help-0

[tribal valve]

its not

[oak perch]

I canceled s-r

[tribal valve]

oh

ohhh

how can we be sure that $2r^2+2s^2+5rs \neq 0$

[ocean seal]

dabbingpotato

[oak perch]

So r^2+s^2+5rs/2 =0 then (r+s)^2=-rs/2

I am assuming arithmetic sequence meaning arithmetic sequence of real numbers

so I took the middle root z, minimal root z-r, maximal root z+s

r,s are positive and r doesn’t equal s

[oak perch]

Positive=negative, impossible

[tribal valve]

okay but what if r is positive and s is negative

[oak perch]

No r and s being positive are given

Three real roots

I set the middle root z

minimal z-r, so r>0

Maximal root z+s so s>0

I mean r,s>=0 but not both being 0

If both zero, three roots equal, arithmetic

So r,s>=0, and not both zero

This is rigorous

[tribal valve]

god damn

but couldn't r and s be used interchangebly

if the roots were supposedly something like 10, 13, 19

z = 13

[oak perch]

r=3

s=6

[tribal valve]

could r be -6

and s be -3

tho

[oak perch]

No

[tribal valve]

and it would still work

[oak perch]

I let the minimal root be z-r

And I let the maximal root be z+s

[tribal valve]

how do you decide

why does it have to be like htat

[oak perch]

Because real numbers have order

Given three real number there exist an order on them

[tribal valve]

yes

[oak perch]

Three roots, u,v,w, u<=v<=w

I let z=v

r=v-u>=0

s=w-v>=0

[tribal valve]

okay i get that

[oak perch]

Okay

[tribal valve]

but how did you decided that z + s would be the maximum

[oak perch]

u minimal, w maximal

u=v-(v-u)=v-r

[tribal valve]

alright i get it

[oak perch]

w=v+(w-v)=w+s

[tribal valve]

so the reason they can't be the same

is becaus

e

it would be an arithemetic sequence

if they were both 0

[oak perch]

Because v,v,v is arithmetic

[tribal valve]

so 7 7 7 is an arithmetic sequence

[oak perch]

So one of r,s can be zero, but at least on non-zero

Like (0,pos), (pos,0),(pos,pos)

[oak perch]

So positive =non-positive, contradiction

And s doesn’t equal r is used to cancel s-r both sides on this step

[tribal valve]

ye

so

basically

that means like

1, 1, 1, 1, 1, 1, 1 ... is technically an arithmetic sequence

[oak perch]

I can’t see why this is relevant

I only used: u,v>=0, and can’t both be zero

[tribal valve]

well isn't that what happens when both r and s are 0

[oak perch]

and u doesn’t equal v

Yeah

Okay

[tribal valve]

yeah

so when that happens doesnt it have to be arithmetic

didn't we declare this here

[oak perch]

That’s why we require they can’t both be zero

[tribal valve]

right

so when they are

its arithmetic

[oak perch]

“Not arithmetic”->”that equality fails”

And “not arithmetic” is equivalent to

“r,s>=0, not both zero, r doesn’t equal s”

“That equality fails” becomes “(r+s)^2 doesn’t equal -rs/2

And true, positive on LHS doesn’t equal non-positive ON RHS

[tribal valve]

ok i think i got it

[oak perch]

Good

[tribal valve]

thanks so much

that was super long

[oak perch]

Np

[tribal valve]

ur super smart

i need to level up my math

[oak perch]

Just calculation stuffs

Only patience is required here probably

[tribal valve]

oki tysm

byeeee

[oak perch]

Np

Good day

[tribal valve]

u too

❤️

.close

bot broke

.close

[oak perch]

Bad bot

[tribal valve]

.close

[oak perch]

<@&268886789983436800>

[tribal valve]

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[delicate ibex]

okey so in translation basically it says to calculate the missing angels

how do i do that

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[brisk tree]

For a non-negative random variable $X$ with distribution function $F$, show that $\int_0^{\infty} (1 − F(t))^2 dt \leq (E (\sqrt{X}))^2$, can someone help?

[ocean seal]

Tapi

[tacit arch]

if you don't know the integral definition of the distribution function, then what do you know

[molten pivot]

Maybe expand the square

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[pallid scarab]

Your channel got closed because the discord bot is drunk, allow me to reopen

[pallid scarab]

Write the expectation of sqrt(X) as an integral.
Then, knowing that $P(Event) = \int_{-\infty}^{+\infty}\chi_{(Event)}(t)dt$ (where $\chi_A$ is the indicator function of set A), write F(t) as an integral

[ocean seal]

rafilou2003

[brisk tree]

we've used $P(X \in (-\infty, u])$ to define it, but yea the integral definition is in the notes, i'm trying to use it

[ocean seal]

Tapi

[brisk tree]

ok

[lone heart]

@pallid scarab Has your question been resolved?

[brisk tree]

$\int_0^{\infty} ( 1-\int_{-\infty}^{\infty}\chi_{(P(X \leq t)) )^2 }dt \leq (\int_0^{\infty} \sqrt{x}f(x)dx)^2$

[pallid scarab]

Oh X has a density ?

[brisk tree]

probably not

but im not sure how to write E(sqrt X) in terms of (1-F(.))

