help-36

[tight void]

ok

did u get h as 3.7...

or something around it

[undone agate]

i don't really know hot to do

[tight void]

oh sry sry

[undone agate]

let me check the answer

it's ok

[tight void]

is it the ans?

If yes then I can guide u

[undone agate]

yup

[tight void]

OKay then

lemme show

[undone agate]

okok

[tight void]

Have a look at this

[undone agate]

okok

[tight void]

u got it?

[undone agate]

still looking

[tight void]

ok

Sry for bad hand writing tho -_-

[undone agate]

it's fine i know what u are writing

haha

[tight void]

Okay then -_-

[undone agate]

I GOT It

thanksssss

[tight void]

welcome

[undone agate]

HELPPPPPPPP

[viral oriole]

6^x = (2 * 3)^x = 2^x * 3^x

3^x cancels out

@undone agate

[undone agate]

oh hi

[viral oriole]

Is it clear now?

[undone agate]

yesss

thankssss

[viral oriole]

yw

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[odd sleet]

Hi

[odd sleet]

I got a question in automata theory

The question is; given a monotonic decidable partial function N -> N, is the set of all values of f decidable

whatever it means

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[mild cloud]

i understand this is cos(A+B) but how do you get A and B from 7pi/12

[odd sleet]

3/12 and 4/12 could work

[mild cloud]

but i need to use one that its in the unit circle

[odd sleet]

what's 3/12 and 4/12 simplified?

[mild cloud]

1/4 and 1/3

Hmmm

i see

what if its something like this

[odd sleet]

3 - 4 = -1

[mild cloud]

that makes sense

tyty

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[quick spire]

what should i do , here ?

[odd sleet]

!status

[final saddle]

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

[quick spire]

1

[coral prawn]

You should probably begin by writing out the n=2 case

You can let z1=a1+b1*i and z2=a2+b2*i for instance

On the left-hand side of the inequality you get (a1+a2)^2 + (b1+b2)^2 (first sum up the components and then take the sum of their squares)

which you can expand like you normally would

On the right-hand side you get (1/k) * (a1^2 + a2^2 + b1^2 + b2^2)

which is basically the same as what you have on the LHS but without the extra garbage that you get from expanding the binomials

Then you should see quite easily that the best you can do is k=1

The rest, as you can probably imagine, is a proof by induction

(most likely; I haven't actually fully solved it myself)

[final saddle]

@quick spire Has your question been resolved?

[quick spire]

i just exactly did that , but i just don't know how to go further than that

like how can i generalize this , for any n

[quick spire]

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[rustic wedge]

<@&268886789983436800>

[sturdy ocean]

.close

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[vale reef]

not sure what correct diagram would be for part c

[cedar ivy]

Just a chord in circle

[keen forge]

I need help with this 2x^2 – 36x + 156 = 1,2x -17

is this even correct

[vale reef]

nvm i got

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[unborn basalt]

6. no idea
7. A/B
8. not C (what does B even mean)
9. A
   ?

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[unborn basalt]

6. no idea
7. A/B
8. not C (what does B even mean)
9. A
   ?

[delicate halo]

If you solve for x²-2 how many zeros do you get

[unborn basalt]

Sry I don't get u

[delicate halo]

x²-2 = 0

How many solutions of x

Solve for x

[unborn basalt]

2

[delicate halo]

Exactly

[unborn basalt]

x = +/- sqrt2

[delicate halo]

Correct

[unborn basalt]

Wdym by 'zeros'

[delicate halo]

That’s what zeros mean

x values that make it 0

[unborn basalt]

So this is 2 zeros?

[delicate halo]

Yes

[unborn basalt]

Ah...

K imma google that ltr

For 7

Both A and B seems good to me

[delicate halo]

No look at the graph for A

Oh wait

[dire delta]

In that case A can't be the function you are looking for

Because it asks for an intercept at x = 0 specifically

[delicate halo]

It does have a intercept at x = 0

[dire delta]

Yes but it has one at x = 1 too

[delicate halo]

I think it’s B because A isn’t even

[unborn basalt]

What does it mean by 'even'

[delicate halo]

If you do f(-x) does the function sign change?

Plug in -x everywhere you see x

If it changes it’s odd if it doesn’t it’s even

[unborn basalt]

A changes so it's not even ahh

[delicate halo]

Yea

[unborn basalt]

Then for 6

[dire delta]

For 6 you need to look at what characteristics are given to you

[unborn basalt]

I know that the function is sth like
1 / (x-5)
If x =/ 5

[dire delta]

You know that f(5) = 0

And you know that for every x != 5 f(x) != 0

So theres a particular x that cannot be used in g(x) = 1/f(x)

You see why?

[unborn basalt]

Kinda feel like it's C

[dire delta]

The thing is: Why do you feel like its C

[unborn basalt]

Cuz x=/ 5

And it's a continuous function

[dire delta]

At the moment you are picturing how the equation for the function may look and what characteristics this function has but this is not needed here

The simple reason as to why C must be correct is because f(5) = 0

So 1/f(5) is 1/0

[unborn basalt]

Ye

[dire delta]

which is undefined and as such cant be part of the domain

And for everything else we know f(x) isnt 0 so it has a defined value

[unborn basalt]

Can I check if
8. B
9. A (cannot be C/D)
?

