help-19

[amber schooner]

yea these don’t look like simply sinx and cosx

they have a shift tho

what was the transformation

oh no that’s at 2

not zero

nvm

[lament stone]

The red curve is the sine wave, and the blue curve is the cosine wave.

I know that it's an integral with a lower bound of pi/4 and an upper bound of 5*pi/4, but I'm unsure what to put inside the integral.

This is a better image:

Red curve is sine wave, and blue curve is cosine wave. Need to find the geometric area of the purple region.

I don't know where to begin.

Should I do this integral?

$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}\left(\sin\left(x\right)-\cos\left(x\right)\right)dx$

[clever fjord]

Vanouper

[lament stone]

Don't know if I should do this for solving geometric area.

[fluid tundra]

integral from a to b of |f(x) - g(x)| finds the area between f and g on that interval

in this case, the integral you have is correct

[lament stone]

Okay, thanks for the help, that's all I needed to know.

.close

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[fervent trench]

What's up?

[fervent trench]

Can someone breakdown this formula |pn| = sqrt( (a1*an)^n )

geometric progression

What is this used for?

[cold sage]

what is pn

[fervent trench]

|Pn|

[cold sage]

thanks, i understand now

[quasi sparrow]

!original

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

[fervent trench]

I do not have the original problem

it's just a formula that I never understood.

[odd edge]

@fervent trench Has your question been resolved?

[quasi sparrow]

Show the formula and its entire context

[odd edge]

@fervent trench Has your question been resolved?

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[graceful rose]

how do i find x and y so i can do the jacobian matrix

[graceful rose]

?

from the examples i've done, x and y are usually provided

[forest sky]

hint: there are two expressions in terms of x and y that show up a bunch of times

[sharp oak]

"find x and y"?

Like, the intersections?

[graceful rose]

an equation x and an equation y

[graceful rose]

ah i think i see it

[graceful rose]

ok wait so i tried to isolate the x and the y and set them equal to each other then solve it for that but that didnt work

[forest sky]

you want 2 new variables here. define them in terms of x and y, then you can solve for x and y in terms of those variables

[graceful rose]

.close

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[mystic saffron]

Hey so if the original function is a one to one, then the inverse is also a one to one, or is it just an ordinary function?

[hasty dome]

Well, what does one to one mean?

[mystic saffron]

A one to one function is one that passes both the horizontal line test and the vertical line test

[mystic saffron]

bijections are invertible, yes

[hasty dome]

OK, so another definition is that each input has a different output. No two inputs have the same output. That's the horizontal line test.

Does that make sense?

[mystic saffron]

Yes

[hasty dome]

OK, so the inputs and outputs are kind of paired up.

[mystic saffron]

Yes

[hasty dome]

Which means that those same pairs apply when you have the inverse function.

So, that means that the inverse will also be one to one, since each input to the inverse has only one output.

Sorry, each has a unique output.

[mystic saffron]

But in order for a function to have an inverse, it must be a one to one function

So does that mean that any function that has an inverse, the inverse is a one to one?

[hasty dome]

Yes.

Each input in the original function is paired with a unique output.

Like f(x) = 2x.

(3, 6) is one pair.

Nothing else has 3 as the first element of the tuple.

[mystic saffron]

Okay

[hasty dome]

Nothing else has 6.

For the inverse, you have (6, 3).

[mystic saffron]

Also, is this graph I drew wrong?

[hasty dome]

Yes.

You change the x intercept to a y intercept.

Then you use the reciprocal of the slope.

So, that's f(x) = 1/2 x - 1.

[mystic saffron]

Wait

But I use a different method

I just find points and reverse them

[hasty dome]

Right.

The inverse of a line will also be a line.

[mystic saffron]

Idk where I got this point from

I make a lot of silly mistakes for some reason, like just brain farts

[hasty dome]

It's easy to make mistakes in calculations and small things.

[mystic saffron]

But it happens to me so often

Idk y

Anyways, I think that’s the only mistake I made

[hasty dome]

Yes, that looks fine after it becomes a line.

[mystic saffron]

Aside from the red point my graph seems right, yeah?

[hasty dome]

Yes.

[mystic saffron]

Yeah

Are these graphs I drew correct? (Last image is the answer key)

If they are wrong, I think I know why… I inferred these points

[hasty dome]

a looks good.

b looks good.

c looks good.

So, those all seem fine.

It's easier to notice that they're piecewise linear functions.

So, you only need to handle the x values where at least one of the lines changes slope.

And the leftmost and rightmost points.

If you get those points correct and then draw lines between them, that will work.

[mystic saffron]

So I did this right

Someone told me that I should do more points than I think I need so I just did as many as possible

[hasty dome]

You can just do x = -6, -3, 0, 3, and 6.

[mystic saffron]

The thing that I was confused if I did right was thst the most right and most left values in my table, I inferred. The line didn’t reach it yet but I knew it would so I did them

Is that ok?

-6 and 6

[hasty dome]

Yes, those are fine.

