help-17

[inner osprey]

precisely

and that's the answer to your question

[atomic pine]

you alr have that since m^2-n^2 is positive.

[kind light]

833 4913

[inner osprey]

fair point

[inner osprey]

sounds about right

[rare swift]

2023 ² = 7× 17²
So m+n = 119
m- n = 17
I am getting m = 68

[atomic pine]

2023^2=7^2 times 17^4

[rare swift]

Ohh fck

Nvm I fcked up my calculations as usual

[kind light]

alr ty yaal

.close

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[rare swift]

Is the answer m = 2023?

Just a guess tbh

[atomic pine]

n is positive

2023^2-0^2=2023^2 but n=0

[rare swift]

[kind light]

n cant be 0

everybody going gangsta until $n\in\mathbb{I}$

[twin meteor]

Skill_Issue

[atomic pine]

whats mathbb I

[rare swift]

I stood by your name skill issue

[kind light]

is it not imaginary

[atomic pine]

never used that ever

[kind light]

huh

is it C? idrk

[atomic pine]

thats the complex numbers

i think u can just do $$i\mathbb R$$ lol

[twin meteor]

qwertytrewq

[atomic pine]

if what u r referring to imaginary is what i think it is

[kind light]

doesent look nice :(

$:\mathbb{m}$

[twin meteor]

Skill_Issue

[civic otter]

But I think it's correct

[atomic pine]

ull see that a lot often in pure math

[kind light]

$e\mathbb{R}^2$

[twin meteor]

Skill_Issue

[kind light]

err

[atomic pine]

we use $\mathrm{Im}(x)$ to denote the imaginery part of $x$ but thats kinda different

[civic otter]

The same way as the even numbers are $2\mathbb{Z}$

[twin meteor]

qwertytrewq

Alberto Z.

[civic otter]

@kind light

[kind light]

ok

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[livid grotto]

The limit of a function?

Yeah that's the limit of a function, you'll need to be a little more specific on what you need to know

Sorry?

That's a bit like asking what the uses of addition are

Thou art requesting information of a personal persuasion, which contradicts the rules of this institution

<@&268886789983436800> trolling

[dense cedar]

🤣

Bro just turned Shakespeare

[mild flower]

what...

thy

(compare my)

[dense cedar]

What thee conceiveth as regular text may seemeth humeth'rous to oth'rs. P'rspectives is what matt'rs

[worthy citrus]

Thou shalt ask a specific question

[dense cedar]

I appreciate thee

[mild flower]

thee - me
thy - my
thou - i

anyway yes do you have a specific question?

[dense cedar]

F'r thy specific questioneth i shall suggesteth thee to behold at the page mention'd below
https://brilliant.org/wiki/limits-of-functions/

[mild flower]

lim isn't really a function on its own and explaining how limits work is kind of about half of a semester's worth of material

[dense cedar]

7

[mild flower]

mmk well if you have a class called pre calculus then that will include it

[kind light]

woo same

[jagged cargo]

when one doth consider a function, there may come a time when the value of the function draweth nearer to a number

,w plot 1/x

[twin meteor]

Results provided by WolframAlpha

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

Requested by fungusdesu

[jagged cargo]

verify, shall thou substituteth x = infinity, thou yield 0

[dense cedar]

Let's consider a case , suppose you drop a ball from a specific height h , then after first drop it rises to a height half of the initial. This pattern continues

[jagged cargo]

in simple words, if thou wishest to find the limit of a function, thou shalt observe the value that the function approacheth

[dense cedar]

If n tends to infinity , what will be the height after n'th drop ?

Nah

[kind light]

it should be 1/8, and thats summation

[dense cedar]

It's (1/2)^n

As n approaches infinity

The denominator gets quite high

[kind light]

$\lim_{n\to\infty}(\frac{1}{2})^n$

[dense cedar]

It's a number , which is like 0.0000000000000000000000000000000001

[kind light]

thats $\sum^{\infty}_{m=1}(\frac{1}{2})^n$

[dense cedar]

In that case we assume the value of the function to be zero

[twin meteor]

Skill_Issue

[dense cedar]

You just applied the limit here

[twin meteor]

Skill_Issue

[dense cedar]

Since n tends to infinity our , height after n'th drop is 0

[kind light]

ok 😭

[dense cedar]

w > cube root of unity ?

[kind light]

heres your epsilon-delta limit definition

[dense cedar]

What omega do you refer to here ?

Which context of mathematics

Ah ok

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[atomic pine]

thats not quite number theory also cardinality (alephs) and ordinals (omegas) are somewhat different.

[atomic pine]

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[small python]

I'm not able to understand the rewritten equations in the explanation.

[small python]

How does it turn into (50/8)^2 ?

[gentle thicket]

they completed square

[dense cedar]

Y = 4x²-50x +126
Or y-126 = 4x(x-50/4)

So Max value at (50/4,126)

You can also use calculus

[small python]

ok thanks

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[potent stirrup]

Can you give me a hint for part c)?

I have the answer right below, but here looking for a hint.

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[lavish nebula]

confused about d, any help?

[vocal sleet]

@lavish nebula Has your question been resolved?

[lavish nebula]

<@&286206848099549185>

[gentle thicket]

is it still not solved?

