help-13

[sharp harness]

yeah you are right

[cedar kiln]

@haughty yoke Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[deft forum]

How do i begin with

Ohh

Nvm

Silly me

.close

[cedar kiln]

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[safe yew]

Hello is there something wrong with this solution ?? The answers r wrong and I can’t figure out what wrong with it can u do a fast review pls xd ?

[glad kestrel]

well i have no idea what the original equation actually is

your paper is very difficult to read

and that's not how squaring things works

[safe yew]

Ye I realized the squaring thingy xd

It should be 4x^2 I fixed it and wanna see if it fix the problem

[glad kestrel]

what

[safe yew]

The (2x-2)^2

[glad kestrel]

yeah there's like 4 different things wrong with this

[safe yew]

😂

Can u tell me where the mistakes xd

[glad kestrel]

line 1: no idea what the equation is supposed to be.. what's that sqrt off to the side?
line 2: did you multiply by the sqrt? if so, show that. what's up with the random 2 off to the side?

line 3: not how squaring works. $(a+b)^2\neq a^2+b^2$

[wraith dagger]

a disappointing son

[glad kestrel]

line 4: calculation errors

[safe yew]

;-;

[safe yew]

Ye I fixed this xd

[faint dirge]

given ab≠0

[safe yew]

Sec lemme fix this really quick and see what I get

[safe yew]

It’s 0= 3 + 2x-2 / root x^2 -2x

[glad kestrel]

[safe yew]

So I multiply with the root first line

[glad kestrel]

okay

[safe yew]

So I get 3 * the root + 2x-2

So I do square to remove the root

And got 9(x^2 -2x) + (2x-2)^2

= 0

Is there something wrong so far ?

[glad kestrel]

yes

[glad kestrel]

.

[safe yew]

I didn’t do that yet xd

[glad kestrel]

you just did

[glad kestrel]

here

[safe yew]

The (2x-2)^2 ??

Or the other 1

[glad kestrel]

the entire thing

that's not how square roots work

[safe yew]

How it work then xd

[glad kestrel]

take it with real numbers... you're doing the equivalent of saying (2+3)^2 = 2^2+3^2... which will get you 25=14

which ain't right

you square it as you'd square any other polynomial

[safe yew]

😅

Sec lemme try something difference

[glad kestrel]

you have two terms... $3\sqrt{x^2-2x}$ and $(2x-2)$

[safe yew]

Okay I have no idea what to do

[wraith dagger]

a disappointing son

[glad kestrel]

it's just distribution

FOIL, if you prefer

[safe yew]

Foil ?

[glad kestrel]

hm actually there's a much easier way to solve this

[safe yew]

How

[glad kestrel]

try putting the sqrt on the other side of the equation

[safe yew]

What ;-;

Didn’t understand

[glad kestrel]

you have $0=3\sqrt{x^2-2x}+2x-2$

[wraith dagger]

a disappointing son

[glad kestrel]

move the sqrt over

[safe yew]

The root ;-; ?

-3 root x^2 -2x = 2x -2 ?

[glad kestrel]

yes

though

now you can apply what you were doing before

squaring both sides

[safe yew]

9 (x^2-2x) = 4x^2 - 8x+4 ?

So what the difference between those 2 ways ?

[glad kestrel]

you were not squaring things properly

the first way

so this way makes it easier

[safe yew]

But those r the same answers xd

But not in the same place

Ok let’s say I’m in exam and I did the wrong squaring is there a way for me to know that I’m doing it wrong xd ?

And is moving 1 of them to the other side always help ?

Or this is special situation

[glad kestrel]

you're doing it wrong if you're not squaring things correctly

as i already stated

[glad kestrel]

.

you kept applying this

when it's not how squaring works

exponents are not distributive

[safe yew]

Where I did that xd ?

[glad kestrel]

here

[glad kestrel]

here

[safe yew]

Hold up

I think I understand what u mean

If I square all that

It became

(3 root (x^2 -2x) + (2x-2))^2

[safe yew]

And I square them both like I did here

Is that what u mean ?

[glad kestrel]

that's what i'm telling you is incorrect to do

because exponents are not distributive

[safe yew]

I see

Never noticed that

[glad kestrel]

$(2 + 3)^5\neq 2^5+3^5$

[wraith dagger]

a disappointing son

[safe yew]

Just question how old r u if I may ask ?

[glad kestrel]

18

[safe yew]

Ok not much older

So

When I’m studying math there is always that question where I feel I solve it right but someone else do it other way and get correct answers and me wrong xd

So in exams I never do that

Can’t check for right answers

And never know if my way is right or wrong

How I can solve this problem ?

[glad kestrel]

you can't

we all make mistakes

you'll just have to learn from them

[safe yew]

🥲

Ok thx u

.close

[cedar kiln]

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[grim lake]

x belongs to N, y belongs to N and 3x – y = 12}.```
Find the domain and range of the relation R

[grim lake]

its answer is 0 right

i mean

domain = {}
range = {}

no

how to ddo this qustion help

i just saw that if x is 5 we can take y as 3 and it can go till infinity

[cedar kiln]

@grim lake Has your question been resolved?

