help-42

[pale blade]

there's a very easy way to do it

using the fact that sqrt6 is irrational

hint: the square of a rational number is rational

[weak silo]

Okay, so I don’t need to go squa ring both sides and all that

[pale blade]

you sqaure something, but not this

[weak silo]

Hm, could you give me another hint

[pale blade]

square sqrt(3)+sqrt(2)

[calm coral]

@weak silo Has your question been resolved?

[calm coral]

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[calm coral]

@grizzled temple Has your question been resolved?

[grizzled temple]

<@&286206848099549185>

[cunning veldt]

notice that (-1, 0) and (1, 0) are both points of the unit circle
is it possible for x to be both -1 and 1 when y is 0 ? what happens with these two points ?

[grizzled temple]

ohhhh

so when y = 0

wait

doesnt the graph look like this?

so um why can't x be both -1 and 1 when y is 0?

[mint verge]

it can as you can see

[grizzled temple]

yeah exactly

so im a bit confused

😭

[mint verge]

about what

[marsh summit]

for what value of t is x = 1?

[mint verge]

no value

[marsh summit]

well there you go then, there's no value of t that gets you (1,0)

[mint verge]

i was talking about graph

[grizzled temple]

ohhh

[mint verge]

i didnt see this

lol

[grizzled temple]

wait but how do u

even know what to test?

[mint verge]

(t²+1)/(t²-1) = 1

[grizzled temple]

this is the question btw

[mint verge]

t²+1 = t²-1

1 = -1

[marsh summit]

i kind of just guessed

[grizzled temple]

ohh

so u just sort of

gave x the value and see if it works or not?

[mint verge]

yes

[marsh summit]

i mean (1,0) is a fairly special point, you'd expect it to be one of the extreme ones ((1,0) (-1,0) (0,1) (0,-1)) and not something random like (1/2,sqrt(3)/2)

[grizzled temple]

ohh

and then

u just

oh

[marsh summit]

so then yeah you just check all of those

[grizzled temple]

hmm

alright then so with this sort of quesiton

should i just sort of test those more important values

and see if it works?

[marsh summit]

i'm... not entirely sure what "this sort of question" would be...?

[grizzled temple]

oh well

i sort of have another one

thats a bit more complicated

but i havent tried it yet

looks like this tho

[marsh summit]

ah hmm

[cunning veldt]

(0, 1)

y can't be 1

[mint verge]

check the values of x²+y² = 1

[marsh summit]

oh yep you're right

alright so yep the same trick works again

[grizzled temple]

ohhh

so again just test each one

using values liek 1

[marsh summit]

or alternatively: let t approach infinity and see if that's the missing point

[grizzled temple]

and seeing if it fits

[marsh summit]

in this case, for very large t, x goes to 0 and y goes to 1

[grizzled temple]

ohhh

[marsh summit]

with the idea being that to get exactly (0,1) you would need t to be infinity but that's not actually a number so that's why (0,1) is missing

[grizzled temple]

wait thast kinda cool

OHHH

so apply limits

wait thats lwk so cool

okok thank u very much

[mint verge]

why did we use limits tho?

for find the undefined values

[marsh summit]

...i'm not sure i really have a precise explanation, it's just like

hm

if you take [0,1] and wrap it around a circle

in a way where the two endpoints get mapped to the same point

[mint verge]

yes

[marsh summit]

if you then look at (0,1), that point where the endpoints went to, is now the one missing point

[mint verge]

how

[marsh summit]

well nothing other than the endpoints was mapped there

assuming you didn't like, wrap it around multiple times

[mint verge]

i dont understand

wrap?

[marsh summit]

like this

the circle, and then [0,1] around the edge

with the two endpoints mapping to the same point on the circle

(the part where it's not quite on the circle is just to make it easier to see what's going on)

[mint verge]

what is [0,1]

[marsh summit]

more explicitly this is the function $(\cos(2\pi t), \sin(2\pi t))$

[potent lotus]

bee [it/its]

[marsh summit]

${x \in \mathbb{R} : 0 \leq x \leq 1}$

[mint verge]

okay but

[marsh summit]

it's the interval from 0 to 1, and including 0 and 1

[potent lotus]

bee [it/its]

[mint verge]

Is this something that can be wrapped?

[marsh summit]

yes, $f(t) = (\cos(2\pi t),\sin(2\pi t))$

[potent lotus]

bee [it/its]

[marsh summit]

calling that "wrapping" was just like, geometric intuition

[mint verge]

ok but i cant think of it lol

[mint verge]

yes this is a circle?

[marsh summit]

it's a function from [0,1] to the circle

[mint verge]

then

what will we do

with this

[marsh summit]

well if you look at the image of $f$ on $(0,1)$ (everything except the endpoints)

[potent lotus]

bee [it/its]

[marsh summit]

(...ok thinking about it that's an annoying notational clash with how i'm writing coordinates)

[mint verge]

what are endpoints

[marsh summit]

0 and 1

so we're looking at the set ${x \in \mathbb{R} : 0 < x < 1}$

[potent lotus]

bee [it/its]

[mint verge]

okay

you mean (1,0)?