[mystic saffron]

Okay

@hasty dome does this one look good

[hasty dome]

Yes, that looks fine. You just need to do the endpoints.

So, (-6, -4) becomes (-4, -6) and (6, 2) becomes (2, 6).

Then you just connect the points with a line.

[mystic saffron]

Oh so I’m being extra

[odd edge]

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[pastel roost]

Can anyone help figure out the order of steps?

[pastel roost]

.close

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[zealous meteor]

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[granite ruin]

hi

[granite ruin]

can someone please

HELP

and this

[wanton bison]

any idea what f(x-2) means?

[granite ruin]

doesnt it mean shift down

[wanton bison]

no

it means shift to the right

If f(x) = 1/x then f(x-2) = ?

[granite ruin]

shift to the right 2 units

[wanton bison]

yea but i mean if you plug in x-2 for x

you get 1/(x-2)...

so g(x) = 1/(x-2) + 2

[granite ruin]

uhh

see thats where im stuck

how do u get that

[wanton bison]

you can plug in thinggggs for x

[clever fjord]

adonhs

[granite ruin]

OHHH

im stupi

that was

😭

[wanton bison]

yea you simply plug it in

hence now you know g(x) and should be able to fill up the table

For the g(x) = 3f(x)

you can determine first f(x)

and then multiply it by 3

[granite ruin]

how would u graph that?

[wanton bison]

or easier

take the point (-2,1)

[granite ruin]

no cuz when i graph it- it doesnt join the line

[granite ruin]

okay

[wanton bison]

what happens if you multiply the function value by 3?

[granite ruin]

.

[wanton bison]

.

f(x+4) + 3

that's something total different

i bet you used the old g(x)

[granite ruin]

no i didnt

lemme send a pic

[odd edge]

@granite ruin Has your question been resolved?

[wanton bison]

,rotate

[wanton bison]

huh

f(x) = 1/x i assume

so what is f(x+4) + 3?

[granite ruin]

imma say this rn

i don’t get it

[wanton bison]

so i assume this reaction was fake lol

[granite ruin]

no i just confused myself again

😭

[wanton bison]

so you dont get that you can plug in things into a function

x -> input
f(x) -> output

say f(x) = x²

if i plug in a for x then i get f(a) = a²

if i plug in x+1 for x then i get f(x+1) = (x+1)²

@granite ruin

[granite ruin]

it’s just confusing cuz u have one then u see all these other numbers

i get it a lil now

thanks

[wanton bison]

"then you see all the other numbers" ok

i just dont understand what you struggle with

so i can help out

[granite ruin]

it’s okay it’s late anyways it’s probably because of that

i’ve been out all day 😢

i’m sorry

.close

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[unkempt aurora]

Can someone help me on number 9

<@&286206848099549185>

[sand night]

You cut off the top right

[quasi sparrow]

!15m

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[sand night]

Plug in t=0

[unkempt aurora]

For g

[sand night]

Into the equation

[unkempt aurora]

I did f and e

[sand night]

And then 180

[unkempt aurora]

I’m confused on what it means when it says stop

[sand night]

For part g

Wait do you assume you get on at t=0 or t=6

I would just plug in 180

[unkempt aurora]

For g?

[sand night]

Ye

Plug in t= 180

[unkempt aurora]

Alright thanks

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@unkempt aurora Has your question been resolved?

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[cobalt forge]

What kind of triangle is TAO?

[mystic saffron]

It is not specified

[cobalt forge]

But you can figure it out

What is the measure of angle TAO?

Remember, TA is a tangent line and AO is a radius

[mystic saffron]

Is it 90 degrees

[cobalt forge]

Yes

Therefore TAO is what sort of triangle

[mystic saffron]

Right angle triangle

[cobalt forge]

Yes

What part of the triangle is AO, relative to angle x?

[mystic saffron]

Oh now I can use trigonometry

[cobalt forge]

Indeed

[mystic saffron]

Oh thanks

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@mystic saffron Has your question been resolved?

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[cunning dust]

RRR GGG BBB

what are the chances of selecting 3 Greens from this list ^

[cunning dust]

its combinatorics i believe

[west merlin]

3 greens in a row? with or without replacement?

[cunning dust]

3 greens in a row yes, wdym by your second quesstion

okay well so my actual question asks me to calculate EVERY individual possible out come

so not 9C3

but like

show calculations of each possibilty

[vernal pilot]

His question is, after selecting a letter, do you put it back in the list when you make your second sellection

[west merlin]

when i take a green out, is there RRR GG BBB left?

[cunning dust]

no

wait

when selecting 3 yeah ofc

but

[cunning dust]

like in the overall question of calculating EVERY single possibility

no

so like

every time you choose 3

it goes back in list

and choose again

do yk what i mean?

[blissful tree]

so just a 1/3 chance of getting green each selection?

