help-38

[torn sorrel]

Are you following?

[vast bay]

Yes

Cosx =√3/2

[torn sorrel]

Yea

[vast bay]

Ok thanks

[torn sorrel]

Yw

You can do this from here

[vast bay]

Yes

[torn sorrel]

Right

So what will be the solution?

[vast bay]

30°

[torn sorrel]

Yea

[vast bay]

And 330?

[torn sorrel]

Should write jn radian form though

[vast bay]

Ok

π\6

[torn sorrel]

How many degrees is one quadrant?

[torn sorrel]

Yea

[vast bay]

90

[torn sorrel]

So to go from 1st to 4th

How many degrees you add?

[vast bay]

270?

[torn sorrel]

Yea

[vast bay]

So 300° then

[torn sorrel]

In radian form?

[torn sorrel]

Yea

[vast bay]

5π/3

[torn sorrel]

Yea

[vast bay]

I thought to get the answer in another quadrant that you took the answer away from the quadrant

Like 360-30

[torn sorrel]

Now you got yhe idea right

I mean how to do it

[torn sorrel]

This is not how it's done

[vast bay]

So you just add 90 for how many quadrants it's going up

[torn sorrel]

Yeap

[vast bay]

So this is wrong then?

[torn sorrel]

Yea it's wrong

[vast bay]

It would be 120 and 30°

[torn sorrel]

Yeap

Depending upon the domain

[vast bay]

Same domain

[torn sorrel]

Yea then thiss is true

[vast bay]

Thankyou

[torn sorrel]

Yw

[vast bay]

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[tidal pagoda]

Uh

[tidal pagoda]

Im struggling w some simple trig 😭

I just dont really understand this picture

[clear cloud]

can you rotate ?

[tidal pagoda]

Yeah

[torn sorrel]

,rotate

[tidal pagoda]

Oh

Idk if i even drew it correctly

I struggle with interpreting questions like this

[wraith hinge]

The distance away from the tower stays constant

[tidal pagoda]

Oh

[wraith hinge]

your diagram implies he moves backwards

[tidal pagoda]

Here let me try changing it

[clear cloud]

it should be like this

[wraith hinge]

no

The towers taller than 45m

First measurement measures 45m mark

2nd measures top of the tower

[tidal pagoda]

Like this

[wraith hinge]

and both measurements are taken from 1.3m above ground

[tidal pagoda]

Sorry the paper is so dirty, its been a headscratcher for a bit

[tidal pagoda]

O

Oh youre right

I fixed the diagram but

How would i solve for the tower height

[wraith hinge]

Given the first measurement, what can you work out?

You have the angle and the height

[tidal pagoda]

So with angle and height i could either figure out the adjacent side or the hypo side

[wraith hinge]

Which would help with finding the height of the other measurement

[tidal pagoda]

I was thinking hypo

But idk if it worked

[wraith hinge]

how would the hypotenuse help?

[tidal pagoda]

Im getting the adjacent for the other triangle

Which i could use to math the other x

With the angle and adjacent angle

[wraith hinge]

Yes

[tidal pagoda]

Since it would be Opposite divided by adjacent

[wraith hinge]

thats right

so work out the adjacent

[tidal pagoda]

I did it before but it just didnt seem right

[wraith hinge]

Did you minus 1.3 from 45?

[tidal pagoda]

Let me retry it rq

Gimme a sec

89*

[wraith hinge]

Whyd you use sin?

[tidal pagoda]

Oh

Isnt sin O/H

[wraith hinge]

It is

but you should only have to use tan

[tidal pagoda]

I used sin for the A for the second triangle

Of the*

[wraith hinge]

The a of the second triangle should be the same as the a for the first triangle

the difference is in the o

[tidal pagoda]

Wasnt i supposed to find the H for the first triangle then use that as the A of the second triangle

In order to find the O

OH WAIT

I GET IT

Give me one second

Mb

[wraith hinge]

noo you want to find how far away the observer is 😭

horizontal distance from the tower

[tidal pagoda]

YEAH I JST REALIXED IM SO SLOW

ITS 75M

AYO

Im like

Super sliw

I SWEAR IM SMART(cap) ITS JUST LIKE 3AM FOR ME

I was doing my assignment 😔 cuz its due

ANYWAY TY

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[fast tulip]

Why is 2/5 x 1/4 2/20

Shouldn’t it be 8/20 x 5/20

[amber spruce]

That's right

8/20 × 5/20 will be 40/400
Which as same as 2/20

[fast tulip]

My teachers answer is 2/20

Oh

[amber spruce]

It's 1/10 anyway

Hope this helps 😁

[fast tulip]

Alright

Thank you so much

[amber spruce]

👍

[fast tulip]

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[fast tulip]

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[frank haven]

How would you go about solving this?

[frank haven]

That was how I went about it as well, actually. I was wondering if there was some quicker way that involved finding a common multiple of 8 and 3? There was some approach that worked as such

Ah thank you.

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[frank merlin]

i cant find the sum of b and c

[worldly wing]

Remember the Taylor series for sin x

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@frank merlin Has your question been resolved?

