help-17

[short pilot]

For a polarized sphere it should be distributed

[lime smelt]

but that's bounded charge

isn't the whole point of using electric displacement instead of the electric field to not talk about that

[short pilot]

Oof

Sry sir

I am not that much aware bout the displacement method

I thought it was gauss law question

[lime smelt]

alright

i'll just figure it out on my own then

.close

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[short pilot]

@lime smelt

U can ask in the physics server

You can join it

By visiting #old-network

[lime smelt]

alright

thanks

[short pilot]

Welcome

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[stoic light]

need help with this

[stoic light]

give sequence

[tawny nacelle]

what have you tried so far?

[stoic light]

i dunno how to give sequence

so i know that it must be 6 jordan blocks

then make a sequence that adds up to 10?

[tawny nacelle]

perhaps it is a good idea to calculate the size of the largest block?

can you do that?

[stoic light]

is 6

[tawny nacelle]

it will necessarily be part of the sequence, and it's gonna force the possible sequence options to be smaller

[tawny nacelle]

the largest block is of size 6?

[stoic light]

yeah

[tawny nacelle]

how did you get that?

[stoic light]

oh no

[tawny nacelle]

note that if the largest block is of size 6, then even if the rest are of size 1, we can't make a sum equal to 10 because 6 + 1 + 1 + 1 + 1 + 1 = 11

[stoic light]

is it 5

[tawny nacelle]

there's 6 blocks

[stoic light]

the largest

[tawny nacelle]

what is your method for getting these numbers?

[stoic light]

5,1,1,1,1,1

[tawny nacelle]

I would really want you to answer this...

[stoic light]

like what numers

the largest size of the block

[tawny nacelle]

the largest block size

[stoic light]

?

[tawny nacelle]

yeah

[stoic light]

actually i dont know

[tawny nacelle]

how are you getting 5?

oh, hm

so there's a theorem stating that the size k of the largest Jordan block is going to be the smallest k such that rank((T - lambda * I)^k) = rank((T - lambda * I)^k+1)

you can use this to deduce what the largest block size is going to be

what k satisfies this property?

[stoic light]

k = 1?

[tawny nacelle]

not quite

hint: look at your sequence of nullities

there should be rank-nullity alarm bells going off

[stoic light]

i mean cuz i dont quite know how to do the rank

the non zero vector of the matrix

[tawny nacelle]

can you state rank-nullity for me?

[stoic light]

null is number of non_leading col

and rank is number of non zero row

?

[tawny nacelle]

nono, the rank-nullity theorem

have you learned it?

[stoic light]

The rank-nullity theorem states that the rank and the nullity (the dimension of the kernel) sum to the number of columns in a given matrix

[tawny nacelle]

sure

that's good enough (although, I prefer writing this as rank(T) + nullity(T) = dimV but w/e)

okay, then we automatically have that dimV - nullity(T) = rank(T), yeah?

in our case, dimV = 10, and T = (A - 2I)^k

[stoic light]

yeah

[tawny nacelle]

so 10 - nullity((A - 2I)^k) = rank((A - 2I)^k), we agree?

[stoic light]

yep

[tawny nacelle]

okay, so now we can see that rank((A - 2I)^k) changes if and only if nullity((A - 2I)^k) changes

since 10 is a constant

[stoic light]

yeah

[tawny nacelle]

now, this says that if rank((A - 2I)^k) = rank((A - 2I)^k+1), then k is the size of our largest block

[stoic light]

ok

[tawny nacelle]

so we just need to look for the k such that rank((A - 2I)^k), and thus nullity((A - 2I)^k), stops changing

what is that k?

reposting for convenience

when does nullity((A - 2I)^k) stop changing values?

[stoic light]

4

[tawny nacelle]

!

done, that is our largest block

[stoic light]

ohhh ok

[tawny nacelle]

now, what're our options?

[stoic light]

4,2,1,1,1,1

[tawny nacelle]

that's a valid choice. are there any others?

[stoic light]

4,1,2,1,1,1

[tawny nacelle]

that are not permutations, that is

is there any way to cut up 10 into 6 pieces, one of them being a 4, other than this way?

(this is not a trick question)

[stoic light]

no

[tawny nacelle]

good

so this should be your sequence

[stoic light]

ohh ok

[woeful stream]

you can also use 8 and 9 to double check your sequence

[tawny nacelle]

I didn't double check myself

so I'm slightly worried I'm wrong, but I don't suspect I am

okay, I checked

it seems like we should indeed have 4 blocks of size 1

so this is good

[stoic light]

does your method works in this example

[woeful stream]

yeah this one needs the other numbers

iirc the sequence has to be concave

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[vast shale]

hi

[short pilot]

@vast shale

[vocal sleet]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[vast shale]

yea wait a little pls🫸🏻🫷🏻

[short pilot]

Okok

[vast shale]

bruh my discords tripping 🙏🏻🙏🏻

[short pilot]

You got

[short pilot]

6(b cross c)

Now express b as a or c

[vast shale]

what am i doing wrong

[short pilot]

and u get the answer

b=-3c-a/2

[vast shale]

hold up

[vast shale]

lemme do this

[short pilot]

Ok

Cz see in answer be have

a cross b

Or c cross a

Not b cross c

[vast shale]

true

[short pilot]

Try

[river kettle]

Punpij

[vast shale]

i got 3(c cross a) 😭😭

[vast shale]

😮‍💨

ok lmao i got it, we had to replace c vectoe

thanks i was so close man 😭😭

.close

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[blissful panther]

Consider a mines game with n^2 = 25 where n is the amount of mines/gems.

m = amount of mines in the grid
g = (n^2) - m. The amount of gems in the grid
r as revealed gems

When m = 3, g= 22 and revealed gems are 5. The payout multiplier is 2

When m = 24, g= 1, and 1 gem is revealed, the payout multiplier is 24.75

Find the equation for the payout muliplier considering m, g and r.