[cunning dust]

honestly idk

the question is

Calculate the chances of EACH possbility happening with the following: 3 green, 3 blue, 3 red

and like to start it off i wanted to calculate the chances of selecting 3 greens

but

idk how to do iit

[blissful tree]

weird question

[cunning dust]

yeah

so

overall

theres 9C3 different possibilities

correct?

[blissful tree]

no?

[cunning dust]

so 84 different combinations

[cunning dust]

oh?

what would it be

[blissful tree]

depends

[cunning dust]

on?

[blissful tree]

is GGB different than BGG for example?

[cunning dust]

no

same thinng

i know what you mean

it would be different if it was permutations where order does matter

do correct me if im wrong please

👍 im not very confident at this

[blissful tree]

ok how 9C3 then?

[cunning dust]

I think it would be 9C3 because we are selecting 3 colours right? out of 9 possible options

and order does NOT matter

so

bbg is same as gbb

bbg and gbb is ONE combination of colours

[blissful tree]

it would be 9 choose 3 if it were 9 distinct colors

[cunning dust]

oh

[blissful tree]

but it doesn't matter if you choose the first G or the second G

they are the same

[cunning dust]

how would i go about doing it then?

[blissful tree]

you can think of it like...

there are better ways than this but this is pretty straightforward: you can have 0, 1, 2, or 3 Rs. in each of those cases you can count how many ways you can make combinations with Gs and Bs

but that's a naive way, there is a good formula for this

but also i'm curious what is this question from

[cunning dust]

So would I have to count through calculations or would I like list them down

[cunning dust]

It's a "challenging" question the teacher gave us as homework

I thought it was easy at first

But there's a lot of possible outcomes

So I'd need a lot of calculations to calculate all of them

[blissful tree]

ok and what exactly is the question? you've posted a lot of fragments of it

do you have a pic of it?

[cunning dust]

Well she didn't give us the question, she just read it out loud and I wrote it down

I have written down

[blissful tree]

💀

[cunning dust]

Calculate the chances of EACH possibility happening with the following: 3 green, 3 blue, 3 red

And I can only select 3 at a time

[blissful tree]

tbh are you sure you are selecting with replacement

as in are you sure letters don't disappear when you select them

it's a little weird to have 3 green, 3 blue, 3 red if so

[blissful tree]

also don't you need to list them all out anyway to state the probability for every one of them

[cunning dust]

Okay so when selecting 3 colours from RRRGGGBBB

Say I select Red as the first one

Then I'll have RRGGGBBB to choose from

I misunderstood what you guys meant earlier but yeah they disappear when I'm selecting but once I've selected my 3 so like say red green blue, that's one possibility so I start over again to look at other possibilities

[cunning dust]

Yeah

[blissful tree]

lol ok

[blissful tree]

you can use this to list them all out then

[cunning dust]

Green Green green
Green green blue
Green green red

Red red red
Red red blue
Red red green

Blue blue blue
Blue blue red
Blue blue green

Green red blue

Wait

Oh shit I did it wrong one sec there's a lot more

Wait did I

💀

[blissful tree]

i think there should be 10

[cunning dust]

Yeah I counted 10 too

[blissful tree]

and yea that list is right i think

[cunning dust]

How would I go about finding the possibility of each combination

[blissful tree]

GGG, RRR and BBB are the easiest ones

let's take GGG

we need to get a green, then another green, then another green

know how to calculate the probability for that?

[cunning dust]

Would it be like

9!/3!3! Or smthn like that

[blissful tree]

definitely not, that's greater than 1

[cunning dust]

Oh

[blissful tree]

and syntax is kinda bad

doesn't matter bc both are wrong but is that (9!/3!)*3! 9!/(3!*3!)

[cunning dust]

Wait wdym

[blissful tree]

it doesn't matter lol but i was just recommending not writing 9!/3!3!

it's hard to interpret what that is

[cunning dust]

Ohh

Mb

okay so

uh

the chance of getting GGG

is less than 1

right?

[blissful tree]

yea definitely

[cunning dust]

could you like give me a hint

nothings coming to mind rn

😭

[blissful tree]

what's the prob we get green on the first selection?

[cunning dust]

i mean theres 3 greens

so theres 3 possible greens that I can have in the first selection

[cunning dust]

1/3?

[blissful tree]

yep

[cunning dust]

and in second box

1/4?

cause it would be 2/8

cause other green is gone

[blissful tree]

yea

[cunning dust]

and then

last one

1/7

[blissful tree]

yep

[cunning dust]

,calc 1/3 * 1/4 * 1/7

[clever fjord]

Result:

0.011904761904762

[cunning dust]

so thats the chance

of

green green green

,calc 0.0119*100

[clever fjord]

Result:

1.19

[cunning dust]

thats the %?

