help-43

[quartz yoke]

can someone please check my work

[quartz yoke]

<@&286206848099549185>

[compact pewter]

@quartz yoke Has your question been resolved?

[feral flicker]

Looks good so far. You still need to evaluate the results for c).

[quartz yoke]

im just having trouble understanding fully what the initial matrix is

in there do i have the probabilities or the population

[feral flicker]

Actually, I was mistaken. For part c), you multiply the state matrix by the population, [1400 600]^T.

On the first assignment, 1400 students received an A and 600 students did not receive an A.

[quartz yoke]

why did you set it up like that

Look how my book does it

[feral flicker]

$\begin{bmatrix}0.7&0.1\0.3&0.9\end{bmatrix}\cdot \begin{bmatrix}1400\600\end{bmatrix}$

[boreal girder]

Kookiemon

[quartz yoke]

But look

@feral flicker

[feral flicker]

In that example, the population is given as a percentage. Each iteration will return a percentage.

In your problem, the population is given as integers and will return integers, meaning you will need to round the results.

[quartz yoke]

Well the amount of subscribers are integers

I don’t understand what you mean

[feral flicker]

That's an odd way of doing that. I'm almost inclined to believe that was a major typo by the author/editor.

[quartz yoke]

Really?

Bro I’m so confused

[feral flicker]

You may want to check the website of the book to see if there is any errata.

[quartz yoke]

I’ve been stuck on this for days now just trying to figure out the initial matrix

[feral flicker]

In the example, x_0 should be [15000 20000 65000]^T, and not written as decimals.

[quartz yoke]

idek at this point

People didnt tell me that before

[feral flicker]

You could also ask your instructor if there was a typo in the text in Example 2.

[quartz yoke]

my instructor did it the same way

[feral flicker]

In this example, all of the given population data works out to percentages so you can technically convert everything to a percentage.

15000/100000 = 0.15, 20000/100000 = 0.20, 65000/100000 = 0.65

[quartz yoke]

and my work?

Why is does it not work

[feral flicker]

If you are looking to do the same for part c), you could use [1400/2000, 600/2000]^T. That will give you the percentage of the student population that got an A and did not get an A.

That would return percentages of the population.

[quartz yoke]

Oh

[feral flicker]

It seems odd to me because Example 2 specifically asked "How many subscribers will each company have in one year?" which seems like it is asking for an actual value, not a percentage. 🤔

[quartz yoke]

Btw why are you doing it horizontally

Isn’t it supposed to be vertical

[feral flicker]

The ^T indicates that the matrix should be transposed, ie. read as vertical. It is a common notation that you will see in books.

It's something that will almost certainly have been mentioned in the introduction or chapter 1 of your linear algebra book.

[quartz yoke]

Yeah it was

You’re right

But why not just write it vertical in the first place

[feral flicker]

Because that would require latex and is unnecessary for what I was writing.

[quartz yoke]

So is my work correct though? Just the percentage thing?

[feral flicker]

,calc 1400/2000

[boreal girder]

Result:

0.7

[feral flicker]

,calc 600/2000

[boreal girder]

Result:

0.3

[feral flicker]

Yes.

Part c) ask for the distribution so percentages would be an appropriate answer.

[compact pewter]

@quartz yoke Has your question been resolved?

[compact pewter]

@quartz yoke Has your question been resolved?

[quartz yoke]

<@&286206848099549185> is anyone free

[quartz yoke]

@latent lodge can someone please let me know if im doing this right

[remote bane]

markov chains?

[compact pewter]

@quartz yoke Has your question been resolved?

[compact pewter]

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[mighty gate]

focus on the the velocity v time graph in green; it is my professor’s key to the problem posed in black

shouldn’t the velocity end at 4 m/s at 7s???

rise over run to find acceleration from -2m/s to 4 m/s:

(6m/s) / (3s) = 2 m/s²

the velocity would increase at the same rate as it did when a = -2m/s^2

[lavish gull]

correct. (unless your prof has some random hidden assumption or smth)

[mighty gate]

you’re positive?

why would my professor slip up like this

[rigid perch]

,calc 2 + -22 + 23

[boreal girder]

Result:

4

[mighty gate]

that’s so annoying

[lavish gull]

if you are asking if i am positive you are right, yes

[mighty gate]

what was this about?

[lavish gull]

i mean, just look at your prof's velocity-time graph at t = 5s

it's not even an integer

[rigid perch]

just confirming the calculation

the problem is that the professor seems to have accidentally made the slope of the graph +/- 1 m/s^2 rather than +/- 2 m/s^2

[lavish gull]

makes no sense either

because at t = 5s the velocity is not an integer (it's somewhere in between 0 and -1m/s)

plus if it was, then the graph should have ended at 1m/s, not 2

[mighty gate]

lol yes

[lavish gull]

to go from -2 to 2 in 3s means the slope is 1.33m/s^2, which.... I have no idea how

[mighty gate]

yet the position v time graph looks correct

I was so worried. thought my intuition failed me big time

[lavish gull]

time to punt the question back to your prof

[mighty gate]

lmao

thanks

.close

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[dense dagger]

What are you trying to do with that?

Solve what

9r^2+72r-25=0?

I asked "solve what" since 9r^2 + 72r - 25 isn't a complete question. It's just an expression, I don't know what you want to do with it if you don't tell me. Anyways, to factor it, you can use the ac method. a = 9, b = 72, c = -25 (verify this). Find two numbers that multiply to a*c AND add up to b

[neat ravine]

u get r = -75 or 3

[dense dagger]

!noans but you get 75 and -3, not -75 and 3

[compact pewter]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[twilit flax]

.close

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[cloud spruce]

Could someone please help me with identifying and classifying discontinuities in graphs?

