help-13

[wraith dagger]

Zybikron

[modern compass]

@azure magnet

[azure magnet]

so

the answer would be 1 and 9 then

[modern compass]

yes

[azure magnet]

p = 1

q = 9

u sure buddy, if so TY very much

last question, what IF I had to do it in reverse i.e

I get something like 3e^x + 15e^-x

and u had to do it in reverse if u know what I mean

[modern compass]

You'd basically follow this but backwards

[azure magnet]

ok

OHHH

still use the identies

[modern compass]

$3e^x + 15e^-x = A(e^x+e^{-x}) - B(e^x-e^{-x})$

[wraith dagger]

Zybikron

[azure magnet]

yeah

oh

[modern compass]

So A-B = 3 and A+B=15

[azure magnet]

how much time u got bud

I have one final question to ask

which is non-related and is lengthy

[modern compass]

not that much

[azure magnet]

rip

thank for your help man

.close

[cedar kiln]

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[cedar kiln]

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[gusty lotus]

Hi

[gusty lotus]

Uh so for this I think I'd know how to do it if the derivatives of the functions were linear

but they're both squared functions

so how can they be parallel if they're not linear?

[cosmic tusk]

the tangent lines need to be parallel

tangent lines are always linear

[gusty lotus]

wait what

but the derivative of f(x)

is 6x^2 +2x

[violet flume]

its a root finding problem

fren

for a quadratic

[slim hawk]

And if you plug in any X you will find the slope of the tangent line at that x value

[cosmic tusk]

^

[violet flume]

set the slopes equal

u will see

[slim hawk]

So you want to find where the derivative at f(x) is equal to the derivative at g(x)

[violet flume]

you have two quadratics

[gusty lotus]

wait so I set the derivatives equal to each other?

[violet flume]

so we expect maybe 2

or 1

or maybe none

but probably 1 or 2

[gusty lotus]

so just to make sure i got 6x^2 +2x for the f prime of x

[violet flume]

then maybe we check that the outputs of g(x_1) and f(x_1) and same for x_2 if we find two roots are different

if your problem implies that a line isnt parallel to itself

[slim hawk]

Yes

[gusty lotus]

and -6x^2+2x+48 for g prime of x

[violet flume]

yup

[gusty lotus]

ok and then set them equal to each other

and the reason is to find what?

[violet flume]

well what happens when the derivatives are equal

whats the interpretation

for the same x, we have two equal derivatives

they are changing at the same rate

what does this mean about their tangent lines?

[gusty lotus]

oh so when we solve for the x when we set them equal to each other, we'll be finding what x values make them equal/give them the same value right?

[violet flume]

so we set them equal and solve for x

i can spoil that its a quadratic

the derivatives are equal when 6x^2-48=0

[gusty lotus]

I got

[violet flume]

this gives us two x values

[gusty lotus]

+-2

[violet flume]

call em x1 and x2

so lets imagine the tangent line to f and g

at x1

whats going on with these tangent lines

the derivatives are equal, so we have two possibilities

[gusty lotus]

isnt it 12x^2

[violet flume]

ah, youre right

mb

[gusty lotus]

oh ok

[violet flume]

so

the derivatives are equal

[gusty lotus]

yes

[violet flume]

two possibilities for the tangent lines

f* and g* call em

[gusty lotus]

ok

[violet flume]

what are they

the possibilities

[gusty lotus]

+-2

[violet flume]

okay

at x=2

the tangent lines have the same slope

are they parallel?

are they distinct?

are they necessarily parallel? are they necessarily distinct?

[gusty lotus]

uh parallel is when they have the same slope but different b value

[violet flume]

yea

what gives us our b value of a tangent line

[violet flume]

so necessarily parallel if distinct, we need to check if they are distinct

[violet flume]

how would you do this without desmos

[gusty lotus]

how would we do that algebraically

hmm

[violet flume]

whats the equation for a tangent line

given f'

for some f

say

[gusty lotus]

we plug them back in?

[violet flume]

$T_f (x) = f'(x) \cdot x + f(x)$

[wraith dagger]

jan Niku (Shuri for Honorable)

[violet flume]

you said parallel if slope is equal and b is unequal, yea?

what makes b

[gusty lotus]

the constant

?

[violet flume]

yup

[gusty lotus]

yay

[violet flume]

and in the tangent line equation?

(slope) * x + (constant)

whats the constant

[gusty lotus]

48

[violet flume]

no

in the equation

[gusty lotus]

this one?