[blissful tree]

yes

[cunning dust]

oh ok

Green Green green - 3/9 * 2/8 * 1/7
Green green blue
Green green red

Red red red - 3/9 * 2/8 * 1/7
Red red blue
Red red green

Blue blue blue - 3/9 * 2/8 * 1/7
Blue blue red
Blue blue green

Green red blue

how would i go about finding the other ones

[blissful tree]

those are trickier

take ggb for example

with order considered you could draw ggb, gbg, or bgg

and all those count towards ggb

try to calculate the probability of each those

they should all be the same

see if you notice why when you calculate them

i will brb

[cunning dust]

Green Green green - 3/9 * 2/8 * 1/7
Green green blue - 3/9 * 2/8 * 3/7
Green green red - 3/9 * 2/8 * 3/7

Red red red - 3/9 * 2/8 * 1/7
Red red blue - 3/9 * 2/8 * 3/7
Red red green - 3/9 * 2/8 * 3/7

Blue blue blue - 3/9 * 2/8 * 1/7
Blue blue red - 3/9 * 2/8 * 3/7
Blue blue green - 3/9 * 2/8 * 3/7

Green red blue - 3/9 * 3/8 * 3/7

i think

i have done it all

[blissful tree]

well nvm my brb then

ok so

[cunning dust]

that was pretty easy unless i did it all wrong

🔥

[blissful tree]

not quite done yet

those might be "right" though

[blissful tree]

did you understand what i was saying here?

[cunning dust]

yeah

ggb is 3/9 * 2/8 * 3/7
gbg is 3/9 * 3/8 * 2/7

same thing

gives same result

[blissful tree]

yea

but

the probability of ggb (as in the probability of, with order, ggb, gbg, or bgg) is 3(3/9 * 2/8 * 3/7)

[cunning dust]

oh?

[blissful tree]

prob of ggb is (3/9 * 2/8 * 3/7), prob of gbg is (3/9 * 2/8 * 3/7), prob of bgg is (3/9 * 2/8 * 3/7)

[cunning dust]

yep

[blissful tree]

so you need to count how many arrangements there are for each combination

now i'm actually gonna brb

[cunning dust]

mmm alr

ill have a think about it and continue doing it, ill let you know if i need more help

[odd edge]

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[odd edge]

@near parrot Has your question been resolved?

[near parrot]

<@&286206848099549185>

[quasi sparrow]

do you have to use jensen

[near parrot]

No, at least the exercise does not refer to it.

But I think that the study of the function in question one must be useful for something.

[candid ruin]

why is this so hard

[quasi sparrow]

just expand the inequality for a few small n and try to find a pattern

n=2, 3

[near parrot]

Ok I'll try thanks

[near parrot]

I have no idea how to prove it correctly

[quasi sparrow]

did you expand it for n=2 and 3

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@near parrot Has your question been resolved?

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[near parrot]

,tex Let ( f: [0,1] \rightarrow \mathbb{R} ) be a function that is twice differentiable, and such that for all ( x \in [0,1] ), ( f''(x) \geq 1 ). Show that:

[ f(0) - 2f\left(\frac{1}{2}\right) + f(1) \geq \frac{1}{4} ]

[clever fjord]

Wakizashi

[near parrot]

I have no idea how to start

,tex Since ( f''(x) > 0 ) for all ( x \in [0,1] ), it follows that ( f'(x) ) is increasing on this interval. Consequently, ( f(x) ) is convex on ([0,1]).

[clever fjord]

Wakizashi

[near parrot]

Only thing that i know

[proud ibex]

Try integrating both sides of the inequality twice to get one including f(x)

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[smoky lodge]

When we ask

[smoky lodge]

Prove Something = something + O(some function)

Is it essentially equivalent to saying

Prove Something ≤ something + some function

[grave fiber]

?

is this a question or...

[mortal mirage]

why not consult the literal definition

[grave fiber]

to prove something one must assume something thy has already been proved and consult it how ever thy likes

[smoky lodge]

The definition doens't make it clear

Definition simply says if f(n) is O(g(n)) then f(n) ≤ cg(n)

So what I figured is

[mortal mirage]

eventually, yeah

[smoky lodge]

Something = something + O(some function) is the same as saying Something - something ≤ c*(some function)

So we must simply prove Something ≤ something+ c*(some function)

?

I suppose that makes sense

Thanks

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[ocean hamlet]

Why is this wrong?

[odd edge]

Please don't occupy multiple help channels.

[static totem]

8^88 is 4

just do it again, it's like a bad luck thing

[ocean hamlet]

yea but what is wrong in my calculation?

[static totem]

the middle of the left side

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[plush meteor]

is this correct?

[plush meteor]

taking a while to load

there it is

[vivid bloom]

wait u shud be getting log|secx+tanx| +c

idk how u get it i just remeber the formula

[plush meteor]

yeah

maybe the 2 are equal

cuz i dont see any mistakes in my working

if there isnt i found a nice way to make a formula for tan^-1(z)

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@plush meteor Has your question been resolved?

[plush meteor]

idk

idts

but thanks for asking bot

[plush meteor]

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[winter zodiac]

Hi, can someone help me with this partial derivative.

[stone cipher]

what did you get

[winter zodiac]

It's sending...

C^2/3(a+b)^2

Since the photo it's not sending

[stone cipher]

that's not right

you have a in the top and the bottom

or

it doesn't feel right

hmm

[winter zodiac]

Any idea how should I begin?