[ocean seal]

Tapi

[pallid scarab]

Let's assume it does have a density

Then instead of writing F as integral of indicator of event, write it simply as $F(x) = \int_{-\infty}^x f(t)dt = \int_0^xf(t)dt$ since X is non-negative

[ocean seal]

rafilou2003

[brisk tree]

ok

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@pallid scarab Has your question been resolved?

[lone heart]

@pallid scarab Has your question been resolved?

[lone heart]

@pallid scarab Has your question been resolved?

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[blissful hound]

is cosa+sena=1?

[fallen verge]

no, cos^2(a)+sen^2(a)=1

[blissful hound]

and where can i find all trygonometric identities?

[blissful hound]

AHHHHHHHHHHHHHHHHHHHH OKAY

[fallen verge]

google it, there should be many resources

[blissful hound]

i googled them but it's normally co/ca=tag

and so on

are there any key words?

or do i just need to keep looking?

THANK YOU GOOD SIR

hmmmm,i'll keep that in mind,thx

that was all,thanks for helping me both of you c:

.close

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[hallow forge]

Hi, I'd like help with this question. I know that in total there are 20 marbles. Do I multiply 20 by 1/4 and 3/5 and subtract the sum by 20?

[hallow forge]

Great, but I'm confused on how to multiply fractions with integers.

I converted 1/4 to 25/100
And 3/5 to 60/100

Do I multiply them by 20/1? Do I also convert it to 2000/100? I don't think so...

I can't use a calculator for this

Yes

Oh I see... let me try...

Okay, so now I got 5/20 and 12/20 but... the sum of the two would be 17/20

Therefore, the answer is 3

Cool thx

.close

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[noble schooner]

Hi I need help finding an x value for a curve with inflection points

[noble schooner]

How to get 5pi/3?

Ohh okay

so there is a way to get that value on my calculator?

[fleet widget]

actually i dont understand why its 5pi/6 tho

it should be 11pi/6 ??

[noble schooner]

Yea

Divided by 2

[fleet widget]

idk i was thinking about the trigonometric circle

[noble schooner]

I still am confused how to get the value

What chart 😅?

I’ll show u the calculator I have

[fleet widget]

its bcs i was thinking about that

[noble schooner]

I thought graphing it might help

But I’ll try to use the chart to help me get my answer

[fleet widget]

you can use arcos by doing shift+cos on ur calculator

if you do arcos 1/2 you'll find pi/3

but you wont find the other

[noble schooner]

Yea I can find pi/3

But not 5pi/3

[fleet widget]

you can also use that if that easier to you

[noble schooner]

Thank you I’ve tried and it was almost correct but not quite 😭

[fleet widget]

you understood how the circle works?

[noble schooner]

Oh wait

that did help

okay I got you now

[alpine sable]

inverse cosine has range 0,pi so you wont get 5/3)pi there ig

[fleet widget]

ye

[noble schooner]

I was supposed to look at the positive quadrant of cos

Is there a way to remember this circle without having to memorise it manually?

[fleet widget]

for all the angles i just know them by heart idk

[noble schooner]

Aw okay

I’ll take a sc then but thank you I finally understand

[alpine sable]

hmm you can kinda

[noble schooner]

How?

[fleet widget]

there are ways of remembering

but you'll need to draw it

[alpine sable]

you need like 3-4 values from first quadrant

then apply them to the other quadrants

[fleet widget]

you must know pi/6

pi/4

pi/3

pi/2

and from that you can do something

[alpine sable]

ig thats all the trivial ones you need

[noble schooner]

Okay

[alpine sable]

there are some formulas

like cos(x+y) =.. that could help u derive some

[noble schooner]

I feel like that’ll be a bit confusing to figure out

But now I know how it works

[alpine sable]

yay

[noble schooner]

Thank you guys 🙏

.close

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[subtle gate]

i need an explaination on how

[subtle gate]

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

[naive valley]

well it's not true in general, it's true for some specific x

[subtle gate]

i think i got5 it

yeah alright i did 💀

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[hushed grail]

is it possible to check the continuity of an entire function? ```
my math professor wants us to prove whether a function such as f(x) = x^2-cos(x)/3+cos(x) is continuous or not using calculus 2.4 theorems

but from my understanding these theorems can only prove if a function is continuous at a point?

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@hushed grail Has your question been resolved?

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@hushed grail Has your question been resolved?

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@hushed grail Has your question been resolved?

[tardy stag]

yeah you prove it's continuous at every point

[vale wigeon]

when we say that a function as a whole is continuous we mean that it is continuous at every point in its domain

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[young bramble]

an easy way to do these kinds of problems could be to just use a general length, such as setting the length = 1 and breadth = 1 (a square is just a special rectangle) and simulating the situation

can you go from there?

or you could do the same but with pronumerals

[near lion]

Umm no sry, I suck at maths

[young bramble]

alright

so using the first situation

if length = 1, and breadth = 1, what will both be if you increase both by 10% each?

[near lion]

1.1

[young bramble]

yes

what's the original area using length = 1 and breadth = 1?

and then, what's the new area when length = breadth = 1.1?

[near lion]

Yes

[young bramble]

so the original area would be 1x1
and the new area would be 1.1 x 1.1

and then yuo can compare those results to get the increase of the new area from the OG area.

[near lion]

It's 0.21

Soo

21%?

The answer is c?