[lavish nebula]

https://tenor.com/view/sisyphus-rael-gif-943215023730044045

[lavish nebula]

no

[gentle thicket]

kk

lemme see

[gentle thicket]

okay, so our goal is to find the equation of tangent at x=4 on h(x) right?

[lavish nebula]

yes

[gentle thicket]

do you agree that equation of tangent will be linear?

[lavish nebula]

yes tangents are linear

[gentle thicket]

fr

so tell me the general form for the equation of a straight line

[lavish nebula]

y = mx + b

[gentle thicket]

perfect

but let's talk about slope intercept form

which is (y-y_0) = m(x-x_0)

where y_0 and x_0 are y and x intercepts respectively

now we know the value of x_0

which is 4
since we have to form an equation with intercept of x=4

[lavish nebula]

right

[gentle thicket]

so now the remaining values are of y_0 and m

now y_0 will be the y intercept when x_0 is plugged into h(x)

got it?

[lavish nebula]

so y_0 is 23

[gentle thicket]

yup

now all we want is the value of slope at x=4

you got any idea how to find it?

[lavish nebula]

(y-23) = m(x-4) and you plug in y and x values? Don’t think i’ve ever used a slope intercept form so idk

[gentle thicket]

yeah that's correct

but we want to find the value of slope too

see

you know what derivative is?

[lavish nebula]

yes

do you need to find the second derivative of h? fuck i hate back to school hw i learnt this like 6 months ago

[gentle thicket]

uh i don't think so

only first derivative is fine here

[lavish nebula]

can you remind me how derivatives relate to tangents of a curve

[gentle thicket]

see derivative is nothing but slope of the curve right?

[lavish nebula]

just for context

[gentle thicket]

like that's literally how derivative is defined

being the slope of the curve

[lavish nebula]

right

[gentle thicket]

so we want slope at x=4

of the tangent on h

[lavish nebula]

so it’s just 12x - 18

[gentle thicket]

yeah

exactly

that's the general equation of the slope at any point on the curve of h

[lavish nebula]

so you plug 4 into the x of the derivative

is that true

[gentle thicket]

yeah

so we would get value of slope at x=4

which will be m

[lavish nebula]

and what does this number represent, is it just the gradient?

right

[gentle thicket]

yeah slope is the other name of gradient

it is same thing

[lavish nebula]

30 is the m value, we have x0 and y0

y - 23 = 30 (x - 4)

so y = 30x - 120 -23 ?

would that be the equation of the tangent

[gentle thicket]

but 23 with plus sign

[lavish nebula]

lol

fuck

4 hours of math mybad

ty for the help love you🙏

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[near hearth]

what exactly is a "combinatorial proof"? (expanding out would imply an "algebraic proof" perhaps?)

[peak matrix]

I suppose that you will have to do some two-way counting. Meaning find some problem, answer to which is both the LHS and RHS

[near hearth]

ok thank you will give it a shot

[peak matrix]

np

in case you needed a hint, lmk

[near hearth]

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[velvet zephyr]

huh

[placid wren]

i forgot to add 1 to a term in the series

[vocal sleet]

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[rare swift]

What does Taylor theorem have to do with Lagranges mean value theorem

[vocal sleet]

@rare swift Has your question been resolved?

[woeful igloo]

LMVT is baby version of Taylor series

while LMVT is not generalized and only specifies the existance of at least a singular value c that satisfies the formula, taylors series, by keeping the remainder function vague enough, applies all over the region in the neighbourhood of a point x

so, if you reduce the difference a,b in LMVT to be small enough, you land in h-neighbourhood of the point x. At that point, you can approximate taylor series expansion to LMVT by capping the remainder function to first order

[rare swift]

I don't understand how is the Taylor theorem the genralized version of lmvt when both of them do diff things lmvt says that for f(x) in interval (a,b) satisfying all the conditions there exists a point c for which f'(c) will have the same slope as the cord ab
But Taylor theorem approximates a function near some value how are these 2 related

Also I'm dumb so can u explain using some examples

[woeful igloo]

LMVT merely guarentees the existance of a single such point within (a,b)

[rare swift]

Atleast one point

[woeful igloo]

yeah

for the sake of the argument, lets go with one of such possible points

so, for a point small distance dx away from c, the function is not going to be satisfying LMVT

[rare swift]

Ok

[rare swift]

Yes

[woeful igloo]

so, if we reduce the interval (a,b) to a very small region around a, you can write b as b = a + h

so interval becomes (a, a+h)

Now, if you use taylor series on this interval, you get:
f(a+h) = f(a) + h f'(a) + ...

now lets say all the higher order terms are pretty small

so, you can rearrange this to:
f(a+h) - f(a) / ((a+h) - a) = f'(a)

thats LMVT

[rare swift]

wait right
I GET IT

[rare swift]

Wait so is lmvt wrong for close intervals

[woeful igloo]

No, by LMVT, there is at least one such point

but taylors series manipulats it by adding a remainder function

so, for a small enough error, the single point expands to a small set of points

[rare swift]

But we clearly assumed the higher order derivaties to be small

[woeful igloo]

Yes

otherwise there would be an exactly single point

[rare swift]

So there should be some error

[woeful igloo]

By inducing this error, we can trace back to LMVT, because LMVT only tells us one point exists, but not about the specific location of this point

[rare swift]

Ohh ok i get it in lmvt we say there exists some point c
But in Taylor series we are doing it for all points regardless the fact if it's c Or a

[rare swift]

Where does the error occur
While ignoring the remainder?

