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[midnight plank]

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[raw vector]

From the numbers {1;2;3;4}, how many natural numbers with 6 digits can be made so that 1 is not next to 2?

[raw vector]

What i got so far:

n(Ω)=4⁶
Treat "12" or "21" as a digit.
•There are 5.4⁴ numbers which have 1 "12" or "21"

Then idk about if the number has 2 or 3 of those "12" or "21"

Also idk if my approach is the easiest one

[shy leaf]

my guess would be to go total - ones where 1 and 2 are next to each other

same as here but idk if the numbers work out

total is 4^6

nah theres repeats its annoying this way

[white gate]

i dont think approaching for the opposite is needed

[raw vector]

Actually, the cases where the number has 2 or 3 of those "12" or "21" are included in those 5.4⁵

[shy leaf]

i was gonna say

5x4^5 is larger than 4^6

[white gate]

the easiest situation is when"

• there's no 1 and 2
• there's only 1
• there's only 2

[raw vector]

Oh yeah...

[shy leaf]

you right target

[raw vector]

I messed up there

[shy leaf]

so theres 2^6 with just 3s and 4s

theres 3^6 x 2 for 3s and 4s with the inclusion of 1 or 2

[white gate]

wait nvm this is more troublesome than i think

lol

[shy leaf]

no you are totally right

now we need the cases where there are both 1 and 2

[white gate]

that sounds reasonable until you involve both 1 and 2 in

[shy leaf]

but arent together

this means you can place the 1s and 2s in spots 1/3-6, 2/4-6, 3/5-6, 4-6

which is 10 situations

[raw vector]

Because there can be multiple 1's and 2's

[shy leaf]

multiply by 2 to allocate for 1/2 and 2/1

[white gate]

yeah.

[shy leaf]

the multiplicity fks me

yeah

maybe we do cases with the opposite

if we then look at cases with 5 spaces

two being condensed into 12 or 21

we then need to dodge the cases like a 121 or 212

[mellow sand]

this

as soon as there is a 12 or 21 the number is bad

[shy leaf]

im aware

[mellow sand]

so we just use complementary counting with that, no?

[shy leaf]

im looking at the total - cases where there is 21 or 12

which is why im looking like that

[white gate]

i dont even know how opposite would work because if you combine digits then there's lots of combinations to do

[mellow sand]

2! • 5 • 4^4

no?

[white gate]

12
121
1212
12121
121212

[mellow sand]

that’s the number of 6 digit numbers with 1 and 2 next to each other

[white gate]

yea

[mellow sand]

then there is 4^6 in total

unless i’m mistaking something

[shy leaf]

you are

cases like 124124 or 124121

cases with repeats of doubles

[mellow sand]

this is counting that

i left all 4 digits for the remaining spaces

[white gate]

that is just counting the "12" case

[shy leaf]

then halve it to include 21 cases

[mellow sand]

2! counts the 21 case

oh wait

i see now

damn that is weird

[white gate]

ikr

[shy leaf]

this is strange

[mellow sand]

maybe constructive counting?

eh still a lot of case work

[raw vector]

Yeah...

[white gate]

hm, im thinking of this way

instead of counting how many numbers there are, why dont we consider the probabilities of individual digits instead

so for example:
let the number be abcdeg
if a = 1, 2, then do 3 works:

• check if b = 1
• check if b = 2
• check if b = 3, 4

its still lots of work but i think its easier to manage...?

if a = 3, 4, skip the checking and randomly choose b

then the cycle repeats until digit g

[raw vector]

Still too much work, i think we just try to count the 6 digit numbers where 1 and 2 are adjacent

[white gate]

... which is still lots of work :L

[raw vector]

Treat "12" or "21" as the digit x

There are 3 cases

1 x, 2 x's, or 3 x's

[white gate]

the number 211111 goes brr brr

[raw vector]

Fair enough

There is just 1 case

There is x

5 places for x and 4 slots for other numbers

So 5.6⁴

I made the mistake of counting ut as 5.4⁶

So i think the answer is 4⁶-5.6⁴

[white gate]

why 6^4

[raw vector]

Which is negative

Fuck

Oh

4⁴

I got confused

So 4⁶-5.4⁴ = 2816

[white gate]

that's still not enough because 211111

[raw vector]

That's included

[white gate]

no you didnt

[raw vector]

I included the digits 1 and 2 in the 4⁴ part

[white gate]

wait

[raw vector]

Otherwise it would have been 2⁴

So the answer is 4⁶-5.4⁴=2816

[white gate]

you did, but then you overcount the case where it goes 212222

[raw vector]

How?

I didn't care if there were multiple x's intertwined or whatever, i only had 2 cases. There exists no x's or there exists x(s)

[white gate]

isn't the "4^4" just throwing 1 out of 4 digits in 1 out of 4 spots?

which means you can accidentally create 212 without knowing

[raw vector]

But that doesn't matter

No?

I was counting the cases where there existed x

Also self correction, i forgot to multiply by 2! because it can be "12" or "21"

So 4⁶-5.4⁴.2

1536

Eh...

I'm definitely not sure about it

[white gate]

huh, it really doesnt

but what about case like 212133

which is also a case from 4^4

[mellow sand]

the problem is say we place xx12xx
there is a chance in those x spots we get x212xx

this will be found in another case tho

so we are over counting

[raw vector]

I definitely overcounted the cases were there were "12" or "21", i gotta add by something

[white gate]

honestly i dont think there are even more than 1000 numbers that satisfies this problem

ugh this is such a hassle

[feral sedge]

count the number of ways to end in each digit

[raw vector]

I'm a 9th grader and an 8th grader asked this, i feel behind damn

[feral sedge]

i.e., starting from 1 digit

ways to end in 1 is 1 way
ways to end in 2 is 1 way
ways to end in 3 is 1 way
ways to end in 4 is 1 way

& continue until you get to 6 digits

[raw vector]

What

How does that have anything to do with the problem?