[frank merlin]

why is it divided by 3 tho

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[urban wasp]

would the answer be: A

[wraith arch]

yes

[urban wasp]

k ty

how do i close

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[urban wasp]

.reopen

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✅

[urban wasp]

is this right?

and for question 2 all the options are wrong

[solemn thorn]

really?

for question 2 all the options are wrong?

[urban wasp]

i think

[maiden hare]

q4 doesn't seem right

[solemn thorn]

double check the problem

[urban wasp]

nvm i got it ty

its (12,3)

[brittle barn]

^

[urban wasp]

no im sure its right i double checked wit my friend

[maiden hare]

And I'm sure it's wrong

[urban wasp]

and im sure ur wrong

[solemn thorn]

check segment lengths again

[urban wasp]

bruhhh

[brittle barn]

im sure its wrong

as well

[urban wasp]

my friend otistic man

he said its right

[brittle barn]

LOOOOOOL

[urban wasp]

i checked to

and i thought it was right lol

[solemn thorn]

i thought you double checked WITH your friend

[brittle barn]

check $AC$ and $BC$

[solid kiln]

nameless individual

[urban wasp]

its scaline right

[brittle barn]

no its not

[urban wasp]

scalene*

[brittle barn]

no its not

[urban wasp]

nvm i got it its a right isosceles triangle

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[urban wasp]

.reopen

wait is it?

[trim jolt]

✅

[urban wasp]

i wanna be sure

[brittle barn]

it is

[solemn thorn]

make sure to talk to your friend about this

[brittle barn]

nice username btw

[urban wasp]

ok thanks

can u guys just check everything again

[brittle barn]

its all correct afaik

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[fair forge]

why is this valid?

[fair forge]

i think it's probably (a-b)(a+b)=a^2-b^2 but i just dont see it

[amber python]

,, a - b = \f {a^2 - b^2} {a + b}

[fair forge]

yes ty very much

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[wooden topaz]

Given to random variables. Does it hold that if X is independent of Y, that X² is also independent of Y?

[wooden topaz]

And if how do you prove it?

[lilac flame]

yes this is true

unravel definitions

[wooden topaz]

Unravel definitions? Im stuck at the following:

E(X)*E(Y)= E(XY); And E(X²*Y)= E(X*X*Y)= E(X*(XY))= Cov(X,XY)-E(X)E(XY)=Cov(X,XY)-E(X)E(X)E(Y)

[wooden topaz]

Cause what I want to achieve is something to use the Expected Value with

[digital bolt]

try not using expected values

just work with the probability that the variables take a given value

[wooden topaz]

So P(XeA, Y²eA)=P(XeA YeSqrt(A))=P(XeA)*P(YeSqrt(A))=P(XeA)*P(Y²eA)?

[digital bolt]

this is literally illegible

[wooden topaz]

X element A, Y element B is the notation our prof uses sorry. It means the set of w for which holds: X(w) is element of A.

[digital bolt]

ah

broadly correct but sqrt(A) is the sqrt of a set here which isn't actually defined by default

so you should define just like, B

where B = {b st. b^2 = a for some a in A}

[wooden topaz]

okay thank you!

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[urban copper]

how to find the number of real solutions for the equation $$4x + cos^2(x) = 0$$

[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

someone is saying that using bolzano theorem we can justify that the root exist

and showing that is strictly increasing we can argue that the solution is unique

but idk

how to do that

[marsh forum]

I'd first find the limits at $-\infty$ and $\infty$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[urban copper]

find the horizontal asymptotes

?

[marsh forum]

no

the key here is to see that there's ATLEAST one real root

from the limits alone

[urban copper]

wdym

[marsh forum]

what's the limit at -infty

[urban copper]

I wanted to ask, is this finding the vertical asymptotes or

well

,, \lim_{n \to \pm \infty} 4x + \cos^2(x)

[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

cos is bounded

[marsh forum]

I don't know much about asymptotes

yes but it's roughly -\infty

and \infty

right?

[urban copper]

sure

[marsh forum]

that means?

[urban copper]

isnt cos oscilating between -1 and 1 or

[marsh forum]

it is

[urban copper]

is that sin or

[urban copper]

not sure tbh

[marsh forum]

there's one zero between -\infty and \infty

[urban copper]

sure

changes sign

[marsh forum]

now check if it's monotonic

[urban copper]

well

we can use lagrange theorem

the mvt no?

im not sure

[marsh forum]

IVT

right

[urban copper]

we need to differentiate

[marsh forum]

yes

4-sin(2x)

is the differential

right?

[urban copper]

exactly

well

[marsh forum]

do what can you infer from that?

[urban copper]

,w differentiate cos^2(x)

[solid kiln]

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[urban copper]

well

not quite

[marsh forum]

what's sin(2x)

[urban copper]

,w sin(2x) = 2sin(x)cos(x)

[solid kiln]

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[urban copper]

I didnt know

[marsh forum]

this is an important formula, you better learn it $sin(2x)=2sin(x)cos(x)$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[urban copper]

sure

how to proceed

[marsh forum]

so what's the differential ?

