help-42

[worthy comet]

Is that correct?

To make x equal r we add plus 1 to to cot

[hasty kite]

u dont need to specify tht

[worthy comet]

but did i do it correct?

[hasty kite]

idk wht r u saying

[worthy comet]

like was my logic correct

[hasty kite]

idk

but all u need to know is these 3 identities

theres is only one identity u can proove in trignometry and thats sin2 + cos2 if u want i can show u how

[worthy comet]

sure

The more i know the bettetr

better

[worthy comet]

thx

[hasty kite]

O is opposite side

H is hypotenuse

A is base

[worthy comet]

Thought a stood for adjacent

[hasty kite]

same thing

out of topic but which grade question is this

[worthy comet]

uh pre cal

sorry it took long for me to respond @hasty kite was doing some other stuffs

.close

[calm coral]

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[midnight relic]

I'm doing this problem for Stokes Theorem and I did the cross product first, but after that I was confused and looked at the answer key and I was really confused where it said n = k, I was mainly confused on how one would get that, can someone please help me understand this? Thank you

[swift laurel]

we can integrate over any surface bounded by that particular curve. the simplest such surface would actually be a region in the xy-plane, which has a unit normal either = k or -k depending on the direction of circulation (right hand rule)

note that the special case where we choose the surface as a region in the xy-plane actually makes stokes' theorem equivalent to green's theorem

[midnight relic]

ohhhh ic now thank you

.close

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[toxic haven]

i don’t understand

how they got 2/12

.close

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[remote mural]

I need help

Please

<@&286206848099549185>

[merry python]

do you know the binomial formula it's talking about?

[remote mural]

No

[merry python]

i take it this is from a book. can you please take a quick look if there's any mention of the formula before this page?

[merry python]

ok good. so there's the answer to the first question right there on step 1

[remote mural]

That’s the answer

[merry python]

to the first question (the one about special product)

for the binomial formula, you need to use the $(a-b)^2$ formula they wrote in step 2

[potent lotus]

TooManyCooks

[remote mural]

Yeah I only need help on special Product port

Part

[merry python]

can you identify what a and b are in your problem?

[remote mural]

No idk what a and b are in the problem

[merry python]

so you have a formula there that says $(a-b)^2 = a^2 -2ab + b^2$

[potent lotus]

TooManyCooks

[merry python]

your problem is $(6x-2)^2$

[potent lotus]

TooManyCooks

[merry python]

can you identify which terms a and b would be given that information

[remote mural]

6 is A. And 2 is B

[merry python]

not quite correct on a

it's not just 6

[remote mural]

Brah

[merry python]

bruh

do you see the problem?

about why it's notjust 6

[remote mural]

Idk This is Hard

[merry python]

you can do this. i just need you to match the stuff on the formulas

so you have $(6x-2)^2$ and you want it to look like $(a-b)^2$

[potent lotus]

TooManyCooks

[merry python]

you chose b correctly

[remote mural]

(A-B)^2?

Right?

[merry python]

right, but what's A

[remote mural]

6x

[merry python]

good!

ok so now you can use the formula

[remote mural]

Let’s Fukin go Brah

I love U

[merry python]

the book says $(a-b)^2 = a^2 -2ab + b^2$

[potent lotus]

TooManyCooks

[merry python]

plug in A and B into that

[remote mural]

Kay

[remote mural]

Do I put it where it’s special product

[merry python]

the first question?

[remote mural]

Ya

[merry python]

That's what I was mentioning earlier. This was discussed in step 1 in the picture you sent

"This type of binomial is known as a square of a binomial difference"

the stuff we're working on right now is the stuff that goes below "solve using the binomial formula"

[remote mural]

Ok

Now what about check using the foil method

[merry python]

before that, what did you get

[remote mural]

Oh I have to solve that

[merry python]

yes

you have to plug it in the formula, you already have a and b

[remote mural]

(
a
−
b
)
+

a
+
−
a
b
+
b
+

Oh shit

Holduo

[merry python]

[remote mural]

(a−b)+=a+−ab+b+

That what I got

[merry python]

i need you to plug in a = 6x and b = 2 like you mentioned

[remote mural]

At the end

?

[merry python]

into this one $(a-b)^2 = a^2 -2ab + b^2$

[potent lotus]

TooManyCooks

[merry python]

that's what the binomial formula question is asking for. it's just asking you to plug in a and b and get an answer

[remote mural]

.close

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[merry python]

[calm coral]

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[thin radish]

yo is this proof a good idea?

