help-10

[vagrant spruce]

f(x) = 2 x 3^x

[somber mural]

yes

[vagrant spruce]

thx

[keen jetty]

Btw @somber mural ik it wasnt my question but thx anyway ive got my a level later today and exponential graphs is on it too xD

[somber mural]

same

I do A level maths too

[keen jetty]

Ooh gl dude ^^

[somber mural]

same to you

[vagrant spruce]

im strugglign with the last part i underlined it

[torn wave]

Getting transformation of functions down and a good understanding of how base functions look/work is very helpful

[somber mural]

what type of graph have you drawn

[vagrant spruce]

graph showing the cycles

x axis is seconds

[somber mural]

yes

[vagrant spruce]

y is feet

[somber mural]

but you can see it is a certain type of graph

[torn wave]

graph good but function is off

[somber mural]

a certain trigonometric one

[vagrant spruce]

its a degree function

thats prob why

[torn wave]

wait that's true

my bad I'm more used to seeing in radians

function correct

[vagrant spruce]

how do i solve that problem

the ome i underlined

[somber mural]

can you answer this

@vagrant spruce you have written your function

so you want to find the 2 x values that give the height as 90

so set y to 90

and solve for x

[vagrant spruce]

oh ok thanks

wait 2 x values

wont i get 1 if i solve it

[somber mural]

you have drawn the graph

in one cycle, how many times does the graph y value equal 90

[vagrant spruce]

once

oh twice

[somber mural]

when you do the inverse cosine, you need to make sure you account for both values

then solve for x with both those values

[torn wave]

to elaborate, calculating inverse cosine will give you the reference angle for theta. There are two quadrants where this reference angle would be positive for cosine (1 and 4). Finding the standard angles for both will give you two thetas, and therefore two final x values (for time)

[obtuse pebble]

@vagrant spruce Has your question been resolved?

[vagrant spruce]

im confused

how do i solve thris

[somber mural]

$y=49cos(12(x-15))+50$

[warm shale]

IntelligentCake

[somber mural]

you want to find when y is 90

so

$90=49cos(12(x-15))+50$

[warm shale]

IntelligentCake

[somber mural]

subtract 50 from both sides

$40=49cos(12(x-15))$

[warm shale]

IntelligentCake

[somber mural]

$\frac{40}{49} = cos(12(x-15))$

[warm shale]

IntelligentCake

[somber mural]

do you agree @vagrant spruce

[vagrant spruce]

do u leep the fraction

[somber mural]

leep?

[vagrant spruce]

keep

[somber mural]

what do you mean keep

you now have cos(12(x-15)) = 40/49

now do inverse cosine

so 12(x-15) = angle

[vagrant spruce]

inverse cos on 40/49?

[somber mural]

yes

so you get rid of cos on the left hand side

[vagrant spruce]

35.31

[somber mural]

ok but there is also another solution

at 360 - the angle you just found

[vagrant spruce]

wdym

[somber mural]

this is a graph of cosx

notice how it is symmetrical

so there are 2 solutions that give 40/49

[vagrant spruce]

idk

i have to go soon

[somber mural]

ok well I've given you the other angle to calculate

once you have that

you can solve for x for both equations

[vagrant spruce]

ok i did all that and got 17.9

how do i get 2nd time

idk abt the 2nd one

[somber mural]

you solved 12(x-15) = 35.31

not you need to solve 12(x-15) = 360 - 35.51

[vagrant spruce]

why 360

[somber mural]

I have already expalined

this is cosx

from 0 to 360

it equals 1 twice yes?

at 0

and at 360

so 2 solutions

you are trying to find when it is 40/49

so that is at 2 x values

[vagrant spruce]

i solved ot

42

and 17

[somber mural]

ok

[vagrant spruce]

how do i graph this

[somber mural]

I haven't checked but you can always put x back into original

[vagrant spruce]

how do i get the points

is there a vid i can watch like a yt vid

[somber mural]

sorry I must go now, you will need to wait for someone else

[vagrant spruce]

is there a vid abt it

idk what to search

.close

[obtuse pebble]

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[gentle flower]

help

[gentle flower]

last set before i sleep so i have trouble

with these three

[gentle flower]

this is -1

[obtuse pebble]

@gentle flower Has your question been resolved?

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[peak hamlet]

I have found r=-10 and p=3

I can't find q

I grouped together the parts of the equation depending on whether they had a matching degree

which is how i found p and r

[timid silo]

Do the same for q

[peak hamlet]

I did but i cant seem to get it

[timid silo]

You will get a system of equations

[peak hamlet]

(r-5q)x^2 and (-5q+4p)x^2

[timid silo]

There are more than that

[peak hamlet]

There are three parts of the equation that share the degree of 2

[timid silo]

Multiply every term with (x-1)(x-4)

[timid silo]

Yes, group all the different parts with the same degrees

Together

And then make them equal to -63
Do the same for third degrees and first degrees

[timid silo]

Okay so compare

what is the x^4 value equal to

[peak hamlet]

3

[timid silo]

And what about no x?

