help-13

[crimson sedge]

I guess it's against the rules to speak in any language other than English. I mean, yeah, I could do it, but I'm not doing your homework.

Plus you didn't show any work, you literally want people to go from 0 to the solution without doing any effort yourself.

[cedar kiln]

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[sacred grail]

,w expand x^2((x-2)^2 + 4) - 12(x-2)^2 when x=y+1

[wild spear]

damn🤙🏻

[wraith dagger]

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[sacred grail]

can probably solve for a factorisation of this into two quadratics

[crimson sedge]

Wolfram is truly magical. My favourite.

[sacred grail]

depress the quartic then factorise

[cedar kiln]

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[grand heath]

I want to ask 8b isn't y is a member if real number such that y is greater than -4

[south tundra]

y doesn't have to be greater than -4

As you can see on the graph

You can use 8a for 8b

[grand heath]

so it can be -infinity and + infinity so y is a member of all real no.

am a right?

I*

[hollow minnow]

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[toxic moat]

hi

the graph is a line

whats the eq of a line

[light shuttle]

no clue

[toxic moat]

y=mx+c

try to figure out m and c

[quartz frost]

do you know what m and c mean?

[light shuttle]

no

[toxic moat]

u can do some sub

eg

x=0

y= 0+c = c

aka its the y intercept

[light shuttle]

that didnt explain it

im confused still

[toxic moat]

by which bit

[light shuttle]

all of it

[toxic moat]

so

a graph takes up a range of values

so it has to give some val of y

for every x we put in

in this case we put in x=0

we get some y

its given by

y=mx+c

so when x=0 we get

y=m(0)+c=c

[light shuttle]

can i get da answer

[toxic moat]

lol

no

we r tryna understand c first

o well

or u can do m

imma gtg

💕

[light shuttle]

i really dont understand any of this

yall just confusing me more

(y2 - y1)/(x2 - x1)

is that it

the slope

so its not the slope

how about 1/2 x + 2

[livid hound]

how are you getting that

[light shuttle]

dont worry bout it

am i right

[livid hound]

no

[light shuttle]

damn it

[livid hound]

were you guessing?

[light shuttle]

maybe

i dont get this shit

[livid hound]

to apply the slope formula, you'd need to first identify two points on that line

[cedar kiln]

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[cedar kiln]

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[cedar kiln]

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[winter lion]

A playlist consist of songs from two Broadway musicals: Les Miserables and Mamma Mia. I have 40 Les Mis tracks saved on Spotify, and 19 from Mamma Mia. Order matters, repetition allowed and I prefer to keep answer in form n^k or a combination.
Questions and what I have tried so far:

• Suppose my playlist contains 7 songs. How many playlists could I create? 59^7
• Suppose my 7-song playlist contains songs from one show or the other but not both. Now how many playlists are possible? 40^7 + 19^7
• What if my playlist could contain 7 to 10 songs, but they’re all from Les Mis. Now how many playlists are possible? 40^7 + 40^8 + 40^9 + 40^10
• What if Kayla sticks to 7 songs in her playlist, but wants to play at least one from each show. Repeated songs are still OK. Now how many playlists are possible? 59^7 - 40^7 - 19^7

[winter lion]

for the last question, I am unsure if it is what I got or the other way I tried with a combination which was (40 1) * (19 1) * 59^5

[cedar kiln]

@winter lion Has your question been resolved?

[winter lion]

<@&286206848099549185>

[thick cipher]

I think there might be a sum there

In those cases write down each different case

[winter lion]

what would each case be

wouldn't that be a lot of cases

[thick cipher]

1 song from Les Mis 6 songs from Mama Mia etc

[winter lion]

so I would need to do like 12-14 cases?

[thick cipher]

Yes

[winter lion]

are parts a, b, c correct?

[thick cipher]

Yes

[winter lion]

is there a simpler way to do part d without having to sum 12-14 cases

is there a way to use combinational instead

because my thinking was that 59^7 - 40^7 - 19^7 represents all playlists with a mix of the two musicals

since 40^7 is all of only one song and 19^7 is all of only the other song

@thick cipher

[thick cipher]

Perhaps with sums and the binomial formula

[winter lion]

oh yeah i dont know how to do it then

the two ways i tried i thought would work but they both had different results

values*

[thick cipher]

Sometimes it's the same result but you need to simplify

[winter lion]

I am confused on the question i posted above

for the last part

@blissful glade

[cedar kiln]

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[crimson sedge]

IS this correct?

[crimson sedge]

Or is D correct on number 8?

[winter lion]

7 is correct since it is reflected over BD line so BD and (BD)' would be in the same spot

[crimson sedge]

yeah

[winter lion]

for 8, A and B are correct but it seems you are missing another true choice

[crimson sedge]

D?

[winter lion]

yes

[crimson sedge]

ty

can you also help me with another one please @winter lion

Im kinda confused on the tranformation rule

[winter lion]

i am confused. is it not already reflected?

