help-36

[proven chasm]

If the plane is parallel to the y-axis, does that mean that the slope in ax + by + c is 0 for b

so it'd look like (-5)(1) + (-2)(0) + c = 7

c = 12?

[molten basin]

The general form for the equation of a plane in 3D space is:

Ax + By + Cz + D = 0

[proven chasm]

That's what I used

idk about cz though

I used ax + by + c = z

but that doesn't really answer my question

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[rotund badge]

You can basically ignore the y here I guess, just solve for z = ax + b

[proven chasm]

ye that's waht i did

ty

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[solid ice]

Instructions: Write out sample space (S) and find the number of outcomes n(S) for each of the following experiments

Question:Selecting randomly two delegates from a group of 15 girls.

[pliant shore]

!status

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What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[solid ice]

I don't know where to begin

[pliant shore]

The sample space is just all the possible outcomes that can happen

So it's a description in words

For the number of outcomes, which one do you need
Permutations or combinations?

[solid ice]

I don't know any of that sorry

This is my first statistics lesson

[pliant shore]

Oh ok

[solid ice]

I wanna know how to get the sample space and the number of outcomes

[pliant shore]

Yeah, so for the number of outcomes

Does the order matter?

[solid ice]

Yes

[pliant shore]

Like if I pick girl A first then girl B, will I get the same group as girl B then girl A?

[solid ice]

Wait, I don't get it

[solid ice]

Yes you will, I think?

[pliant shore]

Yeah, so in this case you use combinations

Permutations are when the order matters

Say if I have 8 runners in a race

There are 8 people who can be 1st

7 people who can be 2nd

And 6 people who can be 3rd

So there are 8 * 7 * 6 different ways for the top 3 finishers

[solid ice]

Ohh

That's the example of permutations?

[pliant shore]

Yep

[solid ice]

How about combinations

[pliant shore]

So if you're just choosing 3 of the people from the 8 runners

There are 6 different orderings, 3 * 2 * 1

ABC, ACB, BAC, BCA, CAB, and CBA

For each one of those orderings, you still have the same 3 people

So you need to divide by 3! = 6

8 * 7 * 6 / 3!

Or with different notation, ${8 \choose 3}$ or $^8C_3$

[soft zealot]

south

[solid ice]

What..

[pliant shore]

It's just notation for combinations

[solid ice]

Could you give me an example similar to the question I gave? I really don't understand

[pliant shore]

So if you have just 4 girls for example

There are 4 ways to choose the first girl and 3 ways to choose the 2nd girl

But the order doesn't matter: you will still pick the same two girls

The possible orders are AB and BA

[pliant shore]

So what number do you have to divide by here?

[solid ice]

2?

[pliant shore]

Yep

Yeah so if it's 4 girls it's just 4 * 3 / 2

[solid ice]

So 2 divided by 4??

[pliant shore]

No

[solid ice]

Why is there a fraction? They weren't any fraction teached to us in our lecture

[pliant shore]

It's not a fraction
This is guaranteed to give you a whole number

[solid ice]

My lesson is about Fundamental Counting Principle

[pliant shore]

Ah so it's just the multiplication rule and stuff like that right

E.g if I have 3 starters, 2 main courses, and 4 desserts

[solid ice]

YES

[pliant shore]

I have 3 * 2 * 4 meals in total

[solid ice]

Yes

[pliant shore]

Yeah sorry I skipped ahead, but yep

The question you sent me doesn't use the material in your lecture then

[solid ice]

How come?

[pliant shore]

It uses combinations which isn't in your lecture

[solid ice]

Then

[solid ice]

How would I able to solve it?

[pliant shore]

So it would be just 15 * 14 / 2

Using the reasoning above

For the sample space, give each girl a number

Then it would be {(1, 2), (1, 3), (1, 4), .... (2, 3), (2, 4).... , (14, 15)}

[solid ice]

How about the number of outcomes?

[pliant shore]

Yeah the number of outcomes is 15 * 14 / 2

[solid ice]

What, I don't get it...

I'm looking for a whole number

[pliant shore]

Yes that is a whole number

15 * 14/2 = 15 * 7

[solid ice]

Do I have to solve this?

[pliant shore]

Yeah you need to simplify it ofc

[solid ice]

Ohh okay

105?

Ohhh

So the outcome is 105

[solid ice]

I don't get this, do I have to make 105 pairs?

I mean write down over 105 pairs

[pliant shore]

I'm just listing the outcomes

[pliant shore]

No

You can just use dots

[solid ice]

Okay, I guess my teacher will understand

Okay so Then it would be as sample space {(1, 2), (1, 3), (1, 4), .... (2, 3), (2, 4).... , (14, 15)} and 105 as outcomes?

[pliant shore]

Yeah I feel a written description also works

All unordered pairs of numbers such that each number is an integer chosen from between 1 and 15

[solid ice]

I think my problem is solved now

[pliant shore]

Yeah no worries

[solid ice]

THANK YOU SO MUCH

I really appreciate you

Your help

Tyyy

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[tranquil pine]

From what I understand, cos(-pi/2) = 0.9996242169.

[tranquil pine]

here its said cos(-pi/2 ) = 0.

how please?

[pliant shore]

No, cause this is in radians

You're evaluating cos(-pi/2 degrees)

[tranquil pine]

that was quick lol

ah I get it

because cos(-90) = 0.

perfect, thanks! 🙏

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[tranquil pine]

can

[tranquil pine]

someone

explain

if two waves are superpositioned

i can add them like vectors to get the resultant amplitude

why

[warm ether]

you mean physically?