[woeful igloo]

yes, and as a penalty for generalizing, we dont get an exact function but a remainder that goes on

[woeful igloo]

at each higher derivative term, you get infinitsimally closer to the actual value of the function at x+h

[rare swift]

That n+1 fold integral right?
Which can be represented in lagranges form

Correct me if I'm wrong
In lmvt we find one point c such that the slope of function at the point is equal to the slope of chord
In Taylor series expansion we find a function such that the function matches the slope of our original function at all points by doing this we induce a error which is of the remainder

[woeful igloo]

the sum of taylors series terms can be taken as a convergent series. Each higher order of derivative gets you closer to the answer

[woeful igloo]

we cant 'find' the point in LMVT. LMVT simply guarentees its existence

[rare swift]

ok

[rare swift]

Ohh yes mb

One last question
Can we prove taylors theorem using lmvt

@woeful igloo

Wait nvm I get it we can

Thanks alot
❤❤

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[nocturne badger]

How did they distribute x(x-1)?

[outer warren]

to each term inside

a common factor will cancel for each fraction/term

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[granite dragon]

hey I'm stuck with a trigonometry question. find the 4th root of a complex number/polar form. (I have no idea what the topic of the question is)

[gusty karma]

do you know $e^{i\theta}$

[twin meteor]

loafin around

[granite dragon]

what is that

I know there's a form for this

[gusty karma]

https://en.wikipedia.org/wiki/Euler's_formula

[granite dragon]

r^1/2(cos(theta/h+2pik/s)+isin(theta/n+2pik/n))

[gusty karma]

thats really all there is to the problem lol

[gusty karma]

?

[granite dragon]

there's an explanation to the question

and they give me a formula

let me send you

but I still get it wrong lol

[gusty karma]

it's a lot easier through euler's formula lol

[granite dragon]

how do I get r^(1/4) to sqrt(2)

[gusty karma]

because the fourth root of 4 is swrt2

[granite dragon]

okay

[gusty karma]

sqrt2

$4^{\frac{1}{4}} = \sqrt{2}$

[twin meteor]

loafin around

[granite dragon]

well I got it in decimals on the calc

[gusty karma]

sqrt of 4 is 2, sqrt 2 is sqrt 2?

i don't get what you're missing

[granite dragon]

btw what is that form called? polar form? complex numbers?

[granite dragon]

ohh okay

thanks

[gusty karma]

it's the expansion of complex number polar form

polar is (r, angle)

and it translates into cartesian

[granite dragon]

isn't cartesian normal coordinates

[gusty karma]

yes

ex: polar (1, pi) -> cartesian(1cos(pi), 1sin(pi) = (-1, 0)

[granite dragon]

hmm, with the imaginary number?

[gusty karma]

imaginary number uses euler's formula in polar

[granite dragon]

now how do I use euler's formula to solve that question

can you send me a video tutorial

idk what to type to search for that

[gusty karma]

you translate z into e^i\theta

just search euler's formula complex number

something like that

here the euler's formula expression would be $4e^{\frac{4\pi i}{3}}$

[twin meteor]

loafin around

[granite dragon]

😭

okay

[gusty karma]

4 is the radius

4pi/3 is the angle

[granite dragon]

so would it be like this

I meant -2-2sqrt(3)i

[granite dragon]

ignore the fraction of 2

no it's wrong😭

[gusty karma]

what's the fourth root of $4e^{\frac{4\pi i}{3}}$

[twin meteor]

loafin around

[granite dragon]

yeah what is it

[gusty karma]

i don't know you solve it

basic exponents

[granite dragon]

I tried to write the formula

[gusty karma]

solve it in the exponent form

[granite dragon]

do you mean cube root of e to the power of 4pi*i

[gusty karma]

the problem asks you for the fourth root

what's the fourth root of the above expression

exponent rules in case you forgot

[granite dragon]

is it sqrt(2)cuberoot(e^(4pii))

[gusty karma]

no

use the power of a power rule

what is $(4e^{\frac{4\pi i}{3}})^{\frac{1}{4}}$

[twin meteor]

loafin around

[granite dragon]

you mean this

[gusty karma]

yes what you circled

but not the thing on the bottom

[granite dragon]

done taht form the start

but then what

[gusty karma]

let's walk through this

$(4e^{\frac{4\pi i}{3}})^{\frac{1}{4}} =4^{\frac{1}{4}}*(e^{\frac{4\pi i}{3}})^{\frac{1}{4}}$

[twin meteor]

loafin around

[granite dragon]

so it's sqrt(2)*e^(pii/3)

[gusty karma]

yes

do you remember how to expand $e^{\frac{\pi i}{3}}$

[twin meteor]

loafin around

[granite dragon]

cube root of e^(pii)