[blissful panther]

I've been struggling with this and reaching nowhere

I reached an approximation that multiplier payout = 1/P
where P is the probablity of revealing amount of gems in n^2

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[past badge]

What does this mean?

<@&286206848099549185>

[blissful sentinel]

Please don't ping Helpers until it's been at least 15 minutes

[past badge]

My bad

its been 15 min now

<@&286206848099549185>

[vocal sleet]

@past badge Has your question been resolved?

[dusky mural]

What do you have to graph if your function has +3 at the end?

[past badge]

3 up

[dusky mural]

But in this case it’s 3 to the right

So you sketch the smiley at (8, 1)

For the second one, how many degrees is pi/4

[past badge]

45

[past badge]

why right

[past badge]

but is it clockwise or anticlockwise

[dusky mural]

Counterclockwise

[dusky mural]

Because adding a number to z on the complex plane moves it along the x axis

[past badge]

answers

did clockwise

[dusky mural]

Oh I don’t know then. My bad

[past badge]

nw

[dusky mural]

Wait this is divided by e right?

Because when divided by euler’s identity, then it rotates clockwise

When z is multiplied then it rotates counterclockwise

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[rancid gust]

is this factorable? if so, how?

[gentle thicket]

kinda yeah

[rancid gust]

how could i go about factoring it?

[gentle thicket]

try to sub sqrtx = u

[rancid gust]

ok, thanks

.close

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[smoky pebble]

how is possible for in right side inequality it doesnt make sense |z1-z2|=|z1+z2|

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[dry pecan]

In this diagram, xy and x’y’ are parallel tangents and we have to prove that <AOB is 90 by proving that the square of side AB is equal to the sum of squares of sides BO and AO I.e we have to use PGT to prove that the angle is 90.

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@dry pecan Has your question been resolved?

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@dry pecan Has your question been resolved?

[hallow crescent]

OC is the height of the triangle ABO?

also is AB tangent to the circle at C?

PO = PQ = OC := 1 (lets set it to 1 for easier computations)

$\angle OQB = 90 \$
$=> OB = \sqrt{(1+QB^2)} \$

$\ \angle APO = 90 \$
$=> AO = \sqrt{(1+AP^2)} \$

[twin meteor]

woomy

[hallow crescent]

use pythagoras to find W and Z, and then show that (z+w)^2 = the sum of the other 2 sides squared

@dry pecan

[dry pecan]

Can u elaborate it further a bit?

What is AP and QB here?

[hallow crescent]

this is AP

and this QB

QB is a side of the right triangle OQB, Q is the 90degrees angle and OB is the hypothenuse

OQB

what we do is calculate OB using pythagoras

since Q is 90degreese

OB^2 = OQ^2 + QB^2

AP is the side of the triangle APO, P is 90 degrees, so AO = AP^2 + PO^2

sorry i was at lunch

@dry pecan

[dry pecan]

Thanks a ton !

[hallow crescent]

did you manage to prove the question?

if you have any other questions or still dont manage, feel free to ask :)

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[wary moon]

Say 1a=3b
And
ax=½b

[wary moon]

I think it's written like this

since the original one is 1tablespoon=3teaspoon but what is _tablespoon=½teaspoons

[onyx juniper]

Substitute 3b into the second equation to get 3bx = b/2

[wary moon]

So bx=b/6?

[onyx juniper]

Yes

[wary moon]

Now wat

[onyx juniper]

You can cancel the b on both sides to get x = 1/6

[wary moon]

So ½teaspon=⅙tablespoon?

[onyx juniper]

You can check by plugging in x= 1/6 and see if it matches with the original equation

I'm not familiar with teaspoons and tablespoons but if a=3b is correct then it should be right

[wary moon]

so a⅙=½b

if ax=½b

[onyx juniper]

Yeah, and then if you multiply everything by 6, you get a=3b which checks out

[wary moon]

Oh damn ty

.close

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[meager shoal]

is

that

supposed to be an double integral

@untold arrow

u can use polar coordinate to get easy answer

yes

can i see how u set up the integral

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[vast shale]

What is the difference between |f(x)| and f(|x|)? How would I go about graphing them?

[meager shoal]

|f(x)| is u get the y first and then make y positive

for f(|x|) make the function even by making the function the reflection of the positive x-axis values

[vast shale]

Just for clarification, by “get the y first” do you mean to make the output positive?

