help-36

[formal trail]

plug in your function for f(x) then use any limit solving techniques you can to simplify and take the limit

[faint rock]

I don't fully understand what you mean, the only function I have is f(x)=2x-5

At first i thought that f(x+) became 2x-5+2h-5 but that doesn't seem to be the case

[formal trail]

f(x+h) means substitute (x + h) for every instance of x

similarly to how f(2) means substitute 2 for every instance of x

[faint rock]

Okay I think I understand what you mean, in this case it would then become 2(x+h)-5?

[formal trail]

yes

[faint rock]

Alright, thank you so much man, have a great day 🙂

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[odd tinsel]

I retried the problem and got it incorrect- why

[odd tinsel]

I suspect something to do with revolution of y axis and x given bounds, (which would then be y- bounds of 8, 10) but that answer is also incorrect

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[quiet garden]

How do I find these

[odd tinsel]

i think ur right, you're just missing the "-"s because its in q3

[quiet garden]

For what

For both?

[odd tinsel]

yeah, I havent been in trig or whatever for years but from what I remember the ‘-‘ in your 5pi/4 is indicating that you are ‘going back’, so you look where 5pi/4 is on the unit circle and you can see it should terminate at -sqrt2/sqrt2 for both x,y

[quiet garden]

Isn’t that for 5pi/4

[odd tinsel]

thats your t, yes

[quiet garden]

Mine has - in front tho

[odd tinsel]

right, I believe that indicates that its going backwards, idr

Maybe wait until a trig specialist comes but thats what I recall

Its either that or you negate -5pi/4 from the start of the unit circle and see what the new position is

So 360-225*

idr tho irs been years

[quiet garden]

Well I put it on both and it still says wrong

[odd tinsel]

yeah idk

wait 15mins from your post and then ping Helpers

[quiet garden]

I changed it to - in the x coordinate and it worked

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[desert juniper]

Further

[desert juniper]

How do I simplify this

[cerulean igloo]

ln(a)-ln(b)=ln(a/b)

[desert juniper]

I’m looking to solve in terms of v

[cerulean igloo]

Replace a with v and b with t

And also, if ln(x)=y, then x=e^y

[desert juniper]

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[desert juniper]

Thanks

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[zenith oxide]

I'm trying to make a formula where I put in two variables and get a third variable out so like r*p=v but I want r to have more importance but v has to be between 0 and 100 and I'm just going in circles

[zenith oxide]

I should clarify

R is an unknown number from 1 to 10 inclusive
P is an unknown number from 1 to 10 inclusive
And v is an unknown number from 1 to 100 inclusive

And when I say r should have more important I mean like
r=5 p=10 leads to a larger number than r=10 and p = 5

If r = 5 it's worth more than if p = 5

Sorry if that doesn't make sense I'm a bit "impaired"

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[sly quest]

Quick question. Why can you let a variable be equal to sine or cosine of a different variable? Doesnt that only output values between -1 and 1?

[tranquil pine]

in what context?

[rustic sinew]

Yes you can and yes it does

[manic herald]

well then that variable will only take on values between -1 and 1

[sly quest]

I was integrating sqrt(a^2 - x^2) where it is suggested to let x = asinb

[sly quest]

But what if the variable doesnt

Am I not integrating x generally?

[tranquil pine]

if you let x = asinb then x will take values between -a and a and thats when the square root is defined

if x was any other value the output wouldnt be a real number

[sly quest]

Sorry im not really understanding. Let's say a is 16. Then I can only make such substitution as long as x is between negative and positive 16?

Ooooooohh

I got that

Thank you

I have no idea how to close a room lol

[sweet summit]

.close 🙂

[sly quest]

Thx

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[placid laurel]

I did the question but my book doesnt have answers

so it would be lit if someone could solve this and lmk when they got

I got root 12

nvm ig

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[radiant reef]

how do you find area under curve for parametric?

[vital crag]

solve for y in terms of x

[radiant reef]

cos^-1(x/2) = t?

then do i plug it into the y equation

[vital crag]

use pythagorean theorem

[radiant reef]

sorry im confused

where

[vital crag]

$\sin^2 x + \cos^2 x = 1$

[soft zealot]

riemann

[radiant reef]

do i just add them together then

cos^2(t) + sin^2(t) = (x^2)/4 + y/3?

im kinda confused tho like why can i just add them together

[tulip coyote]

[as an aside, there is another way, in that you can notice what the curve looks like, and use parametric integration]

[tulip coyote]

You know that sin^2 + cos^2 = 1 is an identity, and the fact that you have x = 2cos(t) and y = 3sin^2(t)

[radiant reef]

and i could just add them?

[tulip coyote]

so having (x/2)^2 gets you cos^2(t) and y/3 = sin^2(t), so adding those gets you 1

[radiant reef]

ah yes

then solve for y

[tulip coyote]

Yep, then that one you can e.g. integrate directly

[radiant reef]

area formula?

[tulip coyote]

do you have one in mind?

[radiant reef]

A = int from a to b (top-bottom)

[tulip coyote]

Yep yep that's it!

Worth sketching the curve out too!

