help-10

[flat anvil]

Okay then ask then

[gleaming locust]

its a couple questiopns im rlly stuck on

[flat anvil]

Then send them

Cuz idk the question cuz you haven’t sent it

[leaden ibex]

that's...the same concept

[gleaming locust]

what?

yes but idk how to work backwards

[leaden ibex]

Do the same thing, but instead of having a number for the height you have a unknown

and then solve the equation you get for y

[gleaming locust]

im stuck

help me with it

[tacit quest]

Want me to help

@gleaming locust

[obtuse pebble]

@gleaming locust Has your question been resolved?

[leaden ibex]

Well, what is the formula for the area of a trapezium

[obtuse pebble]

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[latent adder]

Hi

[latent adder]

I'm trying to find the kernel of this matrix below:

[spiral maple]

<@&268886789983436800>

[latent adder]

$\begin{array}{ccccc}
1 & -1 & 0 & 2 & 0 \
0 & 0 & 1 & 1 & 0 \
0 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 0 \
\end{array}$

[warm shale]

omega

[glossy yew]

Thanks

[latent adder]

I understand that $x_2$ and $x_4$ are free variables

[warm shale]

omega

[latent adder]

but after that I'm having some trouble

<@&286206848099549185>

Hello?

anybody do linear algebra here?

[obtuse pebble]

@latent adder Has your question been resolved?

[obtuse pebble]

@latent adder Has your question been resolved?

[robust sleet]

did you rref it?

[obtuse pebble]

@latent adder Has your question been resolved?

[obtuse pebble]

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[green saffron]

How do I find a polynomial function with a degree of 3, leading coefficient 2, and zeroes 14, 1/2, and 9/2?

[idle thunder]

uh

[leaden ibex]

aw, you gave him the answer :/

[gloomy valve]

rip

[green saffron]

I didn't see it

[idle thunder]

good

[idle thunder]

we can help explain it now!

[green saffron]

Also it's she/her please

[leaden ibex]

Do you remember how to represent polynomials in terms of their roots?

"we"

[green saffron]

No

I don't remember any of this, I missed school because I had covid

[leaden ibex]

Alright, let's take a polynomial of degree 1. Would you agree that we can represent any linear function as $y = c(x-a)$ where c and a are some real numbers

[warm shale]

Remavas

[leaden ibex]

(And do you know polynomial division, because that would make the explanation even easier)

[green saffron]

I don't know I never learned any of this

[leaden ibex]

Well ok, we can represent any linear function in this way. You can do some easy algebra to get to the usual y = mx + b form

And this linear function has a root at x = a

[green saffron]

Oh yeah that's slop intercept form isn't it

[leaden ibex]

mhm

[green saffron]

Yeah

[leaden ibex]

If we now go to second-degree polynomials

If we have a 2nd degree polynomial with two roots, we can write it as $y = a(x-b)(x-c)$

[warm shale]

Remavas

[green saffron]

Yeah

[leaden ibex]

where b and c are the roots

[green saffron]

Si

[leaden ibex]

There is a whole theorem on this you can prove with polynomial division, but since you don't know that, we'll resist that urge

[green saffron]

Thank you 🙏

[leaden ibex]

So what do you think, what will that form look like for a 3rd degree polynomial with 3 real roots?

[green saffron]

Uh

Y = a(x-b)(x-c)(x-d)?

[leaden ibex]

exactly

[green saffron]

Sorry it autocorrected y to yeah for some reason

[leaden ibex]

a happens to be the leading coefficient since after we use FOIL on the product

we'll get $ax^3 + \cdots$

[warm shale]

Remavas

[green saffron]

That makes sense yeah

[leaden ibex]

and b,c,d are your zeroes

[green saffron]

Hm

Alright that makes sense

[leaden ibex]

In general, any polynomial can be expressed as $P(x) = a(x-b)^i\cdot (x-c)^j \cdot \cdots \cdot (x^2 + px +q)^l \cdot (x^2 + rx + s)^m \cdot \cdots$

[warm shale]

Remavas

[green saffron]

Long equation

[leaden ibex]

Well yeah, but you won't need it for your task

[leaden ibex]

this is more than sufficient^

[green saffron]

Ah thank freyja

[leaden ibex]

this also becomes much simpler in the complex plane but that is another story

[green saffron]

I'll probably learn that later on in the year lol

Or next year

So b would be 14, c would be 1/2, and d would be 9/2

[leaden ibex]

yup

[green saffron]

Groovy

So y =(x-14)x(x-1/2)x(x-9/2)

[leaden ibex]

you forgot something

[green saffron]

What did I forget?

[leaden ibex]

the a

[green saffron]

Ah yeah

My a key is broken sorry

Have to press hella hard

[leaden ibex]

well we can replace it with a number directly

[leaden ibex]

^

and you're given the leading coefficient

[green saffron]

So a =2

[leaden ibex]

And your polynomial becomes $y = 2(x-14)(x-1/2)(x-9/2)$

[warm shale]

Remavas

[green saffron]

Yeah!

And then you have to multiply it, right?

[leaden ibex]

Do they ask you to simplify it or put it into a canonical form?

[green saffron]

No but would you mind showing me how anyway?