[meager shoal]

yes

[vast shale]

Oh okay then, thank you very much. That’s all i was wondering

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[stark sparrow]

I need some help creating the graph

[stark sparrow]

Helllo

Yea I just forgot it

Like for me I just somehow can’t make it

I forgot a lot

wdym?

Okok

VA is a discontinuity

Yea

Is this alright for now?

A non removable is a jump right?

This is AP calculus AB, but school just started so it’s just review

Nice

so there’s two holes?

What about this?

Oh

Should I get someone else maybe?

Okok

like there could be anything, it’s not specific since it does say any zeros

But what about the relative maximum I feel like it doesn’t seem right

Ill try number 7

But isn’t their two holes? You said a non removable is a hole too?

So it’s a jump?

But there isn’t a point to jump to

Oh ok

What does it mean by continuos for all x

Is this number seven

Even it’s like where it does change if reflected on the y axis

Wait it says increasing 0 to 2 not 0.2

Oh lol

How should it look like then

Ok thanks

<@&286206848099549185> is number 7 correct?

.close

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[acoustic patrol]

Hey I'd like to ask how to make an equation to find the slightly dark triangles figure 8

[acoustic patrol]

An example is number 7

[weary sparrow]

Hmmm

It changes based on weather the number is even or odd

Maybe try going from there

(I'm just guessing)

[acoustic patrol]

I don't know where to start. For number 7 I got the equation because there were 5 slightly dark triangles surrounding the 3 dark triangles

And there were 3 groups so 3 • (3 +n) would be the equation

N equaled 3 because figure size was 3 so if you plugged that in you would get the exact slightly dark triangles

[weary sparrow]

Ok

For even

Example 2 in the figure

First we do 3 x 2

For the sides

3n

Now idk how to go about the triangle tbh

But n+1 will be the side

[acoustic patrol]

Wait which one are we doing?

Which figure

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@acoustic patrol Has your question been resolved?

[weak mason]

the first one

no 8

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@acoustic patrol Has your question been resolved?

[acoustic patrol]

3(n+1) ?

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[untold arrow]

what's r:(a-c,b-c)? i cannot understand

[dull bear]

r is a function from the open interval (a - c, b - c) to the reals

[untold arrow]

why do they make such a function

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@untold arrow Has your question been resolved?

[granite flicker]

it's a way to make it feasible to represent the derivative without a limit. often makes calculations easier. it basically represents an error function wuch that when you divide by h and take the limit it would vanish leaving the derivative being A.

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[warm horizon]

the RHS is (1+x)^{(n+1)choose2}

but how does 1 + 2 + 3 + ... + n equal to n+1 choose 2?

[viral shell]

do you know what the formula for sum of first n natural numbers is?

[warm horizon]

its like

n/2 * (first term + last term)

i think

[viral shell]

yes thats the general arithmetic sum formula i think

for first n numbers the first term is 1 and the last term is n

so n(n+1)/2

[warm horizon]

yeah

[viral shell]

if you write out (n+1) choose 2, the (n-1)! cancels out and you're left with n(n+1)/2

[warm horizon]

oh ok

is part (b) an equating coefficients question ?

so like using part a) equate coefficients of x^{(n+1choose2)-1} on both sides?

[viral shell]

they're adding the coefficients of x^1

[warm horizon]

oh

its the same thing though right

[viral shell]

ig this works better

[warm horizon]

idk abt that

equating that makes me more confused more

x^1 makes sense tho i can just equate coeffs of x^1 on both sides

and then use symmetry identity to get it to the thing in the question

[viral shell]

problem with x^1 is that multiply it n times would give you x^n

and there's multiple ways to get x^n like say x^n of (1+x)^n and 1's of all other terms

so multiplying x^(n-1)th terms would be better

[warm horizon]

o ok

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[warm horizon]

thank u

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[dire comet]

I have this hyper specific question which I can't find a way to solve, although I have not tried to solve it, I just know I can't because I don't know how to solve in case of something is ever-growing rate especially with a lot of things to consider. chatGPT didn't provide accurate answers and I don't know what else I can do other than posting it here. So here's the question

Let's say there exists an abandoned village where all of the village's buildings are already built with only 1 member, he one day decides to recruit more from an outside nation.

He meets x number of people every day and recruites 75% of them

Out of the new members that join, 50% of them decide to not recruit more. 30% who decide to recruit but aren't good, meeting 30% less people each day and recruiting only 50% of them, the remaining 20% decide to recruit and are as good as the original member.

In the span of a year, they need a population of 1200 (1000 adults + 200 children (below 15))

Children are easier to recruit as compared to adults with a recruitment rate of 90% for people that are not that good at recruiting and 100% for those who are.
But they also recruit less children as compared to adults so that the given adult to children ratio is satisfied.

Find the value of x.

[hard atlas]

are the children also going to recruit others?

[dire comet]

No

[hard atlas]

so they belong to the 50% who decide not to recruit?

[dire comet]

Yes

You can also decide to alter some value to make calculations easier if you want to

[hard atlas]

1200 people after a year means that they on average only need <4 new people per day. so thats basically nothing

unless they truly fuck up they will get a lot more

[dire comet]

That was a lot simpler than i thought

[hard atlas]

so the true result would probably be something like x=1.something. which in the context of the model makes no sense

even if just your original guy meets 5 people per day and recruits 75% you get enough. and thats ignoring all the other people who could start recruiting

[dire comet]

So let's change the question a bit (for private reasons)

Let's say value of x is 5. Then what is the amount of people that will be recruited in a year?