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@winter zodiac Has your question been resolved?

[winter zodiac]

<@&286206848099549185>

Already solved it

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[rotund burrow]

i got a bad result using law of sines to find A after finding b. a triangle solver online got the same result for b and B so i'm really confused as to why law of sines didn't work

[rotund burrow]

c ~= 6.62 so sin(A) should ~= 11sin(35)/6.62

the online solver uses law of cosines instead and i can do that but i'm confused how this is wrong

[zenith jasper]

how did you get c

[rotund burrow]

c^2 = 11^+7^2 - (2*7*11)cos(35)

and just followed that down

[zenith jasper]

ok wait

[rotund burrow]

triangle calculator has the same result as me for that part and for angle B, but for A it gives a weird number

[zenith jasper]

did it give you two possibilities

because sin(x) = sin(180-x)

[rotund burrow]

which, my calculation or the triangle solver?

[zenith jasper]

triangle solver

[rotund burrow]

nope, only the one

[zenith jasper]

wait

i think i know why

[rotund burrow]

yeah?

[zenith jasper]

i need to calculate it with real value not 6.62 hold on

i think i know why

[rotund burrow]

real value also got me roughly the same weird result

[zenith jasper]

i think it's because A is supposed to be the largest angle (since it's opposite of 11), but if you simply use law of sine you get around 72.32 degrees

agree?

[zenith jasper]

did you get this for A

[rotund burrow]

ohhh law of sine is failing because A>90deg

yeah

[zenith jasper]

right

but

if u do 180-35-A for B, B>A

which shouldnt happen, as A is supposed to be the biggest angle

so that leaves us with the only other choice, 180-72.32 deg ~ 107.68 deg

[rotund burrow]

that makes sense

[zenith jasper]

you need to be very careful of what the other angles would be after you do law of sine

[rotund burrow]

so if i used law of sines for all of them, is there any way to know which angle is wrong beyond running law of cosines on all of them?

[zenith jasper]

hold on i think my textbook has this

[rotund burrow]

i mean, A and B

[zenith jasper]

yeah

[rotund burrow]

well i guess like you said i can use 180 degree check on both and since i know a>b i know A>B

[zenith jasper]

check if sin theta < 1, angle sum of triangle check

that's all i can think of rn

and the 180-x

there might be more

[rotund burrow]

yeah

thanks a lot

really threw me for a loop there, i'm just now taking notes in this class for the first time and it definitely helps but i also definitely need to get better at it haha

[zenith jasper]

you're welcome, you can always use law of cosines to double-check

[rotund burrow]

yeah, i mostly just needed to know why sines didnt work

and i got it, thanks!

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[viscid sage]

Unsure how to start

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[distant umbra]

we basically have to show that t_n>t_(n+1) which means its dicreasing which means its derivative is less then 0

i took the derivative and got

1/(n^2+2n+1) - 1/(n^2+n)

surely the denominator n^2+n is smaller than the other denominator so their difference gives a negative value thus the f' is negative hence dicreasing

is my method right?

because the solution did a little more complicated method

in the solution they integrated tn-tn+1 from 1 to infinity

and got the answer as somewhat 0.307

and then argued this

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[junior ether]

$\lim_{x\to0}\frac{0}{x}$

[clever fjord]

Jash

[junior ether]

0 or indeterminate?

wdym what do i have

[stark elbow]

It's ||0||

You only care about a neighorhood around x = 0

[zinc glacier]

oh my god <@&268886789983436800> im not dealing with bullshit today

anyway

[mystic saffron]

its lim as x -> 0 0/x

[zinc glacier]

you know that 0/x is defined except for x=0

[warped grove]

alright clam down everyone

if you're not helping, get out

[rotund hawk]

What are your explanations for these two?

[junior ether]

well i was gonna say indeterminate cuz when u do direct sub u get 0/0 which is indeterminate but if u evaluate the function first which is 0 and then put the limit u get 0

but oops helped me in dms so its fine

[leaden widget]

you can use l'hopitals as an option

[warped grove]

L'Hospital's is pretty overkill here

But yeah you could

[leaden widget]

or also think about it intuitively, the value of that expression is 0 for all x except at x = 0, so that includes any open interval containing 0 you could come up with, meaning that if you just think about delta epsilon stuff, you can make the value of the expression arbirarily close to 0 no matter how close to x = 0 you get

[junior ether]

ye

0/x is just a piecewise function i think

which equals 0 for x != 0

[leaden widget]

you can also think of it as a removable discontinuity

its basically the line y = 0 with a removable discontinuity at x = 0

[junior ether]

ye ty

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[faint peak]

I need help with the following problem: A wholesale supplier of fertilizer has 3 types of garden fertilizer: the first type X with 5% nitrogen, of which 4000kg is in stock; the second type Y with 10% nitrogen, of which 1500kg is in stock; the third type Z with 20% nitrogen, of which 8000kg is in stock. There is an order for 1000kg of fertilizer that must have a nitrogen content of 15%. Therefore, the three types of fertilizer are mixed. Due to the stock, it is desired to use twice as much of type Z as type X. How many kilograms of each type should be mixed?