[remote mural]

this version of Lagrange's theorem only works for finite groups

you've indicated that G is not finite

@thin radish

[thin radish]

yeah

i saw that mistake

ty

[remote mural]

no worries

:)

For finite groups, though, I could walk you through a proof (although I think you've got the hang of it)

[thin radish]

i though to shot it with homomorphism or something like that

[remote mural]

hm

[thin radish]

G ismorphic to something that it is p-group

something like that

[remote mural]

Define a map $\phi : G \to G/H$ by ...

[potent lotus]

maxim

[thin radish]

yeah something like that

[remote mural]

or H

[thin radish]

it could be a good idea

[remote mural]

i mean, if G is isomorphic to a p-group then it is also a p-group

[thin radish]

yeah

[remote mural]

well

are you still looking to prove this for infinite G?

if so yeah a homomorphism is probably our best bet

[thin radish]

if G is finite then its a piece of cake

i am trying for G infinite

[remote mural]

yeah if $G$ is finite then we just use that version of Lagrange's theorem

[potent lotus]

maxim

[thin radish]

exactly

[remote mural]

and use the fact that $|G/H| = |G|/|H|$

[potent lotus]

maxim

[remote mural]

anyway

gimmie a second to think

[thin radish]

i am trying to solve it myself i wanted an opinion about my proof

if you find a solution just dont write it to me

i hope you understand

ty for your help

[remote mural]

no worries

if you need further help let me know

I think I have an idea of how to proceed

[thin radish]

you know what just write your idea here and i am gonna see it if i am rly stuck

[calm coral]

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[fleet echo]

how did it happen

i didn’t understand

[compact venture]

So

What if we said y=x+2

[fleet echo]

okay

[compact venture]

Then the line above the white arrow would look like 9x^2y-4y=0

Because I said y=x+2

Now

Is it more obvious that you can factor out y to get (9x^2-4)y=0?

[fleet echo]

wait how did you get 2y-4y?

[compact venture]

.

Just a sec

$9x^2y-4y=0$

[potent lotus]

caliπnia

[compact venture]

Was my intention

[fleet echo]

okay okay

[compact venture]

It can be seem you can factor out y , right?

[fleet echo]

yea

[compact venture]

So when you do that you get

[fleet echo]

(9x^2-4)y

[knotty pumice]

you can question why would it be valid for a binomial

[compact venture]

$(9x^2-4)y=0$

[potent lotus]

caliπnia

[compact venture]

But

y in this case is (x+2)

[fleet echo]

yup

[fleet echo]

so how do u get that answer?

[knotty pumice]

replace it back

[fleet echo]

explain

pls

[knotty pumice]

(9x² - 4)(x+2) {since y = x+2}

you know its valid for a monomial?

[fleet echo]

it is?

[knotty pumice]

so its a good enough explanation

[remote mural]

What in exact didn't you understand 'bout this?

[fleet echo]

so u just combine the x+2 into the other and the outside of the parenthesis you combine it?

[knotty pumice]

not sure i followed

[remote mural]

This is all 'bout taking commons and decomposing the equation

Into factors

[fleet echo]

i dont understand, im sorry😔

[remote mural]

Alright lemme explain

[remote mural]

First of all they wrote the entire equation in standard form which is like variable with highest power to variable with lowest power

[fleet echo]

thats right

[remote mural]

How did they do it?
Answer: The constant with the highest power is x^3 which is with coefficient 9

[fleet echo]

yes

[remote mural]

It descends to variable with lowest power x^0=1. The coefficient of x^0 is -8

[remote mural]

Now this stuff is factorised

[fleet echo]

well, the thing that i dont understand is how did it factored into (9x^2-4)(x+2)

[remote mural]

Well, 9x^2 is taken as common and -4 is taken as common

Look at it carefully

These things are like the highest common you can take for simplification

Grabbing commons from both of the equations in first line

We end up with x+2 and 9x^2-4

And for further simplification and decomposition

9x^2-4 can be taken to the formula a^2-b^2

Which is like (3x)^2-2^2

Now you can write it as (3x+2) (3x-2)

Therefore (x+2)=0 (3x+2)=0 (3x-2)=0 are the required factors

[fleet echo]

what is a common?

[knotty pumice]

multiplicity is associative over addition

[fleet echo]

oh okay i get it now

thank u guys

.close

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[terse swan]

Can someone help me?

[calm coral]

Please don't occupy multiple help channels.

[past zinc]

you need to evaluate derivative P' ( I) and then solve equation P' ( I) = 0

when you find soem zerors

then check, where is maximum

[terse swan]

Can you walk me through it, if you don't mind

[past zinc]

🙂 i give you a hint then

[terse swan]

sure

I just don't know where to start

Do I take the derivative of both the top and bottom?

