help-10

[crimson hill]

okay in the case that x<0, x would always be < -(1/n), so i get that it wouldn;t be included
in the case that x is 0 it would be undefined so that's whatever
in the case that x is 1 it would be > -(1/n) ,same with any x > 1
so it would be included by the first count only
and by the second count it would exceed 1+ 1/n

ok

things make sense now

world is clear future is bright

[haughty coyote]

"in the case that x is 0 it would be undefined so that's whatever " no, 0 > -1/n for all n

0 is in the intersection

[crimson hill]

i meant that -(1/0) is undefined

[haughty coyote]

this to me feels like taking things out of context

[crimson hill]

but i see what you're saying

[haughty coyote]

because I only defined n as such for x < 0, so it shouldn't be expected to work for x = 0

[crimson hill]

the 0 thing specifically or my approach as a whole ?

[crimson hill]

i see that, yeah i shouldn't have done anything with 0 in the first place you're right

(x=0 that is)

....surely there's a better way to approach this than what i did
surely the induction doesn't have to be like that

[haughty coyote]

what induction ?

[crimson hill]

like if you were to prove that the notation is equiavalent to the set would you genuinely have to exhaust x > and x < 0

[haughty coyote]

these 3 messages are my proof

[crimson hill]

complete induction i mean, as in going through each x option

[haughty coyote]

pls don't call that induction

[crimson hill]

between 0 and 1
i guess that's not very complete at all is it

yeah no i regret that

[haughty coyote]

don't use names that already have a meaning to call things that aren't that thing

[crimson hill]

my apologies

[crimson hill]

no i genuinely dont know how id prove this
like would playing w x<0 and x>1 be sufficient ?

[haughty coyote]

as I said, here's a short proof

not as well written as it could have been, but well enough

[crimson hill]

wait that counts as a proof?
that's great actually
i thought i'd have to use some other notation i don't know of
also what is that type of proof called ?

[haughty coyote]

in this case double inclusion

[crimson hill]

(so i avoid calling it "an induction/exhaustion except we only exhaust 2 things")

ohh

i see

thanks !!!!

[haughty coyote]

based on the principle that 2 sets are equal iff they're included in each other

[crimson hill]

got it
ty

that was rly helpful! thanks so much!

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[short aurora]

if f:R_>R is odd, solve f(x) + f⁻¹(x) = eˣ - 1

[short aurora]

This one has been melting my brain for the past hour

[sage geode]

If f is odd, then so if f^(-1), right?

[short aurora]

yes

i have proved that

[sage geode]

So f + f^(-1) is odd

But on the right hand side you have e^x - 1, which is definitely not odd

So no such function exists

[short aurora]

e^x - 1 not being odd doesn't mean it can't have common points with the graph of f(x)+f^-1(x)

[sage geode]

Ah, did the problem ask to solve for x instead of f?

[short aurora]

yeah for x of course

x is always the one missing

[sage geode]

Well since f and f^(-1) are odd, then we have f(-x) + f^-1(-x) = e^(-x) - 1
-f(x) - f^-1(x) = e^(-x) - 1
f(x) + f^-1(x) = 1 - e^(-x)

So e^x - 1 = 1 - e^(-x)

[short aurora]

uhm

i dont get the last line

[sage geode]

Initially we had f(x) + f^-1(x) = e^x - 1

And now we also showed that f(x) + f^-1(x) = 1 - e^(-x)

So, since a = b and a = c means that b = c

Similarly e^x - 1 = 1 - e^(-x) in this case

[short aurora]

oh right

okay let me try to solve it now

[short aurora]

im a bit skeptical about this line , since you're using the inital equation as a given fact and not as something we have to solve. I dont think im allowed to just replace x with -x in it , or with anything for that matter. But correct me if im wrong

[sage geode]

On the other thought yeah an equation for x being true doesn't imply that it's true for -x

But there's another argument to prove a solution

For every odd function, f(0) = 0

So f(0) + f^-1(0) = 0

And e^x - 1 is also 0 at x = 0

So x = 0 is a solution

Other solutions can't be determined unless f is given

[short aurora]

what does that even mean

[sage geode]

A typo, mb

[short aurora]

oh

this is true only if f is continuous though

[sage geode]

That f(0) = 0?

[short aurora]

it could be defined in something like (-a,0)u(0,a) and still be odd

yea

[sage geode]

Well it's given that f: R -> R so I guess 0 should be in the domain

[short aurora]

oh well

[sage geode]

Continuity isn't necessary for f(0) = 0 to be true tho, the only requirement is 0 to be in the domain

It just so happens that continuity at 0 also requires that

[short aurora]

yeah youre right

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[wanton kraken]

"A point lies on the hypothenuse of a right-angled triangle, and it divides the hypothenuse by parts, equal to 30 cm and 40 cm.

That point is equally distant from the catheti. Find the catheti"

[wanton kraken]

so

I can easily get to conclusions that the 4-angled figure is a rectangle at least

and since the problem states that the point is equally distant from the catheti, I assume that all sides are equal, making the rectangle a square

but what do I do to proceed?

so I marked some segments

am I correct thus far?

slight edit to correct a mistake

[obtuse pebble]

@wanton kraken Has your question been resolved?

[wanton kraken]

<@&286206848099549185>

[wanton kraken]

welp

that's as much as I can get from a server full of thousands dedicated to mathematics, including people with university degrees

welp

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[agile pendant]

Hello
Can anyone help me find the relationship between a and b so that the limit of f(x) when x approaches infinity is 9/2 ?