[leaden ibex]

You just use FOIL

[green saffron]

What is foil?

[balmy mortar]

$$(a+b)(c+d)$$

[warm shale]

Shuri2060

[balmy mortar]

instructions on multiplying this out

[green saffron]

Ah thank you

[leaden ibex]

I honestly don't remember myself what the abbreviation exactly means.

[leaden ibex]

but you should be able to do it^

[green saffron]

So it is $$(2+14)x(1/2+9/2)$$

[warm shale]

Modern_Valkyrie

[leaden ibex]

woah what what

We went from a third degree polynomial to a first degree polynomial

[green saffron]

I'm confused sorry

[green saffron]

y = 2(x-14)(x-1/2)(x-9/2)

[balmy mortar]

$$abc = (ab)c = a(bc)$$

[warm shale]

Shuri2060

[green saffron]

I jsut pasted that to read it easier

My teacher said that we don't need to simplify btw

[balmy mortar]

I have no idea what the original question was

in which case, yes

why would you multiply out lol

🤦

[green saffron]

polynomial function with a degree of 3, leading coefficient 2, and zeroes 14, 1/2, and 9/2

[balmy mortar]

factorised form is simplified form

[leaden ibex]

Finding a polynomial of degree 3 with given leading coeff and 3 roots

[leaden ibex]

So why do it

[balmy mortar]

this answer is fine as is (get rid of the fractions if you like)

[green saffron]

I just saw that it said that

[leaden ibex]

Any form is correct, and this one is very useful too

[green saffron]

It was way at the bottom of the page

[leaden ibex]

since it immediately shows you the roots of the polynomial

[green saffron]

Yeah

[balmy mortar]

$y = 2(x-14)\left(x-\frac12\right)\left(x-\frac92\right)$

[warm shale]

Shuri2060

[balmy mortar]

You can make this look nicer with a few simple steps though

$y = \frac12(x-14)\left(2x-1\right)\left(2x-9\right)$

[warm shale]

Shuri2060

[balmy mortar]

I divide everything by 1/4 and multiply each bracket with a fraction by 2

[green saffron]

I should do that?

[balmy mortar]

You could. I think it looks nicer.

[balmy mortar]

But this answer is perfectly fine

[green saffron]

I think I'll just stick with the original answer, I don't want to confuse myself lol

Thank you for the help thogh

[leaden ibex]

Just don't put it into $ax^3 + bx^2 + cx + d$ form unless you really need to

[warm shale]

Remavas

[leaden ibex]

It will be menial work with little to no point

[green saffron]

Alright

Thank yu

[hushed moat]

Alternatively use vieta if you want it in that form

[obtuse pebble]

@green saffron Has your question been resolved?

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[obtuse pebble]

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[weak grove]

I am to show this is true for all integers x

Could I have a hint please?

I think its probably pretty easy to do by induction if the exponent wasnt so large

(expanding (x+1)^42 isnt fun)

[balmy mortar]

fermat might help.

40 === -1 might also help

[obtuse pebble]

@weak grove Has your question been resolved?

[weak grove]

40 == -1 might help indeed

ill give it another shout

thx

[obtuse pebble]

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[untold ravine]

hello

[untold ravine]

https://gyazo.com/41fa800e0be511f8595ed3d4834f59c3

for this question i used the sin area rule

sin^-1(12.5/(5x6x1/2))

i got 56.4

but its not obtuse

[high lily]

there are two solutions for x to
sin(x) = c
for 0<c<1 and 0<x<180°

using arcsin directly only gives one of them
apply properties of the unit circle to get the other

[untold ravine]

hhmm

we ahvent learnt this yet

[high lily]

you should have

[untold ravine]

but algebraically its correct i dont know why its wrong

i used sin area rule

[high lily]

have you done questions like
find all solutions to sin(t) = 1/2
for 0 < t < 360°

[untold ravine]

no

its a gcse question i think

this one

[high lily]

are you familiar with the supplementary identity for sine?

[untold ravine]

ah no

i know sin and sin^-1

[high lily]

sin(180°-x) = sin(x)

[untold ravine]

hmm

we havent got this far yet

[leaden ibex]

Let me draw you something

[untold ravine]

this is grade 10

[high lily]

it seems you're the knowledge for this question

usually its something they'd give you after teaching you a decent amount about the unit circle

[untold ravine]

oh

[leaden ibex]

crude but works

[untold ravine]

i see

[leaden ibex]

Have you already learnt about the unit circle?

If not, this probably means nothing to you

[untold ravine]

no

yeah unforunately

https://gyazo.com/c9f946bf346fb9833c768d4c74cc9649

i just rearranged this

to get sinC

and do sin^-1(C)

but its wrong because the angle has to be obtuse

[leaden ibex]

Well, Ramonov already told you something crucial

$\sin \alpha = \sin(180^{\circ} - \alpha)$

[warm shale]

Remavas

[untold ravine]

hmm

so i cant just just re arrange the equation

: /

[high lily]

rearranging to sin(C) = stuff is fine

you just need to be aware of the limitations of the inverse sine function

[untold ravine]

i see

it cant work for triangles?

obtuse angles**

i assume

[high lily]

you'd need additional work

and for that you'd consider properties of the unit circle and/or the identifed that come from it

[untold ravine]

yes

[high lily]

specifically stuff like
sin(180°-x) = sin(x)

and in this case,

the angle you want will be
180°-sin^-1(12.5/(5x6x1/2))

[untold ravine]

ah

i see

thanks

[leaden ibex]

I'd recommend you learn the unit circle

It's a powerful tool

[untold ravine]

yes

[obtuse pebble]

@untold ravine Has your question been resolved?