[dire delta]

Well for 8 you can rule out C as you have a polynomial and polynomials are always continuous

You can also rule out D as the function is obviously not decreasing on the given intervall

I guess this R set there is including every y value that the function can take?

[unborn basalt]

Ruling out A cuz Y >= vertex

[unborn basalt]

Ig so

[dire delta]

if so you can rule A out as well as obviously negative y values must be part of this set

For x = 0, y = -2

[unborn basalt]

Mmm

[dire delta]

So the only thing that can be true is B which can easily be checked with the midnight formula or the p-q-formula

idk how you call it in english tbh

[delicate halo]

Rational root theorem I think

[dire delta]

but its (b +- sqrt(b^2 - 4ac))/(2a) or -p/2 +- sqrt((p/2)^2 - q)

The left formula is easier to use as it is applicable for equations ax^2 + bx + c = 0, while the right equation can only be used on equations of the form x^2 + px + q = 0

Meaning that if you have an equation of the form ax^2 + px + q you must divide the entire term by a in order to use the p-q-formula

Looking at x^2 - 2 we got a = 1, b = 0, c = -2 you can just replace those and check for yourself

you'll get 2 values x_1 and x_2 for which f(x_1) = f(x_2) = 0 so B is true

[unborn basalt]

Last ques

I hv no idea what I'm looking at

[dire delta]

I have to think about that for a moment

So first you need to be aware of the fact that y = f(x) is included in y = 1/2 f(2x) - 1

This means y = 1/2 f(2x) - 1 is a modification of the original function.

Now we know that f(2) = 4, so our first objective is, to get f(2) into the equation of 1/2 f(2x) - 1, giving us the corresponding x coordinate

We do that by simply solving for x in the equation 2x = 2

[unborn basalt]

x=1

[dire delta]

indeed that is our corresponding x coordinate

Now for the y value you already know how that works

you know f(2) = 4, so you just solve 1/2 * 4 - 1

[unborn basalt]

1

B

[dire delta]

yup

[unborn basalt]

Ok so that's how u do that-

[dire delta]

So for reference if these types of questions are asked in the future

The corresponding x value is determined by solving the equation inside the function term. If f(5) = 0 in one function than f(2x - 4) must be solved so that 2x - 4 = 5, giving you the corresponding x value. The corresponding y value than arises by simply solving the equation

if that makes sense

[unborn basalt]

Yep got it

Thanks a lot u 2!

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[next pagoda]

I've tried a few u substitutions here

but I haven't figured it out yet

first I tried u = x^2 + 44

then I tried u = x^2

but I need to get du to be x^3 dx somehow

[manic herald]

no you don't

$x^3 = x^2 \cdot x$

[junior radish]

You can do that

[soft zealot]

tushar

[next pagoda]

?

are you subbing x^3 for u?

[final tangle]

no

[manic herald]

keep your first substitution

just solve for x^2

[final tangle]

did you consider the conversion from dx to du?

[next pagoda]

my first sub is x^2 +44

[manic herald]

yes

[steep marsh]

/

[next pagoda]

the derive of x^2 + 44 is 2x dx

[manic herald]

yes

you have an x^3

$x^3 = \frac{x^2}{2} \cdot 2x$

[soft zealot]

tushar

[manic herald]

can you express x^2 in terms of u?

[next pagoda]

oh wait

are you just rewriting the original expression? Specifically x^3 then?

[manic herald]

yes

[next pagoda]

hmm

then I'd have indef integral (x^2 + 44)^1/2 * x^2 / 2 *2x dx

hmm

[manic herald]

yes

and u = x^2 + 44 and du = 2x dx

[next pagoda]

right

[manic herald]

so perform the change of variables

all you really have left to do is express x^2 in terms of u

you have u = x^2 + 44

[west walrus]

[manic herald]

so what is x^2

[next pagoda]

x^2 = u- 44

[manic herald]

what

yes

[next pagoda]

indef integral u^1/2 * (u-44) / 2 du ?

[manic herald]

yes

should be able to proceed from here with power rule

[next pagoda]

yeah

alrighty

ty somuch

❤️

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[stoic crow]

Can someone help me with this?

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@stoic crow Has your question been resolved?

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[tardy vortex]

hi

[tardy vortex]

could someone explain

what a conjugate diameter of an ellipse is

ive tried googling it but i dont get it

i just need a simple explanation, nothing in depth or anything

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@tardy vortex Has your question been resolved?

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@tardy vortex Has your question been resolved?

[vital crag]

show where you read this and the entire context

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[upper elk]

The figure below consists of a rectangle, two circles and a triangle. The circles are tangent to three sides
the rectangle and has its center at S and P. The corners A and B of the triangle ABP lie at the point of intersection
between the two circles. The rectangle has length 15 and width 8.
a)
b)
pp
Explain why the distance between S and P is 7.
Find the area of ​​triangle ABP.

