help-10

[kind hawk]

yes

[viral blade]

you could do that yeah

[timid silo]

Oh gosh, you cant just say n-1 now because you cant just break the piece in half lmao

yeah fair enough i suppose

[viral blade]

Yeah lol it can get kicked back arbitrarily far

this reminds me of the problem of how many ways to tile an nxn square with 2x1 tiles

[timid silo]

how do you even tackle such problems then?

it does seem very

pretty difficult

[viral blade]

maybe look at the set of all shaped endings that can be created by recursively removing the right-most tile

[kind hawk]

examples. try building smaller blocks. see how things fit together

[viral blade]

So like

O...(n)...O
O...(n)...OO
O...(n)...OOO

for example

[kind hawk]

what are the basic combinations that you can build and into which you can split everything

[timid silo]

hm should i start drawing again

it probably gives me a better idea

i guess same as before, i will try doing a(1), a(2), a(3)?

and see how that works out

in fact

[kind hawk]

it does. thats the point

when you have no clue, start doing examples

thats how you learn

[timid silo]

Got it. Let's do it

brb

btw

in general, how would you even recognise how many initial values you need to have?

keep building up ur cases until they all start repeating the previous ones?

[viral blade]

until every possibility is covered by one

the tricky part is choosing them so that the number of possibilities each case can be written in terms of smaller cases

That's why I like the "removing the right most block" setup

Your set of cases is the set of shapes you could get by starting with a 3x∞ block and removing right-most blocks

[timid silo]

okay i probably missed a few but writing a_3 seems insane so i wont do that 💀

[viral blade]

Are we throwing in 2x1 blocks too now?

And 3x1?

[timid silo]

does that not work?

like

we can have any square can we not

[viral blade]

Well there's definitely a lot of cases missing if you're doing that now

Wym any square?

[timid silo]

i was thinking we were following those conditions?

[viral blade]

the problem is that one 2x1 and two adjacent 1x1 blocks are different things

[timid silo]

oh yeah fair enough

i will separate

[viral blade]

hint for a_2: ||it's the same as a_3 from the previous problem||

[timid silo]

okay i didnt miss any of them did i

[viral blade]

I don't think so, no

[timid silo]

okay so a_1 = 1, a_2 = 7

[timid silo]

i will open this if it gets desperate lmao

anyways

[viral blade]

You might as well now it doesn't matter

[timid silo]

oh okay

hmmm the same method

[viral blade]

No the same answer

[timid silo]

okay lets just do the good ol strat

[viral blade]

It's just rotated 90°

2x3 block still

[timid silo]

oh waity

then

i did miss some

there you go

[viral blade]

Not so far as I know

[timid silo]

i hope this is it

[viral blade]

Ah ok

[timid silo]

this is painful

[viral blade]

probably is yeah

[timid silo]

but anyways a_1 = 1, a_2 = 11 (probably; hopefully)

anyways so

i guess we can apply the same logic on the easy ones first?

[viral blade]

This is where it's going to become pain

[timid silo]

so like, if it ends with three squares then it is simple a(n) = a(n-1) + ...

So thats good

[viral blade]

I think you're going to run into some problems classifying everything like this

[timid silo]

what do you mean

[viral blade]

You're not going to be able to find classes the way you're looking for

[timid silo]

yeah because again u cant just split open the thing

uhhhh

[viral blade]

You need to have a bunch of cases that are all written in terms of each other

So like

a_n is written in terms of a_n-1 and b_n-1

b_n is written in terms of a_n-1

[timid silo]

yeah i see

whats a better way to think about this then

[viral blade]

And only one of those cases is going to be the case you need

Well this was denascites idea maybe he'll help

[viral blade]

I tried to explain my idea here

[viral blade]

So this would be an example, maybe like b_n

Maybe b_n+3, whatevers needed

And then the rectangular block would be like a_n

Then you'd have other shapes be other cases

[kind hawk]

I am confused. which pieces do you have?

[viral blade]

Square and L

[kind hawk]

but how big is the L

[viral blade]

same as before

[kind hawk]

oh same as before? oh boi

[viral blade]

We could make it bigger but then it's pretty much just the exact same problem again as before

[kind hawk]

well yes. that's why I suggested it

its a little bit different but not too much

[timid silo]

oh

i misinterpreted you then 💀

[timid silo]

i thought u wanted it like this

[viral blade]

Probably pfires problem is best if you want a harder problem along the same lines

[kind hawk]

that runs into this problem. which sucks

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

anyways thanks a lot @viral blade @kind hawk !

i think this really helped

[obtuse pebble]

Channel closed

Closed by @brave kite

Use .reopen if this was a mistake.

[viral blade]

no problem

[kind hawk]

youre welcome

[obtuse pebble]

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.

[maiden quail]

This book asks me to factor these really complicated polynomials, how am i supposed to do this stuff by hand?