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[near lion]

Yes

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[calm bough]

What are the chances that you lose 4 times in a row if you have a win chance of 87%

[modern sedge]

So it's a chance that you don't win and don't win and don't win and don't win

assuming that your chance is independent on previous result, do you know how to compute that? How to compute P(A and B and C and D)?

[modern sedge]

Are you online?

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[half vale]

What is 3+5?

[echo socket]

,w 3 + 5

[ocean seal]

Results provided by WolframAlpha

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

Requested by alonelybean

[echo socket]

.close

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[echo socket]

Don't troll

[half vale]

who trolloed

[echo socket]

You

[half vale]

.close

.close

[echo socket]

The channel is already closing

[half vale]

Idk

Ok

3n+2=7

[echo socket]

.reopen

[lone heart]

✅

[echo socket]

Do you have an actual question to ask or no?

[half vale]

I have a question

3n+2=7

n=1.66?

<@&286206848099549185>

[echo socket]

Approximately speaking, yes

[echo socket]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[echo socket]

The exact form for n would be 5/3 though

[half vale]

Ok

Yeah

Thank you

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[fair condor]

.

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[alpine sable]

hello the one on the left side is the correct anwser with three question marks benief it the right side is my anwser clearly im messing up my algrebra somehwere i just have no idea kinda suck at radicals and stuff

[verbal ermine]

its the same

just written in a different form

[alpine sable]

ah can you help me understand the math to do that cause im just staring at it and yeah

[paper arrow]

determine whether the function f(x) = x/(3+tan(x)) is continuous on interval

[verbal ermine]

yep

[alpine sable]

ty you so much i appreciate it

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[somber spoke]

For the last part

I’ve just drawn the graph and said it doesn’t take values between the turning points

[upbeat gorge]

no, those are values x can’t take

[granite badger]

sorry i misread

[upbeat gorge]

ngl that was my first impression too, then I was like “hold on”

[lime bobcat]

What is your definition of "turning point"?

[somber spoke]

Is this okay or is there like an algebra method theyre after

[somber spoke]

I’ll show u what the graph looks like 1 sec

[granite badger]

i would do partial fractions

[somber spoke]

There’s the minima at (1,1)

And a maxima at (5,1/9)

And clearly the graph doesn’t take values between 1 and 1/9

But I’m not sure if that’s what they’re after

By prove I think they want me to do some algebra or something

[lime bobcat]

The steps for sketching are:

1. Maximum domain of definition
2. X- and Y- intersects
3. Even/odd symmetry or similar easy-to-see symmetry axes
4. Vertical assymptotes
5. Horizontal/Oblicuous assymptotes
6. Monotony
7. Relative extrema
8. Curvature
9. Inflection points

[somber spoke]

Already done that but

Bit

[lime bobcat]

↑ This are the basic things to calculate for sketching a function by hand

[somber spoke]

I was looking at the last part

[lime bobcat]

If you mean the relative extrema. The candidates are x such that f'(x)=0

If you mean the inflection points. The candidates are x such that f''(x)=0.

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@somber spoke Has your question been resolved?

[lone heart]

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[topaz loom]

Hey guys sorry for the mess

I would just like to know If I messed up anywhere during the calculus

Because I don't think I have the right answer, I could be mistaken though

[somber spoke]

Did u check on wolframalpha

[topaz loom]

Yeah it gives me this:

I don't see a ln there

[wind cloak]

Are you using an x for multiplication

[topaz loom]

No

Never

[ocean seal]

NEON

[wind cloak]

Is this the integral?

[topaz loom]

Yes it is

[wind cloak]

Hmm

What Wolfram gave you is right

Why is yours so convoluted

[topaz loom]

Oh wow wait my mistake

Didn't mean to make you think so much

[wind cloak]

Wait a minute did you distribute multiplication over addition

[topaz loom]

Actually no you are right I wrote the integral wrong wait

[wind cloak]

$ab(c + d) \neq ac + ad + bc + bd$

[ocean seal]

NEON

[topaz loom]

This is the integral

[wind cloak]

oh

[topaz loom]

I replaced the minuses with a .

My bad

[wind cloak]

Needs to be a -3/x in the second line

[topaz loom]

Yeah and then my answer is almost good I just have some signs problems

Yup

Alright thank you for making me realise

I got it now

.close

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[cobalt ember]

On the figure is for graphs
ONE
and
B
drawn in the same coor-
dinate system. One is the graph of the function
f
, and
the second is the graph of a stem function
F
to
f
.
one.
Explain which is the graph for
f
, and which one
there is a graph of the stem function
F
.

[cobalt ember]

hey, i have a lack of knowledge regarding this topic specifically how to mathematically calculate this question.

[peak bough]

Hvis F er en stamfunktion til f, så gælder at F'(x) = f(x).

Hvad er hældningen af den grønne graf?

[cobalt ember]

ingen info omkring hældningen men visuelt så er A = -x

der er hældningskoefficienten = -1

[peak bough]

Ja, et estimat er fint nok i den her situation

Men ja, hældningen af A vil ca. være -1

Det interessante at bide mærke i her er at hældningen er konstant

[cobalt ember]

men hvad er F'(x) ift. F(x)??

[peak bough]

Altså hvis grafen A var grafen for funktionen F skulle der gælde at den røde graf skulle være konstant, da F'(x) = f(x)

[peak bough]

Så hvad ville du kunne konkludere fra dette?