[crimson sedge]

yes but

this part

I don't know how to like format the tranformation rule

like yk (x, y) > (-y, -x)

[winter lion]

yes

what do you think it is

for 5 and 6

[crimson sedge]

idk tbh im very lost at that one

they didnt teach it in my study

[winter lion]

lets take an example coordinate

for example we have (-2, 6)

[crimson sedge]

yeah

[winter lion]

after the reflection it is at (6,5)

so how exactly did each coordinate change

-2 went to 6 and 6 went to 5

so therefore, it would be changing from (x, y) to (-x+8, x-1)

now try it with the other coordinates

the x would be -x+8 because it is changing from a negative to a positive so it would be -x and the value increases by 8 so +8. Therefore, -x + 8

[crimson sedge]

ah

[winter lion]

not (-x+8, x-1) but (-(x+8), x-1)

u would first have to translate it then reflect it

do u see how that works with all four original coordinates

[crimson sedge]

yeah

[winter lion]

u just have to write out teh original and final and see how they chagne

if it is reflected then it is changing signs

then transformative would be + or -

now try the same with 6 and see what you get

[crimson sedge]

(x, y) - (-2 + 8 = 6, 6 - 1 = 5)

lioke this is how it works

[winter lion]

yes exactly

try the same with problem 6 now

and let me know what u get

[crimson sedge]

kk

G(-4, -2) > G'(-4, -4)
(x, y) > (-x, -y + -2)
(x, y) > (-4, -2 + -2 = -4)

tbh I got no clue what i mdoing

I think I messed up lol

[winter lion]

hmmmm thats not quite it

lets think about it

so one example coordinate is (-4, 2)

for G

sorry i meant G (-4, -2), correct?

that is the original position of G

what is the final position of G aka G'

[crimson sedge]

-4, -4

'

[winter lion]

yes so it is going from (-4, 2) to (-4, -4)

now how is each coordinate changing from original to new

so how is the x coordinate changing and how is the y

[crimson sedge]

wait you mean -2 or

[winter lion]

yes -2 my bad again lol

[crimson sedge]

x coordinate isnt changing and y coordinate is being added by 02

-2

[winter lion]

yes

so now how would you write the equation for how x is changing and y is changing

[winter lion]

it is not being added by 2. -2 to -4 is adding -2, not 2

[crimson sedge]

yeah

[crimson sedge]

-2

[winter lion]

yes

so what would it look like

for the transformation in format (x, y)

[crimson sedge]

(x, y) to (x, y + -2)

[winter lion]

the original is always (x, y) and after the transformation when x is staying the same and y is being add by -2 what would it look like

yes that is correct

but rather try (x, y-2)

since + - is equal to -

[crimson sedge]

ah yeah

[winter lion]

do you understand why the first was that for problem 5

[crimson sedge]

but wouldn't it be different for other cordinates?

Like H

[winter lion]

test it

it shouldn't be

[crimson sedge]

alri 1 sec

H(3, -1)
H'(3, -5)

if you added -2 to y it wouldn't be -5

[winter lion]

oh i apologize

it is reflecting over the x axis you would agree right?

one step we forgot is that it is being reflected over x-axis

so you would flip the sign

so original is (x, y) corect?

i know an easier way for you to understand this

[crimson sedge]

yeah but its being reflected by its own line y = -3

[winter lion]

yes exactly

so it is being reflected over x so it would first take teh format of (x, -y)

[crimson sedge]

yes

[winter lion]

now for an example you have (3,-1) which using the format i sent would become (3, 1)

[crimson sedge]

but it doesnt do that? right? since they are both negative

[winter lion]

how would you make (3, 1) into (3, -5)

[crimson sedge]

add -6

or subtract 6

[winter lion]

yes

so it would go from (x, y) to (x, |y| -6)

now test it witha ll of them

so (-4, 2) would become (-4, 4)

(3, -1) would become (3, -5) for H

(-3, -6) would become (-3, 0) for J

etc.

[cedar kiln]

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[wide lintel]

need help with this word problem

[cosmic steppe]

Their distances are the same

You can represent their time

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[visual light]

Hi, I got this challenge to solve a partial diff problem when I havent learnt of that yet

[visual light]

This is the problem along with my solution

sorry if the quality is low, discord compression

id appreciate a nudge to the right direction if my approach is wrong so plz dont spoil the solution lol

<@&286206848099549185>

[thick cipher]

Here it is a variable separation problem. You need to solve explicitly for $\phi(x)$ and $\phi(y)$

[wraith dagger]

black_couscous

[still barn]

your step where you bring a function inside the differentiation operator and then cancel is very invalid.

If you have f(x) = x and calculate (1/f) df/dx that is 1/x, not d(f/f)/dx = d1/dx = 0.