[violet flume]

you are confusing equations

the one with 48 provided us with x values where the derivatives were equal

[gusty lotus]

im sorry😭

[violet flume]

the slope of either tangent line is from f' and g'

lol ur good

just f'-g'=0 is its own thing

[gusty lotus]

OH

the equation

[violet flume]

yea

[gusty lotus]

when we set them equal

uhh

lemme see

[violet flume]

you can forget that equation now

you have the x's

you dont need it any more

[gusty lotus]

yea

[violet flume]

now heres our question

we got 4 lines

2 pairs of 2

at x=2 we have these two lines:

$T_g (2) = g'(2) \cdot 2 + g(2)$

[wraith dagger]

jan Niku (Shuri for Honorable)

[violet flume]

$T_f(2) = f'(2) \cdot 2 + f(2)$

[wraith dagger]

jan Niku (Shuri for Honorable)

[violet flume]

how can we know if Tf and Tg are distinct at x=2?

we know that f'(2) = g'(2)

we did that work

[gusty lotus]

im just confused on where this rule came from. is this a rule or something?

[violet flume]

This is just how tangent lines work

[gusty lotus]

what is Tf(x)

[violet flume]

f'(x) is the slope of the function f at x, yea?

[gusty lotus]

yea

[violet flume]

tangent to f at x

so if you just do f'(x) * x

you get a line with the correct slope

but

its probably not at the correct place

since its gonna cross the x axis at x

thats probably not where we want it

so we have to shift it up or down

thats what the +f(x) does

[gusty lotus]

oh

so can we do it in terms of this case? Sorry it's just that I learn better by doing examples

so we got the x values where they're equal

the derivative functions

[violet flume]

just check that the outputs are different

of f and g

at those x's

im worried my equation is wrong now bc im tired

[gusty lotus]

oh lol

[violet flume]

well i was trying to be fancy

theres no need to be fancy

you just need to check the outputs

[gusty lotus]

uh so plug them back into the original functions? Not the derivative functions

[violet flume]

yup

[gusty lotus]

cause at the derivative functions they're equal

[violet flume]

the function and tangent line have to agree at the point of tangency

yea im missing a piece bc dumb

[gusty lotus]

it happens

so I just plugged them back in

f(2)=20

f(-2)=-12

g(2)=84

g(-2)=-76

[violet flume]

looks like two distinct lines to me

[gusty lotus]

which means that they're parallel at those points?

right?

[violet flume]

yup 🙂

[gusty lotus]

yay

so if f(2) = g(2)

then I would cross 2 out from my answer

because they're the same exact line? which means that they're not parallel?

[crimson sedge]

[gusty lotus]

is that a good thing or a bad thing

[gusty lotus]

does that mean that what I said is right?

[violet flume]

what what a bad gif

noet gif

[gusty lotus]

lol

[violet flume]

what u call the thing

dont worry

well imagine this

you got two lines

they have the same slope

and we can find a point where they cross

whats this means about the lines

[gusty lotus]

parallel?

[violet flume]

no

consider we start at the point where they cross

and we move forward in x by unit

how much has each line changed?

determined by its slope right

[gusty lotus]

yea

[violet flume]

they were the same at some point

we move forward one unit

line 1 has gone up by its slope

line 2 has gone up by its slope

but they have the same slope

[gusty lotus]

yes

[violet flume]

do you see the problem?

so theyre the same somewhere, and they change at the same rate everywhere

[gusty lotus]

oh

yea

hmm

[cedar kiln]

@gusty lotus Has your question been resolved?

[gusty lotus]

idk what's up with me today. I don't seem to get it lol. Sorry for the trouble. I'll give it some thought and see if I'll get it. Thank you jan

[cedar kiln]

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[rustic scarab]

hey can someone explain this

[cedar kiln]

@rustic scarab Has your question been resolved?

[cedar kiln]

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[dry snow]

how to write this in latex?

[violet flume]

$\begin{bmatrix} \leq \ = \ \geq \end{bmatrix}$

[wraith dagger]

jan Niku (Shuri for Honorable)

[dry snow]

copy + paste of this

[violet flume]

huh?

here

$\begin{bmatrix} \leq \ = \ \geq \end{bmatrix}$

[wraith dagger]

jan Niku (Shuri for Honorable)

$\begin{bmatrix} \leq \\ = \\ \geq \end{bmatrix}$

[dry snow]

all good

[violet flume]

copy the code above the image

[dry snow]

i double the \

[violet flume]

discord eats characters

yes

[dry snow]

thank you so much

[violet flume]

theyre newlines

np

[cedar kiln]

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[zinc pagoda]

can someone please check for me if my answer is correct or not?

[south rover]

can u show how u worked this out?

[zinc pagoda]

o waite

I realized I made a mistake

is my calculation correct?

[soft cloak]

It's correct, but it's not in standard form

[cedar kiln]

@zinc pagoda Has your question been resolved?

[zinc pagoda]

how to change it into standard form?