Also that's just an absurd amount of work for such a simple problem

[feral sedge]

this is an absurd amount of work?

i solved this problem in 2 minutes by hand

[raw vector]

But mainly i don't see the correlation between that and the fact that 1 and 2 are not adjacent

[feral sedge]

from this

going to 2 digits

[raw vector]

And how do you know how many cases are there where 1 and 2 are not adjacent?

[feral sedge]

say you want to compute the number of ways to end in 1, when there are 2 digits

[mellow sand]

can you show your work?

[feral sedge]

from previous, this would be: you add the number of ways to end in 1 + the number of ways to end in 3 + the number of ways to end in 4

[mellow sand]

i’m curious to see what you did

[feral sedge]

do you see?

[raw vector]

And what does that have to do with the problem?

It doesn't have anything about what digit the number ends in

[feral sedge]

it's a way of counting

[raw vector]

n(Ω)=4⁶, now we need to subtract the number of cases where 1 and 2 are adjacent

How do count the latter part?

[feral sedge]

that way is prone to error

ok well you do not believe me and i do not have much time

so i will just post

1 3 11 39 139 495
1 3 11 39 139 495
1 4 14 50 178 634
1 4 14 50 178 634

counting via last digit

the answer is the sum of the 6th column, i.e. 495 + 495 + 634 + 634

which is 2258

[mellow sand]

6?

i really don’t see what you’re doing

[raw vector]

Neither do i

It's not clear

It's not just that the number doesn't end in "12" or "21", it can't be adjacent

You only adressed the cases where it ended in "12" or "21" i think

When 124444 could be a case

Also i don't get those numbers

[feral sedge]

these 2 messages explain the process

if a (n - 1)-digit number ends in 1, then you can only append a 1, or a 3, or a 4, without violating the problem constraint

to get an n-digit number

the idea is you do this for the 4 digits and add them all

[raw vector]

I get the process now, but doesn't that branch out quite a bit?

[feral sedge]

it looks like a lot of work but you’re really just mindlessly adding numbers, so it’s unlikely you make an error

[feral sedge]

no

as you can see in this table there are only 12 numbers

and many of them are less than 3 digits and very small to add and compute

I guarantee you this method is much, much simpler than trying any casework solution to count this

[raw vector]

From 1,3,4 how do you get the 4rd collumn?

[feral sedge]

wdym

[raw vector]

The 39

[feral sedge]

that's 11 + 14 + 14

more formally you write this out using a recurrence

but this intuitive explanation works for a problem like this

[feral sedge]

to interpret,
11 is the number of 3-digit numbers that end in 1
14 is the number of 3-digit numbers that end in 3
14 is the number of 3-digit numbers that end in 4

[raw vector]

So like a "Pascal-triangle-like" process?

[feral sedge]

uh, sure

[raw vector]

Does this have a formal name?

[feral sedge]

you can just call it a recurrence

[raw vector]

Also how do i formally write this out?

[feral sedge]

like

a_1 = 1, b_1 = 1, c_1 = 1, d_1 = 1

define the recurrences
a_n = a_{n - 1} + c_{n - 1} + d_{n - 1}
…etc

you can notice that the values of a/b and c/d will always coincide because 1/2 and 3/4 are essentially the same as far as this problem is concerned

[white gate]

aha, i knew it, it's all about going from left to right/right to left

[feral sedge]

so you can do this more simply as

a_1 = 1, b_1 = 1

define the recurrence
a_n = a_{n - 1} + 2b_{n - 1}
b_n = 2a_{n - 1} + 2b_{n - 1}

and your goal will be to compute 2a_6 + 2b_6

[raw vector]

The question was for 8th graders...

I'm in 9th grade and still haven't learnt that yet

[feral sedge]

that’s why i preferred this intuitive explanation

[feral sedge]

this is clearly a contest style math question, you will never “learn” it in a school setting

you only study this on your own, or have a tutor, or smth, etc

[raw vector]

Yes, it's a contest question, but the solution can only use the knowledge taught in 8th grade

I'm assuming the intended solution is something similar to this though

[feral sedge]

this is essentially adding numbers but in a clever way

[raw vector]

What's a and b? a for 1 or 2 and b for 3 or 4 or sth?

[feral sedge]

yes

it is sort of an ad-hoc construction that works based on the problem

[raw vector]

I did study on my own and learn about recurrence sequence but never thought to use multiple sequences like this

I don't get the 2b_{n-1} part though

[feral sedge]

.

wait

2b_{n - 1} indicates that you add b_{n - 1} numbers for the (n - 1)-digit numbers that end in 3, and another b_{n - 1} numbers for the (n - 1)-digit numbers that end in 4

b_n = 2a_{n - 1} + 2b_{n - 1} is the number of n-digit numbers that end in 3 (it is also the number of n-digit numbers that end in 4)

[feral sedge]

keep in mind i simplified it from this, so it might make more sense to view this system first before looking at the other

this interpretation is more natural than having only a and b (but you wouldn’t write this out fully on a test)

[feral sedge]

it is essentially what’s going on here, you will notice the 1st / 2nd rows coincide (and the 3rd / 4th rows)

extra note: this style of recursion is common in high school level combinatorics questions https://artofproblemsolving.com/wiki/index.php/Recursion

where there are many problems that doing them without this type of recursion will be nearly impossible

but for the problem you have it is sort of possible, since it is possible to use casework to break up the 4^6 possibilities, it is just maybe slow

for example here

i don't think anyone wants to do this problem without recursion

again the approach is very similar, you count by listing the last digit and setting up a recurrence relation (on an exam, you often do not need to write this out explicitly, once you have the right idea, you are free to add the numbers etc)

[raw vector]

How do i find the recursion relation?