[maiden quail]

j and k are the real problems im having issues with

[knotty crow]

,rccw

[knotty crow]

j) try grouping

in fact same thing in k)

[maiden quail]

Yeah but how do i know how many grouped factors im going to have? How do i know how many terms each factor will have?

some of the previous questions had two term and three term factors for a polynomial

[knotty crow]

I just look at the coefficients (usually when there are 4 terms then we group them in pairs 2 x 2)

let's say we consider j)

4 terms

and see that

[maiden quail]

highest power 4

[knotty crow]

1 and 2 (2 is two times bigger than 1)
8 and 16 (16 is two times bigger than 8)

so it should pay your attention

[maiden quail]

Would it be two 2-term factors with x^2 as the first term? Or is it a single monomial times a 3-term factor? That's that part im having trouble with

i can factor normal quadratics just fine but these higher order and multivariable ones really throw me for a loop

[knotty crow]

2 - term factors with x^3

x^4 + 2x^3 = x^3 * (x + 2)

try to do something with -8x - 16 to get (x+2)

somewhere

[maiden quail]

So wait, (x^2 + 2)(x^2 - 4)

No nvm

[knotty crow]

Sorry, I meant 2 terms in the bracket and x^3 in the front of it

[maiden quail]

Wait but you cant factor out x^3

[knotty crow]

why not

[maiden quail]

The last term doesnt even have an x

[knotty crow]

Do you know concept of gruping the terms?

Thing is we don't have to factor it from all the terms

just a few

[maiden quail]

ya, but you said to put x^3 as a monomial infront of a binomial factor, right?

[knotty crow]

yes

and I did that:
x^3 * (x + 2)
x^3 - monomial in front of the (x+2)
x+2 - binomial

[maiden quail]

That only works for the first two terms, they cant be pulled out of an expression like that

x^3 would have to be a common factor for the entire expression to factor it out

[knotty crow]

yes, only for the first two terms and that's fine

so far

[maiden quail]

We cant consider the first two terms seperate from the expression, sums cant be broken up like products can

[knotty crow]

we can, look at the simple example:
2 + 4 + 3 + 6 = 2(1 + 2) + 3(1 + 2)

,calc 2 + 4 + 3 + 6

[warm shale]

Result:

15

[knotty crow]

,calc 2(1 + 2) + 3(1 + 2)

[warm shale]

Result:

15

[knotty crow]

This is how factorization by grouping works

[maiden quail]

huh, i never knew that

Was always made to factor the whole thing all at once

[knotty crow]

I guess you've never seen that method before

You only knew taking out from the whole expression

[maiden quail]

No, my school never taught us anything besides factoring quadratics

so the second part is -8(x+4) ?

[knotty crow]

-8 * 4 = -32

and it should be -16

[maiden quail]

oh my bad -8(x+2)

[knotty crow]

yes

and now look carefully at this

[maiden quail]

x^3(x+2) + -8(x+2) == (x^3 -8)(x+2)?

[knotty crow]

yes

[maiden quail]

oh, wow

[knotty crow]

you can even factor it further

[maiden quail]

You mean that first factor that's power 3?

[knotty crow]

I don't know if you're familiar with the short multiplication formulas

[maiden quail]

No, im not u ufortunately

[knotty crow]

e.g. a^2 - b^2 = (a-b)(a+b)

stuff like this

[maiden quail]

Huh, where do i find a listing of algebraic rules like that?

[knotty crow]

I'd say there are five worth memorizing

(I mean these short multiplication patterns)

[maiden quail]

Are those just for exponents or in general?

[knotty crow]

for exponents, but they may be different, doesn't matter

[maiden quail]

ok, thank you so much btw

[knotty crow]

the last one

is your case

where a = x and b = 2 (since 2^3 = 8)

[maiden quail]

Oh that's incredible, ive never even considered using roots to plug integers into these kinds of formulas before!

(x-2)(x^2 + 2x + 4)?

[knotty crow]

yes, now it's factored completely

In fact

[maiden quail]

(x + 2)(x -2)(x^2 + 2x +4) is final right?

[knotty crow]

every polynomial can be factored (over the real numbers) into a product of linear factors and irreducible quadratic factors

[maiden quail]

(x^2 + 2x +4) factors into (x+2)(x-2) right?

[knotty crow]

that is, if you have 3rd degree be certain it can be factored further (sometimes it's not easy, even impossible)

[maiden quail]

so its really (x+2)^2 times (x-2)^2 ?

[knotty crow]

nope, that's it

[maiden quail]

oh wait

Nvm my b

[knotty crow]

exactly

,w (x + 2)(x -2)(x^2 + 2x +4) simplify

[warm shale]

Results provided by WolframAlpha

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

Requested by modus7591

[maiden quail]

Wow, can't believe I never learned that. Thanks again for your help, means alot to me

[knotty crow]

My pleasure, good luck!

[maiden quail]

Same to you!

.close

[obtuse pebble]

Channel closed

Closed by @maiden quail

Use .reopen if this was a mistake.

[obtuse pebble]

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.

[grizzled condor]

Hey, I'm working on some calc 2 homework. We're doing partial fraction decomposition. My question is, where am I going wrong? I'm getting A = 1 and C = -1, but apparently it's A = 21/4 and C = -21/4. I'll link the exact HW question in a second.

[tardy epoch]

you should just find the partial fraction decomposition for 1/(x^2-1)^2 then multiply it by 21 at the end

[grizzled condor]

alright, i'll do that

[tardy epoch]

,w partial fraction decomposition 1/(x^2-1)^2

[warm shale]

Results provided by WolframAlpha

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

Requested by rie.mann

[grizzled condor]

im trying to not use wolfram alpha rn so i actually learn, instead of just getting the answer yk

and i was doing it this way because the "help me solve this" was telling me to

[nocturne minnow]

How did you get A - C = 2 from A - C + 21/2 = 21?

[grizzled condor]

dividing 21/2 from both sides

wait im supposed to subtract it :(

that's my error

pretty sure

because then 2a = 21/2 and A = 21/4

ty dldh06

.close

[obtuse pebble]

Channel closed

Closed by @grizzled condor

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.