Here are answers provided but my teacher but I am genuinely confused on what to do 😭

[cloud spruce]

I understand where the first 2 lines come from, but I don't get the As x -> -2 part

[mighty gate]

why are these solutions??

[compact pewter]

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[plain marlin]

Hey can anyone please help with 2nd part i agree that their answer is correct but, im getting 6^(n-1) from my method can anyone tell what i have done wrong

[plain marlin]

It is certain that one roll will be 6

So that leaves us with n-1 rolls which are random

So wouldn't no. Of outcomes be 6^(n-1) ?

[kind viper]

nope

what you're actually counting with 6^(n-1) is the outcomes where specifically the first roll is 6

[cobalt owl]

i was also just solving cengage rn lol

[plain marlin]

Lol I just started this ch

[kind viper]

imagine this for n=2

your count would give only 6 possibilities: 61, 62, 63, ..., 66

and you miss 16, 26, 36, 46, 56

[plain marlin]

Ohh I got it thank you very much

So we cant just fix a certain position for 6 then

[kind viper]

indeed we can't

[plain marlin]

Okk so do I close the channel now ?

[kind viper]

if you got nothing else to ask then yes

[plain marlin]

.close

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[shadow scroll]

this is very short and simple
10 people sit in a row of seats 7 adults and 3 kids how many possible ways could they sit in those seats in a way that 2 or more kids cant be beside each other

[shadow scroll]

my idea so far

is

[silver flume]

6-7 or something

[shadow scroll]

to get the total ammount without the restriction then to subtract with possible ways 2 kids or 3 kids sit togehet

[shadow scroll]

GET OUR

[chilly basalt]

good insight

Have you tried working towards this direction?

[shadow scroll]

yes

i did

[chilly basalt]

Might want to share it w us

[shadow scroll]

10! - (8!×3!) -(9!×2!)

this was my original method but i can feel i know its flawed

idk how to

convey fhe 2nd one

where only 2 kids sit beside each other

but also excluding all 3 sitting together

A K A K A K A

hm

ok now im tryna visualize the possible

ways

K A K A K A A A A technically also works

[elfin finch]

One way to think of it would be putting the kids between the adults ( or on the 2 sides as well)

[shadow scroll]

without inidvidually like

trying how do i

do that

[elfin finch]

Why would you need to convey that?

[shadow scroll]

bc at thhat orginal way im only account for 2 sitting together and the rest which is 1 kid and 7 adults

there could be a possibility that the 1 kid sits beside the griup of 2 kids

[elfin finch]

Ok lets think of the spaces between the adults (or on the 2 sides)
—A—A—A—A—A—A—A—

[elfin finch]

Yeah thats already in the orginal way

[shadow scroll]

mhm

[elfin finch]

Now you need to pick 3 spots to put the kids in right

[shadow scroll]

yes

8 spots open

8×7×6???

for the kids

that and

THE ADULTS

which is just like

8!×8!?

[elfin finch]

No? There are 7 adults and they switch places with eachother

So 7! For the adults

[shadow scroll]

oh wait yeah my bad

8!×7! then?

[compact pewter]

@shadow scroll Has your question been resolved?

[elfin finch]

You mean 8×7×6×7!

[quartz yoke]

is that the answer?

@elfin finch

[elfin finch]

I mean its what i would answer

[quartz yoke]

could you explain it?

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[ebon wharf]

I got this wrong but I dont understand why?

[kind viper]

spotted it

one moment

[ebon wharf]

okayy

[ebon wharf]

OHHHHHHHHHHHH

😭

the method to do it is right though right?

finding the equation of the tangent to the graph and then substituting the point to solve for constant k

[kind viper]

your planning is ok, the execution contains at least one error

[ebon wharf]

i c

thank you so much

is it fine if i keep this open and reattempt the question then post my new working?

[kind viper]

sure ig

[ebon wharf]

can someone check this working pls

this is the question

nvm i checked it

thanks

.close

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[humble kraken]

im having trouble figuring out sequence B
i have no idea where to even start

[spark folio]

1/2 is same as 5/10

[humble kraken]

yeah

[spark folio]

does that help?

[humble kraken]

its adding one to the numerator and denominator

[spark folio]

yep

[humble kraken]

but i cant add 1/1 thats not gonna work

[spark folio]

consider the numerator and denominator as separate sequences

you got 1,2,3,4,5,...

[humble kraken]

ohh

so

sequence of numerator over sequence of denominator?

[spark folio]

well, nth term of sequence of numerators over the nth term of the sequence of denominators

[humble kraken]

so how would the sequence look like??

[spark folio]

what does 1,2,3,4,... look like? what would be its nth term?

and what does 6,7,8,9,... look like? what would be the nth term for that?

[humble kraken]

thats just n i think

[spark folio]

yep

[humble kraken]

this is n+5

so n/(n+5)?

[spark folio]

right, so what would the fraction that has these two terms would look like?

[spark folio]

its a fraction, one is numerator and other is denominator

[cursive current]

Hi

[spark folio]

its not a product

[humble kraken]

i fixed it mb

[humble kraken]

hello

[spark folio]

yea, n/(n+5)

[humble kraken]

thats the answer?

[spark folio]

yep

[humble kraken]

oh okay

ty!

.close

[cursive current]

Everyone my maths exam overrer

[compact pewter]

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[spark folio]

!redir

[compact pewter]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[cursive current]

Means

[spark folio]

go to #discussion

this category of channels is for helping ppl

[compact pewter]

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[delicate musk]

I dont understand where the contradiction is...