I see that it's gotta be 01...10 but that's about it

[feral sedge]

a hint is

[raw vector]

Do i just set up the recursion for the middle 15 numbers?

[feral sedge]

the problem you had is simple in the sense that it you only needed to consider the previous last digit to make the recurrence

sometimes you will have to consider the previous last two, or last three, etc digits

these would appear as terms like a_{n - 2} or a_{n - 3} (or something similar) in a recurrence you set up

[raw vector]

Doesn't the "no 3 consecutive 1's" mean i gotta consider the last 2?

(the one you gave)

[raw vector]

This is the key thing i'm missing from the method

[feral sedge]

you can deduce a bit more actually

the key point is every sequence has to start or end with zero

to get yourself to a point where you can effectively set it up, the thing you're recurring on has to conform with the problem

[raw vector]

Uhmm...

So

Starting from the third digit

u_{1}=2

u_{n} = u{n-1} + 2u{...} - ...u{n-2}

Idk

Actually

If the digit is 0, the next is 1

So that's u{n-2}?

And if it's 11 then it's 0, so that's u{n- something}

And if it's 01 then there's 2 cases so that's 2 u{n-something}

@feral sedge please help here

[feral sedge]

you might not want to consider like last 2 digits, or roughly that perspective

try to force the last digit to be 0

[raw vector]

If i try to force it to 0, the sequence ends in 110, but there's also the ...010 case, wdym "force it"?

Am i trying to count the number of valid sequences length n or am i trying to count the possible cases for the first n digits in the 19-digit sequence?

Since earlier you mentioned you didn't have much time so i guess i'll find that out on my own

Thank you nonetheless

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[regal barn]

The value of a for which the system of equations x + y + z = 1 x + 2y + 4z = lambda x + 4y + 10z = lambda ^ 2 is consistent, are given :
(a) 1,2
(c) -1,3
(b) 1,-2

[regal barn]

so i got two values of lambada

so at these values the equations will be consistent am I right?

The value of ( a ) for which the system of equations ( x + y + z = 1 ), ( x + 2y + 4z = \lambda ), ( x + 4y + 10z = \lambda^2 ) is consistent, are given :
(a) 1,2
(c) -1,3
(b) 1,-2

[grand pond]

Tom

[lethal steeple]

Yes you are right.

[coarse ocean]

Yeah you're right dude

[midnight plank]

@regal barn Has your question been resolved?

[regal barn]

thanks

[midnight plank]

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[regal barn]

i used cos angle formula

[midnight plank]

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[regal barn]

.reopen

[midnight plank]

✅

[tough shale]

What does it mean by B has solution?

@regal barn

[pale vapor]

yeah correct

[regal barn]

i don't know

actually

[slender walrus]

its asking for how many possible values B could be

cos rule also not ideal here

[tough shale]

If it means c, then it has 2 solutions

[regal barn]

hn

[regal barn]

what should I use then?

[slender walrus]

sine rule

[tough shale]

Ohh

Ok

[tough shale]

Is the answer 1?

1 solution

[slender walrus]

no

[oak barn]

What was sine rule again

[regal barn]

sinB=8/14

=4/7

[tough shale]

Is it area = ab/2 sin(C)?

Sine rule

I kinda forgot

Oh that

Ok

[slender walrus]

no, that's not sine rule, thats trig formula for area of a triangle

[tough shale]

Nvm, I remember now

Yeah, thx

[regal barn]

...

[slender walrus]

continue your work

[regal barn]

no idea what should I do with angle sinB

[slender walrus]

sinB isn't the angle

B is the angle

use inverse trig (and supplementary property of sine) to determine the solutions to that equation for 0<B<180°

[regal barn]

B=arcsin(4/7)

what next?

,w arcsin(4/7) in degree

[grand pond]

Results provided by WolframAlpha

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

Requested by tom018546

[slender walrus]

(and supplementary property of sine)

[regal barn]

which one?

[slender walrus]

there's only one thing called the supplementary property of sine

[regal barn]

i don't think so

in my book there are many i meant i can convert sin into any trig

sin(pi-theta)?

[slender walrus]

yes

[regal barn]

so C will be around 26 degree

i am not getting what is our final goal

and what if wolfram alpha is not there

because we need to find arcsin(4/7)

[slender walrus]

so C will be around 26 degree
where's that coming from

[regal barn]

yes true

pi-(30+34)

[slender walrus]

isn't 26

[regal barn]

approx]

[slender walrus]

is nowhere near 26

[molten bay]

180°-34.85°-30°

[regal barn]

what do you mean?

nannii

[regal barn]

this is what i was saying

banned what?

[slender walrus]

who?

[regal barn]

why sending requests to me

i don't know

i see

[slender walrus]

I have never been banned here before, and why are you as a new member here making such accusations?

If you have an issue with me don't interrupt other people's channels and
dm @shadow scaffold

[slender walrus]

it appears you mixed up the value of pi and pi/2 (or 180° with 90°)

[regal barn]

sum of all angles of triangle is pi

so for C i substract

[slender walrus]

also why are you mixing radians with degrees in your work

what's the angle sum of a triangle in degrees?