[past zinc]

minute plz 🙂

[terse swan]

alright

[past zinc]

\text{Let's assume that: }\text{ }\text{ }P\left( I \right)=\frac{g\left( I \right)}{h\left( I \right)}\text{ }\text{ }\text{ then:}\\P'\left( I \right)=\frac{g'\left( I \right)\cdot h\left( I \right)-g\left( I \right)\cdot h'\left( I \right)}{h^{2}\left( I \right)}

$\text{Let's assume that: }\text{ }\text{ }P\left( I \right)=\frac{g\left( I \right)}{h\left( I \right)}\text{ }\text{ }\text{ then:}\\P'\left( I \right)=\frac{g'\left( I \right)\cdot h\left( I \right)-g\left( I \right)\cdot h'\left( I \right)}{h^{2}\left( I \right)}$

[potent lotus]

Joanna Angel

[past zinc]

that is formula

for derivative of the quotient

and in yoru case

you must compute it

so you were right but you have to use this formula

[terse swan]

Oh quotient rule?

what do you after this?

[past zinc]

when you get the ersult of this derivative

you need to equate it to zero

i.e.

[terse swan]

Can we work on this one please? I lost the other one

[past zinc]

P' ( I) = 0

[terse swan]

I'm going to do the quotient rule

I will send what I get

[past zinc]

yes

when you ready publish phot or pdf

[terse swan]

I got (i^2+i+4)(100)-(100i)(2i+1)/(i^2+i+4)^2

what do I do now?

[past zinc]

denominator must be sqaured

due to formula

but fo rnext oepraitosn only nomiantor si important

plz make order in yoru nominator

and equate it to zero

[terse swan]

So bascially I just get the derivative set it equal to 0 and solve?

[past zinc]

i told it ) yes

[terse swan]

Ok I know how to do this now

can you help me with this one

[past zinc]

i cansee it

cant

make it bigger

[terse swan]

Also, why is the numerator only important? why does the bottom not matter

[terse swan]

Hows this?

[calm coral]

@terse swan Has your question been resolved?

[calm coral]

@terse swan Has your question been resolved?

[terse swan]

@past zinc Find dimensions of a rectangle with area 216. Please help

[old falcon]

can you restate

how is the problem you posted above and this problem related

[terse swan]

I figured out how to do both of them

[calm coral]

@terse swan Has your question been resolved?

[calm coral]

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[sharp depot]

help

[sharp depot]

why does the foci have -2

in there

[glacial elk]

by foci you just mean the two points labelled F?

why do you think there should not be a -2 in there?

[sharp depot]

nvm i got it

.close

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[fiery lagoon]

hello

[pure kayak]

greetings

[fiery lagoon]

how to prove lhs = rhs

[pure kayak]

tan^2+1=sec^2

[fiery lagoon]

there is no tan²x +1

it just tan²x

i tried sec²-1

[pure kayak]

and?

[pure kayak]

this is the key really
if you want to work from the LHS, try factoring out sin^2

if you want to work from the RHS, use tan^2=sec^2-1

[wooden gyro]

sin^2/cos^2 - sin^2
sin^2 ( (1/cos^2 ) - 1)
sin^2tan^2

[fiery lagoon]

why (1/cos²)-1

[wooden gyro]

what u mean why?

[fiery lagoon]

isnt it cos²=1-sin²

[wooden gyro]

?

bro tan = sin/cos

tan^2 = sin^2/cos^2

[fiery lagoon]

omg

thx ill try

[wooden gyro]

already solved

.-;.

[fiery lagoon]

man i feel dump this morning

[wooden gyro]

me too

[fiery lagoon]

it got solved thx bro

[wooden gyro]

.close

[fiery lagoon]

.close

[calm coral]

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[indigo wren]

so I've solved all of them

but want to know why problem 4 can't be modelled with poisson

by saying that there are an average number of 4 in 5 to 7

and then do poisson 1 for 1

but bayes rule gives a diff answer

[calm coral]

@indigo wren Has your question been resolved?

[calm coral]

@indigo wren Has your question been resolved?

[calm coral]

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[crisp spear]

what are the steps to get this ?

[calm coral]

Show your work, and if possible, explain where you are stuck.

[native wasp]

! status

[calm coral]

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

[native wasp]

Write that's root stuff like √((2x)^2 - 9^2)

[crisp spear]

the sqrt symbol really need to stay?

[native wasp]

What

[crisp spear]

i mean that can be simplified?

like 4x - 9

remove the exponent and the root

[reef nimbus]

how can you "remove the root"?

[crisp spear]

with the exponent of 2

both term are perfect square right?

[calm coral]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[native wasp]

?

[native wasp]

It's on individual terms

The exponents

Not as a whole

[crisp spear]

oh i see mb

[native wasp]

Put x=(9/2)cosTHETA

@crisp spear try this

Bruh

[native wasp]

U there?