[peak hamlet]

and 4r to -40 so r =-10

[timid silo]

Yes

You have p and r

Combine the terms with second degrees after substituting them

And find q

Could also combine the third degree terms

[peak hamlet]

ok

[obtuse pebble]

@peak hamlet Has your question been resolved?

[peak hamlet]

q=115

boom, thanks

[obtuse pebble]

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[proven zephyr]

$\frac{2x+5}{x+5} = \frac{x-20}{x^2+x-20}$

[warm shale]

MarveI

[proven zephyr]

i saw someone asking this question and i still dont understand

[timid silo]

What dont you understand

[proven zephyr]

okay wait ima type where im stuck at

$\frac{(2x+5)(x-4)}{(x+5)(x-4)} = \frac{x-20}{(x+5)(x-4)}$

[warm shale]

MarveI

[proven zephyr]

ok i already made both have the same numerator

wait nvm i got it now

im dumb

[scarlet locust]

yes now can cancel the denominators

keeping in mind that if you get x=-5 or x=4 later on then they arent solutions

[proven zephyr]

x = 2 right?

$(2x + 5)(x - 4) = x - 20$

[scarlet locust]

thats 1 solution

there will be another

[proven zephyr]

and 0

[warm shale]

MarveI

[scarlet locust]

yup

those are correct

[timid silo]

Gg

[proven zephyr]

$2x^2 + 5x - 8x - 20 = x - 20$

[warm shale]

MarveI

[proven zephyr]

$2x^2 + 2x - 40= 0$

[timid silo]

2 and 0 are right

[warm shale]

MarveI

[proven zephyr]

oh ok

[timid silo]

No

You made a mistake here

[proven zephyr]

+20

yes

[timid silo]

It becomes 2x^2 -4x

[proven zephyr]

wait

what

[timid silo]

-8x + 5x -x

[proven zephyr]

oh yeah i m dumb

[timid silo]

Thats -4x

[proven zephyr]

$2x^2 - 4x = 0$

[warm shale]

MarveI

[proven zephyr]

and then just ax + bx + c = 0 thing

ok ty

.close

[obtuse pebble]

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[meager fiber]

i need a explanation on this:

[meager fiber]

i am trying to learn integration, and was finding the proof of integration of a^x dx

this part was verification, (derivative of a^x/lna)

[obtuse pebble]

@meager fiber Has your question been resolved?

[timid silo]

what do you need explained

[obtuse pebble]

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[modest eagle]

how many natural solutions does x+y+z=8 have

[brave bramble]

See "stars and bars method"
https://brilliant.org/wiki/integer-equations-star-and-bars/

[modest eagle]

hm

close

[obtuse pebble]

@modest eagle Has your question been resolved?

[obtuse pebble]

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[warped fulcrum]

I could use help with this

I tried with generating functions

But i seem to be getting nowhere

[jolly ginkgo]

Do you have to find a_n

[warm shale]

fäf

[warped fulcrum]

Huh?

I'll show you what i had so far

[jolly ginkgo]

What's the doubt there?

[restive ridge]

@warped fulcrum what's the question

you posted a random set of equations

[jolly ginkgo]

Hello there

[warped fulcrum]

Ah mb. Solving the recurrence relation

[restive ridge]

what do you want

[jolly ginkgo]

Use this @warped fulcrum

[jolly ginkgo]

Was just saying hello

[pseudo swift]

I mean generating functions are 1000x overkill here, it's pretty much a geometric series we got here

[warped fulcrum]

This tells me exactly nothing but ok

[timid silo]

what do u even want?

[warped fulcrum]

Yes but the homework specifically asks for it to get familiar with it

[jolly ginkgo]

Well use the knowledge that $a_n - a_{n-1} = 3^{n-1}$

[warm shale]

fäf

[warped fulcrum]

Solving the recurrence relation like I've stated above. Please read the convo first. Wasn't that high up lol

[jolly ginkgo]

You gonna get $a_n - a_0 = 3^{n-1} + 3^{n-2} + \dots 3^{1} + 1$

[warm shale]

fäf

[pseudo swift]

Ah okay

I'm a bit rusty on those

[warped fulcrum]

.Close

.close

[obtuse pebble]

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[sharp mist]

I'm familiar with solving these types of questions when the power of x is one.
I got the answer as (c) by substituting values but I don't really know how to start this one.

[obtuse pebble]

@sharp mist Has your question been resolved?

[sharp mist]

.close

[obtuse pebble]

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[timid silo]

How did they got that red from blue

I am always getting 6.31*10^33

[pine sail]

You can't have 10^33

[timid silo]

There is minus before 33

[pine sail]

Lol.

What you got is correct.

[timid silo]

Is it my mistake calculating or ...

Then the text is wrong?

[pine sail]

But it's exactly same as what they got too.

