help-23

[dusk gate]

But a is not part of the integral

[hard mortar]

it will be in the end

[mint flint]

like imagine F(x) is the antiderivative of the integrand, all it's actually asking is lim[a->inf] F(a+1)-F(a)

do you see what im saying

[hard mortar]

^

[dusk gate]

Yes, if you were analyzing this with trig substitutions

[hard mortar]

I don't even know what that is

I really am not supposed to even know how to solve that integral

[dusk gate]

1/sqrt(a^2 + x^2) is a trig substitution that relies on the Pythagorean Theorem to sub in known ratios, similar to u-substitutions

But yeah, that's kinda beyond the scope of this question

a^2 in this case would be 1, since 1^2 = 1

But we're getting off the rails here

[mint flint]

can you outline the suggested method again

[hard mortar]

yeah you substitude x=atan(u) right?

[mint flint]

that one about inequalities

[hard mortar]

I am familiar with that but am not supposed to know it

anyway

[dusk gate]

pfft.. you can know it.. but you might not be allowed to use it to solve the problem 😉

[hard mortar]

so essentially you have to find some sort of inequality from the function inside the integral

like, if the function was 3sin(x), then it is true that -3<-3sin(x)<=3

then you can integrate each part of the inequality

and you find something

[mint flint]

so squeeze theorem in essense ig

[hard mortar]

not really, you can find anything

it will just be (number) <= integral <= (other number)

but

if lim(number) <= lim(integral) and lim(number) = infinity

then lim(integral) = infinity too

is that too confusing?

wish I could draw here or something

[dusk gate]

https://www.wolframalpha.com/input?i=lim+as+a+goes+to+infinity+(x)

The answer is just the integral itself

[mint flint]

well we know that the function is greater than 1 for all x>0

[hard mortar]

a is part of the integral

[dusk gate]

Where?

[hard mortar]

here

[dusk gate]

Right, but where is that in the function

a can do whatever it wants

how does x change in relation?

We're not told

[hard mortar]

it doesn't have to change, but what changes is the end result, because the integral, like smork said is equal to F(a+1)-F(a) with F being the antiderivative

the integral is a number after all

it can't be a function

[mint flint]

right, by definition the integral is the difference of 2 functions of a

so a is definitely a part of the integral

[hard mortar]

I mean the limit here is obviously equal to infinity

I just havee to find a way to show that

[mint flint]

rly, i graphed it and it seems to be 1

[hard mortar]

I don't see how my inequality method can work otherwise

maybe squeeze theorem

[mint flint]

probably squeeze theorem yeah

[hard mortar]

ok so the original function is greater or equal to 1?

is it lesser or equal to something else?

that would probably solve it

[mint flint]

are you allowed to differentiate?

[hard mortar]

yeah I am going to test something like that to find the extremums

[mint flint]

i was gonna say that yeah

[dusk gate]

I could very well be wholly off base... but if a + 3 = 10, how does that prove that x = 12?

[mint flint]

check my work it should be less than equal to sqrt(2)

[hard mortar]

yeah seems to be right

one moment I will try something

[mint flint]

wait what

[hard mortar]

ah I don't know

I can ask my teacher next time at the end of the day

I will close now because I got tired and this is one exercise from the 20 more I got to do

thanks for helping

.close

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[umbral swan]

Has there been discovered a way to assign a numerical value to the asymptotic growth of functions?

[umbral swan]

Such that the growth value of any exponential function is larger than the growth value of any polynomial

meaning we could compare the absolute growth values of two arbitrary functions and declare which grows faster

and thereby also essentially plot all possible 1D functions on a scale according to their growth

[dusk gate]

isn't that just the degree of the function?

x^3 grows faster than x^2, etc.

[umbral swan]

that would only be applicable for polynomials

[dusk gate]

Got an example of what you're trying to describe?

Because... when it comes down to it.. it sounds like you're describing the limit of a fraction, where the numerator and denominator have different degrees

[umbral swan]

not just degrees, I refer to arbitrary functions

the capability of assigning an arbitrary function a numerical value regarding its growth

[dusk gate]

But you have to have some function in mind to include in the comparison

[umbral swan]

not as a necessity

[dusk gate]

That's... the degree of the function

isn't it?

[umbral swan]

we could declare the function f(x) = x a growth value of 1

[dusk gate]

ax^4 + something has an ultimate growth rate of x^4

[umbral swan]

and then assign all other growth values relative to it, but does that growth function exist?

[umbral swan]

As said before only works for the specific case of polynomials

[dusk gate]

Which is why I'm curious about an example you might have in mind

i.e., where would that not be the case

[umbral swan]

any arbitrary other function?

f(x) = x!

[dusk gate]

thinking

[umbral swan]

f(x) = ln(sin(floor(x^cos(x))))

whatever, polynomial comparisons are trivial of course

[dusk gate]

ehh.. sin and cos should be right out... they're oscillating functions, going up to a max, going down to a min, and repeating

They don't grow infinitely

[umbral swan]

it was a sample

regardless, I'm not certain whether a growth assigning function exists

[dusk gate]

Well, it's always relative to another function

hence the limit of a fraction where the numerator and denominator grow at different rates

[umbral swan]

The relativity of such a function wouldn't be an issue as said

since we can declare the value of any function to be 1

which would infer the values of all other growths

[dusk gate]

Then the value would be 1, and it wouldn't grow at all

[umbral swan]

The growth value is 1, not the function itself

[dusk gate]

Try graphing what you're talking about

See if the visual cues help?