[visual light]

Oh ok I see

Ive done so assuming that t = 1/s and thus substituting t with 1/s in the original term dst/dx^2 was

same for the other term but with t and s switched

alright thanks for the help guys

.close

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[crimson sedge]

x such that x/10 is an integer

or 10x where x is an integera

and then define the bounds for x

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[twin frigate]

What is this "P" thing? https://cdn.discordapp.com/attachments/780519481377423401/1033180339234947122/Screenshot_20221021_064909.png

[hollow minnow]

Power set

[twin frigate]

ty

.close

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[sand mica]

I was wondering if I could get help with this problem because I have the answer but I'm confused as to how to get there. I tried using the generalized binomial theorem to get the expansion 1 - (5C1)(-2x)^1 + (6C2)(-2x)^2...+(8C4)(-2x)^4. The answer is (-2)^4(-5C4) which is derived using the binomial theorem (I think). Why would I use the binomial theorem over the generalized binomial theorem?

[cedar kiln]

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@sand mica Has your question been resolved?

[young star]

@sand mica do you know how to expand the fraction on the bottom

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[sacred tangle]

.close

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[swift spoke]

what manipulation makes A^2 B^2 - ( A^2 B^2 Cos^2 g ) = A^2 B^2 (1 - Cos^2 g) ?

[red pumice]

you just factor out A^2B^2

[swift spoke]

if you factor A^2 B^2 in the R.H.S how does it still remain out? don't you factor all terms? like in - ( A^2 B^2 cos^2 g ) / A^2 B^2 it gives (1 - Cos^2 g) but A^2 B^2 still exists out there without being factored too

[swift spoke]

how does it still remain? don't you divide all terms by the common factor

[red pumice]

a^2 - a = a(a - 1)

a^2 - a is not a - 1

[swift spoke]

but here you said a^2 - a = a(a - 1) and in the equation a^2 is still there

why didn't A^2 B^2 go from A^2B^2 (1-cos^2 g)

[red pumice]

its just an example of factoring a variable out

[swift spoke]

shouldn't it be (1- cos^2 g) alone

[red pumice]

why should it be

[swift spoke]

show me the steps of that factoring process in my example

i am a bit confused

[red pumice]

okay suppose you have the equation

x - xcos^2(x)

then you could also write that as x(1 - cos^2(x))

[swift spoke]

no use the same equation

[red pumice]

cant just give you the answer

but do you understand the concept?

of factoring out a variable

[swift spoke]

this isn't an exam or whatever it is an answer already but I can't understand it

it is an example with an answer

[red pumice]

multiply out A^2 B^2 (1 - cos^2(scripttheta))

youll see that it equals the A^2 B^2 - A^2B^2 * cos^2(scripttheta)

[swift spoke]

so basically it was multiplied

[red pumice]

its just rewritten

[swift spoke]

so I get that a^2 (1-x) would equal a^2 - a^2 x

they're just the same form

[red pumice]

they are

[swift spoke]

thank you a lot

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[hollow minnow]

$x^5 + x^4 + 1 = p^y$

[wraith dagger]

Pure

Pure

[hollow minnow]

Find all pairs (x,y,p)

Well I notice that (1,1,3) and (2,2,7) are solutions

And ah…

[crimson sedge]

u meant (x, y, p) I think?

[hollow minnow]

Yes

I also notice this factorisation

$x^5 +x^4 + 1 = x^5 +x^4 + x^3 -(x^3 -1) = (x^2+x+1)(x^3-x+1)$

[wraith dagger]

Pure

[hollow minnow]

But not much from here unfortunately

[crimson sedge]

why not

oh

ya

[hollow minnow]

Lemme just

,w solve x^5 +x^4 +1=p^y for integer x y and prime number p

Come on

You’ve failed me

[opal basin]

Cant you subtract 1 and factor both sides?

[hollow minnow]

Why would that help?

[wraith dagger]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by Pure#9389

[hollow minnow]

Would help to know if there are other pairs or just these two

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[hollow minnow]

.reopen

[cedar kiln]

✅

[hollow minnow]

mb

[cedar kiln]

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[hollow minnow]

.reopen

[cedar kiln]

✅

[cedar kiln]

@hollow minnow Has your question been resolved?

[cedar kiln]

@hollow minnow Has your question been resolved?

[hollow minnow]

Anyone got an idea?

Let me just bring this to the top don’t mind me

.close

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[hollow minnow]

.reopen

[cedar kiln]

✅

[long matrix]

Can we take the factor a different prime exponents each

[crimson delta]

so we need that $x^2+x+1=p^k$ and $x^3-x+1=p^n$ for some naturals $n, k$. if $x$ is large enough (dont wanna check how large) then $x^3-x+1$ is bigger than $x^2+x+1$, so we need that $x^2+x+1$ is a divisor of $x^3-x+1$. then we can use polynomial division and the result has to be an integer

[wraith dagger]

Denascite

[crimson delta]

I think that works. haven't checked all the details tho

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[hushed badger]

@sacred grail

[cedar kiln]

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[crimson sedge]

what's the result of this infinite series:
tan(20°) - (tan(20°))^3 + (tan(20°))^5 + .....

[solid juniper]

that looks like it's probably supposed to be sin(tan20°)

the + at the end is a little confusing

[sacred grail]

doesn't taylor series for sin have coefficients tho

[solid juniper]

oh yes nvm

probably not that

[sacred grail]

maybe geometric series

[crimson sedge]

nothing wrong with the + at the end

[sacred grail]

[crimson sedge]

but the - is boggling me

[solid juniper]

summation pattern seems unclear

[sacred grail]

i never know whether to put a + ... or a - ... when the signs are alternating lol

if its alternating signs then its a geometric series

[crimson sedge]

it's not alternating

[sacred grail]

it isnt?

well

then i got no clue what the pattern could be

[crimson sedge]

Is this right?