[urban copper]

well

4 - sin(2x)

[marsh forum]

yes

[urban copper]

sin oscillates between 1 and -1

so f'(x) its always positive

[marsh forum]

yes

which means?

[urban copper]

idk

there is only one sol

[marsh forum]

yes!

[urban copper]

but idk howto prove it

or show

wtf

how did ivt is used

I got overly confused

[marsh forum]

the function is continuous on $\R$, yes?

[solid kiln]

ƒ(Why am. I here)=I don't Know

[urban copper]

well

cosine is bounded

4x is a polynomial of degree 1 coefficient 4

should be continuous AFAIK

[marsh forum]

yes

which means we can use IVT

right?

[urban copper]

well

in order to use ivt we need f to be continuous in a closed interval lowkey

one of the corollaries from ivt is bolzano theorem

[marsh forum]

https://math.stackexchange.com/questions/938121/does-intermediate-value-theorem-apply-to-continuity-on-open-intervals

[urban copper]

well assuming we can use IVT on open intervals, what would be the procedure?

I am getting confused a little bit

[marsh forum]

honsetly I've only studied the IVT informally

so I don't think I should help

sorry

[urban copper]

I am studying it formally and I still dunno whats going on here

well

problem is to show that the root is unique

well in order to show the function it is monotonically decreasing we need to inspect the derivative 4 - sin(2x)

I mean

we need to find the critical points as well

I mean at the very least absolute maxima and absolute minima

idk this exercise is too complicated, please ping with hints and ideas

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@urban copper Has your question been resolved?

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@urban copper Has your question been resolved?

[clear cloud]

If youre function is monotonically decreasing or increasing, continious on
[a, b] and the root is in [f(a), f(b)] (or [f(b),f(a)] ) then the root is unique

[clear cloud]

Yes its how to deal with the question

[clear cloud]

You can use the limit aswell

Since 0 is in (-inf,inf) you can argue that there is only a unique root to this function

[trim jolt]

@urban copper Has your question been resolved?

[urban copper]

why

[clear cloud]

Cuz 0 is in the interval

[clear cloud]

Its a bijection due to strictly monotonic behaviour of the function

[urban copper]

which interval

[urban copper]

injection not bijection

injection can be implied from the monotonic increasing functino

but wdym bijection, how

I am getting confused

[clear cloud]

(-inf, +inf)

[urban copper]

okay

[clear cloud]

Yes bijection cuz 0 is in the interval of the limit

Its a corollary of IVT

Due to monotonic behaviour

Dont know if there is a special name in english too, but its also called "bijection theorem"

[clear cloud]

Added to all the other things i said its a bijection

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[urban copper]

.solved

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[merry scaffold]

dang i always use this without proving

[wraith arch]

!status

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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

[sour tree]

do you know induction?

[urban copper]

i do but i still need handholding

maybe I simply dont?

[clear cloud]

Let this be True and show that its True for ( n+1)

[clear cloud]

Crazy

[livid thunder]

why does that seem crazy?

[merry scaffold]

here ig..

,rotate

[sour tree]

I cant really explain induction, but in this case its pretty straightforwaed

1. for n=1 we have 1 = (1+1)/2 (this step is easy, always crucial, sometimes you do like n=5 depends on a problem)
2. Suppose statement is true for some n. Lets show rhats its true for n+1.
   In this case you should do it that way:
   1+…+n+1 = 1+…+n+n+1 and keep going from there

[urban copper]

keep going from there?

[merry scaffold]

ye becuz jsut like in the picture (n+1) is continously goin in nth number times

[sour tree]

well because statement is true for n then 1+…+n+n+1= n(n+1)/2 + n+1

[solid kiln]

y0shi

[wraith arch]

using the formula ofc

[sour tree]

i remember this was my first ever proof by induction i did, like my 1st year in high school

i couldnt understand why does it work

[clear cloud]

The legend say that its 10yo gauss that discovered this formula

[frozen plover]

he really did

[clear cloud]

Not like fermat and his great theorem kekw

[urban copper]

can someone hand hold me hard for this

idk induction very well

[clear cloud]

Show that its works for n=0 first since n is positive integers its the initialisation

Then assume that the statement given is true

And show that its true for n+1

So you write the formula

You add both sides n+1

So in the right side you have n(n+1)/2 + n+1

You put under common denominator

[wraith hinge]

ohh Vietnamese?

[merry scaffold]

how u know 🤔🤔

[clear cloud]

It will give you [n(n+1)+2n+2]/2

[clear cloud]

Top of paper

[merry scaffold]

i mean the language 🤔🤔🤔🤔🤔🤔🤔

[clear cloud]

Then you factor, so its (n+1)(n+2)/2

And job is done cuz if we have Sn = n(n+1)/2
Sn+1 = (n+1)(n+2)/2

So its true for all n, natural numbers

Imagine induction like dominos (not pizza) you push the nth dominos so the (n+1)th fall too

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[random niche]

What is wrong with this?
$\int \sin^3{x} \cos^3{x} dx \$
$\int \sin^2{x} \cos^2{x} \sin x dx \$
$\int (1- \cos^2 x) \cos ^2 {x} \sin dx \$
$u = \cos x \$
$du = -\sin x dx \$
$-du = \sin x dx \$
$-\int (1-u^2) u^2 du \$
$-\int u^2 - u^4 du$
like is this correct so far?