[upper elk]

how do i solve this

[dry light]

If the circles are only tangent to 3 sides, logically the radius is gonna be half the height

[weak dawn]

that means that the side legnth of the rectangle would have to be ||4 times the radius|| if both circles were tangent, giving that the the distance of the inside region where the circles intersect should be ||15-4r|| ||r is 4 and the distance in between the intersections is 1. giving that SP=7||

[upper elk]

before i sent this i figured out the radius and all that but i dont understand what to do next

[silent tinsel]

Maybe this helps

[upper elk]

wait why is hypotenus 4

[silent tinsel]

The radius of the circle is 4

[weak dawn]

the radius is 4

yeah

[upper elk]

ah

i thought that 4 and 3.5 switched place

now 16 minus 12.25?

3.75 is the shortest catheus

[silent tinsel]

Square root of that

[upper elk]

yes

so the area is 6.7

[silent tinsel]

6.8 if you round it?

[upper elk]

yes

[upper elk]

but why is that line the length of the radius, im just making sure i understand everything

pA

[upper elk]

.

[silent tinsel]

Height is 8 which is twice the radius

[upper elk]

ah i just didnt see it

ty

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[teal scroll]

how do i show that if the set of zero divisors is an ideal, then its a prime ideal?

[wise berry]

To prove that Z(R) is a prime ideal, you need to show that for any two elements a, b in R, if their product ab lies in Z(R), then at least one of them must lie in Z(R). That is, you need to show that if ab has a non-zero divisor c, then either a or b has a non-zero divisor.

[final saddle]

@teal scroll Has your question been resolved?

[teal scroll]

i get that part, but im not sure how to show it

[wise berry]

Oh, I see

let a, b be two elements of R such that ab has a non-zero divisor c. This means that c is not equal to zero, and there exist non-zero elements x, y in R such that cx = 0 and yb = 0. Then, we have:

(ab)(cx) = a(bc)x = 0
and
(ab)(yb) = (ay)(b^2) = 0.

Since ab is a zero divisor, at least one of (bc)x and (ay)(b^2) must be non-zero. You can continue this reasoning?

[teal scroll]

how did you get that cx = yb = 0

[wise berry]

Sorry, I think I made a wrong assumption

Let me rewrite it

Now, I dont really know how to explain it. If someone else wants to help, please feel free to do so. I will give it some thought.

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[tranquil pine]

Why the hell is he rearranging this

[muted hull]

I'm confused what the question is?

[tranquil pine]

is it not meant to be that

[tranquil pine]

wdym the question is up there

[muted hull]

Is that the end of his writing?

If so, yes, he's wrong

[tranquil pine]

huh

he is manipulating the formula and idk why

you first do the 2 brackets and expand it

idek where he got +5x from

[muted hull]

Multiplication is commutative so it doesn't matter which order you use to multiply

5x comes from +9x - 4x

[tranquil pine]

oh

wait

[muted hull]

Yeah :)

[tranquil pine]

nvm ty

[muted hull]

Np

[tranquil pine]

.close

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[upper elk]

Calculate.
A brick foundation wall is 12 m long, 5 m wide and
0.5 m thick. See sketch.
What will be the area of ​​the area that the stones in
foundation cover?

[upper elk]

how do i solve tghis

[muted hull]

Personally I'd fine the area that all of it covers then subtract the white rectangle in the center

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[stray stump]

Not sure why they set r = 0?

and how they got 4cos3 theta = 0

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@stray stump Has your question been resolved?

[granite tapir]

it looks like daisy petals

[vital crag]

That's the start of the curve

[granite tapir]

r=0 represents the points that are zero distance from the origin

it is where the curve intersects or touches the origin

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[gilded marten]

can someone help me with this

[dry light]

You didn't even answer the question

[gilded marten]

i dont know how...

i wouldnt be here if i knew how to answer the question

[dreamy canopy]

I would guess differentiate, substitute in values and try solve.
You know f' is a maximum at 1/3, so f'(1/3) = 0

[gilded marten]

f'(x) = (abx + a)e^(bx)

[dreamy canopy]

And hopefully you get some useful result from f(1/3) which you can use to cancel stuff out

[gilded marten]

is the critical point 1/3

[dreamy canopy]

yep

[gilded marten]

so f(1/3) = (ab(1) + a)e^(b(1))

[dreamy canopy]

not sure where the /3 went

[gilded marten]

f(1/3) = 1

do i plug in 1/3 and then make that equation equal to 1?

[dreamy canopy]

thats what i would do, although as a disclaimer im a bit rusty

[gilded marten]

im supposed to set f'(x) = 0 for critical points and solve for x

idk how this guy is doing it

i just looked at their answers and guessed and got it right

i just put 3e for a and -3 for b

[dreamy canopy]

well if you want i can show you how to get b

[gilded marten]

i dont know how to get either

[dreamy canopy]

give me a mine to write it out

f(x) = axe^bx
f^(x) = (abx + a)e^bx = abxe^bx + ae^bx
f(1/3) = 1 (Given in question) ---> a(1/3)e^b(1/3) = 1 *
f'(1/3) = 0 (As its a maxima) ---> ab(1/3)e^b(1/3) + ae^b(1/3) = 0 **

[gilded marten]

i see

[dreamy canopy]

ok wow im starting to realise how messy this looks

it looks much nicer by hand

now you can rearrange * and substitute into ** get rid of all the exponentials to find b

[gilded marten]

did you get the derivative in your head

i had to use a calculator

[dreamy canopy]

i just trusted you on the derivative haha, but product rule would be the method

a(1/3)e^b(1/3) = 1 ---> ae^(1/3)b = 3

slap that into ** and it should work out quite nicely

[final saddle]

@gilded marten Has your question been resolved?