[tulip coyote]

[if you'd like, I can explain this way too if you would like after? If you haven't seen it before]

[radiant reef]

yes pls

[radiant reef]

there's no bottom curve tho, so would it just be the y

[tulip coyote]

Yea - if you sketch it out somehow, you should find something like this, so the bottom is just the x axis, y=0

[radiant reef]

if you didn't know what it looked like, would you just assumed the second ones is x-axis

[tulip coyote]

Well it might depend on the curve I think, but something like this, seeing they say "area under the curve", it's safe to assume the bottom is just the x axis

[radiant reef]

ohhh okay

if they give you parametrics where one is like sin^2t or cos^2t and then sint or cost, is it safe to just assume you can use pythagorean and add them together

[tulip coyote]

Yea the idea is that you want to try to make use of trig identities where you can (maybe alongside some restrictions on t) to find an explicit equation
So say something like x = sin(t), y = sin(2t) for t between -pi/2 and pi/2, you could use double angle formula and then the fact that for those values that cos(t) = sqrt{1 - sin^2(t)} to get y = 2x * sqrt{1 - x^2}

[radiant reef]

ahhh i get it

i think i messed up somewhere

😭

i plugged in 0 and pi for a and b in the integral

am i supposed to plug it into x and y values

i realized its t

[tulip coyote]

You want the limits as x values, so find the corresponding x values to those

[radiant reef]

ohhh right

so 2 and -2

[tulip coyote]

[also just realised that the parametric added in the restriction 0 <= t <= 1, it should really be this for t between 0 and pi]

[radiant reef]

so would it be a negative integral then since -2 or x(pi) is supposed to be on the bottom?

[tulip coyote]

Well you want -2 at the bottom and 2 at the top

You want the lower x value for the bottom and the higher one as the top

[radiant reef]

yeah but like

pi > 0

but the value when plugged into x

is like -2 and 2

like for ftc before u plug it into u it could be like int from 2 to 3

but after plugging into u its like from -1 to -5

and then u have to turn integral into negative to flip the values?

[tulip coyote]

Note that for the integral here, you already want the smaller x value as the bottom limit and the higher one as the top limit

[tulip coyote]

Say you ignore the whole parametric equations they gave you and that I tell you to find the area under the curve y = (3/4)(4 - x^2) that was sketched here

You'd want to find the integral $\int_{-2}^2 \frac34\pqty{4 - x^2} \dd x$, right?

[soft zealot]

@tulip coyote

[radiant reef]

yeah

[tulip coyote]

And here you aren't really doing any substitutions really (other than the rearraging and finding the x values)

[radiant reef]

yeah okay

thank u

[tulip coyote]

[tulip coyote]

Anyways, this should be not too bad to work out I hope

[radiant reef]

its equal to 8?

units^2

[tulip coyote]

,w int (3/4) * (4 - x^2), x from -2 to 2

[soft zealot]

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[tulip coyote]

Yep

[radiant reef]

thanks

i think i have another question i'll need help with

[tulip coyote]

Sure thing

[radiant reef]

i missed the lesson on this one, but i understand how to find dy/dx

but idk how to find the point

[tulip coyote]

I mean, if they give you theta, you can find what r is from there

[tulip coyote]

What did you find as dy/dx?

[radiant reef]

im still working on it

sorry 😭

[tulip coyote]

Don't worry thought you already had it, take your time

[and ping me when you're done! ]

[radiant reef]

@tulip coyote

[tulip coyote]

Looking good to me!

[tulip coyote]

Happy with that now?

[radiant reef]

yes

how would you find points?

[tulip coyote]

As in the radius r given the theta, or the x and y coordinates?

[radiant reef]

it says to find the equation of the line tangent to the polar

so how would i find the points for the equation

[tulip coyote]

Ohhh, I see

Do they want it in polar form or Cartesian form?

[radiant reef]

i don't know

[tulip coyote]

if they want it in polar form, I'd probably mess it up tbh

[radiant reef]

maybe cartesian form?

[tulip coyote]

I would guess it's polar form they want it in but if Cartesian, then for that one you just know that when theta is 3pi/2, that r is 6pi and from there you can find x and y

You have dy/dx, the gradient from there, so line equation should be fine from there

[radiant reef]

ohhh so plug the theta into x and y equation i set

[tulip coyote]

Yep

[radiant reef]

so (0, -6pi)

y+6pi = -2/3pi x

ok

i have one more sorry 😭

do i just integrate this?

to find position

ftc

its supposed to say dx/dt = sqrt(t^2 + 3t + 1)

[tulip coyote]

those don't look too easy to integrate

[radiant reef]

can use a calculator

[tulip coyote]

In theory integrating the dx/dt and dy/dt would be fine

[tulip coyote]

hmmm, maybe those might help I think if a calc can do it for you, but nonetheless those functions' antiderivatives are a bit something

[radiant reef]

should i integrate them separately

like integrate dx/dt from 0 to 2 = x(2) - x(0)

and do the same for the y

to get the position coord?

[tulip coyote]

That would work out at least theoretically worth trying!