[tranquil pine]

well mathematically physically wouldn't it mean same lol

we dont just write equations in physics for fun

like sound waves

or travelling waves on string

if two of them are superpositioned

we can get resultant amplitude by vector sum

why is that

[rocky carbon]

Yeah because we add their individual particle displacements vectorially

[tranquil pine]

where angle between vectors is the phase difference

[warm ether]

waves are made by forces moving particles from their equilibrium position that propagate through a medium. So when two waves are superimposing these forces will either add to each other or detract from one another, you add forces like vectors, then since the amplitude is proportional to the forces, the amplitudes add in the same fashion

[tranquil pine]

is there a proof for proportionality of amplitude to the forces

i mean this would make sense to me if it was just oscillation not a travelling wave

like forces air resistance may cause oscillation to become damped

affecting their amplitude

but how can we say same for waves

how do forces affect amplitude of travelling waves

it would make sense only if wave was simple harmonic

then the restoring forces would add up vectorially

right ?

thats why we can also find resultant amplitude of super imposed simple harmonic motions

thats what i think after reading your explanation

@rocky carbon@warm ether

[rocky carbon]

Forces add up vectorially in any case

[tranquil pine]

yes

[rocky carbon]

It doesnt need to be Simple harmonic

[tranquil pine]

but its not intuitive to me in case of waves

it is intuitve to me in case of a simple harmonic motion or any kind of oscillation

[rocky carbon]

You can prove it mathematically for Simple harmonic oscillations

[tranquil pine]

like you can consider a spring connected to a block and connected to a fixed end then forces in nature will affect its amplitude

the differential equation so formed will be different

[tranquil pine]

yes i have

[rocky carbon]

No we mean the restoring forces not the external forces

restoring forces in two different waves

[tranquil pine]

oh

how

i dont think ive done that

[rocky carbon]

Its not forces in nature

[tranquil pine]

yeah

can you tell me by an example

[rocky carbon]

Like in a spring block system the force which the spring exerts is what we are talking about

not something like friction

if you mean forces in nature its damped oscillations

[tranquil pine]

yes yes i get you mean the restoring force

which is causing the simple harmonic motion

[rocky carbon]

Yeah

[tranquil pine]

like f = -kx in case of spring mass systems

[rocky carbon]

yes

But in any shm displacement is a function of sin or cos

So you can add two waves

and get the resultant amplitude

[tranquil pine]

so we write equation of shm then find second derivatives then multiply by mass then add the force vectorially

you mean this ?

[rocky carbon]

why not directly add the displacements?

displacement is a sin function right

[tranquil pine]

yes we can add them

[rocky carbon]

like y = A Sin(wt) and x=A sin (wt+φ)

[tranquil pine]

but its not necessary that this will always work

[rocky carbon]

so if we find x+y and do a bit of trigonometry we get the amplitude of the resultant wave

[tranquil pine]

you are taking w to be same

in both case

what if theyre different

[rocky carbon]

w has to be same for superposition

wait

i think it can be different

[tranquil pine]

yes

i think so too

[rocky carbon]

Wait im wrong it has to be same for superposition

[tranquil pine]

no it dosen't

[rocky carbon]

Oh

if w is different its gonna be complicated

[tranquil pine]

but

if its different

resultant motion would not be simple harmonic

[rocky carbon]

yeah it wont

[tranquil pine]

you can take example of asinwt + bcos2wt

resultant won't satisfy shm equation

[rocky carbon]

It wont

I have no idea about that

[tranquil pine]

hmm thanks though your explanation is bearable

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[rocky carbon]

Ive only studied simple harmonic so idk about those

[tranquil pine]

same

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[native valve]

guys i solved this sum but is there a way to solve it using componendo and dividendo?

[native valve]

please tell

someone

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@everyone

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[native valve]

@everyone

@everyone

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[robust linden]

can someone walk me through how i work this out?

[tiny gorge]

you could write it as two equations

one for each row

[robust linden]

like simultaneous ?

[tiny gorge]

yep

one of them only involves a, so you can solve for a

and plug into the other equation to find b

[robust linden]

how do i make it a simultaneous equation?

[tiny gorge]

just read off each row - this is just two simultaneous equations in disguise
the first row is:
5a - 2(-1) = 17

what's the second row?

[robust linden]

what did you do with the 5 before the bracket ?

where the b go?

[tiny gorge]

there's no b in the first row

[robust linden]

would it be

[tiny gorge]

what if i rewrite it like this:
$$\begin{pmatrix}5a \ 25\end{pmatrix} + \begin{pmatrix}2 \ -2b\end{pmatrix} = \begin{pmatrix}17 \ a\end{pmatrix}$$

[soft zealot]

Bungo

[robust linden]

25 -2b = a

[tiny gorge]

where all i did was carry out the scalar multiplication

[tiny gorge]

yep correct

[robust linden]

ohhh

thank you

i'm still a bit confused

is A = 3?

[tiny gorge]

yea, that's correct

[robust linden]

you know in this

[tiny gorge]

now you can use that in the second equation to find b

[robust linden]

how did u write it like that

[robust linden]

i'm not on about this btw

[tiny gorge]

i took the scalar 5 in front of the first vector and just moved it inside (it means, multiply each component of the vector by 5)

and similarly for the -2

[robust linden]

why is it addition between them not subtraction?