Because if y = 1, then you get a horizontal line with no slope

[umbral swan]

I don't think visualisations would help :/

[dusk gate]

Well, you're talking about the overall slope of a function, but the slope changes depending on the x value you're looking at

That's... the whole point of derivatives

which is, by definition, the rate of change in y compared to the rate of change in x

[umbral swan]

which would declare the function f:R->R; f:x|->1 to have a growth value of 0 for instance

[dusk gate]

so.. again.. by definition it's going to be relative

[umbral swan]

which is not an issue

[dusk gate]

I think the answer to your question, then, is "no"

No such value exists

[umbral swan]

why

[dusk gate]

Because you can't isolate the slope of a function above degree=1

once you hit a degree of 2, the slope varies constantly

[umbral swan]

we can categorize asymptotically, e.g.
f element Theta(n)
f element Theta(logn)
why wouldn't we be able to assign such sets a numerical value relative to each other

[dusk gate]

I've been saying it's all relative from one function to another the whole time, and you've disagreed with me

So

[umbral swan]

I appreciate the help, but why are you talking about slopes :D

[dusk gate]

Because slope is the rate of change

It is the growth rate

[umbral swan]

I seek a way to numerically measure asymptotic growth, not compare the slopes of polynomials

[umbral swan]

no not derivative

the asymptotic growth

[dusk gate]

Define asymptotic growth

I don't know what you're talking about

a rate is one value compared to another.. y/x

that is slope

if the correlation between growth rate and slope is lost on you... I don't know where else to go

[umbral swan]

in relative notion a function f grows faster asymptotically towards ->inf than g if there exists a c>0 such that from a x0 onwards (x≥x0) the inequation f(x)≥g(x) holds

[umbral swan]

np, it's an analysis 1 topic 🦆

[dusk gate]

No.

I guess? 🤣

[umbral swan]

I'd rather not have a guess

[dusk gate]

Could be me not understanding exactly what's being asked, since I don't have a way to visualize what you're asking

Could be me just... being right, since growth rates depend on the x-value of a given function

and since x values are variable, growth rates are subsequently variable

[umbral swan]

it's fine if you aren't yet familiar with the topic, but thanks for the attempt :)

[dusk gate]

hence, the overall (average?) growth rate might be undefined

mm... thanks 😉

anywho... have fun finding Jimmy Hoffa!

(no seriously... let me know if you find an answer)

(would be SO helpful with evaluating limits.... )

[umbral swan]

Has there been discovered a way to assign a numerical value to the asymptotic growth of functions?
Such that the growth value of any exponential function is larger than the growth value of any polynomial
meaning we could compare the absolute growth values of two arbitrary functions and declare which grows faster
and thereby also essentially plot all possible 1D functions on a scale according to their growth

[cold relic]

Are you asking if there is an injective mapping from equivalence classes over asymptotic growth to real numbers?

[umbral swan]

hm if the growth comparison implies injectivity then yes

from equivalence classes to real numbers yes, injective

[cold relic]

I would very much assume there is, theoretically at least… though I doubt it is practical

[umbral swan]

hm it'd be quite intriguing as we could essentially map out all growths we have discovered so far and fairly quickly determine which functions grow faster than others

[cold relic]

But if [x^r] represents a unique equivalence class for every unique r in R, then is such a system really useful?

[umbral swan]

hm I'd argue it could be

I stumbled upon it during Theoretical Comp Sci

eg if we were to assign the basic linear function f:R->R, f(x)=x the value 1

and all growths above constant to be ≥0

then all logarithmic functions must be between 0 and 1

yet only occupy a small enough space

In fact arbitrarily small

to leave room for infinitely many other slower and faster growing functions

with values between 0 and 1

the realization of that principle is probably where it might collapse

but not sure

[cold relic]

The problem seems to be that we can’t define it intuitively, because mapping [x^r] to r for example makes it impossible to make the mapping injective

I suppose we could prove the existence of an injective mapping by proving that the cardinality of the domain is equal to the cardinality of the set of real numbers

[umbral swan]

hm then potentially instead onto a different set with higher cardinality yet also an ordering?

[cold relic]

Kinda like a mapping to itself? 🙂

[umbral swan]

yeah

could be larger as well though

[cold relic]

I think the main problem with the numeric approach is that it seems more difficult to figure out the numeric value that it is to just compare the growth of the two functions

It’s much easier to prove that 2^x grows faster than x^2 than it is to define a mapping that meets the criteria and then figure out what the values assigned to 2^x and x^2 are

[umbral swan]

My younger brother sits next to me and asked why can't we measure which function reaches infinity first and measure the distance :]

[umbral swan]

yeah, I think I'll resign to it for now

[cold relic]

lol that’s sweet

[umbral swan]

thx

[cold relic]

Even if we find the numeric values they’re likely given by limits, and comparing those limits will be the same as comparing the original functions

[cold relic]

If you find something practical let us know, maybe it’ll make you rich 🙂

[cold relic]

It was a pleasure

[umbral swan]

Maybe I could suggest it as an addition to the millennium ptoblems

.close

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[fresh owl]

How do I even do this

Tried u sub below but it doesn't seem to help

[devout shale]

e^x-1 = (xe^x+1)+e^x-1-(xe^x+1)

use that

[fresh owl]

And then break everything apart?

The fraction?

Oh wait

No

Ill just get to the same thinf

[devout shale]

not everything

but breaking apart the fraction somehow

will help after you rewrite that way

[fresh owl]

Alright hold on

Im right here

[devout shale]

okay you made a sign mistake

$$\frac{e^x-1}{xe^x+1}=\frac{(xe^x+1)+e^x-1-(xe^x+1)}{xe^x+1}=\frac{(xe^x+1)+e^x-1}{xe^x+1}-\frac{xe^x+1}{xe^x+1}$$

[flat frigate]

Austin

[fresh owl]

Ohhhh so I do it that way

[fresh owl]

Omg yeee

Thank you

Would have never seen that if it wasn't for you!!!