[faint peak]

I can't find all of the equations.

I have 3x + y = 1000

And can't really find the other 2.

Maybe 0.05x + 0.10y + 0.2z = 0.15 but I'm not sure about that one.

[rigid dragon]

I don’t see how you got this equation

[rigid dragon]

This makes sense I believe

And you should also have 2z=x

Since you need two times the amount z as x

[faint peak]

Yeah I tried to apply that to this one

Did it incorrectly though

[rigid dragon]

Oh well that one equation you would only need two then

And yeah it makes sense that x+y+z=1000

[faint peak]

Lemme put it in my calculator rq

[rigid dragon]

So you have x+y+z=1000, 2z=x and 0.05x+0.1y+0.2z=0.15*1000

[faint peak]

yeah i included the times 1000 as well

but

[rigid dragon]

15% of total mass

Should be nitrogen

[faint peak]

x+y+z= 1000, x-2z= 0 and 0.05x+0.1y+0.2z=0.15*1000 right??

[rigid dragon]

Yeah

Seems right

3 equations 3 unknowns

Solve the matrix or use substitution or something

[faint peak]

It's a conflicting system so I think somethings still wrong

[rigid dragon]

Conflicting how so?

[faint peak]

0 =1

when i solve it

[rigid dragon]

So are there no solutions?

Or infinitely many?

[faint peak]

no solutions

but according to my book there is one solution

hmm

[rigid dragon]

Stop using a calculator

And row reduce the matrix by hand

[faint peak]

😦

[rigid dragon]

Or use substitution

That’s my first advice

[faint peak]

shouldnt rref always work?

on calc

[rigid dragon]

I mean are you actually plugging in a 3x4 matrix and getting rref?

Or are you in a system of equations solver

[faint peak]

yes

im doing it with matrices

[rigid dragon]

Then what’s the resulting matrix

[faint peak]

hold on let me log into this acc on my phone and take a picture

@rigid dragon

[rigid dragon]

Please don’t ping

[faint peak]

Mb

[rigid dragon]

I see what you mean

I think that maybe removing z might help

Just putting it as 2x a

[faint peak]

I got the right answer by indeed removing the Z and doing 3x + y = 1000

[rigid dragon]

Since I think the third equation is the issue

Yeah just reduce it to two unknowns

[faint peak]

thanks for the help

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[arctic jackal]

its not exactly a problem per se but can someone explain this equation? like what does each thing mean

[arctic jackal]

context

[quasi sparrow]

x_0 = x_i
x_f = x_i + delta x

[arctic jackal]

.close thank you :)

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[mystic saffron]

Solving this ODE with an integrating factor, did I mess something up

[mystic saffron]

on symbolab it has a -1/2x^-2 instead of just x^-2

[quasi sparrow]

It's just a constant so it doesn't matter

You have C

-C/2 is still a constant

[mystic saffron]

righhttt, so this is fine as is

i didnt think of it like that

thankyou

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[dawn quiver]

help? using squeeze theorem

[dawn quiver]

<@&286206848099549185>

<@&286206848099549185>

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[tardy lagoon]

do you know how to find intersections?

[thin wraith]

i think so

[tardy lagoon]

try setting up an equation for finding intersections

[thin wraith]

what is that?

do i just make the equations equal each other?

[tardy lagoon]

yes

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[young sedge]

I have to solve for multiple equations on [0,2pi)

sec(x/2)=2 still 2 cycles?

i found 4 values 🤷

pi/3, 5pi/3, 7pi/3, 11pi/3

multiply by 2 to isolate x

2pi/3, 10pi/3, 14pi/3, 22pi/3

but 2pi/3 is the only one that fits in the parameters so is that my answer?

[olive needle]

Yes.

[wanton bison]

Yes.

[young sedge]

thanks

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[gloomy holly]

how do I factor this??
x^2 + 5x - 6

[quasi sparrow]

6 + 1 = ?

[gloomy holly]

ok fixed the question

help out please

[copper quarry]

Hello

[gloomy holly]

hi

I need help factoring this

[copper quarry]

Okay

Can you think of two numbers that multiply to -6 but add to +5?

[gloomy holly]

hm

no

oh yes

1- 5

sorry -1 5

[west merlin]

they multiply to make -6

and add them to get +5

[remote reef]

-1 and 5 multiply to make -5

[gloomy holly]

-1 6

[west merlin]

there you go

[remote reef]

yes

[gloomy holly]

now what

[copper quarry]

now you want to put them in the form of (x-a)(x+b)

You know a and b

[gloomy holly]

(x-1) (x+6)

[copper quarry]

correct

now we can expand this to make sure it’s correct

try doing it

[gloomy holly]

kk

I got

x^2 + 6x - x - 6

simplified its

x^2 + 5x -6

ok thanks

how about x^4 - 9

[simple zealot]

you can rewrite it as x^4 + 0x^2 - 9

we know the brackets to factor this is going to be (x^2 + A) (x^2 + B)

where A + B = 0 (the coefficient of the middle value, coefficient x^2 for this)
and A * B = -9 (the coefficient of x^0)

this should be a general guide on how to find it out for yourself, if u need help with using it i can help

you can just find it using trial and error to a certain degree, you just get better at seeing the correct numbers that fit in after some practice

[gloomy holly]

if its only 2 terms

you add 0x^2?