[karmic hedge]

https://i.imgur.com/mncsl14.png

[karmic hedge]

In order to get the critical values I have to get the factors, right?

How do I get the factors of sin(x/2)

(I know I can just look at the graph but if it wasnt there how would I figure this out)

Am I allowed to set it = 0

So that I can do sin^-1(0)?

Cause otherwise idk how to solve this one

Oke I got the right answer so I guess I can haha

.close

[obtuse pebble]

Channel closed

Closed by @karmic hedge

Use .reopen if this was a mistake.

[obtuse pebble]

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.

[tranquil sonnet]

Is there a way to transfomr a^3-2a-1=0 into a^2-a-1=0

[tight thunder]

Factor out a+1

[tranquil sonnet]

wow

dont know how i didnt see that thanmks

.coose

.close

[obtuse pebble]

Channel closed

Closed by @tranquil sonnet

Use .reopen if this was a mistake.

[obtuse pebble]

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.

[timid junco]

I don't know how many base cases I need

I do know that it starts at 14 though

so k cents of postage for 14<=n<=k using 3 and 8 cent stamps

Can I do k-2 cents of postage, by adding a 3 cent stamp we can have k-2+3=k+1?

does that work

[obtuse pebble]

@timid junco Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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.

[tranquil sonnet]

chicken mcnugget theorem

[tranquil sonnet]

.close

[obtuse pebble]

Channel closed

Closed by @tranquil sonnet

Use .reopen if this was a mistake.

[obtuse pebble]

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.

[alpine dew]

how do i do this

[alpine dew]

please

[polar fossil]

!show

[obtuse pebble]

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

[alpine dew]

here let me show it

so i was doing this on pc

(20m + 16) + (-10m + -14)

10m + 2

i got 10m + 2 but that isnt shown as a answer

[polar fossil]

you're right, it's not shown directly as an answer

you may need to simplify some of the answer choices as well

[alpine dew]

ive attempted but im not the best at this subject so it remains difficult

i know it cant be answer choice 3 or 1

or its either 2 or 4

[polar fossil]

you did that one well!

[alpine dew]

some of my answers up to this point have been lucky choices lol

i feel like its answer num 2 but i dont know how to verify it

[polar fossil]

hmm well how about you simplify answer 2 the same way you simplified the initial expression?

[clear field]

simplifying the expression is the long way

[alpine dew]

(16m + 20m) +

[clear field]

look at the similarities in the answers with the question

[alpine dew]

idk if . -2 means multiply negative two

[clear field]

yes it does

[alpine dew]

so thats the same thing as -2(5m + 7)

[clear field]

yes

[alpine dew]

so its the same question just written a bit differently;

[clear field]

correct

[alpine dew]

bruh

the question changed bc i havent answered in so long

😭

ill try to solve the new one

(-4n + 14) + 2n

-2n + 14

yessssss

i finished

i actually learned something while doing ixl

thank you guys 4 the help

ill prolly be back tmrw lol

.close

[obtuse pebble]

Channel closed

Closed by @alpine dew

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.

[untold trail]

need second opinion

[untold trail]

are tehse the points of an exponential function

it looks exponential to me but im not sure

[tight thunder]

do you have their exact coordinates

[brave bramble]

Could also be linear haha

But it looks a bit expo

[tight thunder]

Ye it looks a bit curved at the beginning

[tranquil sonnet]

The increased spacing suggests exponential

Assuming the x increment is constant

[tight thunder]

good observation

[candid yarrow]

have you tried plotting them on a log graph

then find the best fit line on that graph

´´line´

[untold trail]

yup

[candid yarrow]

could you try this

[untold trail]

lemme try

but it doesnt look like log to me

[tight thunder]

No a log graph

Not

a

um

As in a log-scale graph

[untold trail]

oh wait

sorry

like this?

[tight thunder]

Hm.

I don't know if it's clea

clear from the graph

Why don't you try analyzing the numbers themselves

They don't seem exponential to me though

But not linear either

[tough bolt]

Maybe demos's regression function will help

[novel grotto]

if this is desmos, you can use the regression function

[tough bolt]

This probably isn't an exponential function if the numbers you gave is exact

[untold trail]

i dont think i used it right 😅

[untold trail]

ohh okay yeah the values are rounded to the tenths place

thank you!

.close

[obtuse pebble]

Channel closed

Closed by @untold trail

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.

[viscid saddle]

Hello, this is based on a computing question, but I am having trouble understanding why the formula looks like this. In the example, the number is 12 and it is factored into 2 parts, 2^2 and 3^1. The number of divisors is 6 (they are 1, 2, 3, 4, 6, 12) and the product of divisors is 1x2x3x4x6x12 = 1728. When plugging into the formula that i = 2, it becomes (8^2)(3^1^3) = 1728. I have no idea how the formula relates to n, x and k.

[obtuse pebble]

@viscid saddle Has your question been resolved?

[viscid saddle]

Uh I'm gonna go to bed actually, I'll try to work on it more on my own then maybe come back if I'm still lost.

.close

[obtuse pebble]

Channel closed

Closed by @viscid saddle

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.

[neon wolf]

i have solution but dont know how its working

can anybody halp me

<@&286206848099549185>

[obtuse pebble]

@neon wolf Has your question been resolved?

[obtuse pebble]

@neon wolf Has your question been resolved?

[obtuse pebble]

@neon wolf Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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.