@soft cloak

[soft cloak]

Look at what standard form is

[zinc pagoda]

5*10^4

.close

[cedar kiln]

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[zinc pagoda]

.reopen

[cedar kiln]

✅

[zinc pagoda]

thank you guys

.close

[cedar kiln]

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[versed lava]

Im not sure how to approach questions like this

[celest ledge]

Replace 2x-1 with 2x+3-4

[fallen heath]

u = 2x + 3

[versed lava]

yea i get 2

[versed lava]

Whats the thought process involved to reach that conclusion

that u have to do that

[eternal comet]

^I second this

Split the fraction

it becomes trivial

[versed lava]

Yea i know how to solve this now, but if it comes in an exam, how do ik that i need to split the question in this manner

or add and subtract 4

[celest ledge]

To rewrite any rational function as a polynomial plus linear combination of functions of the form x^j/(x-r)^k plus linear combination of functions of the form x^j/((x+a)^2+b^2))^k

[eternal comet]

^partial fraction decomposition

[versed lava]

0_o

[eternal comet]

The first step is polynomial division. In this case it can be done quite easily in your head.

[versed lava]

ohhh

ohhhh

okay

[eternal comet]

(If the degree of the numerator is equal to or higher to that of the denominator)

[versed lava]

I am an idiot

I forgot about this concept

thanks :)

.close

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[zinc pagoda]

Is my calculation correct?

[zealous crystal]

Yeah

[zinc pagoda]

thanks

.close

[cedar kiln]

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[shy oriole]

.reopen

@zinc pagoda question says give answer in standard form

[cedar kiln]

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[spring jolt]

Hello. I have a curve that is y = 2x-x^3, it has a tangent where x = 1 which is parallel to y = 2 - x. What is the equation for the tangent.

[crimson sedge]

what have you tried so far?

[spring jolt]

I know that the K value is the same for the tangent because it is parallell.

But i cannot go any further than this

No clue what they mean by y = 2 - x

what is 2 and what is x?

[livid hound]

y = 2 - x
that's just an equation of a straight line

[spring jolt]

alright

[livid hound]

you can determine the slope of your tangent line you want from that

you should also try finding the point of tangency

and with a slope and a point, you have the required info to get the desired equation

[spring jolt]

what does a slope mean

[livid hound]

gradient

[spring jolt]

i am not very familiar with english mathematics

oh

alright

I will try now with the information you gave me

So the gradient of the tangent should be -x

[livid hound]

no

the gradient should be a numerical value and does not contain x

[spring jolt]

The equation y = 2 -x, this means that the gradient of this one is -1

[livid hound]

yes

[spring jolt]

and that is the same for the tangent

now i need to figure out the M-value right?

in order to finalize the equation

[livid hound]

you use
y = kx + m
where you're from right?

[spring jolt]

yes

[livid hound]

yeh

[spring jolt]

i know that the tangent is where X = 1, so i can just put that in the equation, but what about the Y value, where do i find that?

[livid hound]

use the equation you're given

to find the point of tangency

you're interested in the tangent at the point where x=1 of

y = 2x-x^3
what's the y value when x=1?

[spring jolt]

right, it must be 1 then

[livid hound]

yes

[spring jolt]

thank you for this help man

appreciate it

.close

[cedar kiln]

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[lime lava]

let (a_n)n>=1 an arithmetic progression with strictly positive terms. Calculate according to a_1 and in the following amounts:

[lime lava]

I don't know what to start with

[celest ledge]

$\frac{1}{a_{j}a_{j+1}}=\frac{\frac{1}{a_j}-\frac{1}{a_{j+1}}}{d}$

[wraith dagger]

Cogwheels of the mind

[celest ledge]

Where $d=a_{j+1}-a_{j}$ for any j

[wraith dagger]

Cogwheels of the mind

[lime lava]

Ok ty

[cedar kiln]

@lime lava Has your question been resolved?

[lime lava]

determine an arithmetic progression with 4 terms, knowing that the product of the permits of odd rank is 16, and the product of those of even rank is 55

@celest ledge

[celest ledge]

I don’t understand the question

[celest ledge]

Can you explain it? What does it mean by saying the product of the permits of odd rank?

[lime lava]

Idk 2, but ig

a_1*a_3 = 16

And

a_2*a_4=55

[celest ledge]

That’s it?

You let a_n=pn+q

[lime lava]

I tried this

but it's not good

[lime lava]

I didn't do that at school

a_n= a_1+(n-1)r

only this formula

[celest ledge]

(p+q)(3p+q)=16
(2p+q)(4p+q)=55, so together we have that 7p^2-q^2=62

Replace p with sqrt((62+q^2)/7) and put it in the first equation

We will have a quadratic function of q^2

Then we can solve p and q

[celest ledge]

=rn+(a_1-r) so p=r,q=a_1-r

[lime lava]

Yes, but I didn't learn that

the solution

but I still don't understand

Why a_1 = 3r, a_2 = a-r, a_3 = a+r and a_4 = a+3r

[celest ledge]

You can set them this way

Basically the same, a quadratic function of a^2

But it seems like it wants you to guess what a and r are instead of calculating them…

[lime lava]

uh

[celest ledge]

So a^2-3r^2=71/2 and ar=39/4

[lime lava]

Y'know what

I'll ask my math teacher

[celest ledge]

So a^2r^2=1521/16, replace a^2 with 3r^2+71/2, then you have 3r^4+(71/2)r^2-1521/16=0. I won’t know that 9604 is a square without online calculator, r^2=(-71+/-sqrt(9604))/12.so r^2=27/12=9/4, r=3/2

Therefore a=13/2

So we need to use online calculator to find that sqrt(9604)=98

If calculator isn’t allowed, I don’t know how to solve it in exam

[lime lava]

Uh

Tysm

[cedar kiln]

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[limpid grail]

Any thoughts on how you would start to simplify this into a fraction?