[cedar kiln]

@grim lake Has your question been resolved?

[grim lake]

<@&286206848099549185>

[grim lake]

fucking

.close

[cedar kiln]

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[grim lake]

useless

[cedar kiln]

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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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[blazing pecan]

So anyway, I have this optimization problem. We have a pie with dimensions of 11cm radius 2.5 cm height inside of a cardboard box with dimensions 24x24x4. We want to optimize the box using derivatives so that we get the most use of the cardboard. So the plan is to turn the box into a cylinder to fit into the pie and not waste any excess material.

I want to find out how much to reduce in this box, is my solution of getting the volume of the box and then the volume of the pie inside and the subtracting the two?

After getting the volume,of the remaining space in the box, i converted it into percent and got the new volume. Is this even correct?

[blazing pecan]

I actually had two people help but then the channel closed...

And i thought after getting the dimensions of the box,I found that now the pie wont event fit,so I turned the volume of the box and converted it into the volume of a cylinder with around the exact height I wanted.

[cedar kiln]

@blazing pecan Has your question been resolved?

[blazing pecan]

<@&286206848099549185>

[blazing pecan]

Can someone at least help with the question on the bottom right?

I'm not really sure what to do to minimize there..

[glass blade]

It's total surface area to me

[blazing pecan]

I am aware that the height requirement should be of 3,and according to this we should have volume of 200

but how do I found length and width?

[glass blade]

It should be reduced by 25% of its volume
So it means V - 25%V

[blazing pecan]

so we have to reduce 25% of 200?

Ok so that means its 150 instead?

150=l x w x 3

so now,how do I find the remaining dimensions?

math really does suck..

[glass blade]

Hmm so lw becomes 50

So area becomes A = 50 + 3l + 3w

[blazing pecan]

wait bro,how did u figure out it was 50

[glass blade]

.

3lw = 150
Implies lw = 50

[blazing pecan]

what if the length and width are different?

OH WAIT OK

[glass blade]

Ok now w+2x = 9
w = 9-2x
And l=24-2x

[blazing pecan]

thats l and w

[glass blade]

Yes

[blazing pecan]

ok these ones,how did u figure that out

[toxic hatch]

How can we send our doubts here?

[glass blade]

You need to go to an empty channel and ask for help currently it's occupied haha

[glass blade]

Oh that one yes

[blazing pecan]

u got this bro,gl

[toxic hatch]

Thank You :)

[blazing pecan]

OHH OK I SEE IT

its in the slide bottom left

[glass blade]

Like this

Oh wait length was 24cm

It will be 24-2x

But yes the idea is the same

Because the box is same

[blazing pecan]

so now do i just substitute the values of x into the remaining equations?

[glass blade]

So the original length and width of the box is retained
So 2x+ l = 24 => l = 24-2x
And similarly width = 9-2x

[glass blade]

Yes

[blazing pecan]

the x is 3 right?

because that is what the desired height is?

[glass blade]

Yeah

[blazing pecan]

so l is 18
w is 3

correct?

but 18x3x3=162, which is not equal to 150

[glass blade]

It will be (24-2x)(9-2x) = 50

[blazing pecan]

ohhhh

ok im stupid

[blazing pecan]

but this equals 54 is it not? if we have x=3

[glass blade]

Actually it was lw= 50
I just put the value of l and w

Still idk if we put the value of x why isn't it satisfying

[blazing pecan]

maybe because it has a new volume?

[glass blade]

Nah that shouldn't be an issue lol

[blazing pecan]

i hate that the teacher left it as a practice exercise,

this example is exactly what i need for my project

what are we missing?

[glass blade]

I am checking from my side

Haha

Oh hmm

It's matching if x = 2

[cedar kiln]

@blazing pecan Has your question been resolved?

[glass blade]

Nah I can't match the answer of this question

Smh

[blazing pecan]

It's ok,guess I just have to find another solution

The lesson is about optimization and derivatives if that helps

[glass blade]

Tag helpers one more time

[glass blade]

I think someone can help you

Ofc mention the question as well

[blazing pecan]

Oh ok

<@&286206848099549185>

Bottom right problem

[cedar kiln]

@blazing pecan Has your question been resolved?

[blazing pecan]

<@&286206848099549185>

[glass blade]

[cedar kiln]

@blazing pecan Has your question been resolved?

[blazing pecan]

rip

nobody can tacklemy hard ass proble

[warped oxide]

it's too difficult to read the problem

[cedar kiln]

@blazing pecan Has your question been resolved?

[upper abyss]

"Suppose the same box is then to be reduced by 25% of its volume but with depth 3cm what are the ideal dimensions"
Can you interpret this for me @blazing pecan ?

Oh wait, I think I understand it.