Now I'm stuck

[sacred grail]

it should be a geometric series from the very first term

you don't need to separate it out

[crimson sedge]

the question maybe bugged

[bold hinge]

nah it's just simple manipulation of power series

[crimson sedge]

so how?

[bold hinge]

and that last sum should give us:
x - x^3 + x^5 + ...

[crimson sedge]

how did you get the i?

[bold hinge]

we need the square root of -1 here

because the powers of x differ by 2

so we need a number a such that:
a^1 = a^5 = a^9 = a
a^3 = a^7 = a^11 = -a

[crimson sedge]

can we introduce the tan 20 to the equation now?

[bold hinge]

i pretty much just solved it for you lmao

just let x = tan(20) and you're done

[crimson sedge]

thank you

.close

[cedar kiln]

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[merry tinsel]

could someone explain

why they did

x/0.25=h/0.5

or how they got that

[thick cipher]

Thales

[merry tinsel]

?

[thick cipher]

It might be Thales theorem

[merry tinsel]

Idk what that is

h is not 0.5 btw

[thick cipher]

Yes

[merry tinsel]

idk what thales thorem is

[thick cipher]

I can't draw it here so can you Google it please?

[merry tinsel]

but we are not in a circle?

[thick cipher]

Not that one

The one in a triangle

[merry tinsel]

i mean

we havent learnt this

so i doubt this was used

this was the full quesiton

[thick cipher]

If you have two parallel segments in a triangle you have a relationship between the lengths in the triangle

You might have learnt it with another name

[merry tinsel]

i still dont understand

how tf that was done

how is x=h

@thick cipher

[thick cipher]

It's a proportionality relationship that you have in triangles

But x is not h

[merry tinsel]

in fortnite terms please

[cedar kiln]

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[zinc vortex]

The r is 6, no?

I used the ratio test.

[cedar kiln]

@zinc vortex Has your question been resolved?

[cedar kiln]

@zinc vortex Has your question been resolved?

[obsidian tinsel]

seems like you are right

oh wait a ratio of 6 means it diverges, right?

@zinc vortex if r>1, diverges right?

[cedar kiln]

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[torn atlas]

think this part here is what I'm struggling with. I don't understand how he has gotten x(u,v)y(u,v) and so on? because we just had xyz, so I don't know how he obtained the right turn them all into functions? I was trying to parameterise it but its just xyz so I don't know how to isolate for any variable

[torn atlas]

here is the actual question (what i sent above was the solution provided)

[sacred grail]

<@&268886789983436800> what is this spam?

[torn atlas]

wait what?

[sacred grail]

dw about it

[torn atlas]

were you referring to me?

[sacred grail]

nah someone was spamming

[torn atlas]

oh okay

yeah so i'm just really confused with that bit in blue there

[sacred grail]

like

in general once you parametrise your surface

your surface will be represented in terms of the two parameters u and v

but like

you want to integrate the quantities x, y, z which are presumably like

the cartesian coordinates

so like

for a particular u, v

you need to know where that actually is in space

thats why its written as x(u, v) etc

its just saying

x(u, v) is the x coordinate in space on the surface corresponding to this particular u, v when using this particular parametrisation of the surface

[torn atlas]

okay so am i just subbing in the corresponding x value given by my parameterisation?

[sacred grail]

your parametrisation of the surface will tell you what the surface is

so then like

for a particular u, v

you'll get a position in space

on the surface

that position will be the (x, y, z) triple

so the x in that triple is x(u, v)

[torn atlas]

okay one sec let me try this out

[torn atlas]

okay so it looks like they've just subbed in the correspond x,y and z from the parameterisation ?

[sacred grail]

yeah

you just sub in

[cedar kiln]

@torn atlas Has your question been resolved?

[torn atlas]

tysm cant believe i missed that lmfaoo

.close

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[tropic knot]

I am trying to get the reflection of a line in a plane. First i wanted to intersect the line and the plane. But I cannot solve this.

[tropic knot]

so this is what i have done so far, but i get 0t on the left side while i was trying to solve for t. i guess that means there is no intersection point but then i don't know how to reflect the line in the plane.

[cedar kiln]

@tropic knot Has your question been resolved?

[tropic knot]

<@&286206848099549185>

[cedar kiln]

@tropic knot Has your question been resolved?

[tropic knot]

.close

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[ebon apex]

how do you find x2, y2 with only the equation of the line and x1, y1

[ruby lion]

example?

[ebon apex]

so point a's coodinates are (1, 4)

[ruby lion]

wdym by x2 y2, a specific x2 y2? or just any x2 y2 that satisfies the equation????

[ebon apex]

and the equation of the line is y=-3/2x + 11/2

i need to find the x and y coordinates of b

this is like the second half of the question actually

shall i start from the beginnning

[ruby lion]

do u have pic of the question?