[gilded marten]

thanks

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[tranquil pine]

Hey

[tranquil pine]

Write as a single trig function

3cos pi/5 sin pi/5

[final saddle]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[tranquil pine]

Huh

[gaunt niche]

trig identity

sin(2x)=2sin(x)cos(x)

you can manipulate that by setting x as pi/5 and multiplying both sides by 3/2 to make it 3cos(pi/5)sin(pi/5)

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[tranquil pine]

answer is second ss

[tranquil pine]

we are integrating with respect to x, how did they get that on the right

[swift flicker]

I guess typing error

[tranquil pine]

ok

it is an easy integral if it were dr

[swift flicker]

Yes

[tranquil pine]

.close

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[tranquil pine]

.reopen

[final saddle]

✅

[tranquil pine]

hey

where did the ^1/16 come from

ok i figured it out

it was a log rules thing

.close

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[gleaming geode]

Why does the cross-term here vanish? and shouldn't y^2 be -y^2

[gleaming geode]

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[onyx peak]

When exactly am I allowed to make substitutions with infinite expressions? $x=1+\frac{1}{1+\frac{1}{1 + \frac{1}{1} \dots}}$ after substitution I get $x=1+\frac{1}{x}$. The second equation produces 2 solutions while the infinite fraction produces just 1. What do I do to determine what value does the infinite fraction have when I get 2 solutions? In this case it's quite easy, since one of the solutions is negative. But what if it wasn't?

[soft zealot]

MathIsAlwaysRight

[onyx peak]

nvm, got it

.close

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[dim laurel]

So um like do you know addition

2x + x = 3x

[formal quest]

umm

[tranquil pine]

Bro how can you even post that

[formal quest]

do you know what 2+2 is?

anyways

its 13x

just add it just like any other number

for example

2+2=4

so

2x+2x=4

i hope your problem is fixed now

[final saddle]

@rough pebble Has your question been resolved?

[tranquil pine]

Factor out the variables, you'll be able to solve every question of this kind

[tranquil pine]

(3-9+7)ab

1×ab = ab

[rough pebble]

so then 16-3?

[tranquil pine]

It's 7-9 not 7+9

[rough pebble]

then why doesn't it say that?

[tranquil pine]

It does

What's the difference between 7-9 and -9+7?

[rough pebble]

oh

so what do i do with the 3, add it?

It doesn't tell me to add it though @tranquil pine

I get it now

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[tranquil pine]

Wow

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[rough pebble]

this is so easy I did this in like 4th grade lowkey embarrassed i sent thiss.. thanks tho @tranquil pine

[rough pebble]

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[eternal wharf]

What's is this symbol between b and c

[final saddle]

@eternal wharf Has your question been resolved?

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@eternal wharf Has your question been resolved?

[final saddle]

@eternal wharf Has your question been resolved?

[granite tapir]

http://detexify.kirelabs.org/classify.html

$\oplus$

[soft zealot]

attilavjda

[granite tapir]

@eternal wharf it is symmetric difference, it means the elements that are either one of the sets, but not in both. It seems to be similar to "XOR"

[eternal wharf]

How to solve this divide?

[sturdy cypress]

yeah what's division mean why not use subtraction if it's that what

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[brazen sedge]

3x3 grid, 9 slots total
Calculate how many different possible positions without repetition (ie flipped along x-y axis and same pattern but moved on the x-y axis)

Pretty sure it would be 9^3 to start off with. Then divide by 7. 3(for each rotation) + 2(movement on the x-axis) + 2(movement on the y-axis) = 7 (the value that we chose to divide by)

The only issue is that it doesn't return a whole number, and it fails to factor in ones that CANT move along the x-y axis

had to move in here because some kid sent a message in a channel while I was typing in it and stole it 🫡

[bleak granite]

can you elaborate on what you mean by:

"Calculate how many different possible positions without repetition"?

[brazen sedge]

Same pattern but moved along the X-Y axis, or Flipped on the X/Y axis

As well as rotated (Z axis?)*

[bleak granite]

Are you filling in the squares with numbers? are you moving the sqaures, I need some context 😅

[brazen sedge]

Just filling in the squares, 3 different positions

Effectively trying to figure out an equation, which would work if we used 4/5/6 different combinations etc

[bleak granite]

Something like this where it's reflective along diagonals/cardinals?
0 0 x
0 x 0
x 0 0

[brazen sedge]

yes

[bleak granite]

essentially how many possible combinations there are and for any amount of x's

1 x, or 9 x's

[brazen sedge]

yes

[bleak granite]

are you counting a square with 0 x's?

because that would hold true to the reflective properties

0 0 0
0 0 0
0 0 0

[brazen sedge]

Not really, as we would consider that as just empty

As I have img perms I can prob give an example

[bleak granite]

ok so your definition then does not include an empty square?

[brazen sedge]

Yes

so this would be considered one pattern

[bleak granite]

wait, are you not considering reflectivity along the diagonals?