[marsh summit]

no
i meant (0,1) as in the notation for open intervals, not as coordinates

i had never noticed before now that those two things are written the same way

[mint verge]

yes

lol

[marsh summit]

ok let's just ignore that

the image of $f$ on ${x \in \mathbb{R} : 0 < x < 1}$ is the circle, except missing one point

[potent lotus]

bee [it/its]

[marsh summit]

the one point it's missing is $(1,0)$, which is $f(0)$ and $f(1)$

[potent lotus]

bee [it/its]

[mint verge]

ok but why isnt it cos(t) and sin(t)

[marsh summit]

and we could find that out by taking the limit $\lim_{x \to 1} f(x) = (1,0)$

[potent lotus]

bee [it/its]

[mint verge]

why cos(2πt) and sin(2πt)

[marsh summit]

that just rescales it so that [0,1] does the entire circle

otherwise instead of a circle we would have ended up with just this

[mint verge]

Why does it look like 2/3 quarter circumference?

[marsh summit]

because it's exactly 1/(2pi) of the circumference, and that's roughly 1/6 which is 1/4 * 2/3

(because pi is around 3)

[mint verge]

yes i noticed it

but why 1/2pi

[marsh summit]

because the circumference of the unit circle is 2pi

[mint verge]

yes

[marsh summit]

or you can look at it as just, that's what the period of sin and cos are

$(\cos(2\pi),\sin(2\pi)) = (\cos(0),\sin(0))$

[potent lotus]

bee [it/its]

[marsh summit]

so it takes $2\pi$ for it to go all the way around and get back to where it started

[potent lotus]

bee [it/its]

[mint verge]

yes

but why are we examining the 1 radian part?

[marsh summit]

we aren't

[mint verge]

then let it go and it becomes a full circle

or whatever the function is

[marsh summit]

i mentioned it because you asked why we didn't use (cos(t),sin(t)) and the answer is that if we did that we would end up with just 1 radian

[marsh summit]

well we're looking at just [0,1]

so i rescaled it by 2pi to make [0,1] cover the entire circle instead of just one radian

[mint verge]

It should be at least half

circle

...

[marsh summit]

anyway this is the important point
we work out what the missing point is by it being the value at the end of the interval, and we work that out by a limit

then in the case of $\mathbb{R}$ it's... sort of the same thing if you squint a bit

[potent lotus]

bee [it/its]

[marsh summit]

the set of real numbers is the open interval from -infinity to infinity

so to find the missing point we want to evaluate $f(\infty)$

[potent lotus]

bee [it/its]

[marsh summit]

and to do that we take a limit $\lim_{x\to\infty} f(x)$

[potent lotus]

bee [it/its]

[grizzled temple]

wait guys sorry i just have 1 question i know that because x cannot be equal to 1, y cannot be equal to 0 but um why cant y be equal to 0?

[marsh summit]

obviously this... isn't really true, but it's close enough that it works out

[marsh summit]

y can be equal to 0

if t = 0

[grizzled temple]

oh

[marsh summit]

that gets you the point (-1,0)

[grizzled temple]

OHHH

OH MY GOD

OHHHH

😭😭

wait that makes so much sense now

wait u might be a genius

[mint verge]

@marsh summit i still dont understand btw

[grizzled temple]

or maybe im just dumb 💀

[marsh summit]

$\mathbb{R}$ is homeomorphic to an open interval, and $\mathbb{R} \cup {\infty, -\infty}$ is homeomorphic to a closed interval

[potent lotus]

bee [it/its]

[marsh summit]

and the main thing we're actually using here is the topology, the notion of a limit, so that is actually enough to make this correspondence make sense

[mint verge]

all i know is x = (2t+1)/(2t²+2t+1), y = (2t²+2t)/(2t²+2t+1)

[marsh summit]

yep, and that defines a function from $\mathbb{R}$ to the circle $$f(t) = (\frac{2t+1}{2t^2+2t+1},\frac{2t^2+2t}{2t^2+2t+1})$$

[potent lotus]

bee [it/its]

[mint verge]

okay then there is no problem

why is it undefined at some points

@marsh summit

because y ≠ 1

ahahahaha

looks funny

or i'm psycho

i dont understand..

[marsh summit]

i mean tbh it is kind of just, vague intuition,

but basically the issue is that we're missing the endpoints of $\mathbb{R}$

[potent lotus]

bee [it/its]

[mint verge]

what are endpoints

[marsh summit]

points at the end

[mint verge]

so (1/2,√3/2)

[marsh summit]

...no

[mint verge]

so (√2/2,√2/2)

[marsh summit]

...no

that's not the kind of "point" i meant

for instance the endpoints of $[0,1]$ are $0$ and $1$

[potent lotus]

bee [it/its]

[marsh summit]

those are the real numbers that are where the interval ends

[mint verge]

what's wrong with them

i love them

they are important

looks beautiful

and nothing more

[marsh summit]

well the issue is that we forgot to include them

the endpoints of $\mathbb{R}$ are $\infty$ and $-\infty$ which \textit{aren't real numbers}

[potent lotus]

bee [it/its]

[mint verge]

yes

[marsh summit]

so when you look at the values of $f$ just on $\mathbb{R}$, the value $f(\infty)$ won't be included in that

[potent lotus]

bee [it/its]

[mint verge]

yes

[marsh summit]

so that's what the missing point is

[mint verge]

for x/x there is one more for x = 0

[mint verge]

and how do we know it has an asymptote

horizontal

[marsh summit]

...we don't really

all of this logic is kind of just, why this might work

and why it works in the cases where it works

[mint verge]

and

couldnt it have been y = 1 at a certain point

and perhaps it had an asymptote at infinity

[mint verge]

after this

[marsh summit]

yeah that's just another part of the "this might work"

if you had like, $f(t) = \left(\cos(\frac1t),\sin(\frac1t)\right)$, on all positive real numbers