[solid kiln]

Tomi

[random niche]

<@&286206848099549185>

[austere cedar]

Looks like a cos went missing after the first line

[random niche]

so it should have been u to the power of 3, right?

[lavish sequoia]

hi

[random niche]

also in the solution sin is the value of u

@austere cedar

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[outer cradle]

this is more of a conceptual question. today i spent some time with my physics teacher and he asked me a question that is mostly conceptual and theoretical, i have been thinking about it but havent gotten anywhere so any hints or guiding me in the right direction would be amazing.

why do we differenciate functions? its more like what purpose does this serve in the real life (physics application). like obv to find the slope or rate of change, but thats a mathematical answer. whats the real reasons equations in the real world use differentiation

[outer cradle]

<@&286206848099549185>

[stuck elm]

yes what's up how can I help

[clear cloud]

As many things in mathematics, this is the reasonning we are searching for, obviously to buy bread at bakery you wont use derivatives, but the man who invented the atm where you withdraw money do.

Devolopping something may find an apply but not neccessarly, we use derivatives in financial to find out the most benefit of a production for example

Get meteo previsions also

Earthquake preventive too

[outer cradle]

possibly some context will make answering this question easier. he taught us a condition under which you can determine if something is a wave, and it was for function f(x, t), after double partial differentiation, (d^2y)/(dx)^2 = k (d^2y)/(dt)^2. i then asked him why double differentiation and he said to answer tthat you need to understand why we differentiate things in the first place

[clear cloud]

Oh, wouldnt he wanted to bring the initial purpose which is to find out variation ?

[outer cradle]

what?

im sorry i dont understand

[clear cloud]

The first derivative is to find the variation of the function

The second derivative is to find the variations of the first derivatives, so when the slope of tangent are increasing or decreasing

[outer cradle]

yeah

i get the maths behind it

but under what situation in real life

would you think of using derivatives

how does it manifest in the physical world

[clear cloud]

Hum..

For common people, they dont need derivatives in the absolute

But all science does use it

Even sociology

[sour tree]

My Real Analysis teacher said that we basically try to study complicated things and understand them by approximating them and getting information about them from simpler functions.

Thats what derivatives do

But yeah its not for a regular person, regular person does nit need derivatives

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[outer cradle]

.reopen

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✅

[outer cradle]

so its useful for approximation and understanding complex functions?

what about it allowing us to understand things that might be very fast

and taking derivative gives info about that, which we might not have been able to do otherwise

[sour tree]

Thats what he told us

[sour tree]

what do you mean very fast

[outer cradle]

hmm

im just thinking

at very fast and ig even very slow speeds for example

it might be hard to visually make out the difference

but its easier and more clear if you differentiate it

oh also i was just thinking

technically wouldn't differenciation also reduce the error in our results

cause if we have a function f(t) where t is like +- 0.1 and f(t) = t^4

the error will be smaller if you work with the differenciated function

[rough goblet]

look

derivatives and integrals constitute real analysis

which is a very important field

they help us find ROCs and net changes of functions, and how much they change

any serious application will require a lot of complicated maths

e.g. in chemistry and physics we use a lot of differential equations to figure stuff out

how something moves or what the force on something is going to be

e.g. what the impact force on a car is going to be at a point, and thus what parts we need to reinforce

or aerodynamics

[outer cradle]

hmm

[rough goblet]

or neural networking

[outer cradle]

okay

[rough goblet]

etc

[outer cradle]

cool

[rough goblet]

ye

[outer cradle]

thanks for this stuff guys

[rough goblet]

np

[outer cradle]

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[uncut cedar]

can someone explain why this is wrong? do i need to show my working out

[digital bison]

Not sure, looks like there's a hole in the graph at x=4 and no other defined point, so f(4) should not exist.

[real gale]

Can f(4) exist and be below -3?

[cosmic meadow]

Can you put a wider image with all the context?

[wraith arch]

maybe the teacher wanted you to write undefined?

could be being picky on DNE and undefined

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[cosmic meadow]

If there is not any other relevant information then, what you should have answered is “Undefined” as y0shi said, and not DNE

On the other hand, if f(4) is defined somewhere else in the exercise, then you should find it there.

But, according to the graph, it is just undefined.

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[bold rune]

43 pls

How do i solve these types?

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@bold rune Has your question been resolved?

[digital bison]

construct a system of equations.
You know the total number of tickets sold was 350, which creates one equation, and the total revenue of 12520 combined with the cost per ticket gives you a second equation

[bold rune]

Nah

[bold rune]

Ok wait

Well know i know x=5

[digital bison]

what is x?

[bold rune]

Wait

Im now wrong

I really dont know where to go

[bold rune]

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[bold rune]

I kinda did it

Thanks

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[faint fulcrum]

Hello, how would you demonstrate this property of divisibility?