[gusty karma]

no, using euler's formula

pi/3 is the angle

[granite dragon]

idk euler's formula, never learnt it

oh wait

[gusty karma]

i sent a wiki article about it

[granite dragon]

yes I'm using that

[gusty karma]

$re^{i\theta} = r\cos(\theta) + ri\sin(\theta)$

[twin meteor]

loafin around

[gusty karma]

yes

and now multiply by sqrt2

there's no i in the angle though

[granite dragon]

oh really

[gusty karma]

look at the form of the equation above

[granite dragon]

yuh

[gusty karma]

and this answer matches up with the answer found in the solution

[granite dragon]

yeah

[gusty karma]

👍

[granite dragon]

just gotta change sqrt(2)*sqrt(3) to sqrt(6)

now what if k=1 or k=2

[granite dragon]

thank you

[gusty karma]

remember how sine waves and cosine waves repeat

k is just that

[granite dragon]

yea

[granite dragon]

which one

what even is k

[gusty karma]

k just describes how many times the wave repeats

it does not effect the outcome really

ex sin(0) = sin(pi) = sin(2pi)

[granite dragon]

okay

but the answers are different

[gusty karma]

oh

yeah

because the waves repeat every two pi

$(4e^{\frac{4\pi i}{3}})^{\frac{1}{4}} = (4e^{\frac{10\pi i}{3}})^{\frac{1}{4}} $

@twin meteor pls?

[granite dragon]

i have no idea how to read that

[gusty karma]

the bot is down

you can add 2 pi to the angle in e^4pi/3

and it will still be the same

but the fourth root will be different

you can add 2 pi infinitely many times

so e^10pi/3 also works and you can solve with that

in retrospect this isn't too much simpler with euler's formula

but it's a nice tool to have

[granite dragon]

like this?

[gusty karma]

no

before you take the fourth root

$(4e^{\frac{10\pi i}{3}})^{\frac{1}{4}}$

[twin meteor]

loafin around

[gusty karma]

10/3 is two more than 4/3

[granite dragon]

oh ok

[gusty karma]

and then you can repeat the process

[granite dragon]

where do you put k to get 2pi

oh i'ts kpi/2

wait no

[gusty karma]

$(4e^{(\frac{4}{3}+2k)i\pi})^{\frac{1}{4}}$

[granite dragon]

that doesn't make sense

[gusty karma]

oops

it's a little scuffed

[granite dragon]

oh

with the imaginary number?

ah ok

[twin meteor]

loafin around

[gusty karma]

yep

that's gonna be the euler's form expression

it gets a lot faster once you're used to euler's form

[granite dragon]

i got a wrong answer

[gusty karma]

show your work

the angle is 10/3

divide by 4 its 5/6

[gusty karma]

you need to take the fourth root

[granite dragon]

did I write the formula wrong

it is multiplied with sqrt(2)

[gusty karma]

^

take the fourth root of the exponential

[granite dragon]

ohhhh

[gusty karma]

and you get 5/6 for the angle

[granite dragon]

yes yes

[gusty karma]

and you get $\sqrt{2}\cos(5\pi/6) ...$

[twin meteor]

loafin around

[gusty karma]

too lazy to write the rest

but you get the gist

[granite dragon]

now I got the answer right

thank you

[gusty karma]

and you repeat this for k = 2

but you know the pattern and it's just adding 1/2 pi to the resulting angle

[granite dragon]

yup we good

[gusty karma]

.close

you have to close it

[vocal sleet]

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[vocal sleet]

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[vocal sleet]

Please don't occupy multiple help channels.

[stone wadi]

<@&286206848099549185>

[vocal sleet]

@stone wadi Has your question been resolved?

[stone wadi]

no

[hallow crescent]

isnt x(y/y) just x(1)

so you have

$\frac{x(y/1)}{x(1)} \geq \frac{x(1)}{x(1/y)}$

[twin meteor]

woomy

[stone wadi]

yes

[hallow crescent]

lets define our function x(y) to be ay^2 + by + c (where a, b and c are positive)

[stone wadi]

ya

[hallow crescent]

oh and y/1 is just y so

$\frac{x(y)}{x(1)} \geq \frac{x(1)}{x(1/y)}$

[twin meteor]

woomy

[hallow crescent]

there probably is a nicer way to solve it but now just sub in the values in the quadratic, so:
$\ \frac{ay^2 + by + c}{a + b + c} \geq \frac{a + b + c}{\frac{a}{y^2} + \frac{b}{y} + c}$

[twin meteor]

woomy

[stone wadi]

i did that..

[hallow crescent]

oh sorry, next time post your status on the question so we can start from where you reached

[stone wadi]

oh ok

[hallow crescent]

so a + b + c is always going to be a positive number

hm wait let me try to think of waht to do from here

[stone wadi]

i got aby + bcy + ac(y^2)+ (ab+bc)/y + ac/(y^2)

[twin meteor]

anonymoushuman

anonymoushuman

[hallow crescent]

$\geq$

[twin meteor]

woomy

[stone wadi]

thanks

[hallow crescent]

wait what side of the equation is that?