[simple zealot]

0x^2 is the same as 0, they just didnt write it down because it's 0

im just writing it down for you so i can explain the concept of how to figure it out

it's a difference of two squares, if you know it by that name

[gloomy holly]

ok but I still don't understanf

how would you solve this with 0

[simple zealot]

the values A and B are so they add up to 0, and multiply to give -9

it's just one of the ways to find out the numbers

i can explain concept more if you'd like

this is the proof of the concept

so now you can use it to understand how to do this i hope

[gloomy holly]

idk

whats A and B

how would it equal 0

[simple zealot]

it's just something u give it a thought, in your head, what two numbers when added together give 0, and when multiplied by each other give -9?

[gloomy holly]

9 and -9

no

3 and -3

[simple zealot]

AYYY

u got it

gj

ur gonna get better at it and be quicker with more practice

[gloomy holly]

I have a unit test trmw and I'm just reviewing

am I cooked???

[simple zealot]

nah u good

take it slow and just focus

dont panic

[gloomy holly]

ok im pretty much good

thank you alot for the help have a goodnight brother

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[simple zealot]

np and good luck 🫡

[gloomy holly]

🫡

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[worldly sleet]

I’m doing task 2117 and I’ve been stuck on it for a while

[weary pelican]

how are binary numbers less than 256 represented?

[worldly sleet]

1 0 0 0 0 0 0 0 0

This is = to 256

The number furthest to the left must be 0 always

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[orchid torrent]

Principle of inclusion and exclusion

$|A \cup B|=|A|+|B|-|A \cap B|$

[clever fjord]

Civil Service Pigeon

[worldly sleet]

What does this tell me?

[orchid torrent]

Yk how to read the notation right

[worldly sleet]

Not really, I haven’t gone through it in class

[orchid torrent]

Basically you take the number of numbers that satisfy each condition

And subtract the number that satisfy both (since this is counted twice - once for each individual condition)

[worldly sleet]

Okey let me try

Binary system has 9 numbers and if the first digit is 1 it means it’s equal to 256

Then to have two digits as a 1 we have to take the combinations 1x1

Then we’re left with 9-3 = 6 digits that can either be a two or one

2^6

We can do that two times

And then I don’t know what to subtract with

[orchid torrent]

Binary is 0s and 1s

But it’s clear that it’s at most 8 digits

So let’s consider it starting with two 1s

Consider how many numbers there are if there’s 3, 4, 5, …, 8 digits

You should notice a pattern

[worldly sleet]

It’s it’s 3 digits and it start with two 1s you can have 2 combinations

4 digits gives 2x2 combinations

5 digits gives 2x2x2

6 digits give 2^4 etc

8 digits give 2^6 combinations no?

If we do that but with the last two digits we get 2^6 too

and if both two last digits and first digits are 1s

We get 2^4 combinations

Add that together I get 144 possible combinations and the wrong answer

[orchid torrent]

That’s only for 8 digits

I’ll brb but see if you can figure it out now given this

[worldly sleet]

Thanks for the help

[orchid torrent]

Alright I’m back

Did you figure it out

[worldly sleet]

no

256 has 9 digits right

[orchid torrent]

Actually there’s a few issues here

[orchid torrent]

You’re supposed to subtract the intersection, not add it

Read what I said earlier to see why this is true

And as I said, this is only for 8 digit numbers

So what about for 2,3,4,5,6,7 digits?

[worldly sleet]

We don’t care about 234567

[orchid torrent]

$111_2$ be like:

[clever fjord]

Civil Service Pigeon

[worldly sleet]

Does it matter that there’s a text that says “ten” in the bottom right corner of 256?

Like the 2 is for 111 in the image above

[orchid torrent]

That means less than 256 base 10

Basically the subscript denotes your base

[worldly sleet]

And that means we start at 1?

1 2 4 8 16 32 …

[orchid torrent]

Yes, you need to consider the binary representations up until that of $255_{10}$

[clever fjord]

Civil Service Pigeon

[worldly sleet]

Yes and that’s 9 digits no?

[orchid torrent]

256 base 10 is 100000000 in binary

So you consider everything less than that

Which is obviously any binary number with no more than 8 digits

[orchid torrent]

Such as this

[worldly sleet]

Yes right

[orchid torrent]

So do you now see why you need to consider some binary numbers other than those with 8 digits

[worldly sleet]

No

Either digit is either a 0 or a 1. If the two first digits are 1s, that leaves us 6 digits that can be either 0 or 1. That gives us a combination of (2^6) x 1 x 1

[orchid torrent]

Does my example of 111_2 not mean anything

[worldly sleet]

And if the last two digits of the 8 digits are two 1s you get the same thing

[worldly sleet]

What does it mean explain

[orchid torrent]

It satisfies the condition of starting (and ending) with two 1s.