[steel urchin]

if i have 1/2[sin(pi-pix) - sin(pi+pix)] can i take the pi and go okay whenever i subtract x pi off its going to be -sin(pix) and on the same token when i add xpi to pi i can get -sin(pi+pix)] = sin(pix) right?

[steel urchin]

thank you bog!

.close

[obtuse pebble]

Channel closed

Closed by @steel urchin

Use .reopen if this was a mistake.

[steel urchin]

.reopen

[obtuse pebble]

✅

[steel urchin]

is this some sort of identity?

or the fact its an odd function?

thats what i was doing in my head!

'okay pi - pix is negative direction

and pi plus pix is the positive

.close

[obtuse pebble]

Channel closed

Closed by @steel urchin

Use .reopen if this was a mistake.

[obtuse pebble]

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.

[long root]

so im trying to work this out

[royal basin]

insufficient info

[long root]

and this is part of the solution

[long root]

this is all i was given

how do they get from that to the integral of -1 to -4

[royal basin]

-4 to -1,

but they withhold it from you actually

it's impossible as stated

[long root]

this is what they said

look at the bottom

but i dont understand how they got the line with -1 to -

4

[royal basin]

u-substitution

i guess

this question is confusing.

[long root]

aye

[royal basin]

did they INTEND for you to answer with "not enough info"???

[long root]

idk even know lol

[timber fox]

can you just plainly reason there is a horizontal shift of 4 to the left

[long root]

im just going through schools past papers for upcoming exam and didnt know how to answer this

[long root]

how?

[timber fox]

f(x+a) is a horizontal translation by some value a

[obtuse pebble]

@long root Has your question been resolved?

Channel closed

Closed by @long root

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.

[naive flare]

help pls

[wanton hull]

What is $\sin \theta$

[warm shale]

Bettim

[wanton hull]

Do you know how these ratios work?

[naive flare]

not 100%

[wanton hull]

Well sine is opposite by hypotenuse

You know that right

[naive flare]

yes

[wanton hull]

What is the opposite for theta in that triangle

[naive flare]

?

[oblique nebula]

I think he meant: What is the lenght of the perpendicular opposite to theta

[naive flare]

Ik you have to divide just don’t know what

[oblique nebula]

sin $\theta$ is the lenght of the perpendicular opposite to $\theta$ divided by the lenght of the hypotenuse

[high lily]

do you know the definitions of adjacent,opposite,hypotenuse in a right triangle

[warm shale]

themathboi #2137

[naive flare]

so i need to find which one is which

[oblique nebula]

Maybe answer Ramonov's question first

[high lily]

!nosols

[obtuse pebble]

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.

[stuck pasture]

Oh my bad

[naive flare]

Which equals 0.32552301255

but I put 0.3 in and it says wrong

[hidden compass]

But they ask for only two decimal places, hence what is the answer?

[high lily]

0.3 is to 1dp

you didn't round to the desired precision

[naive flare]

how do I round to the desired precision

[high lily]

count decimal places

you are told they want

to 2dp

[naive flare]

and divide it but that?

[high lily]

what divide

[naive flare]

I’m confused

[high lily]

you are told they want
to 2dp

your final answer should have 2 digits after the decimal point

[naive flare]

so 0.32?

[high lily]

not quite

[naive flare]

0.325?

[high lily]

no

you should be following standard rounding rules, and not truncating nor give more than 2 digits

[naive flare]

0.33?

[high lily]

yes

[naive flare]

thanks for ur help 😀

[obtuse pebble]

@naive flare Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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.

[woven wharf]

It should be giving 412, or I am blind

[woven wharf]

I think this solution is wrong

What do you think

<@&286206848099549185>

[fervent nexus]

Send the question

[woven wharf]

it's there

[fervent nexus]

Like 407 is part of the first AP, 412 will not be part of the first AP

[woven wharf]

Oh I get it

Thank you

.close

[obtuse pebble]

Channel closed

Closed by @woven wharf

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.

[golden cypress]

#help-7｜zen1thxyz message help me with this question please

[novel grotto]

send again so i can rotate the picture

[golden cypress]

q 30

@novel grotto

[novel grotto]

hold on a while

[golden cypress]

sure @novel grotto

[errant lark]

Which one?

[golden cypress]

q30

[novel grotto]

$\sin{\theta}(1+\sin^2{\theta})=1-\sin^2{\theta}$

[warm shale]

orthogonal

[novel grotto]

right?

now the RHS is cos theta multiplied by 1+sin^2theta=2-cos^2theta

@golden cypress do you understand? i don't really know how to explain why we do this but it's a bit weird

[golden cypress]

no i dont understand this sorry

[novel grotto]

i made a typo sorry

[golden cypress]

how is the lhs cos theta?

[novel grotto]

wait i didn't typo at all

what am i saying

[novel grotto]

sorry for confusing you

[novel grotto]

from here, $1-\cos^2{\theta}=\sin^2{\theta}$

[warm shale]

orthogonal

[golden cypress]

1-sin^2theta=cos^2thetq?

theta

[novel grotto]

yes

from trigonometric form of pythag

[errant lark]

But how does that make LHS cos(theta) ?

Shouldn't it be cos^2(theta) ?