.close

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[fallen pier]

What do i do to solve b

,rotate

[fallen pier]

i found the product

but idk how to solve

do i need to use an identity to solve cos 3x=0

so pi/6

and doesnt cos 270 =0

so cos 6pi/4

so my answer is 6pi/4 pi/2 and pi/6

is that right

wait do cos 5pi over 6 work

when subbed into cos 3x

ok ty

whats k

ohh i get it now

tysm

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[hard thistle]

i need help with what to do to find the third vector

[magic junco]

[hard thistle]

i found the first and second eigenvector, but i dont know how to find the third one

i know the eigenvalue -1 has multiplicity two

[icy lance]

why are you so sure theres a third eigenvector

considering you have 2 eigenvalues

[hard thistle]

because there are three eigenvalues

-1 is a repeated eigenvalue

[icy lance]

that doesnt mean anything

they have the same set of eigenvectors

[hard thistle]

then how would i find the general solution then

[icy lance]

each eigenspace would be the span of your eigenvectors

their union is the set of solutions

you could express those algebraically

(a,-a,a) is one general solution for the -1 value

[hard thistle]

it has to be in the form C1e^At + C2e^At+…

[hard thistle]

i have that

this is the solution

i dont know where the second term comes from

[icy lance]

interesting, ive never seen this form before

[hard thistle]

its #17 bte

btw

if an eigenvector has a multiplicity higher than 1, then the generalized eigen vector is (A-lambda*I)^m * a vector u = 0

it has multiplicity 2

when i row reduced i got two free variables

im confused on what to do after

[icy lance]

im truly not sure, im not familiar with this sorry

[hard thistle]

do i @ the helpers then?

[magic junco]

Sure

[hard thistle]

<@&286206848099549185>

[magic junco]

You have to wait for 15 minutes before the next ping

Remember

[hard thistle]

so i can ping every 15 minutes?

[magic junco]

yep

[hard thistle]

can it be whatever i want?

[magic junco]

You can only ping helpers

Pinging individuals for help is prohibited

[hard thistle]

i only did it once so far

i didnt ping any individuals

[magic junco]

yep, as long as you wait for 15 minutes

You can ping

[hard thistle]

does replying count as pinging then?

[magic junco]

Nope

[hard thistle]

<@&286206848099549185>

[safe radish]

@hard thistle Has your question been resolved?

[hard thistle]

<@&286206848099549185>

[safe radish]

@hard thistle Has your question been resolved?

[hard thistle]

<@&286206848099549185>

[hard thistle]

<@&286206848099549185>

[magic junco]

No way, no one comes here

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[hard thistle]

<@&286206848099549185>

[hard thistle]

<@&286206848099549185>

[clear agate]

?

[hard thistle]

i need help finding the eigenvectors of a matrix whos eigenvalues is of multiplicity greater than 1

[hard thistle]

^

[clear agate]

im sorry thats too advanced for me

[hard thistle]

;-;

ok

[clear agate]

<@&286206848099549185>

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@hard thistle Has your question been resolved?

[hard thistle]

<@&286206848099549185>

[magic junco]

What…5 hours

[lost plume]

5

[safe radish]

@hard thistle Has your question been resolved?

[drowsy citrus]

Dude, what's your doubt in this ?

@hard thistle

The question seems answering itself.

[safe radish]

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[stray urchin]

Hi. I’m struggling to figure the question out “What conclusion can you make from this table?” I haven’t done this before this year so I’m new to trigonometry and I really don’t want to lose any marks.

[stray urchin]

<@&286206848099549185>

[last crater]

The ratio of the sides is dependent on said angle

[stray urchin]

These are the angles that they gave

So would it be All ratios are the same?

.close

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[last crater]

You can see in your table they are of samr value

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[compact pelican]

Are you having trouble with 7?

[alpine hornet]

I am hoping anyone can maybe help me check my work , confirm if I have done a correct job on 5 & 6

[compact pelican]

Oh, have you tried wolframalpha... not sure it'll do it, but worth a shot?

[alpine hornet]

I have not tried that

I tried it’s too confusing mso

[compact pelican]

give me a sec

https://www.wolframalpha.com/input?i=sum+-10%2B4*n+for+n%3D0+to+49

But that might not be obvious

[safe radish]

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@alpine hornet Has your question been resolved?

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[lean otter]

where did i go wrong

[lean otter]

im new to partial derivatives

check with online calculator

now im confused

also i dont know what the fz(1, 1, 1) notation means

is that just in respect to z, then plugging in values 1, 1, 1

[desert juniper]

derivative with respect to z at point (1, 1, 1)

[lean otter]

right okay

can you tell me where i went wrong for the first part

[desert juniper]

i dont see an issue on the first part, other than the notation

on the bottom left you're not differentiating f with respect to x, you're differentiating the differentiate of f

[lean otter]

oh shoot

ur right

how do i denote this

[desert juniper]

i'd write (hold):

$\frac{d}{dx}(\frac{df}{dy})=3x^2z^2$

[flat frigate]

LordFelix

[desert juniper]

but there's many ways to write it

(replace d with delta, it's just very annoying to write that on discord)

[lean otter]

thank you i appreciate it

.close

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[keen pebble]

Yuh

[keen pebble]

I have these equations

And im uspposed to do this

i can translate it real qucik

Investigate whether there is a value for a such that the line with the equation y=2x-54 is tangent to the graph G at the point P(6/f(6)).

And i really dont know what im supposed to do yk

i just thought of someting

i would try to get the tangential equation of P x=6 and see how it is attached to a

So now i put 6 into fa(6) so i got 174-18a

and now i try to get the gradient of 6

and then assemble a tangential function

daimn my approach turned out to be fully right and i got it without help

wo

w

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[modest owl]

please help, when solving recurrence relations like a_n = 3a_(n-1) - 5a_(n-2), i know to solve this we have to first make a characteristic equation for it and then we have to factor out the least n term like n-1 or in this case n-2 and after that we get r^n -3^n-1 + 5r^n-2 = 0 =>> r^2 - 3n + 5 = 0 and then so on, but i want to ask what if we have a_n = 2a_n-1 + 2n^2 how do i factor out n-1 from this?

[safe radish]

@modest owl Has your question been resolved?