Both are equal

,w 0.63 * 10^{-32} = 6.3*10^{-33}

[warm shale]

Results provided by WolframAlpha

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

Requested by Yours truly#2323

[pine sail]

See.

Equal.

[timid silo]

Wait my head is processing

[pine sail]

200 = 2 * 10^2 = 0.2 * 10^3

Right?

[timid silo]

Yep that "-power" got me wrong

Thx buddy

.close

[obtuse pebble]

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[pine sail]

$$

[tight bison]

can someone please explain why $|\vec{a}|^2 * |\vec{b}|^2$ can be written as just $\vec{a}^2 * \vec{b}^2$

[warm shale]

magima

[pine sail]

Do you by any chance mean, magnitude in the second case also?

[devout solar]

if you square something its positive

[pine sail]

Yeah.

So it doesn't matter if you take the magnitude.

(5)^2 = (-5)^2 = |+-5|^2 for example.

[tight bison]

Oh okay thx I understand that

[timid silo]

|| is used for absolute ... Always returns positive value

[tight bison]

So absolute and length of the vector are the same when taking the square?

[obtuse pebble]

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[empty spruce]

What are the conditions for $y$ that:
$${ a + by | a, b \in \mathbb{Z} }$$
to be a subring of $\mathbb{R}$

I know $\sqrt{2}$ works, but it seems $\sqrt[3]{2}$ doesn't work.
So $\sqrt{2}$ works because, $\sqrt{2} \cdot \sqrt{2} \in \mathbb{Z}$
However, that's not the case for $\sqrt[3]{2}$, and the fact that it happens at 3 seems to cause an issue.

[warm shale]

ryep

[empty spruce]

I guess the natural question is what about $\sqrt[4]{2}$, and my guess is that it'd work.

[warm shale]

ryep

[warm canopy]

You should be checking the axioms of a subring

[empty spruce]

Oh ok

[warm canopy]

Iirc one of the axioms will put a requirement on y

Your testing might give you an indication of the requirement

[empty spruce]

Aight.

Ok, so $\sqrt[4]{2}$ isn't possible.

My guess is that:
$y^n = a + by$

Is a requirement.
e.g $y = \pi$, then:
$$\pi^2 = a + b\pi \text{ is not possible}$$
$$\sqrt[4]{4} = a + b\sqrt[4]{2} \text{ is not possible}$$
And $y^2$ must exist in the ring, by multiplication.

Idk if this is right: It's kinda nuts, it means $\sqrt{n}$ and $n$ are the only possible $y$s for $${ a + by | a, b \in \mathbb{Z} }$$ to be a subring of $\mathbb{R}$.

[warm shale]

ryep

[empty spruce]

Not even rationals can be y..., wtf.

[obtuse pebble]

@empty spruce Has your question been resolved?

[obtuse pebble]

@empty spruce Has your question been resolved?

[obtuse pebble]

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[twin torrent]

I would like help with this

I have first split this guy into 8 equal volumes by stating its symmetry about x=y, x=-y, and z=0, then integrated as following but I got a bit stuck with integration

[twin torrent]

so I would seek to know if my starting integral is already conceptually wrong, or my integration is trash, or both

and I don't think integrating the bottom integral gives 16 - 8 sqrt(2), so I stopped

[indigo blade]

seems right to me

[twin torrent]

ahahahahah

so the left term becomes -2/3 sin(x) tan^2(x), which for some reason I don't know how to integrate, and integrating the right term is probably not 16 because there's a 3 in the denominator?

is that right?

[indigo blade]

write sinx tan^2(x) in form ((1-cos^x)/cos^x)x sin x then substitute cos x = t to integrate it

What is the numerator in the right term?

is it 2?

if yes then it in integration should be 32/3

[obtuse pebble]

@twin torrent Has your question been resolved?

[obtuse pebble]

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[timid silo]

I am embarrassingly blanking on a super simple 3rd grade math problem, and I need the answer ASAP

If I earned $46.95 after working 3h8m, what is the hourly rate?

I'm wanting to do 46.95 / 3.8, but I know that's not right, 8 minutes is not 80% of 60 minutes. 8 minutes is 7.5% of 60 minutes. so is it 46.95 / 3.075?

[restive ridge]

and I need the answer ASAP
this won't get you help any faster lol

[somber mural]

are you sure 8 minutes is 7.5% of 60

[tardy epoch]

,calc 8/60

[warm shale]

Result:

0.13333333333333

[tardy epoch]

,calc 7.5%

[warm shale]

Result:

0.075

[timid silo]

the point in me saying that

is that I don't have time to go watch a math tutorial lol

im filling out a government applicaation and am asking help

[fickle turret]

8/60 = 1/7,5 not 7,5%

[timid silo]

so what you could do, is tell me the answer, and the method by which you used

[somber mural]

seems kinda lazy ngl

[timid silo]

lol insane

[somber mural]

no is going to tell you the answer

this is for help

not answer giving

[timid silo]

"I am genuinely blanking on how to do this and am filling out an urgent government document"

you are definitely insane if you think I am lazy

that's a totally valid reason for asking for quick help /explanation

[somber mural]

^

[timid silo]

"yeah sorry time sensitive government doc, lemme go pull up math antix real quick"

[somber mural]

in the time you have spent asking, you could have done a lot more

[restive ridge]

developers are supposed to know math

[timid silo]

I said in my message I am blanking on what the calculation even is to do

lol i dunno why people on this server are such self righteous assholes

[somber mural]

you presume that we just give answers

no

[timid silo]

I didn't ask for the answer

[restive ridge]

lmao you understand that we aren't getting paid

[timid silo]

im asking for the formula

[restive ridge]

and that you are not entitled to help

[timid silo]

then get off a help server?