[regal barn]

180-34.85-30

[slender walrus]

,calc 180-34.85-30

[grand pond]

Result:

115.15

[slender walrus]

that'll be C for the first value of B (and you have a valid triangle here)

[regal barn]

oh yes

[slender walrus]

now

(and supplementary property of sine)
sin(180° - B) = sin(B)
to get the other potential value for B

[midnight plank]

@regal barn Has your question been resolved?

[midnight plank]

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[earnest forge]

How do I solve this?

[earnest forge]

This can be rewritten as, $$\ln{(2)} \cdot \lim_{x \to \infty}\frac{ln{(x+1)}}{\ln(x)}$$

[grand pond]

waffle

[earnest forge]

What next?

[lethal steeple]

Use l'hospital

[earnest forge]

No

[lethal steeple]

?

[earnest forge]

Let's not use it.

[lethal steeple]

Okay then you can use ln(1+x) expansion

[fallow scarab]

try x = e^t

or factor an x out of x+1 and then use log rules

[tacit lance]

nvm

I see where I messed up

[earnest forge]

That would give lim x to infinity ln(x)

[lethal steeple]

What would?

[earnest forge]

furyolen's deleted idea

[lethal steeple]

Oh

[tacit lance]

ah yeah mb about that messed up my properties

[visual tiger]

ln(x+1)/ln(x) = 1 + ln(1 + 1/x)/ln(x). ln(1+1/x) -> 0, ln(x) -> infinity

[earnest forge]

This gives, $$\ln{(2)} \cdot \lim_{x \to \infty}\left(1+\frac{\ln{\left(1+\frac{1}{x}\right)}}{\ln{x}}\right)$$

[rustic tartan]

use ln(1/2) = ln(1) - ln(2)

[visual tiger]

almost

[grand pond]

waffle

[earnest forge]

(fixed)

[visual tiger]

yes

now 0/infinity is determinate

[earnest forge]

That's just ln(2)

Hmm

Thanks

[rustic tartan]

No

0/inf = 0

[visual tiger]

yes

That's why I said it's determinate

not indeterminate

[rustic tartan]

I thought u said in*

Nice!

!done

[midnight plank]

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[earnest forge]

.close

[midnight plank]

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[lethal steeple]

L'hospital

[visual tiger]

It's L'Hôpital

squirms in french

[midnight plank]

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[tall lily]

am i allowed to define functions like this

[tall lily]

does it make sense syntaxically

[wise gulch]

I guess

Is that not just a series then tho

@tall lily

[tall lily]

i suppose

is that important?

[wise gulch]

I mean I dont think theres an issue with defining it that way

doesn't seem wrong per se

[tall lily]

okidoki

.close

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[alpine blade]

i need some help on A) and D) on this question, id really appreciate someone who can fact check me since i cant find any help online!!

[alpine blade]

here is the plain text
for a) i did poissonpdf(0.4,0) > 0.706

and for d) im still pretty confused on but i got 37 weeks which im not super confident about

hello???

<@&286206848099549185> s

okay i see

.close

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[tough shale]

If you didn't get it, you don't have to close it. You'll eventually get an answer

[alpine blade]

sorry i was just getting a little inpatient 😞

i just decided to try and figure it out myself lol

[tough shale]

I see

Fyi, the current oldest one is 3d ago, and when I joined, there was one which was over 1 week old

[alpine blade]

oh okay then i guess its just normal for it to take a while

when i first came my questions got answered pretty quickly so i just assumed that was the case

[tough shale]

If it's a harder question, we'll have to wait for someone who is good in that field of math

Just wait for a while, the all time expert Ann will soon get online

[alpine blade]

mk ill open up another channel then ^_^

[midnight plank]

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[tranquil jungle]

I do not know what is wrong

[tranquil jungle]

are my bounds incorrect? I also tried using 0 to pi and got the same result

I think xbar is 0 because x-axis symmetry

[midnight plank]

@tranquil jungle Has your question been resolved?

[tranquil jungle]

.close

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[brisk crescent]

.help

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[acoustic owl]

21 please

[acoustic owl]

I’m having trouble visualizing the process and don’t know where to start

I get the first integral is from x^2 to 1-z but after that I’m stuck

[midnight plank]

@acoustic owl Has your question been resolved?

[stoic spoke]

Because the order is dz dx, you have to visualize what this object looks like when projected onto the z-x plane

[midnight plank]

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[vernal igloo]

help me with this question

[vernal igloo]

can you help me with this question

Im stuck

[golden comet]

ok I can help

[vernal igloo]

thank

[golden comet]

now, three of them lied, only 1 told the truth

[vernal igloo]

yeah

[golden comet]

so if you say, you have two pirate give you the same number, they cannot both be telling the truth, therefore they lied, right?

now look at Tom

if Tom was telling the truth

he would have 10 gold coins

because there is a total of 30

but Pit also said there was 10 gold coins

this would imply that Tom is lying

Using this same logic, you should be able to find out who told the truth

[midnight plank]

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[vernal igloo]

hey

There are
10
more truth-tellers than liars in a room. Everyone in the room was asked, ‘Are you a truth-teller?’ and everyone gave an answer. A total of
20
people answered, ‘Yes.’

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[vernal igloo]

Yuan wants to fill each box with a different prime number less than
20
so that the value of
A
is an integer.

?+?+?+?+?+?+?
---------------- = A
?

What is the largest possible value of
A
?

[tawny apex]

to calculate the largest number of A, we should have the numerator the largest sum possible and the denominator to be the least value of a prime number.

