help-13

[tepid ginkgo]

how you solve it exactly?

[dull oxide]

Try following this probability first:
P(3 TIMES GETTING SOMETHING YELLOW FROM 4 TRYS)

Find it in terms of your unknown, N

[slow thicket]

yakir do you want to talk in Hebrew

would that be easier for you

[tepid ginkgo]

do you speak hebrew?

[slow thicket]

yes

[tepid ginkgo]

תעזור לי בבקשה

[slow thicket]

can you call

I cant type in Hebrew

[slow thicket]

oh that’s why the English was bad

which part

[tepid ginkgo]

א

[slow thicket]

can you call

it would be easier for me to explain

I’m not so fluent in reading/writing

[tepid ginkgo]

I called you

[tepid ginkgo]

.

[slow thicket]

I was on dnd

[tepid ginkgo]

dnd?

[slow thicket]

mic issues one second

do not disturb

[dull oxide]

Try #discussion

Dms against rules

!help

[cedar kiln]

Please read #❓how-to-get-help

@tepid ginkgo Has your question been resolved?

[cedar kiln]

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[winter socket]

hi there, i need help evaluating this integral. i am confused on what to do since the upper bound number is smaller than the bottom bound number. would i integrate as normal?

[winter socket]

i was told the largest number always has to be upper bound

[dull oxide]

this is not necessary

It's standard, but not necessary

[winter socket]

oh alright

[winter socket]

yeah i was never taught that. lol

[dull oxide]

Also, why are you integrating all the way to 5?

The bounds of your integral are on the same side of the absolute value

[winter socket]

sorry, i don’t understand what you mean. i thought you sub in what you get for x in the bound

on each side

[dull oxide]

I don't follow

[cedar kiln]

@winter socket Has your question been resolved?

[winter socket]

sorry that i took a while, i think this will be a more appropriate question.

for the question to the left, i got the c value and plugged it into the appropriate bound. my question i why i cannot do this to the question on the right. i am getting the wrong answer.

<@&286206848099549185>

.close

[cedar kiln]

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[cedar kiln]

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[nova snow]

What is the “golden ratio” I don’t quite understand it

[bright bridge]

wdym

[nova snow]

What is it?

[bright bridge]

(1+√5)/2

[nova snow]

Yeah but what does it mean

[boreal epoch]

did you research on google?

[dull oxide]

Wikipedia is a good resource when in doubt

[nova snow]

yeah I don’t understand their explanation

[dull oxide]

Wikis explanation?

[boreal epoch]

which part did you read that makes no sense @nova snow

[cedar kiln]

@nova snow Has your question been resolved?

[cedar kiln]

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[sweet forge]

Let G be a finite group, with subgroup H. Suppose for some x in G that G - H = [x] the conjugacy class of x.

I need to show that for all g ≠ e:
gg = e ⟺ g ∉ H

[sweet forge]

i can see that |H| = |[x]| = |G| / 2

but what does that tell me

[humble marsh]

where does this come from?

[sweet forge]

hmm, because the order of a conjugacy class [x] must divide the order of G, and the order of H must as well

and [x] union H = G

so |[x]| + |H| = |G|

[humble marsh]

ooh, G\H, not G/H

sorry, misread

[sweet forge]

oh yeah

[severe linden]

I extracted this information from the question
f(x(x=1))=27
f(x)=1/3 f (x-1)

what would be the sequence

the answer i got went into fractions

<@&286206848099549185>

[humble marsh]

#❓how-to-get-help

somebody else is asking a question here

[severe linden]

omg sorry cuh

sorry gang

[sweet forge]

so g ∉ H means g ∈ [x]

[humble marsh]

if g ∈ H, write g = hxh⁻¹ and show that g² ≠ e

that shows it in one direction

[sweet forge]

hxh⁻¹ is in H?

but x is not in H

xhx⁻¹ maybe?

[cedar kiln]

@sweet forge Has your question been resolved?

[drifting violet]

notice that,
g not in H => [g] = G \ H => Z(g) = {e, g} => g^2 = e

can you do the converse now?

[sweet forge]

why is Z(g) = {e, g} ?
what is Z(g)?

[drifting violet]

centralizer

and that's because you noticed that size of the orbit [g] is exactly |G|/2

so size of centralizer must be 2

and clearly it contains e and g (which are different)

[sweet forge]

i get it. i think. because of the orbit stabilizer theorem

[drifting violet]

yep :3

[sweet forge]

but i dont follow with the last step

[drifting violet]

so the centralizer has to contain <g> as g commutes with its powers, in particular g^2 is also in Z(g)

so you're forced to have g^2 = e

[sweet forge]

ah yes, of course, thanks !

[drifting violet]

[cedar kiln]

@sweet forge Has your question been resolved?

[cedar kiln]

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[nimble veldt]

are you trying to cheat a test?

[modest wind]

Help

[nimble veldt]

@modest wind read #❓how-to-get-help

[modest wind]

How to solve this question

[nimble veldt]

do not occupy another ones help-channel, read #❓how-to-get-help

[cedar kiln]

@fluid glade Has your question been resolved?

[cedar kiln]

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[cedar kiln]

Please don't occupy multiple help channels.

[crimson sedge]

Is the bottom wrong or right

[sand orbit]

wrong, it would be -1 and how did u solve the powers?

[cold briar]

i think that was multiplication, not powers

[sand orbit]

yeah

[cold briar]

he was showing what he multiplied both the numerator and denominator with

[sand orbit]

so, what abt the powers raised up there?