[ebon apex]

yeah iill send

[ruby lion]

huh.. what other conditions do they give

[ebon apex]

heres the question

[ruby lion]

well that helps

do yk what perpendicular bisector means

[ebon apex]

line that cuts the other line in half at 90 degree angle

[ruby lion]

sure

what does it mean for the slope/gradient of the line then

if we have two lines and they are perpendicular what is the relationship between their slopes

[ebon apex]

isnt it like the reciprocal

[ruby lion]

kinda

well ye

[ebon apex]

negative**

[ruby lion]

mhm

so.

u have gradient

and and a point.

[ebon apex]

do i need to find the gradient of the other line

[ruby lion]

uh i now do not know what im doing

this question is

bad

[ebon apex]

its worded a bit funny

is there a way though to find the second set of coodinates of the line with just a gradient and the first set of points?

[ruby lion]

any point across the blue line expect for (1, 4) would be a valid b coordiante

u sure thats the whole question

[ruby lion]

uh kinda.

u find the equation of the other line as in AB.

[ebon apex]

that is the whole question

[ruby lion]

but seeing as they dont give u anything else to indicate the length of AB i would assume its any point on the blue line

[ebon apex]

i mean if we have (1,4) to the bisector that half of the line is it not

since a bisector cuts a line in half?

[ruby lion]

oh right

mbv

*mb

so it would be.

solvable

[ebon apex]

do you have to use the point gradient formula?

no actually nvm

[ruby lion]

yh

[ebon apex]

i have the gradient of AB

[ruby lion]

which is?

[ebon apex]

y = -3/2x + 11/2 right?

[ruby lion]

ye

[ebon apex]

im just confused on how to find the second set of points

at B

[ruby lion]

o

uhh

i would say....

actually idk

like i know how to do it idk how to explain it

[low bronze]

uh hi guys

[ruby lion]

#❓how-to-get-help

[ebon apex]

is there a formula?

[ruby lion]

man idk

[ebon apex]

maybe i just move on

[ruby lion]

maybe ask someone else

close this

and open a new one with same question

idk

[ebon apex]

hmmm

someones helping me actually now

so i should be alright

thanks for the help though

[ruby lion]

alr..

wait tell me how they did it tho.

[ebon apex]

they are saying to use

nvm they are confused too

[ruby lion]

what did they say

[ebon apex]

i think to find point of intersection with simultaneous equations

which is midpoint

and work backwards from there

[ruby lion]

yea thats what i was thinking

using the point of intersection find the distance from A to the intersection point, using that find the point B

but idk how to apply that

[ebon apex]

with the simultaneous part

nvm

i got that

i think i got the minndpoint

??

x = 35/13

y = 37/39

??

[ruby lion]

looks ok

[ebon apex]

ok so then 35/13 , 37/39 = (5/2 , x2 + y2 / 2)?

[ruby lion]

not quite

it would be

35/13 which is ur x coordiante

35/13 = 1+x2/2

and 37/39 = 4+y2/2

[ruby lion]

oh my lord so thats how apply it. oh my lordy lord im so stupid i overcomplicated it so much

[ruby lion]

making it neater

$$\frac{35}{13}=\frac{1+x2}{2}$$

[ebon apex]

wait so is mine right still?

[wraith dagger]

Mothy

[ruby lion]

right idea

wrong execution

[ebon apex]

ah ok

so from that find x

[ruby lion]

mhm

and do the same to find y

[ebon apex]

ok cool

can you do that funny thing with the y one too

[ruby lion]

lmao ok

$$\frac{19}{13}=\frac{4+y_2}{2}$$

because we have point a,

(1, 4)

we have midpoint

we use the midpoint as reference (x, y) to solve for unkowns

[ebon apex]

yeah that makes sense

ok my y cord must be wrong because the answer says its different

but the x is right

[ruby lion]

hm

check the midpoint

for y coordiante

[ebon apex]

got it

its 19/13

[ruby lion]

alr cool

as in y2

?

[ebon apex]

for midpoint

[ruby lion]

ah so just apply that to the equation

[ebon apex]

y2 is -14/13

which is correct

[wraith dagger]

Mothy

[ruby lion]

gjgj

[ebon apex]

so b's points are 57/13 , -14/13

i have a test on tuesday

if a question like that is in it

i might die

[ruby lion]

good luck lol

just do prac questions idk man

[ebon apex]

yeah i will

i mean ionce i understand the general concept of smth i can apply it reasonably easily

thanks for your help!!

[ruby lion]

np

[ebon apex]

alright shall i close the channel

how do you do that

[crimson sedge]

write '.close'

[ebon apex]

.close

[cedar kiln]

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[subtle sphinx]

A 24 in. piece of string is cut into two pieces. One piece is used to form a circle while the other is used to form a square. How should the string be cut so that the sum of the areas is a minimum?

[ruby lion]

idk

test

areas of each and see if there is a pattern?

[potent tinsel]

this seems like a tedious algebra lols

if you can use calculus it would be faster

[cedar kiln]

@subtle sphinx Has your question been resolved?