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[runic carbon]

A craft store has 25 assorted grab bags on sale for ₱150 each. Fifteen of the bags contains ₱150 worth of merchandise, six contain ₱100 worth, two contain ₱250 worth of merchandise, and there are one each containing ₱500 and ₱1000 worth of merchandise. Suppose that you purchase one bag; what is your expected gain or loss?

[runic carbon]

I really need the process of getting this coz i don't get the lesson itself xd

<@&286206848099549185>

.close

[obtuse pebble]

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[subtle topaz]

Can someone help this is AP Calculus AB Related Rates

[subtle topaz]

someone?

[tall knot]

I'm trying

[subtle topaz]

ight thx

any luck?

sorry don't wanna sound impatient

<@&286206848099549185>

good to know nobody can help with shit here

WHAT THE ACTUAL FUCK

[short spruce]

calm down lmfao

[short spruce]

you succeeded

[tall knot]

Do you have answer? 😬

[subtle topaz]

400 ft/hr

just the answer

but how did they get it

[tall knot]

Got it

[tall knot]

Sorry

Ok so

[subtle topaz]

nono it's fine I apologize for getting aggressive as well

[tall knot]

Do you know chain rule?

[subtle topaz]

yeah

[tall knot]

Ok so assume x is the length of the shadow

[subtle topaz]

mk

[tall knot]

We now have to find the dx/dt

[subtle topaz]

yeah

[tall knot]

The question had given one information

$$\frac{d \theta}{dt}=-0.25 rad h^{-1}$$

[warm shale]

skypirate

[tall knot]

So how can you find dx/dt by using chain rule?

[subtle topaz]

this equation doesn't have x tho

[tall knot]

You can make one

Use triangle

[subtle topaz]

I did

But what is radh

radh^-1

[tall knot]

That's just radian per hour

[subtle topaz]

oh ok

yea so I got tan theta = 400/x

derivative of that is d theta/dt sec^2theta = (-400/x^2) dx/dt

[tall knot]

Hmmmm

Actually we want the derivative dx/d(theta)

[subtle topaz]

here cud u walk me thru step by step

[tall knot]

Hmmm?

[subtle topaz]

how to solve this

or cud u show me ur work if u wrote it on paper?

[tall knot]

Did you get the dx/d(theta)?

[subtle topaz]

Ohhhhh i c what I did wrong

is x=400cot theta?

[tall knot]

Yeahh

[subtle topaz]

ok ok worked it out myself

got 400 ft/hr

tysm

.close

[obtuse pebble]

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[safe spade]

area is $s^2$

[junior inlet]

sqrt is on your calculator

[warm shale]

Vladilena Milizé

[safe spade]

so just square root that number and round

[spare flare]

the problemis

I DONT KNOW HOW TO DO MATH\

[obtuse pebble]

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[gloomy valve]

what did you try?

[inner raft]

I figured that you add e^1 on top and subtract e^1

Im not really sure how to get 8 to 10 and 6 to 10

[tall knot]

I figured out the bottom, the top not yet

[gloomy valve]

look at the numbers on the top: 10, 11, 12, ...

[inner raft]

add 1

[gloomy valve]

so the first one should have 9

[inner raft]

but its 8/6

[gloomy valve]

yeah its simplified

[tall knot]

Ahhhh

[gloomy valve]

you can write it as 9 - e^0

[tall knot]

And then (9+1)

9+2

9+3

[gloomy valve]

same for the denominator

[tall knot]

I can't type out the answer 😬

But I get it

[inner raft]

wait so how did u get 8/6 to 9-e0/9+e0

[tall knot]

Look at the 6

It can be written as 5(1) + e^0

Then the next one can be 5(2)+e^1

[inner raft]

oh what about the top?

[tall knot]

And for the 8, we can write it as 9 +(0) -e^0

[inner raft]

maybe im wrong about this

[tall knot]

And the next we can write as 9+1-e^1

[inner raft]

but i got n-e^n-1 / n+e^n-1

or should n be 9

[tall knot]

$$\frac{9+(0)-e^0}{5(1)+e^0}, \frac{9+(1)-e^1}{5(2)+e^1},\frac{9+(2)-e^2}{5(3)+e^2},...$$

[warm shale]

skypirate

[tall knot]

@inner raft get the pattern?

[inner raft]

9 + (n-1) - e^n-1 / 5(n-1) + e^n-1 ?

[tall knot]

I think is 5n

[inner raft]

i think ur right

but is the other parts okay?

[tall knot]

yeah, at least I think so

[sick pelican]

think you can make 9+(n-1) into 8+n

[tall knot]

but you can simplify the 9+(n-1)

[sick pelican]

oh lol nice timing

[tall knot]

hahaha

[inner raft]

so
8+n- e^n-1 / 5(n-1) + e^n-1 ?