[crisp spear]

still here

what will be the dx?

[native wasp]

I mean just put it

And look at expression

First

[native wasp]

-(9/2) sintheta dtheta

[calm coral]

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[balmy knoll]

Why does k disappear here?

[balmy knoll]

0.66 is wrong btw its supposed to be 0.6

Here it disappears as well

wait I just realized I times K = J

J times K = I

and K times K = 0

wtf why

[calm coral]

@balmy knoll Has your question been resolved?

[balmy knoll]

<@&286206848099549185>

[dim shadow]

These are quaternions right?

[balmy knoll]

I don't know what a quaternion is

this chapter is about torques

[balmy knoll]

Oh its kind of like quaternions

can you help me?

[dim shadow]

Not exactly I don’t know enough but I can tell you the most likely reason they disappear is the sine function next to the k

Also I think K*K would be 1

Not 0

[balmy knoll]

but why would it even become a constant number

why wouldn't it be k²

[dim shadow]

Have you looked at the times table for these

K*K is either 1 or -1 can’t remember

[balmy knoll]

There wasn't a times table that I found

[dim shadow]

i*i is -1

[balmy knoll]

what would j * i be?

[dim shadow]

K or -K

[balmy knoll]

can u send me the times table?

[dim shadow]

Let me try and find it

[balmy knoll]

Ok thank you alot

this is very annoying

[dim shadow]

Found it

[simple plover]

Do you want the maths reason, or the torques reason for k disappearing?
Because in the y-direction the two forces equal each other.

[dim shadow]

Oooo this is a bigger version

[simple plover]

It's just vectors and torques.

[dim shadow]

Could you give me the math reason I always wondered why the times table worked like this

[balmy knoll]

But k is in the z direction though right?

[balmy knoll]

also positive k times positive k

does it make a negative j?

[dim shadow]

No cause -i * i is not -1

[balmy knoll]

I don't understand how to read this table it gives me confusion

How do I read it? diagonally? horizontally?

[dim shadow]

It’s all the top row multiplied by the left most column

[simple plover]

For torque around O in the y-direction the forces cancel

[balmy knoll]

So is it like J times i?

[dim shadow]

-k

[balmy knoll]

Ah I see

that's right

[balmy knoll]

This is the same times table right

[dim shadow]

Ye

Just bigger

Cause octonions go brr

[balmy knoll]

this times table teaches us that k times k = -1, but -1 just disappears in this excersise right

the forces cancel eachother but do they always disappear?

what if A was 30lbs and B was 25lbs

would it still have disappeared?

[dim shadow]

This is where I can’t help

[balmy knoll]

damnit

thank you for bringing the knowledge that such a times table exists tho

that'll be helpful for things like this

[balmy knoll]

@simple plover do you know the answer to this question?

[calm coral]

@balmy knoll Has your question been resolved?

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[floral mango]

Biggest integer that can be represented with n bits? is it n^2?

[humble ivy]

Write out some small values

Its not n^2, you're close though

[blazing fractal]

@floral mango
what are bits first of all, major hint to the solution

and how so they store values and how are they represented

[floral mango]

hmmm.. Sorry, I got to restart my brain. It keeps flashing critical error.

[next hound]

depends what encoding scheme you are using

theres infinite valid answers, but two that are the best

[floral mango]

is it? ((2^n)-1)

or at least one of the best.

[next hound]

this is a good answer yes

this is for unsigned integers

[floral mango]

Thanks for your help bunny!

I have one more that I am not sure about then I'll be done! Worst-case number of swaps needed to sort an array of length n if you are
only allowed to swap adjacent elements. I think n^2 is the answer.

[calm coral]

@floral mango Has your question been resolved?

[floral mango]

some of my notes say bubblesort does 2*n-1 /2 comparisons worst case

[calm coral]

@floral mango Has your question been resolved?

[clear delta]

bubble sort is definitely O(n²) so that can't be right

[calm coral]

@floral mango Has your question been resolved?

[calm coral]

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[calm coral]

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

[tidal grotto]

@raven pier

[raven pier]

this one actually

[raven pier]

2

[simple plover]

How did you solve 1 (i)? Seems similar.

[raven pier]

i subbed y in the equation

got quadratic and solved discriminant to get 0, meaning it is a tangent

but for 2 i didnt get 0?

[simple plover]

Understood.

Maybe an arithmetical mistake?

Shows only one solution.

One mistake in the algebra will do it.

[calm coral]

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[remote mural]

ok so i dont get why my method is wrong

[calm coral]

Show your work, and if possible, explain where you are stuck.

@remote mural Has your question been resolved?

[remote mural]

basically i pulled made 1/2 into 4x(1/8x)
and then log4x 4x(1/8x)

like this

[calm coral]

Show your work, and if possible, explain where you are stuck.