[faint fulcrum]

I tried this

But I don't know how to continue

help pls

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[digital bison]

what is the mcd?

[digital bison]

greatest common divisor?

[faint fulcrum]

yes

but I was able to do it anyway

thanks anyway

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[faint fulcrum]

Hi, I need help to show if this property is valid or invalid please

mcd is gcd

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[faint fulcrum]

z

[nova spire]

what happens when a = c

[faint fulcrum]

gcd is a in tesis?

[nova spire]

gcd of a and c will be |a| yes

so is it always d?

[faint fulcrum]

yes

[nova spire]

really?

can you prove it?

[faint fulcrum]

ah no

gcd(3,3) = 3

gcd(5,5) = 5

3 is not 5

[faint fulcrum]

nope

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[frozen plover]

I've defined $x_{n+1} = \s{2 + x_n}$, I just need to figure out how I'd show the sequence is bounded

[solid kiln]

argon

[worldly wing]

$x_n<???\implies x_{n+1}<???$

[solid kiln]

kheerii

[worldly wing]

proof by induction esentially

[frozen plover]

the sequence increases

[worldly wing]

yes but it converges to a particular value

[sullen mural]

I’m not so sure about induction tbh. There’s a more clever way to solve this and ofc once we find a solution, there’s a way to verify it rigorously with the definition of convergence

[worldly wing]

what's that value?

[frozen plover]

Yeah I mean I could do that but

This exercise is on the monotone convergence theorem

So I guess I should do that 😓

[worldly wing]

$x_n<2\implies x_{n+1}<2$

[frozen plover]

Done quite a few eps-N proofs already

[solid kiln]

kheerii

[worldly wing]

this is what I meant

[frozen plover]

uh

why is that necessarily true

[worldly wing]

use this definition

[frozen plover]

x_1 = sqrt(2)

[worldly wing]

it still satisfies x_n<2

$x_n<2\implies 2+x_n<4\implies \sqrt{2+x_n}<2\implies x_{n+1}<2$

[solid kiln]

kheerii

[worldly wing]

so you can say that all the terms must be less than 2

by induction

[frozen plover]

Ah okay fair

alright thanks :))

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[worldly wing]

then just monotone converge proves this right?

[worldly wing]

just prove x_{n+1}>x_n from here

[dull thorn]

Looks like 2*cos pi/2^n

[worldly wing]

uhh

[frozen plover]

yep that works too I think

I did something similar a few years ago

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[wraith hinge]

Help please

[unique pine]

do you know the formula for the volume of a cylinder?

[wraith hinge]

Yeah

I think

[unique pine]

and it is?

[wraith hinge]

Pie x l x w?

[unique pine]

wrong

[wraith hinge]

Then idk...

[unique pine]

πr²h

[wraith hinge]

oh

[unique pine]

do you want to know how that came? it always comes in handy

[wraith hinge]

ok

[unique pine]

what is the area of a circle?

[wraith hinge]

Pie Radius Square

[unique pine]

correct

now, could you imagine a cylinder as multiple circles stacked up on each other?

[wraith hinge]

yh

[unique pine]

and, the height of your cylinder is h, so basically the pi * radius squared is stacked up until it covers a length of h

[wraith hinge]

oh

[unique pine]

anyway

lets do the question now

what is the height of your cylinder?

[wraith hinge]

Someoe got it for me

What abt this?

[unique pine]

pi * r^2 is given as 18

so it is just 18 * h

[wraith hinge]

Ok

But idk the height

[unique pine]

make the cylinder stand on the circle

assume it is vertical

the height is 16

[wraith hinge]

ok

so 18x16?

[unique pine]

yes

[wraith hinge]

288

[unique pine]

correct

[wraith hinge]

So is that the answer?

[brittle barn]

yep

[unique pine]

yes

[wraith hinge]

tysm!

[unique pine]

no problem

[brittle barn]

.close plz

[wraith hinge]

AYY!

[brittle barn]

please type .close

[unique pine]

@wraith hinge could you type .close? ^

[wraith hinge]

Ok

.close

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[terse haven]

is this Truth Table correct

[junior saffron]

No. I would recommend splitting it further since you jump from $p\rightarrow q$ to $(p\land (p\rightarrow q))\rightarrow q$ directly, which makes it more likely to commit mistakes. You can use the layout in this image.

[solid kiln]

Crystopher

[terse haven]

not to be rude or anything but i cant really see what you wrote

[junior saffron]

Understandable, how about now?

[terse haven]

ya i can see it

how about now

@junior saffron

[junior saffron]

yes, it looks correct, you see that the statement was in fact a tautology.

[terse haven]

i have some other problems if you dont mind checking them out

should probably clarify financial security/gambling is 5, new taxes is 6, godzilla is 7

[junior saffron]

in 5 it is hard to see how you 'translated' the statements, can you elaborate?
6 seems correct
7 seems wrong at first sight to me. I also have some doubts about the interpretation of the statements

[terse haven]

lets start with 5

wdym how i translated

[junior saffron]

I know you are trying to prove $p$ there, but how did you symbolically interpret statements 1 and 2?