[lyric charm]

can you list all the prime numbers less than 20, for a start?

[vernal igloo]

2,3,5,7,11,13,17,19

[tawny apex]

and how many prime numbers are there?

and how many spaces you have to fill?

[vernal igloo]

7 up 1 down

[tawny apex]

yes

and you have 8 prime numbers

[vernal igloo]

yep

[tawny apex]

now for the largest possible value of A, you should have the least value possible in denominator,right?

and the largest sum possible in the numerator

[pine thicket]

dont forget that you need to guarantee A is an integer

[tawny apex]

oh shit yeah

[vernal igloo]

ok

hey

@pine thicket

or

@tawny apex

[tawny apex]

hey

[vernal igloo]

i cant do this question

The letters p q r s t represent five consecutive positive integers, though not necessarily in that order. The sum of p and q is 69 and the sum of s and t is 72

What is the value of r
?

[tawny apex]

okay okay

first we'll average it out

so $(p+q)/2 = 69/2$

[grand pond]

Aria

[vernal igloo]

umm

no

[tawny apex]

why??

p+q=69

[vernal igloo]

yeah

[tawny apex]

if we divide it by 2 on both sides, it will give this

[vernal igloo]

um

p+q is 69 and

[tawny apex]

thus $(p+q)/2=34.5$

[grand pond]

Aria

[vernal igloo]

s+t is 72

[tawny apex]

yes

[vernal igloo]

any order

ANY ORDER

[tawny apex]

yeah

[vernal igloo]

33+36 is 69

[tawny apex]

ik

[vernal igloo]

s+t=35+37

thus

r is

[tawny apex]

and how would you write it formallly tho?

[vernal igloo]

34

it thinking problem

[tawny apex]

you have to show your work

[tawny apex]

oh

WHY DIDNT YOU TELL ME EARLIER

[vernal igloo]

ummm

It literally is a thinking problem

the question is literally shoutign itselfff

[tawny apex]

but btw if you ever get questions like these and you have to show the work, you first average the numbers

so that it would give us the range

and then solve it

[vernal igloo]

ok

Btw

that was test easiest question : D

Im fucked

The test too hard

[tawny apex]

ohh goodluck lol

[midnight plank]

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[calm quest]

can someone check my working out for part ii:

[thin tree]

,rccw

[blissful spear]

i is 360
ii is 1200

[lunar pivot]

group all the vowels together
PNTGN(EAO)

now you can think of (EAO) as one object

so now you essentially have to permute 6 objects where two of them are identical

6!/2!

we kept the order of vowels fixed but they could be permuted amongst themselves as well (EAO), (EOA), (AEO), ...

so multiply by 3!

so i) should be 6!3!/2! = 2160

[lunar pivot]

for ii) you'll have to make two cases i'll let you figure that out yourself

[calm quest]

can u have a look at my working

im not sure if its right

[calm quest]

how is ii) 1200?

[lunar pivot]

is that the working for the first or second part

[blissful spear]

5!/2!=60
60x4x5=1200

[lunar pivot]

that's not how you do it

what if you choose two Ns

[blissful spear]

Oh wait

[lunar pivot]

you'll have to consider two cases

[midnight plank]

@calm quest Has your question been resolved?

[calm quest]

yeah

its for the second part

i wanted help on that part

[lunar pivot]

oh alright

you got 2160 for the first part yea?

yeah ii) seems fine

[midnight plank]

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[remote bone]

I'm so lost on part B, where do I even start?

[last slate]

how old are u

[north robin]

Bro what does that have to do with anything 😭

[last slate]

so i wanna know why u do not solve this. i think bcz u are too young

[remote bone]

i don't see how that's relevant my man

[midnight plank]

@remote bone Has your question been resolved?

[normal prawn]

U need to write the equation of equilibrium by writing net torque = 0

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[novel flax]

for it to be reflexive, (1,1),(2,2) (3,3) must be there in all the relations possible

for it to be transitive, only (1,3) can come

so is the number of relations=1?

[lethal steeple]

I can still add any one symmetric like (2,1)

It won't make the whole thing symmetric

[novel flax]

oh

icic

.close

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[small geyser]

I need help finding the angle of v here. I will upload an image

[small geyser]

Well i have tried finding bc

Where i got 24 as the length

[cedar mason]

okay

[small geyser]

I need help getting further

[cedar mason]

you dont know trig yet-?

[small geyser]

I do, but im in the starting phrase

[cedar mason]

well

without trig

just write down some angles

write down sums and stuff

see if you shake something loose

[small geyser]

Is there a trigonmetric way to solve it? Like can i use the law of sines or cosines here?

[cedar mason]

now im wondering if im stupid

we have the hypotenuse

we have another side length

go figure

[small geyser]

Is it 11,54??

.close

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[vocal briar]

how do i integrate this?

without integration by parts

i tried u sub with secx

i got stuck though after getting $\frac{du}{tanx} = secx\cdot dx$

[grand pond]

gamer75431

[exotic basalt]

its a product

u have to use ibp

if its just sec^2 x its fine

[vocal briar]

we haven't learned this tho

are you sure its only possible by ibp

[exotic basalt]

what would u write the x in front as then

if i were to do it i would use u=tanx

cuz du=sec^2 x dx

hey its in there

and if u=tanx

then x=inverse tan u

so ur just integrating inverse tan u with respect to u

the only thing i can think of without integrating by parts

does that help or nah

[vocal briar]

uhh

we havne't learned that either lmao

so ig maybe she assigned this by mistake

[exotic basalt]

yeah

u have learned u sub right

just not integrating inverse tan

so the first half make sense to u but the second half dont?