[delicate musk]

.close

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[delicate musk]

nvm

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[rocky path]

A fair die is rolled repeatedly until the sum first exceeds 6. Find the probability that the sum is exactly 7, given that the number of rolls are even

[cursive harbor]

Well lucky for us the sample space is small enough to write explicitly

[rocky path]

The real question was "A fair die is rolled repeatedly until the sum first exceeds 12.Find the probability that the sum is exactly 13,given that the number of rolls taken is even."
I just want to know the method because in this one I can't just count the cases

[chilly basalt]

It’s alright, good to begin with a smaller sample

[rocky path]

[rocky path]

The sample space in this question is 161 for me, but this was just pure counting.

Anybody😭

[pine thicket]

Well tell your method

[vivid breach]

surely you can list all 161 of the sample spaces

[rocky path]

Considering rolls=2 we can say that x+y is less than equal to 6 where x and y belongs to (1, 6)

Subtract it from the total cases

[pine thicket]

And then?

[rocky path]

You get 21 cases which exceeds 6

Similarly for 4 rolls and 6 rolls and then you have to stop

[pine thicket]

So you count the number of cases where the sum of rolls is less that or equal to 6

And then subtract them from total caes?

[rocky path]

Yes, I want to know any alternative which can tackle cases which are huge like 4 rolls

[pine thicket]

But this means you count cases like 5, 5

The sum is 10 which is greater than 6

[rocky path]

Yes

That's just the sample space

[pine thicket]

But how and when do you narrow down to sum = 7

[rocky path]

That's the problem

What would you do in this question?

[pine thicket]

I guess counting all the cases would be the easiest

I would recommend going in an order like 6 to 1

That would make it easier

Like let's say the first role is 6

So now sum should be 7 so now we can only do 1

As we rolled 2 times this is a valid case

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[rocky path]

.reopen

[compact pewter]

✅ Original question: #help-43 message

[rocky path]

Even if the data was 12?

[pine thicket]

Yup

Many cases will start to overlap

[rocky path]

I'm done

[pine thicket]

Lmao

[rocky path]

.close

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[vivid breach]

@rocky path for 4 rolls, split it apart into two pairs of rolls then consider the cases for each

[pine thicket]

.reopen

[compact pewter]

✅ Original question: #help-43 message

[olive bear]

is it 2/7?

[compact pewter]

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[robust robin]

hii i have a question regarding newton's third law

soo i always assume that the first action is positive
and the second one will be negative (especially when pushed back)

however idk if this is really correct? bcz when i tried searching abt the principles it just says that it's up to the person to assign the signs..

but then the answer will be completely different so im rlly confused 🙏 help please

[upper bane]

!original, please (also, while allowed, do note that there is a physics server in #old-network you can use)

[compact pewter]

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

[robust robin]

ohhh thank you i will join that server! alsooo here is an example of a practice problem A 40 g superball strikes a wall with a velocity of 10 m/s that is normal to the wall. It bounces away at a velocity of 7 m/s, still normal to the wall. What is the ball’s change in momentum? If the bounce lasted 0.1 s, what is the force between the ball and the wall?

10 would be positive and 7 would be negative?

[upper bane]

a picture would be nice for next time

if you define "towards the wall" as positive and "away from the wall" as negative, you are right

[robust robin]

thank youuu it's always like this right and i shouldn't interchange the signss?

[upper bane]

stick with one definition for the whole problem. you can of course decide that "away from the wall" is positive, but you cannot assume one thing on one line and another thing the next

if you decide on a convention, stick with it until the problem is completely done with, and that means every last subquestion to do with the problem

[rocky path]

I don't have the answer, nor it's anywhere online

[robust robin]

the answer is always in absolute valueee?

[upper bane]

depends on context

[robust robin]

waittt husufhudhgufg;fg;; in the worksheet it gives a negative answer -0.68 kg*m/s; -0.68 N

so liek u have to be precise with the signs?

[upper bane]

force is a vector, so yes

[robust robin]

its #9 on this pdf https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://www.sfponline.org/uploads/71/momentum_and_impulse.pdf&ved=2ahUKEwj73bbik5SQAxWWfvUHHdJcJJMQFnoECA8QAQ&sqi=2&usg=AOvVaw3hhS_FXV2gqtgJfH2aVbeH

[upper bane]

!nopdfs, please

[compact pewter]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

[robust robin]

okay22

[robust robin]

im sorry im self learning so just one last question, then doesnt that mean that the signs from earlier has to be +10 and -7, because otherwise it will give positive 0.68 and will be wrong

[upper bane]

it will be wrong if you did not state that you are defining "away from the wall" as positive

[compact pewter]

@robust robin Has your question been resolved?

[robust robin]

okayy! slr i went to cook. thank you!!!!!

[compact pewter]

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[brisk fox]

hiiiii

[kind viper]

hello and welcome to the server

you have just claimed a help channel

do you have a math question you want help with?

if not, you should close this and go to #discussion or #chill

[compact pewter]

@brisk fox Has your question been resolved?

[brisk fox]

@compact pewter leave

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[steep oak]

help

[young raft]

send your problem directly

[steep oak]

ok wait

[young raft]

which question do you need help with

[steep oak]

everything

[young raft]

i knew you were gonna say that lol

no one will solve the entire homework for you

but one question is fine

[young raft]

are you scheme a or scheme b

[compact pewter]

@steep oak Has your question been resolved?

[steep oak]

both

[steep oak]

I want to know how to solve it

not the answer in jtself

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[serene coral]

Can someone help me revise on systems of equations?

[chilly basalt]

just post :))

[serene coral]

like in general

it's like i can't see the pattern

[chilly basalt]

Do you have an example?