[brazen sedge]

this would not because its repetitive

[brazen sedge]

we're not considering empty squares. only whats filled in and the pattern that it makes

[bleak granite]

ohhh you're considering rotation

[brazen sedge]

rotation and having it flipped along the xy axis

as well as movement along the xy axis

[bleak granite]

so then:

0 0 0
0 x x
0 0 0

and

0 0 0
x x 0
0 0 0

Would be considered repeats?

[brazen sedge]

yes

same with
0x0
0x0
000

000
0x0
0x0

[bleak granite]

I see

ok

well then first I believe your initial computations were incorrect

[brazen sedge]

x00
x00
000 would also be considered a repeat as its the same thing but moved along the x-axis

[bleak granite]

A square with one x, only has 9 possible combinations, I hope that should be trivial

A square with two x's and 9 combinations for the first but then 8 for the second and so 9*8

A square with three x's has 9 for the first, 8 for the second, and 7 for the third so 9*8*7

Hence, we get 9+9*8+9*8*7+9*8*7*6+9*8*7*6*5+9*8*7*6*5*4+9*8*7*6*5*4*3+9*8*7*6*5*4*3*2+9*8*7*6*5*4*3*2*1=986409

986409 combos not 9^3

before any repeats are considered

[brazen sedge]

i think the guy i was working with originally had repeats in mind but we just couldnt get any further

but this works for the system we're trying to use

[bleak granite]

wait I just realized for a square—based on your definition—with only one x, everything is repeats

so there's only one possible choice

[brazen sedge]

yeah

[bleak granite]

would that be true for two x's?

wait no

nvm

just me being tired

[brazen sedge]

not for two

as whatsthetermforit uhh

diagonal isnt counted as a rotation

rotations in this case are full 90 deg turns

[bleak granite]

I did some thonking

and it could be wrong

but

just gimmie a sec to type this boio out

[brazen sedge]

just for a little bit of context behind this, we consider quantum mapping to be off 3x3 grid so anything thats not 90deg rotations would be considered quantum maps

[bleak granite]

for 2 x's there are five cases after repeats:

0 0 0
0 x 0
0 x 0

0 0 0
0 x 0
x 0 0

x 0 0
0 0 0
0 0 x

x 0 0
0 0 0
x 0 0

x 0 0
0 0 0
0 x 0

nope not factorial

just me being over excited

[bleak granite]

I think those are the only possible cases though for 2 x's

[brazen sedge]

rotations remove that bottom one

[bleak granite]

wait there's only two

yea just realized

only two cases

sorry just type-thinking

[brazen sedge]

the only reason why we dont count diagonals as a rotation is for cases where 3 corners are used, they cant be rotated anything but 90/180/240

[bleak granite]

For three x's there are nine cases after repeats are considered:

0 0 0
x x x
0 0 0

0 0 x
0 x 0
x 0 0

0 0 0
0 x 0
x 0 x

0 0 0
0 x 0
x x 0

0 0 x
x x 0
0 0 0

0 0 0
x 0 x
x 0 0

0 x 0
x 0 0
0 0 x

x 0 x
0 0 0
0 x 0

0 0 x
x 0 0
x 0 0

that look right?

[brazen sedge]

X's dont have to be connected

[bleak granite]

oh wait

ur right

[brazen sedge]

000
x0x
x00 would work

00x
x00
x00 etc

[bleak granite]

up to nine for three

[brazen sedge]

this is weird because its not even school hw its literally just for a rhythm game i play

[bleak granite]

lol

yk honestly

i dont wanna get too excited again but

1 case for 1 x

4 cases for 2xs

9 cases for 3xs

i mean there's gotta be lol

[brazen sedge]

thats why i was thinking its an exponential but

idk

[bleak granite]

i found a 5th case for 2xs

def not quadratic

[brazen sedge]

i just wanna figure out an equation to where we can set X to a number of values we want, which can be 1-9, and if we make pattern repeats legal (ex: two x's in one square in the grid) we can make that value go to 20+

[bleak granite]

yea logically an algorithm would be easier

but again its really late for me

actually thats a lie

im just really tired

[brazen sedge]

lol

[bleak granite]

Have you considered doing a 2x2 and trying to find any patterns?

0 0
x 0

0 x
x 0

0 x
0 x

i mean idk

[brazen sedge]

wouldnt that just be 16

if repeats were allowed

[bleak granite]

if repeats were allowed there are three combos

0 0
x 0

0 x
x 0

0 x
0 x

[brazen sedge]

thats only combos if repetition isnt allowed

also

[bleak granite]

oh my god

[brazen sedge]

0x
xx counts as one

[bleak granite]

i just realized

[brazen sedge]

lol

or would that be 13

[bleak granite]

ok i am really sleep deprived, im sorry couldn't be of much help brother but there are people who are much more awake and much better at math than me 💀🙏 so feel free to ping helpers, im sorry again m8

[brazen sedge]

as

xx
xx theres only 1 variation

[bleak granite]

yea only 1 possible combo for 4xs

try building a table?

[brazen sedge]

could but i just dont want to manually input that many patterns into a 3x3 grid

[brazen sedge]

do i just ping the entire role...?