Let the other two dimensions be x and y. The volume is some known value V

Write the volume and surface area functions, then use them to express the surface area with only one unknown. Differentiate that, set to 0.

[blazing pecan]

Ok let me try that

[worldly chasm]

@blazing pecan are you still having trouble? If so, can you show me your work so far?

[cedar kiln]

Channel closed

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• 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>.
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• Be polite and have a nice day!

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[proper briar]

this one is 10

[proper briar]

mean = average

[upper garnet]

do u know how to find mean

[proper briar]

yeah, but average is already found

I got d)

[upper garnet]

(5+1+4+x+6+7+9)/7=6

[proper briar]

yep\

i sol3ed thatn and got 6

10*

[upper garnet]

,w (32+x)/7=6

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Grubert#2374

[upper garnet]

ys it u

[proper briar]

nice

[upper garnet]

yes it is di

[crimson sedge]

helpage

[shut reef]

This is luigis help

[upper garnet]

ger your own channel

[proper briar]

make ur own channel

[upper garnet]

yeah man tell him

[proper briar]

this one is 1/2?

D)

[upper garnet]

,w 1/(2^5)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Grubert#2374

[upper garnet]

Let’s first look at what this is asking us. The key phrase is ‘at least’. In order to solve at least problems we know that this is actually saying P(1 tail, 2 tails, 3 tails… 5 tails). In other words P(T >= 1). This is the same as 1 - P(T = 0).

[proper briar]

what does P and t stand for

[shut reef]

Probability and tails

I believe

[upper garnet]

proobability

[proper briar]

so that means that there's a 3% chance/

[shut reef]

no

how would your chances of getting atleast 1 on tails with more throws go down

[proper briar]

it won't

i guess

so then it's 97%?

[upper garnet]

jahahahhaahahahahhah

[shut reef]

There is a 3% chance of hitting all 5

[upper garnet]

forgot to say

[proper briar]

oh

[upper garnet]

100-(1/(2^5)

[proper briar]

i don't get it

how'd you make the formula P(T>= 1)

[upper garnet]

a coin has two sides, one flip is either 50 chance head or tail

[proper briar]

yes

[upper garnet]

flipping 5 times

how many

outcomes will there be

[proper briar]

50/2 = 25
25/2 = 12.5
12.5/2 = 6.25
=3.125

wait

can you do 50/0,5

[upper garnet]

what did u do

[proper briar]

nvm

[upper garnet]

its just 32 outcomes

2^5

[proper briar]

my brain

look

you have a 50% change to get tails every flip

that's 1/2

so then 50% get's halfed every time

that's sounded wrong

mb

[shut reef]

yes but thats if you had to hit tails each time

[proper briar]

what

you have 50% change to hit either side

[shut reef]

yes

[proper briar]

wait

[upper garnet]

If you see the complement of at least 1 tails means 0 tails or 5 heads, then we must surely see 1 outcome, e.g. all heads. Then it must follow at least one tails has 31 outcomes,e.g.32–1=31. To find the probability will be just 31/32. or 1-(1/32).

[proper briar]

can you explain what complement means

google isn't really helpful

[proper briar]

so in every situation where tails doesn't occur there are 5 heads that do occur right?

[shut reef]

yes

[upper garnet]

yes if in our dimension

[proper briar]

so now how does atleast one tail have 31 outcome? where'd you get 32 from and 31

[shut reef]

there is only 1 outcome where all are heads

32-1 is31

tadaa

[proper briar]

so how'd you get 32?

[upper garnet]

total outcomes

[shut reef]

2^5

[proper briar]

why 2^5 though?

[shut reef]

5 flips

[proper briar]

5 flips

and where is the 2 from?

[upper garnet]

two outcomes for one fap

flip

[shut reef]

1/2

yes

[upper garnet]

lets say u can only answer

yes,no,fuck u

and there are 35 questions

it would be

3^35

[rustic siren]

yes, no, maybe *

[upper garnet]

same thing

[shut reef]

4^35 actually cause you can always noy answer

[upper garnet]

u have to answer one

or else ur sent to gulag

[shut reef]

I'd like thay

that

[rustic siren]

hey

[proper briar]

either relapse or continue the streak

[upper garnet]

?

[proper briar]

nothing

wait

1 sx

bro

[shut reef]

My guy more focused on a typo than math

[proper briar]

so

there is 5 situations that you get

[proper briar]

so there are 2 sides that's why 2 right?

[shut reef]

yes

well it could land on its side but thats so small of a chance that it wont matter really

[proper briar]

okay now i don't understand why 2^5

[shut reef]

so

you have 5 throws and you have a 50% chance of heads right

[proper briar]

yes

[shut reef]

$/frac{1}{2}/frac{1}{2}/frac{1}{2}/frac{1}{2}/frac{1}{2}$

[wraith dagger]

.,..

[shut reef]

damn that failed

but

[upper garnet]

create a tree

[shut reef]

1/21/21/21/21/2

[upper abyss]

🌲

[upper garnet]

this gfice taif

this fie times

give

five

[proper briar]

there's 6 outcomes

[upper garnet]

?