[vocal briar]

yeah but we also dont deal with reciprocal trig functions

for my math course

[exotic basalt]

ur integrating without using recip trig?

[vocal briar]

would this be possible

[vocal briar]

no we dont learn that

its for the HL curriculumn

[exotic basalt]

but u know what they mean right?

[vocal briar]

ibp? no

i know what cot and stuff means tho

[exotic basalt]

no not ibp i meant sec csc cot

[exotic basalt]

ah ok

[vocal briar]

yeah yeah

we dont integrate reciprocals tho

[exotic basalt]

we'll need to use ibp to integrate cot sadly

BUT

usually we have a definite answer for integrating cotx

ln|sinx|

just remember that if u cant do ibp for now

but hey we have a 2 inside the function

it aint cotx it cot(2x)

so what would the answer be???? think of chain rule

[vocal briar]

wait ya i can do this one

cuz its cos(2x)/sin(2x)

[exotic basalt]

theoretically maybe u can use the double angle formulae for this

not sure tho

[vocal briar]

can we do this

and set u = sin(2x)

[exotic basalt]

yes then find du?

[vocal briar]

can't you do without double angle tho

yeah

ok let me try

[exotic basalt]

i have the answer key right here

tell me when ur done

[vocal briar]

oh yeah i got it

1/2 ln(sin(2x)) + C

[exotic basalt]

correct

[vocal briar]

ok nice

thakn you

[exotic basalt]

i believe u can just use that in ur exam

[vocal briar]

u sub?

ya

[exotic basalt]

u dont need to derive all that just memorize the result if u can

[vocal briar]

oh yeah

but like my test is on integration lol

prob have to show work

but yeah

thanks

.close

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[umbral sparrow]

am i right with B here?

[umbral sparrow]

is this correct or no

[limber grove]

yep

[umbral sparrow]

goat

[somber rune]

yeah that’s definitely irrelevant to asking for help

[midnight plank]

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[umbral sparrow]

okay so i have three parts to this. im on part A right now

[umbral sparrow]

i just want to know if im doing everything correctly so far

[twilit jetty]

looks good to me

nicely formatted btw

[umbral sparrow]

thanks man can u just quick check everything once i finish

[twilit jetty]

alr

[umbral sparrow]

@twilit jetty i just wanted to make sure. did i calculate that right? is it 0.088388?

i calculted the numerator first then denominator then just divided both numbers

[twilit jetty]

thats not the correct amount

avoid using the calculator, you can just simplify the number

[umbral sparrow]

when i typed the whole thing in desmos it didnt work

[twilit jetty]

well I just did

[umbral sparrow]

oh bet okay

[twilit jetty]

again avoid using the calculator

[remote quiver]

how do you even solve for that 😭

[twilit jetty]

lets do it step by step

oh wait youre not zeeko

[umbral sparrow]

okay 8squareroot 2 is 11.3137

[twilit jetty]

thats still using the calculator

what you need here is an exact answer

[twilit jetty]

do you want to do it step by step instead

[umbral sparrow]

wait so why is my teacher using a calculator

im following a similar vid with a similar question and she uses calc

[twilit jetty]

ideally you dont need a calculator to solve the problem

this is one of those problems

[umbral sparrow]

oh okay

[umbral sparrow]

so how would i start

[twilit jetty]

so first step, do you see anything here that should be easy to get rid of

[umbral sparrow]

yeah 2(8) = 16

[twilit jetty]

I was thinking about these but alr

[umbral sparrow]

okay yeah those two

so

[twilit jetty]

so your first step is this

second step is this

now what can you do next

[umbral sparrow]

so 64 + 8squareroot2)^2 - 64

[umbral sparrow]

16(8squareroot2) right

[twilit jetty]

Ill say this again,

we began here

first step is to multiply 2 * 8 = 16 on the denominator

second step is that 8^2 - 8^2 is just 0, because youre subtracting something with itself

you dont need to write 8^2 as 64

[umbral sparrow]

ohhhh ok

[twilit jetty]

given that you didnt see that, Ill be more lenient for what you can do next

[twilit jetty]

do you see anything that can simplify this further

[umbral sparrow]

so now divide both 8√ 2?

[twilit jetty]

thats good, thats what I wouldve done

the top has 8 sqrt 2 but squared

so when you divide both by 8 sqrt 2, you get

does that check out for you

[umbral sparrow]

yes okay so now divide those two

[twilit jetty]

going to assume you mean 8 / 16,

so as a result you get sqrt(2)/2

now you have that cos B = sqrt(2)/2

does that remind you of anything that B has to be

[umbral sparrow]

so that would be 0.7071?

[twilit jetty]

dont do that

again, no calculator

you should be able to recognize what B gets you sqrt(2)/2

[umbral sparrow]

🤦‍♂️

my abad

[umbral sparrow]

uhh wait

okay so its just sqrt(2)/2 then?

[twilit jetty]

cosine of what angle gives you sqrt(2)/2?

this is something you remember

[umbral sparrow]

B

[twilit jetty]

ok lets try something else

do you recognize the number 1/sqrt(2)?

[umbral sparrow]

why that?

[twilit jetty]

this is a yes or no question

[umbral sparrow]

no

[twilit jetty]

have you learned special right triangles before?

these are specific right triangles that you need to remember the sides and angles of

@umbral sparrow yes or no question, you can quickly respond to this one

[umbral sparrow]

i think so yeah

[umbral sparrow]

sorry was looking at soemthing

[twilit jetty]

can you tell me the special right triangles you remember

[umbral sparrow]

theres the 45-45-90 triangle and the 30-60-90 triangle

[twilit jetty]

thats good

now matching with that are special angles

these are angles of which you need to remember the sine and cosine of

have you learned those in class yet?