[radiant reef]

okk

[tulip coyote]

Just so it's neater for me haha, $\dv{x}{t} = \sqrt{t^2 + 3t + 1}$ and $\dv{y}{t} = e^{t^2 - 1}$

[soft zealot]

@tulip coyote

[radiant reef]

@tulip coyote

this is what I got

[tulip coyote]

Gonna take your word for it, you have the right method at least

Interesting they give it like this, would have expected something a bit nicer and exact

But otherwise is that all good?

[radiant reef]

yes

thank you SO SO MUCH

i understand everything sm better now

[tulip coyote]

Perfect alsooo do you want that parametric integration thingy explained too?

[radiant reef]

yes sure

[tulip coyote]

Cool cool, so basically the idea for parametric integration is that it's basically like doing a substitution, you have that (ignoring limits for a second) $\int y \dd x = \int y \dv{x}{t} \dd t$

[soft zealot]

@tulip coyote

[tulip coyote]

The first integral, you have the limits being in increasing order of x

The second one, the limits correspond with the t values that get you those

So going back to our example, $x = 2\cos(t)$ and $y = 3\sin^2(t)$, then that is basically saying (putting the limits back in)
[
\int_{-2}^2 y \dd x = \int_{\pi}^0 d \dv{x}{t} \dd t = \int_{\pi}^0 3\sin^2(t) \cdot -2\sin(t) \dd t
]

[soft zealot]

@tulip coyote

[tulip coyote]

(with the pi being at the bottom because that's what corresponds to the smaller x value, so it looks "the wrong way around", if that makes any sense and I've explained it properly?)

[radiant reef]

uhhhh

yes

because pi is supposed to be larger than 0

[tulip coyote]

Basically you know like with some substitutions, where if you put them in, like the sizes of the limits change

[dusky lily]

how are u getting the bot to do that

[tulip coyote]

So like if you had some integral, say $\int_0^1 f(x) \dd x$, if you chose (just for demonstration purposes) $u = 1 - x$, that when $x=0$ you have $u= 1$, and when $x=1$ you have $u = 0$, so then that gets you
[
\int_0^1 f(x) \dd x = \int_1^0 -f(u) \dd u
]

[soft zealot]

@tulip coyote

[tulip coyote]

If you type LaTeX out, the bot will recognise it and render it for you (basically the stuff between dollar signs in my messages, if you're not used to it - you can see #latex-help for some resourses on it )

[dusky lily]

thanks

[radiant reef]

oh yeah i remember that

[tulip coyote]

it's similar here, it's like the parametric substitution means that when x = -2, t = pi and when x = 2, t = 0

[radiant reef]

ohhhh okay

i think 0 to pi

[tulip coyote]

Of course it's kinda like the other way around in that you have the "substitution integral" first, so in general you want to check the order of the limits and then order them according to that(!)

Anyways, back to the above for a moment-

Of course, with a bit of manipulation and making use of that negative, that integral is $\int_0^{\pi} 6\sin^3(t) \dd t$, you can then use e.g. power reducing to get that as
[
\frac64 \int_0^{\pi} 3\sin(t) - \sin(3t) \dd t
]
and then evaluating that should hopefully get the same thing, evaluating $\frac12 \pqty{-9\cos(t) + \cos(3t)}$ between 0 and $\pi$

[soft zealot]

@tulip coyote

[tulip coyote]

Of course, I'm gonna be lazy and just check with the bot

[radiant reef]

wait

ok now im a bit confused

where did sin 3t come from

[tulip coyote]

That's using the power reduction formula

,w int 6sin^3(t), t from 0 to pi

[soft zealot]

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[tulip coyote]

And it does work out to the same thing

[radiant reef]

ohhhh

interesting

[tulip coyote]

[in this case, the parametric way is a bit more painful than the direct way, but in other cases the parametric way is much easier]

And I guess it makes a nice example too, cause you can see both ways

[radiant reef]

yeaaa thank u

i think i'll stick to pythagorean tho

its interesting to see the other way

[tulip coyote]

Yea, it's just to see another way, and maybe hopefully might be useful if they give you other examples where it's easier to do it like this

Hopefully everything was useful!

[radiant reef]

it was!!

thank u so much

have a good night

🫶

[tulip coyote]

You too

[radiant reef]

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[coarse badger]

I don’t get it

[coarse badger]

Someone help pls

[tardy sigil]

when things are similar

their ratio is same

think

corresponding similar sides

have equal ratio

@coarse badger

[coarse badger]

Yeah, but idk how to find the scale factor

@tardy sigil

[vital surge]

you find scale factor by comparing 2 sides on the triangles you know correspond to each other

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[tranquil pine]

Whats the point of recursive definitions? like ie for addition, how do you typically find them?

[vernal monolith]

what kind of recursive definitions?

[tranquil pine]

like for addition, multiplication

exponent

factorial

etc

[vernal monolith]

you mean how addition is defined in foundational mathematics?

[tranquil pine]

maybe? like for multiplication, m * (n + 1 ) = (m * n) + m

[worldly vale]

how else would you define it?

[tranquil pine]

But like, how do you think of these definitions intuitively

[worldly vale]

its capturing the notion of "multiplication is repeated addition"

intuitively you know m*(n+1) is adding m to itself n+1 times, well that definition is just saying "add it to itself n times" and then add one more m

and then you follow the tower all the way down and finally use m * 1 = m

by the time you've reached the bottom you really have just added m to itself n+1 times

[tranquil pine]

okay thank you so much!