[cold briar]

where?

[sand orbit]

x/2^-3

[crimson sedge]

Can you show it on a sheet

What u mean by that

[cold briar]

^

[crimson sedge]

Like where did I do it wrong

[cold briar]

it's not a power, it's just showing what he did

[sand orbit]

uh okay

[cold briar]

he basically meant
x/2 * 3/3 = 3x / 6

[sand orbit]

2/2 = -2x/2

[cold briar]

in a very ambiguous way

[sand orbit]

oh

[cold briar]

yes?

[crimson sedge]

I dont understand

Can you show me on the paper

?

[cold briar]

oh he's talking about the part
2 = -2x

you divided both side by 2 right?

[crimson sedge]

Yes

[cold briar]

let me ask you this...

what is -2 / 2

[crimson sedge]

-1?

Do you mean this bottom part?

[cold briar]

yes

[crimson sedge]

So because X is -2X

On the right

I have to divide 2 by -2 on the left?

[cold briar]

yes

[crimson sedge]

Where do I put the -

On the top or bottom

[cold briar]

wait...

wdym

[crimson sedge]

Like on the left side

[cold briar]

ok so basically

2 = -2x
right?

and you want to divide by -2

to get x

[crimson sedge]

Do you want me to put the - here

[cold briar]

no no no

wait lemme just explain it slowly

[crimson sedge]

Ok

[cold briar]

2 = -2x right?

[crimson sedge]

Yup

[cold briar]

so what you did was divide both sides by 2

[crimson sedge]

Yes

[cold briar]

2 / 2 = 1
and -2 x / 2 = -x right?

[crimson sedge]

Yes

Except

I didnt write -x

But yea

[cold briar]

yes

exactly that's where you went wrong

[crimson sedge]

Ah

And what now

X is -

Is that it

[cold briar]

no

you need to find x

there is a couple of ways to do it

[crimson sedge]

Can I do it like this

[cedar kiln]

@crimson sedge Has your question been resolved?

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[dull oxide]

This drawing is wildly not to scale

[pulsar fox]

Yeah, it actually does not makes any sense.
(According to me)

[ionic thistle]

i really want an answer for it

[pulsar fox]

@ionic thistle Have you been taught any methods in your school or university regarding these types of problems ?_?

[ionic thistle]

i'm studying "yos" by my self

for university

[pulsar fox]

Hmmmmmm

[ionic thistle]

where can i find a community expert in this kind of question?

[jagged spire]

Looks like the pictures used in intelligence test, so I think not a geometry question but a find the pattern question

[pulsar fox]

Maybe you can try pinging @Helpers for advance help

[dull oxide]

There's no correct answer

[zealous compass]

If nothing is in proportion the answer is whatever I want it to be

[dull oxide]

K must be 14 and L>0, so no answer you can choose will be right

[ionic thistle]

according to what?

[dull oxide]

Assuming the interiors are rectangles

[ionic thistle]

make sense

[dull oxide]

This is a bad problem

It's not to scale and nothing is mentioned about proportion

Whoever gave you this problem is betraying your education

[ionic thistle]

i have question

what do u call this kind of question??

[ionic thistle]

.

[dull oxide]

I would call this problem a lot of things:
Bad
Not thought out
Ill formed
A poor representation of university mathematics
A waste of time

[vague rapids]

We may try to look for a pattern

instead of geometery

[dull oxide]

My guess is its supposed to be a geometry problem, but it betrays every idea of geometry

If someone gave you this problem to prepare you for university, they are wasting your time.

[vague rapids]

I don't think that's a geometry problem since those area would be almost impossible to calculate

[dull oxide]

Having said that, maybe beard will surprise us with an interesting approach

[ionic thistle]

i mean specific name to search about this kind of question?

[dull oxide]

It may be a logic question

[vague rapids]

Either geometry or Mental Ability

Most probably not geometry

[dull oxide]

Maybe the outer triangles must sum to the inner squares

It's definitely not geometry

[vague rapids]

Number figure 'relation'

[pulsar fox]

@ionic thistle If you dont mind me asking, where did you found this question ?_?

[ionic thistle]

like i said i'm studying "yos" in turkey for university

[pulsar fox]

Bruuh

[dull oxide]

Yos?

[pulsar fox]

What does Yos stands for ?_?

[vague rapids]

Do you have the answer with you?

[ionic thistle]

Algebra

and geometry

[pulsar fox]

I don't think my man has an answer for it. It surely is a weird question.

[pulsar fox]

[ionic thistle]

no i dont

[vague rapids]

You're studying for geometry. So a random mental ability question shouldn't pop up

[ionic thistle]

and intelligence quotient

[vague rapids]

Oh this

[vague rapids]

Well, mental ability is in this category

[ionic thistle]

okey

thank u everyone

[dull oxide]

Yes but who or what assigned the problem to you?

Was it a school? A company?

[ionic thistle]

no . we have books that have different questions to practice on and that was one of them

[dull oxide]

who gave you the book?

who recommended you the book?

[ionic thistle]

it name is metropol intelligence quotient

its like the best

[vague rapids]

Yeah then definitely a mental ability question

[dull oxide]

colleges will not care about this stuff

Maybe turkey is different.

But your time is better spent studying what they actually teach in college

[ionic thistle]

i mean it has an exam for any one came from outside the turkey

and they call it yos

without it there is no university

[dull oxide]

what's this book called?

[ionic thistle]

metropol intelligence quotient

[cedar kiln]

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[strange fulcrum]

Why is there not a maxima and minima?