[golden cypress]

yea that's what I am asking

[novel grotto]

bruh lemme clear up

$\sin{\theta}(1+\sin^2{\theta})=1-\sin^2{\theta}$

[warm shale]

orthogonal

[golden cypress]

check this please

[novel grotto]

$\sin{\theta}(2-\cos^2{\theta})=\cos^2{\theta}$

[warm shale]

orthogonal

[novel grotto]

it has too many typos because i am tired sorry

just ignore it

[golden cypress]

alright

[novel grotto]

this should be correct however

refer to this

[golden cypress]

yes understood that

[novel grotto]

square both sides

of this thing

ping me when you are done

[golden cypress]

done

[novel grotto]

show your work

[golden cypress]

how do you use latex?

[novel grotto]

\sin{\theta} is for sin theta
^ is for exponent
put a $ sign at the ends of your expression

[golden cypress]

sin²x(4+cos⁴x-4cos²x)=cos⁴x

[novel grotto]

ill replace theta with x ok?

[golden cypress]

just put x instead of theta

ok sure

[novel grotto]

$\sin^2{x}(4+\cos^4{x}-4\cos^2{x})=\cos^4{x}$

[warm shale]

orthogonal

[novel grotto]

$(1-\cos^2{x})(4+\cos^4{x}-4\cos^2{x})=\cos^4{x}$

[warm shale]

orthogonal

[novel grotto]

expand LHS

@golden cypress

[golden cypress]

done

@novel grotto -cos^6x+4cos⁴x-8cos²x+4

[novel grotto]

yes, shift all the terms except 4 to the RHS

what do you get

[golden cypress]

Oh wow, I got the answer 😂

[novel grotto]

yes

[golden cypress]

@novel grotto love you, you're a great teacher. Thanks for helping me out.

[obtuse pebble]

@golden cypress Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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.

[harsh remnant]

Did I do this correctly?

[ruby path]

no

$x^2 \leq 4 \implies |x| \leq 2$

[warm shale]

NEON

[harsh remnant]

Ohh

Does that mean

-2 ≤ x ≤ 2 ?

[ruby path]

yes precisely

[harsh remnant]

Ooo

I see

Thank you!

.close

[obtuse pebble]

Channel closed

Closed by @harsh remnant

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.

[tired skiff]

I need help with this proof

[tired skiff]

,rccw

[tired skiff]

question 2d specifically

[obtuse pebble]

@tired skiff Has your question been resolved?

[obtuse pebble]

@tired skiff Has your question been resolved?

[tired skiff]

<@&286206848099549185>

[tired skiff]

anyone?

[errant lark]

$e^{\frac{i\theta}{n}} \neq \frac{e^{i\theta}}{e^{n}}$

[warm shale]

Enemagneto

[tired skiff]

i solved it

.close

[obtuse pebble]

Channel closed

Closed by @tired skiff

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.

[astral aurora]

I have solved this simple multiply

But i want to know any other method

[sage geode]

What do you mean by "simple multiply"?

There are like 2022 matrices to multiply by one another

Did you diagonalize?

[astral aurora]

Nope

No eigen values

[sage geode]

[astral aurora]

Is there any helpful solution?

[ornate oar]

Wth did you do then

[astral aurora]

A^2022 =I+2022B

[royal basin]

what?

what's B?

[sage geode]

What's B and how do you know such B exists?

[astral aurora]

1,1
-1,-1

[warm canopy]

its not diagonalisable but it has a nice jordan form

[sage geode]

Okay such B may exist but how do you find it

[royal basin]

and from where do you get that A^2022 = I + 2022B, with this B?

[astral aurora]

I multiplied A×A

[royal basin]

not obvious to me, that's for sure...

[sage geode]

,w multiply {{5/2, 3/2}, {-3/2, -1/2}} by {{5/2, 3/2}, {-3/2, -1/2}}

[warm shale]

Results provided by WolframAlpha

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

Requested by alonelybean

[sage geode]

Then?

[royal basin]

uh wait

you sure WA didnt accidentally do pointwise multiplication

,w {{5/2, 3/2}, {-3/2, -1/2}} * {{5/2, 3/2}, {-3/2, -1/2}}

[warm shale]

Results provided by WolframAlpha

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

Requested by ann_dec

[royal basin]

this is more like it...

@astral aurora is this what you got for A^2?

[sage geode]

Oh, so that's why it used to give me the wrong matrices unless I was on the website

Thanks for letting me know

[astral aurora]

Yes same

Now took common I and 3/2

[royal basin]

ok, so sticking with $B = \bmqty{1 & 1 \ -1 & -1}$ you noticed that $A = I + \frac{3}{2}B$ and $A^2 = I + 3B$...?

[warm shale]

Ann

[royal basin]

i guess B is nilpotent with index 2 yeah

ok

so then you would have $A^n = I + \frac{3}{2}nB$ though

[warm shale]

Ann

[royal basin]

so I + 3033B not I+2022B

(or your notation was inconsistent, whatever)

ok, yeah, this works.

[warm canopy]

from the very stack exchange website

[royal basin]

ok so arjunn used B for a different matrix than here

[astral aurora]

They multiplied it with 3/2

[royal basin]

i guess eigenstuff won't be of much help here, since A-I is nilpotent

so the eigenvalues of A are 1 (alg mult 2), but it is not diagonalizable

[warm canopy]

im not really sure what you want here arjuun, there are many methods in the stack exchange post

[ornate oar]

Jordan decomp is then the systematic way I guess?

[astral aurora]

What?