The question is to simplify ,so I'm assuming that it goes into some sort of neat expression ,but I'm not even sure when to begin or how to approach it

[cedar kiln]

@limpid grail Has your question been resolved?

[upper garnet]

factor common terms

[limpid grail]

im struggling to see what you did here

[upper garnet]

simple factorization

[limpid grail]

what did you factor out

[upper garnet]

common terms

like the constant, the x and (x-1)

[limpid grail]

what constant

im confused how you can subtract these to become one statement

[upper garnet]

1/6 and 1/12

[limpid grail]

<@&286206848099549185>

Im still confused about how I can simplify this into a single fraction

[cedar kiln]

@limpid grail Has your question been resolved?

[zealous crystal]

Same way, you would do a/6+b/12

[cedar kiln]

@limpid grail Has your question been resolved?

[upper garnet]

^

[limpid grail]

I know you said to use factorisation ,but I still don't know what i'm supposed to factorise out. If I knew how to factorise this I wouldn't be asking this question.

sorry for being stupid btw

im kinda bad at math so even this is causing me alot of issues

[safe abyss]

here's a little something you could do

for getting a single fraction, like you said

@limpid grail

[limpid grail]

yes

[safe abyss]

do you know how to do the rest to get it as one fraction?

[limpid grail]

I can time the first part by 2/2

[safe abyss]

yesyes

[limpid grail]

If i pulled (x-1)^-1/2 and (x-1)^-3/2

Would it be 12 × (x-1)^-1/2 × (x-1)^-3/2

*to the denominator

i think it wants me to remove the negative powers also

.close

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[bold vine]

Read this^^

[violet rapids]

wait

[bold vine]

[violet rapids]

How old are you?

... can you post a direct link to the video please

[upper garnet]

bruh

no

are u trolling

havent learnt it

[bold vine]

Just post a link to the video

Directly

[upper garnet]

suspicious link and has nothing to do with math

yes

-_-

[dense wing]

Just use a calculator if you want to be lazy

[upper garnet]

is that even high school math

[violet rapids]

no lol

[upper garnet]

and it seems like u just found it from google

[violet rapids]

this is way below MS curriculum

[upper garnet]

'free math worksheets'

[dense wing]

K, close the channel then

Or ask an actual question.

[bold vine]

Just report them or something

[dense wing]

Are you having troubles reading the stuff you posted?

Cause clearly x isn't 0

[bold vine]

They're trolling

[dense wing]

im aware

[upper garnet]

next post is gonna be something about topology bet

[bold vine]

Just remove them from the server if they aren't gonna be serious

[upper garnet]

ok bye

<@&268886789983436800>

[dense wing]

<@&268886789983436800>

[upper garnet]

Trolling and posting a suspicous link

[vernal palm]

modinat0rs

[bold vine]

Wow - you're so funny I almost laughed

[dense wing]

Concatenation jokes, funny ell oh ell

[vernal palm]

concatenation!!!!

[sharp widget]

.close

[cedar kiln]

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[odd wharf]

I need int integrate int [x^2 sqrt(x+1)] dx with u-sub as u=x+1

[odd wharf]

so int [(u-1)^2 sqrt(u)] du

[bold vine]

$\int x^2 \sqrt{x+1} dx$

[wraith dagger]

Shen

[bold vine]

$u = x + 1$

[wraith dagger]

Shen

[bold vine]

$\int (u-1)^2 \sqrt{u} du$

[wraith dagger]

Shen

[bold vine]

Yup

[odd wharf]

wait let me try the next step

$\int {(u)^(1/2) (u^2)-(2u)+1} du$

[bold vine]

$

[wraith dagger]

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

[odd wharf]

eh whatever

we get this

[bold vine]

There you go

[odd wharf]

\int_{0}^{x}\left(u^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}}\right)dz

does that work?