[modest owl]

<@&286206848099549185>

[tough sigil]

sorry can’t help

i’m getting my own help done

actually

i can help

[modest owl]

please do

[tough sigil]

sure thing dude

let me just solve the equation

and then i can help

for the given recurrence relation you can approach it by using the method of undetermined coefficients to find a particular solution. the form of the particular solution will be a polynomial of the same degree as the non-homogeneous term, which is 2n^2

assuming a particular solution of the form p(n)= An^2 + Bn + C, substitute this into the recurrence relation:

[ a_n = 2a_{n-1} + 2n^2 ]

[flat frigate]

Rick

[tough sigil]

[ An^2 + Bn + C = 2(A(n-1)^2 + B(n-1) + C) + 2n^2 ]

[flat frigate]

Rick

[tough sigil]

after simplifying you can solve for the coefficients a, b, and c. once you have the particular solution, add it to the solution of the homogeneous part of the recurrence relation to get the general solution

hopefully that can help you solve the question better

[uncut gate]

if its an arithmetic series you can write a_n-1 as a_n - r

[modest owl]

i am trying to solv it this way:

a_n = 2a_(n-1) + 2n^2
r^n - 2r^(n-1) - 2n^2 = 0
where do i go from here?

@tough sigil

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[exotic charm]

help

idk how to do the first part of part c

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[exotic charm]

<@&286206848099549185>

[still crag]

Hello

[exotic charm]

hi

how to do part c i have no clue

[drowsy citrus]

I think $f_x(x,y)$ is the derivative with respect to x. Then you put x =0. Then you differentiate wrt t.

[flat frigate]

Solomaniac

[drowsy citrus]

That should give you $f_{xy}(0,t)$. Put t=0

[flat frigate]

Solomaniac

[exotic charm]

i did this and got -1

[drowsy citrus]

Is it incorrect ?

[exotic charm]

idk if it's incorrect but im not sure what to do from there

like when u sub x = 0 u get -y

sub y = t to get -t

differentiate wrt t gives -1

no clue what to do from there

[exotic charm]

i dont get how differentiating wrt t gives me fxy

is what im asking

[plucky elk]

t and y play the same role in f

[exotic charm]

why

[plucky elk]

it's just the dummy variable of the second argument of f

f(x, y), f(a, b), f(c, d) are all the same things if x=a=c and y=b=d

[exotic charm]

i get that but like

when you find fxy dont you have to differentiate the whole partial derivative of x with respect to y

here they are subbing x = 0 and then differentiating it with respect to y?

@plucky elk

[plucky elk]

What is "here"

[exotic charm]

in that example

i dont get why subbing x = 0 gives me fxy

normally when u do fxy u differentiate the whole of fx with respect to y?

[exotic charm]

<@&286206848099549185>

[tough sigil]

!15mins

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[exotic charm]

it has been longer than 15 mins

[tough sigil]

sure lol

dw i’m js playing

[exotic charm]

<@&286206848099549185> plsss

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@exotic charm Has your question been resolved?

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[devout steeple]

I solved this karnaugh map, and compared it to a KMap calculator online

[devout steeple]

is my solution less correct than theirs, or entirely invalid?

sorry if this isn't the right place for this... I couldn't find a good fit for it in computer science servers either.

[edgy patrol]

@devout steeple

your solution is also correct

but KMap is more simple than your expression

you must add several block as possible as big

your method is 2 2 4

KMap is 2 4 4

[devout steeple]

ohhhh i see what you are saying

i need to make the purple group include the cells above it

[edgy patrol]

yes

you are correct

[devout steeple]

thank you!

[edgy patrol]

you are welcome

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[dim needle]

1 and 2. I understand the concept but I don’t know where to start lol

[vapid cypress]

find p(x) by multiplyng x+2 and 3x^2 + x + 3 and add 2

a = bq + r

for 2 use remainder theorem

[dim needle]

is this for one?

[vapid cypress]

assume the pollynomial is f(x) find f(-2) and equate to -7

[vapid cypress]

yea

[dim needle]

Ohh okay

is 5 basically similar to the ones above?

[vapid cypress]

yea nearly

for y intercept x= 0

[dim needle]

same with 8 and 9?

same concept as 1 and 2?

[vapid cypress]

yea

[dim needle]

Alright im overthinking this.

Thank you i appreciate it

.close

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[lean otter]

can someone please help with me with question 83

[lean otter]

here is my work thusfar

but the answer is wrong

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

fuck no

[fathom jewel]

chain rule

you forgot u'

[lean otter]

is that because u is a function

[fathom jewel]

a function of x yes

[lean otter]

so the derivative of a constant x a function is just the constant times a function

[fathom jewel]

u is defined as g(x) = ...

[lean otter]

so it should be pi/10 u'

[fathom jewel]

yes

or g'(x) whatever

[lean otter]

ok

thx

i get it now

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[junior wagon]

Is there a quick way (without computer) to compute $8^8 (\mod 9)$?

[flat frigate]

WhoTao

[junior wagon]

yes!

[solar hazel]

8^2 is 1 mod 9

[junior wagon]

right. So then (8^2)^4 = 8^2 = 1 mod 9

[solar hazel]

yes

[junior wagon]

ty

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[jolly pasture]

hii rly struggling here

[jolly pasture]

the new angle a measurement should be 143

but that gives me the same angle measurement as c1

which like. shouldn’t be the answer

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@jolly pasture Has your question been resolved?

[jolly pasture]

<@&286206848099549185>

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@jolly pasture Has your question been resolved?

[plucky elk]

question got cut off

[jolly pasture]

just had to find an SSA triangle

i got it though

thanks

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[fair forge]

can someone explain to me why

the range is [0, infinite)

how they got it thank you

[devout shale]

can the square root of something be negative

@fair forge

[fair forge]

no

[devout shale]

Can the square root of something be 0?

[fair forge]

im not sure

[devout shale]

sqrt(0) = ?

[fair forge]

0

[devout shale]

so, yes the square root of 0 can be 0

Now more specifically, if f(x)=sqrt(2x-6)

Can 2x-6=0 ?

[fair forge]

x=3\

[devout shale]

Alright

so far we've shown the range cannot include anything negative

and does include 0

Now we want to know, does it go alll the way to infinity?

What do you think about that

[fair forge]

yes because its not -

[devout shale]

What do you mean by that?

