help-23

[hazy elbow]

From here three cases
x<-3
-3=<x<1/2
x>=1/2

[blazing roost]

oh this marking critical points

on number line

oh yes ik this

[hazy elbow]

I'd say use this

[blazing roost]

if i use this

then this

and this

is very long

i am asking that can we do it by a short methos

[hazy elbow]

I don't understand that process

[blazing roost]

if can you solve this question then pls tell is it long or not

[hazy elbow]

Pretty short if you try
Just three inequalities and their union

[blazing roost]

can you pls send pic of solution

[old grail]

i guess its 4 inequalites, we should try... if we consider +/- for both

[blazing roost]

true

[old grail]

at first keep x-3 as it is the 2x-1

[blazing roost]

?

[old grail]

then go with -(2x-1)

[hazy elbow]

can you write all 4 inequalities

[hazy elbow]

That is not allowed

[old grail]

wait let me write ...then u shall check it

[blazing roost]

honestly

[old grail]

x-3 >2x-1 equation 1

[blazing roost]

i was looking for someone who could tell me a shorter approach

i can make that cases too

and waste time

thanks

@old grail

@hazy elbow

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[lusty spear]

hi everyone. I have a quick question on discrete random variables. Am I correct with saying that for a discrete random variable X and k in X (Omega) with Omega being the set of all results,
if P(X=k) = 0 then k not in X(Omega) ?
I have this feeling that this statement is not true since k could be in X(Omega) , but it wont ever happen. But then I think like we could put everything in X(Omega) and then X(Omega) would be infinitely big consisting of a bunch of useless k's that will never happen
I'm not sure so maybe someone can help me with this. Thanks in advance

[safe radish]

@lusty spear Has your question been resolved?

[lusty spear]

<@&286206848099549185> I'd appreciate your help a lot

[quasi bison]

if P(X=k) = 0 then k not in X(Omega) ?
no

you can only say that if you know Omega does not have any nonempty null events

[lusty spear]

what do you mean by Omega does not have any nonempty null events?

[quasi bison]

a null event is an event whose probability is 0

[lusty spear]

but in Omega there would only be the results right ?

[quasi bison]

ok, i should probably have not said Omega

instead i should have said whatever letter your class uses for the event sigma-algebra

since that is where the events "live"

[lusty spear]

We used whatever letter

we just say like

A subset of Omega

[quasi bison]

no, not a particular event

but the collection of all subsets of Omega that are events

[lusty spear]

I cant remember any particular letter for that in our course

wouldnt that simply be the power set of Omega ?

[quasi bison]

no, not always.

[lusty spear]

alright , and you say that only if I am sure that there are no empty events possible , the initial statement is true ?

wait but nonempty null event confuses me

[quasi bison]

❌

a nonempty null event is an event that is null (has probability 0) but isn't empty

[lusty spear]

oh I see

[quasi bison]

example: you generate a random real number uniformly distributed between 0 and 1. what's the probability you end up with exactly 0.42069?

[lusty spear]

I understand

hmm alright, I'll take that. thanks a lot !!

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[lean otter]

I have to show that this funciton is uniform continous

For 0 its pretty obvious

But for the x>=0 I don`t know what kind of an estimation I should use

[viral loom]

What is the definition of uniform continuity?

[lean otter]

|x-y|<Delta --> |g(x)-g(y)|<epsilon

and if i can find such delta then it is proven

[viral loom]

Can you describe it with your own words?

[lean otter]

I can show u an example i did before

[viral loom]

I'd like first to know that you know the real meaning of uniform continuity, that's why I asked it.

[lean otter]

ok wait i have to trnslate

I dont know if u asked this but normal continuity has like only one key for one door and uniform continuity has one key for all doors metaphorically said

[viral loom]

Which one would be continuous here and which one uniformly continuous?

[lean otter]

I suppose red is uniform

[viral loom]

Ok. Can you compute the derivative of g(x)=x(e^x) for x≥0?

[rapid igloo]

there's no way this is uniformly continuous

[lean otter]

We cant use that

We havent introduced derivates

[viral loom]

It is, the red one.

[rapid igloo]

sure, but not x -> x e^x

[lean otter]

Now im confused

blue or red?

[viral loom]

Well, I'm asking questions first. Solve the exercise later.

Red is uniformly continuous, blue is not uniformly.

[lean otter]

derivate (1+x)e^x

wait

nooo

[viral loom]

Take your time.

[lean otter]

now first one was wright

[viral loom]

Yes, that's the derivative.

What does the derivative tell you about the behavior of the function?

[lean otter]

extremal

[viral loom]

Explain.

[lean otter]

x=-1 must be extremal point

[viral loom]

Can you use the derivative to show that the function is uniformly continuous for x≥0?

[lean otter]

No I think not

[viral loom]

x=1 is not an extremal point for this function. The derivative g′(x)=(1+x)e^x is always increasing for ≥ 0
x≥0, meaning the function g(x)=xe^x does not have a maximum value for x ≥ 0. the maximization operation shows that the maximum is not attained at any point.