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[grim glen]

I’m trying to figure out the relationship between these numbers, the total pot last year was around 1700 and the payouts were as follows:

700 - 41.1%
375 - 22.05%
260 - 15.3%
210 - 12.4%
155 - 9.1 %

Are there any straightforward methods i can use to figure out why these were weighted the way they were?

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.reopen

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✅

[fossil geyser]

Those percents seem to be just the straight proportion of the total

700/1700 = 0.411 = 41.1%

etc

In other words, 700 is literally 41.1% of the pot, if the total pot is 1700

@grim glen

or are you asking why those specific percentages were used instead of something else?

[grim glen]

Yeah

I cant seem to find a real reason why they were used that way

Is there a common pattern im missing?

[fossil geyser]

I don't think so. It seems pretty arbitrary

[grim glen]

Yeah im looking through all other values and it doesnt seem to be based off anything i can see atm

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[dense badge]

hey so if I wanna find out if a function is homogeneous, I count the degree of each term, and if they're all equal the equation is homogeneous right? okay so what about trig functions like sin(x), what's the degree of x in this case and what about if x is under a root, will it be 1/2?

[tiny gorge]

degree is only defined for polynomials, sin(x) is not a polynomial

[dense badge]

so it would have a degree of 0?

[tiny gorge]

undefined, i'd say

[dense badge]

so the function wouldnt be homogeneous

right?

[tiny gorge]

i don't think so, unless there's some more general definition of homogeneous

if so, i'm unaware of it

[dense badge]

okay what about under a sqrt

[tiny gorge]

sqrt is also not a polynomial

[hybrid heath]

!original

[final saddle]

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.

[dense badge]

well okay

it wasn't a problem is was just a question

thank you

.close

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[tiny gorge]

yea that looks valid

sure, why not?

P(A) = P(A intersect B) + P(A intersect B^c), right?

by disjoint additivity

you're just using a rearranged version of that

(twice)

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[hoary oxide]

okie so

y' = e^(x^3) * 3x^2

wouldn't this mean the derivative is INDEPENDENT of y, meaning that for the slope fields, it shouldn't vary for y?

And because the derivative isn't symmetrical when you do x & -x, doesn't that mean it shouldn't reflect?

I got the answer correct (B) BUT IDK WHY

[final saddle]

@hoary oxide Has your question been resolved?

[fossil geyser]

This is one differential equation for which y=e^(x^3) is a solution

but it is not the only one

for example y' = y * 3x^2

is also satisfied by y=e^(x^3)

[hoary oxide]

ahhh gotchya

thank uuuu

'rep

idk how to d it

+rep

t!rep

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[hoary oxide]

there we go

wut

[fossil geyser]

Don't worry about it lol

[hoary oxide]

sorry : (

[fossil geyser]

It doesn't really do anything, I didn't even realize it still existed lol

but thanks anyway

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[raw sundial]

is my answer correct?

are you for sure

can you help me with more

can you help me agian

ok

is this right?

thanks

.close

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[calm tangle]

can someone explain how they make this approximation to get to the bottom line?

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[vital crag]

try using taylor/geometric series expansion in s/r

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[proper yoke]

write in parameter form an equation for the line (in the plane) determined by : 2x - y = 5.

[proper yoke]

I understand why y becomes 2t -5 but why does x become t?

[spring haven]

how does y become 2t-5?

without x becoming t

[proper yoke]

I don’t know but I thought that I was supposed to solve out x and get the answer

Like why can’t I switch out y to a t and solve for x?

[spring haven]

you can

it will give you a different parametisation

[proper yoke]

Oh okey but is that a valid answer to this question?

[spring haven]

yes it is

there are multiple parametisations that are valid

x=t and y=2t-5 is one of them

[proper yoke]

X = (5-t)/2?

[spring haven]

yes (with y=t)

you have to specify both coordinates

[proper yoke]

Okay, in my book they only gave the first answer where x=t so I was confused

Would this question be solved differently if the question said “write in parameter form an equation for the line (in the room) determined by : 2x - y = 5.”?

[spring haven]

"in the room"?

what does that mean

[proper yoke]

The book talks about lines in the room and lines in the plane

I think the room is when you have three variables

Like this next question where 2x + y - z +3 = 0

Is one answer to that question z = 2t + s + 3, x = t and y = s?

.close

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[slow belfry]

could anyone please give me an idea on how to solve the fifth question in the second part? (integrals aren't allowed in that specific question)

[slow belfry]

I have:

\[f_n(0) = 0\]
\[F_n(1) = 0\]
\[f_n(e) = 0\]
\[F_n(e) = U_n\]

[soft zealot]

ᑎᗩᑕᖇᗴᝪᑌᔑᗞᗩᗯᑎ596

[slow belfry]

<@&286206848099549185>

[peak tusk]

could you translate the question

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[slow belfry]

okie, 1min

[slow belfry]

prove that Un+1 = ...that stuff

or you mean the translation of the whole exercice?

[peak tusk]

no just translate the relevant stuff

like given information

etc

[slow belfry]

you have these

[ashen ice]

Can u help me

Pls

[slow belfry]

U_n is a decreasing serie

[ashen ice]

Where do i request help

[blissful meadow]

Read #❓how-to-get-help , this channel is occupied.