[tall knot]

hmmm still the 5(n-1) huh?

[inner raft]

uhh

5(n)

?

[tall knot]

yup

you can plug in n=1 and try

see whether it's true or not

[inner raft]

BINGO

its goes to 4/3

which is 8/6

i got it. thanks for the help everyone

.close

[obtuse pebble]

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[swift river]

How those two numbers are similar

Isn’t it supposed to be (3/sqrt of 2)

[spiral maple]

1/sqrt(2)=sqrt(2)/2

[swift river]

Like this

[sick pelican]

rationalized version

they basically just multiply sqrt 2 on both the numerator and the denominator

[swift river]

Oh

dang

I missed that

Thnx

.close

[obtuse pebble]

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[tawdry slate]

how would you solve that

[tawdry slate]

using u sub

[obtuse pebble]

@tawdry slate Has your question been resolved?

[leaden ibex]

what substitution have you tried

[obtuse pebble]

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[stable prairie]

how to find the 9 elements

[royal basin]

every polynomial in Z3[x] is congruent modulo x^2 + 1 to some linear polynomial

[stable prairie]

i can find the elements using that?

[obtuse pebble]

@stable prairie Has your question been resolved?

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[cunning cliff]

May i know if my sokution is exactly right?

[broken bridge]

i will do the same thing if i were you

[distant moth]

Both correct

[obtuse pebble]

@cunning cliff Has your question been resolved?

[cunning cliff]

.close

[obtuse pebble]

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[autumn patrol]

Is my methodology correct? Im practicing for a quiz in a few hours and kind of lost

[obtuse pebble]

@autumn patrol Has your question been resolved?

[obtuse pebble]

@autumn patrol Has your question been resolved?

[autumn patrol]

Need help <@&286206848099549185>

[rugged kite]

you got something wrong

n⁵+3 ≤ n⁵ is most certainly not correct

As for the methodology, recall that Θ is not about always being bounded by constant multiples of n², just that being true for large enough n

[autumn patrol]

I took the whole fraction as a whole, maybe that was my mistake

what if I add any constant to make it larger?

[rugged kite]

Have you studied limits yet?

[autumn patrol]

n^5 + 3 <= 4n^5 for ex

[rugged kite]

that does work

[autumn patrol]

Yes but im prompted to use ad hoc calculations not limit theorem

like manually test each Big-O and Big-Omega instead of lim f(x)/g(x)

[rugged kite]

I see

Anyway, n^5+3 ≤ 4n^5 works

[autumn patrol]

and i can make n^3 + 9 <= 4n^3 so I get n^2

[rugged kite]

Although, for the second inequality i don't get what you did

[autumn patrol]

4n^5/4n^3 = n^2

[autumn patrol]

which one

[rugged kite]

You (up to a few inaccuracies) proved f(n) ≤ n² and n²/2 ≤ n²

But that does absolutely not imply f(n) ≥ n²/2

[autumn patrol]

I must use a different approach to find Big-Omega then

My idea was, since I previously proved n^2 from 4n^5/4n^3 is >= n^5+3/n^3+9 I can just replace it

so n^2 >= Cn^2

Both proofs must have the same constant C and starting n?

[rugged kite]

Again, a>b and a>c does not mean b>c

nor b<c for that matter

Honestly you can just do the same thing you did for the first direction

i trust you to find a reasonable lower bound for n^5+3

As for 1/(n^3+9), that means finding an upper bound for n^3+9, which you can do similarly

[autumn patrol]

May I ask why n^5+3 needs a lower bound and n^3+9 needs an upper bound?

Why not find lower bound of both for ex

[autumn patrol]

and this^

For big omega:
h(n) = a/b = 4n^2

since h(x) >= n^5 +3/n^3 +9 I can just use h(n) instead

n^5 +3 <= 8n^5 (a)
n^3 +9 <= 2n^3 (b)

so at the end im gonna get

4n^2 >= 2n^2 >= n^2 with C = 2, n > 1
f(n) belongs to big theta(n^2)

[rugged kite]

Because 1/a > 1/b if and only if a < b

1/2 > 1/3 because 2 < 3

[obtuse pebble]

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[timid silo]

anyone help with set in discrete math

[compact shadow]

Pleased don’t ask to ask…

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

that's not even a definite question. If you want to discuss this go to #math-discussion

@timid silo if its not a question what is it ?

[compact shadow]

#❓how-to-get-help

It says , don’t ask permission to ask, just ask.

If you still don’t want to give you question, better close this channel for others.

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

let me do a favor . i betta quit the server.

[compact shadow]

Yeah you better

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[acoustic pawn]

"The integration here is over y, along the wire, with x treated as a constant. We must now write theta as a function of y, or y as a function of theta"
why is that?