[lean otter]

I corrected it to -1

[lean otter]

Yes

No extremal for x>=0

[viral loom]

Since the derivative is increasing, the function is monotically increasing. This means that for any two points x and y with x<y, we have g(x)≤g(y). This property can be used to show that the function is uniformly continuous.

[lean otter]

I understand

[viral loom]

You can use the definition of uniformly continuous for x ≥ 0 to show that

Do you have any other question about this exercise or uniformly continuity?

[lean otter]

So I just show it by putting the fuctions in the definition of uniformly cont

[viral loom]

That's correct.

[lean otter]

And then find an estimate

[lean otter]

Well I did that but I cant find any estimate

[viral loom]

Let ϵ>0 be given. We want to find a δ>0 such that if ∣x−y∣<δ for any x,y≥0, then ∣g(x)−g(y)∣<ϵ.

Note that g(x)=xe^x is differentiable for x≥0, and its derivative (x)=(1+x)e^x is always increasing for x≥0. This means that the function g(x) is monotonically increasing for x≥0. (This is what we had).

for any x,y≥0 with x<y, we have g(x)≤g(y). This means that ∣g(x)−g(y)∣=g(y)−g(x).

Now, consider the difference g(y)−g(x)=ye^y − xe^x . By the Mean Value Theorem, there exists a c between x and y such that g(y)−g(x)=g′(c)(y−x).

Since g′(c)=(1+c)e^c is always increasing for c≥0, we can bound g′(c) by g′(y)=(1+y)e^y for x<y. Therefore, g(y)−g(x)≤g′(y)(y−x).

We want this to be less than ϵ, so we choose δ=ϵ/g′(y) for y≥0. Then, if ∣x−y∣<δ, we have ∣g(x)−g(y)∣≤g′(y)∣x−y∣<g′(y)δ=ϵ.

Therefore, for any ϵ>0, we have found a δ>0 such that if ∣x−y∣<δ for any x,y≥0, then ∣g(x)−g(y)∣<ϵ. This shows that g(x)=xe^x is uniformly continuous for x≥0.

Take note of this and read slowly and with time, no need to rush.

[rapid igloo]

δ can't depend on y

[lean otter]

Hmmm but we dont know derivates

We cant use them

My idea was doing it buy an estimate

[viral loom]

That's true, too much writting for nothing

[rapid igloo]

in fact if a function f is uniformly continuous on [0, inf) you can show that there exists A, B such that |f(x)| <= Ax + B

and the issue here is e^x

[lean otter]

Can u explain pls

like i had a function before

I used estimate in purple

My question is

Is there any estimation so i can extract the x and y from the exponential terms so that I can generate a Delta

Like |xe^x - y e^y| --> |x-y| ...

Ok I've got it thank you @viral loom and @rapid igloo

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[gray sable]

6 - 2/3(x+5) = 4x find the value of x

[gray sable]

. Close

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[wind kite]

why am i still single

[worthy hemlock]

Not the purpose of a help channel

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[velvet kettle]

This is actually one of the millennium problems which are yet to be solved

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[deep creek]

looks like a gre question lol

anyway what did you try?

[tulip niche]

Looks like something from the west definitely

[green flare]

Yeah its Gre style questions

[devout shale]

find the percentage of measurements below 60, and also the percentage of measurements below 62. Then take the difference. For (a)

(b) is the same method

and then compare

[deep creek]

actually, from your work i see that you might've been confused on what the mean is in this context

it's not an arithmetic average like you wrote out

[green flare]

Yes I did

Okay I see

[deep creek]

if you know what a normal distribution is and how to draw a quick graph, you can answer this easily

[green flare]

Oh it says normally distributed

Got it

[stoic dune]

[green flare]

Did I draw this correctly? I guess it should be A then because it falls in 13.5% which is greater than. 2.35%

[deep creek]

yeah, nice job

you don't need to compute the area between the two values

intuitively 60-62 is bigger since it's closer to the mean

[green flare]

Okay cool

Yeah

Sounds good

Thank you!!

[deep creek]

np, glad i could help

[green flare]

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[marble obsidian]

(\forall x (i(x) \to {(\exists y \overline{i(y)}) \land R(x,y)}))

[flat frigate]

metnal

[marble obsidian]

why are the curly brackets there? do they denote a set in this case?

cuz i dont think the expression inside them generates multiple values?

its just a boolean value

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[formal plume]

Hello! I'm learning about cartesian vectors in class and I'm not quite sure if I answered the question correctly. I've attached an image of my work below and I'd like feedback if possible

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[lean otter]

hello i did the first question but i cannot do the second or third one i thought the second was ab^24 but that is not correct

[green bane]

!show

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[lean otter]

4 times 2 is 8 times 3 is 24

ab^24

[green bane]

B?

[lean otter]

for B yes

[green bane]

you do the power not togeter

so its like 4^3 a^3 (b^2)^3

[lean otter]

so the answer is 12 a^3 b^6?