[fair forge]

infinity numbers are not nevative numbers

[devout shale]

That doesn't really have to do with our function so it isn't a good reason

f(x)=sqrt(2x-6)

we know the range so far is

[0,

and we want to know what the other side should be

Do you have any ideas for that?

[fair forge]

get it from the domain?

[devout shale]

Good idea

So we can notice a few things now

sqrt(x) is an increasing function

Right?

As x gets bigger, sqrt(x) gets bigger

[fair forge]

ok

[devout shale]

In our case, we have sqrt(2x-6)

2x-6 is also an increasing function, do you agree?

[fair forge]

yes because we cant have a negative

[devout shale]

That is not a good reason

as x gets bigger, does 2x-6 get bigger?

[fair forge]

since its positive its gonna go up

[devout shale]

Since the coefficient on the x is positive

yes

Okay so, since sqrt is increasing, and 2x-6 is increasing

as x gets bigger

so does sqrt(2x-6)

right?

[fair forge]

wouldnt it be lower since it has square root

[devout shale]

We just said earlier that sqrt(x) is increasing

[fair forge]

ok

its increasing

[devout shale]

so now we look to our domain

as x gets bigger, our range gets larger

but how big can x get in our domain?

[fair forge]

all real numbers

[devout shale]

so how big can our range be?

[fair forge]

all real numbers

[devout shale]

but it can't be negative so that cannot be true

[fair forge]

so include 0

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[fair forge]

thank u @devout shale ill try to understand it more

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[charred jetty]

Is there properties for sqrts ? like $\sqrt{a + b} \le \sqrt{a} + \sqrt{b}$\
And how do I prove them?

[flat frigate]

Roman_Garland

[lean otter]

this is true

you can prove by squaring both sides

[halcyon root]

ah, nice

[charred jetty]

Is there a rule

or theorem?

With inequalities?

[lean otter]

this is used often enough that some books might call it a theorem

but it's basic enough

as the proof is one line lol

[charred jetty]

What about $\sqrt{a - b} \ge \sqrt{a} - \sqrt{b}, a, b > 0, a > b$

[halcyon root]

I think they're asking for more theorems

left side is gte

[lean otter]

that's false

[drowsy citrus]

You need condition on a and b

[lean otter]

or like you can't say in general

[blissful lance]

can someone help me

[charred jetty]

Am I right now?

[blissful lance]

pls

[drowsy citrus]

Still, can't comment.

Need a condition like a,b >0

Because what if a=-2 b=-4

RHS can't be commented

[flat frigate]

Roman_Garland

[charred jetty]

Now ?

[drowsy citrus]

Just square both sides. You'll understand.

[charred jetty]

If $a, b > 0$ and $a > b$
Then $\sqrt{a - b} \ge \sqrt{a} - \sqrt{b}$\
Proof:
$\sqrt{a - b} \ge \sqrt{a} - \sqrt{b}$\
$a - b \ge a + b - 2\sqrt{ab}$\
$- 2b \ge - 2\sqrt{ab}$\
$ b \ge \sqrt{ab}$\
Oh.... that is wrong, right?

[flat frigate]

Roman_Garland

[charred jetty]

.close

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[ocean raptor]

How many 3-digit numbers less than 150 can be formed from the digits 0, 3, 4, 6, and 9? is it 241?

[potent seal]

How did you get that

[lean otter]

umm zero right?

unless a leading 0 counts as a 3rd digit

[ocean raptor]

fundamental principle of couting

[pliant egret]

tell me any number less than 150 formed with those digits

[ocean raptor]

039

does that count or

its just 0

[lean otter]

if those count

then lets see

[ocean raptor]

yes those count

[pliant egret]

039 is just 39 no? and 39 is a 2digit number

[lean otter]

1 digit numbers: 5

2 digit numbers: 4*5 = 20

so total 25

[lean otter]

padding with zeros counts apparently lol

[halcyon root]

feels like a pnc problem

[lean otter]

its just 25

first get the 1 digit numbers

then get the 2 digit numbers which dont have the first digit being 0

im curious which grade level problems is this that leading 0 counts

[halcyon root]

It's 4x5 because the four non-zero numbers = 4x4 and once with the zeroes is 4 more?

[ocean raptor]

I just found the question on the internet

I was curious

[lean otter]

sure that's one way

i think of it as

__

first digit 4 options

second digit 5 options

[ocean raptor]

they are broad

anyways thanks

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[lean otter]

sure thing

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[dull stratus]

if we have three vectors that are coplanar right then they will all lie on the same plane, but that means the plane is non-specific right? just needs to have same normal vector but they can be any plane (series of parallel planes) right? cuz vectors are translatable in some space

[hot thistle]

is this related to some other question

if you are told 3 vectors are coplanar, then yes that is not sufficient to find a unique plane (in general)

[dull stratus]

alright cool, just making sure i got the right idea

not related to a specific question just making sure im understanding y'know

[hot thistle]

if you are familiar with linear algebra, you get at least 1 degree of freedom from "three coplanar vectors," where the degree is the translation of the plane in space
if you know that their span is only 1 dimensional (so all 3 vectors are parallel) then you get 2 degrees of freedom, where one is the translation in space, and another is the rotation of that plane about a line created by the vectors
if all vectors are the 0 vector, you get 3 degrees of freedom, so any plane will do

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[keen steppe]

,tex
Let
\begin{flalign*}
& S_1 = \left{ x \in \mathbb{R}^4 \mid 2x_1 - x_2 + x_3 - x_4 = 0 ; \ x_1 - 3x_3 + x_4 = 0 \right}; & \
& S_2 = \left{ x \in \mathbb{R}^4 \mid 2x_1 - x_2 - x_3 + x_4 = 0 \right}; & \
& T_1 = \left\langle (1,0,1), (0,1,1) \right\rangle; & \
& T_2 = \left\langle (2,1,3), (0,0,1) \right\rangle. &
\end{flalign*}

Find a linear transformation ( f : \mathbb{R}^4 \rightarrow \mathbb{R}^3 ) that simultaneously satisfies:
\begin{flalign*}
& f(S_1) \subseteq T_1; & \
& f(S_2) \subseteq T_2; & \
& \dim \mathrm{Nu} \ f = 1. &
\end{flalign*}

[flat frigate]

renato

[wooden forum]

!status

[safe radish]

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

@keen steppe Has your question been resolved?