[stone wadi]

2ab+2bc+2ca

[hallow crescent]

inequality*

oh perfect thankjs

[stone wadi]

i cross multiplied

[hallow crescent]

oh yea i forgot about that

$aby + bcy + acy^2 + \frac{(ab+bc)}{y} + \frac{ac}{y^2} \geq 2ab + 2bc + 2ca$

[twin meteor]

woomy

[hallow crescent]

correct?

[stone wadi]

ya

[hallow crescent]

factor out a y:
$y(ab + bc) + acy^2 + \frac{(ab+bc)}{y} + \frac{ac}{y^2} \geq 2ab + 2bc + 2ca$

[twin meteor]

woomy

[hallow crescent]

now you can factor out ab+bc

$acy^2 + (ab+bc)(y + \frac{1}{y}) + \frac{ac}{y^2} \geq 2ab + 2bc + 2ca$

[twin meteor]

woomy

[hallow crescent]

now its easier to see that you can factor out an ac

$(ab+bc)(y + \frac{1}{y}) + ac(y^2 + \frac{1}{y^2}) \geq 2ab + 2bc + 2ca$

[twin meteor]

woomy

[hallow crescent]

hmm now

oh

lets factor the second side of the inequality to look more like the first: $\ (ab+bc)(y + \frac{1}{y}) + ac(y^2 + \frac{1}{y^2}) \geq 2(ab + bc) + 2ac$

[twin meteor]

woomy

[hallow crescent]

now its clear

ill leave the rest to you

try to confront the terms of (ab+bc) and of (ac) and notice an inequality that arises

@stone wadi do you need more help? i can continue if you wish

[stone wadi]

could, you wait, a minute?

[hallow crescent]

yep, no problem

[stone wadi]

yeah, pls continue

[hallow crescent]

rearange so that you have everything with ab+bc on one side

and everything with ac on the other

[stone wadi]

$(ab+bc)(y+(1/y)-2) \geq ac(2- (y^2)-(1/y^2))$

[hallow crescent]

put the $ at the begging and end of the message

not before and after \geq

[twin meteor]

anonymoushuman

[hallow crescent]

not quite correct

we are doing (ab+bc)(something) - 2(ab+bc)

[stone wadi]

=(ab+bc)(something - 2)

[hallow crescent]

sorry

yea your correct

im dumb

[stone wadi]

np

[hallow crescent]

yea thats correct mb lol

so: $\ (ab+bc)(y + \frac{1}{y}) - 2(ab + bc)\geq 2ac - ac(y^2 + \frac{1}{y^2})$

[twin meteor]

woomy

[hallow crescent]

$\ (ab+bc)(-2 + y + \frac{1}{y}))\geq ac(2 - y^2 + \frac{1}{y^2})$

[twin meteor]

woomy

[stone wadi]

ya

[hallow crescent]

im thinking about isolating the y's on one side and the constants on the other

but im not sure how that will help show that the inequality holds for all a,b,c

[stone wadi]

$\frac{-1}{y^2}$

[hallow crescent]

$\frac{top}{bottom}$ remember the ${}$

[twin meteor]

anonymoushuman

[stone wadi]

in the end

[hallow crescent]

whered you get that?

[stone wadi]

here

[stone wadi]

into this

[hallow crescent]

could you maybe circle or screenshot the specific part

[hallow crescent]

oh

ohh yea

the minus multiplies through

$\ (ab+bc)(-2 + y + \frac{1}{y}))\geq ac(2 - y^2 - \frac{1}{y^2})$

[twin meteor]

woomy

[stone wadi]

ya

[hallow crescent]

im not so sure what we can do here

[stone wadi]

well, i gtg, gn

[hallow crescent]

byee i also gtg to s leep

[stone wadi]

bye

[vocal sleet]

@stone wadi Has your question been resolved?

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[dapper herald]

help

ii just need the answer

[odd helm]

what's the det of A

[lyric relic]

$det(5A^{-1}B^t)=det(5Id)det(A^{-1})det(B^t)$

[twin meteor]

everg

[oak magnet]

Roblox game ?

[lyric relic]

not actually

[oak magnet]

I thought

[lyric relic]

np

[oak magnet]

As he just need the answer and that flash blue on top of picture, just as every others questions from roblox game

[vocal sleet]

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[chilly spire]

how do i change any % out of X to % out of 100%?

[chilly spire]

for example i have 77% out of 85%

how do i change it to X out of 100%

[pallid forge]

wouldn't that just be the answer to 77% of 85%

[chilly spire]

what will?

idk the X

but i think i managed to get help from chat GPT

[loud grove]

77/85

[chilly spire]

then times 100

[loud grove]

yes

[chilly spire]

great

thank you

[vocal sleet]

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[austere crow]

Hi, this is part of my homework that isnt graded however it just didnt make sense for me

[urban edge]

usually we write that as |x|

[austere crow]

ill try that

it worked!

thank you

.close

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[viral walrus]

how to solve?

sorry

oops

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[paper mica]

hello my name is Rachmir and im studying too be a md and wiskunde (mathmatics)
i dont understand this if someone could help me out would mean the world

[pale perch]

,rccw

[pale perch]

would it be possible to translate?

is a to complete the table?

[paper mica]

ye

its dutch

[pale perch]

b saying to find the value of k?

im just making educated guesses

[paper mica]

ye sadly

im new in this

[pale perch]

what are you struggling with?