$\int_{0}^{x}\left(u^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}}\right)dz$

[bold vine]

no bounds and wrong differential but ye

[wraith dagger]

SauceBauce

[odd wharf]

swag

ok so

now I can just

[bold vine]

Use reverse power rule on each term

[odd wharf]

$\int_{0}^{x}\left(u^{\frac{5}{2}}\right)dz-\int_{0}^{x}\left(2u^{\frac{3}{2}}\right)dz+\int_{0}^{x}\left(u^{\frac{1}{2}}\right)dz$

[wraith dagger]

SauceBauce

[odd wharf]

but I think the answer is wrong

[bold vine]

Wrong differential and keep it an indefinite integral

du

[odd wharf]

yeah sorry I'm copying and pasting from desmos

[bold vine]

If it's definite, then change the bounds. If it isn't, then replace u with x + 1 at the end and add an arbitrary constant

[odd wharf]

$\int \left(u^{\frac{5}{2}}\right)du-\int \left(2u^{\frac{3}{2}}\right)du+\int \left(u^{\frac{1}{2}}\right)du$

[wraith dagger]

SauceBauce

[odd wharf]

ok there

now we have

$\frac{u^{\frac{7}{2}}}{\left(\frac{7}{2}\right)}-\frac{u^{\frac{5}{2}}}{\left(\frac{5}{2}\right)}+\frac{u^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}$

[wraith dagger]

SauceBauce

[odd wharf]

or

$\frac{2}{7}u^{\frac{7}{2}}-\frac{2}{5}u^{\frac{5}{2}}+\frac{2}{3}u^{\frac{3}{2}}$

[wraith dagger]

SauceBauce

[odd wharf]

which is wrong.

well also because I forgot to replace u but ehhh

[gaunt hamlet]

What happened to the coefficient on the second term?

It's 2u^(3/2)

But you lost the 2

[odd wharf]

oh right

$\frac{2}{7}u^{\frac{7}{2}}-\frac{12}{5}u^{\frac{5}{2}}+\frac{2}{3}u^{\frac{3}{2}}$

[wraith dagger]

SauceBauce

[odd wharf]

there we go

[gaunt hamlet]

Why 12/5?

[odd wharf]

cause 2 = 10/5

so

2/5 + 10/5 = 12/5

[gaunt hamlet]

Why are you adding them

[odd wharf]

oh

right

$\frac{2}{7}u^{\frac{7}{2}}-\frac{4}{5}u^{\frac{5}{2}}+\frac{2}{3}u^{\frac{3}{2}}$

[wraith dagger]

SauceBauce

[gaunt hamlet]

Now you have to sub back for u and add + C

[odd wharf]

...

alright imma just trust the math

.close

[cedar kiln]

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[crimson sedge]

in page 3 of this link what does "m(b-a)" exactly mean? m is supposed to be a number as implied its defined as a suprenum while in page 3 its treated as a function:

(integral calculus)
https://www.math.ucdavis.edu/~hunter/m125b/ch1.pdf

[vagrant breach]

Aren't they the sizes of the rectangles

so a and b are the integral limits, m the inf of the function, so m(a-b) is going to be a rectangle that bounds the function from below and M(a-b) from above

[crimson sedge]

ohh i get it now

thnx

.close

[cedar kiln]

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[slow sparrow]

Can I get help on this

[dense wing]

what's the question?

[slow sparrow]

Expand and evaluate

[dense wing]

🤨

You can't evaluate that so...

[slow sparrow]

This is my very first time seeing a question like this so I have no clue what to do

[dense wing]

Hint: When do infinite geometric series converge?

[rigid ridge]

using i as an iterator when using it as an exponent with e is very suspicious

[dense wing]

It isn't

[distant swan]

^^^^

[dense wing]

Didn't know I needed a bunch of arrows with a ping

[slow sparrow]

-1 < r<1?

[rigid ridge]

it goes against all intuitions about e

[dense wing]

yes

[dense wing]

what's r in your question?

[distant swan]

Is it possible to get a finite number from this sum?

Figure that out

[dense wing]

@distant swan can you not?

[slow sparrow]

3

[dense wing]

no

[distant swan]

What

[dense wing]

cause then you'd have 3^i

$\sum_{i=1}^\infty ar^i$ is a geometric series

[wraith dagger]

Mosh

[dense wing]

what's r going to be?

[slow sparrow]

No clue

[dense wing]

Use your eyes

and compare your sum

to the general form I wrote

[slow sparrow]

1?

[dense wing]

Compare $3e^i$ and $ar^i$

[wraith dagger]

Mosh

[slow sparrow]

e

[dense wing]

yes, finally

[slow sparrow]

Sorry

[dense wing]

now is $|e|<1$?

[wraith dagger]

Mosh

[slow sparrow]

Yes

[dense wing]

???

,w euler's constant e

[rigid ridge]

is e defined otherwise in the question maybe?

LOL

RIP

uwu w approximate e

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Lance#8018

[slow sparrow]

So it’s diverging?

[dense wing]

yes

|e| isn't less than 1

[slow sparrow]

I haven’t learnt about eulers number yet

Which is why I was confused

.close

[cedar kiln]

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[cedar kiln]

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[restive viper]

hey

[restive viper]

i need help

<@&286206848099549185>

I forgot how to find the area of a paralelagram of a square that has 5 numbers.