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

@remote mural Has your question been resolved?

[remote mural]

<@&286206848099549185> hi can someone help me wi5h this question?

[old falcon]

i think using some substitutions helps

but i havent finished the problem out

use $\log _{4x} (0.5) = \frac{\log _x (0.5) }{ \log _x (4x) }$

[potent lotus]

jan Niku

[old falcon]

we can simplify this quite a bit

to i think $\frac{-2\log _x 2 }{ 2 \log _x 2 + 1 }$

[potent lotus]

jan Niku

[old falcon]

you can arrive at a similar term for the $\log _{0.5x}(0.5)$ piece

[potent lotus]

jan Niku

[old falcon]

likewise, $\log _x 8$ is just $3 \log _x 2$

[potent lotus]

jan Niku

[old falcon]

at which point you have everything written in terms of log _x 2

ping if u come back

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

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[arctic mauve]

does anyone know how to tackle 13.7?

i dont know all

[calm coral]

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[arctic mauve]

.close

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[soft orbit]

Minimum value of (2√5 cosx+ 4 sinx +7 )
How can we find it through derivative method?

[dark wagon]

find the first derivative of the function

equate that to zero

from that you will get a certain value of x

thne first the second derivative of the function

and see if the value which you found out earlier gets you negative value or positive value when you put the value in the second derivative equation of the main function

if negative then maxima

if positive then minima

did you understand?
@soft orbit

[soft orbit]

2√5 sinx=4cosx

tanx=2/√5

[dark wagon]

yes good

[soft orbit]

what should I do next?

[dark wagon]

now find the second derivative of the function and put the value in it

and see if it comes negative of positive

[soft orbit]

sec^2x =0 ?

[dark wagon]

noo

the main function

[soft orbit]

I see

[dark wagon]

2root5 cosx + 4sinx + 7

diffn this this two times

[soft orbit]

2√5 cos x +4sinx

[dark wagon]

no

-2sqrt 5 sinx + 4cosx

this is first derivative

diffn this again and put the value and check

[soft orbit]

-2√5 cosx-4sinx

[dark wagon]

correct

now put the value of x = tan^-1 2qsrt5

which you found earlier

[soft orbit]

It looks very messy

@dark wagon

[dark wagon]

yeah it is

just use basic inverse trigonometry

make a triangle and solve

its easy

[soft orbit]

Ohh i got it

It is one

.close

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[verbal lintel]

Whats the formula for the last question? Idk how to solve it at all

[calm coral]

@verbal lintel Has your question been resolved?

[swift laurel]

this corresponds to the inverse normal function (which takes a percentage and gives a z-score)

[verbal lintel]

How do I do that?

Take the z score as 0.75? @swift laurel

[swift laurel]

what sort of calculator are you using?

[calm coral]

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[mental cosmos]

hey how does one go about solving this?

[leaden thunder]

Use the definitions of continuous and differentiable

[mental cosmos]

do i just set their differentials at x=1 equal to each other. And then both their values at x=1 equal to each other and solve the two equations?

thats the best i could come up with as i dont understand the definitions too well

.close

[calm coral]

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[ember zinc]

im really sorry, but i really dont understand this equation, ik i should probably be finding common factors but i geniunely dk where to start w the top line, any help would be really appreciated, sorry

[dull wagon]

factorisation by groupingg

[ember zinc]

how would i go about grouping the top line?

[minor gust]

With those

So see there is 4 terms

You can factor the first two terms

And then the second two

[ember zinc]

ohhh

so like

x(x+3y) -5(x - 15) or is it plus 15 bc minus minus or

[minor gust]

Other one should be (x+3y)

[ember zinc]

i forgot to divide 15 😭😭

ty

[minor gust]

Then

On the bottom

Take out the GCF

[ember zinc]

2x right?

[minor gust]

Ye

[ember zinc]

2x(x-5)?

[minor gust]

Ye

And now what’s ur next step u think

[ember zinc]

cant remember what u call it but i think u take out one of the (x+3y)

[minor gust]

You have a (x-5) on the top and bottom

[ember zinc]

ohh

[minor gust]

Btw when you do the thing where you spilt it in half

[ember zinc]

so i take them both out or just the (x-5)?

[ember zinc]

huh?

[minor gust]

You get x(x+3y) -5(x+3y)

Right

[ember zinc]

yeah

[minor gust]

That it (x-5)(x+3y)

Just btw

[ember zinc]

ohhh

right

bc there were two (x+3y)s right?