[potent lotus]

bee [it/its]

[marsh summit]

then that does get you every point on the circle

and $\lim_{t\to\infty} f(t) = (1,0)$ but that's also on the circle at for instance $t = \frac1{2\pi}$

[potent lotus]

bee [it/its]

[mint verge]

yes

[marsh summit]

the idea was just that the value at infty is a reasonably intelligent guess

[mint verge]

well yes

but i still dont understand maybe

[marsh summit]

the basic reason for the guess is just visual intuition on what happens if you map [0,1] to the circle, + viewing the entire real line as an open interval

[mint verge]

i still dont understand

ahaha

[marsh summit]

you have something like this, with f(0) = f(1) and every other point reached exactly once

if you remove the endpoints, 0 and 1, then you end up with this gap, there's no value between 0 and 1 with f(x) = (1,0)

to work out where that gap is, you take f(1)

and if you only know how f works on the open interval, you can get f(1) by taking a limit

[mint verge]

i dont know what to say

i only see a pan to crack eggs

and there is some kind of wire around it

[marsh summit]

that's a circle

and the interval [0,1], wrapped around the edge of the circle

i.e. a function from [0,1] to the circle

[mint verge]

okay

then radius is 1/2π

[marsh summit]

not necessarily

[mint verge]

why

[marsh summit]

we don't really know much about this function other than that it's continuous

so it might just have a scaling factor of 2pi in it

[mint verge]

then why did we wrap the [0,1] thing around

[marsh summit]

because that's the original setup of the problem

[mint verge]

how

[marsh summit]

we are given a function from [0,1] to the circle

well

we are given a function from {x : 0 < x < 1} to the circle

and told that it misses exactly one point, and asked to find what that point is

[mint verge]

one point?

[marsh summit]

yes, one point on the circle

[mint verge]

does it need to say this?

maybe two

maybe three

Cant we know this?

already

with the function we have

with using the function we have

[marsh summit]

https://cdn.discordapp.com/attachments/948770026914213909/1220248091215790221/image.png?ex=660e3f98&is=65fbca98&hm=7ea220eaab475cc65180806b1da1468878bf03a35629a63e06f93faa56080c8d&

What point is missing?

point, not points

there are a lot of functions from [0,1] to the unit circle that miss multiple points, and if you use this argument on them it will return complete nonsense, and we don't care about that

[mint verge]

ok

[marsh summit]

this argument only works on a very particular class of functions from {x : 0 < x < 1} to the unit circle

and it requires the unjustified guess that probably the function we were given is in that class

if it isn't, then it won't work, but that's fine because we can check the answer afterwards, this is just a heuristic method to find which point we should be interested in in the first place

[mint verge]

i dont understand still..

and probably i wont understand

[calm coral]

@grizzled temple Has your question been resolved?

[mint verge]

@grizzled temple you can close here

if there is no more question

[grizzled temple]

oh yes

.close

[calm coral]

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[narrow hamlet]

if these r solutions

[narrow hamlet]

how did they get 117.4 - 45i
like... 1.96^3 = ~7
not 117
where -3 - 4i are the solutions btw

Also #2

shouldnt this be pi - 53/ 180 -53
since theta from -pi to pi

😄

[calm coral]

@narrow hamlet Has your question been resolved?

[calm coral]

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[solar marsh]

Hey, I don't understand this... :)
I understand how you could come up with this, but none of these formulas fit...

In the diagram, we see four squares, with the entire configuration resting on a horizontal straight line. The smaller squares have side lengths a, b and c. The vertices A and C of two small squares
coincide with diagonally opposite vertices of the large square. The corner point B of the third small square lies on one side
of the large square. Which of the following expressions gives the side length of the large square?

[cedar carbon]

@exotic falcon my sir, I am confused and in need of help. How to approach these fun wacky formulae?

@graceful dust my sir responded to slowly. What should I do?

[solar marsh]

[solar marsh]

I did

[graceful dust]

your sir has responded i will stay on reserve

[exotic falcon]

I see

Well

It is a square so it is fine if we just find one of its sides

[solar marsh]

[exotic falcon]

Do you agree

[solar marsh]

Yea

[exotic falcon]

Let’s try finding the leftmost side

[cedar carbon]

You are right. Pardon me for taking some of your time Madm.

[graceful dust]

it is no problem. you could not have known when your sir would arrive

[exotic falcon]

Can you both get the fuck out of cosmos doubt channel?

We’re trying to solve this

[solar marsh]

Hahahah

[cedar carbon]

Pardon me my sir. Please, continue.

[exotic falcon]

So yes proceeed by instead trying to find the length of the leftmost side of that square

[solar marsh]

Need this as my nickname here

[solar marsh]

Ok ok

I know the formula a^2=c*p

So p as the corner point up to the height on the side of the square at the hypotenuse

Does that help me?

Hello?

?

<@&286206848099549185>

[jagged latch]

what is teh question

[solar marsh]

Hey, I don't understand this... :)
I understand how you could come up with this, but none of these formulas fit...

In the diagram, we see four squares, with the entire configuration resting on a horizontal straight line. The smaller squares have side lengths a, b and c. The vertices A and C of two small squares
coincide with diagonally opposite vertices of the large square. The corner point B of the third small square lies on one side
of the large square. Which of the following expressions gives the side length of the large square?