[jagged spire]

Use secondary condition, x+y=1 to make function have only one variable. Then you can use derivatives or a key fact about polynomial functions

[strange fulcrum]

x=1-y -----> x'= -1?

[jagged spire]

No, don't derive yet because this is only a condition but they are asking you about the function f(x,y).

You can make a new function with the x that you have defined, for example. f(x,y) = f(1 - y, y) = ?

[strange fulcrum]

oh yeah ofc

but i don't understand why there is not a maximi or minimi?

[jagged spire]

I can show you a graph of the function, would that help to understand?

[strange fulcrum]

Yea please

[jagged spire]

ok

I must have misunderstood the translation, I see the problem now.

Sorry for wasting your time.

[strange fulcrum]

no problem

How do I solve these equation?

[cedar kiln]

@strange fulcrum Has your question been resolved?

[lethal coral]

By using the first one and second you can get z in function of y, and x in function of y

Then you can substitute them into the last equation

I think

[strange fulcrum]

So z=2y-x and the second x=4-3z-2y?

[jagged spire]

plug in z into the second

[strange fulcrum]

Oh yeah then I get a function x in y

[jagged spire]

can put this back into the first equation (for z) to get everything in y or x

[lethal coral]

no, You need to get rid of x

[strange fulcrum]

If I plug in z in the second equation I get
X+8y=7

Which means x=7-8y

[lethal coral]

or you can do this
(x+2y+3z)-(x-3y+z)=4-0

[strange fulcrum]

We get 5y+2z=4

[lethal coral]

and (x+2y+3z)-(3(x-3y+z))=4-(3*0)

[strange fulcrum]

What should I do now

[lethal coral]

(x+2y+3z)-(3(x-3y+z))=4-(3*0)

(x+2y+3z)-(3x-9y+3z))=4

[strange fulcrum]

Where did 3x-9y+3z come from

[lethal coral]

OK
c=d ==> a.c = a.d

[strange fulcrum]

Yea

But the first method seems easier

[strange fulcrum]

Like this

[lethal coral]

you need both

[strange fulcrum]

Oh

[lethal coral]

the first for get z,y and second to get y,x

[strange fulcrum]

I got z =2y-x

Should I plug it in the second equation?

I’m so confused right now

What is the first step??

[jagged spire]

Since $z=2y-x$, $$x=4-3z-2y=4-3(2y-x)-2y=4-6y+3x$$ So then, $6y-4 = 2x$ and thus $3y-2=x$. Can plug back into $z$ to get $z=2y-(3y-2)=2y-3y+2=-y+2$. Now you have $z=2-y$ and $x=3y-2$. Can plug this into $x^2+y^2+z^2=83/7$ to get quadratic equation in $y$.

[wraith dagger]

Ara

[lethal coral]

hhhhh, My method is not the same way as Ara
Of course, there is more than one way to solve

[strange fulcrum]

Bro right now I don’t know what to do

I don’t understand either

I just understand z=2y-x

Then I don’t know what to do

[jagged spire]

You also have x=4-6y+3x

[strange fulcrum]

Where did that come from

[jagged spire]

oh I see

sorry, I confused the equations

[jagged spire]

Let me go back to this then

Plug the z=2y-x into x=4-3z-2y

[strange fulcrum]

Okay

I got that

[jagged spire]

To make the equation easier to solve, we want to make the equation have as few different letters as possible. In this case we just picked y, so try to put everything in terms of y. Then continue this process for the third equation

To start with, x is in terms of z and y, but we want x only in terms of y, so plug in z into x=4-3z-2y to try to get rid of anything that's different from y. In the end it should look like x = only y's

[strange fulcrum]

Okay after I plugged z=2y-x inside my second equation I got x=4y-2

Should I plug x into z now

[jagged spire]

yes, to make z only in terms of y

we want everything in terms of y, so this is the next step before we turn the third equation into all y's

[strange fulcrum]

Okay now I got x=4y-2 and z=-2y+2

[jagged spire]

ok now do the same method to third equation

[strange fulcrum]

I got y1=-1/7 and y2=9/7

But here in the book they called it t

I don’t know why

Oh they did same, they just used y like t

[jagged spire]

Ah, this is a different way of solving called parametric

[strange fulcrum]

Its same thing they just determined from begging y=t

But thanks so much

I just have one question

Could we say x=t

Or z=t would it work?

Why they chooses y=t

And why did you also choose y?

Or could you choose any variable?

[strange fulcrum]

@jagged spire

[jagged spire]

I chose y for two reasons. The other helper chose y, so I stick with that, but also I think they chose y because you had two equations:

"So z=2y-x and the second x=4-3z-2y?"

the first is in terms of x and y, the second is in terms of y and z. pick what they have in common = y

[strange fulcrum]

Okay thanks so much

I appreciate the help so much. I understand so much better now

[jagged spire]

You're welcome. Good luck with your math

[cedar kiln]

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[tribal sphinx]

"The two polynomials in the unknown x, ax^3-2x^2-x+3 and (a+b)x^2+cx+6 are proportional for: A) a=0,b=-2,c=-1 B)no value of a,b,c C) a=0 and any value of b and c D)a=0,b=-4,c=-2 E)every value of a,b,c". I think the answer is B since the first polynomial has x^3 and and the second x^2 so no matter what's the value of a,b,c they'll never be proportional. Does this make sense? Or am i wrong?

[cedar kiln]

@tribal sphinx Has your question been resolved?