[proper briar]

okay 2^5

then why just -1?

[shut reef]

what

so you understand why it is 2^5?

[proper briar]

no

[shut reef]

ok so

[proper briar]

okay

you flip once

you have a 50% chance

[wraith dagger]

.,..

[shut reef]

hes

yes

[proper briar]

hm

so then 1/32

[shut reef]

does that help

yes

[proper briar]

so 2 is the amount of possible outcomse

outcomes

[shut reef]

32

[proper briar]

and then the 5^ is the times you do it

[shut reef]

yes

so in total 32

[proper briar]

yes

then why -1?

why not -16

[shut reef]

there is a 1/32 probability of hitting all heads and in all of the other outcomes there is atleast 1 tail

[proper briar]

in all of the other outcomes?

in the other outcomes there could be 2 tails that doesn't make sense

[shut reef]

yes but the question is 1 or more

[proper briar]

so that's why it's -1?

[upper garnet]

sexy tree

[proper briar]

i don't understand the numbers

[rustic siren]

there are only one row of flips which doesnt contain a tail

what numbers

on the side?

[proper briar]

on the side and the ones below

[upper garnet]

each number presents outcomes

first one 2^1 second 2^2 third 2^3

[rustic siren]

its the possible outcomes after each flip

[proper briar]

that part makes sense

what about hte ones down

[rustic siren]

down its the bad/good outcomes

[proper briar]

why is one bad and the rest good?

it'

it's a serious question

[upper garnet]

don't worry coin sides don't have feelings

[shut reef]

$P(T\geq1)=\frac{32}{32}-\frac{1}{32}=\frac{31}{32}$

[wraith dagger]

.,..

[proper briar]

p= probability
t= tails?

[shut reef]

yes

[proper briar]

okay lemme think for a second

so
50%x5>=1?

[shut reef]

what

thats the solution not a equation you have to solve

[upper garnet]

=1 just means atleast once

[rustic siren]

P(we have at least one tail)= ...

[proper briar]

okay let's skip this one

cause i'm not relaly getting it

and i have 49 mins left so i'd be smarter to go on other exercises

[shut reef]

I see

[upper garnet]

P(T=3) means prob of 3 tails, P(T<2) means prob of tails less than 2 P(3<=T<=10) means prob between 3 and 10

[proper briar]

so then this means probability of tails is less than 1

?

[shut reef]

<less than

more than
<=less or the same than
= more or the same than

[proper briar]

yeahi know the signs

so less or same as 1?

[rustic siren]

so its a test?

[shut reef]

No the test is tomorrow

[proper briar]

the test is tommorow

io have 49 more mins to study

[shut reef]

he would have failed on time already

[proper briar]

that's against rules right?

[shut reef]

yes

[proper briar]

okay

let's continue on other ones

fuck probability

[shut reef]

ok

[proper briar]

P=? in cilinder

[upper garnet]

lol

[shut reef]

You will have to figure it out during the test then and I dont have much hope of that happening

[upper garnet]

just

Google

[proper briar]

it's not guaranteed that i will get it on the test

[rustic siren]

me neither

[upper garnet]

volume of cilinder

[proper briar]

oh volume

[rustic siren]

cuz this one is probably way more simple than the one in the test

[proper briar]

these are past entrance exams

[upper garnet]

entrance exams for what

[shut reef]

some school

[upper garnet]

university?

[proper briar]

it's a private school

high schol

not university lol

[upper garnet]

imagine loo

[shut reef]

yeaahhh

[upper garnet]

finding volume of a cilinder with enough info on an entrance test

[shut reef]

do you know how to find the volume

[upper garnet]

he does he deleted the question

[shut reef]

oh

[rustic siren]

so back to the probability

imma try one more time

[proper briar]

ok

[rustic siren]

so there are 32 outcomes right?

[proper briar]

yes

[shut reef]

Luigi what is the probability of all 5 landing on tails

[proper briar]

3%

somehow

[shut reef]

so what is the probability of all landing on heads

[rustic siren]

for example HHTHT means head head tail head tail, ok?

[proper briar]

3% somehow

[proper briar]

okay

[shut reef]

so what is the other 97% when all of them cant be heads

[rustic siren]

write down the question now from the exercise please

[proper briar]

1 sec

https://cdn.discordapp.com/attachments/903430334681608232/962419191082782810/unknown.png

[rustic siren]

read it out loud

[proper briar]

okay

done

[rustic siren]

we are asking how many of the 32 contains at least one T, right?

[proper briar]

yeah

waiyt

wait

wait

2^5 because 2 sides and 5 times

okay

got it

continue

[rustic siren]

so which of the outcomes doesnt suffice that sentence?

[proper briar]

pls

[proper briar]

1 = 100% which is impossible

[rustic siren]

where there are no tails, right?