[minor gust]

no

You spilt the thing in half

If it’s factorabke

You will get the same thing inside of the parenthesis

It’s still just one entity

But now you have (x-5)(x+3y) as yours factors

Idk how to explain it

[ember zinc]

ur explaining it way better than my teacher either way lol

[minor gust]

There’s other ways to do it but I’ve done it this way my entire life and it has worked

It’s super easy

So now you have (x-5)(x+3y) over 2x(x-5)

Do you know what to do next

[ember zinc]

do i just cancel out the (x-5)

[minor gust]

Ye

[ember zinc]

alr so i should have (x+3y) ober 2x?

[minor gust]

Ye

[ember zinc]

over

[minor gust]

And that’s ur answer

[ember zinc]

tysm

geniunely ur an absolute lifesaver

[minor gust]

Anytime

[calm coral]

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[normal wasp]

ok so in a club there are 320 seats and there are rows after they added 4 seats to each row and added 1 more row there was 420 seats in the club i need to find how many rows are in the club now

[calm coral]

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[remote mural]

I wanted to ask, how would you find the area of a triangle?

[remote mural]

I'm trying to calculate for "r", I have theta but need area

say my point was at (3,4)

meaning the base is 3

the height is 4

making the hypotenuse 5

[dull wagon]

if you have base and height,
simple area formula

[remote mural]

area = b*h?

and this is true for triangles?

[dull wagon]

thats for a rectangle

[remote mural]

oh

would it be

1/2b*h

then

[dull wagon]

yeh

[remote mural]

oh okay, thanks so much

.close

[calm coral]

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[remote mural]

.reopen

[calm coral]

✅

[remote mural]

sorry to bother, but I'm getting an incorrect answer.. I'm taking my base/height*1/2 and dividing it by my RA and getting it incorrect

the actual question's point is at (6,8)

I found theta to be 0.93 rad

which is correct ^

area is 6*8 x 1/2 = 24

[dull wagon]

what's the original question asking for?

[remote mural]

24/0.93 = 25.8

apparently it's 10 tho

[remote mural]

everything I stated

[dull wagon]

the r value would just be the hypotenuse

[remote mural]

wait raelly?

[dull wagon]

you don't need anything to do with area here

[remote mural]

really?*

it isn't radius?

[dull wagon]

radius of the circle centred around the origin that the point lies on

[remote mural]

oh you're right

because height is technically radius

hypotenuse*

2* hypotenuse is diameter

[dull wagon]

"height"/ relative altitude depends on the base you're using

[remote mural]

makes so much more sense lo

l

thanks again

.close

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[junior fern]

what would $\gcd(0,2)$ be?

[potent lotus]

Joshii

[junior fern]

would it be 2?

[nimble badge]

Yeah that looks right, 0/2=0 and 2/2=1 anything bigger isn’t an integer

[cloud topaz]

Yes

[calm coral]

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[spark adder]

Hello everyone I have a question about graph theory

[pure breach]

!da2a

[calm coral]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[spark adder]

so basically here is the question

Here is the way I tried to solve it, is A the answer?

[calm coral]

@spark adder Has your question been resolved?

[spark adder]

<@&286206848099549185>

[vestal ivy]

clockwise

yeah you're correct

[spark adder]

Thank you

.close

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[swift marsh]

https://gyazo.com/91a3d0855d784c2a12f15080ab876d52

[swift marsh]

need to create a function

that looks like this

[velvet osprey]

what are you going to use this function for, and why does it have to have this shape?

or are you just doing this for fun and/or shits & giggles?

[swift marsh]

Well basically

shits and giggles

Just any function that looks like this generically

but doesnt go below the x axis

anything with a nice dfeined curve

[velvet osprey]

hm

[swift marsh]

You see

[velvet osprey]

i was gonna suggest a cubic but then you said it can't go below the x-axis.

[swift marsh]

Im trying to do a reimann sum

[velvet osprey]

how's it supposed to look to the left of the portion you've sk-

!original

[calm coral]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[swift marsh]

THere is no problem

Im like

Making a question

one could say

[lilac void]

so you want a formula for something like this?

(src: calcworkshop)

[swift marsh]

and the question is gonna be like 'using a riemann sum with x number of rectangles

Yes

something like that

[velvet osprey]

so you are composing a question of your own

[swift marsh]

YES

[velvet osprey]

that will ask for an approximation of an integral with a riemann sum

ok, alright...

[swift marsh]

Yea

But i have no idea what function looks like that

[velvet osprey]

i mean the thing is

that diagram's not to be taken too literally

[swift marsh]

Yea i know

but it looks so clean

[velvet osprey]

the function can be anything, not just that sort of bendy thing

[lilac void]

you actually propose a really good question, what is a function that graphs that?