[somber mural]

what does this say lol

[timid silo]

and the method by which you used

[restive ridge]

wat

[somber mural]

we already helped you

[restive ridge]

lmao

[timid silo]

this response

was in response to illuminator

saying "we dont care if its urgent"

[somber mural]

you have wasted far more time arguing than just taking the help

so yes, seems kinda lazy

[timid silo]

Sure I have a hard time not making time to argue against behavior i find riduclous

thanks for the help.

[somber mural]

you putting a reason for the help changes nothing. you are the same as others. wait until someone comes to help

[keen jetty]

[timid silo]

There are two common approaches to helping people work through problems.

One is to take this wise kung fu mentor approach and tell the student to go spend 5-30 minutes thinking about it and come to it on their own. This is usually a good approach, but you guys over do it ( I think for ego reasons)

The other approach, more reasonable in this context, is "Yeah sure lemme help you out, all we gotta do is do this step, and then this step, and then voila here's the answer."

"thanks man, that makes sense and now I have the answer to my urgent matter"

[keen jetty]

[timid silo]

but instead you wanted an excuse to feel superior wise kung fu whatever

[restive ridge]

@timid silo since I assume you're a dev, give this so meta post and the approved answer a read https://meta.stackoverflow.com/questions/326569/under-what-circumstances-may-i-add-urgent-or-other-similar-phrases-to-my-quest

[somber mural]

lmao

[restive ridge]

indeed

redirecting to SO is always the solution

[timid silo]

Yeah I mean I didn't ask an esoteric software question. I asked a 3rd grade math problem that I'm blanking on

if this were any other discord server

I feel like you'd just

i dunno

be a normal person

and help a brother out

[somber mural]

for some "urgent application" you sure are spending a lot of time arguing

just close the channel and move on

[timid silo]

sure I'll close the channel, just wanted to make it clear that i don't think the behavior is reasonable

how do I close without it asking me?

!resolved

[tardy epoch]

.close

[obtuse pebble]

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[timid silo]

ty

[obtuse pebble]

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[dense ingot]

Part c, the mark scheme splits the integral around pi, why?

[dense ingot]

graphing f(x) there are no asymtotes

[obtuse pebble]

@dense ingot Has your question been resolved?

[dense ingot]

<@&286206848099549185>

t = tan(x/2) so t is undefined when x = pi

.close

[obtuse pebble]

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[main cedar]

when solving $2x+4\equiv_5 3$ I did
$$2x\equiv_5-1\equiv_54$$
$$x\equiv_52$$

However, how can I guarantee that this is all of the possible solutions?

[warm shale]

lirmirit

[kind hawk]

Z_5 is a field, so 2 is invertible and there is only one solution to 2x=b for all b

or you can check all options

note that if you multiply 2x=4 by 3 on both sides you get x=2

that's because 2^-1 = 3 in Z_5

[main cedar]

wait, i thought Z_5 was a ring

are rings and fields similar?

[main cedar]

oh yea, this is a much simpler way to solve it

[kind hawk]

fields are commutative rings in which each nonzero element has a multiplicative inverse

Z_n in general is a ring, Z_p with p a prime is also a field

[main cedar]

ah, this makes sense

ty once again

.close

[obtuse pebble]

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[stuck compass]

so at the derivative of f'(3) would g'(x) equal 1/2

because that would be the slope at x=3

[warm canopy]

You certain on 1/2?

[stuck compass]

over one, up 2? I believe so

unless the open dot means the function is undefined

[warm canopy]

What's the definition of slope of a line

[stuck compass]

rise over run?

so 2

sorry

[warm canopy]

Yes

Now also when calculating f'(x) you're gonna need to use the product rule, not sure if you've done/doing that or not

[stuck compass]

g'(x) * f(x) + g(x) * f'(x)?

[warm canopy]

Not f(x)

That doesn't appear in the definition of f(x)

[stuck compass]

oh right just the equation

[warm canopy]

Yeah you can say h(x) =(5x-4)² and replace f with h in what you wrote

[stuck compass]

okay yes that makes sense

then we need to look at g(3)?

the equation would be g(3) * 10(5x-4) + (5x-4)^2 * g'(3)

[warm canopy]

Yes but remember you need to replace ALL x with 3

[stuck compass]

ahhh yes good point

g(3) = 2 and so does g'(3)

[warm canopy]

I would agree with you

[stuck compass]

i am getting f'(3)=462

for part b im thinking the non differentiable ones are the open dots and the pointed edge at x=-2

so the non differentiable values of g(x) would be 2 and -2

[warm canopy]

You are correct, but why?