[bleak granite]

,tex \begin{tabular}{|c|c|}
\hline
\textbf{Numbers of $x$'s} & \textbf{Number of Combos} \
\hline
1 & 4 & \
\hline
2 & 6 & \
\hline
3 & 4 & \
\hline
4 & 1
\end{tabular}

[soft zealot]

XxMrFancyu2xX
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[bleak granite]

meh close enough

[brazen sedge]

lmao

[bleak granite]

consider making a table

and just try to find any patterns

[brazen sedge]

i'll try eventually

[bleak granite]

looks like pascals triangle honestly

[brazen sedge]

never heard

[bleak granite]

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

[brazen sedge]

ah

[bleak granite]

yk that might be it lmao

cause for 0 x's there only one combo 👀

wait

it is exponential...

[brazen sedge]

I WAS RIGHT AHA

LFGGGG

[bleak granite]

$\sum_{k=0}^{n}\binom{n}{k}=2^n$

[soft zealot]

XxMrFancyu2xX

[bleak granite]

that's just a theory though...

and that's before repeats

so take it with a GRAIN of salt

[brazen sedge]

true

but imma enjoy that happiness rq as well so

[bleak granite]

lmao well at least you know it is before repeats 😂

well gn man

[brazen sedge]

GN

idk if i should leave this open if im going off to eat rq

as im gonna continue it

[bleak granite]

if you want to

but it might close on you

or a mod/helper may come by to close it insted

ok i really need sleep gn now fr

[brazen sedge]

alrighty

going to eat and coming back ~20 mins

[bleak granite]

Ok I fix table:

\begin{tabular}{||c|c||}
\hline
\textbf{Numbers of $x$'s} & \textbf{Number of Combos} \
\hline
1 & 4 \
\hline
2 & 6 \
\hline
3 & 4 \
\hline
4 & 1 \
\hline
\end{tabular}

[soft zealot]

XxMrFancyu2xX

[final saddle]

@brazen sedge Has your question been resolved?

[brazen sedge]

thats only w/ 2x2 grid + repetitive

(assuming ur asleep)

.close

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[next pagoda]

How could I rewrite e^7x?

[clear hearth]

Just use L'hopital

[next pagoda]

For this problem, I'm not supposed to

[clear hearth]

Ok

[next pagoda]

I can only use factoring methods

and trig identities and such

[clear hearth]

You can divide both by 7x

[next pagoda]

(this is for my study guide for calc 1

wait

can you explain that

are you using ln ?

[clear hearth]

No

Wait please

We will use this identity

Here x=7X

That's why we divide by 7x

Got it?

[next pagoda]

no

[clear hearth]

What's the matter?

[next pagoda]

I'm just not following

[clear hearth]

Ok

I'll just show you the answer

It'll be much better

And you will get an idea of what I mean

Do you understand what to do now?

[next pagoda]

why you are allowed to do that

what property is this/?

[clear hearth]

I just took -1 common

And multipled 1/7x on both numerator and denominator

So they cancel out

Any doubts?

Don't be shy

[next pagoda]

I'm not getting it yet

still thinking

what do you mean you took -1 common?

[clear hearth]

Which part specifically?

Ok wait

We have (x+y)

We take -1 common

-1(-x-y)

Got it?

[next pagoda]

yes

[clear hearth]

I did the same thing on 1-e^7x

[next pagoda]

why?

[clear hearth]

I took minus 1 common

To get the form e^x-1

To attain this form

[next pagoda]

ohhhh

So, you tried to make the problem above look like that?

[clear hearth]

Yes yes

Exactly

So that I can apply the theorem

[next pagoda]

is there a proof for that? Idk why l^x - 1 / x = 1 exactly (ignore this question for now)

okay gotcha

[clear hearth]

There are so many proofs online

You can search them in your free time

[next pagoda]

okay

[clear hearth]

So are we clear now?

[next pagoda]

on that yeah

[clear hearth]

Ok shall I continue?

[next pagoda]

Yes please

I'd like to know how you're dividing by 7x

[clear hearth]

Or will you do it yourself

Ok ok

Clear?

[next pagoda]

no 😦

[clear hearth]

What's the matter

[next pagoda]

hold on

okay yeah

I see what you did

still unclear where the 7x is coming from.

Like idk that property

But I see where you got the first term in the denom

and you use a lim property

[clear hearth]

Yes yes

[next pagoda]

:0

[next pagoda]

I remember this

[clear hearth]

Ok next step?

[next pagoda]

umm

the denom will be 3/7

[next pagoda]

because of this

[clear hearth]

?

[next pagoda]

what

[clear hearth]

Isn't that what I wrote?

[next pagoda]

yeah

but like, when you take the limit for the denom, it will be come just 3/7 no?

[clear hearth]

Yes

[next pagoda]

maybe that's not the next step you were looking for

[clear hearth]

That's the next step

[next pagoda]

oh ok

there's a problem in the num if we take the limit though

[clear hearth]

Clear?

[next pagoda]

yup

looks good so far

[clear hearth]

Woila

Confused about the numerator limit?

[next pagoda]

what is the upper limit?

the numerator?

if so, no

I saw the prop you mentioned above

both will be 1

[clear hearth]

Yeah

[next pagoda]

when you take the lim as x appr 0

wow that question was involved lol

I doubt my professor will include that on our final

thank you so much for powering through that with me

[clear hearth]

he will

[next pagoda]

given I'm so slow

you think so?