[cerulean sail]

“…since the first polynomial has x^3…”
Careful with that reasoning: what if a was zero?

Technically the second one has x^3 in it too, just that its coefficient is zero

[tribal sphinx]

right, i didn't think that far lol

okay, i think it could be D but i will have to divide each element by 2 to get the same on both sides

just tried again and A could also be the answer

[cerulean sail]

Right, you have that $a x^{3} - 2x^{2} - x + 3$ and $(a + b) x^{2} + cx + 6$ are proportional, meaning there’s some $k$ such that
[
a x^{3} - 2x^{2} - x + 3 = k(a + b) x^{2} + ck x + 6k
]

[wraith dagger]

@cerulean sail

[cerulean sail]

By equating coefficients that’s how I got that a=0, maybe try following through from there to find out what k, b and c are?

[cerulean sail]

(Spoiler: I don’t agree with this)

[tribal sphinx]

where does k come from?

[cerulean sail]

Definition of proportionality

[tribal sphinx]

could it be option D?

i know im getting this wrong but is k=2?

[cerulean sail]

How are you getting this?

[tribal sphinx]

if i plug in the values from option D, i get -4x^2-2x+6 which divided by 2 gives me -2x^2-x+3

[cerulean sail]

Try to do it without looking at the options first

[cerulean sail]

In particular, notice the constant terms here

You have that 3 = 6k, so then what does that give you as k?

(The whole point of doing it like that is that it helps you find the option it ends up being without needing to do trial and error)

[tribal sphinx]

1/2?

[cerulean sail]

Yep! Now from there, consider the coefficients of x and of x^2

[cerulean sail]

We already concluded that a = 0 and now we have k = 1/2

What does that give you as b and c?

[crimson sedge]

clenchedcoochieflaps

[tribal sphinx]

idk if i understand but i have to find a number that multiplied by k would give me -2 and -1?

[cerulean sail]

Yea kinda, basically you have that $k(a + b) = -2, ck = -1$ and that $a=0$ and $k = \frac{1}{2}$

[wraith dagger]

@cerulean sail

[cerulean sail]

From there you can find b and c if you put in the values of a and k and then solve

[tribal sphinx]

okay, then i think that b=-4 and c=-2?

[cerulean sail]

It is(!)

See the react there haha

[tribal sphinx]

so D is correct?

omg thank you so much!!

and thank you for being patient lol!!

[cerulean sail]

It’s all good

Lovely to work with you

[cedar kiln]

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[crimson sedge]

hello are these highlighted arithmetic sequence i dont seem to figure out its common difference thingy

[crimson sedge]

also how do i do the last question i dont seem to understand that too

this is a weird question

that i have ever come across

what even is happening in this

its crazy as ever

<@&286206848099549185> can you please put some thought into it

this is a weird question ever

[vague rapids]

What do you want to find?

[vague rapids]

The differences are in AP not the numbers

6-2,12-6, 20-12, p-20 are in AP

[crimson sedge]

i dont get it the common difference thingy in those highlighted

how do i do the last question

[crimson sedge]

i need to find values for q,r and s

[vague rapids]

for p it is in AP so you can easily find it

[crimson sedge]

ohh

ap is i dont get

p is an arithematic sequence and its common difference is 2

[vague rapids]

yes

[vague rapids]

For q notice this

[crimson sedge]

i dont know what is AP

[vague rapids]

An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference from any succeeding term to its preceding term remains constant throughout the sequence

[vague rapids]

It means for example

4, 8, 12, 16, 20

The same number 4 is added to every new term

[crimson sedge]

oh we used to call it arithematic

[vague rapids]

Arithmetic Progression in short is AP

[crimson sedge]

ohh

i understood

[vague rapids]

can you find p now?

[crimson sedge]

hmm

i cant find q i dont understand its sequence

[vague rapids]

Yeah lemme explain it

[crimson sedge]

oka

[vague rapids]

The sequence is 2, 6, 12, 20, p

sequence of Difference of two consecutive numbers goes 4, 6, 8, p-20

I mean 6-2, 12-6, 20-12, p-20

[crimson sedge]

is there like a name for this specific sequence

[vague rapids]

No

It's like the differences are in AP

[crimson sedge]

so i would i remember this type of the sequence

i probably should have

oh

[vague rapids]

You don't need to remember just visualise

[crimson sedge]

okay @vague rapids can you like

help me with my last question

[nova geyser]

math isn't like science u don't memorize

[vague rapids]

Okay

[crimson sedge]

its something i dont understand too

last question

[nova geyser]

c?

[crimson sedge]

its this one last question

[crimson sedge]

i know

[nova geyser]

i would like u see from how is odd and even number

if n=odd so n =
if n=even so n =

[crimson sedge]

right

[nova geyser]

@crimson sedge find this first

[crimson sedge]

i dont get you @nova geyser

[vague rapids]

Notice if n is odd the middle number is n+1 (n means row number)
If n is even the middle number is n itself

[crimson sedge]

i need to understand please let them explain

[nova geyser]

so when n=2 the middle number is 2
when n=4 the middle number is 4

so when n=even the result will be the same

[crimson sedge]

im sorry i myself is confused

[vague rapids]

Look at the question and find middle number of first 4 rows

You'll understand yourself

[crimson sedge]

alright

thaks

.close

[cedar kiln]

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[eternal agate]

How do I graph semi circles? More specifically, how do I know which side is missing?