[hard atlas]

probably more than on earth

[dire comet]

Lol

[hard atlas]

exponential increase is quite crazy

[dire comet]

Yeah

I'm always intimidated by scary looking questions

I need to change that

[hard atlas]

lets say at some point you have 100 people. lets simplify even more and say that only 50 of them recruit and only have a 50% chance. so 50 people go recruit, find 250 people and convince 125 of them to stay. so just after one day you are at 225. then 112 of those go and recruit, meet roughly 560 people and convince 280 of them, so you are at 500 already

so 2 days means a factor of 5

and thats with worse recruiting skills

so over a full year, thats roughly 5^180 = fucking lul

[dire comet]

That's crazy

Thanks bud

Can I close now?

[hard atlas]

sure

[dire comet]

.close

[hard atlas]

its your channel

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[indigo sigil]

.reopen

[wheat zodiac]

x² - 6x - 41 = 0

[wheat zodiac]

hi

[vast shale]

wt about this

[wheat zodiac]

help me solve this pls

x² - 6x - 41 = 0

[vast shale]

okay

do you know discriminat formula

[wheat zodiac]

no

[vast shale]

then what formulas do you know for this

or what have you tried

[fervent wasp]

@wheat zodiac what have you tried?

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[astral cairn]

i need to integrate and uh

calc gives this but idk how

[dull dagger]

The rule is
Integration of sin(ax)=(-1/a)cos(ax) +c

[desert hornet]

!nosols

[vocal sleet]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[vast shale]

do you know about u substitution?

[dull dagger]

It's not answer

[astral cairn]

u substitution?

[vast shale]

yeah

[astral cairn]

ive been 2hours to integral so i dont think so

[vast shale]

oh

well id advise you to learn u sub

but you can also solve it with simple inspection

if you need to integrate -sin(3x), then what do you need to differentiate to get that?

you should know chain rule

[astral cairn]

-sin comes from cos

[vast shale]

yes

but we need 3x and not x

so we need to differentiate cos(3x)

what do you get when you differentiate that

[astral cairn]

wait do i need to solve cos(3x) with uh

is it this or Dg( f (x))= g′( f (x)) ⋅ f ′(x)

[vast shale]

yeah

thats chain rule

[astral cairn]

ye i know it

[vast shale]

so using that what do you get

[astral cairn]

my english bad related to math

wait

3cos(3x)

[vast shale]

thats not right

we're differentiating cos how can you get cos again

[astral cairn]

oh yea cos -sin

-3sin(3x)

[vast shale]

yes

[astral cairn]

but thats not done i think

[vast shale]

but we need to get -sin 3x not -3sin 3x

[astral cairn]

isnt g(x)=cosx f(x) 3x

[vast shale]

no no that part is right

but we need to differentiate something and get -sin(3x) instead of -3sin(3x)

so instead of just differentiating cos(3x), we can differentiate (1/3) cos(3x)

this way, when we differentiate, the 3 and 1/3 cancel out

[astral cairn]

oh yea I didn't even think about it

[vast shale]

but you really shouldnt be using inspection to do integration

[astral cairn]

inspection?

[vast shale]

u should solve it using u substitution

[vast shale]

like basically guessing the answer

the way we did

[astral cairn]

i think u substitution comes later im in school

first integrate lesson

[vast shale]

we guessed cos 3x, but that ddint work

so we modified our guess to 1/3 cos 3x

inspection is basically just guessing

[vast shale]

oh

in that case yeah i guess u do have to use inspection

[astral cairn]

yea for now

[vast shale]

theyll teach it soon enough its a pretty basic technique

[astral cairn]

thanks for the help still

[vast shale]

np

[astral cairn]

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[heady quest]

Hey so we took multivariable but i have no idea what these type of questions are... Does anyone know the exact name of this subject

[chrome raptor]

Ordinary Differential Equations

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[violet flare]

hi, in a problem where a parametric equation of a line is given and an equation of plane too ,is asked for the projection of the line to the plane , i found the perpendicular to the plane vector and the direction vector of the line and then i substituted this to the formula with the inner product of the projection of a vector to another , the problem is that the result isn't the same as the solutions one , i checked in chatgpt and the right result came up by substracting what i found in the direction vector , is what i did right ? why the right vector came with the substraction ?(i have solve it with other ways i am intersting fo this one )

[spiral turtle]

!show

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Show your work, and if possible, explain where you are stuck.

[violet flare]

that the equation of line and plane that are given respectively ε:(x,y,z)=(15,-12,17)+t(15,-15,11) to
π:13x-9y+16z-69=0 , the direction vector is u=(15,-15,11) and the perpendicular to the plane is (13,-9,+16) , they arent parallel cause they arent multiple of eachother , i found one point od intersection and in order to find the equation of the line that ε projects to π i want to find the direction vector by applying the proj(u->v)=(<u,v>/<v,v> )v , by that i find proj(u->v)= (13,-9,+16) which dont match with (2,-6,-5) the solution but i search for that in chatgpr and when i substract from (15,-15,11) the proj(u->v) i get the result why this happens ? is just a coincidence ? is my method valid ?