[lone heart]

@cobalt ember Has your question been resolved?

[lone heart]

@cobalt ember Has your question been resolved?

[lone heart]

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[marsh drum]

Wouldn't they be independent though?

[wise dirge]

do you know what independent means?

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[stable hill]

umm.. how do I solve this?

[proven leaf]

well, this one looks pretty ugly, firstly maybe solve for x in your first equation?

[lethal belfry]

I assume you know how to simplify the first eqn

For the second one using all the properties of logs

[stable hill]

I think I made the equation harder

[pallid scarab]

For the second one put everything as 8^(...)

[proven leaf]

no that's actually pretty good, the equation on the right you cancel logs

[alpine sable]

assuming you know all the properties
convert base to 2 in the second eqn
write 3 as log8 base 2
and you can the take log on both sides to get an equation without log

[lethal belfry]

For instance the rha/s would be 3+ log(x/y)

[alpine sable]

wont be helpful if you dont know the logarithms properties

[stable hill]

is this valid?

idk what to do after this equation

[proven leaf]

$\log_{8}((x^2y)^2)=\log_{8}\left(\frac{512x}{y}\right)\implies(x^2y)^2=\frac{512x}{y}\implies(x^2(x-2))^2=\frac{512x}{x-2}$

[ocean seal]

MrFancy

[proven leaf]

that's the equation you should end up with I believe

[stable hill]

hmm i will try

what should I do after this?

I haven't learn how to solve this ax³+bx²+cx+d equation yet

[long axle]

I wouldn’t even do all that ngl

[stable hill]

is theres any other solutions?

[long axle]

Yea

I just did it

[stable hill]

Can you show?

[long axle]

I can guide u thru it ig

[stable hill]

sure

[long axle]

Aight sorry I’m good now

Aight so

Let’s start at

[stable hill]

No worries

[long axle]

$(x^2 (x-2))^2 = \frac {512x}{x-2}$

[ocean seal]

Stephen

[long axle]

First distribute that exponent

Don’t expand anything

Lmk what u get

[stable hill]

is it like this?

[long axle]

Yea I mean u can simplify it

$x^2 \cdot (x-2) \cdot x^2 \cdot (x-2)$

[ocean seal]

Stephen

[long axle]

U can combine terms

[long axle]

Hm

What happened to the other x^2

What’s x^2 • x^2

[stable hill]

x⁴

[long axle]

Ok and what about (x-2)(x-2)

Don’t expand

[stable hill]

(x-2)²

[long axle]

Yes

So what do we have in the left side

[stable hill]

512x/x-2

[long axle]

That’s the right side

[stable hill]

ahh my language failed sorry

x⁴(x-2)²

[long axle]

Good

Now

$x^4 (x-2)^2 = \frac {512x}{x-2}$

[ocean seal]

Stephen

[long axle]

Now get all terms with an x on one side

Then simplify without expanding

Lmk what u get

[stable hill]

is it like this?

[long axle]

Yes

Nice

Now u can simplify the left a little more

U can get rid of the denominator how?

[stable hill]

I see

after that I do cube root right?

[long axle]

Yea

[stable hill]

alright

and I get X=4 and X=-2

ignore the X=-2 yes?

since that log -a is invalid

[long axle]

Wait what did u get for (x,y) before eliminating any solutions

[stable hill]

hmm

Sorry, I do it by myself afterwards

[long axle]

So yea, the -2 doesn’t work out with the log, but what y-value did u get for x = 4

[stable hill]

y=2

[long axle]

Yup, now plug that solution back into both equations to see if it still works

[stable hill]

mmm yes

solved then

Thank you very much for your help

[long axle]

Np

[stable hill]

.close

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[vague iris]

What do I need to do here? Just get the derivative of the function then plug in the x-value?

[fathom grove]

yes

that’s exactly what you need to do

[vague iris]

okay tyy

is the slope the y-value I get after plugging in the x?

[fathom grove]

after taking the derivative, yes

[vague iris]

okay thanks!

.close

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[upbeat eagle]

Question

[upbeat eagle]

Now is it enough to throw it in the calcualtor , and it tells you 2 Imaginary ROOTS

Does that guarantee that it means the X axis is -5, +4 and 0

[median oar]

why 0

[upbeat eagle]

Or do you have to do

[upbeat eagle]

im just using the rule from x(x+5)(X-4)

[median oar]

where is the x

[upbeat eagle]

x = 0 x = -5 x=+4

Im just hoping its the same thing, and it can be applied here

Ig not

[median oar]

there's no x

why can u apply it

[upbeat eagle]

Ive never gotten an imaginary number before

So i thought it just means x = 0

[median oar]

that would imply x^2 + 1 = 0 has the x = 0 solution

but clearly that's not true

[upbeat eagle]

yes

[median oar]

0^2 + 1 = 1 ≠ 0

[upbeat eagle]

ok, so I would have to make it a b^2 - 4ac

If i ever get imaginary answers

[median oar]

calculator telling you there's 2 imaginary roots and using the discriminant is the same thing

[upbeat eagle]

ok?