[serene coral]

For example: y^2 +xy^2=6x^2, (xy)^2+1=5x^2

[covert creek]

$\begin{cases} y^2+xy^2=6x^2 \ (xy)^2+1=5x^2 \end{cases}$

[boreal girder]

Fionna The Unemployed

[paper anchor]

Well, isolate y^2 in the first one

find its value in terms of x

Then substitute it into y^2 in the second one

Then evaluate until you get the value of x

Then substitute it into the first one and evaluate that to find y

[serene coral]

so i have $y^2 = \frac{6x^2}{1+x}$

[boreal girder]

Thomas

[paper anchor]

Yes

[serene coral]

so we have $x^2 \frac{6x^2}{1+x} + 1 = 5x^2$

[boreal girder]

Thomas

[paper anchor]

Yes

[serene coral]

look ugly

[paper anchor]

It does ngl

Where did you source it from?

[serene coral]

multiplying them all by the denominator

any better ways?

[paper anchor]

No ig, but where did you get it from

This is giving a quartic equation, no way anybody would be able to solve it unless it factors to a quadratic, or they're a madlad

[serene coral]

yeah, i just feel discouraged when a quartic or higher comes up

[paper anchor]

I mean if it was ax^4 + bx^2 + c = 0, then we could have done it

[serene coral]

well let's try, shall we?

$6x^4+x+1=5x^3+5x^2$

[boreal girder]

Thomas

[paper anchor]

The only way I see where we could solve it is if it factors to a quadratic

There's no other formula

Hit and trial maybe, but that should be the last resort

[serene coral]

,w rational roots theorem

[boreal girder]

Sending query to Wolfram Alpha, please wait.

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[serene coral]

is the bot broken?

[paper anchor]

I don't think it'd give that, or maybe it does, not sure

(3x^2 + ax + 1)(2x^2 + bx + 1)

[serene coral]

but nevertheless does that theorem help?

[paper anchor]

Not sure mate, that's honestly the first time I've heard of it

Or use graph, that's the easiest

[serene coral]

let me cook w rrt

so the possible rational roots are factors of 1 over factors of 6: $\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}$

[boreal girder]

Thomas

[serene coral]

so now we try out which one works

[paper anchor]

There is a value there, yes

[serene coral]

after some testing, i've found that 1/2 is indeed a value

[paper anchor]

Yes, that one is correct

There is one more

[serene coral]

so now we divide by x - 1/2 or 2x-1

which one's better?

[paper anchor]

2x - 1 sounds good

[serene coral]

any other ways except long division?

[paper anchor]

I don't think so

[serene coral]

welp -_-

okay so the quotient is $3x^3-x^2-3x-1$

[boreal girder]

Thomas

[paper anchor]

Okay, can you divide it once more?

[serene coral]

by 2x-1?

[paper anchor]

yes

[serene coral]

prob not but i'll try

i mean 3/2 isn't a great start

[paper anchor]

Alright, that'd be easier to solve

[serene coral]

so how do we factor this?

[paper anchor]

Wait let me check

,w (6x^4 -5x^3 -5x^2+x+1)/(2x - 1)

[boreal girder]

Results provided by WolframAlpha

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Requested by doctorstrangejr

[paper anchor]

Well, that's x

[serene coral]

looks...not so enticing

[paper anchor]

But it is indeed correct, I checked with graph

[serene coral]

,w cubic formula

[boreal girder]

Sending query to Wolfram Alpha, please wait.

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[serene coral]

why me everytime

doesn't let me send queries

[paper anchor]

Dude you sure you picked this question from book, like what's your level of linear?

[serene coral]

wdym "level of linear"? (english isn't my first language, so bear w me)

[paper anchor]

Like are you in school, or did you start college, or are you studying for advanced linears

[serene coral]

am in 9th grade in my curriculum

what does 9th grade teach in yours?

[paper anchor]

Okay dude no way they'd ask you to solve a quartic

[paper anchor]

Similar stuff, you Indian by any chance?

[serene coral]

no but we get the same level of "supposed smartness" as them (maybe less but yeah)

[paper anchor]

Ahhh okay, well we are taught similar stuff here, but cubic is the maximum we go

[serene coral]

where are you?

[paper anchor]

Maybe you chose the question from the internet, can you send one from your book where you're stuck at

[paper anchor]

I am, in fact, indian

[serene coral]

not rlly, i got this problem from a friend's book, and i only wrote down that and a few other questions

[paper anchor]

Well it is pretty advanced for grade 9

And that's the easiest method among the ones there are to solve this

And it leads to a quartic

Maybe for olympiads it checks, but they don't ask stuff like this ig

[serene coral]

so how do we solve this

lemme try again w the query

,w cubic formula

[boreal girder]

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Requested by thomas2406

[paper anchor]

Well you need a root for solving cubics

And divide the cubic by (x - your root)

And you tried that

[serene coral]

well i found this:

[paper anchor]

Dude

IG that's the Quadratic formula for cubes, and there's no way anybody uses or remember this

[serene coral]

so is there any other way that doesn't involve my brain exploding?

[paper anchor]

Nope, with cubic I don't think so

The best way is dividing by a factor

[toxic cipher]

cardano's method

you make the cubic depressed which makes it easier to solve

[serene coral]

,w cardano's method

[boreal girder]

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[paper anchor]

,w solve [3x^3 - x^2 - 3x - 1, x]

[boreal girder]

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[paper anchor]

Whoops

[paper anchor]

Well, nobody likes you cubic, boooooo

[serene coral]

,w cardano's method

[boreal girder]

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[cursive harbor]

The cubic is dead, hope you're happy

[paper anchor]

Ehhhh, I mean Cardano would maybe too advanced for your level

[cursive harbor]

In grade 9? Definitely

[paper anchor]

omg no I did too much, he didn't specify how much😭

[toxic cipher]

i mean it isn't that hard to just memorise and apply it

[cursive harbor]

The way to solve cubics is to guess one root and make it a quadratic

[serene coral]

the other 2 imaginary roots also look like sh..