[ashen ice]

Aight

[slow belfry]

and also convergent

fn+1 - fn <= 0 in [1; e]

ig that's what we have so far

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@slow belfry Has your question been resolved?

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@slow belfry Has your question been resolved?

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@slow belfry Has your question been resolved?

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[flint perch]

Im not even sure what i+j = k means

i forgot lol

its i+j modulo k right? so the remainder when i+j /k?

if and only if m is prime, can I just translate that to m does not divide by 2

but that would be saying m is odd

and not all odd numbers are prime?

well that doesnt work

cause 2 is prime

[tulip coyote]

m=2 also forms a field, a field with two elements (it's pretty easy to show that's a field)

The idea is that you want to show that along with other ring axioms, you can find an inverse for all the nonzero elements if m is prime

And that if m is composite, you want to somehow break a field axiom

hint that might help: ||all fields are integral domains||

@flint perch

[flint perch]

im still here

trying to understand how to do this

[tulip coyote]

As in actually showing the field axioms?

[flint perch]

yea , this is all i have down rn

[tulip coyote]

stolen from the other one

[tulip coyote]

Worth of course noting down the field axioms if you have them

[flint perch]

yup

[tulip coyote]

Hopefully it's only 8 that you need to think about (but if you haven't shown the others elsewhere in general, we can do those?)

Some of these are basically automatic for you anyway

[flint perch]

sure, Im stuck on this whole thing :(( but a few examples and I'll prob start getting the gist of it

[tulip coyote]

Awww well I'm not exactly sure how much they actually want you to do

But for now, let's assume that all of those, with the exception of 8, are true, and that we have m being prime

Then of course, all the nonzero elements 1, 2, ..., (m - 1), what can we say about them and m?

[flint perch]

if the set is A, there is are two x in A and x =/= 1 or m-1 such that x times x is m-1

x times b in A

[tulip coyote]

Not necessarily (in fact you need m to be 1 mod 4, considering m prime, if I recall right)

[tulip coyote]

May be worth e.g. comparing each of these separately: m is prime, so 1 and m are...?

[flint perch]

you mean like factors of m

[tulip coyote]

Yep, compare the factors of m and each of the elements 1, 2, ..., (m - 1)

[flint perch]

wait i’ll be back in 15 min

[tulip coyote]

Alright sure! Ping me when you're back

[final saddle]

@flint perch Has your question been resolved?

[flint perch]

@tulip coyote

[tulip coyote]

you're back

[flint perch]

why would you compare factors of m? because m is not in the set. also, you mean like this? for the comparision?

i’m very lost

[tulip coyote]

The idea is that you want to work with the knowledge about how each of those and m relate to each other (the word I was looking for was "coprime/relatively prime") - in other words, their greatest common divisor is 1 (does that hint at something you can do from there?)

Remember that the conclusion we would like is that, say that p is in {1, 2, ..., (m-1)}, you want to be able to find an b such that p * b = 1 (remembering that means that p * b has remainder 1)

[flint perch]

Im so screwed

i dont know what to do

the tetbook doesnt have solutions I can look to

i am so so screwed

we are just startring this course too

i dnt know what to do anymore

[tulip coyote]

awwww

Well I mean, we can help out as much as possible

[flint perch]

I thought I was gonna get like half done today😭

😭 😭 😱

this course is a nightmare

[tulip coyote]

Awwww when do you have to get everything done by?

It is quite a bit

[flint perch]

no its just for practice

we have an assignment due in a week ish but I dont understand anything so thats why Im going through the textbook

this is so stupid, if theres no solutions in the textbook how are you gonna learn? or if you get stuck you cant do anything

[tulip coyote]

Awww I see

[tulip coyote]

Where is the book from? How comes they don't come with solutions(?)

I guess the expectation is that you'd have either someone helping you, or if they're particularly not great, they expect you should be able to figure it out

[flint perch]

M. Laczkovich, V. Sos, Real analysis. Foundations and functions of one variable.
Published by Springer, 2015.

[tulip coyote]

Like the one we're working on does require a bit of thought and remembering stuff

[flint perch]

i went through two weeks of calc content in a day

linear algebras not too bad as well(I havent started but looks doable)

but this stuff

i do like 1 question a day🥲

[tulip coyote]

some topics do be a bit like that

[tulip coyote]

[anyways, to help you out, Beizout's lemma might be useful to remember here ]

[flint perch]

i dont know how gcd(a,b)=d and ax+by=d is gonna help

[tulip coyote]

Well, letting, say, n be any integer between 1 and (m-1), you know that the gcd of m and n is 1, so you have an a and b such that an + bm = 1

Remember you wanted n to have a corresponding something such that when you multiply by it, you get a remainder of 1 modulo m

Does that make it any clearer?