[median oar]

what for

[upbeat eagle]

But then how do u know x = 0

[median oar]

you dont

x = 0 is not true

[upbeat eagle]

oh x > -31

[median oar]

??

no

[upbeat eagle]

no like discrimant there is 2 roots, 1 root and no roots

x>0
x=0
X<0

[median oar]

there is

• 2 roots
• 1 repeated root
• 2 imaginary roots

[upbeat eagle]

ok

Lets start from the beggining.

[median oar]

this is wrong

it is not x that is on the left

it is ∆

∆ denotes the discriminant

[upbeat eagle]

I have a question, is it possible to get 2 real roots as imaginary numbers?

[median oar]

what does this even mean

[upbeat eagle]

Umm so you know how i got 2 imaginary numbers

[median oar]

yeah?

[upbeat eagle]

Like if you throw that into the calc can you get imaginary numbers and have B^2 -4ac >0

like get a value above 0

[median oar]

what are you talking about

[upbeat eagle]

Ok never mind

It doesnt happen

[median oar]

what are you throwing into the calculator

real numbers are imaginary numbers

[upbeat eagle]

OH

Ok

[median oar]

every real number is an imaginary number as well

but not all imaginary numbers are real

only some of them are real

[upbeat eagle]

ok i think i follow

[median oar]

roots also come in "pairs"

but they "reflect" over the real axis

[upbeat eagle]

Yes

Ok

So for that question

Could you explain to me why you would want to use b^2 - 4ac

[median oar]

it tells you if the quadratic has roots or not

[upbeat eagle]

oh ok

[median oar]

(we are only interested in real roots at this point)

[upbeat eagle]

Ah ok

I understand

I also remember doing quadratics which wouldnt work in the calculator as B^2 -4ac

prior to this

Ok Ive understood

haha

thank you

WAITT WAITT WAITT WAITT WAITT WAITT WAITT WAITT WAITT WAITT WAITT.

.

one more question

So in that question , usually if you throw in a quadratic it gives you real numbers

But in this case it doesnt

Does that mean you can assume if you get 2 imaginary numbers no REAL roots exist therfore x DOES not Exist ALWAYS?

[median oar]

Where are you getting x= 0 from

[upbeat eagle]

OH I MEAN x doesnt exist

Like I have x=5 x=-4, and the other quadratic gives two imaginary roots therfore its ONLY 2 roots, 5 and -4 which are x

Sorry for bad wording

[median oar]

What you mean is

The only 2 real values of x that satisfy this equation is 5 and -4

And yes

[upbeat eagle]

Ok and the follow up question I have

is that whenever you recieve imaginary roots from a quadratic or anything X^3 x^4

whatever

It means that x does not exist

as in there are no solutions for it

therfore (IF I had to draw a graph) i wouldnt draw anything intercepting x

[median oar]

Well

[upbeat eagle]

Or at least in graph scenario

[median oar]

x³ always has 1 root

[upbeat eagle]

[median oar]

At least

[upbeat eagle]

Ok X^2

then

[median oar]

Every odd power has at least 1 real root

[upbeat eagle]

good to know for the futre ill write that down

[median oar]

(We are considering only polynomial with real coefficients atm)

[upbeat eagle]

Yes

Polynomial is?

intigers?

[median oar]

aₙxⁿ + … + a₁x¹ + a₀

[upbeat eagle]

Ah ok

but im doing ax^3 bx^2 cx d

or ax^4 ect

[median oar]

That’s a polynomial of degree 3

That’s degree 4

[upbeat eagle]

ah ok

[median oar]

Think about it

When x is super negative

Odd power makes it negative

[upbeat eagle]

yes

[median oar]

When x is very positive big

The polynomial is positive and big

[upbeat eagle]

yes...

[median oar]

Somewhere in the middle it has to cross the x axis

So it can go from very negative to very positive

[upbeat eagle]

does it have to?

[median oar]

Yes

Polynomials are continuous

[upbeat eagle]

Cant it just start otnop of x axist and continue in straight line

like x+5 or something

[median oar]

Nope

[median oar]

This has to go very negative on the left side

So it will be under the x axis

[upbeat eagle]

ok y= 5

[median oar]

The to the right it’ll be very positive

That has even degree

Not odd degree

[upbeat eagle]

Ok i think ive understoon

Also polynomial

Does that mean like curves

[median oar]

No

[upbeat eagle]

so would like something y=5 not be a poolynomail

[median oar]

Polynomial refers strictly to these things

[median oar]

And no

That is a polynomial

[upbeat eagle]

oh ok

[median oar]

It is 0xⁿ + … + 0x¹ + 5

[upbeat eagle]

GOT IT!

Ok that was nice of you

[median oar]

👍

I’m actually about to fall asleep haha

[upbeat eagle]

haha well i sure hope you have an amazing night

sleeping

Becuase u def deserve that

I apprecaite it

a lot

.close

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[alpine sable]

.close

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[vast sonnet]

Consider the triangle ABC, with a right-angled at A, AC = 6 and tg C = rad3 . Show that the area of the triangle ABC is equal to 18rad3 .

[vast sonnet]

how would I go about drawing this?

because afterwards i think i can solve it

[plush crow]

can you show ur attempts?