[tiny gorge]

for the -2?

[robust linden]

between the vectors

[tiny gorge]

i just viewed it as plus (-2 times the vector)

instead of minus (2 times the vector)

so i moved the -2 inside

you can do it either way

[robust linden]

hm

wait let me try answer it now

[robust linden]

would the other way be harder?

[tiny gorge]

nope

[robust linden]

okok

[robust linden]

is this the same as

5a + 2 = 17
25 - 2b = a

b = 12?

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[brisk lynx]

all angles are right angles+quadrilateral(four sides)

[rain stirrup]

i got it

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[fathom walrus]

@rain stirrup what are ur thoughts

[rain stirrup]

i dont know

[fathom walrus]

if i have a bag with a green ball in it

and i reach in and randomly grab a ball

whats the probability i get a green ball?

[rain stirrup]

100%?

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[violet hull]

could someone explain to me how to solve this?

[bitter sparrow]

,rotate 270

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[violet hull]

<@&286206848099549185>

[tulip vector]

@violet hull what is your current topic

Like do you have an exact task?

[violet hull]

No, I just need like the answer to the equation below

im unsure like where to start with it

[tulip vector]

Also what does the 1,(6) mean? Is it a rounded result?

[violet hull]

yeah so like it can also be 15/9

[tulip vector]

Is 7a-ab-ac=1,6
Equal to 7(a+0,2)-a(b+c)?
Or are they unrelated

[violet hull]

Unrelated

[tulip vector]

Ok..

[violet hull]

i have to say what 7(a+0,2)-a(b+c) is qual to based on the information given in the first equation i guess

[tulip vector]

?

[violet hull]

Okay so

it gave me the first equation

7a-ab-ac=1,(6)

[tulip vector]

Ok

[violet hull]

i have to solve the second one based on the first one

[tulip vector]

Ah ok give me a minute

I think I have the solution

Just have a struggle with my tablet

Sec

[violet hull]

Alright no worries

[tulip vector]

So if you have 7a-ab-ac = 1,6 and the other 7a-ab-ac= -1,4, then I guess ? Would be 3, because 3-1,4 =1,6

[violet hull]

Oh thank you

[tulip vector]

Was it helpful?

Btw the third line is unnecessary

[violet hull]

At the end?

[tulip vector]

The 7(a+0,2)=a(b+c) lol

I just wanted to see how the equation works and my IPad is weird rn. I have struggles with my Pen💀

[violet hull]

No worries

[tulip vector]

I mean it's technically correct, except I forgot to write the question mark at the end

Do you understand why or do you have any questions left?

[violet hull]

Thanks I understand now. I was just confused abt the end part but now I got it

[tulip vector]

Oki. I think that you have to write like

I hope this helps

If not, feel free to ask

Or smth like cutting away 7a-ab-ac, cause it's there in both equations

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[atomic kite]

Can someone help me with this question

[atomic kite]

Do you just do the root test and then as it approaches infinity the value reaches 0<1 so it’s convergent

[polar solstice]

does anyone know year 10 algebra
and is willing to explain it to my class in a call

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[fossil geyser]

Yeah this works

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[tranquil pine]

why is ∫[0,L]asin((nπ/L)x)sin((mπ/L)x) = 0 when m ≠ n

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[silent dock]

Shouldn't it be just 5?

[jolly tapir]

you're right

sqrts cannot spit out negative numbers lol

[silent dock]

yes that's what I was saying but

it can be (-5)^2

[royal gust]

Putting negatives under the square root, then using an algebraic identity is also very sus

[silent dock]

in this situation it's only asking about the positive number

[silent dock]

imaginary numbers

it's -50/-2 so not a big deal

[royal gust]

Applebee's

[silent dock]

Do you want the menu?

[royal gust]

Okay. Yes imaginary numbers exist. But this is very dangerous

[silent dock]

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[royal gust]

Like, this tier of mistake can just happen

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[sonic crystal]

The red graph is the shape I thought f(x) would look like given by the sign testing which isn’t all that off but the blue shape is the correct one, how would I figure this out tho

[sonic crystal]

Btw should say - inf on the numberline for the left side

[formal trail]

you should also test the second derivative if you want the concavity

[sonic crystal]

But I only have one stationary point which would only tell me it’s concave up at x=0 but I figured that out from the sign testing, or is there some other thing I’m missing

[formal trail]

there are inflection points that you are not accounting for

[sonic crystal]

Is that for the second derivative = 0

[robust fulcrum]

yes

[sonic crystal]

okay

[robust fulcrum]

try that and check if the inflection occurs somewhere close to that correct graph's inflection points

[sonic crystal]

Btw is a minimum and maximum a point of inflection technically I’ve been wondering this for awhile

[formal trail]

a point of inflection is where the second derivative changes signs

so it generally can't be a minimum or maximum, although it can also have f'(x) = 0

[sonic crystal]

So mins and max don’t count

[robust fulcrum]

so f'(x)=0 is not the necessary condition for an inflection point?

[formal trail]

it's not, a simple example is sin(x)

every x-intercept is also an inflection point

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[earnest shale]

For this question ik how to do the first 15 kg block

for the second half

do I take the entire thing as 25 kg?

and then do 25x9.8xsin50

or do I do smth else?

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[craggy bane]

I'm stuck at part (b)(ii)

Can someone teach me how to do this ?