[umbral sparrow]

not really

but we went over it

[twilit jetty]

so nothing about "what is sin(30 degrees)" or "sin(45 degrees)"?

[umbral sparrow]

oh wait yeah we did that

[twilit jetty]

and what is cos(45 degrees)?

[umbral sparrow]

1/sqrt2 right

[twilit jetty]

yea

and 1/sqrt(2) is the same number as sqrt(2)/2, right?

[umbral sparrow]

ohh yeah

[twilit jetty]

so it looks like you did recognize 1/sqrt(2) after all

since cos(B) = sqrt(2)/2,

B is 45 degrees

you are looking at a 45-45-90 triangle the whole time

you can tell because the sides are 8, 8, 8 sqrt2

which match the 1, 1, sqrt2 that the 45-45-90 triangle normally has (but 8 times bigger)

[umbral sparrow]

okay i see it now

[twilit jetty]

this is why you didnt need a calculator for this specific question, the teacher gave you a special triangle for you to find the cosine of

for most questions, its almost guaranteed that you do need a calculator

I think its been proven that if the sides are all rational numbers, you will always need a calculator to find out the angles

only exception is an equilateral triangle where the angles are all 60 degrees

[umbral sparrow]

got it okay im just wondering why my teacher didn't just do it like this and resorted to calculator 😭

but i get it

[twilit jetty]

well its a good habit to avoid reducing numbers to decimals which you cant really work with

you use decimals when you need to see how large the number is

but other than that, math mostly leaves numbers as-is since its exact that way

[umbral sparrow]

my teachers doing a similar one and i was tryna follow it ig

[twilit jetty]

other than building that habit, you dont necessarily need to shun a calculator

[umbral sparrow]

interesting im understanding it now

[twilit jetty]

so youre still allowed to use one for these questions, its just harder to notice the patterns

1/sqrt(2) and sqrt(2)/2 are easier to recognize than 0.7071

[umbral sparrow]

true

yeah i see what u mean

[twilit jetty]

yea it wasnt really a kind move to make you go through the hoops like this, but here its easier to recognize the patterns

for question 9, did you notice its a 30-60-90 triangle?

(the image doesnt help very much, its just a general image of any triangle and isnt to scale or anything)

[umbral sparrow]

yea i kinda did now when i search it up it clearly says it

[twilit jetty]

thats good

usually the sides are 1, sqrt3, 2

so this is a 5x larger version of that

[umbral sparrow]

i think i need to draw a 45-45-90 triangle and label my lengths too kinda like how she did with hers

[twilit jetty]

just some triangle with ABC and abc, you dont need to draw it to scale

[twilit jetty]

the triangle here is supposed to be 30-60-90 but you can see its very loosely drawn to just "a triangle"

if you want to use that the triangle is special, you draw it that way

[umbral sparrow]

oh yeahh ur right okay

[twilit jetty]

since youre just using the calculator on all of them, you dont need the to-scale

[umbral sparrow]

is this good

[twilit jetty]

yea sure

wild seeing the 8 sqrt(2) not be the diagonal side but thats how it be

[umbral sparrow]

man i still have two more parts left and this literally took forever

[midnight plank]

@umbral sparrow Has your question been resolved?

[midnight plank]

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[lost sphinx]

for this function called max(), is the stuff we put into the () a range/set instead of 1 variable/term/expression?

[gaunt plume]

It depends on how it's used

Normally it's used as a shorthand for the reader rather than as a formal thing

But if you did want to define it properly then you would probably do it on sets

So a function that takes a set as its single argument (from a certain viewpoint, every function takes exactly one argument but let's not go there), and outputs the maximum value of that set

But then you get problems with it not necessarily being defined, hence why it's normally just a shorthand

[lost sphinx]

alr

so id write max{set} instead of max(var1,var2,var3)?

[gaunt plume]

Theoretically

Although even in that context, the set brackets are often just implied

Don't overthink it

[lost sphinx]

alr thx

.close

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[inland patio]

Working a true or false exercise. Consider the following claim: \

Any solution to a homogenous linear differential equation with constant coefficients is of the form $ae^{ct}$ or $at^ke^{ct}$, where $a$ and $c$ are complex numbers and $k$ is a positive integer. \

This is false, however, I wonder if it's true that any basis vector of the solution space is of this form?

[grand pond]

psie

[runic hamlet]

bases can look wild

so still false

e^t+te^t can still be a basis vector

[inland patio]

yeah...you're right. Ok, hmm.

.close

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[inland patio]

True or false:\

For any homogeneous linear differential equation with constant coefficients having auxilary polynomial $p(t)$, if $c_1,c_2,\ldots,c_k$ are the distinct zeros of $p(t)$, then ${e^{c_1t},e^{c_2t},\ldots,e^{c_kt}}$ is a basis for the solution space of the given differential equation. \

This is false, however, it is true if the differential equation is of order $k$, right?

[grand pond]

psie

[midnight plank]

@inland patio Has your question been resolved?

[spring wave]

you are right

[inland patio]

thank you

.close

[midnight plank]

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[midnight plank]

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[chilly cobalt]

ABCDEF is a hexagon with all equal sides. There exists a circle passing through A,B and is tangent to DE. Let BC insersect the circle at P, if CP = 9, what's the area of the hexagon?