[broken mist]

The area of a parallelogram can be found by multiplying its base length by its height. I'm not sure what you mean by a square that has 5 numbers

[restive viper]

oh

its like uh

a square right

outside it has 4 numbers

and like

inside

there is one.

<@&286206848099549185>

[jaunty mural]

Did I not tell you not 10 mins ago NOT to spam the ping

The poor people who assign themselves that tag didn't ask for this.

[restive viper]

https://tenor.com/view/bro-ayo-wtf-bruh-moment-gif-21632234

[jaunty mural]

like once every 15 mins (and NOT immediately when you post) at most please.

[restive viper]

but like

are you a upper rank

[slim jasper]

He is right, that's what matters.

[restive viper]

can you help me

[dire geode]

Definitely more upper rank than you

[restive viper]

ok.

can you help me tho.

thats your rank

a helper.

thats your job

so you should be able

[dire geode]

That's your answer

[restive viper]

"im not sure what you mean by a square that has 5 numbers"'

[slim jasper]

Nobody here has any job; it's a server people hang out in in their free time.

[restive viper]

thats not my answe tho

[dire geode]

The beginning is your answer

Show a specific question if you want specific help

[restive viper]

happy

[dire geode]

Don't need your attitude. Bye

[restive viper]

ok

<@&286206848099549185>

lol

[fallow steppe]

i have muted you, so you have time to familiarize yourself with our rules @restive viper

.close

[cedar kiln]

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[cedar kiln]

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[simple geyser]

Heyooo guys

[simple geyser]

I need some help understanding the basics of generating functions when it comes to manipulating A(x)

anyone who has knowledge on this topic is very appreciated to help

I don't understand how the line I marked in red is achieved

[upper abyss]

There's two lines in that line, haha. Which of the two is giving you grief?

Does it make sense that (1 - 2x)A(x) = B(x)?

Note the third line on this page is very important here. It looks like the constants are not independent

[simple geyser]

that's the one troubling me

I don't get how we can get that equation from any above equation

[upper abyss]

The line above the one you marked in red is detailing that
2xA(x) + B(x) = A(x)

[simple geyser]

so the professor was trying to build a gen func for the sequence 0,1,3,7,15

[upper abyss]

Definitely want to understand why that one works

[simple geyser]

$n_k = 2n_{k-1}+1$

[wraith dagger]

Arkid

[simple geyser]

because $a_0 = 0$

[wraith dagger]

Arkid

[simple geyser]

I'm sorry, I wasn't clear enough

[upper abyss]

Take a sec, think this hint out

[simple geyser]

ohhh

so A-2xA = B

;-; finally on the right track

it's subtraction mb

now it all makes sense

tyvm, this was really killing me

can you help me with one more thing?

I've done little of this like 1 year ago and I don't remember how it exactly works

this

so why does A + B =0 and 2A+B=-1

I'll close this and open new tbh just in case

.close

[cedar kiln]

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[high hornet]

How would I find this graph?

[obsidian coral]

,rotate

[obsidian coral]

Is that a test?

[high hornet]

No

It's a practise exam, so my teacher has put the marks on it

[obsidian coral]

My guess, it looks like it has something to do with piecewise functions

[high hornet]

It's to do with absolute functions

[obsidian coral]

An absolute value piecewise function

[high hornet]

Kk

How do you graph them on desmos?

[obsidian coral]

By bounding it, with intervals

[gaunt hamlet]

You do it in curly brackets. For example, y = x {x > 2}

[high hornet]

Ok

So would assume this is correct if I wrote It down in a large y={} form?

.close

[cedar kiln]

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[heavy mauve]

since both have no common divisor larger than 1, doesnt it mean than a and b are two different prime numbers?

[bold vine]

Nope - 9 and 10 fit this description (although they are not the solutions to the equation)

[heavy mauve]

that makes sense

[bold vine]

It may help to compare 5/9 and 4/7 more directly by creating a common denominator between them

[heavy mauve]

so it's about not having any common factor ehh

so at most one of them can be an even number

[heavy mauve]

thanks

[bold vine]

Yeah! That's one restriction that such numbers have 🙂

[bold vine]

np np - gl

[heavy mauve]

.close

[cedar kiln]

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[cedar kiln]

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[warm vector]

Sine law

[warm vector]

Do you know what it is?

[bleak mirage]

55+40=95
180-95=ac

which side of the angle

outer or inter

outer then do 360-95

so 265

[steel canopy]

a/sin(theta)=b/sin(theta)=c/sin(theta)

[steel canopy]

a = Allstown. b = Brimtom and c = Cripton

We know bearing from c to b. We know bearing from c to a. Distance from b to a is 38. We need distance from c to a. a/sin(theta)=c/sin(theta) you can find distance between c to a. From there you should know all angles.

[cedar kiln]

@high bay Has your question been resolved?

[steel canopy]

30?

Ok. Calculate and tell me what u got

[high bay]

?

[steel canopy]

sin94?

[high bay]

my bad

[high bay]

?