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[worldly loom]

my answer was confirmed by 2 AI's so I'm not sure if I'm just going crazy or what

Just asking for confirmation this question is bugged or if I did something wrong

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[stone bough]

what i'd like to ask is there are two different coordinates of R idk why it is considered the (-13,0) neglecting (3,2)

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[zenith pollen]

I get 1264/(4+pi) yea

[worldly loom]

Okay I'll mark it as complete then idk what the answer is

Thanks

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[wide folio]

the answer for c is shown in the box with the red corner (This is directional derivatives)

[wide folio]

so for c

i got +/- 3/5 for u_2

and +/- 4/5 for u_1

how do I know which two numbers will be pairs

originally i had 4/5i - 3/5j, -4/5i + 3/5j

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[plucky scaffold]

is this right?

[mellow axle]

in a) you're missing (1,3) - 1r3 and 3s3

[plucky scaffold]

so 1, 3 isnt 5?

[mellow axle]

wdym?

[plucky scaffold]

1+2=3

then

3<=x

x can be

3, 5

so yes for d

right?

i fixed graph

[mellow axle]

uh not sure about the visual

but look at your part a and b

they contian the same elements

[plucky scaffold]

i have a key

[plucky scaffold]

^^

[mellow axle]

oh yes didn't see that msg

[plucky scaffold]

am i good

?

[mellow axle]

yes, d is yes

[plucky scaffold]

and a b c?

[mellow axle]

a and b look fine, idk c

[plucky scaffold]

ok

thx man

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[fallen reef]

For c, I don't understand why it's not antisymmetric. Sure it has (1,3) and (3,1) but it also has (2,3) and NOT (3,2), wouldn't that make it antisymmetric?

[sturdy cypress]

it would make it not symmetric

that's not enough

[fallen reef]

hmm okay

So if it contains even one instance where two elements are symmetric, it is not antisym?

[sturdy cypress]

right

[fallen reef]

thanks

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[jovial iris]

g(x) = |x - 5| + 1

[jovial iris]

How would I go about finding inverse of this function?

I know that:

But when I try to do it the standard way, it simply doesn't work

I should probably mention:

g: B -> C
Where B = {-2, -1, 2, 5, 14}
C = {1, 4, 7, 8, 10}

[tiny gorge]

well, as a function of a general real variable it doesn't have an inverse

but you have a very restricted domain

[jovial iris]

Yeah, but what about this restricted domain

[tiny gorge]

you can just work out which value in C each value of B gets sent to

[jovial iris]

I have found mapping and it is bijective

[tiny gorge]

and then your inverse just does the opposite

for example, g(-2) = 8

[jovial iris]

Yeah but I have to write it down mathematically using some step by step process, ig

[tiny gorge]

so g^-1(8) = -2

[jovial iris]

Yeah but how do you derive the complete function?

Why can't we just split into two cases: x >= 5 && x < 5

And do the standard thingy

[tiny gorge]

well there will be infinitely many formulas that work given that you only have to pass through 5 points

[jovial iris]

Then combine in the end

[tiny gorge]

because |x-5| + 1 is not invertible in general

[jovial iris]

So you suggest me just case bashing?

[tiny gorge]

dunno, what's the context? is someone demanding a formula?

it's only 5 points, i would probably just brute force it unless there's a natural formula to use

[jovial iris]

You have to provide a formula, yes

And step by step how you got it

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[finite island]

I am trying to solve the following integral:

$\int \frac{2-3x}{\sqrt{1+3x-2x^2}} , dx$

I have solved the first term and found it :

$\frac{3}{2} *\sqrt{1+3x-2x^2} $

and the second term is where I am having trouble. I seem to have the numerator correct

$\arcsin\left(\frac{4x-3}{\sqrt{17}}\right) $

however the denominator is meant to have a sqrt(2) and I have no idea where it gets that from.

[soft zealot]

drinking water

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[elder garden]

i got to

1/4 (lnx+2)^4 + c

but im not sure how i could turn it into the form required?

i tried doing some index rules to turn the brackets into squared twice

[formal trail]

how did u end up in the numerator?

[elder garden]

hang on sorry wdym?

OHHH THANKS

[dense crystal]

your u³ is in the denominator

[elder garden]

yea the u^3 should be under the 1/

[dense crystal]

its should be u^(-3)

[elder garden]

thank you soo much :)

yep. got it now. thank you!!

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[waxen jay]

Just wanna make sure I understand finding the slope of a tangent line:

Basically find the cloest point to the point of the tangent, use slope formula on that point and the tangent point to find the approximation of the tangent slope? Right?

[formal trail]

if you want to approximate it, then yes. if you want an exact answer you take the limit of the slope formula as the distance between points approaches 0

[waxen jay]

ok thank you

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[tough hound]

This is my homework problem. i am using the lumped capacitance method for part a since Bi < .2 i have solved part b and KNOW the steady state temperature. can i use the steady state temperature for my initial temperature (935.92K) even if it says that the thermo couple is at room temperature?

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[glass badger]

According to my teacher's note, the answer should be 11/15, but I keep getting 12/15 (=4/5). Please, help.

[tiny gorge]

you have to make up your mind whether duplicates are allowed

[candid mulch]

You are choosing two numbers from
{1,2,3,4,5,6}

[tiny gorge]

your "6 choose 2" formula assumes no duplicates
but you are listing (1,1) as a possible outcome

[glass badger]

Just a minute. I'll redo it.