[proper briar]

1/32 no way too low

[proper briar]

what?

okay it's 31/32

[shut reef]

So you understand why that is

[proper briar]

because there are high chances that a coin gets on head atleast once if you throw it 5 times

but that's just logic

nothing else

[shut reef]

Luigi

what will you study if you get in

[proper briar]

it's the same as school

but it's the best school in our city

so that's why i want to join

okay let's do this one next

[proper briar]

it's the best because the rest of the schools which are public suck

[rustic siren]

why?

[proper briar]

the teachers just give the class, if you don't understand something it's your fault and they won't help you

you must search it online or smthn

[shut reef]

Hate to break it to you but it's not looking good

hope it goes well tho

[proper briar]

i mean these are just the hard ones

i can do simple math

these are logic

[rustic siren]

not really

[proper briar]

okay let's do this one now

[rustic siren]

okay, do it

[shut reef]

solve E first

[proper briar]

okay

i'll try

i'll be back

[rustic siren]

E as in the table

[shut reef]

yes

[rustic siren]

not abcd E

[shut reef]

heah as in the table

[rustic siren]

the bike row

[shut reef]

only one that can be solved

and b too tbh

[proper briar]

i'll find c and E first

[shut reef]

Thats all you should need

oh wait lol no

[proper briar]

damn

it's relly simple

i hope atleast

E=5

c=7

then D = 9

now i need a and the other

[shut reef]

You only need a anymore

[proper briar]

why not?

it's still needed for the final answer

[shut reef]

do as you wish

[proper briar]

a= 45

[shut reef]

yes

[proper briar]

B= 48+45

is it correct?

E=5
c=7
D = 9
a= 45
b= 48+45= 93

[shut reef]

yes

[proper briar]

nice

finally

[shut reef]

Damn

the other one took you an hour and this a few minutes

its looking better

[proper briar]

lol

i'll try this one next

[shut reef]

okay

[proper briar]

71?

c)

[rustic siren]

why

[shut reef]

yeah why

[proper briar]

well

firsly

the one left to d is 121

because they're the same

you guys know what im talking about right?

[rustic siren]

yea

[proper briar]

okay

then

this part must be 180

then 180-121

is

59

then D=59

D has an oposite that is the same

so 59x2 = 118

no

it's 118

[shut reef]

what

what on earth did you multiply it for

[proper briar]

because there's an opposite

that is on the same line

so it must be the same

so i multiplied

[rustic siren]

oh no, you calculated all angles and then subtracted from 360?

[proper briar]

yes

exactly

(going to the restroom

[rustic siren]

well, its not really time efficient :DD

[shut reef]

assuming that the lines never intercect then it is 180-50-59

[shut reef]

yeah

[proper briar]

i mean 2 mins isn't bad

[rustic siren]

just say that CDJ is 121 too, and its opposite is 121, which is x+50-> x=71

[proper briar]

i'll have 2 hours to do the whole test

https://cdn.discordapp.com/attachments/903430334681608232/962435326910029874/unknown.png

[shut reef]

we have 200 mins/test

[rustic siren]

yeah, but if you take 5times the time on every problem, its gonna cost you

[shut reef]

What do you think it is

fast

[proper briar]

c)

oih

[shut reef]

correct

[proper briar]

there's a simpelr wayt

[shut reef]

next

[proper briar]

though

one sec

121-50

[shut reef]

doesnt matter if its right

[proper briar]

yeah true

okay i'll go next

[rustic siren]

i already told you the simpler way anyway

[proper briar]

i just figured it out lol

[rustic siren]

good job

[proper briar]

this one seems hard

i'll try

but it seems difficult

i got D)
but im not sure at all

it's dumb logic

i've got no idea how to do it

[proper briar]

help boys @shut reef

[rustic siren]

wdym

[proper briar]

i have 13 mins

i can't explain

can u tell me how i can solve it?

[rustic siren]

in 13 mins u write the test?

[proper briar]

no

13 mins

until i go to bed

[rustic siren]

your intuition seems good, but those are negative numbers

[rustic siren]

wow, strict time

[proper briar]

no it's cause i wanna wake up tommorow on time

and prepare a little more before the test

[proper briar]

so my answer is correct/

[rustic siren]

no

[proper briar]

lol

[shut reef]

You only need 4 hours of sleep to survive

[proper briar]

yeah but not for my brain to function correctly

[proper briar]

how did you solve it?

[rustic siren]

so we know that c is bigger than b which is bigger than a, right?

which is your answer

BUT

each one is negative

so if we thought oh 1<2<3

[proper briar]

it's the opposite?

[rustic siren]

it becomes -3<-2<-1

[proper briar]

makes sense

[proper briar]

but what is your logic for this?

[shut reef]

well a=3 b=5 and c=5 if b=2 so a<b<c

[proper briar]

https://cdn.discordapp.com/attachments/903430334681608232/962437677788049418/unknown.png

[rustic siren]

b/c = 2/5 so c is bigger than b

[shut reef]

and reverse

[proper briar]

okay

makes sense

i'll go on past exams

cause that's the last one

do they change anything?

