help-23

[timber atlas]

This is what I think the answer should be

[cosmic grove]

its correct, where is the problem ?

[mortal meteor]

your last line is wrong. $\frac{b}{c}=\frac{a\cdot b}{a \cdot c}=\frac{a}{a}\cdot \frac{b}{c}\neq a\frac{b}{c}$

[flat frigate]

Alexander42

[safe radish]

@timber atlas Has your question been resolved?

[timber atlas]

? I don’t get what you mean

[mortal meteor]

in the last step you take the 2 out of the fraction (if you can call it like that). But you take it out of the denominator and out of the nominator. so insstead of multiplying $\frac{3\sqrt{5}-5}{5}$ with 2, you have to muliply it with $\frac{2}{2}$, because you need to multiply the denominator with 2 and the nominator with 2

[flat frigate]

Alexander42

[timber atlas]

So what is wrong?

[mortal meteor]

$\frac{6\sqrt{5}-10}{10}=\frac{2\cdot(3\sqrt{5}-5)}{2 \cdot 5}=\frac{2}{2}\cdot \frac{3\sqrt{5}-5}{5}=1 \cdot \frac{3\sqrt{5}-5}{5}$ is what you should have done. Your last line is incorrect

[flat frigate]

Alexander42

[timber atlas]

Ok

Thanks

.close

[safe radish]

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[polar valley]

Find the range of real alpha for which the following series converges

[polar valley]

This was the provided solution, I don't understand why the sequence converging for alpha = 3/2 implies the series coverges for alpha - 3/2 < -1

[safe radish]

@polar valley Has your question been resolved?

[polar valley]

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[trail rune]

Try and compute the bounding boxes of the first few stages

[proud hazel]

Making this in paint would probably help

[safe radish]

@lean otter Has your question been resolved?

[proud hazel]

There is some sort of fibonacci involved here

[safe radish]

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[hollow ridge]

Hi how do I handle this question

[split ether]

Consider rewriting tan(2x) as sin(2x)/cos(2x) and tan(3x) as sin(3x)/cos(3x)

[hollow ridge]

then?

[split ether]

So you get limit of (sin(2x)/sin(3x))*(cos(3x)/cos(2x))

The cos(3x)/cos(2x) basically approaches 1

So the limit is equal to the limit of sin(2x)/sin(3x)

Do you need help with evaluating limit of sin(2x)/sin(3x) as well or?

[hollow ridge]

got it

thanks

[split ether]

Feel free to .close

[hollow ridge]

.close

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[weak pumice]

Is there a way to find number pairs in less than O(n^2) time? There was a question asked in my job interview where N is given and I'd have to find pairs (x,y) in sequence from"1 to N".
Conditions for a pair are
1 <= x <= y<= N
Sum of first x-1 numbers == Sum of x+1 till y

[foggy orchid]

yes this is doable in O(n) time

[weak pumice]

I used AP formula to calculate sum for 1 to x-1 and x+1 till y

[weak pumice]

With the equation? I couldn't able to find the pattern

[foggy orchid]

actually it's doable in O(n) time for an arbitrary sequence of increasing numbers

for the numbers from 1 to N it may be doable even faster with the equation

[weak pumice]

I first tried O(n^2) solution then tired to optimise it by using binary search to find y. But it's still slow for larger 1e16 integers

[foggy orchid]

for 1e16 integers even O(n) is slow

[weak pumice]

Yeah

I thought same

[foggy orchid]

anyway

formula for sum of numbers from 1 to n is n(n+1)/2, right?

[weak pumice]

Yeah

[foggy orchid]

well

if you sum the numbers from 1 to b

and from 1 to a

and subtract the two sums

you get the sum of the numbers from a+1 to b

so you're looking for x,y such that (x-1)(x)/2 = y(y+1)/2 - x(x+1)/2 if I understand the problem correctly

[weak pumice]

(6,8) is an valid pair when N=8, because Sum([1...5]) == Sum([7...8])

[foggy orchid]

so I think I understood it correctly

[foggy orchid]

can you verify this formula to be true

[weak pumice]

Let me check, give me 2 min I'll code it quickly

[foggy orchid]

just evaluate it, we're not done simplifying it

[weak pumice]

We don't know value ofy yet

[foggy orchid]

or x

this is an equation that must hold if x and y are solutions (and if it holds then they're solutions)

[weak pumice]

Ah yes

[foggy orchid]

now the problem is how to find x and y that satisfy this equation

[weak pumice]

Yeah

[foggy orchid]

now we simplify

the /2 are unneeded

(x-1)(x) = y(y+1) - x(x+1)

(I multiplied both sides by 2)

move the xs all to the left

(x-1)x + x(x+1) = y(y+1)

factor x out on the left

x(x-1+x+1) = y(y+1)

the 1s cancel out

2x^2 = y(y+1)

and actually

divide by 2 again

x^2 = y(y+1)/2

you're looking for y such that the sum of numbers from 1 to y is a square number

for y < N

[weak pumice]

not 1 to y, we need sum(x+1 to y)

[foggy orchid]

y=8 is the first such number that satisfies this, actually

can you write me a program that checks which numbers satisfy this real quick

[weak pumice]

Yes

i=1 j=1 x=0.0 y=0.0
i=6 j=8 x=15.0 y=15.0
i=35 j=49 x=595.0 y=595.0
i=204 j=288 x=20706.0 y=20706.0

i,j are x,y

I just used my previous code to calculate pairs till 500

[foggy orchid]

interesting

I don't see a pattern at all

I found the sequence on oeis:

1, 8, 49, 288, 1681, 9800, 57121, 332928, 1940449, 11309768, 65918161, 384199200

[weak pumice]

Yeah, I spent 4 hours trying to find pattern

Then I wrote somewhat optimized brute force algorithm

[foggy orchid]

I don't see how this would be doable for N=1e16

there are just

too many pairs here

even just returning the answer would take too long

are you sure it's not finding the largest pair or smth

[weak pumice]

I'm pretty sure because my code was working for first 3 test sets but was failing for larger test cases, I was getting Time Limit Exceeded for those remaining test cases

[foggy orchid]

yes but 1e16 is ridiculous

[weak pumice]

I have taken screenshots of that question, if it's allowed then I can share it here

[foggy orchid]

are you in a job interview rn?

if not then it's fine

[weak pumice]

No, I failed that test

[foggy orchid]

lol

[weak pumice]

Still, I'll delete it after we close our discussion.

That's the question

[foggy orchid]

aaah

determine the number of pairs

not the pairs

[weak pumice]

Can we do it without knowing the (x,y)?

[foggy orchid]

yes, finding the number of answers is very often easier than finding all of the answers

especially when there are this many answers

[weak pumice]

Omg, howw?

[foggy orchid]

well, we have this sequence to analyze

[weak pumice]

yeah, that sequence is correct for y

[foggy orchid]

hmm...

[weak pumice]

x matches A001109 sequence

y matches A001108 sequence

[foggy orchid]

these numbers alternate between being squares and being 2x a square

you know

these number aren't actually that many now that I look at them

so maybe finding all of the pairs isn't slow either

ah!

the oeis has formulas for this

lmao

[weak pumice]

checking all pairs is takes time, so if I count that A001108 sequence till N then that might work?

[foggy orchid]

a(n) = floor( (1/4) * (3+2*sqrt(2))^n )

[weak pumice]

yes

[foggy orchid]

this is actually a really fast formula

so uh

I guess now the question turned into why this formula is correct

[weak pumice]

count numbers till N?

[foggy orchid]

yep

god dammit euler, what have you not solved

were you allowed google on the interview?

I have no idea how one would reinvent this equation on their own in the span of 10 minutes

[weak pumice]

It was online, I could have googled but instead I thought the question was very easy and I wrote O(nLogn) Solution. Later I found out it was not correct.

[foggy orchid]

I can tell you the O(n) solution I immediately thought of at the start tho

it's related to this classic problem: https://leetcode.com/problems/two-sum/

[weak pumice]

I've solved that already using a Hash Map

here is my solution for that problem

[foggy orchid]

here's another solution that only works if the array is sorted:

[weak pumice]

two pointer method

[foggy orchid]

yes

basically it relies on the fact that if you increase x then y needs to be smaller than what it would need to be for the previous x

this problem follows the exact same structure

if you increase x you need to increase y

so the two pointers but both of them starting on the very left works for an arbitrary number array

[weak pumice]

yeah

how would you have solved it ?

[foggy orchid]

using this formula and iterating over every y is also O(n)

[foggy orchid]

I'd probably have gotten to x^2 = y(+1)/2 and then do the O(n) search over y

[weak pumice]

yeah, but it would still take more than 5s to calculate it for 1e16 number

[foggy orchid]

yep

[foggy orchid]

did you verify this works btw?

[weak pumice]

not yet

I'll check it now

[foggy orchid]

it seems the formula was derived by solving a pell equation

which I've never seen in my life

[weak pumice]

me neither

[foggy orchid]

maybe there's a recurrence relation

where you can get the next answer from the previous answer

or maybe the previous two answers

(1,1) and (6,8) are answers...

[weak pumice]

it's working

[foggy orchid]

nice

[foggy orchid]

<@&286206848099549185> can anyone explain this formula for finding the members of sequence A001108 ?

I've only managed to determine that it's related to recurrence relations and pell's equation... probably

no clue how

[weak pumice]

How A001108 and A001109 are related in those pairs?

[foggy orchid]

A001108 is the index of a triangular number that is square

A001109 is the sqrt of such a triangular number

A001108 contains the number 8 and 8*9/2 is 36

A001109 contains 6 and 6*6 is 36

[weak pumice]

I really want to test that equation against real test cases.. for 1e16 numbers

[foggy orchid]

I think for 1e16 the answer is 21

this formula finishes in literally an instant

why did they give you 5 whole seconds if that was the intended route

and if this is not the intended route then what is

[weak pumice]

there were multiple test cases

[foggy orchid]

ah

[weak pumice]

so I/O would take time

[foggy orchid]

5s for the whole suite

[weak pumice]

yeah

[foggy orchid]

sounds about right

[weak pumice]

where did you find that formula on oeis ?

I don't know how they were expecting a fresh CS graduate would remember that formula...

Again thank you very much for explaining the solution...

.close

[safe radish]

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[lean otter]

HEEELP

HEEELP MEEE

BEFORE I DROWN

[pulsar condor]

why are you asking a math question if you're drowing?

[lean otter]

Can you help me with this? @pulsar condor

[pulsar condor]

Yes, however you're drowning, so that should be taking precedent

Just google the definition of a rhombus

[lean otter]

i did

[pulsar condor]

So read it

[lean otter]

read the pic

[pulsar condor]

Yeah, read the definition of a rhombus

That will dispel all issues you're having with it

[lean otter]

all equal in length but then why do the angles differ

if all the sides are equal

[pulsar condor]

cause it doesn't have to be a square

all a rhombus is, by definition, is a quadrilateral with all 4 sides equal.