@spiral turtle

[spiral turtle]

oh, you projected onto the normal vector of the plane

yes, subtracting is the proper way. You found essentially "How far am I from the plane" by projecting onto the normal of the plane

then you can pull yourself to the plane by subtracting it.

@violet flare

[violet flare]

@spiral turtle yeah , thats cool thanks mate have a good one legend

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[iron carbon]

-700000000000000 is answer

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[maiden egret]

is this correct?

[meager shoal]

C and D should be switched

if u hold radius constant, dr/dt should be zero and not be in the final expression

[maiden egret]

ok, thanks

.closed

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[fossil gulch]

Help

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@fossil gulch Has your question been resolved?

[vast shale]

what course is this

calculus level 9999

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[south ember]

Hello, i need help to calculate this limit pls

[south ember]

without using lhopital

[shrewd widget]

why not

that's the easiest way

[south ember]

We did not see it yet

[shrewd widget]

then try expansion

[inner osprey]

i would probably start by using x - 2 = t, just to make the expression a little cleaner

[inner osprey]

yes

now distribute the e^u

[south ember]

ok

then?

[inner osprey]

then u will probably see what to do next

[south ember]

this?

[inner osprey]

yes

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[south ember]

thanks

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[violet flare]

hi , if i have a vector equation of a line , can i find two of the planes that have section this line ?

[timid flint]

In 3d you should be able to. X E.

Are you looking for how?

[violet flare]

@timid flint yes mate ,is the following way right ?
i will substitute the variables x,y,z of the parametric equations of the line in Ax+By+Cz+D=0 analytic form of the wanted plane the i will get t(Aa+Bb+Cc)+(Ax +By +Cz +D)=0 , that gives me a system of 2 equations , is it right to set arbitary non 0 values for 2 of A,B or C and solve for d ?

[timid flint]

Sorry I am no expert, unsure. X E.

[violet flare]

<@&286206848099549185>

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@violet flare Has your question been resolved?

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@violet flare Has your question been resolved?

[maiden nebula]

@violet flare my initial instinct is to construct two planes that are perpendicular and their intersection is that line

suppose the line has equation s(t) = tv + c
so t is the parameter
c is the root point
and v is the direction vector

i would first find some vector w that is perpendicular to v

so for example, if v = (1, 2, 3)
i might choose w = (2, -1, 0)

Now, one of the two planes you’re after has w as a normal vector and c as a root point.

can you see why this plane contains the line s(t)?

[twin meteor]

Ryan

[maiden nebula]

you can use this equation to in fact prove that that plane contains s(t). i encourage you to try to prove that on your own

once you do that, all you have to do to find the second plane is find a third vector u that is perpendicular to both v and w and do the same process again.

(actually, u doesn’t have to be perpendicular to w. u DOES have to be perpendicular to v, but it just needs to not be a multiple of w. nothing in the problem states that the two planes need to be perpendicular, that was just my “initial instinct” from earlier)

and one last note, to actually justify why the intersection of these two planes is s(t),

• both planes contain the line s(t)
• the planes are not parallel (because u is not a multiple of w), meaning their intersection is a line

conclusion: their intersection is exactly the line s(t)

hope this helps. let me know if you have questions

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[violet flare]

.reopen

[vocal sleet]

✅

[violet flare]

@maiden nebula that's cool mate , i could't see the equation of a plane written like the inner product of the perpendicular to himself vector and one other vector that belongs to plane , i think that helped a lot ,as for proof that the plain contain s(t) using this formula is this done by substituting the parameters of the parametric equation of the line to the formula and the cause <v,w> are perpendicular this stands true , am i right ?

[violet flare]

cause v is parallel to the plane and they have a common point ?

[maiden nebula]

correct! good job.

[maiden nebula]

you’re correct they have a common point, but why exactly is v parallel to the plane?

[violet flare]

@maiden nebula what do you think about the way founded i suggesr=ted before ? In an equation of a plane Ax+By+Cz+D=0 does A,B,C take all the combinations of values unless A^2+B^2+C^2!=0 ?

[violet flare]

cause its perpendicular to w as it is the line too ? am i missing sth ?

[maiden nebula]

yes that’s correct. good job

[maiden nebula]

i don’t really understand what you’re going for here

[violet flare]

that's how i think of that :
i have the parametric equations of a line x=x0+at y=y0+bt z=z0+ct and i want to get an equation of this form Ax+By+Cz+D=0 so to be sure that the points of the line are in the plane i substitute the parametric equations to the wanted one and i end up in this t(Aa+Bb+Cc)+(Ax +By +Cz +D)=0 which gives that
Ax +By +Cz +D=0 and Aa+Bb+Cc=0(cause t takes all the real values in order to describe the line ), then cause A,B and C can take all the possible combinations of real values except A,B and C such A^2+B^2+C^2!=0 i choose 2 of them arbitary and then i solve for C and D and find an equation of a plane