[green bane]

4^3

not 12

[lean otter]

it is wrong

[green bane]

you do 4^3

simplify

[lean otter]

is 64

[green bane]

yeah

so

[lean otter]

64a^3b^6?

[green bane]

ye

[lean otter]

alr cool

[green bane]

i hope

[lean otter]

it is still wrong unfortunately

[green bane]

uh

[tall bough]

which one

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

i still do not have the answer

[lean otter]

b or c

[tall bough]

whats your answer

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[static yacht]

Hello, I need to find the inverse of B:
There is a hint: You don't have to calculate a lot

[static yacht]

What is the smart method, that the hint is suggesting? I started by writing it up as an augmented matrix with I on the right, but this take 8 x 4 numbers pr. row operation

If it's any help for context: B is an ortonormal basis

Answer: Since the matrix is ortogonal, the transposed matrix is the inverse.

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[unborn garnet]

uh

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[ruby relic]

Is anyone able to Explain how to solve this problem?

[ruby relic]

A noble villa is secured by an alarm system. In the event of a break-in, there is a 99% chance of an alarm. However, a false alarm must be taken into account with a probability of 1% if no burglary takes place. The probability of a burglary is about 1 in 1000 per night. Indicate the probability that a burglary will actually occur in the event of an alarm.

[tulip niche]

!show

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[tulip niche]

Hint:tree diagram proly

[ruby relic]

I had this but since It was wrong I did thaf

[graceful quail]

Why so mean to your beautiful workings?

[ruby relic]

A: Burglary
B: Alarm

[ruby relic]

Because it's wrong

[graceful quail]

🤮

[ruby relic]

?

[graceful quail]

Why not A- Alarm, B- Burglary

I would get confused if I changed the first digit case by case

[tulip niche]

Don't use formula

First

Use tree diagram first then formula

[ruby relic]

Didn't know how to start so I just did formula

[graceful quail]

If you are struggling with a question that includes conditionals I would always recommend a tree diagram

[ruby relic]

Bc i need to calculate with Bayes that a Burglary happens when the alarm rings

[ruby relic]

I didn't know how to start one

I tried a Four-field table

[graceful quail]

Id make a tree diagram with the first part being Burglary (0.001) and No Burglary (0.999) and then a second layer with Alarm and No Alarm

From there you can do some calculations

[tulip niche]

Bayes theorem basically

Reverse probability

Wherein u go backwards along the tree

Divide the required branch by all branches mentioned

[graceful quail]

Mhm

[ruby relic]

Is that the solution?

I need to calculate the probability that a Burglary happens when the alarm rings

So shouldnt alarm be The Condition

So the first one

[graceful quail]

P( B | A ) = P( A & B ) / P( A )

Is the relevent formula, I believe

[ruby relic]

I know that this is Bayes' theorem

[graceful quail]

Okay

[ruby relic]

We have P(Alarm|Burglary) which is 99%

I have the the solution now but I don't know how to start with the Tree diagram and four-field board

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[ruby relic]

.reopne

.reopen

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✅

[ruby relic]

<@&286206848099549185> I have this now

[vague frost]

!15m

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[ruby relic]

E=Burglary A=Alarm

[ruby relic]

Look up the last time was 20.06

It's 16 Minutes

[vague frost]

u reopened it

so rewait

[ruby relic]

🧐

.close

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[hasty frigate]

what is the process

[lean otter]

do you know the statement of MVT?

[hasty frigate]

no

what is it

[paper veldt]

if a function f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point c in the interval (a,b) such that f'(c) is equal to the function's average rate of change over [a,b]

[tall bough]

if u dont know something it would be wise to google it

[lean otter]

(especially if its part of the question)

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[coarse kelp]

$\vec{R}'(t) = \langle -3 \sin t, 4, 3\cos t \rangle
\\s = \int_{0}^{t} \sqrt{(-3\sin u)^2 + 4^2 + (3 \cos u)^2} du
\\s = \int_{0}^{t} \sqrt{9 \sin^2 u + 16 + 9 \cos^2 u} du
\\s = \int_{0}^{t} 5 du
\\
\\s = 5t
\\
\\t = \frac{s}{5}
\\
\\ \vec{R}^{*}(s) = \langle 3 \cos \frac{s}{5}, \frac{4}{5}s, 3 \sin \frac{s}{5} \rangle
$

This is what I did in letter A

can someone help me in letter b

no idea how to do it

[light shoal]

can you find the value of t such that the arc length from 0 to t is 10? (I assume that by "10 units away" they mean measuring the distance along the arc, not the straight-line distance)

[flat frigate]

steamhahasteamhaha

[coarse kelp]

$s = 10 = 5t$

[light shoal]

right, so t = 2

so just plug that into the formula for R(t)

[flat frigate]

steamhahasteamhaha

[coarse kelp]

ohh

thanks

that is easy

[light shoal]

again, that's assuming "10 units away" means distance measured along the curve, not the straight-line distance between two points in 3d space

which is a reasonable assumption here, but they stated it poorly

[coarse kelp]

btw I got another question

given a point

how do I find out t_0

look at this

ok so its easy to say t_0 = 0

[coarse kelp]

but in here idk

[light shoal]

well in this case hopefully it's clear that t_0 = 1 (just plug in t=1 to verify)

[coarse kelp]

what how

[coarse kelp]

why 1

[light shoal]

what do you get if you plug in t=1 into R(t)?