[paper mica]

b and c

[pale perch]

any idea for b?

[paper mica]

nope

thats why im here sadly

[pale perch]

what does it exactly say?

[heady ibex]

b is saying why can you not choose k to be 1,5

which is 1.5 in english convention

[pale perch]

aha, id probably need the rest of the context to figure the answer to that

[heady ibex]

k is the number of cards sold

like those you send to family when you go on vacation

[paper mica]

OHH

the awnser too be is : you cant send half a card so its impossible

[heady ibex]

yes

[pale perch]

yup

[paper mica]

W helpers

TY BOTH OF YOU

[heady ibex]

and c is related

to b

[paper mica]

ok thank you again

[vocal sleet]

@paper mica Has your question been resolved?

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[sharp comet]

Sorry it is side ways

What do they mean by explain what ks happening on the graph when x=4

I plugged x into the graph of y= -1/2 + 2

And got y=0

What do I do

[heavy yoke]

,rccw

[sharp comet]

Thanks

[lucid bane]

what does this look like visually? I assume it wants you to describe that

[sharp comet]

Idk how to

Wouldn’t it look the same as before if we know x=4

[lucid bane]

Yeah its saying explicitly what's happneing on the 'graph' so I assume it wants you to plot it and graph it. Draw x and y lines, and then plot your line on it

make a table with some values and then plot those points in your drawing

[sharp comet]

Ok

[sharp comet]

It says explain though

Doesn’t that usually mean with words

[lucid bane]

Yes

but i'm saying graphing it will allow you to describe it in words

[sharp comet]

Ok

[sharp comet]

I have another question I need clarification on

[lucid bane]

,rccw

[sharp comet]

,rrcw

[sharp comet]

So when

The points are on the x axis

Are they excluded

From. The function

Like this guy put brackets meaning -1 and 3 are included

I know min and max are always excluded from the function

So do all positive intervals have parentheses?

And negative

<@&286206848099549185>

[sharp comet]

^

[vocal sleet]

@sharp comet Has your question been resolved?

[sharp comet]

[white palm]

tbh, unless the function is going back up to be positive on intervals not plotted in the picture, then i don’t see the error. the previous picture seems to imply that your definition of positive is that f(x) > 0 not f(x) >= 0.

that might be it

i might be pretty tired also

[distant cipher]

still unresolved>

?

[distant cipher]

.

[distant cipher]

unless the answer is asking you to point out that 0 is not negative nor positive

Im stumped

[vocal sleet]

@sharp comet Has your question been resolved?

[untold valve]

it's with parenthesis because x=0 is neither positive nor negative, which is where you went wrong here

[untold valve]

^

[vocal sleet]

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[viscid delta]

I've got a problem I need to finish tonight that I forgot how to solve over the summer. f(x)=6^3 + 7x^2 - 63 + 20. I need a step by step on how to solve this, because I can't group it, and youtube isn't very helpful.

[red bear]

what does it mean by "solve"?

find the roots?

[viscid delta]

Yeah

[red bear]

alright, are you familar with the rational root theorem

[viscid delta]

No

[red bear]

it says that any rational root of the polynomial will be the ratio of two factors of the first and last coefficients

so in this case, it will be some ratio a/b where a is factor of 20 and b is a factor of 6

does that make sense?

[viscid delta]

is it just any factor of 20 or six, is it common, greatest?

[red bear]

it could be any factor

you have to check all of them

[viscid delta]

how do you know which ones are correct?

[red bear]

you can either plug them into f(x) and check that you get 0, or you can divide f(x) by (x-r) where r is the root, and check that you get 0 remainder

because by the root factor theorem, if r is a root, then (x-r) is a factor of f(x)

on the other hand, you can also use the cubic formula since this is a cubic polynomial

but that's really painful

[atomic forum]

note that negatives are allowed here

[red bear]

right, that's important

[red bear]

i've tried asking your mom, and she gave me a very nice doubt clearing

[viscid delta]

I think that makes enough sense. thx.

.close

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[sour cypress]

How does one calculate how long x is without a calculator

[split wind]

Let me assume this is a quarter of a circle, then i think you mean the radius of he circle is 50

Then you may use Pythagoras Theorem to find the green line

[sour cypress]

Not sure what you mean, the radius is 50 indeed but how do you know its a quarter of a circle

The answer is 30 so it makes sense but i dont see how

It just clicked

[split wind]

very true indeed, we cannot just say it's a quarter of a circle from the picture above.
unless it's written

but if it's not given, then we cannot determine x...

[sour cypress]

Yeah no the side with 10 should have writen next to it that the total is 50

50-10 = 40

[split wind]

the orange line is also 50 given that it's part of a circle

[sour cypress]

So it becomes x^2 + 40^2 = 50^2

Makes total sense now

Thank you

[split wind]

of course we might also have to assume the x is perpendicular to the radius..

[sour cypress]

.close

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[plush bloom]

How should I describe this?

[plush bloom]

Partially overlapping circles?

[plush bloom]

This doesn't sound very professional...