[acoustic pawn]

(I posted this here since its more a question about integration then electric fields, tell me if it doesnt belong here and i'll repost it to the physics help channel)

[royal glacier]

Well, the distance from the wire, x is constant

The variable here is y

To integrate, you need to make a substitution, writing y in terms of θ

You see, instead of dealing with square roots and stuff, you can directly write the hypotenuse as y cscθ

Well yea you need to convert it into a function in θ

[acoustic pawn]

I dont get why you would have to write y in terms of theta honestly

[royal glacier]

Wait @acoustic pawn

The components

The electric field is a vector

Thats why we need to convert it into terms of θ

We cannot integrate directly, as its not a scalar

[acoustic pawn]

Hmm okey, ig i'll start working on my calculus bcs idk how to integrate a vector

[obtuse pebble]

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[obtuse pebble]

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[last kindle]

hello

[last kindle]

the question is

the sum of three consecutive multiples of 9 is 162

find the multiples

how will u do dis

[urban kiln]

u know the concept of AP?

[raven spire]

Hey Tornado

[urban kiln]

Arithmetic progression

@last kindle

[raven spire]

how do you write a multiple of 9?

[last kindle]

i thought of doing

x + (x+9) + (x+18)

as 9 and 18 and multiple of 9

[raven spire]

wow, that would help though

3x + 27 = 162

,calc (162 - 27)/3

[warm shale]

Result:

45

[raven spire]

You'll get x = 45

And consequently, the values of x, x + 9, x + 18

[last kindle]

so if x is 45

i do 45+ 9 and 45 + 18 for values right?

close

.close

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[distant horizon]

Hi

[distant horizon]

D correct?

[turbid sleet]

yes

[hybrid gull]

^^

[obtuse pebble]

@distant horizon Has your question been resolved?

[dim cedar]

You devide the speed by 2 couse you have half an hour

Then you do a cross product with the conversion ft -> miles

Do you understand why?

@distant horizon

[obtuse pebble]

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[swift river]

can someone explain this

[swift river]

i didnt get the last part

[short spruce]

splitting the fraction?

[swift river]

yea

i get the first and second operations

but the last got me confused

1+1/x-1

[short spruce]

you're confused on why $\frac{x-1}{x-1}+\frac{1}{x-1}=1+\frac{1}{x-1}$?

[warm shale]

a disappointing son

[swift river]

yes

[short spruce]

well... $\frac{x}{x}=1$...

[warm shale]

a disappointing son

[leaden ibex]

$\frac{\cancelto{1}{x-1}}{\cancelto{1}{x-1}} = 1$

[warm shale]

Remavas

[swift river]

oh

yeah i see

thanks

.close

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[queen sparrow]

How should I go about proving that for multi-variable function f(a,b,c), where c is an arbitrary natural number, you can find number a and b that make the result be a positive integer?

[tardy epoch]

do you have a function? this isn't true in general

[queen sparrow]

yes, let me find the pic for it real quick

[tardy epoch]

FFS

[queen sparrow]

@tardy epoch

[tardy epoch]

if you can prove the numbers underneath the cubic root are positive and p is positive, then you're done?

[queen sparrow]

well, not really. That doesn't make sure it's an integer

[tardy epoch]

oh positive integer. then just make it a cube of some number

$q \pm \sqrt{q^2 + (r-p^2)^3} = n_{\pm}^3$ for some $n_{\pm} \in \N$

[warm shale]

riemann

[queen sparrow]

not in the naturals because it can be a rational number

p does the job of making it natural at the end result

[tardy epoch]

p has to be a multiple of 3 right?

[tardy epoch]

what are you saying can be a rational number

[queen sparrow]

not necessarily, you can have c = 2, a = 5, b = 2

[tardy epoch]

oh i see what you mean

yea i'm just trying to make all the summands integers

[queen sparrow]

we could adapt your strategy to make each of the summands rational, then say that their sum is always an integer

[obtuse pebble]

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[queen sparrow]

thanks riemann

[obtuse pebble]

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[pine pier]

in this definition of compactness, why does it say X = ... and not X ⊂ ... ?

[pine pier]

I thought the definition of open cover only required X to be a subset of the union of open sets

[balmy mortar]

There are 2 versions of defn of cover

So we have a topological space X, and a subset S within it.

We are now interested in possible definitions for an open cover of S

I believe the following are equivalent... (but check):

• An open cover of S is a collection of open sets in X whose union contain S
• An open cover of S is a collection of open sets in the subspace topology S whose union equals S

===
The last definition is what you're seeing on wikipedia

The open sets of the subspace topology S are exactly the open sets of X intersect S, which should lead these 2 defns to being equivalent, basically.

@pine pier ^

Also this elaborates further I think
https://math.stackexchange.com/questions/1733448/difference-in-definition-of-a-cover

[pine pier]

Ah, thank you @balmy mortar

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[royal solar]

do you remember the theorem

[fervent cobalt]

EF= CE+ DE

that’s where i got sticks

stuck

and i put it in photomath and it don’t pop up with anything

[obtuse pebble]

@fervent cobalt Has your question been resolved?

[fervent cobalt]

<@&286206848099549185>

[quartz pecan]

I think CE is supposed to be 2x + 7 instead of x squared plus 7

I suggest you try that

@fervent cobalt

[fervent cobalt]

ty so much

i misunderstood how to add xs smh

.close

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[brazen grail]

https://media.discordapp.net/attachments/645086150729662508/948013478201925692/4BF4F56D-677F-4765-BFBC-A1C003FE8913.jpg?width=507&height=676

[brazen grail]

I dont understand how to solve it

I guessed on it

[high lily]

simplify the given fraction

[brazen grail]

how would i simplify 1/x+2?