[eager sparrow]

just do its absolute value

you will get rid of all the negatives

[eternal agate]

Like I’m the equation:

y = -sqr(4x-x^2)

[eager sparrow]

wait

[eternal agate]

?

[eager sparrow]

that's not a semi circle

is it the sqrt?

[nova geyser]

,w plot y = -sqrt(4x-x^2)

[wraith dagger]

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Requested by JeremyGaming#1700

[nova geyser]

not look a like semicircle

[lyric leaf]

y = sqrt(r^2 - x^2) defines a semicircle 'reflected' over the x axis

[eternal agate]

And my teacher graphed it as a semi circle

[lyric leaf]

• sqrt(r^2 - x^2) means 'above' the x axis,

• sqrt(r^2 - x^2) means 'below' the y axis

[lyric leaf]

likewise

x=sqrt(r^2 - y^2) defines a semicircle 'reflected' over the y axis

with positive being on the right, and - being on the left

[eternal agate]

Ohhhhhhh

So it acts like a parabola

Kinda

Alright thanks

.close

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[cedar kiln]

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[cedar kiln]

@abstract urchin Has your question been resolved?

[iron onyx]

I remember there being some trig identity where a product like that turns into addition of trig functions, let me see if i can find it

yeah these things

looks to be on the wiki list of trig identities - https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities

applying the one with 2 sines one should give you what you're looking for? @abstract urchin

it looks to me like he took the 2 outside the integral and then applied the identity to get the 1/2

which is... weird but i think it's correct/consistent

don't apologize! valid question :P

no problem

[cedar kiln]

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[wooden loom]

I have the distribution function as well

but im not sure what it means when Z = 1/Y

[cedar kiln]

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[grim yacht]

dam was just about to type something

[cedar kiln]

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[silent cove]

can anyone help me with an account on polynomial factorization

[silent cove]

I don’t under

Stand

[radiant topaz]

what is that you don't understand

[silent cove]

Why is x and y variants

And what’s the formula for the count

.close

[cedar kiln]

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[silent cove]

I don’t need help Ty @radiant topaz

[cedar kiln]

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[plush warren]

english?

[forest portal]

22. Condition: Find the greatest whole measure of a, when the expression has a positive solution

[plush warren]

ill try

[forest portal]

Explain how

[radiant topaz]

uh first simplify the equation

[radiant topaz]

don't give no answers

[plush warren]

k

[forest portal]

How?

[radiant topaz]

isolate x

[forest portal]

But I don't get how to simplify expression like this

3ax + 6 = 8x - 12a

I don't know what to do next

[forest portal]

....

@fading birch

@radiant topaz

Explain

Please

[radiant topaz]

How are you even struggling to isolate x?

[forest portal]

It's possible

[radiant topaz]

Do some subtraction

Then factor out x and isolate it

[forest portal]

x = (-12a - 6) / (3a-8)

Right?

3ax - 8x = -12a - 6
x(3a - 8) = -12a - 6
x = (-12a - 6) / (3a-8)

Correct?

[radiant topaz]

Ok

Now you know x>0

So the right side is also >0

Get the values of a by this inequality

[forest portal]

And a cant be equal to 0

[radiant topaz]

You have to solve for it

[forest portal]

3a - 8 cant be equal to 0

?

[radiant topaz]

both numerator and denominator has to be positive

[forest portal]

I have found

the greatest whole a is 2

[cedar kiln]

@forest portal Has your question been resolved?

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[cedar kiln]

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[stone crystal]

Hey guys, could anyone please help me with this one? I'm assuming angle B is the top of the triangle

[livid hound]

what have you tried?

[stone crystal]

I havent been able to figure it out. Been trying to google and to understand how i should think :S So i guess my thought process is the fault here

I'm guessing its either A or C?

[livid hound]

what's your thought process

[stone crystal]

It has to be with the y^2 etc. Because pythagoras says you need to do the ^2 on each side

[livid hound]

go back a step

[stone crystal]

hmm

Im thinking pythagoras, like i have to figure out a-c?

[livid hound]

that's something you'll need

also you should just call that AC

what does pythag state

[stone crystal]

y^2 + x^2 = AC

No wait

AC^2 + x^2 = y?

[livid hound]

no

[stone crystal]

I guess my problem is how to figure out the AC one

[livid hound]

your first problem is that you don't quite know pythag

pythag does NOT say
$$a^2 + b^2 \wthonk c$$

[wraith dagger]

ℝamonov

[stone crystal]

Im a bit confused, isnt that the correct one?

i mean

c^2

I missed the "^2"

[livid hound]

yes, you missed the ^2

[stone crystal]

Yeah, i meant ^2 in my head tho xD My bad!

[livid hound]

and note what a,b,c represent in the theorem

[stone crystal]

Yeah so i guess im confused to figure out what to put on the AC. Since there are no numbers, only Y/x etc.