[clear hearth]

Yes

[next pagoda]

oh boi

[clear hearth]

It'll take 3 seconds for L'hopital

[next pagoda]

I've like never seen this

I'll have to look for it in my notes

[next pagoda]

oh totally

as long as I get a indeterm form

which you do right away

I can use it

but not for this problem

He said don't use it unless he asks for it

[clear hearth]

You should check out the common limits

Holdon

I'll send you

[next pagoda]

I've got a cheat sheet from pauls notes

bet

ty

[clear hearth]

You do? Please share

[next pagoda]

https://tutorial.math.lamar.edu/classes/calci/calci.aspx

this is my favorite website rn

he's even got practice problems with full solutions!

[clear hearth]

Oh boy

That looks complex

[next pagoda]

the cheat sheet?

It's got some of what I need

[clear hearth]

Maybe because I haven't learnt integral calculus yet

Anyways

Have a good day

[next pagoda]

oh really?

thank you so much

[clear hearth]

Yeah

[next pagoda]

You have a good one as well!

.close

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[final saddle]

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• Show your work, and if possible, explain where you are stuck.
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[oak kraken]

Why can a matrix not have only 2 solutions?

[oak kraken]

I am imagining two bell curves where they could intersect at only 2 unique points on different planes?

Not possible? Planes must be flat rectangles only?

Nvm this is not a thing ^

Each plane must go on forever

Unless we are talking about the edge of the universe but that’s prob not Linear Algebra

.close

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[solemn ferry]

i was solving some contest right now and found a question that idk how to solve
x^4-6x-5=0 has 2 real roots a,b. a>b
find (a-b)/(a^4-b^4)

[worldly vale]

Contest?

[solemn ferry]

yes

online

idk how to solve it so i ended it without answer to this

wait

x^4=6x+5

im stupid bruhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

the answer is 1/6

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[toxic inlet]

hi

[toxic inlet]

AC || BD

find CD

what should I do?

I know that AB is 12

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[remote sorrel]

Basically, i don't know how to solve any of these triangles and i need a bit of help on understanding how to solve them.

[solemn ferry]

in 1 sssolve for angle

use lae of cosines

law

[remote sorrel]

Can you explain if thats okay?

[solemn ferry]

What

a2=b2+c2-2b*c*cos A

Nvm

[remote sorrel]

Can you explain a bit further? Like how to solve for x using the law of cosines (with the given triangle(s) )?

[solemn ferry]

you dont need in them

[remote sorrel]

?

[remote sorrel]

Wdym

[solemn ferry]

11/5=(20-x)/x

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@remote sorrel Has your question been resolved?

[remote sorrel]

I see, although i do not fully understand it completely i think i have the idea.
So how about number two?

[solemn ferry]

look for similar triangles

15/x=x/5

x2=15*5

x=5sqrt3

[remote sorrel]

How about for y?

[remote sorrel]

Oh yeah i forgot to be clear about what i don't understand.

It seems that trying to find y and x in this triangle would require me to use the values already given, i don't know which to divide multiply etc.

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@remote sorrel Has your question been resolved?

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@remote sorrel Has your question been resolved?

[solemn ferry]

x = sqrt(5*15)

wait

look at the similartriangles and corresponding sides

[remote sorrel]

Yes, im looking.

what next?

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[sly hamlet]

For 14!/7! Is the answer just 14 x 13 x 12 x 11 x 10 x 9 x 8? Just want to check my answer

[hearty zephyr]

yes

[smoky heath]

Yep

Do you have any more questions?

.close

[sly hamlet]

Also, does anyone know the rule for converting a ^2 number to no exponent for verifying identities

[smoky heath]

I didn't quite understand you

[sly hamlet]

For verifying trig identities

I remember there was a rule to convert let’s say sin^4 to be without exponent

[tiny gorge]

you can convert sin^2 to a double angle form using

and then for sin^4 you can square both sides of that

which will give you a cos^2 on the RHS

to which you can apply the cos^2 identity

[sly hamlet]

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[primal crag]

hello, im a bit stuck on standard form and ordinary numbers, i cant change them into eachother if that makes sense, for example a few questions are:

• write 1.63 x 10 to the power of -3 as an ordinary number

• write 438000 in standard form

please explain how to work them out, thank you!!

[smoky heath]

Ok

The standard form is 1 digit before the decimal point and several after multiplied by 10 to some extent.

[primal crag]

so there will always be 1 digit first?? then a decimal

[smoky heath]

No

[primal crag]

like for example u cant have 26.700000? u can only have 1 digit first before a decimal??

[smoky heath]

The number 1834 in standard form is 1.834 times 10^3

[smoky heath]

Yea

[primal crag]

ohh okay

[smoky heath]

Do the same with your number.

[grim badger]

The idea for standard form is you want the number before the decimal to be between 1 and 10

[primal crag]

ohhh i see

[primal crag]

so then after how do i work out the numbers after the decimal

[primal crag]

u can use this 1st one an an example

[grim badger]

Here
https://www.youtube.com/watch?v=cW8e4KknHS8&ab_channel=Cognito

[smoky heath]

Use my example

[primal crag]

4.38 X 10 to the power of 3 for write 438000 in standard form ??

[grim badger]

Watch this so you understand how to get the exponent part of the number

[primal crag]

ok ty!!