[solar marsh]

;)

[calm coral]

@solar marsh Has your question been resolved?

[calm coral]

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[steel tide]

It should be C?

[inner delta]

Yup

[calm coral]

@steel tide Has your question been resolved?

[calm coral]

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[calm coral]

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[brave snow]

how do i solve

[calm coral]

@brave snow Has your question been resolved?

[calm coral]

@brave snow Has your question been resolved?

[dusky flax]

11th one?

do you know the coordinates of the vertex of a parabola?

[calm coral]

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[strange ivy]

hello, i need help in creating a probability problem from the kerplunk game

[strange ivy]

i cant think of any simple probability problem involving this game 😭

[dry isle]

how many sticks?

[strange ivy]

40 sticks

[dry isle]

how many colors?

bro you gotta specify stuff lmao

[strange ivy]

oh sorry sorry
40 sticks (4 groups so 4 different colors then each group/color has 10sticks of the same color)

the kerplunk we made had different instructions though T-T

[dry isle]

do you want some complec probabilities tho?

[strange ivy]

complec?

[dry isle]

If the player continues to remove sticks until all blue sticks have been removed, what is the probability that the last stick removed is red?

[dry isle]

complex lol

[strange ivy]

looking for something simple

[dry isle]

What is the probability that the stick selected is blue?

[strange ivy]

but can i do something like "a player removes a stick then drops 2 balls, what is the probability that the two balls are both red?"

[dry isle]

yep

If red balls are randomly placed on top of the sticks, what is the probability that both balls dropped after removing a stick are red?

[strange ivy]

ahh ok ty ty for clearing some confusions

.close

[calm coral]

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[austere moth]

For the vectors u,v w in R^[17} it is known that |u| = 2, |v| = 1, |w| = 3, v dot w = 2 and that u is perpendicular to v-w.
Determine the length of |p| for the vector p = u + v - w

[austere moth]

I understand that a length of a vectoirs is the sqrt of the dot product of itself

But since the vector is R^17 that'll be impossible to find out

and most likely the wrong way to go about solving it

[half jolt]

(u + v - w) * (u + v - w)

[half jolt]

That's right

So expand that, but I want you to keep the v - w together.

(u + (v - w)) * (u + (v - w))

You know that u * (v - w) = 0 and you will be able to use that

[undone lake]

Help me pls

[austere moth]

!occupied

[calm coral]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[austere moth]

my initial though is to do something like this \
$|p|^2 = |u+v-w|^2$

[potent lotus]

Merineth

[austere moth]

Since

$|p|^2$ is the dot product of P dot P

[potent lotus]

Merineth

[half jolt]

.

So |p|^2 is (u + v - w) * (u + v - w)

[austere moth]

uh

I'm not sure that is right?

the length of p can't be the vectors (u + v - w)^2

He does it like this

but i have no idea wtf he is doing?

[calm coral]

@austere moth Has your question been resolved?

[austere moth]

<@&286206848099549185>

[calm coral]

@austere moth Has your question been resolved?

[calm coral]

@austere moth Has your question been resolved?

[calm coral]

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[prime radish]

if we take the 90 degree angle, what would be the opposite, adjacent and hypotenuse for that angle?

[quaint sphinx]

the "opposite" would be the hypotenuse
the "adjacent" would be nothing
the "hypotenuse" would be the hypotenuse

[prime radish]

Could you explain with AC, CB and BA please

[quaint sphinx]

do you know what a hypotenuse is?

[prime radish]

Yes, longest in the triangle

[quaint sphinx]

no

hypotenuse is a term exclusive to right triangles

[prime radish]

So longest in a right triangle

[quaint sphinx]

yes

then u can just look at ur diagram to see which side AB, BC, or CA it is

[prime radish]

But if the hypotenuse is the opposite of 90 degree, then what is the hypotenuse in term of trigonometry

[quaint sphinx]

it's still the hypotenuse

u have heard sin() is opposite over hypotenuse right

[prime radish]

Yes

[quaint sphinx]

this is slightly scuffed, but sin(90) works the same way

because sin(90) = opposite over hypotenuse = hypotenuse over hypotenuse = 1

[prime radish]

Oh I see now, thanks

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[dreamy anvil]

If you have a matrix of the form

[dreamy anvil]

[1 n n-1 .... 2
2 1 n ..... 3
3 2 1 ........
... .. .. ...... n
n n-1 n-2 1 ]

Is there a way to quickly compute the inverse function?

hmm

Just like each column is [1,2,3...n], except shifted one down. So second column is [n,1,2..n-1]. An example is

[1 3 2]
[2 1 3]
[3 2 1]

And lets say we want to solve Ax = b, where A is of the form above.

Can we do this quickly? Without just inverting the full matrix?

<@&286206848099549185>

Is there a closed form of the solution in terms of b?

[valid mirage]

You want to find A^-1 ?

[dreamy anvil]

Hmm, its from a programming problem.

I need to figure out if Ax = b has an integral solution in x, and if it has, find the solution.

But the matrix can be really big, so I can't do gauss jordan elimination or anything like that, it takes too much time.

I need a way to read off the solution just from b.

I'm hoping this is possible since A has a very special form.

b is also integral.

[calm coral]

@dreamy anvil Has your question been resolved?

[dreamy anvil]

Hmm.