[lean otter]

if a rhombus is square then it has equal angles right?

[pulsar condor]

Yes

[lean otter]

s

a

[pulsar condor]

cause a square has equal sides and angles

again, by definition

[lean otter]

but if the rhombus is not a square

then it doesnt have equal angles but only equal sides

[pulsar condor]

yes

congrats on reading the definition

[lean otter]

so equal sides doesnt mean equal angles?

[pulsar condor]

no

having equal sides and having equal angles doesn't mean anything in relation to one another

[lean otter]

what makes angle unequal

[pulsar condor]

...

when the angles aren't equal

[lean otter]

why is that

[pulsar condor]

...

why is 2 not equal to 3?

[lean otter]

because 2 is less than 3

[pulsar condor]

because they're different numbers, yes

so 2 angles that are different, won't be the same

[lean otter]

but they have a reason they have diff values so whats the reason for unequal angles

[pulsar condor]

cause their measures are different

which is the exact same reason as 2 not being 3

[lean otter]

why are they differwnt

[pulsar condor]

Cant explain it any simpler

[lean otter]

i

ok

i just

got confused

[pulsar condor]

2 angles aren't equal, when they're different.

Draw 2 angles that are different, they wont measure the same angle

It's not that hard

[lean otter]

ig im just sleepy thanks tho

how do i delete this

bot

.close

[safe radish]

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[lean otter]

dayum

[safe radish]

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[civic forum]

How to solve 8?

[cursive ledge]

Triangle Similarity

[civic forum]

Uh

Yeah I kinda set it up like that

[upbeat swan]

what

[civic forum]

But idk if I’m doing it right

[upbeat swan]

oh

both are right

[civic forum]

Oh really

[upbeat swan]

yeah

[civic forum]

If I solve that I would get X

Correct

?

[upbeat swan]

what x

[civic forum]

The question is

Find the value of x y and z

[upbeat swan]

ahhhh

[civic forum]

So if I solved my equation

I would get X

I think

[upbeat swan]

wait

i thought we were talking about tthat

🤣

[civic forum]

Lol

Naw this

Number 8.

[upbeat swan]

ok

[civic forum]

So first scale factor

I think I set it up

[upbeat swan]

x/x+12 = 11/22

then solve for x

[civic forum]

Ok

[upbeat swan]

x/x+12 = 1/2

[civic forum]

Wait what

How are u getting those

Number

[upbeat swan]

woah

ok

sry again

lmao

[civic forum]

Np

There’s 2 number 8

Just noticed

The first one

[cursive ledge]

0h

LMFAO

[civic forum]

bruh lmao

U both were confused

💀

[upbeat swan]

wait

there are 2 8s

yea

[civic forum]

The first one

Not with the triangle

[upbeat swan]

ok

x/25 = 30/18

i think

= 15/y

[civic forum]

U sure it’s not

25/18

Instead of x/25

Ah nvm ur right

[upbeat swan]

nah

idk if im right

[civic forum]

U have to make sure

Ur using the exact sides

[upbeat swan]

idk i hate these problems where they just use randomly placed shapes

[civic forum]

Yeah I feel u

[upbeat swan]

you cant figure out the correct fractions

[civic forum]

Yeah junk it’s

I think it’s

15/25

Wait nvm…

Ye ur right

Ok

Ok I have C

X

@upbeat swan

[upbeat swan]

yeah

[civic forum]

How do I find y and z

Now

[upbeat swan]

1 sec so i draw it on paint

[civic forum]

Ok

I gotta turn this in

In 8 minutes

No rush

Though

[upbeat swan]

just substitude now

[civic forum]

Yeah with what

Cause there’s no equation

[cursive ledge]

Remember it's a proportion, try reading it as

x is to number a
as number b is to number c

That represents
x --- a
b --- c
or
xc = ab
or
x/b = a/c

Example

If 2 is to x
as 3 is to 6

Notice that 6 is 2 times 3
So x should be 2 times 2

Ofc, this applies because the figures are similar

So they all in same proportion

[upbeat swan]

akz

is this ok or are the shapes wrongly placed

[civic forum]

Damn.

I have to turn it in rn

Like this

Thanks for the help though guys

I’ll get partial credit for it

Instead of 0

[upbeat swan]

sorry i couldnt help. if there were triangles it would've been ez

[civic forum]

Ur good bro

Thanks

[cursive ledge]

Sorry man

Is this an r, m or p tho

[upbeat swan]

m

M

[safe radish]

@civic forum Has your question been resolved?

[safe radish]

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[past apex]

how to do this i tried something but it doesnt work

[feral linden]

What is the definition of x_y ?

[past apex]

yea, we didnt do this in school im studying for compsci competetion

i tried 1*(n^2)^2+1*(n^2)^1+1*(n^2)^0

[feral linden]

I see… sorry for interrupting 😂 Nvm,bad at computer science.

[past apex]

i mean this is connected to math

go on, please, explain what you wanted

[slender coral]

What was the actual question?

[past apex]

find x

ahaha

actually if X_n=111_(n^2), what is X in base n

dont worry, its not a test

[slender coral]

So this is a good first step, what can you do from here?

[past apex]

simplify

i got n^4+n^2+1

and i tried solving it but

[slender coral]

Well, you dont really want to solve it. How would you write that in base n?