[coarse kelp]

oh

ohhhh

[light shoal]

in general you could set the x coordinate of R(t) equal to the x coordinate of the starting point and solve for t

and if that gives you multiple solutions, do the same for the y coordinate etc

until you find a single t value that works

so to illustrate that here, the x coordinate of R(t) is (1/3)t^(3/2), and the x coordinate of the starting point is 1/3

[coarse kelp]

yeahh

[light shoal]

so you'd set $\frac{1}{3}t^{3/2} = \frac{1}{3}$ and solve for t

[flat frigate]

Bungo

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[gritty glacier]

siu

[gritty glacier]

$\int \frac{2-3\sin x}{\cos^2 x} \cdot dx$

[flat frigate]

bettim

[gritty glacier]

how to do this

[thin bridge]

split the fraction
bring in sec and tan

[gritty glacier]

oh

[thin bridge]

(don't tell me that you had sec and tan to begin with and you manipulated that into this)

[gritty glacier]

$2\tan x - 3\sec x + C$?

[flat frigate]

bettim

[gritty glacier]

hehe no no

[thin bridge]

yes

[gritty glacier]

thanks

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[ember bone]

Can anyone help me in solving this? Determine complex numbers α and β which satisfy:
,tex $$
\begin{pmatrix}
1 \ 1
\end{pmatrix} =
\alpha \begin{pmatrix}
\mathit{i} \ 1
\end{pmatrix} + \beta \begin{pmatrix}
\mathit{-i} \ 1
\end{pmatrix}
$$

[flat frigate]

MikoFiko

[quasi bison]

!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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[ember bone]

1. I have no idea where to begin, i plugged it into wolfram but i dont know how it gets the result for the complex solution

[safe radish]

@ember bone Has your question been resolved?

[safe radish]

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[vivid vortex]

hello

[vivid vortex]

so the function above is given

a)prove any primitive is positive

b)

i know how to solve a and b

however,I am stuck at c)

I need to prove this inequality

[safe radish]

@vivid vortex Has your question been resolved?

[vivid vortex]

<@&286206848099549185>

[woven latch]

oh my

im in 8th rn

srry cant

[vivid vortex]

haha,don't even worry

[woven latch]

😐

[vivid vortex]

it gets more and more complex

[woven latch]

i am learning calculas

rn

[vivid vortex]

so enjoy being an 8th grader

[woven latch]

haha ikr

[vivid vortex]

cool

[vivid vortex]

yeye:)))

[woven latch]

not in school

online lmao

[vivid vortex]

that's bad yeah

[woven latch]

just starting to understand functions, derivatives and integrals

[woven latch]

NO i mean form youtube

[vivid vortex]

cool,ngl

[woven latch]

yee

its fun

i like maths

[vivid vortex]

yep

[woven latch]

ok cya i am doin maths rn lol

[vivid vortex]

cool

enjoy

[woven latch]

haha ty

[woven latch]

r is real number

[vivid vortex]

yep

[woven latch]

like pi, e, golden number etc

[vivid vortex]

no,it's the set of real numbers

[woven latch]

OH right

R = {..., -√2, -½, 0, 1, ⅘, pi,….}

[vivid vortex]

exactly

[woven latch]

yeee

[trim swan]

Maybe there's a more elegant solution but couldn't you just evaluate the integral?

Or does it come out in a form that is not obviously between those values

Sorry I don't have paper in front of me, just a suggestion

[vivid vortex]

I understand

i have kinda got to sth

[vivid vortex]

artifices

and such

[vivid vortex]

I am not sure if it's correct

I will send it

@trim swan can u see it?

[safe radish]

@vivid vortex Has your question been resolved?

[summer raft]

if $m\leq f(x)\leq M$ for $a\leq x\leq b$ then $m(b-a)\leq\int_{a}^{b}f(x)\mathrm{d}x\leq M(b-a)$ it is suffice to compare the function with the two given numbers on either side of the inequality on the interval $x\in\left[-3,-1\right]$ @vivid vortex

[flat frigate]

Stumpman

[safe radish]

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[vivid vortex]

Yes, I did sth similar

[vivid vortex]

.close

[safe radish]

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[thorny thorn]

Hello I need help on this

[peak estuary]

we dont help with quizzes

[thorny thorn]

sorry i misspelled

i didnt mean quiz

i accidentally wrote quiz

and no this isnt a quiz

[patent vine]

well in that case

Just use the sum of the angles

[thorny thorn]

you mean like this

70+95+105

like that

[woven swan]

the sum of the angles of a quadrilateral is 360 degrees

because you know three angles you can deduce the last one

[safe radish]

@thorny thorn Has your question been resolved?