[cursive harbor]

You can solve a quadratic using the formula

Also you can say shit it's fine

[serene coral]

okay

[paper anchor]

Well we have one root

[serene coral]

we could use vieta' theorem to construct the quadratic

[toxic cipher]

you have a cubic now right?

okay then try RRT on it

[serene coral]

i used rrt to get this

[toxic cipher]

if im not wrong at least one of the roots of a cubic has to be real

[serene coral]

the original was a quartic

[serene coral]

yes but they are NOT pretty

i calc'd them

[toxic cipher]

have you tried ALL of the root candidates?

oh dang okay

[paper anchor]

There are only 2, I've checked on graph

A strange quartic we have

[toxic cipher]

maybe you did something wrong with getting to here?

[serene coral]

prob not

[toxic cipher]

like a careless mistake or something

[serene coral]

tested it

[paper anchor]

I mean the 1 root he was getting from this was right

[toxic cipher]

well uhh the second real root does not look pretty at all

[serene coral]

where did you get this?

[toxic cipher]

wolfram alpha on this equation

[paper anchor]

Yeah it's 1.299 something

[toxic cipher]

you could try drawing a graph of the cubic 💀

[paper anchor]

They probably used the cubic formula (the one similar to the quadratic formula just 10 times longer)

[toxic cipher]

could get you in the single decimal place ballpark

[paper anchor]

We could just round it off to 1.3

[toxic cipher]

are you allowed to use a calculator for the question?

[paper anchor]

A graphing calculator perchance

[serene coral]

i mean yeah but i still need to find the exact value

[paper anchor]

Also you reckon this is 9th level @toxic cipher ?

[toxic cipher]

i mean RRT and polynomial long division isnt out of the ordinary

but cardano's method or wtv other method you use to solve this cubic most definitely isnt

if this was an olympiad question or a class for advanced students i wouldnt see anything wrong with it though

[paper anchor]

True, the first two are fairly understandable (I guess RRT was called something else here), but there's no other way

[paper anchor]

TBH that's what I am banking at, olympiad question, this one seems odd for olympiads but could be taught in the classes

[serene coral]

so is there any "normal" way to solve this?

[toxic cipher]

the cubic can also written like this:
$x^{2}\left(3x-1\right)-1\left(3x+1\right)$
which is awfully close to being factorisable (by grouping)

[paper anchor]

Nope, divide by factor maybe

[toxic cipher]

maybe this suggests an error in the question?

[toxic cipher]

thing is, the cubic only has one real factor which is a monstrous expression involving cube roots and such

[serene coral]

you can check the original question if you want

[toxic cipher]

well you said here that you transcribed it from a friend

maybe you copied wrongly?

[serene coral]

not rlly, we did some work in class but got nowhere

[toxic cipher]

okay well

in any case, i would suggest cardano's method as your best bet

for exact values of x

which is what you said you wanted earlier. if you would settle for good approximations of x, there are a few methods

[paper anchor]

Could you send an image of the question

[paper anchor]

Yeah but he could keep on dividing by the factor ig, but he says it gets ugly, which is understandable for this kind of answer

[serene coral]

it's just the equation set

[toxic cipher]

wdym "keep dividing by the factor"

[paper anchor]

The insane thing is that this is just the value of x, you're gonna have to solve a quadratic after this to solve y, way too long of a question

[paper anchor]

until it reaches a quadratic, dividing the cubic by (2x - 1)

I mean we got this cubic from a quartic in the first place

[toxic cipher]

the cubic doesnt have (2x-1) as a factor

[paper anchor]

what

Wait a second

[toxic cipher]

$3x^2-x^2-3x-1$

[serene coral]

yeah

[toxic cipher]

this cubic

[serene coral]

i've been telling you that dividing this doesn't end well

[paper anchor]

oh yeah mb

We've already factored out the (2x-1)

which leaves only one value of x left

[serene coral]

and that's the monster we've been focused on

[paper anchor]

Yeah dude no other way to solve this, as far as I know

Cardano's method is your only bet

[toxic cipher]

@serene coral https://brilliant.org/wiki/cardano-method/
https://www.math.ucdavis.edu/~kkreith/tutorials/sample.lesson/cardano.html
read up

[serene coral]

so what to do after depession

[paper anchor]

TBH you could have solved the quartic by depressing the cubic term too, but that would have been a way bigger headache than this, not saying that this isn't one though

[toxic cipher]

if im not wrong depressing the quartic will just give you a cubic equation to solve

like

[serene coral]

what about depressing the cubic?

[toxic cipher]

after you do a bunch of steps

[paper anchor]

Maybe yeah

[toxic cipher]

wdym maybe 😭

[paper anchor]

And then depress it again, we'll reach here only

[paper anchor]

I mean who has experience with solving a quartic, I just saw how it works

[toxic cipher]

i have solved like 3 quartics by hand in my entire life 💀

[paper anchor]

You madlad💀 🫡

[toxic cipher]

not counting solving by grouping factorisation

[serene coral]

so after depressing: $27t^3-28t-\frac{326}{27}=0$ with x = t + 1/9

if i'm correct

[boreal girder]

Thomas

[paper anchor]

Oh yeah I've done them multiple times, but solving a pure quartic is insanity

[serene coral]

so now what?

[paper anchor]

send the constant to the other side

then take t common from LHS

[serene coral]

oh gtg

prob will come back later

.close

[compact pewter]

Channel closed

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[compact pewter]

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[kind viper]

,rccw

[kind viper]

!status

[compact pewter]

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

[pseudo wasp]

No idea how to do

so 1

[kind viper]

there is an assumption you have to make here: each man can produce the same rate of work as any other man, and each boy the same as any other boy

think about work rates rather than completion time

[pseudo wasp]

Well........