[past apex]

oh i got it

10101

[slender coral]

Right idea, but not quite

[past apex]

mb

10101

[slender coral]

Yes

[past apex]

mistyped

yea, that's cool i didnt think of that

you can just add 0s

thank you

[slender coral]

np

[past apex]

also, you seem to know this with bases,is there a better way to do this other than converting everything into 1 base(most likely 10)

[slender coral]

I dont think so, but converting from base 2 to base 16 is pretty easy, so if you can convert everything into base 16, that might work

You also probably need to know how to add, subtract, and divide base 16 numbers, so maybe converting it all to 10 would be best

[past apex]

yea, the thing is i dont know and dont want to learn because it doesnt really seem useful, dividing by base 16 8 2..

[slender coral]

But I dont think you can do it without converting all of the numbers into the same base

[past apex]

substracting and adding is easy

[past apex]

ok thanks

[past apex]

also, sorry, while you're already here would you know if there is a faster way to do this (find all possible Xs) other than writing equations for every character

[slender coral]

Do you know what DeMorgan rules are?

[past apex]

im not sure if this falls under math, but idk where to ask for help

yes

[slender coral]

I think you can use those

so NOT (1011011 AND X)=0100100 OR NOT X

[past apex]

wait you can "move" to the other side

[slender coral]

I dont think so

[past apex]

how did you get not x to the other side then

[slender coral]

This is just using the DeMorgan identity $\lnot(A \wedge B)= \lnot A \vee \lnot B$

[flat frigate]

opfromthestart

[past apex]

oh sorry

[slender coral]

which lets you replace the first by the second

So you should probably "distribute" to get rid of all of the parentheses and then see what you can combine

[past apex]

i ll try in a second i just gotta go to the bathroom

So I got this

[past apex]

correct?

[slender coral]

,rotate

[past apex]

but still i would need to check for each one right>

[slender coral]

I think that is correct

Im not sure how you would solve it other than just doing it bit by bit from this point

[past apex]

yea, thank you <3

[safe radish]

@past apex Has your question been resolved?

[safe radish]

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[sterile saffron]

can someone explain to me (in more detail), how they traversed these steps? For reference: log base is irrelevant, this is solving recurrences using tree recursion

[sterile saffron]

my best attempt is simplifying it into

sum( 1 / ( logn - ilog3))

i am not sure how to go from there to a form that represents geometrc/ harmonic summation

we can also do (1/logn) Sum( 1 / (1- (ilog3 / logn) ))

which is of the form Sum(1/(1-r)), but the bounds do not fall into any of the categories of the summation series

[safe radish]

@sterile saffron Has your question been resolved?

[sterile saffron]

<@&286206848099549185>

[sharp wharf]

Sub in the first few terms of i and also when i=log_3(n) and then you’ll see

[safe radish]

@sterile saffron Has your question been resolved?

[sterile saffron]

thank you, it makes more sense, i am now just confused on where the 1 comes from, i believe it's from i = log3n, but when we plug that in we get 1/0

so we set the boundary of the sum to be upto log3n -1, but what happens when i=log3n?

[safe radish]

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[deft notch]

Can someone check if I set this up right

[safe radish]

@deft notch Has your question been resolved?

[safe radish]

@deft notch Has your question been resolved?

[safe radish]

@deft notch Has your question been resolved?

[safe radish]

@deft notch Has your question been resolved?

[safe radish]

@deft notch Has your question been resolved?

[nocturne spire]

How to solve this question using differentiation?

[safe radish]

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[potent geyser]

I tried a thing

It did not work

I am dead inside now

[unkempt belfry]

i don't know how to do this lol. but im assuming you tried induction

?

[potent geyser]

I have not even tried that, I lost hope while trying to find some patterns

Was a stupid idea

[unkempt belfry]

oh okay, yeah again i have no idea but i would try induction on k

no idea how to show there are infinitely many n for a given k....

[potent geyser]

same

[unkempt belfry]

oh wait!

maybe for each k you can show that there only a finite number of n for which the number of prime

[feral linden]

Contradiction

[unkempt belfry]

and therefore there must be infinitely many n for which the number is composite

[feral linden]

If that’s not the case you will have a j from 1 to k such that

Any n great enough 2^n+3^n-j is prime

[unkempt belfry]

what about the infinitely many n part? sry your solution prbly handles it and im just not seeing it

[feral linden]

If Finitely many n satisfy that, then there exists N, any n>=N, there is a prime in those k numbers

By pigeonhole

There must exist one j from 1 to k such that

2^n+3^n-j is prime for infinity many n

Sorry I said wrong

So not any n great enough but infinitely many n

[unkempt belfry]

ohh ok got it

thanks!

[feral linden]

But I didn’t solve it at all.it’s just an idea that I am forming

I am still developing it

[safe radish]

@potent geyser Has your question been resolved?

[safe radish]

@potent geyser Has your question been resolved?

[feral linden]

I don’t know whether it will help, but I have transformed the question from proving the existence of infinitely many into proving the existence of just one n

If you have a n such that 2^n+3^n-1,…,2^n+3^n-k are all composite numbers, then there exists prime numbers p_1,…,_p_k such that 2^n+3^n-j=0 mod p_j for any 1<=j<=k. but 2^n+3^n-j mod p_j is a periodic sequence, so it has period T_j. Then I let T be a number divisible by any T_j, for example, T=lcm(T_1,…,T_k) or simply T=T_1…T_k, then any n’=n+mT, 2^n’+3^n’-1,…,2^n’+3^n’-k are also all composite numbers

So now we need to prove that any k, there exists (just one is enough) n such that 2^n+3^n-1 ,…,2^n+3^n-k are all composite numbers

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[lean otter]

@broken yew #help-16 message what?