[safe radish]

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[plucky elk]

What inner product

Which is what?

Write a formula

[safe radish]

@lilac totem Has your question been resolved?

[plucky elk]

What?

What part was I rude

Oh are you calling yourself rude? Nah you aren't

Do you know what these $\phi_i$ are?

[flat frigate]

riemann

[plucky elk]

Exactly what though?

Just distinct

Can you show the original problem where this is coming from

Screenshot or picture is best

This inner product does not look right for your vectors

These are usually some integral

This doesn't have anything you're mentioning in your problem. Just show the original non english book

Different books have different notations and you're just making it harder by introducing new resources

[safe radish]

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[minor sphinx]

Is this right so far

[minor sphinx]

<@&286206848099549185>

[hard crest]

I wouldn't bother translating the bounds; just leave them as x=0 to x=4 and put your antiderivative back into terms of x

seems good, now apply that trig identity to simplify the denominator

[vague frost]

u could also j do a substitution of the x^2 +9

[minor sphinx]

wdym

[vague frost]

u=x^2 +9

[minor sphinx]

im not following'

[vague frost]

u=x^2 +9

u can look at the x^3 at the top as x*x^2

[minor sphinx]

alright let me try that

like this?

[plucky elk]

Last line turned wrong

but u= and du= lines are right

[minor sphinx]

forgot the x

[vague frost]

oh ye theres an x missing

[minor sphinx]

would i do IBP now

[vague frost]

x^2 should be turned in terms of u

and ur bounds

[minor sphinx]

so x^2 would u-9

[vague frost]

yeah

[hard crest]

and then just split it

[minor sphinx]

and for the bounds do i plug in x into u=x^2+9

[vague frost]

yea

0 and 4

[minor sphinx]

I’m stuck

[vague frost]

this is wrong

u need to distribute

then if theres a - or a + u can separate

[minor sphinx]

So I can separate it now

@vague frost

[vague frost]

yeah

[minor sphinx]

ty

[vague frost]

np

[minor sphinx]

How does this look?

[vague frost]

looks fine but why do u only have 1 square bracket ?

[hard crest]

are those coefficients right? $\frac12 \int u^{-1/2} \d u = \sqrt{u}$, not $\frac14 \sqrt{u}$

ok I have \d mapped to mathrm{d} but apparently this doesn't

[minor sphinx]

thats what my prof does when we get to the last step

[flat frigate]

kitten.in.a.teacup

[vague frost]

should be a |

not ]

u can use [ ]

[minor sphinx]

oh alright

[minor sphinx]

i multiplied it by 1/2 when i did the antiderivative

[hard crest]

yeah when you take the antiderivative of a polynomial you divide by the (new) exponent

[vague frost]

oh shit yea it should cancel

im blind

[minor sphinx]

wait so it should only be u^1/2 because when you do the reciprocal u cancel out the 1/2

[hard crest]

yeah

if you're ever not sure just take the derivative to check that it's right

[minor sphinx]

alright ty

I got this but after I calculated it I got the wrong answer

[hard crest]

the 9/4 is also wrong for similar reasons

[minor sphinx]

o let me fix that and try

[minor sphinx]

would it just be 9?

[hard crest]

yeah

[minor sphinx]

ty

im still getting the wrong answer its supposed to be 4/5 but im getting 15.1

[safe radish]

@minor sphinx Has your question been resolved?

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[minor sphinx]

.reopen

[safe radish]

✅

[minor sphinx]

@hard crest @vague frost

[hard crest]

oh sorry i got distracted

[vague frost]

what ?

[minor sphinx]

im stuck getting the wrong answer

[vague frost]

show

[minor sphinx]

supposed to be 4/5 but im getting 15.1

[hard crest]

9 and 25 describe values of u, not of x

you don't need to backsubstitute since you already brought your limits forward

[minor sphinx]

omg

thank you

[hard crest]

btw it took me three tries to actually get the right answer as well because i've lost the ability to accurately keep track of negative signs it seems

[minor sphinx]

lmao i have the same issues

i always seem to forget a number somwhere also

but thanks for the help both of you i really appreciate it

.close

[safe radish]

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[safe radish]

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[atomic jackal]

im only getting 1 solution

im supposed to get 2

nvm

.close

[safe radish]

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[vivid vortex]

how would u guys solve this one?

[brazen helm]

hmm let me see

[royal horizon]

I got this from from maxima:

[vivid vortex]

what's that?

[royal horizon]

The solution in terms of x

I don't know if this will help though

[vivid vortex]

these r 3 positive integers

[vivid vortex]

how did u do that?

[royal horizon]

Using a computer

[vivid vortex]

is there a theorem?

oh

[vivid vortex]

that's supposed to come from my brain

idk how much of that i could've reproduced

i was thinking of matrices

[royal horizon]

Matrices will only help if it was a linear equation

[vivid vortex]

oh yeah,right

[royal horizon]

Let's just focus on what $x$'s makes $z=-\frac{589x-6474}{x+1}$ an integer

[flat frigate]

Math Is Fun

[vivid vortex]

maybe,the fact that it's not rational

[royal horizon]

We also know that $10\geq x\geq1$

[flat frigate]

Math Is Fun

[vivid vortex]

how so

[royal horizon]

Because x is a positive integer

and z is positive

[vivid vortex]

why can't it be also 0?