[steel canopy]

Does it tell you to round?

[high bay]

3sf

[steel canopy]

That should be right

Okay

[high bay]

Thank you :)

[steel canopy]

Since everything is good

[high bay]

It's right

[steel canopy]

Also 85 wasn't correct because it's higher than 38 which is hypotenuse

If you have number less than 38 it should be correct

Just think of trigonometry, too. a^2+b^2=c^2

[high bay]

aa yes that makes sense

[steel canopy]

To ensure that's ur answer make sense

[high bay]

So I could use a^2+b^2=c^2 to double check my answer?

[steel canopy]

ye you could. If you know your hypotenuse

[high bay]

ohhhh

got it

[steel canopy]

Since you know both distances must be less than hypotenuse

If you do enough problem you should be able to tell if it's correct or not

[high bay]

ohh I see, I just realised

tkxxx

[cedar kiln]

@high bay Has your question been resolved?

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[analog minnow]

?

[cedar kiln]

Please don't occupy multiple help channels.

[glad kestrel]

what have you tried

[cedar kiln]

@analog minnow Has your question been resolved?

[analog minnow]

No it has not @cedar kiln

This is the one i need resolved

[analog minnow]

Nevermind

[cedar kiln]

@analog minnow Has your question been resolved?

[analog minnow]

Yah by myself @cedar kiln

[gray condor]

have you heard about quadratics?

@analog minnow

[analog minnow]

Yah i got it

[gray condor]

so you have the answer?

[cedar kiln]

@analog minnow Has your question been resolved?

[cedar kiln]

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[upbeat whale]

is somethign wrong there

i did it a different method and got 33 again

says b is wrong

[cedar kiln]

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[potent moth]

A river is flowing downstream at a rate of 2 metres per second. Brendan, who has an average swimming speed of 3 metres per second, decides to go for a swim in the river. He dives into the river and swims downstream to a certain point, then swims back upstream to the starting point. The total time taken is 4 minutes. How far did Brendan swim downstream?

[inland quartz]

Fellow physics man, but np we can do thiss

[potent moth]

I keep getting 1000m…

Oh no this is math question

[inland quartz]

So these type of q's you gotta divide into two parts

THe downstream part, and the upstream part

[potent moth]

Yes

[inland quartz]

You know how to relate speed distance and time ya?

[potent moth]

Yeah

[inland quartz]

In this case time is given to us, so we'll figure out the expression for time up and down in terms of distance and speed, and proceed from there

Alrightt then, let's assume the distance he swims one way is d

[potent moth]

ok…

[inland quartz]

What would be the time taken for downstream?

[potent moth]

5t?

[inland quartz]

Oh no we need it in terms of d

d is what we're tryna solve for ya, that's why

[potent moth]

d = 5t?

Oh

[inland quartz]

What is t here?

[potent moth]

Time

[inland quartz]

Okkk, but only for the downsteam part, so could we say t = d/5?

[potent moth]

Oh yee

S

yes

[inland quartz]

And similarly, for the upstream?

[potent moth]

T=d

[inland quartz]

Now if we add both these times, we get 4 mins as per the question right

?

[inland quartz]

Yeppp

And u can solve for d now

[potent moth]

Yaay!

[inland quartz]

🥳

[potent moth]

d = 600?

[inland quartz]

Are u sure?

[potent moth]

Wait

Let me do it again

Ummm yes?

[inland quartz]

Show your working will ya

[inland quartz]

Oh I see

See, you cannot take 5 directly to the other side cus it doesn't divide the entire left side

Do I make sense>

[potent moth]

Yes

So

Wait

I will write again

Is this correct this time?

Oh so the answer is 200m!!!

Thank you so much!!

I got it!

.close

[cedar kiln]

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[kind wind]

I need help for 2cii

[kind wind]

Idk how to get b

[steady trench]

OT= 1/5(2r+3s)

Since p=3s

OT=1/5(2r+p)

[cedar kiln]

@kind wind Has your question been resolved?

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[kind wind]

. reopen

[kind wind]

I need help for 2cii

[rigid ridge]

could someone here please zoom in on the figure, i cant view it too well from my mobile

[cedar kiln]

@kind wind Has your question been resolved?

[kind wind]

@rigid ridge is this fine

[rigid ridge]

i meant a close up of a diagram :P

[kind wind]

This?@rigid ridge

[rigid ridge]

yeah thx

[kind wind]

I need cii and d thanks

[rigid ridge]

@kind wind are you given that ST=2/5 SR for the entire question?

[kind wind]

Yes

[rigid ridge]

bruh

i was trying to find that the entire time

this is sad

[kind wind]

Its written on the question what

[rigid ridge]

it is a subquestion

and you told me only to look at cii

😔

[kind wind]

Emm my bad

[rigid ridge]

find OTR/QRT using b

this problem is not too fun but the buildup is really good

[kind wind]

How

[rigid ridge]

combine the statement of b with the idea of ci

[kind wind]

i cant get the answer

[rigid ridge]

how have you solved ci?