[swift marsh]

Yes i know but my creativity is so shit

[velvet osprey]

it looks like some kind of cubic

a cubic whose derivative has two real roots, to be more specific

something like this

[swift marsh]

that looks goodf

from x = 0 to x = 2

Thank yoou!!!

.close

[calm coral]

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[torn lily]

hey why is it equals to (5x-2)²?

[torn lily]

the task was to divide 25x²-20x+4 into its factors and one the right is the correct answer, I dont get where the (5x-2)² comes from

[dull wagon]

25 = 5^2

[modern condor]

what is 25 a square of ?

oops

[dull wagon]

$p^2q^2 = (pq)^2$

[potent lotus]

ℝαμΩℕωⅤ

[torn lily]

so i should always multiply the numbers inside the brackets to see if its squareable?

[torn lily]

I get that, but 2 isn't the square of 2/5 though

[calm coral]

@torn lily Has your question been resolved?

[uncut python]

hello

[uncut python]

what is the question

solve for x or show that

[torn lily]

here's the question

the task was to divide 25x²-20x+4 into its factors and one the right is the correct answer, I dont get where the (5x-2)² comes from

[uncut python]

do u need th value of x

[torn lily]

no, i have to divide the equation into ints factors

[uncut python]

ok

[torn lily]

i just dont get where the (5x-2)^2 comes from

see i got it until 25(x-2/5)^2

[uncut python]

ok wait

[torn lily]

OK

[uncut python]

5^2(x-2/5)^2

5 square is 25

so we write 5^2

[torn lily]

yea

[uncut python]

now when the powers are same you can multiply the base

for this both have ^2

now multiply 5 to x and 2/5

[torn lily]

(5x-10/5)^2, right

[uncut python]

yeah

[torn lily]

and thats (5x-2)^2

[uncut python]

yeah

[torn lily]

i get it now thank u sir

[uncut python]

ok

[torn lily]

.close

[calm coral]

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[remote mural]

i dont rly understand why they only took away lambda on the diagonals

arent u supposed to take it from every entry in the matrix?

[torn gorge]

no you arent

it has computed the characteristic polynomial

which is

where I is the diagonal matrix with ones in his entries

[remote mural]

oh wait

yea ur right lol

lol

im very big brain

🤯

[calm coral]

@remote mural Has your question been resolved?

[remote mural]

wait one more question @torn gorge

if A was a 3x3 matrix and it was a matrix such that it only had 2 distinct eigenvalues

why can we not express A as a product of PDP^-1 ?

where P is an invertible matrix and D is a diagonal matrix

[torn gorge]

we can fro example A=D=(1 0 0) (0 1 0) (0 0 2)

it has 2 eigenvalues (one and two)

[remote mural]

would P have two of the same columns then?

cz there are two equal eigenvalues for 2 columns of D?

[calm coral]

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[misty glen]

I saw an example of a visual proof that can be misleading

but I can't figure out what the problem is

[marsh valley]

I think it stems from the two triangles being changed by swapping them

[misty glen]

Ohh okay now I see it

thank you

[fringe reef]

the second shape is not a triangle :)

it's two triangles

[misty glen]

Yeah makes sense

.close

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[untold kernel]

Hi

[untold kernel]

I have this function:

[remote mural]

what do u wanna do with it

[untold kernel]

I get two asymptotes:

Now I don't know which response choose in my test:

b) is continuous and has a vertical asymptote

c) is continuous and has a horizontal asymptote

d) has an oblique asymptote and a vertical asymptote```

That's the graph:

I don't know which option to choose because none of them say: "Two asymptotes, one vertical and one horizontal"

@remote muralThat's my problem 😐

[remote mural]

i think it's d?

i dont work too much with oblique asymptotes

[remote mural]

how do u know it has a horizontal asymptote

oh yea it can pass through horizontal asymptotes

it's not b and c so i assume it's between a nad d?

[untold kernel]

I calculated, I can't paste the full process : ( but I know it's correct bk of the calculator.

My problem is that any of the options fit with my situation

[remote mural]

yea i see what u eman

wait it's d

i think it's d

[tiny eagle]

There is not an oblique asymptote

[untold kernel]

I think it's not bk if I have an horizontal asympt I can't have an oblique

[remote mural]

wait how do u define an oblique asymptote

lol

[untold kernel]

At moment I don't know 🙂 I only know that if i have an horizontal one then it can not have an oblique

[tiny eagle]

An oblique asymptote happens if the degree of the numerator is one above the degree of the denominator. There will be a diagonal asymptote in that case. For example,

[remote mural]

ohh

so in this case cz the degree of the denom is higher

there is no

[tiny eagle]

Oblique asymptotes are also called slant asymptotes sometimes

[remote mural]

what is an asymptotic discontinuity btw

[untold kernel]

oh

[remote mural]

cz i assume it cant be b and c

so it has to be a?