<@&286206848099549185> I've been stuck with this for over a day, pls give me some help

[obtuse pebble]

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[timid silo]

∫f(x)+g(x)dx = ∫f(x)dx + ∫g(x)dx is this always possible?

[untold flax]

yes

[timid silo]

always always?

[kind hawk]

well assuming f and g are integrable

otherwise you might end up with an infinity-infinity situation on the RHS

[timid silo]

makes sense

thanks

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[sage dagger]

I only need help with C and D, how do I find the domain and range for these equations?

[obtuse pebble]

@sage dagger Has your question been resolved?

[obtuse pebble]

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[timid silo]

You are trying to weld two pieces of metal together, but they keep slipping. Your solution is to clamp them together with a normal force of 4000N. The coefficient of the static friction for steel-on-steel is 0.1. How strong is the friction force keeping them from sliding?
I put 400N

[timid silo]

Correct?

Friction= (0.1)(4000)=400N

https://media.discordapp.net/attachments/963825410741768272/1029520454043578378/unknown.png

Ff=uFn

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[sullen wigeon]

question from a statistics 120 assignment i just cant. seem to figure out

[sullen wigeon]

82% of all adults think gambling should be illegal. In a sample of 50 Adults what is the probability that:

P(x=41)
P(x ≥38)
P(x>40)

Calculate the mean and standard deviation of adults who think gambling should be legal

[rigid lintel]

this is all the information youre given?

[sullen wigeon]

yes

[rigid lintel]

so is it binomially distributed with n = 50, p = 0.82

[sullen wigeon]

i literally have no idea

[rigid lintel]

it probably is

lets just assume that

[sullen wigeon]

ok

[rigid lintel]

so just use the formula for the first

[sullen wigeon]

what formula

[sullen wigeon]

i got 0.15

so 15%?

[rigid lintel]

well yeah rounded off that is

[sullen wigeon]

what about the ones with inequalities?

[rigid lintel]

but yes

well if its larger or equal to 38

that means summing the probabilities that its 38, 39, 40, ..., 50

not very nice to do by hand

[sullen wigeon]

is there an online calculator you'd reccomend for that? i dont have my ti-84 on me rn

[rigid lintel]

https://homepage.divms.uiowa.edu/~mbognar/applets/bin.html

its nice and minimalistic

[sullen wigeon]

and for x i have to put in each value from 38 to 50?

oh wait it shows me

thank you so much for your help!

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[mild spoke]

I need some help with a physics problem

[mild spoke]

we are currently learning about free fall

and these are the equations they want us to use

[royal basin]

there are three things you're asked for here, all given as a continuous chunk of text but could be formatted into subproblems like so:

(a) A cart is released from the top of a tower and is in free fall for 2.6 seconds. What is its velocity at the end of its free fall?

(b) After free fall, the cart is slowed down back to rest over a duration of 1 second. What is its rate of deceleration during this phase?

(c) What vertical distance did the cart cover in total during the free fall and slowdown phases?

which of these parts are you having trouble with?

@mild spoke

[obtuse pebble]

@mild spoke Has your question been resolved?

[mild spoke]

For the entire thing, basically @royal basin

[royal basin]

mkay so let's start at the beginning i guess

do you know the acceleration experienced by an object in free fall on Earth?

[obtuse pebble]

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[solar bolt]

a bit sttuck on this one

[obtuse pebble]

@solar bolt Has your question been resolved?

[elfin burrow]

integration by parts

i don't think the given equation helps

[solar bolt]

what is intergration by parts

like splitting?

[elfin burrow]

are you familiar with this formula?

[solar bolt]

no

[elfin burrow]

https://www.khanacademy.org/math/ap-calculus-bc/bc-integration-new#bc-6-11

i'd suggest learning about it

[solar bolt]

thanks

[deft gust]

@solar bolt I think the question intended you to do something like this

And solve for I

[solar bolt]

what is that

[deft gust]

k

[solar bolt]

ok

[deft gust]

and I don't think it wants you to solve that integral

[solar bolt]

yeah its all general

[deft gust]

If you couldn't figure this out on a test you could just take the derivative of each of the answers

[elfin burrow]

lmao

[solar bolt]

how do i close

[elfin burrow]

.close

[solar bolt]

.close

[obtuse pebble]

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[blazing swift]

<@&286206848099549185>

I am having trouble finding this I am a grade nine that hasn’t been to school in a while bc im sick

[viscid violet]

It must be linear function, which is something like $ax + b$.

[warm shale]

Robin S.

[viscid violet]

Just create a function, let's say $f(x) = ax + b$ then plug in those x values and find a and b with those y values.

[warm shale]

Robin S.

[blazing swift]

I have no clue what your saying right now your kinda speaking jibberish can I just get the answer pls

[viscid violet]

No.

[blazing swift]

Well that’s all I was looking for so I’ll be gone

[viscid violet]

I can't say the answer, it is written in the rules of the server and even if I say it, you will not learn anything.

Then see you later!..

[blazing swift]

.close

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[pallid osprey]

Is the answer -1/2 + sqr2/2

[mental plaza]

,w cos((3pi/4) - (5pi/3)) exact

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Springsskateboard#8145

[mental plaza]

I’d that

is that

the same

if it is then it shd be fine

[pallid osprey]

so the answer i have is correct?