[keen steppe]

1

@wooden forum

@wooden forum

@here

[safe radish]

@keen steppe Has your question been resolved?

[hot thistle]

@keen steppe you've had like 5 people help you with this. i find it hard to believe you dont know where to begin

[keen steppe]

i mean we found two basis, one for s1 and one for s2 I think

(yesterday)

[hot thistle]

why not mention that earlier

[safe radish]

@keen steppe Has your question been resolved?

[safe radish]

@keen steppe Has your question been resolved?

[keen steppe]

how do I get this started?

[safe radish]

@keen steppe Has your question been resolved?

[safe radish]

@keen steppe Has your question been resolved?

[narrow mortar]

To find a linear transformation $f: \mathbb{R}^4 \rightarrow \mathbb{R}^3$ that simultaneously satisfies the given conditions, we can start by finding a basis for $S_1$ and $S_2$.

[flat frigate]

Flamey

[narrow mortar]

@keen steppe

For $S_1$, we have the following system of equations:
[
\begin{cases}
2x_1 - x_2 + x_3 - x_4 = 0 \
x_1 - 3x_3 + x_4 = 0
\end{cases}
]

[flat frigate]

Flamey

Flamey

[keen steppe]

I cannot see what it equals in the third equation

[keen steppe]

.

[keen steppe]

.

[keen steppe]

.

@narrow mortar

@narrow mortar

[narrow mortar]

= 0

Let $x_3 = 1$ and $x_4 = 0$. Then $x_2 = 7$. So one basis vector for $S_1$ is $(2, -7, 1, 0)$. Let $x_3 = 0$ and $x_4 = 1$. Then $x_2 = -3$. So another basis vector for $S_1$ is $(1, -3, 0, 1)$. Thus, a basis for $S_1$ is ${(2, -7, 1, 0), (1, -3, 0, 1)}$.

[flat frigate]

Flamey

[narrow mortar]

@keen steppe

[keen steppe]

,w 2x1-x2+x3-x4=0, x1-3x3+x4 = 0

[flat frigate]

Results provided by WolframAlpha

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

Requested by xbz24

[keen steppe]

how did you manage to find the basis vectors for $S_1$?

[flat frigate]

renato

[keen steppe]

,w rref {{2,-1,1,-1,0}, {1, 0, -3, 1,0}}

[flat frigate]

Results provided by WolframAlpha

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Requested by xbz24

[keen steppe]

,, \begin{cases}
x_1-3x_3+x_4 = 0 \
x_2 -7x_3 + 3x_4 = 0
\end{cases}
\begin{cases}
x_1 = 3x_3-x_4 \
x_2 = 7x_3 - 3x_4 \
x_3 = x_3 \
x_4 = x_4
\end{cases}

[flat frigate]

renato

[keen steppe]

,,
\begin{cases}
x_1 = 3s-t \
x_2 = 7s - 3t \
x_3 = s \
x_4 = t
\end{cases}

[flat frigate]

renato

[keen steppe]

,,
\begin{pmatrix}
3s-t \
7s - 3t \
s \
t
\end{pmatrix} =
s \begin{pmatrix}
3\
7 \
1 \
0
\end{pmatrix} +
t \begin{pmatrix}
-1\
-3 \
0 \
1
\end{pmatrix}

[flat frigate]

renato

[keen steppe]

So one basis vector for S1 is (2,-7,1,0),

no its (3,7,1,0), when x3 = 1 and x4 = 0

[keen steppe]

@narrow mortar

[keen steppe]

@narrow mortar .

@here

[keen steppe]

just change x3 for s and x4 for t but the principle still remains, this is the base I got for S1, is this correct? @narrow mortar

[narrow mortar]

yeah, one basis vector for $S_1$ is $(2, -7, 1, 0)$. Let $x_3 = 0$ and $x_4 = 1$. Then $x_2 = -3$. So another basis vector for $S_1$ is $(1, -3, 0, 1)$. Thus, a basis for $S_1$ is ${(2, -7, 1, 0), (1, -3, 0, 1)}$.

[flat frigate]

Flamey

[narrow mortar]

For $S_2$, we have the equation $2x_1 - x_2 - x_3 + x_4 = 0$. Let $x_1 = 1$, $x_2 = 0$, $x_3 = 0$, and $x_4 = 0$. Then we get $2 - 0 - 0 + 0 = 0$, so $(1, 0, 0, 0)$ is a basis vector for $S_2$. Thus, a basis for $S_2$ is ${(1, 0, 0, 0)}$.

[flat frigate]

Flamey

[keen steppe]

did you read the basis that I gave you, for S1 I got a different result

@narrow mortar

@narrow mortar

[narrow mortar]

For $S_1$, we have the following system of equations:
[
\begin{cases}
2x_1 - x_2 + x_3 - x_4 = 0 \
x_1 - 3x_3 + x_4 = 0
\end{cases}
]

From the second equation, we have $x_1 = 3x_3 - x_4$. Substituting this into the first equation, we get:
[
2(3x_3 - x_4) - x_2 + x_3 - x_4 = 0 \implies 6x_3 - 2x_4 - x_2 + x_3 - x_4 = 0 \implies 7x_3 - 3x_4 - x_2 = 0.
]

Let $x_3 = 1$ and $x_4 = 0$. Then $x_2 = 7$.