[stuck compass]

im assuming because they do not go to infinity? There is no slope at 2 or -2?

[warm canopy]

How do we know if a derivative exists at a point?

Maybe rather, what's the definition of the derivative at a point?

[stuck compass]

that there will be an instantaneous rate of change

at that given point

[warm canopy]

Have you not seen the definition in terms of limits?

[stuck compass]

in terms of limits it'd just be the slope right?

[warm canopy]

So you haven't seen a definition in terms of limits?

[stuck compass]

well I know tangent line and all that but not exactly limits i guess

[warm canopy]

Okay, it's a bit easier to explain in this case

Roughly, If the derivative were to exist at a point, then we would expect that as we approach that point from the left and the right, the slope at those points should get closer and closer to the slope we we're interested right?

[stuck compass]

would that be like x--> infinity or x--> 0

[warm canopy]

x --> point we're interested in

[stuck compass]

right yes

[warm canopy]

Look at the graph, if we approached -2 from the left, what's the slope of the function for all x close to -2 but not -2

From the left

[stuck compass]

it'd be 1

then from the right it'd be negative 1?

[warm canopy]

Exactly

These don't match up

So we don't know which we would assign to the derivative at -2

Which is why it doesn't exist

Same thing is happening at 2

[stuck compass]

that makes more sense, the graph definitely helps me see that which is a first

thank you for the help again

[warm canopy]

Youre welcome:)

[stuck compass]

.close

[obtuse pebble]

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[small bloom]

if I have sin x=2/3 how do I calculate this using either 30-60-90 or 45-45-90 triangles?

[devout solar]

wdym

[small bloom]

How do I calculate x?

[tardy epoch]

It's not a special angle so you can't easily do it. You'll need trig identities

Or just a calculator

,calc asin(2/3)

[warm shale]

Result:

0.72972765622697

[small bloom]

I tried haha but my answer key says the answer is 42

[tardy epoch]

,calc sin(42)

[warm shale]

Result:

-0.91652154791563

[tardy epoch]

,calc sin(42 deg)

[warm shale]

Result:

0.66913060635886

[astral dawn]

😮

[tardy epoch]

Kinda close to 2/3, but not exact

[small bloom]

okay!

[astral dawn]

0.72972765622697 i think this is radians or something

yup

[small bloom]

ohhh

okay thank you!!

.close

[obtuse pebble]

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[obtuse pebble]

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[gilded needle]

yes, note that g is invertible

so applying $g^{-1}$ to both sides of $g(g(b)) = -3$ gives $g(b) = g^{-1}(-3)$

[warm shale]

OurBelovedBungo

[gilded needle]

and doing it again gives $b = g^{-1}(g^{-1}(-3))$

[warm shale]

OurBelovedBungo

[gilded needle]

yeah, the function has an inverse, i.e. g(x) = y always has a unique solution x for any given y

correct

so $g^{-1}(-3) = 0$

[warm shale]

OurBelovedBungo

[gilded needle]

and so $b = g^{-1}(0)$

[warm shale]

OurBelovedBungo

[obtuse pebble]

Channel closed

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✅

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[obtuse pebble]

@timid silo Has your question been resolved?

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[uncut hedge]

Substitution

I can’t solve it

I’ve tried for over 30m

[obtuse pebble]

@uncut hedge Has your question been resolved?

[uncut hedge]

Hello?

[pale mortar]

which question?

[uncut hedge]

11

And 10

I tied the equation

Tried

It gave me -200 for 11

And the answer was 200

When I checked the back of the book

I kept re-checking my work

But I couldn’t find the error

[pale mortar]

80+0.22y=z
100+0.12y=z

so you simply do
80+0.22y=100+0.12y

[uncut hedge]

Yup

You got it right

So what I did was

0.22x+y=80

0.12x+y=100

Then I rearranged it

Then I got -200

But the way u did it was right

[pale mortar]

pietro just stays 1 day, thats important info

[uncut hedge]

and

I don’t get it

What does the 1 day have to do wit anything

Oh it equals one day

Per rental

And whatnot

[uncut hedge]

@pale mortar how abt 10

[subtle eagle]

its the same as 11

try write out the equations here

[uncut hedge]

Alright

25x+50y=h

100x+20y=h

@subtle eagle

[subtle eagle]

the material is only a one time

so u can see 25 and 100 as constant

no need times x

[uncut hedge]

So basically

Ohhh

I see

25x+100y=h

Would that be correct

[subtle eagle]

since the material is only one time, u can view it as x=1 if that's easier to understand

[subtle eagle]

this one is correct

[uncut hedge]