[shut reef]

no

We were taught to use those is () were already in use in primary school and this seems like one of those

[shut reef]

Tf is that

[proper briar]

i don't know either

[proper briar]

do you know this one?

[shut reef]

so skip

[proper briar]

yep

i'll do this one

[upper garnet]

A+H=7

H+Y=A

A=7-H

[rustic siren]

yeah

[proper briar]

how?

[upper garnet]

H+Y=7-H

[livid hound]

A+H=7
not necessarily

[rustic siren]

even more simple

[upper garnet]

let's hope H+Y isn't >=10

[rustic siren]

we know from the middle that either A+H = 7 or A+H = 6

and only one solution gives us that

[proper briar]

https://cdn.discordapp.com/attachments/903430334681608232/962440847482306610/unknown.png

why 6?

[upper garnet]

if h+y>=10

u need to add one on left column

[rustic siren]

yea

[proper briar]

oh

k

then?

[proper briar]

which one?

okay i'll see tommorow

good night boys thanks for helping @rustic siren @upper garnet @shut reef @livid hound

.close

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[grim lake]

1. Are the formulas given above correct?
2. is
   tan(1/x) = 1/(tan x) = cotx?

I think they are just different

[rustic siren]

"tan(1/x) = 1/(tan x) = cotx? "

[grim lake]

yes is this correct

[rustic siren]

where do you see it?

[grim lake]

in this video

[rustic siren]

its not on the pic

[grim lake]

Its just inverse

but it is

[cedar kiln]

Channel closed

Closed due to the original message being deleted

[grim lake]

what no

[grim lake]

@cedar kiln

[glad kestrel]

tangent and inverse tangent are not the same function

your picture is correct... what you said is not

[rustic siren]

wdym just inverse XD

[grim lake]

ok

.close

.close

[glad kestrel]

it's already closed

[grim lake]

💀

ok, by

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[wraith beacon]

how can i figure out how many solution the equation: y^2 + 16 + 3y - 10(absolute value of y) = 0 has, the thing is it has an absolute value and im not sure how to handle it..

[wraith beacon]

i hypothesise it has only 1

well i need it to have 1 for my argument for a problem to work

i need to prove that somehow

[random shale]

Split the function for when y < 0 and y>=0

When y < 0 we know |y| = -y and y>=0 we have |y|=y

[wraith beacon]

yeah

[random shale]

You'll get two quadratics, solve both, just remember what range y is in

[wraith beacon]

OH

OHHH

bruhh

thank you

.close

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[upper stirrup]

Is there a website where I can like input an expression with a variable and it'll give me the value of the expression when I input the value of the variable

[upper abyss]

Try excel

That's pretty good for that kind of work

[upper stirrup]

oh

how do I do that?

it says I need to buy it or smth

[upper abyss]

Google has a free one

Check out "Google drive excel"

You likely already have some spreadsheet software on your computer though, try searching for office products?

[upper stirrup]

oh

I looked up a tutorial on google sheets

that works right

hm ok

first time I'm using sheets in like half a year

hm it says parse error

oh the distribution

ok got it to work

ty

[cedar kiln]

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[random relic]

hey guys i might need some help

[random relic]

working on a riddle where i need to coordinate a square from the origin, and im not getting further

[cosmic steppe]

you didnt even post the question

[random relic]

alr, sorry

one min

we calculated all the pins which is 96

this is the 4th chip

[cedar kiln]

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[grim lake]

is
tan2A = sin2A/cos2A?

[cedar kiln]

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[grim lake]

.reopen

[cedar kiln]

✅

[bold hinge]

Let B = 2A

Then does tanB = sinB/cosB

[cedar kiln]

@grim lake Has your question been resolved?

[cedar kiln]

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[elder walrus]

I was watching Primer's most recent video, then I decided to use the "game" and try and find an optimized way to play the game infinitely, and I think I messed up some probability somewhere in here
https://www.toptal.com/developers/hastebin/oxiqawixuc.py
I have been messing with it in python
am I being stupid or did I do the math wrong somewhere at the bottom?

[elder walrus]

the tree diagram would be something like this:

fair blob (.5)    heads (.5) ...
                  tails (.5) ...
cheater blob (.5) heads (.75)...
                  tails (.25) ...

btw don't tell me what it truly is, just tell me first off is all I have here accurate, and if anything is wrong what line it is on

I want to try to figure it out on my own just it feels off that .5333 + .4 < 1 rofl
(from lines 70 and 74)

if I think about it, given one each, the only options are fair or cheater, so it probably should add up exactly to 1?
but it doesn't?

<@&286206848099549185>

fyi, ** in python means to the power of, or ^

[cedar kiln]

@elder walrus Has your question been resolved?