[subtle sphinx]

no lol

do it using algebra not trial and error

[cedar kiln]

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[cedar kiln]

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[frigid depot]

Hi, is anyone able to help me with this question?

[frigid depot]

mainly just interpreting it

im not too sure what it wants

[red pumice]

you create a while loop

such that the while loop increases N while the difference between real pi and the approximation is larger than tol

once your approximation is accurate to level tol (ie |x-pi| < tol) the while loop terminates and returns how large N had to become

[cedar kiln]

@frigid depot Has your question been resolved?

[frigid depot]

What do i increase the value of N by

[red pumice]

'just 1

[frigid depot]

oh so increase it by 1

[red pumice]

every loop you increase by 1

[frigid depot]

okay i give that a go

[red pumice]

the while statement should check wether your approximation is good enough

and if its not

it will make the approximation slightly better by increasing N by 1

[frigid depot]

k = 1; 
x = sqrt(12) * ((-3)^(1-k)/(2*k-1));
N = 0; 
while abs(x - pi) > tol
    k = k + 1; 
    x = x + sqrt(12)*((-3)^(1-k)/(2*k-1)); 
    N= N +1; 
end

this is what my code is doing now

[red pumice]

are you familiar with sigma notation?

[frigid depot]

not too familiar to be honest

doing a comp maths unit without doing any maths since high school was a mistake

[red pumice]

hahah no its good to challenge yourself

you should start N at 1 though

i like how you dont recompute the sum everytime

its almost correct

i think it just undershoots N by 1 because you start it at 0

[frigid depot]

so i just do N = 1?

[red pumice]

yes

[frigid depot]

huh that worked

but i dont quite understand

why

would you be able to explain please

[red pumice]

its just about sigma notation

you cant sum from k = 1 to N = 0

that would be backwards

[frigid depot]

Oh

right

[red pumice]

you can sum from k= 1 to N = 1 though

[frigid depot]

that makes sense

[red pumice]

thats just one step

[frigid depot]

i get that

i have to brush up on my math

before the final

or im doomed

[red pumice]

hahah

sounds like youll be alright

gl

[frigid depot]

anyways, thanks for the help

appreciate

[cedar kiln]

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[dry stump]

I've been trying to solve this optimization problem for a while but I can't rewrite the function solely depending on u. Anyone got any ideas?

[cedar kiln]

@dry stump Has your question been resolved?

[cedar kiln]

@dry stump Has your question been resolved?

[cedar kiln]

@dry stump Has your question been resolved?

[cedar kiln]

@dry stump Has your question been resolved?

[cedar kiln]

@dry stump Has your question been resolved?

[hollow nebula]

<@&286206848099549185>

[obtuse oxide]

just inteested, but whats 178465937438374 * 273927394827394 * pi

[silent sapphire]

Help

[cedar kiln]

@dry stump Has your question been resolved?

[dry stump]

Still looking for the answer...

[crimson sedge]

<@&286206848099549185> bro hes been waiting for a day 😭

[warm folio]

yeah i think it's just the problem that most can't solve

[bitter finch]

^

[fading wigeon]

I can't understand the question forget about solving it

[warm folio]

haha yeah

[fading wigeon]

@dry stump why dont u try asking this in the advanced section I think u might be able to find some help

[warm folio]

yeah this is real math, not the arithmetic that we usually call math

[dry stump]

Ok I will

[cedar kiln]

@dry stump Has your question been resolved?

[balmy apex]

@dry stump do you still need help?

[crimson sedge]

@dry stump still need help ?

[cedar kiln]

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[cursive harbor]

Not really math, more so problem solving. Not sure what to do.

I have unlimited tries but i can't seem to get it

[high spruce]

green always replaces where purple is

so you know where the next green would be

have a thikn where the next purple would be

[cursive harbor]

smart, did not notice that.

guessing purple moves one spot away?

[high spruce]

hm for me there are 2 possiblities

etiehr there is no purple

or it's the same image as teh first one

both would fit a logical pattern

[cursive harbor]

So my answer is Green at the bottom and purple above it

Let me think

I was thinking there was a trend of purple moving one spot down, now its at the bottom it would move one space up

let me see if it works

[cursive harbor]

This worked

What was the logic behind this one

Im guessing since purple reached the bottom, it restarted from the top?

[high spruce]

just look at the purple

[cursive harbor]

Oh

You're right

[high spruce]

and look at the green

the green goes from bottom to top

and then in the middle

and purple follow sthe same pattern just shifted

[cursive harbor]

Got you, it just clicked

Thanks man.