[lean otter]

let me look

[broken yew]

im not gonna find it for you

wtf

[lean otter]

i didnt ask

[broken yew]

It's not like there are an infinite number of cases to check

[lean otter]

tru

.close

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[mighty garden]

I trying study and I couldn't understand how do this problem.

When I look at the hints my brain not understanding what they did in these steps.

[lean otter]

ok, well the first step is basically they took away 3x from both sides of the top equation

[mighty garden]

I see and that how they got 2x in the next step

[lean otter]

you understand what they did with the next step?

[mighty garden]

No

I not sure where the 4 went

[lean otter]

think, what is the reverse of multiplication

[mighty garden]

Division

[lean otter]

so what can you do to both sides to get rid of the 4?

[mighty garden]

Multiply by 1/4 on both sides

[lean otter]

yep, same thing as dividing by 4

[mighty garden]

I thought you would distribute the 4 since it near the parentheses

[lean otter]

that's also another way to do it

but you would still have to divide later

however, most people prefer to just divide

because it's faster than expanding and simplifying

[mighty garden]

Oh alright

[lean otter]

another way to think about it is this

4g=y

you would just divide by 4 if you solved for g

right?

[mighty garden]

Yea

[lean otter]

now image replacing that g with (2-y)

4(2-y)=2x

if you were trying to solve for y, you would just divide by 4

[mighty garden]

Alright

[lean otter]

you understand why they got 1/2 x in the next step?

[mighty garden]

Yep

They divide by 4 on both sides

[lean otter]

yes, now here is were we bring in the other equation

currently, after simplifying the top one, what equations do we have

[mighty garden]

The other equation is 2-y=6x

But 2-y already in other equation

[lean otter]

go on

[mighty garden]

Doesn't it cancel each other out

So it just be 6x

[lean otter]

cancel out isn't the word you would use you here, but your thinking is correct

the word would be substitute

from the bottom equation we get the 2-y=6x

so 2-y can be represented by 6x in the same system

when you go to 2-y=1/2 x, you can sub in 6x for 2-y since they are the same thing

so now, what is the only possible answer for x

after you substitute in 6x

[mighty garden]

I not sure

[lean otter]

well you equation is 6x=1/2 x

so think, what number can be multiplied by 6 and equal half it's value

[mighty garden]

1/2

[lean otter]

1/2*6=3

1/2^1/2=1/4

try thinking logically

what is any number times 0?

[mighty garden]

It be 0

[lean otter]

so applying the logic that any number times 0 is zero

try to do 6x=1/2 x again

you don't need to do math, just think logically

[mighty garden]

So x is 0

[lean otter]

yep

and now you don't even need to solve for y

because what is 0 divided by any number

remember, 0/y can be written as 1/y * 0

[mighty garden]

Oh

So y is also 0

[lean otter]

not exactly

try to solve for y using the 2nd equation

plug in x=0 and solve for y

[mighty garden]

It's -2

[lean otter]

be careful

2-y=0

what is your next step

[mighty garden]

Subtract 2 by both sides

[lean otter]

-y=-2

[mighty garden]

Oh got divide by neg 1 on both sides

So y is 2

[lean otter]

there you go

so what is x/y

(0,2) is your coordinate pair

[mighty garden]

That makes more since now

Thank you

[lean otter]

np

[mighty garden]

.close

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[crisp wren]

I have one more question

[plucky elk]

this is even simpler?

[crisp wren]

oh srry wrong question

What is the probability that a number selected from the numbers (1, 2, 3,..........,15) is a multiple of 4?

there

thats the question

is it 1/5?

[static ledge]

i mean there are only a limited number of numbers out of 1 through 15 so just consider each number and count the number of multiples of 4 , so yeah. 4, 8 and 12 are the only multiples of 4 in the domain of [1,15]

[crisp wren]

one more question

[static ledge]

sure

[crisp wren]

There are n sweets in a bag. 6 of the sweets are orange. The rest of the sweets are yellow. Hannah takes a random sweet out of the bag and she eats it. Hannah then takes at random another sweet from the bag. She eats it.

The probability that Hannah eats two orange sweats is 1/3

obviously

[static ledge]

and so the problem is how many sweets are in the bag?

[crisp wren]

show that n^2-n-90=0

and

solve n^2-n-90=0 to find the value of n

"There are n sweets in a bag"

[static ledge]

sure ill explain what you have to do

You have to reverse engineer the problem.
Note that she takes the sweet and eats it, meaning she takes a sample without replacement. That means that 1/3 is going to equal the product of two probabilities; it's going to be the product of 6/n and (6-1)/(n-1)

[crisp wren]

wait

lemme say my answer

[static ledge]

ok

[crisp wren]

Since no sweet is replaced we can make things easy. The probability that the first sweet is orange is 6/𝑛. Since she eats the first and doesn't replace. The probability of her eating another orange sweet is 5/(𝑛−1).

The probability of her choosing orange and orange is:6/𝑛∗5/(𝑛−1)=1/3.

Now simplify: 30/𝑛(𝑛−1)=1/3
Multiply by 3 and simplify: 90/(𝑛2−𝑛)=1.
Multiply by 𝑛2−𝑛: 90=𝑛2−𝑛.
Take away 90 from both sides: 0=𝑛2−𝑛−90

The key is that the object isn't replaced and its an and question. Oh and the value for n if 10.