[warm canopy]

it has a nice jordan form that is easy to calculate powers of

[royal basin]

aint that what we did essentially

[astral aurora]

There are many methods but I did not understand them

[ornate oar]

Yea looks like it

[royal basin]

best show us the methods and ask about those methods specifically...

otherwise you are forcing us to read your mind

[astral aurora]

I mentioned this

.close

[obtuse pebble]

Channel closed

Closed by @astral aurora

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.

[clear matrix]

what can i find using -Δ/4a? i just heard about it but i cant find anything about it on it in google

[sage geode]

It's the y-coordinate of the vertex of the parabola

[clear matrix]

is it useful? like is it allowed on tests?

[royal basin]

your teacher knows that better than us

[clear matrix]

yea that makes sense

should i use it instead of substituting the x value into the original equation?

[sage geode]

If the coefficients are nice enough, I would rather substitute x = -b/2a

[clear matrix]

oh alr! ty

and btw what does c/a do?

[sage geode]

It's just the product of roots

[clear matrix]

and -b/a?

[clear matrix]

which means?

[sage geode]

The sum of them

[clear matrix]

like if you add them or the amount of them?

[sage geode]

If $x_1, x_2$ are the solutions to $ax^2 + bx + c = 0$ where $a \ne 0$, then [\begin{cases} x_1 + x_2 = -\frac{b}a \ x_1\cdot x_2 = \frac{c}a \end{cases}]

[warm shale]

A Lonely Bean

[clear matrix]

oh i see

tysm!

.close

[obtuse pebble]

Channel closed

Closed by @clear matrix

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.

[forest glade]

they want me to use the distance formula or the midpoint formula

[forest glade]

but what do they want me to say?? like filling in the blanks

what could i say that require either formula?

[urban patrol]

parallel just means they have the same slope but aren't on top of each other

that can be easily proved with (y2-y1)/(x2-x1)

diagonals can be found with the distance formula

for seeing of the diagonals bisect you have to find the midpoints of both diagonals and see if they're equal

[forest glade]

so paralells can be proven with the midpoint formula??

[urban patrol]

no

with the formula I gave you

for slope

[forest glade]

im sorry im a bit lost

[urban patrol]

confused with the formula?

[forest glade]

according to the instructions i can only use the midpoint or distance formula to back up any claims

this means i have no clue what they want me to fill in the blanks with

[urban patrol]

I dont think there's any way to prove they're parallel with those two formulas

[forest glade]

paralell was just my guess

so opposite sides are -------- eachother

[urban patrol]

oh

[forest glade]

using the midpoint or distance formula i havw to figure out what they want me to say i guess

[urban patrol]

thats not how you fill in the question

it just says prove opposite sides are equal to each other

thats just the distance formula

[forest glade]

where does it say that?

am i defining the formulas..?

[urban patrol]

well you added on parallel I assume

[forest glade]

yeah, that was my assumption, wrong

waaaait

im bliiiind

thats not a - its an equals sign

[urban patrol]

yeah

so use distance formula to prove opposite sides are equal and do the same for the diagonals

[forest glade]

okay this makes sense now

[urban patrol]

I explained how to do the third one

[forest glade]

im not filling in any blanks

yeah, thank you!

im gonna go give these a go

actually wait

it says 1 point for formula, 1 point for work, 1 for conclusion

what do they want for work?? or conclusion?

do they want me to explain?

[urban patrol]

plug in values into formula

simplify it

then conlude with your solution

[forest glade]

ohh

oh no i have another question

what values

like for the opposite sides

the length??

[urban patrol]

yes

[forest glade]

okok

thanks

[obtuse pebble]

@forest glade Has your question been resolved?

Channel closed

Closed by @forest glade

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.

[elder stone]

Hii I have trouble understanding this notation of degrees. It would be a massive help if I could be enlightened.

[plain stag]

6 degrees, 37 minutes

60' = 1° if that helps

[elder stone]

minutess really

yes thank you so much

[polar fossil]

you'll also sometimes see seconds, with two apostrophes; those are 1/60 of a minute as you'd expect

[elder stone]

ahh ill jot it down. Thanks for the heads up.

.close

[obtuse pebble]

Channel closed

Closed by @elder stone

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.

[charred topaz]

The number of policies sold by an agent in an insurance company in a week follows a Poisson distribution. The mode of the Poisson distribution is at two points: 2 and 3. The probability that an agent sells more than n policies in a week is less than 20%. Determine the minimum value of n

[charred topaz]

So im interested in Pr(N > n) < 0.20

I have to lamba given the modes at 2 and 3 and I found my lamba value to be 3

So when x = 0 and lamba is three on a poisson dist. I get 0.04979

x = 1 I get 0.14936

I add 0.04979 and 0.14936 to get 0.199

And this value is less than 0.20

[plain stag]

All looks fine, though don't draw the wrong conclusion

[charred topaz]

I was wondering where I am going wrong in this logic

wym?

[plain stag]

Indeed for n=2 P(N < n) < 0.2

But that's not the probability you're looking for

[charred topaz]

I am sorry, im a bit lost

I thought im looking for P(N > n) < 0.2 and what value of n would suffice this statement

[plain stag]

You are

[charred topaz]

So the minimum value is 1 no?

[plain stag]

No

P(N > 1) = 0.801

[charred topaz]

ohhhh

ohhh

ohhh

Yea

wait yea

okay I got it

So I need to keep going till that value goes. below 0.20

I'm just trying to see where my thinking was mistaken

[plain stag]

You need this sum to surpass 0.8

[charred topaz]

because your 100% correct

Yea

i just don't know why i didn't think of that

[plain stag]

You just reversed the inequality within the probability statement, that's all

[charred topaz]

I was fixing it and i was still confused, but now makes sense

wym>

?