Hmm

What abt the other equation then

[subtle eagle]

its the same

x also=1

[uncut hedge]

Ahh

So basically

25x+50y=h

100x+25y=h

[subtle eagle]

since the h is the same as the other equation, u can merge them together

[uncut hedge]

Ahhh

By rearranging

25x+50y=100x+25y

[subtle eagle]

correct

sub x=1, find y

[uncut hedge]

3

Is my y

@subtle eagle

[subtle eagle]

wait sry i just realised u copied the number wrong

its 20 not 25

wait

ye

it says 20 in the question

[uncut hedge]

Ahhhh

Alr

Thx

I understand it

[subtle eagle]

np

[uncut hedge]

.close

[obtuse pebble]

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[hasty perch]

where do I start>

[obtuse pebble]

@hasty perch Has your question been resolved?

[obtuse pebble]

@hasty perch Has your question been resolved?

[drifting wraith]

start by understanding what they mean

what they can look like

what they can't look like

[hasty perch]

So say propositions are a,b,c,d,e

That alone is 5 not logically equivalent formulas

[drifting wraith]

agreed

[hasty perch]

Just dont know where to go from there

[drifting wraith]

i mean can you do ~a | ~b

i used 3 conectives

but they say 2

i don't know xD

[hasty perch]

well no

you can only negate one

[drifting wraith]

okay

so you mean you can also use disjunction at most once?

it gets way too simple in this interpretation imo

[hasty perch]

well twice

right?

[drifting wraith]

sure twice

[hasty perch]

but u cant negate

in that formula

[drifting wraith]

so essentially you have 2 forms ~o | o and o | o | o

[hasty perch]

or o | o

or -o

[drifting wraith]

or o

[hasty perch]

o

[drifting wraith]

or ~o

yep

[hasty perch]

so is there a simple way of figuring out all possible combinations for each form?

[drifting wraith]

of course

you have to choose which ones you use out of 5

[hasty perch]

a b c d e f

[drifting wraith]

e.g. 5 choose 2 = 10

[hasty perch]

sorry I dont get that?\

[drifting wraith]

5 × 4 / 2 = 10

5 × 4 × 3 / 6 = 10 as well

so 10 of " o | o | o " and 10 of "o | o"

hm

[hasty perch]

im so lost

[drifting wraith]

i'm not sure why

[hasty perch]

wdym which ones i use out of 5

cant i just use a b c d e f as the propostions

[drifting wraith]

a b c d e

[hasty perch]

yep and then

isnt it just a matter of figuring all combinations in the forms o | o and o | o | o, etc...

with those letters

[drifting wraith]

yes?

i'm absolutely lost as well

it's just a matter of that

[hasty perch]

so what does this mean?

[drifting wraith]

you have 6 forms or however much we counted

[hasty perch]

yes

[drifting wraith]

we count them individually and add up

o | o | o requires us to pick 3

propositions

[hasty perch]

5 forms btew

btw

[drifting wraith]

and nothing else

like the number of ways to pick 3 out of 5 will not be smaller than what we want, and won't be larger

we're done at 5 choose 3 = 10

[hasty perch]

so there is 10 different combinations to choose 3 from 5

[drifting wraith]

yeah

[hasty perch]

so thats 10 for o | o | o

[drifting wraith]

pretty sure

only ~o | o is left

[hasty perch]

which is 10

no

20

[drifting wraith]

it's probably 20

yes

[hasty perch]

so we have 10 + 20 + 10 + 5 + 5

[drifting wraith]

oh, if you do ~a | a

that's like a new form

it's not equivalent to others

[hasty perch]

true

so that adds another 5?

so in total 55

[drifting wraith]

no 1

[hasty perch]

what about -b | b

[drifting wraith]

that's equivalent to ~a | a

[hasty perch]

oh yeah

always true

what about --a

[drifting wraith]

that's a

already have that

[hasty perch]

ok

so in total 51

[drifting wraith]

i guess

i don't know if there's a more robust way

[hasty perch]

mm

seems logical haha

[drifting wraith]

the task could also mean you can use any number of each

now that's a challenge, i don't know how to do that

we missed 30

[hasty perch]

huh?

[drifting wraith]

let me count again

~o            5
o             5
o | o        10
o | o | o    10
o | ~o       20
a | ~a        1

30 would be 3 connections, doesn't count

well good job

[hasty perch]

so 51 is fine?

[drifting wraith]

i don't know

looks solid, but i'm used to be more certain

[hasty perch]

maybe we can ask someone else

[drifting wraith]

good idea

[hasty perch]

what is this sign?

[drifting wraith]

it's usually followed by mod something

[hasty perch]

yeah mod x

[drifting wraith]

means they are equal mod x

[hasty perch]

so 15y mod x = 6z mod x?