[inner osprey]

you can just say something literal like “two circles with different radii that intersect each other”

if that level of specificity isn’t needed, then “intersecting circles” will suffice

[plush bloom]

That excludes tangencies right

[inner osprey]

no it doesn’t, but you can specify that separately if you’d like

[plush bloom]

.close

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[vast shale]

hello

so

[woeful igloo]

z and and its complex conjugate when added together leave you with only the real part

so a purely real part cant have any complex component

so you get no z that satisfies the equation

[vast shale]

real

[woeful igloo]

just leave the graph blank with only the axes marked on it

[peak matrix]

you can even write "empty graph" under it, so that its completely clear

[empty frigate]

i think "real = complex" is just a bad description

the graph is the set of points that satisfy the equation, so it's empty if no points satisfy the equation

[peak matrix]

and what about it? that's a number, not an equation

[woeful igloo]

3x - iy is not even a number by itself, its just a expression in complex numbers

[peak matrix]

variable?

[empty frigate]

yeah 3x - iy is not a variable

[peak matrix]

it would be an expression

still not equation though

[empty frigate]

the graph of an equation is empty if there are no numbers that satisfy the equation

i don't know what it would mean to graph something that's not an equation

[thin vale]

graph theory is mentioned almost W

[empty frigate]

no

the graph will be empty if no numbers satisfy the equation

[peak matrix]

if the left hand side is real, and the right hand side has non-zero imaginary component, then it's empty

[empty frigate]

if x has to be a real number, then the graph of this is empty, because no real number x satisfies 2x = 1 + 2i

the only complex number that satisfies that is x = 1/2 + i, and that's not a real number

[thin vale]

empty graph is mentioned 🔥 🔥 Peterson graph next up ⁉️

[twin meteor]

Results provided by WolframAlpha

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

Requested by zxzwz

[vast shale]

haha

[vocal sleet]

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[thin valley]

Im having trouble with finding the formula for this and understanding the question. Anybody could help?

[edgy gulch]

,rccw

[thin valley]

Sorry for bad image quality my hand wasn't steady

[edgy gulch]

which one

[thin valley]

Hmm number 4 first

[edgy gulch]

try simplifying those roots first

[thin valley]

Like this? 2√32 = 2√16•2 = 2√16 • √2 = 2•4√2?

Then uhh

8√2

[edgy gulch]

k good

[thin valley]

Lemme write it real quick

Alright done

I couldn't simplify 2√18

[edgy gulch]

you can

18 is 9 * 2

[thin valley]

How?

Oh

So that means 2*3√2?

[edgy gulch]

yes

[thin valley]

So I add 8+6 then?

Idk how to do the division

[edgy gulch]

well you get 14√2 after adding

then you have √2 top and bottom, just cancel it

so you only have to do 14/6

[thin valley]

Alright

Done now number 5 which is the part im most confused

Can 8√6 be simplified?

[edgy gulch]

well you can

but look at all the roots first, what number can they all get reduced to

[thin valley]

√6 means √2x3

√2 can't be simplified or made to Integers

[edgy gulch]

yep that works

all of those numbers in those roots can get reduced to 3

so try to make them all sqrt(3)

[thin valley]

Alright

Could the top roots be 6 and the bottom one be 3?

Meaning the root at the end will be √2

[edgy gulch]

dont think thats the idea here

[thin valley]

How so?

[edgy gulch]

since every root can be reduced to only sqrt(3), you can simplify them by adding or subtracting on top and bottom

and after you simplify it to only one term top and above, you can just cancel the sqrt(3)

[thin valley]

Alright but, uhh according to my teacher if the sqrt is 6 and it becomes √2x3 that means it cant be 8•2√3 as the sqrt of 2 is and irrational number

I'm sorry if I added length to this help

[edgy gulch]

doesnt matter

oh i get what you mean now

24 is 6 * 4

so you wanna simplify sqrt(24) to 2sqrt(6)

[thin valley]

Yeah

[edgy gulch]

yeah thats valid too

[thin valley]

Alright

Ill write that down

Btw the numbers that get Added are the one not in the roots right?

Same with subtraction

Is this correct?

Root 9 is 3 right?

So i should remove the √ in √3?

It should be 2*3 which is 6?

[civic otter]

You said that √9 is 3 ...

So instead of √9 just write 3, as simple as that

[thin valley]

OHHHH I UNDERSTAND NOW

My brain not functioning

So the answer should be 7√2 right?

Tysm now I can get ready for tomorrow's exam

Alright time to close

.close

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[bitter igloo]

Guys

[bitter igloo]

How to do 1ai

To show the radius r of the top side of the cup is 4.5 ?

pls help

pls help with 1 ai

nvn got it

.close

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[bitter igloo]

guys

how to do 2b

how to find upper bound

of that

pls

help

<@&286206848099549185> plss help : (

: (

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[fossil pollen]

1,3,6,10... whats the 50th term

[vast shale]

!status

[vocal sleet]

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

[fossil pollen]

i dont know where to begin

[vast shale]

ok

[fossil pollen]

The common difference are in AP

Thats what i notice

[vast shale]

do you know what a aritmethic progression is

okay

good

what is the common difference

[fossil pollen]

2,3,4...

an-an-1

[vast shale]

is the common difference constant?