[mental plaza]

idk

do you

have

a picture

or

maybe

you can key into your calculator

and see if it’s the same result

[daring rock]

No

[mental plaza]

oo maybe not then

[daring rock]

Did you use the sum/difference formula?

[pallid osprey]

let me take a picture of my paper

[daring rock]

ok

[daring rock]

Oh

The problem says 5pi/3, but you did 5pi/4

[pallid osprey]

oh

i see give me a sec

[daring rock]

Looks good 👍

[pallid osprey]

like this

[daring rock]

Yeah, that should be fine. Sometimes those online programs are really picky about how you input the answer

But that's definitely correct

[pallid osprey]

it was incorrect

lmao

[daring rock]

That's the same lol

Idk, I guess combine the fractions next time

[pallid osprey]

exactly

yeah

i got sin(10 degrees)

[daring rock]

70-80, not 80-70

[pallid osprey]

70-80=-10

[daring rock]

Yeah, sin(-10 deg) is right

[pallid osprey]

wait

its says rewrite the expression

so would that still be right

[daring rock]

Yeah, rewrite it as a trig function of a single number

-10 deg is a single number

So like, don't evaluate it with a calculator, just put sin(-10 deg)

Use a degree sign if you have one

[daring rock]

Looks good

[pallid osprey]

it was right

cos(45 degree)

[daring rock]

yep

[obtuse pebble]

@pallid osprey Has your question been resolved?

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[random depot]

cassie cage 😮

[spark vector]

?

[random depot]

in the question

[spark vector]

Can you explain?

<@&286206848099549185>

[obtuse pebble]

@spark vector Has your question been resolved?

[obtuse pebble]

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[sand light]

This doesnt make any sense

[sand light]

Part a is easy

but ive tested like 5 points for part b

some are positive, some are zero, and some are negative

and the domain is 0 to inifity

So i dont really know how to answer this

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[timid silo]

is 2 root x the same as root x +root x?

[teal turret]

Yes

[signal sinew]

2*rootx is same yes

[timid silo]

ok thx

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[kindred lichen]

15.712121212 , the 12 is repeating, i know how to change simiral simpler questions like this to a ratio of integers, but i cant grasp the way to change this

[kindred lichen]

this is indeed 7th grade mathematics, but i am entering college and have forgot everything

[teal turret]

Change to an improper fraction?

[kindred lichen]

yes

[teal turret]

https://www.calculatorsoup.com/calculators/math/decimal-to-fraction-calculator.php

this site gives a good explanation on how to do it

just type in 15.712 and say 2 repeating decimals

then it shows u how to do it

@kindred lichen

[kindred lichen]

alright thank you

.close

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[tacit briar]

expected value of the sum of two dice.

[vestal marsh]

Just take the probability of a side and multiply it by the number on that side. Do this for all sides and sum them up

There you go

[obtuse pebble]

@tacit briar Has your question been resolved?

[tacit briar]

Ev of a die is 3.5

So 7?

[vestal marsh]

Yes

@tacit briar

[viral blade]

yeah expected value is linear when you're adding together several random variables

So E(X) + E(Y) = E(X+Y)

[obtuse pebble]

@tacit briar Has your question been resolved?

[obtuse pebble]

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[finite matrix]

There is gotta be a trick in this question, i cant find it

[finite matrix]

ping me when you answer

[high lily]

what have you tried

[obtuse pebble]

@finite matrix Has your question been resolved?

[finite matrix]

yea

[high lily]

that wasn't a yes/no question

[finite matrix]

oh

like i tried opening brackets

but it was getting more complex

(a+b)^3 = a^3 + b^3 + 3a^2b + 3b^2a
2022 and 2023 form 4045 or 1 or -1

[high lily]

consider stuff like factorisation of sum/difference of two cubes

[finite matrix]

elaborate please

[high lily]

a^3 - b^3 = ( ? )( ? ?)

[finite matrix]

(a-b)(a^2+b^2+ab)

[finite matrix]

ohh

[high lily]

yeh and also

[finite matrix]

okay

[high lily]

a^3 + b^3 = ( ? )( ? ?)

[finite matrix]

(a+b) * (a^2+b^2-ab)

okay i try it in that way

[finite matrix]

yoo its getting simplified a lot

nicee

i got the answer

thanks

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[heady vessel]

hello

[heady vessel]

these are the questions

here are the answers

why does the answers not include + C ???

[robust sleet]

its primitive

if it was integral it would be + C because it will be the set of all possible primitives

[heady vessel]

oh

okay i understand that

also question C

why is that the answer???

.close

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[full light]

Am struggling with question 12 pls help

[sharp flume]

you know what a parallelogram is?

[river tree]

Try drawing a diagram of a parallelogram with diagonals that meet at right angles. Drawing the diagonals will split it into four triangles, and you can use those triangles to show which angles on the edges of the parallelogram equal each other

[obtuse pebble]

@full light Has your question been resolved?

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[pine fog]

I have p and q but how am i supposed to get the last three

[full light]

like this?

[obtuse pebble]

@pine fog Has your question been resolved?

[high lily]

were you told additional info about a,b,c?

[obtuse pebble]

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[untold furnace]

These 2 questions

Are they correct or not

Sorry for different language

[obtuse pebble]

@untold furnace Has your question been resolved?

[untold furnace]

<@&286206848099549185>

[obtuse pebble]

@untold furnace Has your question been resolved?