[royal horizon]

I tested one by one using a calculator and got x=6

[vivid vortex]

and why not greater than 10?

[royal horizon]

Find all triples x,y,z of positive integers

[vivid vortex]

yeah ,but 9 is positive too

[royal horizon]

I mean, the fact that x is positive implies it's greater than or equal to 1

And the fact that $z$ is positive implies $-\frac{589x-6474}{x+1}\geq1$

[vivid vortex]

0 is positive too,why not?

[flat frigate]

Math Is Fun

[royal horizon]

The set of positive integers is defined as the set of integers such that it's greater than 0

[vivid vortex]

oh,really?

okay

[royal horizon]

A positive number is any number greater than 0. [From https://byjus.com/question-answer/what-is-a-positive-integer/]

[vivid vortex]

yeah

what ab 10 is greater

10

[royal horizon]

Solving the inequality gets what we want, namely x≤10

[vivid vortex]

that's what i was asking ab

why that 10

im sorry if my question's stupid

[distant lichen]

( i got 6,420,9, correct?

[vivid vortex]

idk,i dont have the answers to this

[vivid vortex]

how did u solve it?

[royal horizon]

Yes

[vivid vortex]

im so confused

[distant lichen]

@royal horizon 's method, trying out z's by this

[gaunt garnet]

Is using programs allowed in the question

[vivid vortex]

nah

[distant lichen]

apparently no

[vivid vortex]

why's x<= 10?

[distant lichen]

this

[distant lichen]

and this

[vivid vortex]

why's it an integer for x<=10 then?

im just trying to learn

there r so many things we haven't been taught in school

[royal horizon]

Can you solve the inequality above?

[vivid vortex]

ofc

i just don't understand the <=10

[royal horizon]

you got <=10.9... right?

[vivid vortex]

its origins

is an integer supposed to be <=10??

[royal horizon]

I mean you got x≤10.99151103565365... right?

From the inequality

[distant lichen]

It's not. we're only trying to know the boundary of x so we know what to try

[vivid vortex]

that's what perplexed me

cus i was like,nah

[vivid vortex]

and that boundary,how did u figure it out?

[distant lichen]

solving this inequality

[vivid vortex]

ohhhhhhhh

sorry

yeah,okay

now i saw this one

im so sorry

i just didn't see this phot

photo

okay

and from there on i basically figure out the other values as well

and see which ones fit,right?

[raven heart]

honestly we don't need maxima here

there's much easier than that

[vivid vortex]

please,go ahead

[raven heart]

the first equation is a bit of a pain to use

one thing we see though is that x+1 should divide z

(cause z/(x+1) has to be an integer)

just a little sanity check for later

the second equation is much more interesting

[royal horizon]

Yeah I just noticed that

[raven heart]

xz + 589x + z

that thing is almost factorable

if you add 589 to it, we get xz + 589x + z + 589, which is (x+1)(z+589)

now our 2nd equation becomes (x+1)(z+589) = 6474+589 = 7063

[vivid vortex]

no way,i did exactly that

[raven heart]

everything being integers here

[distant lichen]

That IS a very good method

[vivid vortex]

and got stuck

[raven heart]

luckily 7063 factors in 7*1009

(1009 being prime)

therefore we either have (x+1) = 7 and (z+589)=1009 or the reverse (which can't work since z would be negative)

you get x and z

then you get y from the first eq

[vivid vortex]

yep

thanks so much

thanks to each one of u

im sorry if i caused u a headache

[raven heart]

it's ok was fun lol

[vivid vortex]

yeah,it seemed like at some point i annoyed since i didn't get answers,not u necessarily

i didn't mean to annoy lol

.close

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[full hazel]

Why is it A^-1 times B instead of B times A^-1? Will it always be like this in these types of problems?

[quasi bison]

$AX = B \implies A^{-1}AX = A^{-1}B$

[flat frigate]

Ann (glomed)

[quasi bison]

if you were to postmultiply by A^-1 you could not cancel AXA^-1

[vivid vortex]

the product of two matrices is not necessarily distributive

[raven heart]

commutative but yeah

[vivid vortex]

yeah

sorry

gosh

don't know why i am so defocused

sorry

basically

A^(-1)B doesn't always imply BA(^-1)

[mortal sandal]

They aren't equal

[vivid vortex]

that's why u multiply to the left side or the right side of the matrix,so to speak

[mortal sandal]

Doesn't always equal

[safe radish]

@full hazel Has your question been resolved?

[full hazel]

Thanks @mortal sandal @vivid vortex @raven heart @quasi bison

[safe radish]

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[vivid vortex]

u r welcome

[safe radish]

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[split grail]

How do i do this intersection?