If x = (a,b,c..). then like the first row will be a + nb + (n-1)c ..

so if that is like b_1

nvm

[dreamy anvil]

;close

/close

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[subtle comet]

Question: Firefighters have a triangular prism water container and a cylindrical container. Both have the same height of 10 feet. If the triangular base has sides of 6 feet, 8 feet, and 10 feet, and the cylinder’s volume is twice that of the prism, what is the radius of the cylinder?

[subtle comet]

explanation Explanation:

Step 1:

Calculate the area of the triangular base. Since it’s a right triangle (3-4-5 Pythagorean triple), Area = 12 × 6 × 8 = 24 sq. ft..

Step 2:

Calculate the volume of the prism.
Volume = base area × height = 24 × 10 = 240 cu. Ft.

Step 3:

The cylinder’s volume is twice that of the prism.
Cylinder volume = 2 × 240 = 480 cu. ft.

Step 4:

Using the formula for the volume of a cylinder, V=πr2h,
480 = π r2 × 10
=> r2 = 480/3.14
=> r2 = 15.286
=> r = 15.286
=> r = 3.90 feet

is one of the triangular base side number a typo

the 10 specifically

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[kind wagon]

how would one even think about solving this

just a curiosity, not trying to solve it

where would i start

maybe expand sin(a+b)

and write tan^2 f(x) as sin/cos

sigh

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[calm coral]

Please don't occupy multiple help channels.

[vernal sandal]

i need help on this

im not too sure what it demands

upon sketching the graph i thought there was a line of symmetry across p = 0.5 but that was wrong

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[quartz stump]

Hi. Can someone help me with question 1 please?

[calm coral]

@quartz stump Has your question been resolved?

[torpid vine]

When they say f(x) <= 0, they just mean its below the x axis, from -2 to 1

They use brackets to show it reaches -2, and 1

For the second one it is saying it is below the x-axis from 3 all the way to infinity, which means for the rest of the graph

So the points they mentioned are where the graph intersects with the x axis, so they are the x-intercepts

So start from the left, and check

-4 is not in the intervals that are below the x-axis so it must be above

Same for -3

At -2 it is on the x-axis

So the graph will be decreasing until -2

Because after -2 it is below the x-axis but still need to go back up

So it will be like a little curve below the x-axis at that point in which it hits the x-axis at -1

Down there is a gap between -1 which is the end of the first interval they told us, and the beginning of the second interval which is 3

So in those points it must be above the x-axis once again

So it will make a curve above the x-axis once again at those points

Since the seond interval says it goes to infinity

The rest of the graph must be below the x-axis from now on

Also, since there are brackets, and <= signs you might want to put filled in circles at the points that they mention

Except for infinity

Because that is a paranthesis

Which mean f(x) does not ncessarily reach it

That is it, I hope it helps

[calm coral]

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[fair dirge]

Hi can someone help me with this? Its grade 12 polynomial equations and inequalities

[tight trellis]

Can you recall the factor theorem?

If we assume (x+1) is a factor of P(x), what does this mean?

[fair dirge]

That the equation remainder is 0

[tight trellis]

Right, what would you sub in to P(x) to give 0?

Or to reword it, if (x+1) is a factor, what would be the value of the root of the function corresponding to this factor?

[fair dirge]

-1?

[tight trellis]

Yeah exactly

So, if P(-1)=0, then (x+1) is a factor

[calm coral]

@fair dirge Has your question been resolved?

[fair dirge]

I just did the division part wnd its not a factor, both -1, and 1 arent factors

P(2) is a factor tho

Cuz remainder is 0

Thank you so much for your help!

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[fair dirge]

.reopen

[calm coral]

✅

[fair dirge]

Can someone help with this aswell

[empty quail]

oohhhh fun question uwu

first you have a x-intercepts

and it tells you that it opens down so that will be what?

also don't forget to find y-intercepts

@fair dirge

[fair dirge]

hi im here

ok solving rn

so my three points are at -2,1,3

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[brazen elbow]

what

[empty quail]

<@&268886789983436800>

[torpid brook]

dont abuse help channels

.close

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[icy hazel]

I need help with geometry

[glacial nacelle]

what's the question?

[elder thicket]

hi, can someone help with a problem?

[calm coral]

@icy hazel Has your question been resolved?

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[royal smelt]

I'm trying to find an eigenvector of this matrix that gives me an eigenvalue of (n-1). Do you have any ideas

[honest quiver]

Start by making an equation that involves the eigenvector and the matrix (think definition of eigenvectors and eigenvalues)

Also to clarify, is n the dimension of the matrix?

[royal smelt]

yes

would this work?

[honest quiver]

Actually I believe so

And you can define a whole eigenspace with such eigenvectors

Because you notice so long as the first coordinate is 0 and the others sum to 0, it is in the eigenspace

[royal smelt]

would there be n - 1 of these vectors that work? or is it n - 2

[0 1 -1 0 ... 0], [0 0 1 -1 0 ... 0] etc

i think its n - 2 of them right

[honest quiver]

Should be

[royal smelt]

okay just wanted to make sure

thanks!

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[timid robin]

Its mainly aleks help ive only got the annoying topics left

[low ocean]

have you tried setting up an equation ?

[timid robin]

yes, i dont have discord on my phone so idk how to share that besides microsoft paint or something

i just dont know if it goes like, .02(2000)+0.02x=0.03

[low ocean]

not quite

what does it mean if the overall % from both plants is 3%

how do you calculate overall %? it might help writing that down first

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[light vale]

Yo I need help

[bright hornet]

?