[high lily]

1/(x+2) itself is simplified

[balmy mortar]

$$\frac{1}{\frac{1}{4}+\frac{1}{6}}$$

[warm shale]

Shuri2060

[brazen grail]

so simplified it would be 2/2x+5?

[balmy mortar]

Try this. Can you?

Definitely not.

[high lily]

assuming you mean 2/(2x+5) no

[balmy mortar]

Do you know how to add fractions?

[brazen grail]

ye

[balmy mortar]

I don't see you doing it correctly

[high lily]

$\frac{1}{1} + \frac{1}{1} \neq \frac{1+1}{1+1}$

[warm shale]

ℝamonov

[brazen grail]

10/24?

[balmy mortar]

its 1/(10/24)

[brazen grail]

oh

yeah

[balmy mortar]

24/10

[brazen grail]

i never worked with a fraction over a fraction

besides negative exponents

[high lily]

well take things one step at a time

[balmy mortar]

Try this for your original question

1/(x/y) = y/x

You should hopefully know this, which is needed in the last step.

[brazen grail]

just making sure

cause doing / is confusing for me

[brazen grail]

i dont

[high lily]

if that top fraction line is distinctly longer than the bottom line implying
$$\frac{1}{\br{\frac xy}}$$
then yes

[warm shale]

ℝamonov

[brazen grail]

does there need to be a parenthesis?

[high lily]

not necessarily, depends on how well you write your math
but they do make things a lot clearer

[brazen grail]

oh

going from here, why did they flip the denominator and numerator of the fraction under 1?

[balmy mortar]

you should revise elementary algebra laws for fraction

if you did not know this

[high lily]

a definition of division

[balmy mortar]

1 divided by a half is 2

because 2 times a half = 1

[high lily]

division is multiplication of the reciprocal

you could also consider
$$\frac{1}{\br{\frac xy}} \cdot \frac yy = \frac{y}{x}$$

[warm shale]

ℝamonov

[obtuse pebble]

@brazen grail Has your question been resolved?

[brazen grail]

sorry i was eating dinner

[obtuse pebble]

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[brazen grail]

.reopen

[obtuse pebble]

✅

[brazen grail]

okay

[balmy mortar]

Not sure what is going on here

Why is that 1 underlined

[brazen grail]

where

[balmy mortar]

first line

[brazen grail]

?

[balmy mortar]

It's not clear where your lines separate

yes

[brazen grail]

let me rewrite it

does this make more sense

[balmy mortar]

I don't understand why you have underlined 1

Like make your choice

1 equation or 2 equations

If it's 2 equations I-

wait im confused af

[brazen grail]

were trying to simplifiy 1/(x/y) correct?

[balmy mortar]

like this is not how math works

why are there 2 equals signs

thats just nonsense

[brazen grail]

yeah mb

[balmy mortar]

You were also shown a more elegant way of doing it

Either work but, at least write it properly if you're going to do it like that

[brazen grail]

ok

we have 24/10

where do i go from here?

[balmy mortar]

That's not your original question.

do that

[brazen grail]

?

[balmy mortar]

that was my question

1/(10/24) = 24/10

[brazen grail]

yeah i am lost

https://images-ext-1.discordapp.net/external/ISkUSbzFv_2nM1NxvnjH5jH4gdLQLM-O5XBoZPySyuw/%3Fwidth%3D507%26height%3D676/https/media.discordapp.net/attachments/645086150729662508/948013478201925692/4BF4F56D-677F-4765-BFBC-A1C003FE8913.jpg

simplify right?

[balmy mortar]

Yes, so proceed in the same manner

To add 1/4 and 1/6

you found a common denominator of 24

(note you could have chosen 12, but either works)

[brazen grail]

(x+2)(x+3)?

[balmy mortar]

yes.

[brazen grail]

why is the x>3 there?

like what do i do with it

[balmy mortar]

It tells you x > 3

That's all.

[brazen grail]

its just there to throw me off?

[balmy mortar]

Because x > 3, you know you are never dividing by 0 for any of these, presumably.

[brazen grail]

huh

welp

thanks anyways i guess

.close

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[mellow tide]

Whats the equation to these 2 questions?

[short spruce]

is this a test

[mellow tide]

No

Revision sheet

[short spruce]

are you familiar with soh cah toa?

[mellow tide]

No

[short spruce]

are you familiar with the ratios that trig functions represent?

[mellow tide]

No

[short spruce]

then you may want to know that if you're doing questions about them

perhaps look back at your notes / google

[high lily]

have you learned any trig at all?

[mellow tide]

Okay ty

.close

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[stark drift]

hey i have this geometric series or sequence (i think) and i dont know how to find the sum of n turns.

[stark drift]

hey i have this geometric series or sequence (i think) and i dont know how to find the sum of n turns.

,,k_0 = 10^3\k_1 = 10^3 + (k_0 * r)\k_2 = 10^3 + (k_1 * r)

[warm shale]

Monsieur Gilbært

[stark drift]

and so on

[balmy mortar]

There is a formula.

Seen it?

[stark drift]

yeah

[balmy mortar]

Also this is not geometric?

[timid silo]

that isn't geometric

[balmy mortar]

geometric means a common ratio between terms.