[kind viper]

so let m be the work rate of a man and b the work rate of a boy, both in fractions of the task per day

[pseudo wasp]

I don't get it lmao

[pseudo wasp]

So you want me to get individual?

Like each mans?

work rate

[kind viper]

idk what you mean by "get individual"

[pseudo wasp]

you want me to find

[kind viper]

we're saying all the men work at the same rate as each other

and all the boys work at the same rate as each other

and i wasnt done yet

[pseudo wasp]

Okie.......

https://tenor.com/view/crypto-i-cant-say-i-wont-say-i-wont-tell-anyone-im-not-telling-gif-6012111083505594710

[kind viper]

6 men and 8 boys take 10 days to finish the task, which means that all combined they complete 1/10 of the task per day.

this means $6m + 8b = \frac{1}{10}$.

[boreal girder]

Ann

[kind viper]

do you understand?

[pseudo wasp]

yup

:)

[kind viper]

ok, there's one more equation to write

then you solve that as a system

[pseudo wasp]

Hmmm

[kind viper]

26 men and 48 boys can do the same task in 2 days

[pseudo wasp]

I got it.......

26m+48b =1/2

13m+24b = 1/4

and then I have to solve..... for m and b

right.......?

but I am getting negative work for men

Wait imma send you ma work

[kind viper]

hm

[kind viper]

well, i can tell you i'm 100% confident we translated the problem correctly

,rccw

[pseudo wasp]

Oh wait....

I suck

m=1/20

[kind viper]

26m + 48b = 1/2.

there's also a less algebraic way to do this problem

[pseudo wasp]

Tell me , I am all ears

[kind viper]

6 men and 8 boys require 10 days
multiply the work team by 5 and divide the completion time by 5
30 men and 40 boys require 2 days

so a team of 30 men + 40 boys has the same work output as 26 men + 48 boys

[pseudo wasp]

Ok.......

[kind viper]

you can get the second team from the first if you remove 4 men and add 8 boys

therefore 4 men work as fast as 8 boys

[pseudo wasp]

Oh so you are saying like I will get ratio of men and boys

then put in an eq.?

[kind viper]

rather you get the ratio of the work rates of men and boys.

which is to say one man works the same as two boys

[pseudo wasp]

😭

[kind viper]

what.

what's wrong.

[pseudo wasp]

Yes that's what I mean

Nothing

It's just I suck

so I will put this in the original eq.

?

[kind viper]

beating yourself up helps nobody. not me, not you, not anyone.

[kind viper]

forget about equations at all

if you're gonna listen to me explain a non algebraic solution then you're gonna listen to me explaining a NON ALGEBRAIC SOLUTION

[pseudo wasp]

Okie miss

[kind viper]

1 man has the same work output as 2 boys

so 15 men and 20 boys are equivalent to 15 + 20/2 = 25 men
while the team of 26 men and 48 boys is equivalent to 50 men, and these require 2 days to completion

so a team half the size will require double the time...

[pseudo wasp]

Ohhh so replace all boys with men or vice versa

interesting approach

[kind viper]

yes

basically reduce everything to be in terms of workers of the same type

[pseudo wasp]

.close

[compact pewter]

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[compact pewter]

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[flint cliff]

why isnt my calculation giving this answer

[verbal kiln]

!original

[compact pewter]

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

[bleak dock]

$-(0 - \ln(e^0 + 1)$

[boreal girder]

south

[bleak dock]

you forgot the brackets here, so the double negative = positive

[bleak dock]

woops, there should be ) on the end

[quiet sentinel]

this should be correct im pretty sure

[bleak dock]

,w ln 8 - ln(1 + 8) - 0 + ln(1 + e^0)

[boreal girder]

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[bleak dock]

damn it the bot is broken again

[flint cliff]

omg

i used a few mintes ago

[bleak dock]

same

it's been broken for 2 days at this point

but yeah I disagree with the 1.777...

[flint cliff]

the 1.77 is the answer

[quiet sentinel]

its a mistake most likely unless all 3 of us are idiots

[flint cliff]

Exactly

So my answer is correct

Of 0.5.....

[bleak dock]

yeah I confirmed using Wolfram Alpha

[bleak dock]

it's actually ln(16) - ln(9) if you work it out

[flint cliff]

Okay thanks

[bleak dock]

or 2 (ln 4 - ln 3)

[flint cliff]

.close

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[tender pasture]

questions like this we find the lcm when do find the hcf

[verbal kiln]

Are you asking for help with the question?

also hello fellow white hole

[tender pasture]

yes

[tender pasture]

no

in this case they found the lcm

when do we find the hcf

wht would the question ask

[verbal kiln]

You mean GCF?

greatest common factor?

[tender pasture]

yh

[tender pasture]

yes

[compact pewter]

@tender pasture Has your question been resolved?

[tender pasture]

.close

[compact pewter]

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[rugged berry]

Hey, could someone help me with the following problem:

Let $\phi : U \subset \mathbb{R}^{m - 1} \rightarrow \mathbb{R}^m$ be a hyperplane. Show that the outer normal unit vectors are preserved under transformation of the parameter.

In the screenshots one can see how we defined the outer normal unit vectors of a hyperplane as well as the cross product.

I set up a new hyperplane $\tilde{\phi} = \phi \circ T$ where $T$ is a $C^1$ diffeomorphism. Thus the chain rule yields $D\tilde{\phi}(u) = D\phi{T(u)} DT(u)$. Then I plugged this into the formula for the cross product -- hoping that I can maybe pull out the $DT(u)$ somehow in order to get the cross product for the original surface $\phi$ back. This way I could set up a relation between the normal-vector of $\phi$ and $\tilde{\phi}$ (which only differ by the value of the cross-product).