[spice phoenix]

!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/

[light vale]

I’m send a picture

[final verge]

A

It's kinda obvious -

[calm coral]

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[remote mural]

Help pls hurry

[remote mural]

XD

Im stupid

Its a drill btw

I have the ans

But

I just copied it from my friend

Idk hwo to do it

[cloud nexus]

Ok

How far away is bus B from bus A when bus A starts moving

[remote mural]

Idk

Just the ans

No sol needed

I gtg

[cloud nexus]

Dude

Do the math

[remote mural]

Can u explain in one text

[cloud nexus]

If you want to learn math, stay here

If you just want an answer because “I wont ever use algebra in real life” then I cant help you

[remote mural]

No i need to learn bud

Im doing smth rn

My teacher just called me

[remote mural]

Ok

[cloud nexus]

Alright

Good

Anyways, to solve this problem, we need to find a formula for both the position of truck A and truck B

And then set them equal and do a little algebra

So how would we find the equation for truck B?

[remote mural]

Okay

what equation?

[cloud nexus]

Formula

I mean

[remote mural]

for what

in truck b

[cloud nexus]

For truck b’s position

Where it is in our scenario

[remote mural]

x-2?

Basically just x

[cloud nexus]

If we let the time when truck B starts moving to be time 0 and time goes up in hours, whats the formula for truck B’s position

[remote mural]

+50

Ohh i got it

Tyy

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[tough vector]

i dont get b

[tough vector]

i tried 2*1*26*26 but thats wrong

[fluid flame]

that fixes a Q in position 2

you’re not counting labels such as KAQN

[tough vector]

yeah thats what i thought

but im not sure how to do that

i thought abt using factorials but idk how to implement that

[fluid flame]

count the complement is easier in this case

{number of all possible labels}-{number of labels without a Q}

[tough vector]

how do i find number of labels without a Q then

oh

[fluid flame]

that should be easier to figure out

[tough vector]

its 2*25*25*25

[fluid flame]

seems good

[tough vector]

alr that worked

thanks a lot

[fluid flame]

if the first thing looks convoluted counting complement is usually the second thing to try

[tough vector]

o

thanks

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[brazen elbow]

how to make big curly brackets for simul equaton in latex

[low ocean]

do \left{ and \right} work for you

i think they scale up in height depending on their contents

[brazen elbow]

no, that would only work for 1 line

im looking for one that covers many lines

[left sorrel]

can you write it out?

[low ocean]

really? i thought it would cover multiple lines if you \begin{align} inside the brackets

[winged smelt]

One that covers many lines with a single command?

[brazen elbow]

yes

[swift laurel]

depends on the use case, cases environment, \left\{+ a multiline environment like arrayor split, empheq package

[brazen elbow]

not as in latex code, as in texit $...$

[winged smelt]

How about you define a custom command for just the brackets

So you don't have to make it too complicated

[swift laurel]

texit is latex though...

[winged smelt]

For example

$\br{\f45}$

[potent lotus]

ColdTee

[brazen elbow]

$\left{ax+b \newline cx + d\right.$

[potent lotus]

Fungus 34A05

[brazen elbow]

i tried this but it doesnt work

[swift laurel]

,, \begin{cases}
ax^2 = c \
dx = y
\end{cases}

[median oyster]

use a double backslash

[potent lotus]

cloud

[winged smelt]

$\br{ax+b \newline cx + d}$

[median oyster]

and you need to be in displaystyle

[potent lotus]

ColdTee

[brazen elbow]

newline just refuses to newline

[median oyster]

not inline

[brazen elbow]

display style?

[brazen elbow]

holy shit magic

[median oyster]

single signs $...$ is inline

[potent lotus]

Desync

[median oyster]

as in, it puts in within text

which doesn't allow linebreaks

[winged smelt]

Oh you meant cases

[brazen elbow]

euh

[swift laurel]

yeah don't use single dollar signs in general unless you actually want inline math

[brazen elbow]

oh alright

,tex

\begin{cases}
ax+b \\
cx + d
\end{cases}

[potent lotus]

Fungus 34A05
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[brazen elbow]

euh

why error

[median oyster]

cases needs to be used within a math environment

[winged smelt]

Also this is more like a #latex-help question

[median oyster]

that might be why

[swift laurel]

cases should be in a math environment. the ,, prefix i used wraps in gather*

[brazen elbow]

hmm

,,
\begin{cases}
ax+b \
cx + d
\end{cases}

[potent lotus]

Fungus 34A05

[brazen elbow]

holy ahit yall wizards

alright thanks yall

.close

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[vale mantle]

Can someone please help me with question 1

[vale mantle]

You have to find an expression for X

Better photo

[calm coral]

@vale mantle Has your question been resolved?

[surreal patio]

u can remove the lns on both sides

u can rewrite $80kt$ as $lne^{80kt}$ and using ln rules write the entire right hand side as $ln((e^{80kt})(\frac{1}{15}))$

[potent lotus]

JustToPro

[vale mantle]

alr ill try that

[vale mantle]

what next?

i have x/(80-x)

idk how to chance that to get just x

[surreal patio]

u can multiply both sides with 80-x and rearrange to get x on 1 side i think

[vale mantle]

This is what I got

This is the answer

exercise 3 level 1 question 1a

so I don't think I did it correct

[calm coral]

@vale mantle Has your question been resolved?