[grim lake]

ooooooooooooooooooooooooo

ok

[elder walrus]

;-;

[reef venture]

let me go through your code one sec

[elder walrus]

you are a godsend

https://primerlearning.org/
if it provides more context if you get confused at all, this is the "game" it is in relation to

[reef venture]

no I watched the video yesterday so I am aware

[elder walrus]

hehe

btw don't tell me anything directly for now, just tell me first off if all I have here accurate, and if anything is wrong what line(s) it is on

and I'd like to focus on lines >=55

[reef venture]

what are you choosing randomly? the amount of flips?

[elder walrus]

no flips

just picking at random cheater or fair
or a pattern
whatever

guessing which it is with no info

[reef venture]

alright

[elder walrus]

and just to make sure, you saw this right?

[reef venture]

yes I did

[elder walrus]

good

[reef venture]

say after you flip once, are you still choosing randomly? or are you saying it is a cheater everytime you get a heads and fair if you get tails?

[elder walrus]

no, it says in line 50, though not the best wording

if you only do one flip and determined to do no more, you'd mark heads as cheater and tails as fair

[reef venture]

yes that's what I was asking alright

[elder walrus]

"or are you saying it is a cheater everytime you get a heads and fair if you get tails?"
somehow I didn't see this and read it right rofl

[reef venture]

until 55 its correct then

[elder walrus]

as in you have determined past it false, or that's all you've checked so far

[reef venture]

checked so far

let go through the rest, one sec

[elder walrus]

also btw where it brings up p_[words]
you'll have to scroll up to where the variables are to check

[reef venture]

yeah ok

[elder walrus]

p ofc meaning probability

[reef venture]

i get it

[elder walrus]

👍 just wanted to clarify because I don't always write the most legible code, though for some reason I did label like 90% of the variables at the top for this project

[reef venture]

Hmm everything does look correct to me but like you said P(tails given 1 each) + P(heads given 1 each) should add up to 1

[elder walrus]

rofl

fyi anyone that sees this I think I need additional assistance

[elder walrus]

any ideas as to potential culprits?

wait
maybe the fact in this situation, B on the bottom doesn't equal B on the top?

but like I don't see how it could be done differently

[reef venture]

I don't think that is the problem

[elder walrus]

but if I made B on the bottom equal B on the top, then it would be conditional to A which is probably not right? so idk

[reef venture]

yeah It will cancel out and you'll just get P(A)

[elder walrus]

oh yk you're right

duh

yeah then ofc B on the bottom doesn't perfectly match B on the top

[reef venture]

wait

I see the problem

[elder walrus]

fyi, at the top where I say (two times), I just noticed that implies once is fair once is cheater
not what I meant
if you flip a coin 2 times, there are 4 possibilities
tt
th
ht
hh

I simply applied what is the best option to guesstimate to each set, being unsure about equal

[reef venture]

this is only true if A and B are independent

which here isn't the case, B is dependent on A

[elder walrus]

as in, B is the same if applied to A1 or A2

[reef venture]

yes

P(B | A) != P(B)

[elder walrus]

but we know exactly how they aren't independent

so what is the right way to do it

also probably everything past line 50 is wrong then

cuz I think I did it on the same basis but I could be wrong

I didn't write it the same, but that's because I didn't bring up a wikipedia article to formalize it in my head

[reef venture]

No clue 💀

the only other thing in my mind is P(A) + P(B) - P(A U B)

but we don't know P(A U B)

[elder walrus]

yes we do

we know all of it

[reef venture]

P(cheater or 1 each)?

[elder walrus]

or wait

or yeah

lets simplify it and think of it on the 1 flip case

P(cheater or heads)

wait how do you calculate U

looking on wikipedia to remind myself

[reef venture]

using and 💀

wait I think we can use the bayes theorem here

[elder walrus]

this is what you're thinking of? we know all parts
there is nothing to solve for rofl

or wait

what we are solving for is P(A)

[reef venture]

P(cheater given 1 each) = (P(1 each given cheater) [which we know] + P(cheater))/P(1 each)

[elder walrus]

oh yeah what we're solving is A|B

[elder walrus]

this works regardless if independant?

[reef venture]

nvm what I said

[elder walrus]

is this true?

[reef venture]

I think P(A) and P(B) still need to be independent

💀

yeah that wont work too

Yeah I don't think I can help much, hopefully someone else might be able to

[elder walrus]

P(AnB)=P(A)*P(A|B)?

if this works with dependent events, then you can move it around to mean P(A|B)=P(AnB)/P(A)
which is different from the other formula

but a very odd difference

[elder walrus]

https://stats.stackexchange.com/questions/237103/how-do-you-find-the-intersection-of-two-dependent-events-when-you-dont-have-the
screenie from here

it feels wrong because I doubt A or B can just be interchangeable

[elder walrus]

you swap P(B) with P(A) at the bottom

@reef venture are you still here?