[solid kiln]

Crystopher

[junior saffron]

with tose i mean these

[terse haven]

The first statement asserts that if p (you are a gambler) is true, then ¬q (you do not have financial security) must also be true.

The second statement simply states that ¬q (you do not have financial security) is true.

idk if that helps explain it

not a good explainer when it comes to math

[junior saffron]

so \
$1. p\rightarrow \lnot q$\
$2. \lnot p$

[solid kiln]

Crystopher

[terse haven]

yes

[junior saffron]

What I see in the upper row of this table is the following:\
$p$\
$q$\
$\lnot q$\

[solid kiln]

Crystopher

[junior saffron]

$p\lnot q$\
$\lnot q$\
$p$

[solid kiln]

Crystopher

[junior saffron]

specially, $p\lnot q$ makes no sense.

[solid kiln]

Crystopher

[terse haven]

gotcha

[junior saffron]

was that supposed to mean $p\rightarrow \lnot q$?

[solid kiln]

Crystopher

[terse haven]

ya i think i made a mistake on that

[junior saffron]

in such case, the there's a row wrong in that table.

[terse haven]

could you help me establish the truth table

and then i can do the T's and F's on my own

just so i get the basic form of it i guess is what im trying to say

[junior saffron]

When starting out with these you should make many columns, else you commit mistakes easily. Like you know the truth value of operations like $\lnot p$, $p\land q$, $p\lor q$ and $p\rightarrow q$. These are fairly simple and should be easy to memorize and 'apply' directly. But when you do more complex expressions it is best to break it into several parts (at least until you are more confident).

[solid kiln]

Crystopher

[junior saffron]

for instance, instead of directly finding values of $p\rightarrow \lnot q$ you find $\lnot q$ first, then you don't need to mentally change the truth value of the entries in the $q$ column and directly doing implication. Then, instead of directly trying to see values of $\lnot q \land (p\rightarrow \lnot q) \rightarrow p$ you find first the values of $\lnot q \land (p\rightarrow \lnot q)$. This way makes you write more columns but at the same time the risk of doing something wrong diminishes a lot.

[solid kiln]

Crystopher

[terse haven]

is ~q still FTFT

[junior saffron]

yes

[terse haven]

idk why the image is sideways

[junior saffron]

;rcclw

.rcclw

[terse haven]

what

[junior saffron]

don't remember the command for that

it is possible to rotate the images in Discord using some command

[junior saffron]

last column wrong, once again I recommend you to use the entire table I provided.

[terse haven]

oh wait shoot i missed something

[junior saffron]

you missed the column for $\lnot q \land (p\rightarrow \lnot q)$, which was intended to be done first

[solid kiln]

Crystopher

[terse haven]

ya i noticed

is the last column TTTF

[junior saffron]

yes, looks good

[terse haven]

now what about 7

[junior saffron]

wait, it seems the table itself is correct, but what is concerning is how you interpreted the statements, although these are not so easy to translate symbolically

[terse haven]

so its right?

well lemme see hold on

[junior saffron]

correction, I got different on this row

[terse haven]

so the 5th row is wrong

[junior saffron]

yes, since $q$ is true there, so $(\lnot p \land r)\rightarrow q$ must be true regardless of $p$ and $r$

[solid kiln]

Crystopher

[junior saffron]

But a more pressing matter may be the interpretation, like how did you symbolically represent the statements and the conclusion.

[terse haven]

ok i think i found the problem for 7 on the 5th row

the last letter in the 5th row is a T not a F

[junior saffron]

yes, this being the reason

[terse haven]

alright

so in conclusion 7 is correct

[junior saffron]

if the interpretation is correct, then yes.

[terse haven]

alright

i have one last question if you dont mind taking a look at it

[junior saffron]

what is it?

[terse haven]

pulling it up

[junior saffron]

all cells look good

[terse haven]

you sure?

double check

[junior saffron]

confirmed

[terse haven]

alright well thank you for helping me get this done

.close

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[sonic pilot]

I am thinking of how to deal with this fraction with negative sign
I mean I know I have to apply the negative exponent rule, to me the exponent 3 positive like in the second picture, so then I divide (1/1 / 1/3^3)^-2, but my confusion is, does the 3 should have the negative sign, if the negative sign is for the whole fraction?

[swift light]

$(\frac{a^n}{b^m})^p = \frac{a^{np}}{b^{mp}}$

This is the property you should use I think

[solid kiln]

Josh ♚

[sonic pilot]

mmm :/,

[sonic pilot]

the thing is that the exponent outside the fraction is negative, does that property apply either way?

[wraith hinge]

it does

[sonic pilot]

so the result would be -1/729?

[wraith hinge]

3^6 is 729? if so then yup

yes

[sonic pilot]

thanks

[spark spoke]

Make sure the result is positive.

[sonic pilot]

both the numerator and denominator get the negative sign? or either when doing the division 1/1 1/3^3

[spark spoke]

As $\left( -3^3 \right)^{-2} = \left( -\frac{1}{3^3} \right)^{2} = (-\frac{1}{3^3}) \cdot (-\frac{1}{3^3}) = \frac{1}{3^6}$

[sonic pilot]

like -1/1 1/3^3 vs 1/1 1/-3^3

where did the numerator go?