[high spruce]

np :)

[cursive harbor]

Sorry bro, got another one to look at, does this seem logical

[high spruce]

huh

why'd my message get deleted

[cursive harbor]

Not sure

[high spruce]

lol intelligence quotient isn't allowed

I asked if it's an intelligence quotient test

[cursive harbor]

Yes, doing a practise one as i got a real one coming soon

[high spruce]

ok lemme have a look

[high spruce]

why

it really doesn't matter bro

[cursive harbor]

Just to prepare idk

[high spruce]

ok the purple are correct

look at the green though

it gets reflected horizontally then diagnoally, then repeats

[cursive harbor]

Yep

so i thought it would be horizontal

[high spruce]

hm?

but we just had a horizontal one

so the next would be diagonal

also the one you did was vertical

[cursive harbor]

Not sure if i get it

[high spruce]

ok purples ones, get reflected about that same line each turn

[cursive harbor]

got it

were you referring to the original image being reflected diagonally

[high spruce]

green lines get rotated by the horizontal line below them, and then the diagnoal line that cuts through the figure

it's hard to verbalize with words tbh

[cursive harbor]

I think i get it

so it would be in the top left like the original?

[high spruce]

yeah the green would be like in the original

[cursive harbor]

i assumed it was just going in a square lol

right down left up e.t.c

[high spruce]

I see

wait what I'm hallucnating lmao

sorry I thought your image was the second in the sequence

urm yeah then I'd go for what you did

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

hello help

[crimson sedge]

these the questions right

and these r my answers

but my answers dont@match the multiple choice answers

[modest fossil]

for the first one you just made a mistake in the subtraction and addition

1000 - 300 + 30 isn't equal to 1000 - 330

[crimson sedge]

oh oops but still subtracting@correctly would equal to 730

that’s still not an answer choice

[modest fossil]

but 73 is

divide by 10 (i think) for average rate of change would give 73

[crimson sedge]

ohhh cuz y2-y1 OVER x2-x1

bruhhh that was easy

what abt the second one?

[modest fossil]

i don't know limits that well

idk

[crimson sedge]

😔

[modest fossil]

i think you mixed up cosine of pi/6

it's square root 3 over 2

not square root 2 over 3

which i think would lead to one of the multiple choice answers

[crimson sedge]

hmm let me try that rq

bruhh they were such simple@mistakes

@modest fossil ty!!

[modest fossil]

np

simple mistakes usually go unnoticed

[crimson sedge]

yeah i got 148 other problems like this to go thru too i hate math

bye

.close

[cedar kiln]

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[abstract locust]

Help

[velvet mortar]

cos^2 theta + sin^2 theta = 1 might help

Wait you don't even need that

@abstract locust you almost did it complete

Make (cos ^2 theta - sin^2 theta) into - (... + ...)

Then you can cancel the like terms

And get - cot ^2 theta

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[forest socket]

$2x^2 - 4x(sqrt(6)/2)-8x +6.5 - 4sqrt(6)$

[wraith dagger]

Tomasz

[forest socket]

does the $4xsqrt(6)/2$ go to $-8x$

[wraith dagger]

Tomasz

[forest socket]

$2x^2 - 12x(sqrt(6)/2)- +6.5 - 4sqrt(6)$

[wraith dagger]

Tomasz

[forest socket]

so it looks like this

and where is C is C just (6.5-4sqrt(6))

[turbid bluff]

$2x^2 - 4x(\frac{\sqrt{6}}{2})-8x +6.5 - 4\sqrt{6}$

[wraith dagger]

Wraith

[forest socket]

yes

does the $4x(\frac{\sqrt{6}}{2})$ adds to $-8x$

[wraith dagger]

Tomasz

[turbid bluff]

Quick answer: No

remove x from the equation and you just have 2 constants

those constants are sqrt(6)/2 and -8

[forest socket]

it is quadratic equation

[turbid bluff]

those two numbers are not equal each other

yes

[forest socket]

i need $ax^2 + bx +c$

[wraith dagger]

Tomasz

[forest socket]

i have a

a = 2, b=?, c=?

[turbid bluff]

okay, so you need "coefficients" for each of the terms a, b and c?

were you asked to find the roots of the quadratic?

[forest socket]

i need to calculate delta and x1 and x2

hallo

[turbid bluff]

To your first question, just collect the coefficients together;
So lets take another look at your equation...
You have one term with x^2 in it, that term is 2x^2
So your value for a = 2
You have 2 terms in front of single "x",
Those values are:
-4x(sqrt(6)/2)
and
-8x
We can collect these terms together like this:
-4x(sqrt(6)/2) -8x
= x(-4(sqrt(6)/2) - 8)

Notice how we've removed x as a common factor^^^

This means that (-4(sqrt(6)/2) - 8) is your value for b, but we can simplify it further:
-4 times sqrt(6)/2 is just -2sqrt(6)
So your value for b is -2sqrt(6)-8

Your value for c is just the constants and can stay the same,
Personally, I don't like decimal points and prefer fractions (your teacher should too...)
c = 13/2 -4sqrt(6)

[forest socket]

i think i understood

[turbid bluff]

So, you can rewrite your entire equation as:
ax^2 + bx + c

$2x^2 -(2\sqrt{6}+8)x + \frac{13}{2} -4\sqrt{6}$

[wraith dagger]

Wraith

[turbid bluff]

what is delta, x1 and x2?

Are they root values?