[coral epoch]

for the first one, we got a false proposition (because {2,3} isnt a part of the empty set) which implies something else which is either true or false, so we got either
false ⇒ false
or
false ⇒ true

both are true since something that is false can imply anything, so the answer is waar

for the second one, we got a triangle abc, and we have that ac is shorter than the lengths of ab and cb combined (which is always the case, because there can’t be a triangle with a side longer than the combined length of the other two sides) implying that ac is longer than the lengths of the other two sides combined (which is impossible), we got
true ⇒ false

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[plain sandal]

Sorry, they told me to post my problem here. So here's the problem that I've been trying to solve:

[astral mantle]

what can u tell me about alpha+beta, and alpha*beta

[sage geode]

Also rewrite (c - b)/a into c/a - b/a and therefore into alpha*beta + (alpha + beta) according to the Vieta's theorem

[obtuse pebble]

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[obtuse pebble]

@plain sandal Has your question been resolved?

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[clever nacelle]

Why must one of m or n be an odd number?

to ensure it's rational?

[final thunder]

We assume the greatest common denominator of m and n is 1

If n and m are both even then at least 2 divides both of them

[clever nacelle]

yep I understand that

[clever nacelle]

but it says that If they are divisible by an integer that is greater than 1 then it doesn't change it

[timid silo]

we assume m and n are minimal, but then we can show that they are even so there are even smaller m and n, giving a contradiction

[clever nacelle]

by minimal it means simplified as much as possible right?

[timid silo]

yes

[final thunder]

Yes

[clever nacelle]

I still don't understand why one of them has to be an odd number, is it because if they were both even then we could simplify it and then we would eventually end up with 1 odd number?

[timid silo]

yes

[clever nacelle]

ah ok

this is a complex first example haha

thanks

.close

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[clever nacelle]

??

I don't understand

how can n=n+1

[timid silo]

can it?

what is the set then?

[clever nacelle]

the empty set?

[timid silo]

yes

[clever nacelle]

oh ok

I thought something was suspicious

thanks

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[raven spire]

[obtuse pebble]

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[latent jasper]

25 people with different names (A, B, C, D, ... ) are to be divided into 5 unnumbered (non-empty) groups. In how many ways can this be done if A is not allowed to end up in same group as any of B, C or D?

[latent jasper]

Im supposed to solve this by using stirlings number somehow but I am confused

The total way that the combination can happen is S(25, 5)=2 436 684 974 110 751

But the I have to exclude the cases A is in the same group as 3 of them, or as 2 of them, or with one of them.

I could do if the groups were distinct.

Ok so I just got an idea

if we have case 1 when say all ABCD are the same group which means we can count them as one, so we have 22 people to distribute in 5 non-numbered groups which means we have S(22,5)

WE have four cases where A is with 2 people, or 1 person so we can do the same thing.

Answer = S(25,5) - (S(22,5)+ (3×S(23,5))+(3×S(24,5)))

But is that correct?

<@&286206848099549185>

[drifting flax]

,w C(25,5)

[warm shale]

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[lime jewel]

Can yall help me with this problem?

[drifting flax]

occupied

[latent jasper]

wdm?

[drifting flax]

uh, that's not correct, it's just me testing

[latent jasper]

we kinda have a similar problem, if you solve it, leave a tip

[latent jasper]

aha okey

[obtuse pebble]

@latent jasper Has your question been resolved?

[latent jasper]

Anyone?

[obtuse pebble]

@latent jasper Has your question been resolved?

[obtuse pebble]

@latent jasper Has your question been resolved?

[timid silo]

,calc 6!/e

[warm shale]

Result:

264.87319764344

[timid silo]

265 Ways

in each box has to be a ball

green ball not in green box

This is derangement

"In combinatorial mathematics, a derangement is a permutation of the elements of a set, such that no element appears in its original position."

you can calculate the ways with n!/e but its approximate so you have to round up to the next number or round down

[timid silo]

There are 25 different names?

[obtuse pebble]

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[acoustic apex]

4x - 4 - 2x + 1
How do i simplify this as far as possible?

[restive ridge]

what have you tried so far

[acoustic apex]

nothing

i dont know how to do it at all

[high lily]

combine like terms

you can think of x as 🍎 if that helps

[acoustic apex]

????????????????????

[high lily]

do you know what like terms are?

[acoustic apex]

no sorry

english isnt my first language

[late lodge]

how to do find intersection point

of 2 slopes

[cedar lichen]

#❓how-to-get-help

[acoustic apex]

this is kinda occupied

[late lodge]

😦

[high lily]

if someone gave you 5 apples and 4 bananas, then took back 3 apples and took back 2 bananas as well
represented by the expression
5🍎 + 4🍌 - 3🍎 - 2🍌
how much of each fruit would you still have?

[acoustic apex]

wait

are the apples the xes

im so confused rn

was that just an example

[high lily]

don't worry about the original problem for a minute,
i gave you a relatable real life example to make it easier to recognise the idea of like terms

[acoustic apex]

oooh ok

2 apples and 2 bananas

[high lily]

yes

so you have some sense of what like terms are, same idea can be applied to your question

[acoustic apex]

alright

[high lily]

just like how you didn't mix apples with bananas here, don't mix your terms with x and constants

[acoustic apex]

ok

how do i do it in the original problem

[high lily]

pretty much the same way as just now

just like how you identified and separated terms with apples and bananas,
do the same for terms with x and terms with constants

[acoustic apex]

4x-2x

-4+1

so 2x-3

?