[tardy mango]

find the y coordinate of any point on the line in terms of k

and ||set that equal to 0||

[split grail]

thanks

.close

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[cold saffron]

What type of problem is this and where can I find more like this? I don’t know how to solve it at all

[tardy mango]

What type of problem is this: idk

However, I can help you solve it

[cold saffron]

That’s fine thanks

[karmic hedge]

This is an algebra problem

[sonic geyser]

Geometry problem using first grade equations

[cold saffron]

A=1/2*bh?

[tardy mango]

Consider the fact that these two are both equal

I leave it to you to justify why

But do you see what you can do with this

[sonic geyser]

That's for the triangle

[cold saffron]

I don’t have a clue

[sonic geyser]

You have this

[cold saffron]

The only thing I can think of is pythogaren theorem but that’s only for solving the right triangle

[sonic geyser]

You don't need Pythagoras

Use equations

One side is 5 units long, and that side is also x + 2 + x units long

[cold saffron]

Squared?

[sonic geyser]

No

One side is one thing and another thing as well

It's an equality

An equation in this case

[cold saffron]

X+2+x=5 then solve

[sonic geyser]

Yes

[cold saffron]

X+x=3

X=3

So the area is 3?

[sonic geyser]

No

The solution of the equation is wrong

x + x ≠ x

And the solution of the problem is not the solution of the equation

Think about what you are really doing

[safe radish]

@cold saffron Has your question been resolved?

[safe radish]

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[lean otter]

hello

[safe radish]

Please don't occupy multiple help channels.

[lean otter]

This was my attempt at this question, but someone told me my solution is lacking still

just wanted to confirm if possible?

[safe radish]

@lean otter Has your question been resolved?

[hard crest]

I don't understand this deduction

[timid pasture]

@lean otter genuine question, where are you getting these questions from?

[lean otter]

its from the discrete math book i am studying

its "Discrete Mathematics and Its Applications by Kenneth Rosen"

[lean otter]

hmm i think i made it a bit unclear with my variable picking i will rewrite one sec

[safe radish]

@lean otter Has your question been resolved?

[finite yacht]

Regardless of the details of your proof, it can't be correct because it does not seem to rely on the fact that a and b are irrational. If it were correct, then the property would be true for a = b = 2, which of course is not

So at some point you have to use that hypothesis

[safe radish]

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[wintry mortar]

i get infinity over infinity

, but the key answer is +infinity, how?

[split ether]

You could use l'hopital's rule to show that

[wintry mortar]

i did lhopital it

it's now reversed

n+4/2sqrt(n)

i did it again and it's nsqrtn

does this work here

thanks

.close

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[wintry mortar]

help

.close

[safe radish]

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[lean otter]

What's the problem?

Okay

[wintry mortar]

.reopen

[safe radish]

✅

[lean otter]

Okay...So what are you having trouble with?

Do you know what is the direct comparison test?

[wintry mortar]

An < Bn

idk how to solve for sin(n)/sqrtn

i removed the 7 so i got an

[lean otter]

so you are trying to do it with

[
\f{\map \sin n}{\s n} \le \f{7+\map \sin n}{\s n}
]

?

[wintry mortar]

?

is there another way

[lean otter]

im asking you if thats what you are trying to do?

[wintry mortar]

yes

[lean otter]

that doesnt help you

[wintry mortar]

why

[lean otter]

[
\sum\f{\map \sin n}{\s n} \textsx{converges}
]

[wintry mortar]

how

[lean otter]

can be proven using Dirichlet's test

[wintry mortar]

what is that

[lean otter]

Look it up? Either way your inequality here doesn't do you much because of the convergence of the series sin(n)/sqrt(n)

[safe radish]

@wintry mortar Has your question been resolved?

[kind fractal]

Is it possible to use the fact that sin goes from -1 to 1 only (?

So you can get a smaller numerator
7 - 3 < 7 - sin(n)

Then you can make the denominator bigger

(7-3)/n

which diverges

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[wintry mortar]

.reopen

[safe radish]

✅

[wintry mortar]

my teacher does this but i don't understand it

[obtuse jackal]

You're comparing to a simpler series

(7-sin n)/sqrt(n) >= 6 / sqrt(n), which you know to be a divergent series

[safe radish]

@wintry mortar Has your question been resolved?

[wintry mortar]

does this also work with cos

[obtuse jackal]

Ofc

Same bounds

[obtuse jackal]

This is true but useless here and far more complicated to prove

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[compact cipher]

i need help

[safe radish]

Please don't occupy multiple help channels.

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@compact cipher Has your question been resolved?

[tardy mango]

#help-6

.close

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[limber drift]

Need a walkthrough:
Arccsc(csc(11pi/6))

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@limber drift Has your question been resolved?

[limber drift]

Lets count sheeps while we wait everybody!

1 sheep

2 sheep

3 sheep

4 sheep

5 sheep

6 sheep

7 sheep

im tired fuck sheeps

ping me when help is here.

Hopefully I dont have to wait another 3 hours like last time.

oh wait i see someone

nvm

bro disapear

$arccsc(csc(11pi/6))$

[flat frigate]

Yamada

[limber drift]

need walkthrough for that.