[kind wind]

No

[rigid ridge]

try solving that first

[kind wind]

I cant solve man

[rigid ridge]

hint: for OST/OTR, use the fact that we have the same heights if we take bases ST and TR

[kind wind]

Where same height

[rigid ridge]

the bases lie on the same line

so the distance to the same line is the same

which is the height by definition

[kind wind]

Which one is same height

I am done with ci

How to do cii

@rigid ridge

[rigid ridge]

like I said, find OTR/QRT

[kind wind]

Like i said,i dont know how

[rigid ridge]

use the idea i just described for ci

and combine it with b

[kind wind]

Its same height also?

[rigid ridge]

it's a very similar idea

in fact, the same

[kind wind]

Can give hint?

[cedar kiln]

@kind wind Has your question been resolved?

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[steep meadow]

Being dumb rn. |a-b| = |a+-b| ≤ |a|+|-b| = |a|+|b| is valid right?

[celest ledge]

Yeah

[steep meadow]

Bet thx

.clode

. close

.close

[cedar kiln]

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[long fulcrum]

Hello, am i missing something but doesn't cross product work in the same way as product rule?

[long fulcrum]

From product rule, I obtain this instead

$\frac{\text{d}}{\text{dt}} (\vec{w} \times \vec{r}) = \vec{w} \times \frac{\text{d}\vec{r}}{\text{dt}} + \vec{r} \times \frac{\text{d}\vec{w}}{\text{dt}}$

[wraith dagger]

Serena!

[long fulcrum]

Which isn't the same?

[celest ledge]

Your last term is wrong

You switched the order, you can’t do that cause y cross product x =- x cross product y

[long fulcrum]

Ah wait so this isn't the actual product rule?

[celest ledge]

It is

[long fulcrum]

I mean it tends to work with other functions i guess, but for cross products it wouldn't is what you mean(?)

[celest ledge]

You just don’t switch order randomly

d/dt(a times b)=a’ times b +a times b’

You can’t replace a times b’ with b’ times a like you did

It will change its sign

[long fulcrum]

Yes I'm aware but like i'm confused in the sense that

Is my form not the actual product rule?

Was the second and first term switched in order to make it easier to remember?

Not for cross products i mean in general

[celest ledge]

Why you care about what it’s called, just by definition show that d/dt(a times b)=a’ times b +a times b’ holds but what you think is true isn’t true

[long fulcrum]

I mean isn't the cross product derived off the product rule?

If we follow this definition of the product rule then you'd get it in my form

which is wrong

Why?

[celest ledge]

Yeah a=(x,y,z),b=(u,v,w) a times b=c=
(yw-zv,zu-xw,xv-yu)

So c’ , each term like its first component yw=yw’+y’w, yw’ goes to a times b’, y’w goes to a’ times b

Just show it by definition

Calculate that

[long fulcrum]

Uh yeah i guess i can show it the same way

as i did for dot product which is by expanding and then using the trend

[celest ledge]

I don’t know why we should have issue about that. It’s not something we can’t easily prove

[long fulcrum]

You mean about my question right?

[celest ledge]

Yeah

[long fulcrum]

Yeah that was essentially what i was concerned about

[celest ledge]

You calculate (a times b)’ by definition

You will easily see which one is correct

[long fulcrum]

Yeah so apparently just use a different definition for cross-product altogether then? Lol

[celest ledge]

What

No

[long fulcrum]

I mean i derived the product rule and i got it in the same form as this just to check if in actuality i had the terms reversed or something

[celest ledge]

I don’t know what you don’t understand. c=a times b is a function with value as vectors. You calculate c’ by definition

[long fulcrum]

I've derived the cross product thingy as well the same way you mentioned by decomposing

[celest ledge]

I don’t know why it is so hard to you

[long fulcrum]

Okay whatever then

Thank you

.close

[cedar kiln]

Channel closed

Closed by @long fulcrum

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[astral shard]

does p have to be greater than q in fermants little theorem rsa?

[trail jackal]

which theorem do you mean? Fermats little theorem is, that for every prime $p$ it holds $a^p=a$ modulo $p$. I don't see a q here...

[wraith dagger]

Alexander42

[astral shard]

well its e x d mod phi = 1

wait nvm

thats something else

basically p x q = n
then you calculate phi
then you choose e which needs to be greater than 2
then you make d = inverse of e mod phi

thats my private key generation

and public key

for an rsa

p and q are prime numbers

should of mentioned that

😄

[trail jackal]

RSA is symmetric in the primes p,q. So there is no reason at all for p being greater than q (although it is not good to choose p=q)

[astral shard]

oh okay

[trail jackal]

just name p the smaller of the two primes and q the greater one, it doesn't matter

[astral shard]

ah okay

but does it matter at all

if they are are greater for example 17 x 3

instead of 3 x 17