[high lily]

yes

[acoustic apex]

ok thanks

.close

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[primal idol]

Hello, this is a series problem.

How should i set up my series here?

so for every 6 hour, the amount of antibiotic is half of 200

[obtuse pebble]

@primal idol Has your question been resolved?

[cursive trellis]

Heya

This is a geometric series

The amount of anibiotic left is 50% of what was there 6 hours ago

Do you know how to find r? the common ratio

[obtuse pebble]

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[sinful minnow]

No idea :/

[sinful minnow]

.close

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[misty steeple]

How do I solve this?

I understand that the sum of the roots is 19, and the product of the roots is 216

I'm not sure how I can write the sum of the product pairs in a way which allows me to solve it using the sum (19) and the product (216)

[timid silo]

the geometric sequence depends on two things - the start and the common ratio. You can create a system of two equations in these two variables with 19 and 216

[misty steeple]

What would the system of equations be?

[timid silo]

first write out the sum of the first three terms in a geometric sequence in terms of the start and common ratio

then the product

[misty steeple]

what do you mean the by sum of the first 3 terms?

are these 3 terms alpha + beta + gamma?

[timid silo]

do you know what a geometric sequence looks like?

[misty steeple]

not particularly

[timid silo]

does this ring any bells?

[misty steeple]

Kind of

so i need to find the common ratio of 19 and 216?

[timid silo]

no

the three roots will be a, ar and ar^2

what is the sum of these three roots in terms of a and r?

[misty steeple]

a + ar + ar^2?

[timid silo]

yup

and thats equal to 19

now what is the product of the roots?

[misty steeple]

ohh and that's equal to 216

[timid silo]

yeah exactly

then you can solve for a and r

[misty steeple]

alright thanks

i'll see how far i get

[timid silo]

np

[obtuse pebble]

@misty steeple Has your question been resolved?

[misty steeple]

@timid silo what is the final answer?

i got an answer but i doubt it's right haha

[timid silo]

let's see your working

I havent done it so idk the answer

,w a^3r^3=216, a+ar+ar^2=19

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[misty steeple]

i got it right!!

let's goo

i struggled with the system of equations for a while lol

wait

so from here

i wrote a, ar, ar^2 as product pairs

and plugged in a and r

i got 73.5

[timid silo]

nice!

lets check

,w roots x^3-19x^2+73.5x-216

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[timid silo]

hmm

[misty steeple]

uh oh

is it the right method?

[timid silo]

the last one should be a^2r^3 I think

[misty steeple]

ohh

let me try that

i got 114

,w roots x^3-19x^2+114x-216

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[timid silo]

yup

looks good now

[misty steeple]

so what's 4,6 and 9

i guess the roots

but like where would i find that

previously in my working

[timid silo]

yeah they are the roots

they're a, ar^2 and ar^3

[misty steeple]

ohh i see now

so the sum of that is 19

the product of that is 216

[timid silo]

yup exactly

[misty steeple]

and blah blah blah

tysm bro

[timid silo]

np :)

[misty steeple]

is there a way i can like

give you a 5 star review

[timid silo]

nah lol

im just here as a volunteer

[misty steeple]

alright well thanks again and see u next time :)))

uh oh

@timid silo

my classmate says the answer is 18

not 114

he also got 114 but apparently on the answer sheet it says 18

i don't have the answer sheet

this must be wrong right?

,w roots x^3-19x^2+18x-216

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[misty steeple]

yeah it's definitely not 18

sorry for disturbing you

,w roots x^3-19x^2+54x-216

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[misty steeple]

,w roots x^3-19x^2+56x-216

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[timid silo]

ye definitely not 18

[obtuse pebble]

@misty steeple Has your question been resolved?

[obtuse pebble]

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[mortal gust]

hello

[mortal gust]

its my first day in 9th grade and i missed 2 months of school and i really need help

[marsh sail]

Follow what the table says.

[marsh sail]

Each two cells beside each other add up to the 3rd cell.

[mortal gust]

mb its side ways

but i think thats it

i drew it in before i put it in

[marsh sail]

Does it make sense to you?

If it does, then go for it 😉

[mortal gust]

yeah ima try it

i got it right 🙂

[marsh sail]

nice

[mortal gust]

bruh now i gotta do this

i believe this is correct can you double check?

[marsh sail]

I think it's pretty clear.

Divide each quantity by 200 and multiply the result by 100.

[mortal gust]

ohhh ok

yeah i got it right

[obtuse pebble]

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[obtuse pebble]

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[atomic oasis]

could someone help me with number 1

here’s what i have so far

idk if i’m getting the matrix representation of the projection right

[obtuse pebble]

@atomic oasis Has your question been resolved?

[atomic oasis]

<@&286206848099549185>

[obtuse pebble]

@atomic oasis Has your question been resolved?

[obtuse pebble]

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[timid silo]

(k^2-2k)x^2+(4-6k)x+8=0
all solutions are integers, k is a constant. what are the possible values of k?

[timid silo]

im confused tbh?

[random depot]

1. simplify the terms
2. use quadratic formula and sub in the values

[zenith raft]

does k need to be an integer, or real number?