[stark drift]

thanks for pointing that out

[balmy mortar]

So nevermind the formula.

[hushed moat]

More of a recurrence relation

[stark drift]

yeah im aware, but i dont know whet this is

[balmy mortar]

But try writing out the sum

for a few terms

and try to simplify it.

[stark drift]

ive tried but this is as far as i get

,,i could write something along the lines of this 10^3 (1+r^1 + r^2 + r^3 ...)

but idk where to go from there

,,10^3 (r^0 +r^1 + r^2 + r^3 ...)

[balmy mortar]

Like I said

Try writing our the sum for small n

k0 + k1 + .... + kn

And simplify that

see if you can spot a pattern

which can generalise.

[stark drift]

numerical sum right?

[warm shale]

Monsieur Gilbært

[stark drift]

iim doing something horribly wrong

i will be back

.close

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[brisk arrow]

There’s a bag it has 11 balls inside of it, each balls are being printed with number from 1 to 11, then we pick 4 balls out at once, assume event A is the event when “ball 6” is the second largest in those 4(ex.2,4,6,7 and 4,5,6,10) then the probability of event A is?

[brisk arrow]

Here’s my attempt

On the denominator is the probability of all events

While on the numerator is the probability of event A which i feel uncertain about

1/11 is the probability of getting the ball6. Multiply by 1/5 cuz we want a number that is bigger than 6 so we got 7,8,9,10,11. We choose one of those.

Next we got 1/(5*4) cuz the last 2 balls have to be smaller than 6 right? So 1,2,3,4,5 we pick two out of those

Since we pick those 4 balls out at once, so the order of those balls isn’t matter thus I multiply it by 1/4!

That’s all, is there anything wrong?

Can someone check this out for me

<@&286206848099549185>

[stark drift]

looks correct to me

although its been a while since ive done combinations and probability

[obtuse pebble]

@brisk arrow Has your question been resolved?

[obtuse pebble]

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[blazing rover]

What I've come up with so far: x^2 = y^2 + y + 2

but I don't see how this can be useful

[obtuse pebble]

@blazing rover Has your question been resolved?

[obtuse pebble]

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[timid silo]

why does that hold

[balmy mortar]

Any group of prime order must be cyclic.

This is a consequence of one of the early theorems

Maybe lagrange? idk.

Yes lagrange.

The order of an element must divide the order of the group.

If the order of the group is p, prime. Then the order of all the elements are either 1 or p.

Hence at least one element has order p, so the group is cyclic.

[obtuse pebble]

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[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[lone blade]

how is this homemorphic to K_3, 3

[obtuse pebble]

@lone blade Has your question been resolved?

[compact shadow]

I don’t think they are. The adjacency matrices of them, one’s determinant is zero while the other one is not

[royal basin]

who said this was isomorphic to K_3,3?

this graph you are showing is planar. move h and d inside the rectangle and past each other and see for yourself

[lone blade]

Hm I see

This is the problem

[compact shadow]

hcgad is K_5

[royal basin]

are you sure

@compact shadow where's the path from g to d

[compact shadow]

Then how about hcebd ?

[royal basin]

deg(c)=3

where's the paths from c to b or c to d

cant have both

[compact shadow]

?

What vertices of subgraph is subset of vertices and edges of subgraph is subset of edges right?

[royal basin]

.....

[compact shadow]

Maybe it’s the way of my expression hcgad I mean

[royal basin]

g cannot be part of a K5 minor either

its degree is too low

[compact shadow]

Oh I was being dumb nvm my bad

it doesn’t contain a subgraph hemeomorphic to K_5 or K_3,3

h,a,b,d,e can’t make a K_5 cause there must be edges h—e—b then we can’t find edge connecting e and a or e and d. and all triple of three vertices that are pair wise disconnected are (b,g,f) , (d,g,c) ,(e,g,d) , (f,b,c) can’t choose two of them to make a K_3,3

[lone blade]

Yeah so is it planar then?

I realllly don’t think it is

[compact shadow]

It is by that theorem

The theorem says it’s planar, I don’t know how to draw it though

[obtuse pebble]

@lone blade Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[fleet locust]

If I have L1 as an abritrary (but not specified) finite language, and L2 as an arbitrary (but not specified) non-regular language.

I need to find out that following languages is regular or not and why.

What will it be, if you do L1 - L2 or L2 - L1?

What I know now:

• L1 is a regular, because it's finite language.
• I have to use closure properties of regular languages.
• If L1 and L2 are regular languages then L1-L2 is also regular

I want some hints to find out that, not the answer please, thanks.

[obtuse pebble]

@fleet locust Has your question been resolved?

[fleet locust]

If anyone want the answer, so look at this:

Set Difference operator:
If L and M are regular languages, then so is L – M = strings in L but not M.