[static ledge]

that's absolutely correct! good job

[crisp wren]

les go

[static ledge]

do you have any other questions or is that all?

[crisp wren]

can I give u a question?

[static ledge]

why not

[crisp wren]

A jar contains four marbles: three red, one white. Two marbles are drawn with replacement.
(i.e. A marble is randomly selected, the color noted, the marble replaced in the jar, then a second
marble is drawn.)
List a sample space containing four outcomes.
List a sample space with sixteen outcomes.
Write the probability of each of the four outcomes in (a).
What are the probabilities of the outcomes in (b)?
What is the probability the colors of the two marbles match?
What is the probability the same marble is drawn twice?

its ez

want me to give u a hint?

[static ledge]

no im well versed in probability but thanks

there are quite a bit of subproblems though so it might take me a minute or two to write out some of the answers

[crisp wren]

k

[static ledge]

sample space with four outcomes:
(R,R)(R,W)
(W,R)(W,W)
sample space with sixteen outcomes:
(R,R)(R,R)(R,R)(R,W)
(R,R)(R,R)(R,R)(R,W)
(R,R)(R,R)(R,R)(R,W)
(W,R)(W,R)(W,R)(W,W)
probability of the outcomes in (a):
R,W: 3/16
W,R: 3/16
W,W: 1/16
R,R: 9/16
probability of each individual outcome in (b):
each outcome has 1/16 chance of occurring.
probability that the colors of the two marbles match?
this probability is equal to 9/16+1/16=5/8
probability that the same marble is drawn twice?
this probability is equal to 4/16=1/4

[static ledge]

how did i do?

[crisp wren]

lemme check

[static ledge]

i chose not to go with a tree diagram for the sample spaces because that would be incredibly difficult to pull off in discord

[crisp wren]

For this question, we consider that the first question if we have a sample space containing four outcomes and then we can just define like are robbery, then Read W means white. So we only consider the color and then in this case we have ar ar ar w w r w W. So we have four outcomes. If we only care about the color, then we can have a sample space with four outcomes and then if we can see the each marble differs from others and then we have we have four marbles in total and then we take two marbles intent And then in total we have 16 outcomes. The former roles are represented by 1234. So the outcomes will be (111) 213 14 21 to 2 to 3 to 4 through 132 problem. So in total we have 16 outcomes in this sample space and we get the probability for each outcome and so because we have three red marbles and one white. So for us we have a probability of 3/4 and there are word based 3/4 square, there is nine out of 16. And similarly we have the probability for rw, which is 3/4 times 1/4. And similarly we can get other and two outcomes probabilities and then we have the properties for the first um sample space for a second for each marble. We will have a probability of one or four. Then, for each combination in the sample space, the probability will be 1/16, which is 1/4 square. And then question E probably that kind of of the two models match means that we will have our own this means that we just need to sum up the probability of R and w w, which is 9/16 plus 1/16, which is 10/16 and is 5/8. The last question and it’s a marvelous joint tie. We will have 112233 or 44. And some of these four probabilities From the second question and then we know that it is one or 4. There’s 1/16 times four, so it’s 1/4.

[static ledge]

is this the answer you had for this problem?

[crisp wren]

yessir

[static ledge]

seems like we have similar solutions

[crisp wren]

ye so ur correct

nice job

cya now

.close

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[west coral]

I need help finding out a semester total using the quiz total and final exam total

[narrow pecan]

rogerhub??

idk

or u could j calculate the class weights and do the multiplication and add them together

like do u know the class weightws

[west coral]

I mean it’s really 2 simply numbers that I’m struggling to figure out for some reason

I don’t no

Like my first quizzes I got 85/100 and then my finals I got 83/100

Like how much is the overall of that term

[narrow pecan]

finals are worth more than quizzes though

[west coral]

Yea

[narrow pecan]

right

[west coral]

Quizzes were /20

Finals /50

[narrow pecan]

uhh idk i thought it mostly had to be calculated w class weights/what each category on your overall course summary states

if that makes sense

[west coral]

Ughh this is stressing me lmao

[worthy hemlock]

Then your current grade is (quiz grade times quiz percentage) + (final grade times final percentage)

It should be that

[narrow pecan]

yeah

along w the classwork times classwork percentage etc

[west coral]

Got it

[narrow pecan]

like whatever else u do in class that counts towards ur grade

do u have the syllabus

[west coral]

Uh

[worthy hemlock]

Or you can use a grade calculator

Google one

[narrow pecan]

im p sure all syllabi state their category weights

[west coral]

Projects Homework Classworks

[narrow pecan]

yes is there like any mention of percentages along w those

[west coral]

Not at all

[narrow pecan]

like how much of each category counts towards the 100% of ur grade

[west coral]

Just a grade /100

[narrow pecan]

oh ://

[west coral]

Yea that’s annoying

International school with the dumbest system I’ve ever seen

[narrow pecan]

hahah yeah intl schools have weird grading systems

[worthy hemlock]

Can you post a screenshot of the syllabus? Just the section with the categories

Like hw, exams, etc

[narrow pecan]

or a picture if ur teacher/prof handed it out in person

[west coral]

Sadly not

It’s only posted on their website

[narrow pecan]

uhh screenshot it??

or use snipping tool

[west coral]

Alright one sec

[lyric ingot]

Windows + Shift + s

[west coral]

Yea ik

Just getting on my laptop

What did you want again ?