[wild swallow]

that seems kinda complicated

apply rrt

looking at factors of 8

[zenith raft]

I think it's not too bad if we're searching for integers k

[wild swallow]

if k is real it might get bad

[zenith raft]

yea lol

[random depot]

[wild swallow]

no k must be an integer

otherwise the expression won't be integer

or am i tripping

[zenith raft]

if k is an integer I think a necessary condition is that k^2 + 4k + 4 is a square

so that limits it pretty severely

[zenith raft]

by quadratic formula if I didn't mess up

[wild swallow]

k(k-2) needs to divide 8 if we apply rrt

[zenith raft]

that sounds simpler lol

[wild swallow]

and then k could be 4 or -2 but hmmm

quadratic formula doesn't work if k is 0 or 2

and then 2(2-3k) needs to divide 8

which happens for both

[zenith raft]

oh yea, didn't realize that

[wild swallow]

all we need to check is if there are integer solutions when k=4 or -2

[zenith raft]

or -1?

oh no nvm

[wild swallow]

if k=4 the middle term is -20

if k=-2 the middle term is 16

8x^2 + 16x + 8 = 0 definitely has integer solutions

,w 8x^2 - 20x + 8 = 0

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[wild swallow]

this doesn't

so k can be 0, 2, -2

[wild swallow]

k^2 + 4k + 4 is always a square

it's possible we're still missing a whole bunch of k that aren't integer

[zenith raft]

not a very helpful necessary condition

umm I think you missed k = 1

[wild swallow]

yes you are right

[zenith raft]

there might not be any

desmos is making me think there aren't any

[wild swallow]

quadratic formula seems to do it actually

k=3/2

[random depot]

ez

[wild swallow]

,w (k^2 - 2k)x^2 + (4 - 6k)x + 8 = 0, k=3/2

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[wild swallow]

[zenith raft]

lol

works

[wild swallow]

doesn't

[zenith raft]

I meant that the weird answer format works

[wild swallow]

oh yeah

[obtuse pebble]

@timid silo Has your question been resolved?

[wild swallow]

i think they're all integer

[obtuse pebble]

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[pseudo hatch]

What is the next step

[stable rain]

sub in the val of x

[pseudo hatch]

Is it underestimate?

[stable rain]

idk feel we cant say

not mch info

[pseudo hatch]

Oh

[cedar lichen]

We know the slope is negative around x = 2

Doesn't guarantee anything for x = 2.1 but we can make a reasonable estimate

[stable rain]

is thr a prev part

[pseudo hatch]

No

[stable rain]

ye idt u can say anything

abt over/under est

[pseudo hatch]

Also is there an easy way to simplify x^5+2x^3+x-4

[stable rain]

doesnt seem like?

,w solve x^5+2x^3+x-4=0

[warm shale]

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Requested by Circle#0025

[stable rain]

ye

its weird lol

[pseudo hatch]

Because my teacher did that without showing work

[stable rain]

except that x=1

[pseudo hatch]

How do u get x=1

[stable rain]

well

if u plug in x=1 in the eq

ull c it =0

[pseudo hatch]

Ok

.close

[obtuse pebble]

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[stable rain]

[obtuse pebble]

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[teal smelt]

Solve 2^2x + 2^x +6 =0 using the quadratic formula

[teal smelt]

im very confused as how to approach this

[high lily]

try doing a sub like
u=2^x to get something more recognisable

[teal smelt]

would that re-write the equations as "let u = 2x" 2^u + u -6

[high lily]

no

[teal smelt]

do you mind showing me what you mean?

[high lily]

consider 2^(2x) = (2^x)^2

[teal smelt]

i see

ive never really thought about that relation

@high lily is it okay to ask another question here?

without closing this help channel?

.close

[obtuse pebble]

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[timid silo]

$a, b, c \in \mathbb{N}$

$a, b, c < 10$

$a + b + c = 18$

$7 \mid 100a + 10b + 1c$

$100a + 10b + 1c = ?$ (we need to find the minimum value)

And no, we haven't been taught any rules for divisibility of 7, thus we aren't allowed to use any.

[warm shale]

trololol !

[timid silo]

I have no idea what to do.

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[upper sorrel]

one unpleasant way I see is to use brute force, but that would be very time consumeing

tho think about it again I think brute force isn't too bad in this case

@timid silo

ping me if you need help

[warm shale]

Arya

Arya

Arya

[raven spire]

I suppose you'll be able to do the rest?

@timid silo

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[earnest folio]

sos

whats the ratio

oh dude this is cool

SOS THOUGH

wait what

how do i do this

[trail musk]

find the ratios between consecutive terms

[earnest folio]

its all wildly different

[trail musk]

what is 58/28

[earnest folio]

tried the average and it was wrong

2.07

if i remember right

[trail musk]

what's 90/58

[earnest folio]

:/

oh wait

so u find all that and then u averGe??!?

[trail musk]

yeah that's a method

[earnest folio]

i got 1.5 as the average and it was wrong

it said to round to the .5 lmao

[trail musk]

,w (58/28, 90/58, 119/90)

[warm shale]

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Requested by Mr. Gamer#6658

[earnest folio]

yep i got that already

previous failed attempt

s

[trail musk]

try 2

[earnest folio]

if its wrong i get a 60

are you sure

whatever man im doing it

it wasnt 2 :)

[trail musk]

it's actually 1

damn you already went ahead

[earnest folio]

it wasnt 1 either

it was linear

29-31

thank u for the valiant effort though this server is super cool

https://c.tenor.com/bWUeVRqW9-IAAAAj/fast-cat-cat-excited.gif

[trail musk]

oh i didn't even see that it gave you the choice

[obtuse pebble]

@earnest folio Has your question been resolved?