[fossil pollen]

no

[vast shale]

ok

have you ever heard about triangular numbers

?

[fossil pollen]

huh

not really no

[peak matrix]

and do you know arithmetic series

[fossil pollen]

yes

[peak matrix]

cool

let's analyze it a bit

[fossil pollen]

oka

[peak matrix]

1, 3, 6, 10

so the difference between the first 2 is 2 as you said

[fossil pollen]

yes

[peak matrix]

so we can also write it as

1, 1 + 2, 6, 10

[fossil pollen]

Indeed

[peak matrix]

then the difference between 2nd and 3rd is 3

[fossil pollen]

Oh wait

sus

yes

nth term is sum of n

[peak matrix]

so 1, 1 + 2, 1 + 2 + 3

[fossil pollen]

!1!1!1!

[peak matrix]

yep

sum of first n natural numbers

[fossil pollen]

im soooo dumb

Thanks a lot

[peak matrix]

which can be found throgh arithmetic series

np

[fossil pollen]

what if it started from some random number instead

[peak matrix]

Yeah, I think he already noticed :]

[fossil pollen]

like say 5,7,10...

[peak matrix]

now the differences are 3, 3, ???

did you mean 5, 7, 11

[fossil pollen]

2,3,4..

[peak matrix]

oh wait

7-5 = 2

yeah i cant count lol

in this case, it's 4 + 1, 4 + 1 + 2, 4 + 1 + 2 + 3...

sso the sequence is just an = 4 + (1 + 2 + 3 + 4 + ... n)

it's same as the 1, 3, 6, 10 ... sequence, just shifted by 4

so the 50th term would be larger by 4

[fossil pollen]

oooh just 4 added

Alrighty thanks a lot

[vocal sleet]

@fossil pollen Has your question been resolved?

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[bleak prawn]

sin(x) + cos(2x) = 0

sin(x) = -cos(2x)

cos(pi/2 - x) = cos(pi - 2x)

pi/2 - x = pi - 2x + 2pik

x = pi/2 + 2pik

sin(x) + cos(2x)

sin(pi/2) + cos(2(pi/2))

1 + (-1)

= 0

So my only point is (pi/2, 0)

But the answer sheet says thats wrong? What did I do incorrectly?

[mellow void]

from line 3 to 4

cos is not injective on [0, 2pi]

[bleak prawn]

Hi flip 🤗

[mellow void]

hellu

[bleak prawn]

What does injective mean?

[mellow void]

oh sorry, injective (one-to-one) means that equal outputs implies equal inputs

f is injective if f(x) = f(y) implies x = y

[bleak prawn]

Ohhhhhhh I think I see what you mean

pi/2 - x = pi - 2x + 2pik AND pi/2 - x = 2x - pi + 2pik

right?

[mellow void]

yeah, exactly

[bleak prawn]

Ok got it. 1 moment I will try solve

Hmm I got a different answer to what they wanted again

pi/2 - x = 2x - pi + 2pik

3pi/2 = x + 2pik

x = 3pi/2 - 2pik

so x = pi/2 (which we already had) and x = 3pi/2

[mellow void]

you should have a 3x floating around, I think

[bleak prawn]

Oh oops!

pi/2 - x = 2x - pi + 2pik
3pi/2 = 3x + 2pik
x = pi/2 - 2pik/3
so x = pi/2 (which we already had) and x = pi/6 and x = -5pi/6 and x = -3pi/2

Hmm its still not the answers they wanted

[mellow void]

these (-5pi/6 and -3pi/2) are outside of the interval

[bleak prawn]

Yeah I realized haha my bad

I dont understand how they got those 2 values for x?

[mellow void]

they could be just 2pi + the other numbers you got

unless if they aren't

[bleak prawn]

Yeah theyre not... wtf did I do wrong 😂

Im so confused

[mellow void]

rewriting the solution space for this (writing k as -k) we have x = pi/2 + 2kpi/3

there are more than two solutions generated here

[bleak prawn]

Ohhhhhhh

I see

Ok I got the answers I wanted

Thank you!!

❤️

.close

[vocal sleet]

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[earnest torrent]

How solve it?

And other question. If {an} convergent sequence and bn is positive bounded sequence lim of an * bn = infinity?

I answered in the test no because if an = n and bn = 1/n the multiplication is constant 1

[frozen bobcat]

typically you'd use logarithms to bring the exponent down and then find the limit however possible

[frozen bobcat]

is the question if the product goes to infinity?

[earnest torrent]

Yea

@frozen bobcat

[frozen bobcat]

then yes your answer is fine

[vocal sleet]

@earnest torrent Has your question been resolved?

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[surreal marten]

The number of pairs of positive integers (x, y) satisfying the equation
x2 + y2 + 2xy – 2008x – 2008y – 2009 = 0 is ______.

[surreal marten]

can any1 help with this question

ig the ans is 1 but can anyone crosschk if i m wrong can anyone tell me correct ans

[spiral turtle]

What pair?

[surreal marten]

bcuz sometimes i m getting ans as 2008 somehow

[surreal marten]

pairs of positive inteegers

[spiral turtle]

You said you're getting 1 pair as the answer

[surreal marten]

yep

[spiral turtle]

What pair did you find?