[chilly cobalt]

i have no idea where to begin💔

[cedar mason]

is this diagram accurate

[chilly cobalt]

yea

(well it would be if the sides were actually equal)

[midnight plank]

@chilly cobalt Has your question been resolved?

[lethal path]

let M be the midpoint of ED and let O be the centre of the hexagon

and then yep you can get sides AO, OM, and AM in terms of the side length of the hexagon s also

which allows you to find the circumradius in terms of s also

I have a strong temptation to coord geo bash this since that's what I'm used to, but I will refrain from doing so here

[midnight plank]

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[lethal path]

.reopen

[midnight plank]

✅

[lethal path]

but you should get that BC = 9 * 8 = 72

[midnight plank]

@chilly cobalt Has your question been resolved?

[lethal path]

,w circumradius triangle with sides 1, sqrt(3) + 2, sqrt(3) + 2

[grand pond]

Results provided by WolframAlpha

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Requested by sooouth.

[lethal path]

ok so not that idea

https://www.desmos.com/calculator/ajrjuo8s6q

might as well give you the graph

[midnight plank]

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[twilit field]

so to start , the equation of the tangent is $z-z_0= 2x_0(x-x_0) + 2y_0(y-y_0)$

[lethal path]

yeah then try replacing $\sqrt{x^2 + y^2} - \sqrt{x_0^2 + y_0^2} = 2x_0(x-x_0) + 2y(y-y_0)$

[grand pond]

south

[twilit field]

hmm

okay

[lethal path]

or $\sqrt{x^2 + y^2} + \sqrt{x_0^2 + y_0^2} = 2x_0(x-x_0) + 2y(y-y_0)$ if z0 is negative and so on

[grand pond]

south

What a wonderful world !

[twilit field]

wait

I messed up

[lethal path]

wait this is trippy

[twilit field]

so to start , the equation of the tangent is $z-z_0= \frac{x_0}{z_0}(x-x_0) + \frac{y_0}{z_0}(y-y_0)$

[lethal path]

but the idea is that you can substitute x = 0 and y = 0

[grand pond]

What a wonderful world !

[lethal path]

and then you should be able to force z = 0

[twilit field]

hmm

[lethal path]

yeah so I think this now looks correct, seems to work from here

[twilit field]

so we start by now assuming x=y=0

[lethal path]

yes and then if the plane really doesn't pass through the origin, then you won't have z = 0

it'd pass through some other point with x = 0 and y = 0

which means another z-coord that isn't 0

[twilit field]

so $z-z_0 = \frac{-x_0^2}{z_0} + \frac{-y_0^2}{z_0}$

[lethal path]

fix the typo

[grand pond]

What a wonderful world !

[lethal path]

but well yeah $-x_0^2 - y_0^2$ equals if $x^2 + y^2 = z^2$....

[grand pond]

south

[twilit field]

Wait

but we know $(x_0^2+y_0^2=z_0^2)$

[grand pond]

What a wonderful world !

[lethal path]

perfect

[twilit field]

so $z-z_0=-z_0$, thus $z=0$

cool

[grand pond]

What a wonderful world !

[lethal path]

good catch by the way, because the tangent plane formula only works for z = f(x, y)

[twilit field]

yea

thanks

I have a few more problems, if that's fine

[lethal path]

bruh a uni typed so horribly

[lethal path]

sure

[twilit field]

IMO

Also, that feels wrong

the last bit

[lethal path]

it's based on $\nabla (a, b,c) \cdot ((x,y,z)-(a,b,c)) = 0$

[twilit field]

So to start I find the equation of a tangent plane to the sphere

[grand pond]

south

[lethal path]

yes

you should also realise that the normal vector to a(x - x0) + b(y - y0) + c(z - z0) = d is just (a, b, c)

so you can save the effort actually

[twilit field]

hmm

yea, makes snese

[twilit field]

lol, didn't realise

thanks

[lethal path]

yes so the normal line to a tangent plane is just given by the gradient vector

the line passing through the given point with the gradient vector as its direction vector

then it's self-evident cause you would just set t = -1

so yeah it's more accurate to say the gradient, direction of steepest increase, comes first

and then the tangent plane comes from the dot product of the gradient being 0

[lethal path]

like so, the gradient at a given point (a, b, c)

and the vector (x, y, z) - (a, b, c)

[twilit field]

got it

thanks

.close

[midnight plank]

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[midnight plank]

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[finite jewel]

assuming youre given 2 sides and 1 non included angle, if we dont know if the angle were finding is acute or obtuse, is there any case in which sine rule produces one answer only?

[midnight plank]

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[midnight plank]

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[midnight plank]

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[late palm]

This is how I solved it

But my answer is not in the options

[lyric charm]

it's probably simplifiable to one of the options eventually

but you did yourself a disservice by not simplifying the surds as much as you could have

[late palm]

I can't simplifyit further😭

Idk how

Maybe i should rationalise this?

But still the numerator is in the root

[small gust]

i need help on rule of limits

[pale vapor]

3-2sqrt(2) can be rewritten as (sqrt(2)-1)^2

yeah and take 2sqrt(2) common in denominator

[midnight plank]

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[midnight plank]

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[twilit field]

sorry

gtg

.close

[midnight plank]

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[gloomy scaffold]

u take the integral of a velocity function f(x) for the displacement, but the velocity function is a function of displacement
do you just do tf(x)

[gloomy scaffold]

ie you have a function of velocity based on displacement

oh wait no im clowning you just get a differential equation dont you

[solid iris]

yeah u solve an ODE

$\dot x=f(x)$