[vast sonnet]

i didnt start it cause i dont know how to draw the triangle

but i can do it after i have it drawn out

[plush crow]

understood

so, lets start

you know the triangle have a right angle (90 degree or π)

[vast sonnet]

yes

[plush crow]

make a line to the right and this point will be c

the distance has 6

[vast sonnet]

could you explain the logic behind drawing it like this?

and maybe also provide a rough sketch so that i dont misunderstand something

[plush crow]

first, look the problem and see what it gives to you

make a list for me, only the important things

like: triangle ABC with right angle

[vast sonnet]

right-angled at A, AC = 6 and tg C = rad3

[plush crow]

nice

now, what you wanna prove?

[vast sonnet]

area is equal to 18rad3

[plush crow]

now we start with the basic, drawing what we know

A right angle, having 6 as distance with C

when we have a trignometric function, we can get the others sides of the triangle

so we draw it, we don't know the value, so give a letter, maybe x

understand that we start with what we know, and later complete with what we want know?

The A is right angle, so the AB will be perpendicular to AC

and we have a fig

did you manage to draw?

[vast sonnet]

This is what I drew earlier

And this is what you meant aswell, no?

The thing is

How do you knos

That it can't look like that

Or is it that it can

But the results are the same either way?

[plush crow]

ah

[plush crow]

.

this

[vast sonnet]

I c

Thanks!

.close

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[tall talon]

$$\begin{cases}x\left(y+z\right)=a&\ y\left(x+z\right)=b&\ z\left(x+y\right)=c&\end{cases}$$
$$\begin{cases}x+y=\frac{c}{z}&\ x+z=\frac{b}{y}&\ y+z=\frac{a}{x}&\end{cases}$$
$$\begin{cases}xy+x\left(\frac{a}{x}-y\right)=a&\ xy:+y\left(\frac{b}{y}-x\right)=b&\ x\left(\frac{a}{x}-y\right)+y\left(\frac{b}{y}-x\right)=c&\end{cases}$$
from the last equation:
$$a-xy+b-xy=c$$
$$a+b-c=2xy$$
$$xy=\frac{a+b-c}{2}$$
$$\frac{a+b-c}{2}+xz=a$$
$$xz=\frac{2a}{2}-\frac{a+b-c}{2}$$
$$xz=\frac{a-b+c}{2}$$
repeating the same process we get:
$$\begin{cases}xy=\frac{a+b-c}{2}&\ xz=\frac{a-b+c}{2}&\ yz=\frac{b-a+c}{2}&\end{cases}$$
From which we can deduce:
$$\frac{a+b-c}{2y}=\frac{a-b+c}{2z}$$
$$\frac{a+b-c}{y}=\frac{a-b+c}{z}$$
$$y\left(a-b+c\right)=z\left(a+b-c\right)$$
$$yz\left(a-b+c\right)=z^2\left(a+b-c\right)$$
$$\frac{\left(b-a+c\right)\left(a-b+c\right)}{2}=z^2\left(a+b-c\right)$$
$$\frac{\frac{1}{2}\left(b-a+c\right)\left(a-b+c\right)}{\left(a+b-c\right)}=z^2$$
$$z=\pm :\sqrt{\frac{\frac{1}{2}\left(b-a+c\right)\left(a-b+c\right)}{\left(a+b-c\right)}}$$

[ocean seal]

themathboi #2137
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tall talon]

Hello. I was told that my working is correct, but since a, b, c can be any values, z would be undefined in some cases.

How can I avoid this?

[lone heart]

@tall talon Has your question been resolved?

[tall talon]

<@&286206848099549185>

[fluid venture]

can z be a complex number of has it to be a real number?

[tall talon]

There are no restrictions for z

[fluid venture]

in that case just make sure that a+b doesn't equal c

[tall talon]

This is what I am trying to avoid. A, b and c are random, so I need to have a universal formula for z.

[exotic belfry]

as a, b, c can be any values it is not guaranteed that there is a solution for x,y,z in all cases.

[tall talon]

fair enough,
if I somehow wanted a, b, c to be real numbers, how could I get rid of this square root in my notation?

[exotic belfry]

for example you divide by x, y, z in the first step. what if one of them would be zero?

[tall talon]

That is part of my problem

[exotic belfry]

you are using mathematical oparations (eg. division by something) which are not always defined. you need to handle this cases seperatly. for example. what would happen if z = 0? would mean that c has to be zero. or youre dividing by a+b-c. handle the case a+b-c = 0 separatly.

your solution (last line) is not valid in all cases. add the neccessary conditions. and do the cases where this conditions are not fullfilled separatly.

[tall talon]

thanks, this problem turned out to be way harder than I thought.

[exotic belfry]

i dont think so. your way is right. but you have to do some special cases.

[tall talon]

I'll definitely try

.close

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[lone heart]

Please don't occupy multiple help channels.

[wary stream]

You have #help-5 already

And don't make people download stuff

[spice jungle]

i cant see help 5

[lone heart]

@spice jungle Has your question been resolved?

[spice jungle]

no

help5 is blocked

if can get help here

CAN YOU HELP ME NOW PLEASE @wary stream

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@spice jungle Has your question been resolved?

[spice jungle]

no

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[potent oasis]

Could someone help me solve this? I have tried u substitution but I can’t figure out how to do it because the du would be in the denominator. Could anyone help me figure out where to start?

[fallen verge]

Thats a trig derivative

[potent oasis]

But how do I get rid of the x in the front?