[narrow pecan]

the part w the categories

[west coral]

Alright

[narrow pecan]

and anything mentioning each category if it says something abotu percentages

[west coral]

here it is

nothing mentioning percentages

man them not showing more details is dumb

[worthy hemlock]

That's not it

That's your grades

Or something

[west coral]

Yea it is

[narrow pecan]

uhh can u try clicking on mathematics us adv

which is what ur taking rn right

or what u need to calculate the grade for

[west coral]

Nothing shows up

[narrow pecan]

?????

[west coral]

I guess I do need percentages to calculate it properly

[narrow pecan]

okay what about this part

like whered u find that

[west coral]

it's all here

but it only shows numbers

[worthy hemlock]

What exactly are you trying to calculate?

[west coral]

like the first pic

[narrow pecan]

overall grade i think

[west coral]

the total of the semester

[narrow pecan]

after they took quizzes and a final

[west coral]

^

[worthy hemlock]

For a specific class?

[west coral]

nope

the whole term

[narrow pecan]

wait what

so like all of your classes

[west coral]

yes

[narrow pecan]

huh

[worthy hemlock]

Then that's your grade for the term

It's in there

[west coral]

im very shit at explaining

[worthy hemlock]

The 85.83

[west coral]

well that's what I though

The final meant the final exams

[narrow pecan]

does your school website have like your own personal login

[west coral]

not the final mark for the whole school term

yea that's where im getting these from

[narrow pecan]

oh okay

[west coral]

like look im ashamed to say this but the first term i did very bad

[narrow pecan]

so each of the numbers underneath "ratio" is what you received in each class?

[worthy hemlock]

Explain clearly what on earth you want to do

[narrow pecan]

or is it what you got on the final in each class

[west coral]

yes yes that one

Look ill show you a horrible term i had maybe that could help

[narrow pecan]

okay but you said you got a 83 on ur finals

does that mean u averaged out each of ur course final grades

[narrow pecan]

uhh yeah maybe if you know what each of the grades are (like in each category)

[west coral]

you know what im sorry for the hassle

[narrow pecan]

????

okay look i cant tell u like what exactly u got on the overall term grade but since you have a 85-ish on your final avg and you got around the same on your quizzes im going to assume u turn in ur homework/classwork/projects in on time and j say u have like a very low a or (more likely) a mid-low b

[worthy hemlock]

It is because you are not providing enough details on what you want to do

[safe radish]

@west coral Has your question been resolved?

[safe radish]

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[west hedge]

Those twos are not equivalent

The blue underlined should be 1/2

Am I tripping or the teacher

<@&286206848099549185>

[unreal pasture]

sin²θ+cos²θ=1, and you have another 1, so 1+1=2.

(The teacher wrote the 1 in yellow)

[worthy hemlock]

You should know that you have to wait at least 15 minutes before pinging helpers

[west hedge]

I see

[west hedge]

I’m sorry

The question is solved

.close

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[west hedge]

Thank you castle

[safe radish]

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[brazen kraken]

Hello, here is the pic. If anyone can help, that would be amazing. I've been trying to understand it all day but I've had no luck. Thanks!

[flat frigate]

Andreww

Andreww

[safe radish]

@brazen kraken Has your question been resolved?

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[lean otter]

i tried saying by way of contradiction assume p1>sqrtn and p2>sqrt n

idk where to go with this

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

<@&286206848099549185>

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

no one knows?

[queen parcel]

yup.

Now what can you say about p1 • p2?

You assumed that they are both factors of n

You also assumed that p1 > √n, and that p2 > √n

[lean otter]

p1 p2 > n

?

i dont think u can say that

[queen parcel]

You can

[lean otter]

what if p1 = 0.8 and p2 = 0.9 and n = 0.78

[queen parcel]

Good question: I think you assume that n, p1, and p2 are positive integers

[lean otter]

ohhhh right

thx

imma try it

ok i got it

thanks

[queen parcel]

👍

np np

@lean otter .close

[lean otter]

.close

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[dusk berry]

can i get some help with this

[dusk berry]

not sure how or where to start with this

[spice grove]

Well

Do you know what arithmetic mean is?

For two numbers let's say a and b.

[dusk berry]

cz sin^n + cos^n =/= 1^n right?

or am i in the wrong

[dusk berry]

uh can u give an example

[spice grove]

Basically

We have 5 and 7

Tell their arithmetic mean.

[dusk berry]

i dont understand what u mean

[spice grove]

Do you know what mean is?

Like average.

What's the average of 5 and 7?

[dusk berry]

oh so like bounding?

[spice grove]

The average mate.

Wdym bounding?

The average of 5 and 7 would be 6.

Right?

[dusk berry]

yeah but not sure how that would help

[spice grove]

It surely surely does.

The arithmetic mean of two numbers is?

For example, tell for 11 and 15.

[dusk berry]

13

?

[spice grove]

Good.

We did a+b/2

Right?

Although let's leave this, I'll help you proving some other way.

Alright?

[dusk berry]

that'd be better

[spice grove]

I would explain you this way but you probably don't know geometric mean.

[dusk berry]

i dont think ive taken arithmetics

[spice grove]

So let's try something else

[spice grove]

No we'd use an inequality from there. But leave that.

So

Expand

[(sin^n(x)+cos^n(x)]²

It's as simple as (a+b)^2

[dusk berry]

sin^n2 + cos^n2 + 2cos^n sin^n

[spice grove]

Good.

Now let's take sin^nx as a because it is not looking very cool rn.

Let $sin^n(x)=a$

Alright?