Or how do i get the AC o.O

[livid hound]

focus only on the theorem for what I'm asking in the following question

what does the theorem say a,b,c represent

[stone crystal]

A=x
B= unknown
Y=c

[livid hound]

not reading what I said

[stone crystal]

Sorry im so dumb :S I really appreciate you helping me a lot tho ❤️

[livid hound]

also lowercase

[stone crystal]

hmm

[livid hound]

the response I was expecting was that a,b represent the length of the legs
and c the hypotenuse

[stone crystal]

Yeah, thats what i mean tto

[livid hound]

you've identified that one of the legs is x
the hypotenuse is y

[stone crystal]

Like a is where X is atm

Y is the C "hypotenusa"

[livid hound]

and the remaining keg is AC

[stone crystal]

But we dont have any letter for "AC", thats where im a bit confused

[livid hound]

introduce one yourself or just keep it as AC

[stone crystal]

But if we look at these 4. There are no equation with another variable

[livid hound]

well you'll just use whatever that variable is equivalent to

[stone crystal]

x^2 + AC^2 = y^2 / 2

[livid hound]

applying pythag initially gets you
AC^2 + x^2 = y^2

then solve for AC

[livid hound]

no

you can't just divide one side of the equation by 2

[stone crystal]

hmm

[livid hound]

applying pythag initially gets you
AC^2 + x^2 = y^2
then solve for AC

[stone crystal]

ac^2 = x^2 + y^2? :S

[livid hound]

no

you can't just subtract x^2 from one side while adding x^2 to the other

[stone crystal]

ohh, is this an equation`?

[livid hound]

well yeh

there an equal sign

[stone crystal]

AC^2 = -x^2 + y^2?

[livid hound]

that's valid

and from there
AC = ?

[stone crystal]

hmm

[livid hound]

the absence of numerical values makes little difference to what you'd do

how would you solve
p^2 + 5^2 = 13^2

[stone crystal]

1 sec

p=12?

[livid hound]

how not what

[stone crystal]

p^2 + 25 = 169

p^2 = 144

square root=

p=12

[livid hound]

exactly the same idea for your question

[stone crystal]

Im confused with the x and y and such

[livid hound]

makes no difference to what you'd do

go through the exact same steps you just stated

[stone crystal]

But the answers have an x in front

and x is the short side to the left

[livid hound]

forget about the answer options

[stone crystal]

but short side is already represented in x^2

AC^2 = -x^2 + y^2 ?

[livid hound]

and focus only on what I'm saying

[stone crystal]

solve this?

[livid hound]

and apply the exact same thing you did to get p in that numerical example here

don't overthink

[stone crystal]

so youre thinking

y^2= 2y?

[livid hound]

no

[stone crystal]

in that way

These letters confuse me xD

[livid hound]

you did what was required three minutes ago

[stone crystal]

What i have a problem with. The numbers i could do etc -20 + 80 which is 60. Easy.

But -y +x. I cant just ad a y in to x?

[livid hound]

recall what you did after

p^2 = 144

p + q is always p + q

[stone crystal]

I think the correct answer is C. But i dont know 100% why yet

[livid hound]

don't feel the need to forcibly combine things when you cant

[stone crystal]

So how can i get an answer out of letters?

[livid hound]

leave expressions that can't be combined/simplified further as they are

recall what you did after
p^2 = 144

[stone crystal]

square root yeah

[livid hound]

yes

[stone crystal]

So you mean square root this? AC^2 = -x^2 + y^2

entire

[livid hound]

yes

take the square root of each side

[stone crystal]

so one square over AC. And one large over x+y?

[livid hound]

no

what happened to all those ^2

[stone crystal]

you mean AC = -x + y?

[livid hound]

no

[stone crystal]

xD

[livid hound]

leave expressions that can't be combined/simplified further as they are

like you can literally just chuck a sqrt on each side as the first step

[stone crystal]

Square root of AC^2 = "SQ"AC

dont know how to do a SQ sign

[livid hound]

$\sqrt{AC^2} = \sqrt{y^2 -x^2}$

[wraith dagger]

ℝamonov

[livid hound]

just type sqrt(what you want rooted in parenthesis)

[livid hound]

and no

[stone crystal]

i cant go further than this correct?

[livid hound]

sqrt of that is just AC

the left side simplifies to just AC

[stone crystal]

AC = y^2 - x^2?

no sry

[livid hound]

no

[stone crystal]

AC = y-x

[livid hound]

no

freshman's dream

[stone crystal]

These letters will give me nightmares xD

[livid hound]

$\sqrt{2^2 - 1} \redneq 2-1$

[wraith dagger]

ℝamonov

[livid hound]

letters don't matter, same rules apply

don't feel obligated to forcibly cancel/make things look nicer

if there are no laws that allow for certain actions, don't do anything

leave expressions that can't be combined/simplified further as they are

[stone crystal]

So this is as far as it goes?

[livid hound]

the left side simplifies to AC

can't do much with the right side

[stone crystal]

Ahh yeah true

so

AC^2 = SQ "y^2 - x^2

[livid hound]

no

[stone crystal]

No sry

AC

just plain AC in front

[livid hound]

use () around the entire thing you want rooted

[stone crystal]

AC = sqrt(y^2 - x^2=

[livid hound]

and sqrt not sq

[stone crystal]

AC = sqrt(y^2 - x^2)

[livid hound]

yes

you now have an expression for AC (in terms of x and y)

and from here you can simply apply formula for area of a triangle

[stone crystal]

AC = sqrt(y^2 - x^2) / 2?

shit these letters, fuck :S

So it is C right?

[livid hound]

yes

[livid hound]

no to that though

[stone crystal]

Shoudltn an area for a triangle be AC x AB / 2?

[cedar kiln]

@stone crystal Has your question been resolved?

[cedar kiln]

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[midnight ridge]

solve the equation 3tan(2x + 15) = 4, for 0 <= x<= 180

[lyric narwhal]

!status

[cedar kiln]

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

[midnight ridge]

2!