[solid kiln]

eToThe2iPi

[spark spoke]

There you go 🙂

Sorry about that.

$\frac{1}{\frac{1}{3^3}} = 3^3$

[solid kiln]

eToThe2iPi

[sonic pilot]

ah. I was about to ask if there was a rule to get rid of it
is it 3^3 because it ends up like 3^3/1 = 3^3 right?

[spark spoke]

That's it.

The reciprocal of $\frac{1}{3^3}$ is $3^3$.

[solid kiln]

eToThe2iPi

[sonic pilot]

I see, just switching, but in this case the result was because of the division 1/1 1/3^3, not because we decided to get rid of the two 1s, right?

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[barren crescent]

am i right? little iffy about directrix

[unique minnow]

Yes. The directrix essentially tells you in which direction the parabola opens

[barren crescent]

fantastic

thank you very much!

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[timber bronze]

should i use trig sub for this?

[wraith arch]

would work yeah

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[barren crescent]

am i correct?

[unique minnow]

Looks right !

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[amber wolf]

can anyone check if my answer is correct?

[dull temple]

generally miles is mi, because m means meters

[amber wolf]

oops

besides that anything else wrong

[dull temple]

if you're not confident in your answer, you should show your work

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[echo linden]

Hi, can anyone help me understand this question, thanks

[mighty prism]

,rotate

[wraith hinge]

Me and @wraith hinge Can assist you through this problem. What seems to be troubling you today?

[mighty prism]

?

[mighty prism]

it's asking you to prove this statement

[wraith hinge]

We actually both looked at the problem and we are confused. Sorry.

[echo linden]

What does it mean by draw a perpendicular line AE of length h, does it mean draw a line from A to D and d=h? And from d to D equaling p?

[mighty prism]

no it's saying to drop a perpendicular from A to BC, so a separate line segment

so draw another line from A to BC with the intersection at a new point E and then label the lengths AE=h and ED=p

[echo linden]

So basically a straight line down from A and to B and C

[mighty prism]

right

[mighty prism]

onto the line BC

[wraith hinge]

If you look at the vertex you can see l BD has an equilibrium balance to A (b)C, so therefore in this equation a*^2* = d*^2*

[echo linden]

Is this what you mean?

[mighty prism]

yeah

[echo linden]

BE=BD-ED?

so m-p

[mighty prism]

right, which is m-p

yeah

[echo linden]

So EC=2m-BE?

[mighty prism]

yes

which is m+p

[echo linden]

Oh ok

[mighty prism]

can you see what to do now?

you have two right triangles, so you should be able to just apply pythagorean theorem

[echo linden]

How would I do D?

[mighty prism]

use pythagorean theorem on AED

to get d^2 in terms of p and h

and then you should be able to use algebra to reduce to the identity that you want

[echo linden]

So what would I do from 2h^2+2m^2+2p^2?

How would I deduce it?

[mighty prism]

wait so you have a^2 = (m-p)^2+h^2, b^2 = (m+p)^2 + h^2, and d^2 = p^2 + h^2

so now reduce the equation a^2+b^2=2(d^2+m^2)

so that everything is in terms of p, h and m

and you should have reflexivity

[echo linden]

What happens if the angle at A is a right angle? Would the theorem still work

[mighty prism]

yes, why wouldn't it

we didn't assume anything about the triangle

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@echo linden Has your question been resolved?

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[minor geode]

someone help me?

[minor geode]

??

[livid thunder]

i dont think either is longer?

[minor geode]

well the anwser is LK

[livid thunder]

wait

[minor geode]

but I dont know why

@wraith hinge

[wraith hinge]

is that right angle

or what

[livid thunder]

determine angles LKJ and JLK

the side opposing the larger angle will be longer

is this triangle possible?

[wraith hinge]

wait i made an error

[livid thunder]

yeah u did

[frail heron]

Weird enough

[worthy jacinth]

Let LKJ be x...
LJK be 90-x
L_K be 140-x...
L_J be 40+x
Let the two equal sides be y
y / sin 40 = LK/sin(140-x)
y/sin 50 = LJ/sin(40+x)
sin 50 / sin 40 = LK / LJ
LK > LJ...

[livid thunder]

the triangle cant existy

exist*

the only way for the line to bisect the hypotenuse is if the triangle is isoceles

which implies JL=LK

[wraith hinge]

yeh

[livid thunder]

but if u work out the angles in the triangle

[worthy jacinth]

It didn't say anywhere that the line bisects...

[livid thunder]

which is 40,90,50 it says JL shou ld be longer

[livid thunder]

look at the diagram

[wraith hinge]

but the sides are same

[livid thunder]

the 2 markers imply their lengths are the same

[wraith hinge]

yeh

[worthy jacinth]

2 sides the same not equal the line bisecting the angle.

[livid thunder]

ah right i forgot the angle was 40

ffs

anyways the question answer is still wrong

[worthy jacinth]

It didn't say that the line forms right angle.

[worthy jacinth]

Hmm?