[flat frigate]

Sakata Yaksha

[dusk berry]

ok

[spice grove]

And for cos b.

So tell me

(a-b)^2

What do you about this?

[dusk berry]

a^2 + b^2 - 2ab

[spice grove]

Yes anything special you know about this?

It's range perhaps?

[dusk berry]

wdym range

sry if im not understanding u much but curriculum was reduced due to covid

[spice grove]

Minimum value is for (a-b)^2 is?

[spice grove]

It's completely alright.

Even if you don't know lots of stuff, it's fine at some point no one did.

[spice grove]

[dusk berry]

how do u calc that

[spice grove]

I mean.

It's squared right?

[dusk berry]

yeah

[spice grove]

Let's get back to the basics.

1^2 is?

[dusk berry]

1

[spice grove]

2^2?

[dusk berry]

4

[spice grove]

(-1)^2?

[dusk berry]

1

[spice grove]

Can anything squared be negative?

[dusk berry]

no

[spice grove]

Can it be zero?

[dusk berry]

ye

[spice grove]

So 0 is the minimum value?

[dusk berry]

if its 0

[spice grove]

Well?

[dusk berry]

yea?

[spice grove]

Yes.

So anything squared is either greater than or equal to zero.

Do you agree?

[dusk berry]

ye

[spice grove]

Alright, now we are done.

So

$(a-b)^2 \geq 0$

[flat frigate]

Sakata Yaksha

[spice grove]

Expand.

[dusk berry]

a^2 + b^2 - 2ab > 0

[spice grove]

add 2ab both sides.

[dusk berry]

a^b +b^2 = 2ab

[spice grove]

Woah

≥

No?

[dusk berry]

ye dont have that on my keyboard

[spice grove]

Hm just write geq I'll understand.

[dusk berry]

ok

[spice grove]

Alright so now do one more thing for me.

$$( \sqrt{a}- \sqrt{b})^2 \geq 0$$

[dusk berry]

i think i get wht u mean

ye

[flat frigate]

Sakata Yaksha

[spice grove]

So is it true?

[dusk berry]

ye

[spice grove]

Expand.

[dusk berry]

a + b - 2√a√b

[spice grove]

≥0 yes. Now add 2√a√b both sides.

[dusk berry]

a + b geq 2√a√b

[spice grove]

Well done now what was a originally?

And b?

[dusk berry]

sin^n

[spice grove]

||You should realise by now, we already are done.||

[dusk berry]

yeah thought so

thx for the help

[spice grove]

👍

[dusk berry]

I'll solve then close if i dont have no ques

[spice grove]

As you wish.

[dusk berry]

.close

[safe radish]

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[safe radish]

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[weak pumice]

There is an equation 3x-1y+z = X, how do I find the value of x,y,z..?

[weak pumice]

I studied it in school many years back, I don't know whether that equation is called as linear equation ?

I don't remember how to solve such equations

Could anyone please help me?

[upbeat swan]

what is X

[weak pumice]

X is an number, I wanted to know whether it is possible to get LHS==RHS

[upbeat swan]

its not

at least i think so

you have 3 unknowns and 1 equation

dont think its possible

[weak pumice]

My problem if it's possible to get LHS==RHS then print the x,y,z values

[upbeat swan]

idk what that is

[weak pumice]

I was coding a solution

[upbeat swan]

sry

maybe someone else can help

[weak pumice]

No problem

thank you

[upbeat swan]

np

gl

[weak pumice]

here is the question

x+y+z = N

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[quartz cliff]

Please help me out

Mhm

33

[upper badge]

Oh wait im terribly sorry

I read the question wrong

so there are 9 rows of first class and t seats in each of these rows

how many seats do you think that is?

[quartz cliff]

9 x t

?

[upper badge]

yes

and for economy?

[safe radish]

@quartz cliff Has your question been resolved?

[safe radish]

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• Show your work, and if possible, explain where you are stuck.
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[velvet laurel]

Hello,
Algebra question,
In school we have been learning how to factor polynomials.
An example of my quarry is something like:
x^2 + 10x + 21
And we would be expected to turn this into:
(x + 7) (x + 3)
We would need a number that would add to 10 and multiply to 21.
We were taught to guess and check to find these numbers.
Is there an easier way than guessing, an equation to plug numbers into or something?

[bold prism]

Yes

There is

[smoky epoch]

you could find the roots of your polynomial

[bold prism]

On your calculator would be easier, if you can

Otherwise the formula whic will take a bit

[velvet laurel]

Ok

[bold prism]

^^^

[velvet laurel]

Ok

We can’t really use a calculator all the times

[bold prism]

Here

Change the + or - for both the results

Takes some time remembering but if you cant be bothered guessing then theres that

[velvet laurel]

Kk

I’ll save this and give it some practice

[bold prism]

However be careful

This will give you the answer

So how to make those brackets = 0

[ancient escarp]

on your question of factoring faster, once you get more practice in, you'll be able to see it a lot quicker

[bold prism]

That too ^^ but its nice to learn this formula for future maths!

To find if equations have one root, two or none etc

[velvet laurel]

Yeah we’ve only done really small and easy numbers so far

[bold prism]

Does this help you?

[velvet laurel]

Yes

.close

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• Be polite and have a nice day!

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[lean otter]

Hello can someone give me a hint how to approach this, like which method should I use: disk, washer, slicing, or shell

[lean otter]

I’m thinking of slicing but since it’s only partial of the whole object, I am hesitant to proceed

[glass sonnet]

Have you looked at radians before?