[surreal patio]

sorry i had to go but im back now

then multiply and divide right hand side with $e^{80kt}$

[potent lotus]

JustToPro

[surreal patio]

so u get this
$$x = \frac{e^{80kt}\cdot 80}{15 + e^{80kt}} \cdot \frac{e^{80kt}}{e^{80kt}}$$
$$x=\frac{e^{80kt}\cdot 80}{e^{80kt}} \cdot \frac{e^{80kt}}{15 + e^{80kt}} $$
$$x = 80 \cdot \frac{e^{80kt}}{(e^{80kt}) \frac{15}{e^{80kt}} +1)}$$
$$x = 80\cdot \frac{1}{15e^{-80kt} + 1}$$
$$x=\frac{80}{15e^{-80kt} + 1}$$

[vale mantle]

Oh dang wtf

[surreal patio]

oh wait its +1 in the denominator (i suck)

[surreal patio]

this is also wrong

let me fix it wait

[vale mantle]

Alr

[surreal patio]

now to fix the texit text

i forgot to change the negative into positive when shifting to lhs

[potent lotus]

JustToPro

[surreal patio]

i hope u understand , if not then let me know where u are stuck

[vale mantle]

Ah ok

Thanks for explaining

[surreal patio]

np

[vale mantle]

.close

[calm coral]

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[surreal patio]

i just saw that there is a bracket missing

[calm coral]

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[sinful shuttle]

Does this count as math ?

[surreal patio]

what is cg? centi grams?

[sinful shuttle]

yeh :>

[surreal patio]

i would first convert that into grams , cuz molar mass unit is g/mol

100 centigram = 1 gram
1 centigram = 1/100 centrigram
so divide 0.0500 by 100 to covert into grams

why are there so many spaces for fractions? formula unit is simple
number of moles * avogradros number

[sinful shuttle]

this is something called dimensional analysis, and it's evil

it adds extra steps >.>

but Idk what they want as the end value here

with the magic of process of elimination: I got this far

[surreal patio]

there shouldnt be f.u Na2SO5, avogradro number is a constant it has no unit

[sinful shuttle]

still wrong D:

hrmm

[surreal patio]

maybe just try f.u without the na2so4

[sinful shuttle]

nope :<

[surreal patio]

"units of formula units Na2SO4"

[sinful shuttle]

tryna hunt down the lesson on this @~@ not sure where it'd be tho

[alpine gust]

love chemistry ⚗️

[sinful shuttle]

I'm more of a biology person xD

[surreal patio]

why do units even matter , i only write them at the end of the answer

and im pretty sure thats what everyone does

[sinful shuttle]

to keep track Ig? We lose points if we don't include them :<

[surreal patio]

hm

[surreal patio]

maybe try amu or um

atomic mass unit
unified mass

[sinful shuttle]

OH WAIT I GOT IT

I FOUND THE LESSON

I had to spell out formula units dnjasmklasd

sobbing

ty for your help :D

[surreal patio]

i mean i wasnt much help tho

!done

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[safe moss]

i just need a quick explanation of local maxima's. earlier, someone said max value and local maxima are not the same thing. a max value is the highest while local point also includes other high points as well

[safe moss]

so pretend i got a graph that has 2 peaks. one of them is higher than the other. lets call this peak x. peak x is max value

lets call other peak y

so i say that

both peak x and peak y is local maxima

and only x is the max value

is this correct?

[glass heart]

yes

[safe moss]

alright thanks

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[thick fog]

Could someone tell me which ones are correct and which ones are wrong

[calm coral]

@thick fog Has your question been resolved?

[thick fog]

<@&286206848099549185>

[calm coral]

@thick fog Has your question been resolved?

[dusky flax]

9th one is defo D right

you know how x=1 x=2 and x=3 are parallel in 2d right

same thing in 3d

8th one is E, corrcer

correct

[nimble badge]

4, 5, 6, 8 look right the rest are wrong

[dusky flax]

7th is wrong

6th is correct

wait 5th is kinda weird

1 min I'll try graphing it

5th is correct

4th is correct

3rd is wrong

2nd is wrong

(easy one bruh 😭)

1st one is wrong

so yeah like pixelius said 4, 5, 6, and 8 are the only correct ones

[thick fog]

oh ok thanks for the help

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[frozen orbit]

How do I show this?

[frozen orbit]

Translation: Show that there doesn’t exist $z \in \mathbb{C}$ such that $|z| - z = i$

[brisk pelican]

Seems like you could do it algebraically in a straightforward way

[potent lotus]

Michael

[brisk pelican]

z = a+bi

|z|=sqrt(a^2+b^2)

Since that’s real, b must be 1

[frozen orbit]

yeah I tried, I got $0 = 2xi(y+1) - 2y - 1$, if z = x + iy

[potent lotus]

Michael

[brisk pelican]

So sqrt(a^2+1)-a=0

[brisk pelican]

Maybe idk

[frozen orbit]

Wait, why?