[flint perch]

1 mod m is just m, assuming m != 1 so its a(n) =m...

would it be possible if you just provide a solution to this

and possibly a few others and then I can just look it over and fill in gaps where I dont understand

[tulip coyote]

Well we aren’t really supposed to give solutions

[flint perch]

maybe I should drop out of this course, seeing my level of comprehension how would you say my chances are of passing

[tulip coyote]

I think you’ll be fine, it just might take a bit of time

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[eager junco]

@opal plinth Ive got my final solution

[vital crag]

do you have a question

[eager junco]

yea i did and Nel helped me not too long ago and i mentioned that i would let him know once i was done

[vital crag]

you don't anymore?

[eager junco]

well i just want to verify that ive got it right

[final tangle]

repost, people might not always be available

others may be able to assist

[eager junco]

sure

.close

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[granite bane]

Number 48. I used a = 2 , b = 1, and c = -1

[granite bane]

pls help

you are right how?

[vital surge]

(also, using your example, only F works)

[granite bane]

🗿 I am still confused

Yes

where you get negative from?

3/-1 and 2/-1 ?

oh ok

what you said about the sign flip?

[soft zealot]

swagmafia3000

[granite bane]

oh

because whenever there is negative in inequalites you flip sign?

alright

thanks

.close

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[granite bane]

help pls

[quiet garden]

It’s A

[spare nexus]

Choice A

[granite bane]

is that a formula

[granite bane]

how do I know its going up or down

left or right

[spare nexus]

When k > 0, it goes up. When k < 0, it goes down.

The horizontal shift is a bit trickier to remember. If it's (x + h), it goes left. If it's (x - h), it goes right.

[granite bane]

Oh okay thanks bro

.close

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[dense dune]

Question 22

[marsh temple]

,rccw

[dense dune]

Can I divide the whole equation by 3

[marsh temple]

Yes

[dense dune]

Ok hold on a sec

Is the 2 squared at the end

,rccw

[marsh temple]

No

It's just 2

So the radius of your circle is sqrt(2)

[dense dune]

Oh ok

Thanks

[marsh temple]

Also you didn't complete the square correctly

[dense dune]

Oh

[marsh temple]

y^2 + 2y - 1 =/= (y-1)^2

[dense dune]

Oh

Does that mean it’s A

Wait

[marsh temple]

I have to go so don't keep me waiting too long

[dense dune]

Sorry

I’m not sure what the answer is

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[sour viper]

@rustic sequoia

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[tranquil pine]

how to find a , b ?

[tired walrus]

are you sure you didn't mistype this?

[tranquil pine]

no

i am sure

[tired walrus]

!xy

[final saddle]

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.

[tired walrus]

thing is, since the denominator approaches something nonzero (3+3 = 6), evaluating this limit would have required nothing but direct substitution...

[tranquil pine]

i know

that’s the point

[tired walrus]

and it just becomes $\sqrt{3a + b} - 2 = -\frac{3}{2}$

[soft zealot]

Ann

[tired walrus]

so you only have one equation, but two variables.

you cannot solve for both.

oh, the value of the limit is +1/4, not -1/4.

my point still stands, bc the denom is x plus 3 ...

[tranquil pine]

yes my bad

[tranquil pine]

idk it’s a exam question

[tired walrus]

it's an impossible question, you're literally deprived of info.

are you 120% sure that the denominator is x+3 AND x approaches positive 3?

[tranquil pine]

Yes

[tired walrus]

ok then yeah

it's unsolvable

so if your teacher is holding you at gunpoint about it im so sorry but your life will be in danger then

(but the stakes are probably a lot lower in reality)

[tranquil pine]

i see

thank u

.close

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[vivid tapir]

Are these lemmas of equality of functions equivalent? I have written below that they are not since in the first one we have to equate the images of f and g and in the second and third one we have to equate the codomains. But it need not be the case that the images and codomains are equal, is that the reason why they are not equivalent?

[vivid tapir]

I have to go in a while so can someone respond?

[trail jewel]

can you rephrase the last sentence

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[cursive wave]

Can someone help me with this one?

[rocky tusk]

ok so u can establish a pattern here

u know the original price is 15000

so what do u think the price will be after one year

if it increases by 10%

@cursive wave

[final saddle]

@cursive wave Has your question been resolved?

[cursive wave]

sorry

[rocky tusk]

^

[cursive wave]

1500

I think

NO

[rocky tusk]

but it increased by 10%

so at let’s say t=0 where time is zero years have passed

it’s 15000

so after one year, t=1 what’s the new price

u can think of this is a geometric sequence

if u know what that is

[cursive wave]

yES

YES

Sorry my family keeps calling me

[rocky tusk]

no worries

[cursive wave]

Anyways. so 10% of 15000 is 1500 cuz

So will it increase by

1500

[rocky tusk]

do u know what a geometric sequence is

[cursive wave]

Mhm

[rocky tusk]

ok do u know the formula for finding the nth term

[cursive wave]

When they get multipled by the same thing

Yeah

[rocky tusk]

yes

so what is the formula

an=?

[cursive wave]

sn = t1(r^n - 1) / (r-1)

[rocky tusk]

this is the sum formula

[cursive wave]

OH

Mb

[rocky tusk]

this is saying if u added up the first n terms

[cursive wave]

t1r^n-1

[rocky tusk]

yes

if we start at n=2

[cursive wave]

Yeah, sorry

Mhm

[rocky tusk]

ok so

what’s t1

[cursive wave]

15000

[rocky tusk]

and what’s r

[cursive wave]

.1

[rocky tusk]

no

[cursive wave]

No?