[tardy tapir]

let x = sec(t)

[lone heart]

@potent oasis Has your question been resolved?

[potent oasis]

Why is that though? I don’t understand why

[tardy stag]

because we can use the identity $\sec^2\theta - 1 = \tan^2\theta$

[ocean seal]

hayley!

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[quiet cliff]

Hi i need help factoring polynomials the question is to factor x^2 + 6x

[viral prism]

what is the common factor?

[quiet cliff]

it would be 1 right?

[viral prism]

x

[quiet cliff]

so it would be somethng like (x+) (x+ )

but how would you avoid getting 3 numbers

[viral prism]

write with ``` (start and end with these symbols) to ignore formatting

(x)(x+6)```

do you understand this? @quiet cliff

[quiet cliff]

so you dont have to have a variable + a number

[viral prism]

you mean x^2 + 6x + another number?

or a number with x^2 ?

oh

i get what you mean

well here technically it is (x+0)(x+6) — you can't have any other number.

for these types of expressions

[quiet cliff]

oh that makes sense

thanks

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[foggy sonnet]

I'm a bit confused on the last step of this

y^2-6y-40 becomes (y-10)(y+4)?

[charred summit]

Yes?

What do you want to ask about exactly

[foggy sonnet]

I'm unsure of the step

how does this happen

[charred summit]

You can take (y-10) as common factor and you will get the final expression

[foggy sonnet]

understood. one moment

this would be "greatest common factor", yes?

[charred summit]

Yes it is the greatest common factor

But usually it is not needed to be the greatest common factor to be able to do that
You can do it with any common factor

[foggy sonnet]

I see

how would you solve this problem?

[charred summit]

Taking (y-10) as a common factor?

[foggy sonnet]

ok

[charred summit]

All good?

[foggy sonnet]

for the time, I will look into this further as i don't quite understand. thank you for your time.

.close

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[charred summit]

Wait@foggy sonnet

[foggy sonnet]

?

[charred summit]

Do
.reopen

[foggy sonnet]

.reopen

[lone heart]

✅

[charred summit]

Ok so do you know about what is common factor is

[foggy sonnet]

Honestly I haven't taken a math class in 3 years, algebra in 4

I was once taught it but I don't remember

[charred summit]

So: it is the same thing when we say
$$ax+bx = x(a+b)$$

[ocean seal]

Sherif Player

[charred summit]

Do you remember something like that

[foggy sonnet]

yes

[charred summit]

Here we call x as a common factor
As it is a factor in both
ax and bx

Taking the common factor means we multiply it by all the values in one brackets then divide each one by it

So taking common factor of
(3 + 9)
Be like this
$$3(\frac{3}{3}+\frac{9}{3})$$
$$3(1+3)$$

[ocean seal]

Sherif Player

[foggy sonnet]

yes

I recognize this from similar problems.

[charred summit]

So do you know how to apply that to the question you asked?

[foggy sonnet]

no

[charred summit]

So focus on the two values in the equation
$$y(y-10) + 4(y-10)$$
Don't you see some common factor there

[ocean seal]

Sherif Player

[foggy sonnet]

I see that y^2-6y-40 becomes y-10 y+4
and I see that -40 is from -10(4)
but im unsure on how to get that from the other numbers.

I don't see what to do with the 6y

[charred summit]

You may multiply the brackets then solve it by factorization
But it would be easier if you thought about it in the common factor way
I will explain both of them for you

So focus on the two values in the equation
$$y(y-10) + 4(y-10)$$
Don't you see some common factor there.

[ocean seal]

Sherif Player

[foggy sonnet]

sorry to interupt, but if i go back a step i have y^2-10y+4y-40

[charred summit]

If you can see it
(y-10) is multiplied in the both 2 values
Which means that it is possible to take it as a common factor
Which will result in
$$(y-10)(\frac{y(y-10)}{(y-10)} + \frac{4(y-10)}{(y-10)})$$
$$(y-10)(y+4)$$

[ocean seal]

Sherif Player

[foggy sonnet]

I would never get to y^2-6y-40, would I?

I just took a wrong turn

[charred summit]

This is actually the correct one

[charred summit]

Do you understand this?

[foggy sonnet]

honestly no but i figure i'm just too out of the loop and i eventually will

from y^2-10y+4y-40
are there more actual steps to get to (y-10)(y+4) or is it just a direct thing with the middle 2 numbers sharing the same variable

[charred summit]

So what about we say that
u = (y-10)
So we get the equation
$$yu + 4u$$
So here u can be taken as a common factor
So
$$u(y+4)$$
Substitute u from above to here
$$(y-10)(y+4)$$

[ocean seal]

Sherif Player

[charred summit]

Is that more clear?

[foggy sonnet]

less, honestly. I'm gonna look into this more regardless dude

[charred summit]

So about that there is an actual step called factorization

Which I would not be able to explain to you in just text
I can send a video link about factorization

[foggy sonnet]

I pulled one up from "maths explained"

i'll figure it out eventually, i'm sure.

[charred summit]

https://youtu.be/qeByhTF8WEw?si=ibY-ov1CuCu1SZUh

I think that will help you

[foggy sonnet]

thank you for helping me, I'll get up to speed

[charred summit]

Also I recommend you to watch about the common factor

[foggy sonnet]

will do

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[charred summit]

https://youtu.be/FoTD41xzF5Q?si=-XPq_cJd8HWtjN55

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[quiet cliff]

Need help simplifing the expression 4x-8 over 16

[serene isle]

have you done that kind of question before

or have you just got no idea where to start?