I don't get it. When I use permutations instead of combinations, the answer is 13/30 (listing two (1,1)'s as possible outcomes). When I add +1 to the denominator (adding the element (1,1) to the sample space) while using the combination, the answer is 12/16 = 3/4.

Could you please provide me the solution so that I can analyse?

[tiny gorge]

there's only (1,1), you can't list two of them

and if you use permutations there's more than that

(1,2) and (2,1) are distinct, for example

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[lunar skiff]

yooo, i was about doing it

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[ember karma]

working on this problem and already solved it but in a completely different way than was showed. i don't understand why they put 'a' on the numerators (a/4 and a/3)? that doesn't match up with 1/4 and 1/3 right?

[errant plume]

5x-4=6x+12

[magic sparrow]

!occupied

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[elder garden]

"a" is used because the number of apples is unknown at the start

a quarter of "a" is a*1/4 = a/4

[ember karma]

ohhhhhh i see now

thank you!

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[timid geyser]

.reopen

[misty sequoia]

I don't really understand how to find the antiderivative of 10^(-x)

[steel crescent]

do you know the antiderivative of a^x?

[misty sequoia]

I suppose it's 1/(10^x) => ln(10)/(10^x)

[misty sequoia]

(a^x)/ln(a)

[rough crystal]

What is the derivative of a^x

[steel crescent]

so try using substitution here

u = -x

[misty sequoia]

Can it be resolved without a substitution?

[rough crystal]

I'll take a look.

[steel crescent]

yeah, if you have the basic idea

[misty sequoia]

https://socratic.org/questions/how-do-you-evaluate-the-integral-int10-x-dx

Like that is the source of the solution

[misty sequoia]

Thank you

[steel crescent]

but you still need substitution for it

$\int 10^{-x} dx$

[soft zealot]

Wither

[steel crescent]

u = -x

du = -dx

$-\int 10^{u} du$

[soft zealot]

Wither

[steel crescent]

there you go

[rough crystal]

Still have to solve

the other part, but for question about using u-substitution to undo the composition absolutely.

[rough crystal]

If you want to use the trial and error method sure, but then you have to account for unwinding the chain rule when you find the antiderivative which makes you prone to making more errors

You should solve the other part alone

[steel crescent]

you can also do $\frac{x^2 + 1 + 1}{x^2 + 1}$ for the other part

[rough crystal]

unless you need help on that?

[soft zealot]

Wither

[misty sequoia]

So, it's better to solve the problems like that with substitution methon

[rough crystal]

100%

[misty sequoia]

I'd like want to ask

[rough crystal]

Because you never want to manually account for the trial and error method.

[misty sequoia]

Like if we have something like that:

[rough crystal]

Mhmmm.

[misty sequoia]

Can I split it into several integrals, then write the first term((x^2 + 2x + 1)^3) with substitution and the others without?

[rough crystal]

U-substitution would fail if you tried doing that.

Because when you rewrite an integral in terms of du all the x terms must cancel out and it should only be written in u.

We would notice if we tried applying u-substitution there the x terms wouldn't cancel out and so we would have a u and x term which devoids our whole purpose.

The point of u-substitution is that it undoes a composition.

We can try to make sure though.

[misty sequoia]

Give me a minute

[rough crystal]

You can apply u-substitution multiple times sometimes as there could be multiple compositions so do take that into account.

[misty sequoia]

So, in order to do substitution with multiple composition I need to substitute all the terms?

[rough crystal]

No, no. I'm saying there are cases where you may have to apply u-substition multiple times, but that doesn't mean you'd substitute all the terms.

[misty sequoia]

Could you give an example?

[rough crystal]

Don't know if I can think of an example of the top of my hat, but I'll try.

[misty sequoia]

Also, what I meant is:

[misty sequoia]

And then I substitute the t back to (x + 1)

[rough crystal]

Yep if we look at (x+1)^6 closely we will see that there is no extra composition, so thats why dx = dt

its cause even if we applied the chain rule because this function is wrt to x, the chain rule would just evaluate to 1 multiplied by the other thing.

((x+1)^7) /7 + ln(IxI) - (x^(-2)/-2) + C

Always verify your answer

[misty sequoia]

Yes, that's exactly what I meant

[rough crystal]

7(x+1)^6 / 7 + 1/x + ( -2x^-3 / -2 ) + 0

We've attained the initial funciton (i.e the integrand) therefore our antiderivative is correct.

[misty sequoia]

Like if we have a composition of multiple terms that can't be reduced in a substitution, we can split the integral into a composition of multiple integrals, then make substitution with the one that we need and then substitute the one back

[rough crystal]

No.

Because that doesn't have anything to do regarding the chain rule.

[misty sequoia]

That's what I did right here

[rough crystal]

That is different there isn't multiple compositions.

When I refer to compositions its in regards to the chain rule I'll show an example.

differentiate this tan(sin(x^2))

and tell me how many times you had to apply the chain rule

The whole purpose of u-substitution is that it unwinds the chain rule to rewrite the integrals in terms of its composition thereby making it easier to find the antiderivative.

[misty sequoia]

2 times

[rough crystal]

Write, so essentially what I'm saying is there will be scenarios where you will have to apply u-substitution multiple times to an integral. What I mean by that is that you'll apply it once, and then once its rewritten in terms of du, you will have to apply it again.

I'll show an example.

[misty sequoia]

Hold on, I don't understand how to solve the integral

[rough crystal]

That was just an example of a function that needed to be differentiated multiple times.

Some functions even with u-substitution it will not unwind the chain rule.