[drifting wraith]

yep

[hasty perch]

cool thx

[drifting wraith]

@hasty perch i know what we missed

~(a | b)

10 more

[hasty perch]

ahhh

61

how confident are you with that

[drifting wraith]

i'm more confident that it's 61 than that it's what the task meant

80% confident

[hasty perch]

yeah lol

[drifting wraith]

i'm starting to believe it didn't mean this

it meant you can use any number of connectives

[hasty perch]

what do you think it meant

hm?

[drifting wraith]

yeah

[hasty perch]

it says 2

[drifting wraith]

it shouldn't have said 2

[hasty perch]

whys that

[drifting wraith]

all connectives can be represented with combination of negation and disjunction

it makes just as much sense to ask that

you infer that it means all connectives period

and count all statements

[hasty perch]

should I clarify what the question means

[drifting wraith]

if you can

[hasty perch]

what should i ask?

[drifting wraith]

if you can use more than 2 total

[hasty perch]

ohhh so do you think ur limited to only using 2 negation and disjunction

but can use as many connectives not including those 2?

[drifting wraith]

no

[hasty perch]

is that what I need to find out

o

[drifting wraith]

you can't use other connectives but you're not limited to using 2

[hasty perch]

but it says ur limited to 2

im confused now

[drifting wraith]

it just say 2 as a number

uf it said two as a word

you would be less convinced

[hasty perch]

oh so we arent sure if it means the only two you can use are negation and disjunction

[drifting wraith]

if you can ask, you should ask

no we're sure

we definitely can't use others

we're not limited to 2

[hasty perch]

i see now

[drifting wraith]

it just says 2 as a figure of speech , "i will now list what you're allowed to use"

two: negation and disjunction

it's very plausible, you should ask

[hasty perch]

It means that you can build formulas using only those two logical connectors. Same with the propositions, there are 5 propositions available to build formulas (not 4, not 6) 🙂 So, the logical language has 5 propositions and 2 connectives that one can use to build various formulas.

[drifting wraith]

ok

[hasty perch]

This question has already been asked

[drifting wraith]

language

100% certain we can use a lot

when the language has two vowels, you can use a lot of vowels in a word

that sort of thing

[hasty perch]

so now how do we solve

seems like there can be a lot more possibilities

[drifting wraith]

it's 2^(2^5)

https://math.stackexchange.com/questions/3382747/how-many-logical-connectives-are-possible-involving-n-simple-propositionsp1-p2

if you can use anything, you can build anything, negation and disjunction allow you to build any connective

[hasty perch]

yeah i see

so it is 2^32

[drifting wraith]

4294967296

[hasty perch]

great thank you

do you feel like helping out with anything else?

Ive got 4 more questions Im somewhat stuck on

[drifting wraith]

sure

[hasty perch]

I think i is 0

idk the rest

[drifting wraith]

well powerset has cardinality of 2^n

how many elements in A?

[hasty perch]

4

[drifting wraith]

totally

[hasty perch]

5?

[drifting wraith]

yes

yes 4

xD

[hasty perch]

o

so 2^16

[drifting wraith]

2^16 yeah

2 marks for that?

[hasty perch]

1

[drifting wraith]

m × 2^m = 384

i don't know 🤣

guessable

[hasty perch]

so

is X the cartesian product?

[drifting wraith]

yes

[hasty perch]

and is the cardinality of A X B

|A| X |B|

[drifting wraith]

product of |A| and |B|

[hasty perch]

ok

so what is iii

6

[drifting wraith]

it's the solution to m × 2^m = 384

[hasty perch]

6

[drifting wraith]

absolutely

[hasty perch]

is iv 23?

coz 23 is prime

[drifting wraith]

yes

[hasty perch]

Now this one Im fairly certain A B E F are true

but Im not sure how to solve for C D with the negated quantifiers

[drifting wraith]

i'm bad at notation too

[hasty perch]

oh rip its fine then

[drifting wraith]

well, it's probably only first is negated not the whole thing

[hasty perch]

dont i flip the propositional statement

[drifting wraith]

oh but that's the same result

it's like, how many times do you flip it at the end

even or odd

i think

[hasty perch]

dont worry about that one

[drifting wraith]

sure

[hasty perch]

how are you with big o complexity classes?

[drifting wraith]

im ok :)

pff

[hasty perch]

haha what

easy or nah

[drifting wraith]

D,E,G

[hasty perch]

lmao

[hasty perch]

ahh

i see

[drifting wraith]

ok

so n^7 also works

[hasty perch]

CDEG

[drifting wraith]

you drop the small unnecessary terms

leaving n^6 log(2n)

what does that 2 do

[hasty perch]

alot

more than the others?

[drifting wraith]

n^6(log(2) + log(n))

ok, so eventually it disappears

n^6 log n

[hasty perch]

i see

[drifting wraith]

and then everything above is also "correct"

[hasty perch]

why not F?

[drifting wraith]

it's a constant, the bottomest of all

[hasty perch]

ok is ee

see

so the correct answer is A-G

as the question asks for all correct ones

A - E i mean

[drifting wraith]

no cdeg is correct

[hasty perch]

ah ok

why factorial

thats not apart of the equation?