[quiet cliff]

i have but this one is throwing me off

[serene isle]

its $\frac{4x-8}{16}$ right?

[ocean seal]

Poelymole

[quiet cliff]

yes

[serene isle]

ok so what would your first step be usually

[quiet cliff]

figure out the gcf

[serene isle]

sure

which would be

[quiet cliff]

4

[serene isle]

then what

[quiet cliff]

factor the top

[serene isle]

$\frac{4(x-2)}{16}$

[ocean seal]

Poelymole

[serene isle]

and now?

[quiet cliff]

im stuck here

1/4th

why would that not affect the x-2

[serene isle]

this can be written as $4 \times (x-2) \times \frac{1}{16}$

[ocean seal]

Poelymole

[serene isle]

you with me?

[quiet cliff]

still not really getting it

[serene isle]

you can do multiplication in any order

or write it as $\frac{4}{16}\cdot (x-2)$

[ocean seal]

Poelymole

[quiet cliff]

oh ok

yeah

thanks

[serene isle]

nice

[quiet cliff]

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[tight blaze]

I just need help with part A

[tight blaze]

I got 15.2

x axis 13^2 = 169 ~ y axis 8^2 = 64 = square root of 233 is 15.2

i don't think that's right tho

[maiden glen]

that's right

it says give an exact answer though, so you should leave that as sqrt(233)

[tight blaze]

ah ok ~ thank you

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[subtle blade]

can anyone help me find where i went wrong here? the answer i got in b is too large so i think i did something wrong but im not sure where

[subtle blade]

i think my mistake might have something to do with the ft^3 in the first problem but i have no idea

[lone heart]

@subtle blade Has your question been resolved?

[subtle blade]

<@&286206848099549185>

[lone heart]

@subtle blade Has your question been resolved?

[upbeat gorge]

I’d leave it as pi if they don’t tell you to take pi as some estimate

Otherwise I don’t see any problem with the solution

[subtle blade]

@upbeat gorge my teacher said to estimate pi

18151.44 seconds seems too big of a number right?

thats 304 minutes and idk if it takes that long for something to fill a tank

[upbeat gorge]

Well you would need 16,336-ish pounds of water, which is quite a lot

A cylindrical tank with base radius 4 feet, which means a diameter of 8 feet (roughly one a half people of average height), plus a height of 5 feet (roughly one person)

Plus the sea water is pretty dense

So yeah, it makes sense, but typically reality is one of the last things you check in math problems lmao

[subtle blade]

ohhh

thank you so much i appreciate it!

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[vapid root]

I'm currently trying to prove this. Let y be a positive real number. Then for all n in N, there exist a unique positive real number x such thatx^n = y.

[vapid root]

Intuitively, this make sense. But then, how would this be possible if y is 1.

[naive valley]

what facts do you know about the function f(x) = x^n?

[vapid root]

x^2 = y if y is a prime

[naive valley]

x = 1 would work, no?

[vapid root]

there can't be unique x for all n if y is 1 though

[naive valley]

i don't understand, what goes wrong?

[vapid root]

I need to prove that for all integer n, there is a unique x so that x^n = y. But this can't be true if y= 1. X can only be equal to 1 for all n

[naive valley]

well what's wrong with x = 1? it's a correct solution

1^n = 1 for any integer n (including n=0 and n<0)

[vapid root]

But x wouldn't fit the unique criteria though. Should there be different x for different n

[naive valley]

that's not required

there just has to be exactly one x that works for a given n

no one says that the same x is not allowed to work for multiple n

although that's generally not going to be the case

[vapid root]

Perhaps this is due to wording but doesn't "there exist" implied a different x for a different n?
ex: x= 1 for n=1; x=2 for n=2 so on and so forth?

[naive valley]

no, that is not implied

it will be true for most values of y, but not for y = 1

[vapid root]

i c ic

ty

.closed

[wary stream]

It's .close

[vapid root]

oof

ty

.close

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[alpine sable]

help

[alpine sable]

wouldnt there be an infite number of vectors?

as long as y = -3/4x

it would equal 0

so <x, -3/4x> ?

[lone heart]

@alpine sable Has your question been resolved?

[swift talon]

pretty much but it's good to give your answer in a general form so something like x(1,-3/4) where x is some arbitrary constant

but there's more than 1 answer don't forget

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@alpine sable Has your question been resolved?

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[whole otter]

Product… rule???

[alpine sable]

looks like chain rule?

[whole otter]

How would you solve this

With a monster of an answer

That steps hella long probably

[alpine sable]

derivative (9x²)^x?

[whole otter]

Yes

I don’t see how it is product file

Rule

Maybe chain rule

[alpine sable]

you know the chain rule?

[whole otter]

I can see that possibility

[alpine sable]

me neither

[whole otter]

Yes

I know chain rule

f’(g(x)) * g’(x)

[alpine sable]

lemme think

yea u might actually need product rule later

[whole otter]

So how would you do it

[alpine sable]

find f(x) g(x)

[whole otter]

P obvious what is f and g

(a)^x is f

9x^2 = g

Hello?