And is not 8 digits.

[worldly sleet]

Yes but I thought they asked for starting and endings 1s in the binary system

Don’t they?

[orchid torrent]

Your point is?

Does 111_2 not satisfy that?

[worldly sleet]

01100000000011

[orchid torrent]

Unless stated otherwise

You normally don’t consider leading zeros

Plus I can counter that

Consider 0000011111111_2 = 11111111_2 = 255

By your logic, this would fail

[worldly sleet]

This satisfies to having starting/ending two 1s, it’s one of the combinations

Why would it fail

[orchid torrent]

Yk what

Forget that

[orchid torrent]

Explain in detail why 111_2 would fail by your logic.

[worldly sleet]

It doesn’t

When did I say that

[orchid torrent]

So then why are you saying you don’t need to consider numbers other than those with 8 digits

When there’s clearly a counterexample

Unless you mean you can just shove 0s in front to make it 8 digits

But the method doesn’t account for that so

Take that as you will

[worldly sleet]

You can write 111 with 8 digits in the binary system

[orchid torrent]

Oh so that’s the issue

Well say that you have a 7 digit number you make 8 digits by shoving a zero in front

Then you have 011(…) where (…) is some string of 5 digits

Then there’s only 2^5 combinations, not 2^6.

[worldly sleet]

Yes

But for 256 you have 8 digits to work with and the ninth digit is a leading 0

Then you have 2^6 combinations

[orchid torrent]

It’s clear that whatever I’m saying isn’t being understood

So maybe just wait for someone else to explain

Original question for anyone who is interested:

[worldly sleet]

We’re saying the same thing but you want to use 7 digits and work with 6 digits

Why’s that?

[orchid torrent]

yk what

Let’s just do it like this

1. Start with two 1s
   I. No numbers after (11) -> 1 possibility

II. 1 number after (11) -> 2 possibilities

III. 2 numbers after (11) -> 2^2 = 4 possibilities

[worldly sleet]

Yes

[orchid torrent]

Continue to get 1+2+4+…+64=127

Same logic for ending with two 1s, you get 64

So we have 254 minus over count

I. Two digits (just 11) -> 1 case

II. Three digits (just 111) -> 1 case

III. Four digits (just 1111) -> 1 case

For k>4, if you have k digits in total, there’s (k-4) digits to fill in between the two ending strings, so 2^(k-4) cases

So you have 1+1+1+2+4+8+16=33 over counts

Take this as you will

I’m not engaging in this conversation anymore.

[worldly sleet]

Okey thanks for the help 👍

254 - 33 still gives me the wrong answer

Is the answer sheet wrong?

[worldly sleet]

Can someone explain to me why’s that

[odd edge]

@worldly sleet Has your question been resolved?

[odd edge]

@worldly sleet Has your question been resolved?

[worldly sleet]

<@&286206848099549185>

[worldly sleet]

<@&286206848099549185>

[worldly sleet]

<@&286206848099549185>

[odd edge]

@worldly sleet Has your question been resolved?

[odd edge]

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[shrewd lodge]

How do I solve this?

[torn lotus]

volume of a sphere is $4/3 \pi r^3$

[silent temple]

$\frac{4}{3} \pi r^3$

[clever fjord]

Kihei

[silent temple]

🙂

[clever fjord]

swerriee

[odd edge]

@shrewd lodge Has your question been resolved?

[frail comet]

YO WSG G

[odd edge]

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Closed by @shrewd lodge

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[heady smelt]

Yo wsg

[heady smelt]

we need to evaluate the expression (1/1000)^1/3

[frail comet]

Hey

[subtle depot]

Ok bro

[heady smelt]

oh wait i’m not supposed to ping

i forgot

sorry

sorry

[frail comet]

RICKKKKK

my boyyyyy

[heady smelt]

WSG G

[subtle depot]

This time dont just give the answer

[frail comet]

wha you need help with now?

[subtle depot]

Ur supposed to learn it

[heady smelt]

we need to evaluate the expression (1/1000)^1/3

[frail comet]

what?

[heady smelt]

Really

[frail comet]

are you helping rick or am i?

[heady smelt]

You are

Bbg

[subtle depot]

Ur not helping rick ur just gicing the answers

[frail comet]

no im not?

ask rick

lol

[heady smelt]

he’s showing steps

he’s not

[subtle depot]

Ehat

[heady smelt]

stop truijg to argue since

dunce

[frail comet]

yah?

[subtle depot]

Well if you did teach him

He would knoe, cuz this problem had the same steps as the last

[frail comet]

what is your problem?

bro

[heady smelt]

oh it did

can u explain them