[rocky tusk]

because think about it

[cursive wave]

:so

Help.

Hm?

[rocky tusk]

if u multiplied by 0.1

it would decrease

right

[cursive wave]

Yeah true

[rocky tusk]

the price isn’t decreasing

i saw earlier

[cursive wave]

Yeah

[rocky tusk]

what u did was multiply by 0.1

and added the original well

[cursive wave]

Yeah

[rocky tusk]

think of it like this

if something increases by 10%

[cursive wave]

Yeah

[rocky tusk]

then it’s new price is 1.1 times the original

[cursive wave]

Got it

[rocky tusk]

because 1 times the original is itself

so no increase

same goes if it was decrease

[cursive wave]

So it's 100% + 10%?

[rocky tusk]

u would have 1-0.1=0.9

[cursive wave]

Okay

What if it's already .9? so I take it to negatives?

[rocky tusk]

yes because if we have increase by 10% of x it’s really x+0.1x=x(1+0.1)=1.1x

[rocky tusk]

this is if it decreases by 10%

[cursive wave]

Okay, I get it now

Tyyy

[rocky tusk]

ok so

[cursive wave]

Okay!

[rocky tusk]

what will it be after 10 years

using the formula

[cursive wave]

Hold up lemme do it

8649

waittt

[rocky tusk]

hmm

if u started at 15000

how’d u get less

r=1.1

[cursive wave]

Oh I wrote 100 insteas of 15000

I AM SO SORRY 😭 I DONT MEAN TO TROUBLE YOU

[rocky tusk]

no worries

so what is it

after 10 years

don’t get tricked by the n btw

in the formula

[cursive wave]

576650

if I did not mess up again

I hope

[rocky tusk]

576000?

[cursive wave]

No-

[rocky tusk]

that’s too much

[cursive wave]

Oh

I am doing

[rocky tusk]

what’re u typing in

[cursive wave]

15000(a.5)^10-1

1.5 not a.5

[rocky tusk]

why 1.5?

[cursive wave]

...

I have not slept

[rocky tusk]

where’d u get 1.5

[cursive wave]

In 3 days

I am sorry

[rocky tusk]

and wait

n≠10

[cursive wave]

Yeah

[rocky tusk]

n=11

because

the first term is really year zero

so the second term is year 1

[cursive wave]

Okay

[rocky tusk]

and the tenth term is year 9

so we want the eleventh term

to get year 10

[cursive wave]

Okay, that makes sense

[rocky tusk]

i thought of it like ok increase by 10% so multiply by 1.1 for the first year then again for the second

and u notice

it’s really 1.1^t

where t is what year it is

[cursive wave]

Ohhhh

Yeah

That makes sense

[rocky tusk]

lmk what u get

[cursive wave]

38906

[rocky tusk]

bingo

nice

[cursive wave]

FINALLYYY

I am sorry, idk what I was doing with the 1.5

TYSM

[rocky tusk]

lol no problem

[cursive wave]

.close

[final saddle]

Channel closed

Closed by @cursive wave

Use .reopen if this was a mistake.

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[vivid walrus]

!status

[final saddle]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I 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

[tranquil pine]

1

i think you half the bx value

[tranquil pine]

have u heard of completing the square

yes

[tranquil pine]

yeah do it

perform it here

if i remeber correctly i think you half the b term

to get 3 over 2 x

and u get (x+3/2)-9/4 -20/4

after that point im stuck

wait

u collect like terms

-9-20 is -29

[tranquil pine]

not quite!

oh

what havw i done inncorectly

i think there is a fundamental error in what you think of how completing the square works

do you want me to intuitively explain to you hoe completing the square works or do you want me to jump straight to the formula and just guide you from that

explain how it works

alright would u mind like waiting 5 minutes my phone battery is like 4% so ill get to my pc lol

no i wouldnt ,take your time

[tranquil pine]

wot?

huh

do you still need help or

yes

why

i didnt understand what that sentence meant but its whatever

lets continue

so like

a perfect square trinomial is expressed as [
(a+b)^2 = a^2 + 2ab + b^2
]

[tranquil pine]

the idea behind completing the square is that for SOME quadratic of the form [
ax^2 + bx + c
]
you can do some modifications (completing it) to get something with a perfect square similar to the above

[tranquil pine]

so what we can do is this:

lets start with ax^2 + bx + c

okay

@tranquil pine i want you to try and factor 'a' from all 3 terms

what do u get

(x+a)^2

no

by factoring i mean like
ax + bx -> x(a+b)

oh right

the 3 terms are a b and c right

no

ax^2, bx, c

those are the three terms

x(ax+b) + c

okay great but like

im asking you to factor a not x

a(x^2+bx)

nope

that gets you ax^2 + abx if u distribute it back

is ax^2 + abx what u had to begin with

no

yes

so

you need to get rid of that a that u made appaer in the bx term

consider dividing by a for that term

a(x^2+b/a)

great!

but u missed x

in the second but im guessing that was a mistake

anyways like

a(x^2 +bx/a) + c

doing the same process for c gets you

a(x^2 + bx/a + c/a)

all clear?