That function is not integrable through u-substitution.

Some integrals no matter what are just unsolvable in nature.

[misty sequoia]

It looks like I need to substitute x^2 with t, then sin(t) with u, and so on

[rough crystal]

Don't even worry about trying to integrate that, it won't work. I was just showing you an example that involved the chain rule multiple times.

Now there will be integrals like that involving multiple chain rules (compositions) that are actually solvable, you'd just need to apply u-substitution multiple times.

Anyways, I'm a bit busy, so I gotta head out unless you have any last minute questions?

Mk on that note Ima head out now. Best of luck to you!

[misty sequoia]

I didn't know, thank you

[misty sequoia]

I thought everything could be solved in a progper way

[misty sequoia]

No, I don't thank you a lot for explaining all of that

[misty sequoia]

God bless you

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[ivory vessel]

Given an 8x8 chess board, each board square side is 1 unit. A chessboard square can be a square or a rectangle, 2 squares is a rectangle, and so on. Randomly pick 1 rectangle on the chessboard. Calculate the probability for the chosen rectangle to be a square with a side greater than 4

[ivory vessel]

ok this is kinda an interesting problem but i dont know where to start, i cant even find the amount of rectangles on the board

[barren hound]

that is indeed where you should start

maybe start by counting the number of rectangles on a 2x2 or 3x3 board

[trail mango]

rectangles are uniquely determined by where 2 of their opposite corners are

just find how many ways you can choose those

[ivory vessel]

on a 2x2 there are 9 rectangles

on a 3x3 theres 36

i wonder if theres a definitive formula for this?

[barren hound]

there probably is, but there's also the idea of just having a systematic approach to counting them

[ivory vessel]

yup that should be my next step

i just count the amount of 1x1, 1x2, 2x1, 3x1 and so on

wait, for 3x3 i forgot to count the 3x1s

my prediction is: 4x4 should have 144

[barren hound]

interesting prediction

[ivory vessel]

i notice that for 2x2, there is 3^2 rectangle

for 3x3, there is 6^2 rectangle

in 2x2 there are 3 possible types of rectangle: 1x1, 1x2 and 2x2

for 3x3 its 1x1, 1x2, 1x3, 2x2, 2x3, 3x3

so for 4x4 i predict there will be 12 types

so 12^2?

[barren hound]

that's easy enough to verify

the number of types

[ivory vessel]

i think it should follow the sequence 3, 6, 12, 24?

[barren hound]

a reasonable guess

although i do wonder what happens for 1x1

[ivory vessel]

ah, 1

so 1, 3, 6, ...

[trail mango]

||there are 81 places for corners to be. if you pick one, the amount of rectangles who have that corner is 64. counting up all the rectangles like this counts each rectangle 4 times. so the answer is 81*64/4 me thinks||

[ivory vessel]

i assume behind the spoiler is the answer?

yeah ill give some time to think before opening it

[trail mango]

i hope so

[ivory vessel]

could i get a hint on what this sequence is? im stumped on this

[barren hound]

you should count out the rectangle options for 4x4

[ivory vessel]

for 4x4 its 1x1, 1x2, 1x3, 1x4, 2x2, 2x3, 2x4, 3x3, 3x4, 4x4

10

wait

oH

so given NxN grid, there should be 1+2+...+N amount of rectangle options

so the formula for the amount of rectangle on an NxN grid is

$(\sum_{k=1}^N)^2$

uh

[soft zealot]

FunguDeu

[barren hound]

$\left(\sum_{k=1}^N k\right)^2$

[soft zealot]

lemonsaurus enthusiast

[ivory vessel]

yes

so we got that for 8x8 it should output

,calc(1+2+3+4+5+6+7+8)^2

[soft zealot]

Result:

1296

[barren hound]

one thing you're missing is why this is true

[ivory vessel]

didnt we already do that?

i mean, you could be right, i didnt quite prove it, i just use induction

[barren hound]

you found a pattern, which is good! very helpful

but what's the connection between number of rectangle types and the total number of rectangles? why squared?

[ivory vessel]

ok good question actually

give me some time to think

yeah i cant quite find a connection

i believecit has something to do with squares or combinations?

perhaps, something to do with areas?

like, for NxN grid, the amount of squares is N^2

basically areas

wait i need a whiteboard for this

[ivory vessel]

could i have a hint on the connection?

[barren hound]

hmm. let's look at the 3x3 case

it's (3+2+1)² right?

[ivory vessel]

yeah

[barren hound]

think about what terms are being multiplied into each other

and consider what @trail mango said earlier about corners

also maybe think about "1d rectangles" in a 1d setting

[ivory vessel]

as points?

[barren hound]

this connection took me a while to see

as segments

[ivory vessel]

$3^2+2^2+1^2+212+223+213$

[soft zealot]

FunguDeu

[ivory vessel]

how do you know each corner can have NxN amount of rectangles?

[ivory vessel]

hmm i see your point, for a N length segments there would be N+1 points, which totals to N possible segments from any of that point

and this transfers to 2d too

N=3 segment would have 1+2+3 possible segments, while N=4 would have 1+2+3+4

so when we move up one dimension, we square this

[ivory vessel]

hence, this formula

offtopic but i speculate that the amount of possible upright rectangular prisms in an N^3 cube would have the same exact formula, but cubed?

yeah i can see the connection now

[ivory vessel]

now to get back to the problem, it asks for square, not rectangle

so i reckon i should also derive a formula, but with squares instead

[trail mango]

pick a corner. there are 8 ways to choose a “vertical size” and 8 ways to choose a “horizontal size”

basically just…

well maybe i should draw a picture

just got out of the shower need a min

[ivory vessel]

for square this is pretty simple

$\sum_{k=1}^N k^2$

[soft zealot]

FunguDeu

[ivory vessel]

is this the final answer?