[tiny eagle]

That's the key, an asymptotic discontinuity is a vertical asymptote, lol.

The graph has an asymptote (a vertical one) and is discontinuous at that point, hence an asymptotic discontinuity

[untold kernel]

This is my offial notes (translated to english) that talks about it:

[tiny eagle]

Yeah

That's the same thing as a vertical asymptote, as far as I know

[untold kernel]

Yes

It is the option that makes the most sense.

[tiny eagle]

Yeah, because it is correct. Youve got an asymptotic discontinuity (a vertical asymptote) and a horizontal asymptote, so the answer should be (a) 😃

[untold kernel]

Correct. That's the answer (here the screenshot translated to EN)

Thanks! @tiny eagle@remote mural

.close

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[shut geyser]

post question

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

it is the question

[remote mural]

oh

then yea it's a

lol

[shut geyser]

left triangle is a

uh

idk the english term for it

[zinc pelican]

are you familiar with the law of sines?

[shut geyser]

there's no need

what's that

a

whatever

30* angle in a triangle with a 90* angle

sin 30* is 1/2

do you have an idea how to solve this

you find the side of the right triangle using the left triangle

do you remember the property of the 90-60-30 triangle

no

[drifting seal]

lol

[shut geyser]

that problem is quite easy

you just need to remember the property

[drifting seal]

aol knows it he’s asking if u know it

he does know it

[shut geyser]

side opposite of the 30 angle is half the hypotenuse

i'm sure that property is stated in the video

you can do it from her

here

(i hope you do)

what part of the problem you don't understand?

You need to know sides of the right triangle

to calculate x

[drifting seal]

sin 30 = y/8=1/2

solve for y

then use that to solve for x

let y be the height of the triangle

[zinc pelican]

sine

[drifting seal]

sine?

trig function?

ratio of opposite side/hypotenuse

[zinc pelican]

(sinus is the french name don't worry about it)

[shut geyser]

90-60-30 is introduced earlier than trigonometry

[zinc pelican]

so do you know about 90-60-30?

do you see a triangle in the picture that would fit the properties of a 90-60-30 triangle?

what's a

yeah that's right

now what do you notice

in the right triangle

correct

so since you have A, can you find x?

that's right

i'm not really sure but I can help you

[remote mural]

my god

tan 30° =6/AB

Find AB

[zinc pelican]

he doesn't know trig

[remote mural]

Area = 1/2 (6)(AB)

[zinc pelican]

yeah that's correct

[calm coral]

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[calm coral]

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[junior fern]

why is the highlighted bit important?

also a_0=b_0 is showed i just didnt include it in the screenshot

[calm coral]

@junior fern Has your question been resolved?

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@junior fern Has your question been resolved?

[calm coral]

@junior fern Has your question been resolved?

[junior fern]

.close

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[humble steppe]

yo @exotic falcon

[humble steppe]

If i wanted to prove it was convergent or divergent

would 1/((1/e^1/x) * x^2) be a correct comparison?

no i meant

[exotic falcon]

clearly it converges

you computed its value

that's all you'd need to do

[humble steppe]

but if the question would not have compute in it

[coral osprey]

Well I'd make a substitution and see if the resulting integral is convergent

It's convergent because if you make u = 1/x

[exotic falcon]

(we just did that)

[humble steppe]

1/((1/e^1/x) * x^2) < 1/x^2

1/x^2 converges so 1/((1/e^1/x) * x^2) would to?

[humble steppe]

yeah I know but if the question was to prove it using comparision theorem e.g.

[hushed tartan]

this isnt always true

[humble steppe]

for x>1

[hushed tartan]

yea this should work

mb i forgot about the boundaries of the integral

[humble steppe]

okay thank you!

.close

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[remote mural]

can someone help me?

[calm coral]

Please don't occupy multiple help channels.

@remote mural Has your question been resolved?

[leaden thunder]

.close

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[wheat pewter]

i got the numberline down and i think it is no solution. can someone help me

thats the numberline i have

[graceful dust]

numberline representing what?

[wheat pewter]

this

[graceful dust]

anything greater than 3 doesn’t satisfy the last inequality

so it cannot be a solution

[wheat pewter]

can you elaborate

i thought it was no solution because nothing intersects

[graceful dust]

oh well maybe but it’s confusing what your number line represents then

like e.g. you have all numbers greater than 3 included but those are not solutions

i would make one number line for the equalities before the ‘and’ and another for the inequality after the ‘and’ if you wanted to do that

it shouldn’t be no solution

e.g. 0 is a solution

[calm coral]

@wheat pewter Has your question been resolved?

[calm coral]

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[brittle cape]

Yo how I do this one