[plain stag]

You found P(N < n) when you need P(N > n)

[charred topaz]

Yea if i did the complement that would have worked

idk i just didn't realize what I was solving for ig

Thanks for clearing up the confusion 🙂

.close

[obtuse pebble]

Channel closed

Closed by @charred topaz

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.

[sly moon]

Hi, so I'm trying to figure out how this equation is true

[sly moon]

the second integral (from 0 to inf) diverges'

I asked this question before and realized that I was doing the inside arithmetic wrong

but now doing it right, it still doesn't make sense

you simplify and pull out the 1/2 and you get

$\frac{1}{2}[\int_{0}^{\infty}e^{t(x-1)} dx]$

[warm shale]

bouncingsouls1

[sly moon]

which is equal to

$\frac{e^{x(t-1)}}{t-1}$

[warm shale]

bouncingsouls1

[sly moon]

evaluated from 0 to infinity

which is just infinity

[plain stag]

Not quite

If t is less than 1 then the exponent goes to negative infinity

[sly moon]

but it still doesn't converge then, right?

[sly moon]

like, this isn't true?

[plain stag]

If the exponent goes to negative infinity then the numerator goes to 0, so it does converge

[sly moon]

ooooh

ok

that makes sense

thanks!!

[plain stag]

[sly moon]

.close

[obtuse pebble]

Channel closed

Closed by @sly moon

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.

[cobalt pendant]

I’m stupid

[pallid aurora]

What have you found so far?

[cobalt pendant]

Everything I wrote

[pallid aurora]

Sorry my internet is really bad right now

So it seems like you wrote an area formula, and then manipulated it by multiplying both sides of the equation by 2. Can you identify your height and length? Maybe they have some relation to the circle?

[cobalt pendant]

I believe the height and length might be both 4, and I tried using pythagorean theorem but I still can’t get the answer

[pallid aurora]

You are correct about the height and length being 4, do you understand why?

[obtuse pebble]

@cobalt pendant Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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.

[cobalt pendant]

Yea bc those line are equal, but how do I find answer from that

[cobalt pendant]

Damn 💀

[pallid aurora]

Sorry was away for a minute

So we are looking for arc measure of AB correct

What portion of the circle does arc AB take up

Central angle ACB may help

[obtuse pebble]

@cobalt pendant Has your question been resolved?

[cobalt pendant]

I’m not sure

[pallid aurora]

Take a look at the central angle, how does that relate?

[obtuse pebble]

@cobalt pendant Has your question been resolved?

[cobalt pendant]

The triangle take almost 1/4 of the circle

Idk

[pallid aurora]

Central angle is equal to portion of circle the arc takes

We know central angle is 90

So find circumference and multiple by 1/4

[cobalt pendant]

Ohhhh thanks

[obtuse pebble]

@cobalt pendant Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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.

[pallid drift]

what is a sampling distribution i forgot, when searching it up on google i end up confusing myself

[pallid drift]

would the answer be yes because there's a rather large amount of SRSs?

also would the answer to b) be normal distribution as well as 62.4 or is there a process i have to use to calculate the outliers

[obtuse pebble]

@pallid drift Has your question been resolved?

[pallid drift]

pls anyone

[slim cove]

It looks more or less normal to me 🤷 maybe describe roughly the mean and the stdev

It doesn't really look like there are any outliers to me, but I guess that's a judgment call

I think this question is more qualitative/subjective, so as long as you can justify your answer, you're fine

[pallid drift]

ok then thank u

[slim cove]

No problem!

[pallid drift]

can you please help on a as well or should i just try and guess on it

[slim cove]

What are your thoughts on what a) should be?

[pallid drift]

well going off of what i've read a sampling distribution should cover most possibilities(?) and given that the chart is made of 250 SRSs i think that covers most right

[slim cove]

Yeah, maybe technically it's not the sampling distribution because it's only 250 SRSs rather than every possible SRS, but that seems like nitpicking to me

[pallid drift]

so the answer would be something like "it's not the exact sampling distribution but its a good approximation of one"?

[slim cove]

Yeah

That might be a good way to put it

[pallid drift]

ok!! thank you so much :)

[slim cove]

No problem at all! :)

[pallid drift]

would the answer to c be not normal because n(p) is less than 10?

[pallid drift]

i don't understand 4a

or 4b

[obtuse pebble]

@pallid drift Has your question been resolved?

[pallid drift]

<@&286206848099549185>

[obtuse pebble]

@pallid drift Has your question been resolved?

[obtuse pebble]

@pallid drift Has your question been resolved?

[pallid drift]

thanks guys

i appreciate it

:|

.close

[obtuse pebble]

Channel closed

Closed by @pallid drift

Use .reopen if this was a mistake.

[obtuse pebble]

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.

[tired sierra]

do you guys know any tricks for finding the median when you have a really big range of numbers?

[wooden cipher]

Do you have an example

[tired sierra]

because I vaguely remember seeing one where you add 1 and divide it by 2

but I remember you also add 2 if the number is even or odd or something like that

[high lily]

identify the number of entries,
whether you have odd/even amount, slightly differs

[tired sierra]

theres 24 entries

[wooden cipher]

Is the middle nunber the 10s digit

[high lily]

size of dataset doesn't really matter

[tired sierra]

oh

well its a class of 24 if thats important

so...

when its even you add 2 and then divide by 2?

and when its odd you add 1 and divide by 2...