@trail mango @barren hound

the denominator is squared i forgot to do so

[ivory vessel]

id get 1/9, which seems incorrect

oh wait im dumb

[ivory vessel]

29/216

[trail mango]

here is a different explanation than i originally gave that may be easier to follow: say you look at rectangles that have green as a corner

[ivory vessel]

yes

[trail mango]

you just can't pick the red corners as the opposite corner

everything else is ok

[ivory vessel]

yes

[trail mango]

like this is an ok rectangle

there are 81 - 17 =64 choices (17 is the number of invalid choices, the green dot itself and the 16 red dots)

[ivory vessel]

ah i see, and no matter where the green dot is, it would still be 17

[trail mango]

[ivory vessel]

thats a smart way of approaching the problem holy ahit

[trail mango]

so now take each of the 81 corners and count the amount of rectangles that have that corner

there are 81*64 then

but we're counting each rectangle 4 times

for example, this one will get counted 4 times

[ivory vessel]

yes, since theres 4 corners

[trail mango]

once when we count the rectangles with 1 as a corner, again when we count rectangles with 2 as a corner

and so on

so 81*64/4 is the answer (to how many rectangles there are)

[ivory vessel]

right, thats an innovative way of counting

[ivory vessel]

regarding this, is my logic sound?

[trail mango]

i will need to read the channel idr what the original question even was

[ivory vessel]

i take the total amount of squares divided by the total amount of squares not greater than 4

[ivory vessel]

my math teacher said "this is an insane combinatorics problem yall suck so dont do it" (his exact words but in vietnamese)

[trail mango]

this ain't insane

even i can do it

[ivory vessel]

wtf even you, youre comparing yourself to a class whose average math score is 7/10

either way i digress, hopefully this should be the correct result right? since this one my teacher forgot to put the answer key

[trail mango]

this is supposed to be the number of squares with a side length greater 4?

[ivory vessel]

greater than 4 yes

[trail mango]

oops yea

i was thinking at least 5 in my head

[ivory vessel]

its the amount of squares possible, minus the amount of squares lower than 5

yes

[trail mango]

why?

[ivory vessel]

wdym why

[trail mango]

,calc 1 + 4 + 9 + 16

[soft zealot]

Result:

30

[trail mango]

this should be at least 64 because i can think of 64 squares that have side length less than 5 really easily

(the size 1 squares)

[trail mango]

so i do not believe this counts what you were trying to count

[ivory vessel]

oh

shit

yeah youre right

[ivory vessel]

it should be 5 to 8 instead

[trail mango]

sounds more plausible

now why that?

[ivory vessel]

i forgot its actually 8^2+7^2+6^2...

not 1^2+2^2+3^2

[ivory vessel]

the number of of 1x1 is 8^2, 2x2 is 7^2, the formula i derived actually starts counting from 8x8, not 1x1

hence the error

so i changed to 5 to 8, which would count from 4x4 down to 1x1

[trail mango]

fair enough

[ivory vessel]

does sigma bound work like integral bound?

[trail mango]

what does that mean

[ivory vessel]

like from a to b + from b to c = from a to c

[trail mango]

no

[ivory vessel]

oh ok

wait why

they have same function

[trail mango]

$\sum_{n=1}^{10} f(n) =\sum_{n=1}^{5} f(n) + \sum_{n=6}^{10} f(n)$ for example

[soft zealot]

slayla

[ivory vessel]

yes this waht i meant

[trail mango]

not what you said lol

[ivory vessel]

oh

ok yeah youre right

[barren hound]

tbh summation bounds should be half open

that would make the world a better place

[ivory vessel]

so this is our final result

which we get 5/216

which is right

not sure if right, its present in 1 of 4 options

but yeah if our logic works then this should be it

[trail mango]

i agree with this

[ivory vessel]

nice, thanks layla and hayley!

[trail mango]

hi ing

[dense coral]

hi

[trail mango]

or higher

[ivory vessel]

hing

[trail mango]

hiring

[dense coral]

hgr is probably fine

[dense coral]

lmao

[dense coral]

I wonder if I should use this emote instead of a box or QED when I finish my proof

seems like a good "proof is complete" emote

[trail mango]

that is a wonderful idea

i'm actually doing that

[ivory vessel]

mfw each question in the exercise sheet takes like 2 hours to solve

this is problem 23/67

gonna open a new help channel, i have doubt for problem 24/67

again thanks layla and hayley, very pog explanations

.close

[final saddle]

Channel closed

Closed by @ivory vessel

Use .reopen if this was a mistake.

[trail mango]

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[novel nexus]

need help with this sheet of paper

[final tangle]

,rotate

[final tangle]

!status

[final saddle]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[novel nexus]

1

[final tangle]

write the equation representing
sum of x and y is 9

[novel nexus]

okay

[sonic crystal]

Le epic meme

[final saddle]

@novel nexus Has your question been resolved?

[novel nexus]

!close

[final saddle]

@novel nexus Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[final saddle]

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.