[lyric narwhal]

what have you tried

[midnight ridge]

wait i send my calculations rq

[midnight ridge]

so, tan(2x + 15) = 4/3

basic angle: tan-1 (4/3)

= 53.13 degrees

then idk bcs of the range of x they gave me

i think its in the 1st and 3rd quadrant

[lyric leaf]

just remember tan's periodicity

[midnight ridge]

never saw that word before 🥲

[lyric leaf]

might not be applied correctly

tanx is periodic every pi

or uh in degrees that's 180

because your range is 0 <= x <= 180 there shouldnt be any other x-values to calculate?

[cedar kiln]

@midnight ridge Has your question been resolved?

[cedar kiln]

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[lyric widget]

can someone help me understand or maybe even prove this?

[cedar kiln]

@lyric widget Has your question been resolved?

[cedar kiln]

@lyric widget Has your question been resolved?

[lyric widget]

.close

[cedar kiln]

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[crimson sedge]

can someone help me understand how the support video just randomly got the number 4

[sinful geyser]

?

Wdym

[crimson sedge]

like it says moving 4 steps add 32 and 1 steps adds 8

what do they mean by moving 4 and 1 step

[sinful geyser]

Here's the clue

Do you get it or no

[crimson sedge]

wait so 19-7 is 12

the terms increase by 12 each time?

wait no

[crimson sedge]

yeah i dont

[sinful geyser]

So there are certain 'steps' you need to climb before jumping from 13 to 45

Each step adds a certain number

They have given the condition that every step they take is EQUAL

[crimson sedge]

by equal do you mean each step is the same number?

[sinful geyser]

Yessir

Ok so 4 equal steps right?

[crimson sedge]

yeah

[sinful geyser]

Can we name each step as 'x'?

[crimson sedge]

sure

[sinful geyser]

Ok so 13+4x=45

You're taking 4 steps from 13 to get to 45

Does it make sense?

[crimson sedge]

yes

[sinful geyser]

Solve the equation to get to your answer

[crimson sedge]

ok

ok thanks

.close

[cedar kiln]

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[wintry steeple]

How can you write natural numbers n such that n != x^{1/3}?

I.e. is there a general form of natural numbers which are non cubes? This is the basis of a question I had with Diophantine sets

[muted bear]

isnt every natural number a cube root of another natural number?

[wintry steeple]

7 is not a natural number cubed?

oh wait

maybe i mean n such that there exists no x such that n!=cuberoot(x)

right? cus 7 != cuberoot(x) where x is a natural number

[muted bear]

x=343

[wintry steeple]

or wait no its the cuberoot of 7^3

[muted bear]

yeah

[wintry steeple]

i must have formulated this question wrong

maybe i mean n such that n!=x^3

right cus those would be natural numbers that are non cubes

[muted bear]

yes

theres probably some set theory way to write it

[wintry steeple]

well yh its using diophantine sets im p sure but it requires some algebraic representation of n as a polynomial

like example is non powers of 2 is just (2x+3)y for some x and y

[muted bear]

probably gonna get this wrong
$$n \in \mathbb Z {n | n \neq x^3, x \in \mathbb Z}$$

[wraith dagger]

GarlicBredFries

[wintry steeple]

erm yh that is the set im looking for ofc!

okay to be more specific, looking for a diophantine equation such that the diophantine set is the set of non cubes

[muted bear]

oh

[wintry steeple]

but usually it just requires writing n = non cube

[muted bear]

\lfloor \sqrt(3){x} \rfloor = 0

[wintry steeple]

where 'non cube' is some form of non cubes as a function of any number of variables

nope has to be a polynomial haha

[muted bear]

oh i dont think thats possible since that would require an infinite amount of roots

[wintry steeple]

nah cus as i say, a good example is this

for the record this was an exam question, just can't think of the answer (had the exam today so no solutions)

[muted bear]

oh you just need some of them

not all of them

[wintry steeple]

yes!

wait no

all of them

every non power of 2 can be written like that

cus it has an odd factor >= 3

[muted bear]

oh right

[wintry steeple]

like one thing that came to mind is that all the prime factors of n must have power divisible by 3

NOT divisible by 3

[crimson sedge]

(are you looking at n ≠ x^3 or n ≠ 3^x)

[wintry steeple]

but it seems kinda useless because you end up with a x^(3y+2) or whatever

[wintry steeple]

former

n ≠ x^3

btw how did you get neq sign?

[crimson sedge]

keybinded

you can copy it from ascii tables

here's someone doing it for n ≠ x^2 https://math.stackexchange.com/questions/4438559/does-there-exist-a-polynomial-px-y-which-detects-all-non-squares

[wintry steeple]

ah cool ty!

[wintry steeple]

yh saw this 🙂

[crimson sedge]

seems like the answer would be nontrivial

maybe ask in the number theory channel

[wintry steeple]

thank u for effort as always tho!

<@&286206848099549185>

[crimson sedge]

you could try doing something similar with x^3-ny^3=1

and you can show that if n is a perfect cube, then the only solution is (0,1)

[wintry steeple]

yh was just looking at this but there are like papers analysing this

[crimson sedge]

but I think the fact that there are an infinite number of solutions is still open

or at least not well known

I only found 1 chapter of a book on it

and it was waffling on about geometry

[wintry steeple]

damn for like 6 marks in an exam this must be easier than im thinking

6/100 btw

3 hours lol

[crimson sedge]

are you sure it wasn't n≠3^x lol

[wintry steeple]

yes 100% haha

[crimson sedge]

what a troll lecturer lol

[wintry steeple]

specifically said i.e. set of non cubes

hahahaha honestly

[crimson sedge]

even the nonsquares is nontrivial

[wintry steeple]

definitely

[crimson sedge]

it requires knowing that pell's has an infinite number of solutions, which is a long proof iirc

[wintry steeple]

but there must be an obscure trick somewhere in lectures

[wintry steeple]

and was not mentioned in lectures btw

[crimson sedge]

hmm

[wintry steeple]

btw idk if u saw the previous parts of my question

but the question is specifically

find a diophantine set that is equal to this set

[crimson sedge]

is this like grad level number theory?