[smoky heath]

Nice video

[smoky heath]

No

The number 12345 in standard form is 1.2345 times 10^4

[smoky heath]

Try again

[primal crag]

4.38 X 10 to the power of 5

[smoky heath]

👍

[grim badger]

You're not explaining how to get the exponent part, you're just displaying numbers hence the reason why I sent a video so the OP can see how to get the exponent

[primal crag]

also the videos really helpful thank you i think i understand now at least

go

*give me a random question

and ill try convert it to standard form

[grim badger]

Also this is a good video https://www.youtube.com/watch?v=ZtB0vJMGve4&ab_channel=TheOrganicChemistryTutor

[smoky heath]

This should be explained in voice chat differently for a very long time, so I use examples

[smoky heath]

I Agree

.сlose

[primal crag]

.close

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[latent fog]

Calculate the flow of the field $\vec{F}=(y, x, z-2)$ through the surface $z=1-x^2-y^2$ with $0 \leq z \leq 1 / 2$ oriented with normal $\hat{n}$ such that $\hat{k} \cdot \hat{n}>0$.

[soft zealot]

Lyuka

[latent fog]

I don't understand why the correct value of the normal vector is $$
\hat{n}=\frac{(2 x, 2 y, 1)}{\sqrt{1+4 x^2+4y^2}}
$$

[soft zealot]

Lyuka

[latent fog]

Shouldnt i be using this formula to find it?
$$
\hat{\mathbf{n}}=\frac{(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},-1)}{\sqrt{1+|\nabla f|^2}}
$$

[soft zealot]

Lyuka

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[latent fog]

.reopen

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✅

[latent fog]

<@&286206848099549185>

[tranquil pine]

maybe ask in #multivariable-calculus if you aren't getting a response here

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[old silo]

If a velocity as expressed in polar is R,θ: (a + bt, k + st), what’s the position equation (in polar?)
t is time, the others are constants.

[lapis compass]

I think just integrating both coordinates will work?

someone correct me if I'm wrong

eh, wait maybe I'm wrong

[muted prairie]

Could convert to Cartesian, integrate, and convert back

[old silo]

I don’t think that can be correct. The radius and angle of position would have to influence each other, right?

[lapis compass]

yep, you're right
you can't simply add 2 polar coordinates lol

[muted prairie]

the problem is you don't add polar coordinates by adding their components yeah

[old silo]

I tried that, but the problem is the position values for x and y then give this mess.

And i’m not confident in my ability to convert that back to polar…or that the result would look respectably clean

[muted prairie]

ah yeah

Maybe after applying sin(a+b) and cos(a+b) identities and splitting up the addition in the numerators

But doesn't seem very nice

[old silo]

It does not 😛 I’m not brave enough to try

[muted prairie]

I feel like the answer might just be another spiral

no

[old silo]

It’s definitely a spiral

[muted prairie]

hmm

[old silo]

Though not an archimedean spiral unless a = 0, i think

[lapis compass]

I think you have to integrate (a+bt) cos(k+st) which looks like it will be pretty messy

[old silo]

Could you explain? Is there a formula for integrating in polar?

[old silo]

from memory, I think i already did integrate (a+bt)cos(k+st) and it resulted in the top line of this ^

[lapis compass]

nah, I just converted it to rectangular

[muted prairie]

I'm trying to think if there's some sort of more intuitive way to find average velocity

[old silo]

Yesterday I found a webpage that described differentiating in polar to get the velocity from position, but i couldn’t figure out how to do it in reverse

[muted prairie]

Even if we knew what's the final position for v=(t,t) at t=2π

it might be helpful

[lapis compass]

the answer is super messy if I did it correctly

[muted prairie]

I think we can solve that one by some kind of argument looking at symmetric points

[lapis compass]

I did this

[old silo]

Oh yes?

[muted prairie]

x-component should be 0

y-component is... -2π?

[old silo]

Thinking about it logically, integrating the velocity’s radius would essentially give the ‘total distance travelled’

The ‘arc length’ of the spiral, if you will

[lapis compass]

this is what I got (technically I didn't convert back to polar but you'd get the same curve)

[bleak granite]

that just looks like the Archimedean spiral?

[old silo]

That looks the same as the pic i shared above, yeh (with the constants added)

[old silo]

It’s very close to one. I suspect that it is one when a=0, at least

[bleak granite]

i see

[muted prairie]

Just solve integral of (t,t) maybe and transform from there?

[old silo]

But I still couldn’t figure out its actual equation x.x

[old silo]

I tried but couldn’t get it to match up- Feel free to give it a shot if you want tho

[lapis compass]

both are pretty similar

[muted prairie]

,w integral t*sin(t)

[soft zealot]

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[muted prairie]

,w integral t*cos(t)

[soft zealot]

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[old silo]

Yep. Couldn’t make them match tho

[muted prairie]

,w (sin(t)-tcos(t))^2 + (tsin(t)+cos(t))^2

[soft zealot]

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[muted prairie]

so it's t^2+1

your green spiral is archimedean, it just doesn't start at the origin

If you assume the spiral starts at the origin, you get a spiral centered at (-1,0)

[old silo]

Cool! That makes sense since the velocities are linear, it’s nice to have it proven

I still couldn’t get a position equation tho

[muted prairie]

Seems to be
Move 1 step further from the origin
followed by move 1 step left