[high lily]

when you have an even number,
the median will be the average between the
n/2 and n/2 + 1 value

i.e. in this case the average between the 12th and 13th value

[tired sierra]

ah so its when its even you add 1

what about when its odd

[high lily]

that's not what i said

[tired sierra]

when you have an even number the median will between the average of n/2 and n/2 +1

but then you could just do (n+1)/2 and you get 12.5, where you can remember its between 12 and 13

then average and badabing badaboom

[high lily]

if that works for you, you could think if it like that

[tired sierra]

would it be the same for negative numbers?

[high lily]

wdym

[tired sierra]

not negative

odd numbers

my bad

[high lily]

yeh.

(n+1)/2

[tired sierra]

huh thats it?

[high lily]

yeh

[tired sierra]

I thought you had to add 2 or smthin...

[high lily]

you get 12.5
just don't misinterpret that result

[tired sierra]

I wont, I've been using this for like the past 6 years of my life, I just can't remember the first half lol

[high lily]

don't get any idea that getting a decimal means that the median doesn't exist or that you could pick either 12th or 13th at random

[tired sierra]

alright I won't, thanks for the help

.close

[obtuse pebble]

Channel closed

Closed by @tired sierra

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.

[flint ivy]

I dont really understand b

[flint ivy]

i dont know any of the relations of face edge or vertices to mxn

[finite pivot]

hi can i have help

[flint ivy]

uhh do it in another channel

[finite pivot]

ok

[limber quartz]

There's a formula that comes from binomial theorem

[flint ivy]

can you show me?

i know that 2=v-e+f i think

but like

[royal basin]

???

[limber quartz]

The number of paths through a grid; city block traversal

[royal basin]

you... don't need this here?

[limber quartz]

Ok, nevermind. I got the wrong context.

[limber quartz]

v - e + f = 2

[flint ivy]

ah right

[limber quartz]

v = 12,
e = 17,
f = 6+1=7

Hmm

[flint ivy]

i meant question b

for m times n grid graph

[limber quartz]

I got you but part a isn't working for me

[flint ivy]

f is 7

the 6 grids

plus the one outside

since that technically is a face

[limber quartz]

Is that something conventional when using this formula?

[flint ivy]

according to my instructor

yeah?

i think

[limber quartz]

Ok

For part b you can write down a few cases and see if you can find a pattern

For the number of vertices, edges, faces

[flint ivy]

i tried

but the solutions prposed like a weird answer

[limber quartz]

2x2: 4, 4, 2
2x3: 6, 7, 3
3x3: 9, 12, 5
...

[limber quartz]

Show me

[limber quartz]

What's weird about that

[flint ivy]

was i just meant to figure out the pattern

[limber quartz]

Yeah

[flint ivy]

ah

ok

let me have a go

[flint ivy]

wait waht does this mena

[limber quartz]

Those are some cases

[flint ivy]

it doesnt line up with the answers

[limber quartz]

Vertices, Edges, Faces

[flint ivy]

i dont think so

for a 2x2 theres somehow 3 edges

[limber quartz]

2×2 has 4 verts, 4 edges, 2 faces
2×3 has 6 verts, 7 edges, 3 faces,
etc

[limber quartz]

No

It's a box

4 edges

[flint ivy]

i dont see the pattern

for edges

for vertices i see its mn

[obtuse pebble]

@flint ivy Has your question been resolved?

[flint ivy]

nvm ive been told the answers wrong

my instructor explained it to me now

.close

[obtuse pebble]

Channel closed

Closed by @flint ivy

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.

[tired sierra]

to do this it would just be 1-cosx - sinx right?

[novel grotto]

it is a definite integral

to be exact two

[royal basin]

if you want to do this \textbf{by integration} then it would be $$\int_0^{\pi/2} (1 - \cos(x)) \dd{x} + \int_{\pi/2}^{\pi} \sin(x) \dd{x},$$ but it is \textbf{NOT} ``just'' $1-\cos(x)-\sin(x)$.

[tired sierra]

I know, I don't know how to type up a definite intergral

[warm shale]

Ann

[tired sierra]

wait... why are they seperate

[royal basin]

the shaded region is composed of two parts

[tired sierra]

to find the area under two curves its the top curve - bottom curve

[royal basin]

no, that's BETWEEN two curves.

also your curves switch up who is bottom and who is top halfway through.

[tired sierra]

oh I see

[novel grotto]

if you look at the "top-bounding curve" it switches halfway

so you can't do 1-cosx-sinx

[tired sierra]

I see, that makes more sense now

[royal basin]

i said this just a minute ago

[tired sierra]

I didn't even think about them as x-axis bounded curves, I saw them as two curves

man.

thanks for the help then

[novel grotto]

i was about to finish typing when i saw your message so i sent it anyway

i see no problem with me doing this

[tired sierra]

.close

[obtuse pebble]

Channel closed

Closed by @tired sierra

Use .reopen if this was a mistake.

[obtuse pebble]

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.

[plush girder]

lol

[obtuse pebble]

Please don't occupy multiple help channels.

[plush girder]

.close

[obtuse pebble]

Channel closed

Closed by @plush girder

Use .reopen if this was a mistake.

[plush girder]

.reopen

[obtuse pebble]

✅

[warm canopy]

#help-4

.close

[obtuse pebble]

Channel closed

Closed by @warm canopy

Use .reopen if this was a mistake.

[obtuse pebble]

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.