[maiden nebula]

i see now. yeah i think that works. you have 2 equations and 4 unknowns, so you have 2 degrees of freedom

by the way, my method also has 2 degrees of freedom: one choose of w and one choice of u

[violet flare]

@maiden nebula thanks a lot for everything mate , you are a legend , have a good one

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[rocky oasis]

I need to find the next 2 terms in the sequences and the limits of the sequences

[hybrid flicker]

ok the first one was not easy but

write 3/5 as 6/10

and look at the progression of numerator and denominator separately

so 2,6,12,... is the sequence of numerators

5,10,17,... is the sequence of denominators

the second (an) sequence will follow that same idea (find numerator and denominator sequences)

[rocky oasis]

sounds good but what about the limits

plus i cant relly find the denominator sequence for the first question

[hybrid flicker]

@rocky oasis

[rocky oasis]

yes clear and that made the nomerator sequence 0,2,6,12,20,30

oh so 0,5,7,9,11

[hybrid flicker]

no not 0 for the first one

[rocky oasis]

3

[hybrid flicker]

and the other terms are 5,10,17

wouldn't be 3 either

[rocky oasis]

2,5,10,17,26,37

[hybrid flicker]

yes

can you write it in terms of n?

hint: when the growth between terms is linear (5-2 = 3, 10-5 = 5, 17-10 = 7, etc...), a_n is a quadratic

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[abstract narwhal]

how do i find the vertices for this rotated ellipse

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[hollow orbit]

How do i find the lenght of the line segment CE with the other lengths I am given. I dont know where to start

[bitter pilot]

Well what can we see is given?

[hollow orbit]

this diagram lol

[bitter pilot]

We can say so much that CE is the result of DE and CD together

can you see this?

[hollow orbit]

yes

[bitter pilot]

So we can express it

CE = DE + CD

now we can also see

CE = x+26

makes sense?

[hollow orbit]

yes

[bitter pilot]

So since both are equal to we can equate them

x + 26 = DE + CD

what's DE in the diagram?

[hollow orbit]

10

[bitter pilot]

good

Now the most difficult part is CD

x + 26 = 10 + CD

any idea how we could express the length CD in terms of the other lengths BC and BD?

[hollow orbit]

add them

[bitter pilot]

not quiet

you would then extent it

notice that the blue line

BD

it has BC in it

does that make sense?

[hollow orbit]

yes

[bitter pilot]

now we want only CD

So

Imagine you are at B

if we walked to C

what would be left would be CD

So like

we gotta take BC away from BD

to end up with CD

[hollow orbit]

so subtract them?

[bitter pilot]

yes

Imagine this

BD = 3

BC = 1

how long is CD

[hollow orbit]

2

[bitter pilot]

how did you figure it out?

obviously you subtracted

BD - BC = CD

right?

3-1 = 2

[hollow orbit]

yes

[bitter pilot]

so now

imagine BD the longer length

BD = 27+x
and
BC = 3x+47

In order to get CD we do?

[hollow orbit]

subtract 27+x from 3x+47

[bitter pilot]

yes we subtract BC from BD

we subtract 3x+47 from 27+x

subtract = take it away

So mathematically

CD = BD - BC = 27+x - (3x+47)

so back to our equation

x + 26 = 10 + CD

we plug that in

x + 26 = 10 + BD - BC

x + 26 = 10 + 27+x - (3x+47)

and we solve for x

[hollow orbit]

okay I combined like terms and got x+26=-2x-10. Does it matter where I move the numbers to cancel them out?

[bitter pilot]

no

what matters is if your math is correct

but the how is up to you 😄

looks good so far

[hollow orbit]

Okay thank you

[bitter pilot]

what'd you get?

[hollow orbit]

x=-12

[bitter pilot]

yes nice

[hollow orbit]

do I plug that into x+26 to get CE?

[bitter pilot]

Oh right we want CE

CE = x+26 so yea

[hollow orbit]

Okay i got

[bitter pilot]

nice

[hollow orbit]

Thank you so much for the help

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[snow loom]

for part b

idk how to solve it

i know that im supposed to set x(t) and y(t) equal to x and y

and try to simplify the equation down to a form of sec^2(t) - tan^2(t) = 1

but i dont know how to do that

i tried picking random ponints to plug in, but i couldnt figure it out

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[frail whale]

I need help with a study guide for a Pert math test <@&286206848099549185>

[blissful sentinel]

please don't ping helpers unless your question hasn't been responded to in 15 minutes

also please post your question

[frail whale]

I did

[torn timber]

pert?

[frail whale]

Its like another algebra 1 EOC

[frail whale]

But with just equations

[frail whale]

Specifically I need to find different study material for the PERT MATH test at least something I can use as a study guide the pert math test has algebra in this test

[vast shale]

sure i can help

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@frail whale Has your question been resolved?

[frail whale]

Do you know what the PERT test is

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[brisk badge]

Hii ! can somebody explain why the answer to this question is written like this? (2nd pic is answer)

Im able to do all the dividing and finding the remainder, but shouldnt it be (x+2) +9 or something?

[brisk moss]

you could say 9 is the remainder

but no it should be like that

when you divide 10 by 3, the remainder is 1. i.e., 10/3 = 3 + 1/3

similar thing here

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[worn osprey]

Pls help

Pls help I can't find the area of the squares

[split wind]

Have you learnt anything about ratio of similar figures?