[brisk pelican]

How’d you get that

[brisk pelican]

Oh sorry -1

Cause the imaginary component must come from -z

You see what I mean

[brisk pelican]

But anyway this has no real solutions

[frozen orbit]

$|z| = i + z$, putting in x + iy, squaring both sides to get rid of square root, and just writing it out

[potent lotus]

Michael

[frozen orbit]

That’s what I did

[frozen orbit]

Not really…

[brisk pelican]

Oh I see

I mean even if thag works it’s too much work

[brisk pelican]

Well |z| is real

[frozen orbit]

Yeah, I think it’s unecessary to do what I did

[brisk pelican]

Just think of the imaginary components of both sides

On the right you have i

[frozen orbit]

yeah of course

[brisk pelican]

So on the left you must have i

But it can’t come from |z|

So it must come from z

So z is a-i

If it was anything else it wouldn’t line up

[frozen orbit]

hmm

oh wait actually, that makes sense

Alright I think I get it

[brisk pelican]

🙏

[frozen orbit]

Thanks

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[spark roost]

i keep getting 138/210 but the mark scheme says its 2/5

[inner zealot]

,rotate ccw

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[fossil prairie]

okay so this question isnt related to any homework, but more of a personal question myself and several others havent figured out.
(this question is related to Minecraft..)
how many chunks would be in a 15,000 by 15,000 block area.. if each chunk is 16 by 16 blocks.

[remote mural]

15000/16 no?

937,5 chunks

Supposing theres no half chunks

938

[carmine sorrel]

15000^2/16

,calc 15000^2/16

[potent lotus]

Result:

1.40625e+7

[remote mural]

938x938

[remote mural]

this is how many chunks i suppose

938 by 938

@pearl canyon

[remote mural]

how come u get this awnser

[pearl canyon]

wrong dark

[carmine sorrel]

there's the block area is a 15000 by 15000 square, so there's 15000^2 blocks in the area.
There's 16 blocks in a chunk
so the total number of chunks would be around 15000^2/16
it won't be exact because 15000 isn't divisible by 16, so there will be partial chunks. Up to you if you want to round up or down on that.

[fossil prairie]

would this not be wrong? the chunk are 16 by 16.

I somehow managed to obtain a number of 879,000~ but the answer seems wrong.

[fossil prairie]

256 blocks per chunk

[carmine sorrel]

ah, yeah. fair. 15000^2/256

[cloud nexus]

878906.25 chunks

good luck doing that project man

[fossil prairie]

alright... well I aint the only user getting that answer but it really seems too high 😂 thanks!

[remote mural]

Its not that much

U just don’t wanna do it

Get the motivation to do it

[carmine sorrel]

but again, still not accurate because 15000 isn't divisible by 16. so it's over counting some number of chunks

[cloud nexus]

no its not

thats the exact answer

you can round up or round down

[fossil prairie]

im just after an estimate not perfect numbers :3

[remote mural]

ill@give u ond

alot

[hollow totem]

yeah it's 937.5^2

[carmine sorrel]

15000/16 = 937.5
so starting one whole chunk in the bottom left, then the entire top and right sides of the 15000x15000 area is covered by half chunks.
If you only want whole chunks covered by the square, then those shouldn't be included,
15000^2/256 includes the half chunks.
So it's over counting by adding up all the half chunks to be whole chunks, when they aren't

[cloud nexus]

what do you even need this information for

its like maximum +- 5%

[carmine sorrel]

yeah. so it's not a totally accurate count... like i said.

[fossil prairie]

personal thing, im trying to figure out the amount of debris could be in that area for mc :/

[hollow totem]

so just treat chunks squared like a distance unit

[cloud nexus]

mate, theres no way in hell you are mining out like 8 hundred thousand chunks for ancient debris

[fossil prairie]

im not 😂

[hollow totem]

nothing wrong with a fractional value in that case

[calm coral]

@fossil prairie Has your question been resolved?

[calm coral]

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[novel sonnet]

is this the proper way to find the tension in string A?

[glacial inlet]

it should be $T_{Bx}-T_{Ax}=0$ since the x component of tension B is to the right and that of tension A is to the left

[potent lotus]

y0shi

[civic dirge]

What made you settle on using 35 degrees for the components of A? That looks off

[glacial inlet]

yeah that too

[calm coral]

@novel sonnet Has your question been resolved?

[novel sonnet]

true asf

[novel sonnet]

oh yeah wth its 30 lol

i think i accidentally wrote 35 once and then just repeated it

I'm gonna rewrite it and send again because it still says it's wrong

one moment

[glacial inlet]

you forgot the sin (45) when substituting in T_B

[novel sonnet]

thats right

[glacial inlet]

yeah that looks good

[novel sonnet]

i did this one already.. BUT IT WAS IN RAIDIANS

[glacial inlet]

oh damn

[novel sonnet]

so i got it wrong and erased it

lol

but i thank you

[glacial inlet]

i see

ywyw

[novel sonnet]

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[tribal juniper]

can someone explain why -3 is out of the domain and why 4 isnt?

[tribal juniper]

I was able to reach the factoring part myself, but I wasn't able to understand why 4 is in the domain and -3 isnt

[graceful dust]

there is no x that makes e^x = -3

e^x > 0 for any x

[glacial inlet]

and the domain of ln(x) is from 0 to infinity

not including 0

[graceful dust]

honestly that thing on the bottom right is silly

[tribal juniper]

so just to make sure im getting this right nat log range is -inf to inf but the domain is (0, inf)

[glacial inlet]

yep

[tribal juniper]

alright thanks

.close

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[crystal forge]

graph of ln:

increases slowly, but without bound

[calm coral]

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[shell spruce]

Anyone know how to do 5p

[cloud nexus]

wheres 5p

[upper sparrow]

Assumedly they meant 50 and mistyped

[shell spruce]

It’s 50 💀sorry

[empty quail]

I think they want you to make a parabola equation

for b is just set y=0 and solve for x

actually I see there are two 50 questions

does it belongs to the diagram?

yeah it does lmao stupid for stupid question

[still marlin]

First solve for initial velocity