yes

okay so lets ignore that a thats getting multipled

we will get to it

lets look at [
x^2 + \f{bx}a + \f ca
]

[tranquil pine]

so lets compare it with what we want this to be

which is [
a^2 + 2ab + b^2 = (a+b)^2
]

[tranquil pine]

its a whole different equation

[
\c g{a^2} + \c r{2ab} + \c b{b^2} \
\c g{x^2} + \c r{\f{bx}a} + \c b{\f ca}
]

[tranquil pine]

i tried to colour code it for clarity sake

but do you see how a^2 = x^2 here?

they are both the same in the sense that it is some number squared

so we can also say that a = x

right

okay so i will modify it further a bitr

[
\c g{x^2} + \c r{2xb} + \c b{b^2} \
\c g{x^2} + \c r{\f{bx}a} + \c b{\f ca}
]

[tranquil pine]

so we are starting to get to something that makes this work

now

were did the c term dissapear to

its still there

c/a

we factored a if you remember

oh yh

yuh

so like now

i will actually ignore the c/a part

for now

but we will come back to it

so like

[
\c g{x^2} + \c r{2xb} + \c b{b^2} \
\c g{x^2} + \c r{\f{bx}a}
]

[tranquil pine]

so this is the spicy part

we want to find some term so that like

[
x^2 + \f{bx}a + {???}^2 = (x + {???})^2
]

[tranquil pine]

x+ ab?

we are trying to find what that ??? is

[tranquil pine]

okay so lets try and think of it slowly alright

so like

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

thats what a perfect square is yes?

yes

[tranquil pine]

we want to find the ??? that makes this exactly happen

so so

we said a = x yeah?

so we got the first term a^2 down

but for the 2ab part

how do we compare that with the bx/a term?

so like, there is a 2 missing there

[
a^2 + 2ab + b^2 = (a+b)^2 \
x^2 + \f{bx}a + {???}^2 = (x+ {???})^2
]

[tranquil pine]

b^2

okay i dont think this will work very well if im going to do it this way

how about i show it to you and you can just ask questions okay?

i feel like this is too complicated for me

i will try to make it simple as much as i can

so like

[tranquil pine]

thank you and if u wanna give up i wont take offence

[
\c g {a^2} + 2\c g{a}b + b^2 \
\c g{x^2} + \c g{x}\f ba + {???}^2
]

[tranquil pine]

the greens are like directly together in the comparison we are making

so we want to try and find b

but like

do you notice that there is no 2 in the second one?

we can make a 2 appear by multiplying and dividing by two

so like

[
\c g {a^2} + \c r{2}\c g{a}b + b^2 \
\c g{x^2} + \c r 2\c g{x}\f b{2a} + {???}^2
]

[tranquil pine]

what i did is multiply and divide by 2, which is 2/2, which is equal to 1 so it doesnt change anything

if you notice, the only thing left is the b

so we can conclude that b = b/(2a) and thus the question marks are b/2a

so in conclusion we have

[
\c g {a^2} + \c r{2}\c g{a}\c b b + \c b {b^2} \
\c g{x^2} + \c r 2\c g{x}\c b{\f b{2a}} + \p{\c b{\f b{2a}}}^2
]

[tranquil pine]

i tried to illustrate it the best i could

but from this we get like

[
x^2 + 2x\f b{2a} + \f{b^2}{4a^2} = \p{x+\f b{2a}}^2
]

[tranquil pine]

which is why we did all we did

ok so if u remember we had like [
ax^2 + bx + c =a\p{x^2 + 2x\f b{2a} + \f{b^2}{4a^2} + \f ca}
]

[tranquil pine]

so we can come back to that

[tranquil pine]

with what we did here we added the (b/2a)^2 term

we need to subtract it again to avoid breaking the equalityt

okay sigh i think i really lost you so there is no good way for me to do this with my way of thinking

how about i link you some videos about it instead yuri? @tranquil pine

[tranquil pine]

sure go ahead

[tranquil pine]

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:completing-square-quadratics/v/solving-quadratic-equations-by-completing-the-square

view all of whats being said here

alright

thank you for trying to help

[vital crag]

,tex .cts

[soft zealot]

riemann

[vital crag]

[tranquil pine]

thats all you have to do?

yes

the reason why my thingy is

a lot more fucked up is because we are trying to consider it for any general quadratic ax^2 + bx + c

[tranquil pine]

so to get this, we needed to factor a 'a' from the expression

so if you get like uh

3x^2 + 4x + 6 or something

factor 3 out so you get
3(x^2+ 4/3 x) + 6

and then complete the square on the inside

that makes alot more sense

i get it now😭

wooo

thank you for your time

bye bye

.close

[final saddle]

Channel closed

Closed by @vale hound

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sonic crystal]

A fair die is rolled once. A is the event the score is even and B is the event the score is 5 or 6. Determine, giving reasons, whether or not events A and B are independent.

[sonic crystal]

I don't really remember how to do these types of probability questions, how can I start?

[tranquil pine]

basically, if two events are independent, then there couple of rules, it's pretty easy question, just brush up the rules once

[tired walrus]

you should recall the definition of independent events @sonic crystal

no progress can be made without that