[verbal steppe]

Sketch the surface S after first drawing some of the level curves (section with plane) z = C_i :

[verbal steppe]

If i replace z with C_i i.e -1 it would give me

$-1 = 3|x|+6|y|$

[soft zealot]

Merineth

[verbal steppe]

Isn't this unsolvable ?

or is it solved by doing:

$|y| = \frac{-3|x|-1}{6}$

[soft zealot]

Merineth

[verbal steppe]

Which isn't possible either? since the RHS will always bne nagative and the LHS will always be positive

[final saddle]

@verbal steppe Has your question been resolved?

[verbal steppe]

<@&286206848099549185>

[final saddle]

@verbal steppe Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

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Available help channel!

Send your question here to claim the channel.

Remember:
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• 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.

[south torrent]

test

[south torrent]

.close

[final saddle]

Channel closed

Closed by @south torrent

Use .reopen if this was a mistake.

[south torrent]

.reopen

[final saddle]

✅

[south torrent]

i need help working out this problem idk if its possible so give it a try to help me out!

<@&286206848099549185>

[fathom walrus]

bro thought we wouldnt recognize the quartic zero formula

[south torrent]

Let’s see if u can I need help for this

[hardy yew]

😂 😂

what help do you need?

its a formula, you substitute the values and you get the roots

you've like actually copy pasted the image from wikipedia I think

[south torrent]

Ok so maybe that’s how u do it I guess

Thx

[hardy yew]

what?

ok

[south torrent]

.close

[final saddle]

Channel closed

Closed by @south torrent

Use .reopen if this was a mistake.

[final saddle]

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

[obtuse copper]

Hi! I have trouble understanding how to solve a differential equations, I assume it's a linear first order differential equation?
It goes like this:
(x^2 + 1)y' + 2xy = 4x^2 with initial conditions y(3) = 4,
I know how to solve for the homogenous solution which is a part of the whole solution, but I have no idea how to solve the particular solution? I didn't find anything usefull in my book for this form of differential

[obtuse copper]

I mean besides the fact that the solution is a sum of ci *yi(x), where i is equal to the 'degree' of the differential

[oblique turret]

Write $y'+\dfrac{2x}{x^2+1}y=\dfrac{4x^2}{x^2+1}$

Then find $F(x)$ such that $F'(x)=\dfrac{2x}{x^2+1}$

Multiply by $e^{F(x)}$

$e^{F(x)}y'+e^{F(x)}\dfrac{2x}{x^2+1}y=e^{F(x)}\dfrac{4x^2}{x^2+1}$

You can check that the sum at the left equals $\dfrac{d}{dx}e^{F(x)}y$

Substitute that and solve

Wait there is one thing wrong

I will change it

[soft zealot]

d

[obtuse copper]

ohhh alright, and then you can just do, ummm what is it called

[oblique turret]

Integrate both sides

[obtuse copper]

find the integrating factor

[oblique turret]

Yes

[obtuse copper]

ahh alright

and for higher degree differentials of this form?

or are these way harder to solve?

[oblique turret]

^

There is not a general method, unless the coefficients that multiply the derivatives are constant

In that case it is very easy

[obtuse copper]

then it's just the root method right?

[oblique turret]

Correct

[obtuse copper]

alright, I assume the hardest they can ask on a undergrad exam is to solve a system of differentials with constant coefficients?

or systems of this form

[obtuse copper]

.

[oblique turret]

It depends on what you define as hard

I believe the hardest and longest is to solve general linear systems with constant coefficients

Because you have to find the matrix exponential

[obtuse copper]

yeah I still have to reread through this material, I was so confused during the lecture

[final saddle]

@obtuse copper Has your question been resolved?

Channel closed

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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.

[steep forum]

can anyone solve this kind of question for me ? I only can go upto expanding the first two brackets

[gentle barn]

Alright, how did you do the first two brackets?

[steep forum]

I just expanded (5-x)(2x+3) = 10x+15-2x^2-3x

[steep forum]

and I tried to continue doing the same with the third one which didn't work

[gentle barn]

Let’s take this as example. (5-x)(2x+3)=5(2x+3)-x(2x+3)=-2x^2+7x+15

[steep forum]

Did I do the expanding wrong then ?

[gentle barn]

No, you probably did it in another way. This way is more visualized.

[steep forum]

so are my digits correct ?

(5-x)(2x+3) = 10x+15-2x^2-3x

and if it is how do I continue the question

[storm storm]

simplify the expression

[steep forum]

I don't know how to

[storm storm]

Combine the like terms of 10x+15-2x^2-3x

10x - 3x are like terms

[steep forum]

I don't have any problem with the first two brackets but I don't know how to simplify the third bracket as the hook thing doesn't work on it

[storm storm]

in order to continue the problem, you have to combine the like terms of 10x+15-2x^2-3x

[steep forum]

of the first two brackets right ?

I did

[steep forum]

but I don't know about the third bracket

[jovial mica]

@steep forum

[storm storm]

so after combining 10x+15-2x^2-3x

what did u get

[steep forum]

7x+15-2x^2

[steep forum]

thank you I'll look at it

[storm storm]

correct. next, we'll multiply this result by ( (x + 4) ):
[
(-2x^2 + 7x + 15)(x + 4)
]

[soft zealot]

Flamey

[storm storm]

that’s the next step

[steep forum]

I get it now

thank you guys

[storm storm]

this is the next step , that other image isn’t correct

[steep forum]

I got it now thank you my problem was that after expanding i didn't do the same with (x+4)

[storm storm]

and after u multiply this

combine like terms