[worn osprey]

Literally have never been taught in my school

[split wind]

hmmm

[worn osprey]

That's why I'm so confused about this

[split wind]

I'll try to explain, lemme draw sth rq

[worn osprey]

Okay okay

[split wind]

let's say they are both squares

[worn osprey]

Yeah

[split wind]

side length of 1st square = 1cm
side length of 2nd square = 2cm

[worn osprey]

Uh huh

Wait why is the side length of the 2nd square 1cm too?

[split wind]

typo

[worn osprey]

Oh lmao

[split wind]

area of first square = 1cm×1cm = 1cm²
area of the second square = 2cm×2cm = 4cm²

so

[worn osprey]

Yep still getting it

[worn osprey]

I see

So for volume ratio I assume it'd be in the power of 3?

[split wind]

correct

[worn osprey]

Okay I'm getting it

[split wind]

so, the first sentence implies:
"side length of A = 2×side length of B"

[worn osprey]

Uh huh

So (2x-2) : 2(3x-11)

Is this correct?

[split wind]

nah, its (2x-2)=2(3x-11)

[worn osprey]

Oh okay okay

Why is it 2 times and not 4 times?

[split wind]

because if we have 1:4 (which is (1:2)²) as the area ratio, we shall take the squareroot on the area ratio to get the length ratio

[worn osprey]

Yeah okay

Continue continue

[split wind]

then we can solve this to find x

and hence the area of both squares

[worn osprey]

X = 5

Nice

[split wind]

and then 2x-2=10-2=8,
so side length of A is 8

[worn osprey]

Okay

Side length of B is 3(5) - 11 = 15 - 11 = 4cm

Area of square A = 8 x 8 = 64cm²
Area of square B = 4 x 4 = 16cm²

Wow

Perimeter of square A = 4(8) = 32cm

So the ans is (a) (b) and (c)

[split wind]

yay

done!

[worn osprey]

Thanks!

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[split wind]

cheers!

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[south void]

Hello, i'm self studying mathematics and just took a course about geometry. I have been stuck for several hours in a exercise where i'm supposed to figure out the length of x. The assignment and my current progress is attached as pictures. As a side note we haven't been taught any trigonometry(sin, cos, tan) yet. So this assignment should be solvable by only using the area of the triangle equation. So far I think i have correctly solved the angles but i have no idea where to go from here. Any help is appreciated! PS. the correct answer is x = 100

[real path]

I think it’s like this

Wait no

[steady tide]

I am no helper but maybe use one of the SOH CAH TOA rules?

[real path]

He said it’s solvable without using that

[steady tide]

oh

[edgy gulch]

what does the question translate to

[south void]

"The square is divided into three parts as shown in the image. Determine the length of x"

[real path]

It’s a square?

[south void]

yes its a square

[real path]

Thats weird

Or maybe I did smth wrong

[keen zenith]

the side of the square can't be x

[real path]

Yea can’t be

Wait I think I have smth wrong there

[south void]

So the answer should be X=100. There is also a clue in the answers. It roughly goes like this: "Move the rectangular triangles formed inside the square next to each other and study the angles' magnitudes."

[real path]

Here are some ideas

[south void]

i actually tried that too but couldn't make it work. In that scenario base would be 100, height(side of the square) would be unknown and and both sides would be unknown. Or can you get the x out there?

[real path]

Yea I think the answer is there

X=36+64

Because the two sides are equal

The two angles are 90 degrees

[south void]

lets me try that out! Didn't draw Isosceles triangle like you did when i tried it

yep, that's it! Really appreciate your time and help here

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[plain plume]

how would you factorise 3x^5+5x^4+5x+3 quickly?

because i know (x-1) is a root

but surely theres a way to find the multiplicity more easily right?

[ornate terrace]

differentiation

[short carbon]

synthetic division

after

differentiation

[ornate terrace]

i mean, you can just straight use diffing but ok

[plain plume]

oh right thanks

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[finite hatch]

how come this one has zeros

[finite hatch]

but this one has 1s

i dont ge tit

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[short kraken]

help

[short kraken]

my professor straight up skipped the slides

whats with the absolute values

[vocal sleet]

@short kraken Has your question been resolved?

[gaunt willow]

There’s absolute value because you basically want the distance between x and c and the distance between f(x) and L, not just the number

[short kraken]

you want the distance not just the number

but isnt the distance a number?

[gaunt willow]

Yeah but distance is always positive

[hard atlas]

the point is to get a positive number

the distance from 3 to 5 is 2. but you would get -2 if you just subtract them

[gaunt willow]

The main idea with epsilon delta is you tell me how close you want to get to the limit in terms of distance (epsilon) and I tell you the distance from c you can have to be that close (delta)

[short kraken]

so is it like

wiggle room

[gaunt willow]

And if the limit you have is the real one, then I should be able to choose any arbitrary epsilon

[gaunt willow]

Yeah pretty much

Also notice delta can never be zero

Because we don’t care what’s happening at x=c only what’s happening around it

[short kraken]

cause you cant have negative distance?

oh

[gaunt willow]

That’s why it can’t be less than zero

[short kraken]

so like here

whats happening

i have to find out how close i can get to the limit without touching it?