My guess is, that I can maybe pull the determinant apart in a $\mathop{det}(D\phi(T(u), e_j)$ and a $\mathop{det}DT(u)$ part but I could not find a relation between those.

Thanks!

[rugged berry]

(e_j denotes the j'th canonical basis vector)

(and the comma in the determinant just means, that the arguments are concatenated column-wise)

[boreal girder]

Edlingem

[compact pewter]

@rugged berry Has your question been resolved?

[compact pewter]

@rugged berry Has your question been resolved?

[compact pewter]

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[plain marlin]

Hey isnt there something wrong in this? How can n! Be equal to nⁿ

[plain marlin]

And what does n< or =n even mean here

How can n<n

Be possible

[calm trout]

well that's why it doesn't say n<n

it says n__<__n

because n=n

it means n is less than or equal to n

[plain marlin]

Okk but it will not be less than it so we can remove < sign then right ?

[calm trout]

sure but you're proving an inequality

so it doesn't really make a difference

[plain marlin]

n!=nⁿ this is incorrect then right ?

[calm trout]

well your book doesn't say that either

it says <

[plain marlin]

But how can n! Be equal to nⁿ

[calm trout]

it doesn't say it's equal

it says it's less than or equal

also try n=1

[mortal sentinel]

lulu I think you're misreading the text, it doesn't say $n! = n^n$.

[boreal girder]

bvghfgjfg

[plain marlin]

But it says greater than or equal , which means that equality of them is also possible

[calm trout]

less than or equal

and n=1 is when equality happens

[plain marlin]

Oh right it is possible

Okk i got it

[calm trout]

even if it isn't possible, it doesn't make it false

1 < 2 is a true statement

[plain marlin]

Huh

This was my time seeing a statement like this so it seemed incorrect

Okk thank you very much i got it

[calm trout]

[plain marlin]

.close

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[oblique rapids]

how do u solve b

[smoky junco]

do you know what the definition of the mean is?

[oblique rapids]

yes

the sum of all data values divided by the number of data values

[smoky junco]

okay

now there's a new unknown data point

but we know that the mean is at least 85

since we're trying to find the lowest score, we can assume that the mean will be 85

now you can set up an equation to solve for this unknown value

[oblique rapids]

92+89+94+82+93+X divided by 6= 85

[smoky junco]

yes

now you just gotta solve for X

[oblique rapids]

so 85 times 6= 510, 510-450

so the answer is 60

is that right?

[upper bane]

,calc 92+89+94+82+93

[boreal girder]

Result:

450

[upper bane]

,calc 85*6

[boreal girder]

Result:

510

[upper bane]

correct

[oblique rapids]

yayy

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[carmine garden]

so we have $Z = X-\mu / sigma

Z= (834-800)/40 and Z=(778-800)/40

and I use these Z values in the Z table?

so $P(-0.55<Z<0.85)

?

[eager knot]

Z is here your light bulb distr
X is a N(0,1)

You know the table of a N(0,1)

You're asked P(778 ≤ Z ≤ 834)

which is P(778 ≤ (X-mu)/sigma ≤ 834)

which is P(778 ×sigma + mu≤ X ≤ 834 ×sigma + mu)

and this you know it with a N(0,1) table

[quaint quarry]

Or hard mode:
$$p(778\leq x \leq 834) = \int_{778}^{834}{ \frac{1}{\sqrt{2\pi \sigma ^2}}e^{-\frac{(x-800)^2}{2\sigma^2} }dx$$

[boreal girder]

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

[compact pewter]

@carmine garden Has your question been resolved?

[carmine garden]

so what I did , right

[carmine garden]

We use a N(0,1) table

[compact pewter]

@carmine garden Has your question been resolved?

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[terse linden]

Yo guys i really want to understand implication concept

[terse linden]

Why in direct proof we need to assume that P is true then we proof Q

It doesn’t make all the statements we want to proof built as implication?

For example
When we use proof by contradiction to proof that (2)^1/2 is irrational number
When i ask AI what nature of this statement he said implication

How tf is that implication

I mean there one proposition only 2^1/2 is irrational let it P?

[hushed magnet]

P = "normal axioms"

Q = "sqrt(2) is irrational"

in that sense its an implication P => Q

!nogpt btw

[compact pewter]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

@terse linden Has your question been resolved?

[terse linden]

What the axiom then

[hushed magnet]

the normal rules how numbers work

dont think about it too much

[terse linden]

You mean P in every statement is an introduction (definitions, axioms or theorems)??

[compact pewter]

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[pallid gorge]

where does the +4 go?

[pallid gorge]

oh is it cause it'll be + 4 and -4

so they just dont bother putting it in?

[mild sky]

they used the fact that 4x ln(3x) + 4 has antiderivative 4x ln(3x)

[bleak dock]

yeah, the area under y = 4 is a rectangle

oh wait, that's not relevant

[flint wing]

south with the....clutch...?

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[foggy wing]

21k*21l=6174

[toxic cipher]

what's the original question?

[foggy wing]

How to simplify it correctly because i forgor

[foggy wing]

Multiplication of two numbers is 6174 and their greatest common divisor is 21. Calculate all possibile numbers

[noble sorrel]

what can u divide on the both sides?

[toxic cipher]

Okay thank you

[noble sorrel]

do u know the formula

[solemn matrix]

no no

[foggy wing]

I know for addition

[noble sorrel]

that a.b = gcd . lcm?

[solemn matrix]

6174 is divisible by 21^2

[foggy wing]

Because i did example

[solemn matrix]

divide by 21^2 and you basically simplified the equation

[noble sorrel]

you can get the lcm from this

[foggy wing]

But i forgot how to simplify it correctly

[solemn matrix]

multiplication order

does not matter

what you can do is that turn 21k * 21l into (21 * 21) * kl

and 21 * 21 | 6174 as i just divided before

you should be able to simplify this neatly

[foggy wing]

So k*l=14?