[obtuse pebble]

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[wanton kraken]

help

so im at right triangles which have one angle that equals 30 degrees

the cathetus against the 30 degree angle is half the hypothenuse

and I have to find CL, AL, CD, and BC

Translation:

AL - bisector
CD - height
AL + CD = 21 cm

and find a) CL, AL, CD, BC
b) the distance from point L to AB

sorry for everything that's scuffed

I'm kind of in a hurry

I'm given that AL + CD = 21 cm

but they are two different triangles

that part confuses me a lot

[obtuse pebble]

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[wanton kraken]

help
so im at right triangles which have one angle that equals 30 degrees
the cathetus against the 30 degree angle is half the hypothenuse
and I have to find CL, AL, CD, and BC
Translation:
AL - bisector
CD - height
AL + CD = 21 cm
and find a) CL, AL, CD, BC
b) the distance from point L to AB
sorry for everything that's scuffed
I'm kind of in a hurry
I'm given that AL + CD = 21 cm
but they are two different triangles
that part confuses me a lot

<@&286206848099549185>

Ping is made because the bot even closed this channel

for inactivity

[timid silo]

i can't even see the actual diagram

also i don't understand the lang

[wanton kraken]

yea sorry I just found the angles of all triangles inside to the best of my ability

because it's extra small

I translated the conditions

Or if it's easier:

"Given triangle ABC (alpha : beta : gamma = 2:1:3)
AL - bisector
CD - height
AL + CD = 21 cm
"Find
a)CL, AL, CD, BC (lenghts)
b) the lenght from point L to segment AB"

[obtuse pebble]

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[idle thunder]

What do you need help with? @exotic sparrow

[exotic sparrow]

first let me check the first one

so first I did

a ∫4y^3 dy = 1

Then a [ y^4] = 1

and I put in the values y=3 y = 1

so I got

a [3^4 - 1^4] = 1

80a = 1

a= 1/80

now im on (b)

@idle thunder Mean and Variance

[obtuse pebble]

@exotic sparrow Has your question been resolved?

[novel knoll]

its just applying definition

[obtuse pebble]

@exotic sparrow Has your question been resolved?

[exotic sparrow]

back

ok first you replace a with 1/80

f(y) = 1/20 y^3

Mean = ∫y * f(y) dy

∫ 1/20 y^4

= [1/100 y^5]

sub in 3 and 1

2.43 - 0.01 = 2.42

121/50

E(X) = 121/50 I think

===

VAR(Y) = E(Y^2) - [E(Y)] ^2

E(Y^2) = ∫ x^2 * f(y)

= ∫ 1/20 y^5 = [1/120 y^6] and sub in y = 3 and 1

6.075 - 0.0083 =6.067

VAR(Y) = 6.067

@novel knoll

[novel knoll]

looks fine

[exotic sparrow]

how can we find the distribution function of y

[tardy epoch]

The definition of the distribution function of y is $F(y) = P(Y \le y)$. Calculate the integral

[warm shale]

riemann

[exotic sparrow]

t

back

So the integral of 1/20 y^3

1/80 y^4

[exotic sparrow]

$F(y) = P(Y \le y)$

[warm shale]

FC

[tardy epoch]

Right. Feel free to .close

[exotic sparrow]

so thats the answer?

@tardy epoch

how about the 0 ,otherwise

[tardy epoch]

Plug it into your integral

[exotic sparrow]

=0

[novel knoll]

Its just applying definition

[exotic sparrow]

done

just the median and last one now

done median

just (e)

final one I have no idea how to do it

[novel knoll]

Apply definition again

[obtuse pebble]

@exotic sparrow Has your question been resolved?

[obtuse pebble]

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[leaden ibex]

What do you need to find the area?

[young hinge]

I dont know

[leaden ibex]

Wall what's the formula

[young hinge]

a=root of c2 - b2

[short spruce]

that's the pythagorean theorem, which you do need... but not the formula for area of a triangle

[young hinge]

Then how do I do it?

[short spruce]

you use the formula for area of a triangle

look at your notes

[young hinge]

I need the height for it

and the height is a

[short spruce]

so find a

[young hinge]

what formula do I use to find a?

the one I sent before?

[leaden ibex]

If you want to use the pythagorean theorem

You need to first identify a right triangle with 2 known sides

[young hinge]

I did this

[unique solstice]

Thats a good first step

[young hinge]

isnt it just 2+4 = 6

6 x 3

/2

[unique solstice]

Yeah thatd work

[young hinge]

is that right?

[unique solstice]

Ye

[young hinge]

nice

[obtuse pebble]

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[lone blade]

how to show

[obtuse pebble]

@lone blade Has your question been resolved?

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[untold ravine]

hello

[untold ravine]

https://gyazo.com/c2d310a543ab7dfdc729d4c99930a759

i dont know how to get the range of a function let alone a sin fuction

[short spruce]

if you're unsure of how to identify the range of any function, you shouldn't be starting with trig functions

if you can tell me what range of a function is, that's a start

[hallow star]

The range of Y is simply the difference between the minimum and maximum of the function

[untold ravine]

the maximum and minimum y values that it can reach

yeah

[short spruce]

then what's the trouble here?

[untold ravine]

how do i get it for sin functions

[short spruce]

well you have the function in front of you

i assume you know what the transformations are?

[untold ravine]

yes

[short spruce]

so what's happening to the parent function [sin(x)]?

[untold ravine]

its translated 3 spaces upwards

[short spruce]

no

[untold ravine]

wait

hold on

its being stretched in the positive y direction by 3

and translated upwards by 3

[short spruce]

no

[untold ravine]

5 sorry

couldnt read it