[gaunt willow]

This is a slightly different problem

Then are already giving you epsilon, how close they want to get

You have to tell them what range of x will get you that close

So you need to find a relationship between epsilon and delta

Meaning you need a relationship between |(2x-5)-1| and |x-3|

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[short kraken]

.reopen

[vocal sleet]

✅

[short kraken]

so you set them equal to eachother and then isolate x?

well, set up the inequality not isolate x ig

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[vocal sleet]

@rustic grove Has your question been resolved?

[thin jay]

use the sin rule to find the base of the triangle

and then use trigonometry to find the heigh

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[pastel lion]

I dont understand why its the null space one

[pastel lion]

cuz if we're choosing vectors in column space shouldnt it be the above two?

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[thick shadow]

Is this formula correct

And if it is

[rare swift]

Yes

[fervent wasp]

Correct

[thick shadow]

if $log_{b}a = n$, then is $b^n = a$

[twin meteor]

Randel_

[thick shadow]

I mean

The b just looks a bit too far under the log

[rare swift]

It is

[thick shadow]

That's what bothers me

Thank you very much

Have a nice one

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[rare swift]

?

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[rare swift]

Ok wlc

[thick shadow]

Normally I expect it to be somewhere near it

Like in this photo

That's why I found it a bit confusing

[rare swift]

The position of b is confusing to you?

[thick shadow]

Yeah

[rare swift]

How so?

[thick shadow]

Just the way that it is written

For comparison

[rare swift]

Doesn't matter the way you write it both are same

[thick shadow]

Thanks a lot

Much appreciated

[rare swift]

Wlc once again

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[trim pecan]

i dont even know where to start?

[shrewd widget]

try dividing by (cos theta)²

[trim pecan]

then it would be

[gentle thicket]

yeah

and what's 1/cosine

[trim pecan]

1/cosine?

[gentle thicket]

yeah

what trig function we have defined for it?

[trim pecan]

is this what you mean?

[gentle thicket]

yeah

[trim pecan]

so my integral would

replace the x with theta, right?

[rare swift]

11 won't be multiplied to 3x

[trim pecan]

no it would not I'm saying whatever I get inside would then be multipled my 11, right?

because I took out the 11

[rare swift]

You did a mistake in line 2 by taking 11 out of the complete integral

[rare swift]

It should be 3/11 then

[trim pecan]

so it's not neccesary to take out the 11

[rare swift]

I mean you can be you forgot to divide by it

[rare swift]

Here it should have been 3x/11
Now the lower limit will be 0 and upper limit pi/4
Substitute them to find the final answer

[trim pecan]

ok I got it

11 + 3pi/4

I'll just leave the 11 in there next time I got myself confused at the end

thanks everyone!

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[velvet zephyr]

wait wait wait wait wait wait

[velvet zephyr]

i don't understand how they just conclusded that those are trapeziums and triangles (right angles

To add the area

areas

ping pelase

[tidal dock]

what?

they are right angled because AD is perpendicular to BC

[vocal sleet]

@velvet zephyr Has your question been resolved?

[velvet zephyr]

but they aren't parallel, are they?

to the base

[tidal dock]

who they

[velvet zephyr]

can you please explain the sum of shaded areaAS?

<@&286206848099549185>

.

sans culottes

this also

I am confused betwixt iodised salt and bronze

because bronze is pretty surely homogenous

I searched for iodized salt and the results were again homogenous

I

[limber peak]

Iodized salt is just salt with iodide added which makes it heterogeneous by definition if I am not mistaken

So the answer should be bronze I think

But this isnt my area of expertise I am just voicing how I would look at it

[velvet zephyr]

Om shanti om

thank you

.close

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[velvet zephyr]

Why am I being pinged for help even when I have removed my helpers role?

[gentle thicket]

@velvet zephyr

[lyric relic]

do you still want a proof for (x^3)(y^3)(x^3+y^3)≤2

?

[gentle thicket]

.reopen

[vocal sleet]

✅

[gentle thicket]

what exactly is your concern in the solution?

[velvet zephyr]

iaey

i don't understand how they just conclusded that those are trapeziums and triangles (right angles
To add the area
areas

[velvet zephyr]

because without knowing they are trapeziums or triangles we can't use the formula for areas

[gentle thicket]

why do you think they aren't trapeziums/triangles?

[velvet zephyr]

becaues it ain't given that the lines are parallel ( to the base)

[gentle thicket]

ig, we have to assume that on the basis of the diagram

they also didn't mentioned that diagram is not to scale

so we can assume that it is to scale

[velvet zephyr]

nuh uh

[gentle thicket]

oh that's problem

i thought they gave the diag

[gentle thicket]

but wait

if they didn't gave the diag
how are you supposed to know which area is shaded?

[velvet zephyr]

uh uh

[gentle thicket]

that doesn't make any sense

[velvet zephyr]

so ther eis atleast one flaw

[gentle thicket]

for sure

[velvet zephyr]

hm mh

.clsoe

c..close

.clos

.close

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[gentle thicket]

☠️