[obtuse pebble]

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[hollow iris]

How would I rearrange this equation to find t?

[final thunder]

It’s a quadratic in t

[hollow iris]

Wym

[final thunder]

Compare this to $ax^2 +bx+c=0$

[warm shale]

Pure

[hollow iris]

Oh that

Well i actually know my values in the equation (except for t)

The equation isn’t equal to 0

One sec

[final thunder]

You can make it equal to 0

[hollow iris]

This is what I get after I plug in my values

[final thunder]

Oh

[hollow iris]

Yeah sry I should have explained it more

[final thunder]

Then you can multiply both sides by $\frac{2}{9.8}$ then square root both sides

[warm shale]

Pure

[final thunder]

Take only positive roots as time>0

[hollow iris]

Why do you multiply them by 2/9.8?

[final thunder]

To isolate t

(Get only t on one side)

[hollow iris]

Yeah that’s true. I was wondering why you need to do 2/9.8 in order to isolate for t

Like I get we want t on one side

I just don’t get why we do that in order to isolate for t

[final thunder]

1/2 * 9.8 * 9.8/2 =1

They cancel

[hollow iris]

ah ok ty

.close

[obtuse pebble]

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[random egret]

If I have 255 Green Colors, 255 Red Colors, and 255 Blue Colors - what is the amount of possible colors I can create? Or at least, what would be the equation to find out?

[slim surge]

or $255^3$

[warm shale]

NV

[random egret]

What if the color only changes every 23 digits for each section? This would make the number much smaller I assume, but, how would that be calculated?

[slim surge]

so you mean to say is that there will only be 23 greens, reds and blues?

anytime, if the probabilities are independent of one another, you multiply them
lets say a coin, the probability of getting a heads or tells isn't affected by any other events, so getting heads thrice will have the probability of 1/2 * 1/2 * 1/2

here if the outcomes are independent of one another, the probability will be 23 * 23 * 23 or 23^3

[obtuse pebble]

@random egret Has your question been resolved?

[random egret]

No, every 23 numbers, the color changes in the 255. 255 = White. But, 232 = Azure. So, what I'm saying is more along the lines of 255/23 = 11. So 11 x 11 x 11. I think that would be the correct formula?

[slim surge]

I don't get what you mean exactly
<@&286206848099549185>

[random egret]

255 (White) - 23 = 232 (Azure)

255 (Total Colors) / 23 (The amount of numbers it takes to change a color) = 11 (Total number of character variations)

(Red) 11 x (Blue) 11 x (Green) 11 = 1331

Do you understand? And does this sound correct?

[random egret]

<@&286206848099549185>

[timid silo]

Can we just say 3 colors, 765 balls and of each color 255?

[random egret]

We can't say 765 balls, but, the rest is correct. I'm looking at the RGB Section: https://www.color-blindness.com/color-name-hue/ . Every 22 digits, the color changes, except for Black (0), and White (255). How many colors would there be? I'm having trouble understanding whether I should say every 22 digits (sole color), every 23 digits (sole color and the first digit of the next color), or every 24 digits (sole color, last digit of the previous color, and first digit of the next color). I am deeply sorry if this is confusing.

[exotic minnow]

So essentially every green colors have 252/32 rounded down colors in them and you need to find how many colors can be made off of them?
If yes then you have a set containing total of 33 colors and you need to find all 2-permutations of it. Also to note I am assuming repetition is disallowed so (red,green) is the same as (green,red).
I did a quick check and the formula for this is (n over k) where n is the number of objects in the set and k the number of objects taken (in this case 2). @random egret
I hope I did not mess something up, been a while since I did combinatorics xDD

This might be helpful,
Translation is
Number of choices of k objects from the set of n objects is given by the following table.
Rows are 1.repetition allowed ,2. repetition dissallowed
Columns 1. order important, 2. order not important

[random egret]

Please, hold on while I grab dinner!

[random egret]

"n is the number of objects in the set" so n = 765?

[drifting wraith]

it's 11^3 you got it right

[exotic minnow]

Is it not 33? Like there are 11 green colors, 11 red and 11 blue so that is 33?

[drifting wraith]

as long as it's really 11 and you didn't off by one it

[exotic minnow]

Imma read it again xD

If I am not missing something there is 11 colors in the set of green colors and 11 in red and 11 in blue. To get to how many of all colors there are we add how many are in the individual ones.

[random egret]

"To get to how many of all colors there are we add how many are in the individual ones." Can you explain this

[drifting wraith]

they didn't explain much, your interpretation is equally valid

[random egret]

Also, I just subtracted (255-2) because of black (strictly being 0) and white (strictly being 255), and I got 253 (obviously), but, when divided by 22 (which is how many numbers each color has), it = 11.5

[drifting wraith]

there are 256 numbers, you're supposed to get 254 when you subtract

[exotic minnow]

n is the total number of elements in the set of all colors, which we need to do our formula business. To get how many actual colors are in the set of Green, Blue, Red we do the calculation that you did and get 11 of each. So the total number of any color is 11+11+11.

Also I messed up xD

[random egret]

There are 255 of each R, G, and B: https://www.color-blindness.com/color-name-hue/

[exotic minnow]

You not only need 2 permutation you need all permutations up until 33.

[drifting wraith]

no there's 256

[exotic minnow]

So it would be sum k=2 to 33 (n over k)

[random egret]

So n is what we are looking for? There is actually 22 numbers in each color, my original equation was wrong. 33 numbers in each color doesn't sound right.