Don't presume the triangle is isoscele or that the line bisects the other line into two right angles.

Wait...

Is the triangle even possible?

[worthy jacinth]

Hmm? That's wrong?

JLK forms a right triangle.

[wraith hinge]

AHHHHHHHHH MY BRAINNNN

Lmfao

Oh well

I don't think the angle is gonna matter

[worthy jacinth]

Actually, there's an easy way.

Notice the circle you can draw over the triangles.

With JK being the line through the center.

[livid thunder]

his diagram is incorrect the third line isnt the same length as the other 2

[worthy jacinth]

Isn't it the norm that the diagrams might not be drawn to scale?

[livid thunder]

it is not know that the line that bisects the hypotenuse is the same length as the others

[worthy jacinth]

It's known.

Circle property.

[minor herald]

The assumption was LK≠JL so the bisect segment with JK can't be right angle

[wraith hinge]

oh yeh

true

[worthy jacinth]

Don't you remember the circle property? Draw two lines from the line that passes through the center to cross each other and you got a right triangle exactly at the circle.

[wraith hinge]

This?

[worthy jacinth]

yeah...

[livid thunder]

i barely know it, i never learned geometry

[worthy jacinth]

Uhh...

[wraith hinge]

Yes

[livid thunder]

first time i encountered it was in complex numbers

[worthy jacinth]

This question should've been settled a long time ago and well... somehow it dragged on for so long lol.

[worthy jacinth]

You can use sine law or circle property to solve this.

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@minor geode Has your question been resolved?

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[dreamy sleet]

Consider the sequence: $a_n=6n+\frac{1}{n}$

[solid kiln]

lILi

[dreamy sleet]

I have an idea about it and I want to know if it works

I want to prove that the sequence is increasing by proving that the slope of the corresponding function is always positive on the interval [1, infinity)

I'm doing this by converting a sub n into a function f of x, taking the derivative of that function, turning that derivative into a sequence, and proving that the resulting sequence is bounded below 6 and above 5, and converges to 6

Would that logic be a viable proof that a sub n is increasing?

I'm in calculus 2

[solid kiln]

lILi

[dreamy sleet]

sorry for the adjustments

[quartz cargo]

It suffices to take the derivative and show that it is always positive

Why do you want to bound b_n?

[dreamy sleet]

Oh, fair

[dreamy sleet]

Just wanted to prove that it's always positive but I realise now that doing it in such a way would be overkill

Thank you for confirming though

.close

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[uneven steeple]

isn't N(0,p) open by definition?

[marble wharf]

and?

[uneven steeple]

because for any x in N(0,p)
because N(x,p) is contained in N(0,p)

so union of N(0,p) over all primes is also open

oh nvm, I read the end part incorrectly

.close

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[sudden trout]

How to find the last 5 or 10 or more digits of the number 2^12345 or any large number with 2^n where n is positive integer

[sudden trout]

like 2^32 = 4294967296 and the last 5 digits of it is 67296

[tawny obsidian]

you can do this lightning fast on a computer at least

what we want is like 2^(12345) mod 10^n for the first n digits

[near sapphire]

what we want is like 2^(12345) mod 10^n for the first n digits

[sudden trout]

Bruh I meant

Not brute force

[marble wharf]

well you can do it in roughly log_2(12345) computations. which is like 15 steps. its not that bad

see square-and-multiply

[sudden trout]

And I meant like mathematically not computationally

[marble wharf]

well depending on how big 10^n vs the exponent is, you can do some simplification with eulers theorem

but in the end you will have to compute stuff

[sudden trout]

.close

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[sudden trout]

.close

what

.close

[sacred sapphire]

already closed my bro

[sudden trout]

then why is it in occupied

[sacred sapphire]

it doesnt go away immediately

takes some time

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[deep mountain]

can i have help with question B please

[runic dew]

b) will be zero but assuming it is a not a typo. Otherwise you will need to find number of 40000 dollar payments

[deep mountain]

the awns key said 7

but idk how

like i got 5

[runic dew]

I am having trouble understanding the question. It seems that she is receiving a payment of $40,000. This means the investment will never go to zero

I think it will be good to write down the expression for the couple of terms. That we can see any pattern

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@deep mountain Has your question been resolved?

[deep mountain]

Jesse is investing 200 000 that is compounded

She then is given 40 000 out of that balance per year

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@deep mountain Has your question been resolved?

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[cobalt cloak]

if e1 e2 e3 are eigenvalues of A, is (e1)^n (e2)^n (e3)^n eigenvalues of A^n

[cobalt cloak]

I just wanted to know

[dusty sleet]

!status

[trim jolt]

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

[dusty sleet]

Which of these is your situation?

[cobalt cloak]

I would like to know if that is true or not, i haven't gotten to proving it yet but knowing if its true or not would ease my calculations

[coral ember]

It is

True

[dusty sleet]

It's very easy to prove (btw if you search your question on the Internet you'd find plenty of results)

Start with the definition of eigenvector

Hence, if v is eigenvector for A, it means that ... @cobalt cloak

[cobalt cloak]

Ax=vx