[forest socket]

$delta = b^2-4ac$

[wraith dagger]

Tomasz

[forest socket]

x1 and x2 are places where Y=0 on X axis

$x1 = \frac{-b-\sqrt{delta}}{2a}$

[wraith dagger]

Tomasz

[turbid bluff]

well, there you go!
You know how to do quadratic equations? (also, you know how to multiply with sqrts?

[forest socket]

i can do some quadratic equations but what do you mean multiply sqrt

[turbid bluff]

so b = -(2sqrt(6)+8),
or b = (-2sqrt(6)-8)
Would you know how to square this?

[forest socket]

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

is that it

[wraith dagger]

Tomasz

[turbid bluff]

yes, but remember both parts are minus

so what you want is

[forest socket]

both minus which parts

[turbid bluff]

$(-2\sqrt{6}-8)^2 = (-2\sqrt{6})^2 + 2(-2\sqrt{6} \times -8) + (-8)^2$

[wraith dagger]

Wraith

[turbid bluff]

as in. your (original equation) b value can be split into two parts for that particular "coefficient", both parts of which are negative numbers -2sqrt(6) and -8

[forest socket]

so?

is the formula just

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

[wraith dagger]

Tomasz

[forest socket]

but when there is minus i just add negative number

is that it?

that's why the symbols change between $(a+b)^2$ and $(a-b)^2$

[wraith dagger]

Tomasz

[turbid bluff]

Im getting a little confused myself now :S

[forest socket]

like what you did here

[forest socket]

this is one formula it changes becasue some numbers are negative

[forest socket]

this

[turbid bluff]

whats confusing to me is this:
is
(-a-b)^2
the same as
-(a+b)^2
?

you mind if I call in some helpers?

Unless you're not confused ofc

[forest socket]

those are diffrent i think because minus is out of ()

but i think there is only one formula (a+b)^2 = a^2 + 2ab + b^2

[turbid bluff]

but both were valid as original answer for original b?

wait

[forest socket]

what i am saying now may not be conected to example i gave

[turbid bluff]

it should be (-(a+b))^2

[forest socket]

waht

[turbid bluff]

hang on

ignore me

sorry if I made unnecessarilly confusing

[forest socket]

you made me confusing or confused?

[turbid bluff]

If I made the problem confusing, I hope I havent confused you :S

[forest socket]

$(a-b)^2 = (a+(-b))^2$

[wraith dagger]

Tomasz

[forest socket]

that's what i mean

[turbid bluff]

that is true

[forest socket]

there is one multiplication formula

just some numbers are negative

that's what i meant

[turbid bluff]

okay

I could do the first (more difficult part) apparently but not the bit that should be elementary xd

I don't think I can help you more and I'm scared I'll make it worse

good luck! Don't be afraid to use helpers if needed

[forest socket]

you helped me with moving x out of ()

[turbid bluff]

np

good luck!

[forest socket]

may i ask what is your position

[turbid bluff]

terrible uni student

[forest socket]

are you teacher or to what school are you going to

ok

[turbid bluff]

maths undergrad level

[forest socket]

second grade of highschool me

[turbid bluff]

I'm usually better at helping highschool level than uni lol

usually I need help with my own studies at uni level

if i can help people though with something I'm okay at then that is good for me

[forest socket]

if it is ok

i liked working with you

and could i directly ask you questions if it will be ok

do you agree

[turbid bluff]

nope :P

[forest socket]

ok

[turbid bluff]

there are better mathematicians here and I like being able to do this here and there when I have time

[forest socket]

ok

[turbid bluff]

thank you for the compliment though :v

[forest socket]

ok goodbye

[turbid bluff]

later bud, good luck

[forest socket]

.close

[cedar kiln]

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[hoary charm]

How did they do the calculation in 2.1.16

to get v = (2v1 - v2)u1 + (-v1 +v2)u2

[crimson sedge]

we want $v=au_1+bu_2$, ie $v_1=a+b$, $v_2=a+2b$. Now write $a$ and $b$ in terms of $v_1$, $v_2$

[wraith dagger]

Toby

[hoary charm]

Ohh

ty

[crimson sedge]

np

[hoary charm]

@crimson sedge Does it work the same way for 2.1.17

I always get confused switching between vectors as coordinates and vectors as functions

[crimson sedge]

yup, its the same

here, if you expand the right, you will get precisely p(x)

[hoary charm]

could you walk me through the process

i'm not exactly understanding

even though its supposed to be the same

[crimson sedge]

treat (a_0-a_1)1+... as a normal polynomial

expand out the brackets and cancel terms

[hoary charm]

oh I meant getting to p(x) in terms of the basis

[crimson sedge]

do you prefer working with matrices or linear equations?

[hoary charm]

marrives but for this question

*matrices but for this question Id prefer linear equations

[crimson sedge]

we want p(x) in terms of the basis, so a_2x^2+a_1x+a_0=b*1+c(1+x)+d(1+x+x^2)

since we know {1,x,x^2} is linearly independent, we can compare coefficients

eg a_1=c+d

then solve for b,c,d in terms of a_1,a_2,a_3

you can translate this problem (and the previous one too) into matrices

[hoary charm]

a_1 = c+d?