[wintry steeple]

hence finding a polynomial representing elements such as this n = f

undergrad lol

[crimson sedge]

oof

[wintry steeple]

in whats supposed to be a very easy module may i add

its related to one of hilberts problems

[crimson sedge]

yeah,

and they are very hard

[wintry steeple]

hahahaha yh but what we do based on them is not tbh

well clearly it fucking is

but usually the questions are simple like find a diophantine set for blah blah

but ones that are more obvious in terms of thinking

this one is fucked

[cedar kiln]

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[pale ledge]

Any video that goes into depth on how to solve these?

[pale ledge]

<@&286206848099549185>

[cedar kiln]

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[merry heath]

what is the "x value"

[merry heath]

dividing the first part by (x-2)

.close

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[crimson cape]

Hi how do I see my channel, my help channel

[cedar kiln]

Please don't occupy multiple help channels.

[crimson cape]

hi

what's the Inner Reaistance here Ri

open source

Ri = R1/3*R2/4 divided by R1/3+R2/4 ?

electrical engeeenirnf

engineering

please

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[simple flare]

May I know why in this, the second ODES solution is not what I written using pencil?

[cedar kiln]

@simple flare Has your question been resolved?

[simple flare]

<@&286206848099549185>

[astral bay]

,rccw

[dire geode]

They're equivalent. Use DeMoivre's formula

,tex .demoivre

[wraith dagger]

riemann

[cedar kiln]

@simple flare Has your question been resolved?

[simple flare]

Oh; so actually for +k^2 and - k^2, we can just write in exponential form?

[simple flare]

I still didn’t get why it’s equivalent

Because from what I learnt last time, If k^2 Is positive or negative, they will yield different solution

[dire geode]

Did you use DeMoivre's formula

[dire geode]

.

[simple flare]

No, I just directly apply to the formula from last time

But from here, I watched video from YouTube and refer to textbook, it seems that they consider - and + k^2 can be in the same case

Where cos and sin no need to be involved

[cedar kiln]

@simple flare Has your question been resolved?

[simple flare]

Like how this become the bottom line?

[dire geode]

DeMoivre's formula

[dire geode]

.

[simple flare]

I got this 🙂

[dire geode]

great, that's the same as your answer but with different constants

group cos and sin terms together

[idle glen]

Need help with this problem

[dire geode]

!help

[cedar kiln]

Please read #❓how-to-get-help

[simple flare]

But using grouping it I still got an extra i?

[dire geode]

different constants

[simple flare]

So basically, for - or +, we can just use the same concept?

Should be +i

[simple flare]

I mean iB here

Isn’t this consider complex function?

[dire geode]

iB is just a constant

[dire geode]

no

x is the independent variable

[simple flare]

I don’t understand

What is iB here? Just a constant?

[dire geode]

.

[simple flare]

But isn’t for complex number, we got i?

But why iB here can be a constant?

[dire geode]

when you work with wave functions, your field of scalars is C, not R

[simple flare]

R is what?

[dire geode]

if you haven't taken linear algebra, i strongly recommend it

[simple flare]

That one is in linear algebra what topic?

[dire geode]

R

[simple flare]

Then C is complex?

[dire geode]

C

[dire geode]

https://ocw.mit.edu/courses/18-06-linear-algebra-spring-2010/

pretty common pre-requisite for QM

[simple flare]

Learning linear algebra this sem but at the same time learning QM also

[dire geode]

good luck with that

[simple flare]

So here, because I’m dealing with QM, the iB will just be a constant again, like B?

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[cedar kiln]

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[lunar crystal]

in general are calculators used for questions like this because working out 360*16/20 seems redundant or am I just slow

[rustic siren]

they want the smaller one i think, but yeah, you can get it like that

[steel heart]

$\usepackage{fourier}$
Circumference $= 2\pi r$\
Circumference $= 20\pi$\
$\implies AC+ABC=20\pi$

[wraith dagger]

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

[lunar crystal]

so ac is just 4 pi

[steel heart]

yes

[lunar crystal]

The question asks for it in degrees tho

I was mainly asking if you use generally have calculators in geometry tests, I'm self schooling myself so I was just wondering because it feels redundant to work out things like 360*14/36 something I got in another practice question which results in 5040/36 which I then have to simplify into 140 it just seems like a big time waste

[steel heart]

just a normal day in India

[lunar crystal]

?

[cedar kiln]

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[cedar kiln]

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[rancid cedar]

How do I solve this?

[rancid cedar]

I got the area and perimeter right

[buoyant latch]

part i?

[rancid cedar]

I got (i) right

I cant figure out (ii)

[buoyant latch]

ok show me how you did part i

[steel heart]

find BC
area = 1/2 * b * h
area = (AD*BC)/2
where AD is perpendicular to BC

[rancid cedar]

I'm not near my study atm but i did the Line distance formula to get bc and ab

[buoyant latch]

ok

show me the line distance formula

[rancid cedar]

It's PQ = SQRT ([x2-x1] + [y2-y1])

[rancid cedar]

Or am I wrong?