[drifting wraith]

it's in the chart at the top

O(n!) is larger

and everything larger counts

[hasty perch]

i see

ok cool i get that thank you 🙂

Now...

I have one more

It is a big one

[drifting wraith]

well that scared me

[hasty perch]

i warned you

[drifting wraith]

i can tell 6 is wrong

it's the wrong letter

[hasty perch]

i see that

[drifting wraith]

no

i'm not even sure

sorry, too hard

[hasty perch]

thats fine thanks anyway

[hasty perch]

Is this a proof by cases? @drifting wraith

[drifting wraith]

no

it's a proof by contradiction

oh wait

[hasty perch]

ooo

[drifting wraith]

is thast considered contrapositive

no, it's contradiction

contrapositive is just disproving negation

[hasty perch]

yeah true

A is a subset of B, does a proper subset mean that |B| > |A|

but if A = B then it is not a proper subset

[drifting wraith]

exactly

there's like no standard symbol for it

$\subsetneq$

[warm shale]

frownyfrog

[hasty perch]

⊂

[drifting wraith]

like it would be this but i've never seen it

[hasty perch]

⊂ whats this then

[drifting wraith]

ah true, i guess

[hasty perch]

oh cool

First glance I feel this is true

The proof should be pretty easy

[errant zealot]

https://forms.office.com/pages/responsepage.aspx?id=ErFxx2Bzb0-u95WYFbiKFAhehWPf9UJAjnXjVMvgkxFURU44UEUyQjY5VEc0Vk9XTVVQMDhFS0U0QiQlQCNjPTEu

can someone help me analyze what to do

or answer everything

[hasty perch]

.close

[obtuse pebble]

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[timid silo]

Someone sent this to me

Is this right?

[royal basin]

their handwriting is terrible

also not clear whether they meant cos(x^2) or cos^2(x)

the work is correct only in the latter case

[timid silo]

Its certainly Cos(X^2)

[royal basin]

then they are wrong.

[timid silo]

How so?

[royal basin]

please do not reply-ping me.

$\frac{1+\cos(2x)}{2}$ is $\cos^2(x)$, NOT $\cos(x^2)$.

[warm shale]

Ann

[timid silo]

But identities can certainly be used if they are corect to replace the inside of an integral before integrating

So its simply a case of the identity used was incorrect?

[timid silo]

?

[royal basin]

you just reply-pinged me AGAIN, and in response to the very message in which i asked you not to.

anyway, the identity used is indeed incorrect, and i have already said this in as clear language as i could.

[timid silo]

Btw thanks for this

.close

[obtuse pebble]

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[royal basin]

you're welcome but you really could've just NOT reply-pinged me after i asked you not to and then pointed it out to you a second time

[obtuse pebble]

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[storm rose]

Why is there a l in the final answer for -2?

[summer spire]

you need to give more context

[storm rose]

umm, its about the sign in irrational funtions

this is the exercise

So its looking for the zero points of the equations

[summer spire]

right

[summer spire]

its a asymptote

so it wont reach -2

[storm rose]

ok ty

.close

[obtuse pebble]

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[summer spire]

glad to help

[obtuse pebble]

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[orchid root]

|5-x|+|3x-1|>=x+2

[orchid root]

nope

okay

then

so we take two cases ?

okayy i almost got it

thankyou (:

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[obtuse pebble]

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[indigo oriole]

anyone know what the T is referring to?

[warm canopy]

Transpose

[indigo oriole]

what does that mean, is it to just A x A?

[warm canopy]

Swap rows and columns

[versed turret]

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

[versed turret]

^ A_ij, after transpose = A_ji

[indigo oriole]

oh cheers

got it

for this is it just timesing the 2 like any other matrices?

[somber mural]

mxn X nxp = mxp

[indigo oriole]

so the anwer would be 7 x 3?

[somber mural]

yes

[indigo oriole]

cool

then if you do BA it would be the opposite?

[somber mural]

you cant do BA

it is 5x3 X 7x5

the 3 and 7 are not the same

you can only multiply if columns on left = rows on right

[obtuse pebble]

@indigo oriole Has your question been resolved?

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[obtuse pebble]

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[halcyon stag]

Let S be a nonempty subset of real numbers which is bounded below. Let -S denote the set of all real numbers -x, where x belongs to S. Prove that inf S exists and inf S=-Sup(-S)

[halcyon stag]

This is what I’ve done but I’m not sure if it’s correct

[obtuse pebble]

@halcyon stag Has your question been resolved?

[charred plume]

The ideas all look fine. Maybe one or two minor issues with exposition. For example, right at the beginning of the proof, it would be nice to clarify that $l \le x$ for all $x$ in $S$ -- as is, it almost feels like you're saying $l \le x$ for some specific $x$.

[warm shale]

daveamayombo

[halcyon stag]

Ah yes gotcha

Thanks

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[hasty perch]

What is this in CNF, really stuck