[flat frigate]

Flamey

Flamey

Flamey

[narrow mortar]

We can check that $f$ maps $S_1$ into $T_1$ and $S_2$ into $T_2$:
[
f(2, -7, 1, 0) = (2(2) - (-7) + 1 - 0, 2 - 3(1) + 0, 2(2) - (-7) - 1 + 0) = (10, -1, 10) \in T_1,
]
[
f(1, -3, 0, 1) = (2(1) - (-3) + 0 - 1, 1 - 3(0) + 1, 2(1) - (-3) - 0 + 1) = (4, 2, 6) \in T_1,
]
[
f(1, 0, 0, 0) = (2(1) - 0 + 0 - 0, 1 - 3(0) + 0, 2(1) - 0 - 0 + 0) = (2, 1, 2) \in T_2.
]

Also the null space of $f$ has dimension 1, as required. Thus, the transformation $f$ defined above is the desired linear transformation.

[flat frigate]

Flamey

[keen steppe]

yo but getting the eigenvectors in this matter is not rather informal or something, you coudlnt even know beforehand that dim Basis1 = 2 or that dim Basis2 = 1

[keen steppe]

where did you get the thigns that you plugged to define the transformation f

how are you placing the linear equations you are getting from S1 and S2 in transformation f

how can you tell that the dimension of the nullspace of f = 1?

yo

but $2 - 0 - 0 + 0 \ne 0$

[flat frigate]

renato

[keen steppe]

@narrow mortar

[keen steppe]

@narrow mortar

[keen steppe]

.close

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[jade harbor]

i need help with math

[jade harbor]

i need help with easy math

[pliant egret]

just ask the question

[jade harbor]

can i send a pic of it from my book

[pliant egret]

yes you can

[jade harbor]

1 sec

in need help with 7-19

[pliant egret]

do you know what a scale factor is?

[jade harbor]

No

[pliant egret]

it means the ratio between the sides of those two polygons

[jade harbor]

ok

[pliant egret]

see that you know the same side (left side) of those two

can you tell me what the ratio between them is?

[jade harbor]

uuuu i dont no

[pliant egret]

tell me the length of the left side of the bigger one

[jade harbor]

6ft

[pliant egret]

and the length of the left side of the smaller one?

[jade harbor]

4ft

[pliant egret]

ok so what is a ratio of 6ft and 4 ft

[safe radish]

@jade harbor Has your question been resolved?

[safe radish]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[solid musk]

are my answers correct? I don't think the last question is?

[solid musk]

<@&286206848099549185>

[tough sigil]

Yo

[tough sigil]

I don’t think mini modding is allowed here either

[solid musk]

okay...didn't realise

[tough sigil]

Hey I’d help but I kinda need to get some of my work done

Sorry

If ur still here I’ll come back when I’m done

[versed wave]

what are you talking about lol

[tough sigil]

just don’t start arguments please

thanks

[versed wave]

i assume you want to get your answer checked

[solid musk]

yep...just the last part...find the traffic intensity.

I'm not sure if the formula I should use is Average Packet Transmission Time = Transmission Delay + Propagation Delay and then ρ=150×Average Packet Transmission Time...or if I should do Average Packet Transmission Time = Transmission Delay (and not add Propagation Delay)

I'm able to wait until you're finished work thanks

all good

[jade kernel]

scale factor should be 3:2

[jade kernel]

to find z you take 6/4 = 5.7/z

then you cross multiply

[safe radish]

@solid musk Has your question been resolved?

[safe radish]

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[short geyser]

For a graph G, if G is connected and bipartite is |E| ≥ |V| - 1?

[fathom jewel]

mini modder that'S a cute name

[safe radish]

@short geyser Has your question been resolved?

[short geyser]

<@&286206848099549185>

help u with what 🤔

LMAO WHAT 😭

brotha help ur brotha instead of helping me 😭

[safe radish]

@short geyser Has your question been resolved?

[safe radish]

@short geyser Has your question been resolved?

[safe radish]

@short geyser Has your question been resolved?

[safe radish]

@short geyser Has your question been resolved?

[safe radish]

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[magic tusk]

$\frac{138+x}{3}\geq75$

[flat frigate]

pixel

[magic tusk]

$138+x \geq 225$

[flat frigate]

pixel

[magic tusk]

on the right track?

[dim meteor]

yup

[magic tusk]

$x \geq 87$

[flat frigate]

pixel

[magic tusk]

thats it

🔥

[dim meteor]

perfect

[magic tusk]

$0 > (x-1)(x+2)$

[flat frigate]

pixel

[magic tusk]

x = 1
x = -2

so i have to find values of x so that

0 is greater than

,calc (1-1)(1+2)

[flat frigate]

Result:

0

[magic tusk]

,calc (-2-1)(-2+2)

[flat frigate]

Result:

0

[magic tusk]

ooo

ok

so

$-2 < x < 1$

[flat frigate]

pixel

[magic tusk]

ok uh

mhmm

$y=a(x+\frac{b}{2a})^2 - \frac{b^2-4ac}{4a}$

[flat frigate]

pixel

[magic tusk]

,calc 2/2*1

[flat frigate]

Result:

1

[magic tusk]

calc (2^2-417)/4

,calc (2^2-417)/4

[flat frigate]

Result:

-6

[magic tusk]

mhmm

$y=(x+1)^2 + 6$

[flat frigate]

pixel

[magic tusk]

i completed the square

[past birch]

did u learn differentiation yet

[magic tusk]

no

[past birch]

i think after u complete the square it tells u something about the vertex

k = 6 in this case

and we know that the coefficient of x^2 is positive

so itll be pointing down

[magic tusk]

the question itself is asking me to prove that every value of 'x' put in that equation would give a value greater than or equal to 6?

[past birch]

yes

if u know that 6 is the minimum point, then all other values of x are either equal to 6 or greater

[magic tusk]

yeah

so uh what am i even supposed to do now

😭

[past birch]

you basically proved what they asked???

[magic tusk]

all i did was complete the square

unless im missing something

sorry its late i just wanna finish this question and sleep

[past birch]

yes and that tells you alot

[lean otter]

Doesn't it say in the question that completing the square its self proves what you're looking for?