<@&286206848099549185>

later.

.close

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[potent osprey]

Hello, can someone help me with a complexity theory question I have. In complexity zoo (and several other places), there is a property which says that $ \textnormal{PCP}(r(n), q(n)) \subseteq \textnormal{NTIME}(2^{O(r(n))}q(n) + poly(n)) $ but there doesn't seem to be a proof given. I have found a proof here: https://www.cs.umd.edu/~gasarch/TOPICS/pcp/AS.pdf (p75) but it seems to be wrong since they deterministically go through every possible run of the verifier which should be $\textnormal{NTIME}(2^{O(r(n))} poly(n))$ which may be a bigger set (for example if $q = O(1)$). Anyone know a resource which states a correct proof?

[flat frigate]

DAILI

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@potent osprey Has your question been resolved?

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[timber cave]

Can anyone tell me if I’m correct pls

[hard crest]

an easy way to check these is to just substitute them back into the original equations and check that they equal

you have $4x - 3 = 9$

[flat frigate]

kitten.in.a.teacup

[hard crest]

and somewhere in the kerfuffle you dropped the 4

[timber cave]

One question

When u get the y

U should move to the original equation

Or the one we multiplyed

[hard crest]

any of them will work as long as they have y in them

i usually use one of the original ones

[timber cave]

Si if I want to use20x+15y(-1)=45

it’s correct?

For photo

[hard crest]

y is -1 so you'd just have 20x + 15(-1) = 45

[timber cave]

So now its 20x-15y=45?

[hard crest]

that's your original equation

we know y = (-1)

so rewrite that equation and wherever it says y, write (-1) instead

[timber cave]

ya but if u multiple

it becomes -15y

[hard crest]

what

no

what

start with your original equation, write it down

20x + 15y = 45

okay got it? now, we know that y = (-1)

so replace y with (-1) in that equation

(and write it down for me)

[timber cave]

so x=3

?

[hard crest]

yes x = 3

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[safe radish]

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[delicate apex]

hello

[delicate apex]

i have t his equation

i have v^(2) matrix A and lambda^(2)

however, im not sure what the subscript 2 means on the side

[light shoal]

it means the 2 norm

which is the square root of the sum of squares

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[delicate apex]

ok thanks

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[crude star]

Is this correct

[crude star]

im not really sure

[solar hazel]

no

explain your answers?

[crude star]

for all of them I did if P is true then whats its implying should be true aswell

or should both sides be true

[solar hazel]

i think you have some stuff mixed up then

[crude star]

oh

where did I mess up

[solar hazel]

R implies S is true exactly when (R is false or S is true)

[crude star]

oh shit

since its or

[solar hazel]

like you can take that as definition

[crude star]

one of them being true would work

so like

lets say

R is true, S is true

then it would be true right

[solar hazel]

yes but also this is wrong

[crude star]

ye I get it now

my bad

let me update

ok would this be correct? I added the next part too

i changed both of them

[solar hazel]

closer for b but still not right

[crude star]

by b u mean the first one right

mhm let me review

[solar hazel]

oh yes lol sorry, i mean the one labelled b)

[crude star]

wth

r u sure

it seems perfect

[solar hazel]

also c) is not quite right either

ok explain each row for b)

[crude star]

ok so

OHHHHH

is it

row 2?

oh wait nah

row 1: P is true and P => Q is true aswell

So this is true

row 2: P is true and P => Q is false

so this is false

row 3: P is false and P => Q is true

so this is true

row 4: P is false and P => Q is true

so this is true

and for c last row is incorrect

right

[solar hazel]

oh sorry i messed up on b, you’re right

[crude star]

but for c my answer was wrong indeed

[solar hazel]

yea

[crude star]

here

[solar hazel]

yes

[crude star]

also just want to confirm

some stuff

so since P => Q (P is false or Q is true)

then P => \lnot Q (P is false or Q is false)

and \lnot(P => Q) (P is true and Q is false)

right

[solar hazel]

$\lnot$

[flat frigate]

💜𝓁𝒶𝓎𝓁𝒶💜

[solar hazel]

umm

[crude star]

ye i meant "not"

sorry

[toxic stratus]

$\lnot$

[solar hazel]

’m a little confused what you’re trying to say there

[crude star]

which line

im tryin to understand if each of them refer to what I put in the brackets

[solar hazel]

are you asking if (P implies Q) implies something?

[toxic stratus]

are you saying that's when it's true?

[crude star]

ye

[crude star]

no no

rmbr how earlier u said

P implies Q is true when P is false or Q is true

then does that mean $P \implies \lnot Q$ is true when $P$ is false or $Q$ is false?

[flat frigate]

Calc II Victim

[solar hazel]

yea

[toxic stratus]

[ (p \Rightarrow q) = (\lnot p \lor q) ]

[crude star]

and $\lnot(P \implies Q)$ is true when P is true and Q is false?

[flat frigate]

Calc II Victim

[solar hazel]

yea