[solemn matrix]

ye

[foggy wing]

Ayy thanks that's all

[solemn matrix]

np

but really

what's that for?

!original

[compact pewter]

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

[toxic cipher]

so you have to list down all the factors of 14 now

[bleak dock]

yeah you had the right idea, which is kind of impressive at this stage

it's weird you tripped up on 'basic' algebra

[foggy wing]

Yes

[solemn matrix]

what's the original question?

!original

[compact pewter]

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

[toxic cipher]

@solemn matrix

[bleak dock]

you should congratulate yourself for being able to turn that word problem into algebra

[solemn matrix]

i mean, how do we know that k and l are integers?

[bleak dock]

it's implied here

[solemn matrix]

oh sorry

[bleak dock]

gcd only exists for integers

(it can for polynomials but that's irrelevant)

[toxic cipher]

He wrote down the two numbers in the form $21k$ and $21l$, which is correct

[solemn matrix]

wait

yeah

where $\gcd{k, l} = 1$

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

by looking at the factors of 14

they all have gcd = 1

[solemn matrix]

this one is the crucial one

[foggy wing]

Ok original one is 1.95

[toxic cipher]

so do you know how to solve it now?

and if so, could you tell us what you would do?

[solemn matrix]

positive natural numbers ||(natural numbers are already positive, why do they call it positive natural numbers)||

that makes it easier

[fringe granite]

They call it positive because they are positive

Focus on the solution

[foggy wing]

Well a and b are not equal so it's 21 some letter times 21 another letter

Then i simplify

i had problems with that

[toxic cipher]

hmm well you have kl = 14

so what are the possible values of k and l?

[foggy wing]

Either 1, 2, 7 or 14

[solemn matrix]

and remember to check if those pairs have a gcd of 1

[solemn matrix]

as i said, this is the crucial one, or $\gcd{21k, 21l} \neq 21$

[boreal girder]

1 divided by 0 equals Infinity

[foggy wing]

They do yay

[toxic cipher]

okay so

[solemn matrix]

yeah, every pair does satisfy lol

[toxic cipher]

earlier you said a = 21k and b = 21l right?

[foggy wing]

Ok why

[foggy wing]

Yes

[toxic cipher]

so...?

[bleak dock]

you still need to find the original numbers

[foggy wing]

Ok i found thanks

[solemn matrix]

well, if gcd of k, l are 2 for example, then the gcd of your 2 numbers are 42 now

[toxic cipher]

what @solemn matrix is saying here is that if $\gcd(a, b) = 21$, then yes you can write $a = 21k$ and $b = 21l$, but under the condition that $\gcd(k, l) = 1$.

[toxic cipher]

So suppose that you had a different problem, in which $kl = 28$
The two numbers which multiply to 28 would be
\begin{itemize}
\item (1, 28) ✅ valid
\item (2, 14) ❌ they share a factor of 2
\item (4, 7) ✅
\end{itemize}

You have to reject (k, l) = (2, 14)

[compact pewter]

@foggy wing Has your question been resolved?

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[gusty oyster]

Welp

[gusty oyster]

Am I dumb or do these just have to be distributed on a? I'm a bit rusty on matrices

[kind crane]

which part

[gusty oyster]

B,c,d

[gusty oyster]

B,C,D

[kind crane]

there's a simple formula for inverse of 2x2 matrix

did you learn it?

[gusty oyster]

No

[gusty oyster]

No

[gusty oyster]

Ohhhhh

[gusty oyster]

Thanks

. close

.close

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[marble sleet]

Can someone explain how to do part b?

[steel prawn]

You have drawn a circle in a), you just need to find its equation. you need to find its radius and x and put it in the equation of the circle.

[noble sorrel]

x is shifted by radius units to right

so the equation will be (x-r)2 + y2 = r2

[marble sleet]

So i just need to find r and plug it in?

[noble sorrel]

yes

[marble sleet]

Olay

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[azure edge]

hello! Been trying things in desmos. Why does graphing a set of points for A work, but not L? I can't find the syntax error in L

[azure edge]

.close still a fool

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[scenic yacht]

Hi i am very confused here

[scenic yacht]

so there is a finte set E

and youre trying to work out number of elements

why do you need to do it in steps?

what does it mean to do it in steps?

[hushed magnet]

E is for example all the ways to arrange people in a line

you would do that by first choosing one person to be the first, then one to be the second and so on

so thats separate steps you do

[scenic yacht]

so when we say elemtns in E we mean the number of peipl right?

[hushed magnet]

no

[scenic yacht]

E is a set w elements, why are the ppl not elements?

sorry if im being dumb

am i just not understanding something trivial?

😭

[hushed magnet]

the elements of E are the ways to arrange the people

so one element could be (alice, bob, charlie), another element could be (charlie, alice, bob)

[scenic yacht]

hm i see

so ways to arrange ppl in a line, the number of lines would be the total number of elemens

alr

[scenic yacht]

alright so u have how many people though?

wait

u have n possible lines

with k possible options to make thoselines?

[hushed magnet]

well in this example I had 3 people

[scenic yacht]

yea

[hushed magnet]

try writing out all possible lines

[scenic yacht]

it would just be 3!

[hushed magnet]

yes

for the first position 3 choices

then 2

then 1

[scenic yacht]

yep

[hushed magnet]

thats the counting principle

multiply 3*2*1

[scenic yacht]

yea i get that

but the def makes no sense to me

[hushed magnet]

we have 3 steps to make a line

step 1: first person

step 2: second person

step 3: third person

n_1 = 3

n_2 = 2

n_3 = 1

[scenic yacht]

k successive steps, so 3 steps