[exotic minnow]

Sure if 22 is of each color than 22*3 is n.
33 was total in what I said assuming 11 is correct.

Do note that n is not the answer to your question it is a necessary step to get to the answer of your question.

Well not necessary but necessary to the way we are solving it rn

[drifting wraith]

if you decided that 11 colors exist, it's 11^3

but with 23 per color it doesn't divide neatly, because there are 256 numbers, so 254 after you drop the ends

[random egret]

I believe we have deduced that in order to calculate the amount of possible colors; we need to multiply 12 x 12 x 12. Because I'm starting to get a little lost.

I got 1728

[exotic minnow]

My reasoning is:
There are 22 in each set of colors, blue, green and red which means that we have 22*3 total colors.
We know that repetition is not allowed (aka (green, red) and (red,green) count as 1 not 2) meaning order does not matter.
We know that (red,blue), (blue,red) is the same as (blue,red),(red,blue).
Thus our formula is n over k.
We also know that k goes from 2 to n so the final formula is sum k=2 to n n over k

[drifting wraith]

master oogway decided to count 2-permutations for some reason

[exotic minnow]

The reason k goes from 2 to n is because possible colors can be made with any of i-premutations i e 2,3,4,5,6

[drifting wraith]

like just out of the blue

[exotic minnow]

Well, yeah. You can make a color from 2-permutations that you couldn't with 3,4.... permutations

[random egret]

"There are 22 in each set of colors, blue, green and red which means that we have 22*3 total colors." @exotic minnow No, we need to divide 253 by 22, and then multiply it by 3.

[drifting wraith]

i don't get it

[exotic minnow]

AHA, ok then there are 11*3 total

[random egret]

No, no, no. There are 12 x 3 total.

[exotic minnow]

The question is how many colors are possible to create from named colors which means all non trivial color combinations. Non trivial combinations are all permutations except 1. I think this can also be solved with multiset but eh.

[exotic minnow]

Sure, I suppose xD

[drifting wraith]

ah ok

thanks

so like, 33×22/2

any hue, then any "different" hue, and you get a color like Azure Slate

[random egret]

Just to come to an understanding; I don't understand your vernacular. Secondly, what is your reasoning? (In simplest terms please)

[drifting wraith]

they are making a color name from 2 words

[exotic minnow]

For 11?
I took 252 and divided it by 22 got 11,5 which could be either 11 or 12 depending on how one interprets the line.

Like if after then last "allowed" addition the next is a new color but shortened then it is 12 if it is not a new color then it is 11.

I'm not completely sure which is more reasonable but probably 12.

[random egret]

"Like if after then last "allowed" addition the next is a new color but shortened then it is 12 if it is not a new color then it is 11." This makes no sense to me, please speak in simpler terms.

[drifting wraith]

we can't decide for you how many colors you have

[exotic minnow]

We have 22*11 colors and change. If the change is a new color then 12 is the total if the change is not a new color then it is 11.

change here is 0.5

[drifting wraith]

it's an arbitrary decision if you end up dividing it into 12 or 11

you said you don't want 0 or 255, there's still 254 that you may count as 12th color or not

[random egret]

I still don't understand what you're saying, Master Oogway.

[exotic minnow]

Do you understand what happens after we know if it is 11 or 12?

[random egret]

You exponentiate

[exotic minnow]

For what reason?

[random egret]

Wait @drifting wraith , I forgot to count 0 as a number didn't I?

[drifting wraith]

that was very silly let's forget it happened

[exotic minnow]

Are repetitions allowed here?

They do not seem to be. A combination of (red, green) seems the same as (green, red)

And order does not seem important either.

[drifting wraith]

this is 22 per color

12 rows

[exotic minnow]

so we have 22*3?

Lets say that is n. We then put n into
sum k=2 to n n over k and we get our result!

[drifting wraith]

we don't have anything

[exotic minnow]

[random egret]

Aren't we supposed to round up when it is 0.5, that's what I was taught in gradeschool?

[drifting wraith]

you're deciding how many colors you have, you can't skip the deciding by hiding behind arithmetics

[random egret]

? Okay

[exotic minnow]

I just re read. Color changes every 23 so the change is a new color.

so n is 12*3. We agree on that? @random egret

[drifting wraith]

you mean 12?

[exotic minnow]

yeah sry

fixed

[random egret]

Ultimately I ended up with 1520.875

(1521)

[drifting wraith]

12^3 = 1728

[exotic minnow]

why to the power of 3?

[drifting wraith]

we pick one value for r g and b

[random egret]

• 11.5

I think I found my answer guys.
Thank you to everyone who helped!

[exotic minnow]

we got 12 to be how many different colors of g,r and b are. So total of different colors is g+r+b which is 12+12+12.

[drifting wraith]

you still pick one of each

a color is 3 values of r,g,b

[exotic minnow]

n^k allows for repetition and is order dependent which we do not allow even if n is 12.

[drifting wraith]

yep

[exotic minnow]

So the formula is inappropriate.

[drifting wraith]

repetition and order dependent

just what we want

well, i want

[exotic minnow]

We do?
I thought we count (r,g) and (g,r) as 1 not 2

[drifting wraith]

we never had any consensus on what we want

[exotic minnow]

like combining r and g is the same as combining g and r

[drifting wraith]

we don't really combine colors

we just have numbers for the r,g,b we pick one for each, there are no colors to combine initially

they didn't explain anything, so it's not about misunderstanding