[past birch]

it tells you where the vertex is

and looking at the equation in the form ax^2+bx+c, we know that the coefficient of x^2 is 1, which is positive, therefore the curve is pointing upwards

[magic tusk]

OHHH

yes omg

💀

[past birch]

therefore all other values of x are either equal to 6 or greater

[magic tusk]

yes yes thank you

that makes sense

thank you

.close

[safe radish]

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[past birch]

welcome

[safe radish]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[stiff vine]

How do u even solve these

[lean otter]

for the second one, use L'hopital

[stiff vine]

I didn’t learn that

[rough storm]

factor

[lean otter]

hmm ok

then factorize

[stiff vine]

How

I can only factor bottom

Not too

[lean otter]

you can factor the top

its' a quadratic

after all

it'll just be indeterminate in terms of a

[stiff vine]

What

I don’t see how

[rough storm]

for the first one, horizontial asymptotes is the same as calculating limit as x->inf and lim x->-inf

[stiff vine]

So the answer is 0

[rough storm]

no.

[restive niche]

Factorising will be hard, for the limit to exist, you need a (x+2) to cancel from the numerator and denominator, so make x=-2 a root of the quadratic in the numerator

[rough storm]

you can see what it is by taking out an x^2 inside the square root, i.e. write sqrt(x^2+3) as

[stiff vine]

There is no (x+2) anywhere to factor from numerator

[rough storm]

$\sqrt{x^2\left(1+\f{3}{x^2}\right)}$

[flat frigate]

MSC2020 55N31 (Moosey)

[rough storm]

then you can pull out $\sqrt{x^2}$ as $|x|$

[flat frigate]

MSC2020 55N31 (Moosey)

[rough storm]

so

[stiff vine]

What?? How did u get that

[rough storm]

$\f{\sqrt{x^2+3}}{x+1}=\f{|x|\sqrt{1+\f{3}{x^2}}}{x+1}$

[flat frigate]

MSC2020 55N31 (Moosey)

[rough storm]

this is the same as sqrt(x^2+3)

this is a common trick to calculate limits going to infinity or -infinity

we can do a similar thing to denominator

$\f{|x|\sqrt{1+\f{3}{x^2}}}{x+1}=\f{|x|\sqrt{1+\f{3}{x^2}}}{x(1+\f{1}{x})}$

[flat frigate]

MSC2020 55N31 (Moosey)

[stiff vine]

I didn’t know u could factor square roots

And into the square root itself

[rough storm]

,tex .exp rules

[flat frigate]

MSC2020 55N31 (Moosey)

[rough storm]

distributivity

[stiff vine]

Well I already know those

[rough storm]

yes but you can think of sqrt() as power 1/2

[stiff vine]

Then it should be sqrt(x^2) times sqrt(3)

[rough storm]

i just applied distributivity to this

[stiff vine]

But they’re adding not multiplying

[rough storm]

x^2 * (1+3/x^2)

i'm not distributing over addition

first, pulled out x^2 from x^2+3, and got x^2(1+3/x^2)

then i split the square root applying distributivity sqrt(x^2)sqrt(1+3/x^2)

then, i applied sqrt(x^2)=|x|

[stiff vine]

How do u even pull it out

So sqrt(x^5+2) could become sqrt(x^3)times sqrt(x^2+2)?

[rough storm]

almost

sqrt(x^5+2) could become sqrt(x^3)times sqrt(x^2+2/x^3)?

[stiff vine]

Idk

[rough storm]

wait

there

[rough storm]

note that |x|=x if x is positive

it will be if x->inf

so those cancel as lim x->inf

and 3/x^2 goes to 0, and 1/x goes to 0

so what would the limit be?

[stiff vine]

Infinite

DNE

[rough storm]

...no...

$\lim_{x \to \infty} \f{|x|\sqrt{1+\f{3}{x^2}}}{x\left(1+\f{1}{x}\right)}=\lim_{x \to \infty} \f{\cancel{x}\sqrt{1+\f{3}{x^2}}}{\cancel{x}\left(1+\f{1}{x}\right)}$

[flat frigate]

MSC2020 55N31 (Moosey)

[stiff vine]

I don’t get how the numerato or denominator turn to those

[rough storm]

do you see that $\sqrt{x^2+3}=\sqrt{x^2\left(1+\f{3}{x^2}\right)}$

[flat frigate]

MSC2020 55N31 (Moosey)

[rough storm]

you can think about it as dividing and multiplying by x^2 inside the square root

or just 'pulling out' a factor of x^2

[stiff vine]

I guess

[rough storm]

$\sqrt{x^2\left(1+\f{3}{x^2}\right)}=\sqrt{x^2}\sqrt{1+\f{3}{x^2}}$

[flat frigate]

MSC2020 55N31 (Moosey)

[rough storm]

this is also true. if it helps you can think about it as A=x^2 and B=1+3/x^2

sqrt(AB)=sqrt(A)sqrt(B)

[stiff vine]

Ok so I plug infinite into the x

Which gives 0

@rough storm

[rough storm]

this expression, yes?

both 3/x^2 and 1/x go to zero yes

so what are you left with

@stiff vine

[stiff vine]

Infinite times 1

[rough storm]

$\lim_{x \to \infty} \f{|x|\sqrt{1+\f{3}{x^2}}}{x\left(1+\f{1}{x}\right)}=\lim_{x \to \infty} \f{\cancel{x}\sqrt{1+\f{3}{x^2}}}{\cancel{x}\left(1+\f{1}{x}\right)}=\lim_{x\to \infty} \f{\cancel{x}\sqrt{1+0}}{\cancel{x}\left(1+0\right)}=\lim_{x\to \infty} \f{\sqrt{1+0}}{\left(1+0\right)}$

[flat frigate]

MSC2020 55N31 (Moosey)

[stiff vine]

What happened to the x^2

[rough storm]

what is the limit as 3/x^2 goes to infinity

what about 1/x

[stiff vine]

0

[rough storm]

if you're referring to here, sqrt(x^2) became |x|

[rough storm]

yes. hence they became 0 once you take them to infinity

[stiff vine]

Jow does that even work

So the answer is 1

HA

[rough storm]

yes, that's one of the horizontal asymptotes