[reef venture]

Yes I am

Actually I think what you did is correct, maybe P(A|C) + P(B|C) doesn't need to add up to one always

[elder walrus]

but you literally said the equation doesn't work when dependent

[reef venture]

Yes what I said here is true

but you aren't actually doing
P(cheater)*P(1 each) you are doing P(cheater)*P(1 each given cheater) which is P(cheater and 1 each)

[elder walrus]

I think this is relevant, the top looks dependant

[reef venture]

I don't think they need to add up to 1

[elder walrus]

this is clearly saying Bayes works when dependen right?

[reef venture]

we can't tell if being suspicious is dependent on being an assassin or not

[elder walrus]

this is just wrong though
if there is a probability of being sad or happy given sick, then it would total to 1

If P(A|B) = P(A), then events A and B are said to be independent:

ok
so probability of assassin given suspicious is 3/5
probability of being an assassin is 4/12

probability of suspicious given assasin is 3/4
probability of suspicious is 5/12

it is dependant

I will use bayes formula and try it

adds up the same

[elder walrus]

false

but it looks like it might give same results and you are somehow right about before

it gave the same results

[elder walrus]

I am confident this must be right then because wikipedia says so

well then

so fair is only slightly better then

ok time to figure out how to calculate it for n flips and such

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[fervent notch]

Hey, im stuck trying to prove pq is idempotent in Z(2pq) if p and q are odd

[full karma]

Yo

[fervent notch]

ello

[cedar kiln]

@fervent notch Has your question been resolved?

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[thin roost]

Hi

[thin roost]

I need to find dy/dx of x+y=siny

How do I do it

[wraith dagger]

! बाशुदेव

! बाशुदेव

! बाशुदेव

! बाशुदेव

! बाशुदेव

[thin roost]

Is this correct??

[exotic furnace]

Recheck your derivative of sin y with respect to x

[thin roost]

Ye

??

Isn't this correct

??

Oh shit

It should be

'y' instead of 'x'

That's the mistake right??

[exotic furnace]

I believe that is all

[thin roost]

Is the answer correct??

[exotic furnace]

Show me your final answer

Revised

[thin roost]

Ok

$\frac{1}{cos(y)-1}=\frac{dy}{dx}$

[wraith dagger]

! बाशुदेव

[exotic furnace]

There we go

[thin roost]

Ok

.close

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[tough osprey]

Let $A,B$ be sets. Suppose $A\subset A\cap B$.

Suppose $x\in A$.

By the definition of the subset relation, $x\in A\cap B$.

By the definition of the union, $x\in A$ and $x\in B$.

[wraith dagger]

Trenton

[tough osprey]

I am trying to do part c ii

I stuck at how to relate $A\setminus B$. Intuitively, I think $A\setminus B=\emptyset \implies A=B$.

[wraith dagger]

Trenton

[tough osprey]

But idk how to proceed

[quartz vale]

sorry for interrupting but I've seen \ operator being used between sets and I just cant figure out what that refers to?

[tough osprey]

That is the set minus

You may call it the complement

[quartz vale]

ohk thank you, sorry again (was taught to use -)

[tough osprey]

Ya they are the same thing

[quartz vale]

If $x \in A \setminus B$ then $x \in A$ and $x \notin B$ but there exist not such elements since you have already proven that all elements in A are necessarily present in B, so since there are zero such elements $A \setminus B = \emptyset$

[tough osprey]

If you want to type ||||, you need the ||\setminus|| thing

[wraith dagger]

kinglacto

[chilly lantern]

@tough osprey would probably be better to use definition of intersection here

[tough osprey]

Let $A,B$ be sets. Suppose $A\subset A\cap B$.

Suppose $x\in A$.

By the definition of the subset relation, $x\in A\cap B$.

By the definition of the intersection, $x\in A$ and $x\in B$.

[wraith dagger]

Trenton

[tough osprey]

I made a mistake in my previous message

[chilly lantern]

yep that's right

and then you can conclude that A\B=0 right away

[tough osprey]

Umm wait why?

[chilly lantern]

(unless you somehow forgot the definition of set complements?)

[tough osprey]

Oh

The complement of the set $B$ in the set $A$ is defined to be the set ${x|x\in A$ and $x\notin B}$.

[wraith dagger]

Trenton

[tough osprey]

Let $A,B$ be sets. Suppose $A\subset A\cap B$.

Suppose $x\in A$.

By the definition of the subset relation, $x\in A\cap B$.

By the definition of the intersection, $x\in A$ and $x\in B$.

Since for any $x\in A$, we have $x\in B$.

Hence there does not exist $x\in A$ such that $x\notin B$.

By the definition of complement, there does not exist $x\in A\setminus B$.

Hence $A\setminus B=\emptyset$.

[wraith dagger]

Trenton

[quartz vale]

yes

[tough osprey]

Ok thank you so much @quartz vale @chilly lantern

.close

[cedar kiln]

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[rare osprey]

this is the formula for area of triangle yes?

[tropic oxide]

as stated this only applies to isosceles triangles with θ as the apex angle

the bottom side must also be labeled as having length r for this to work

[rare osprey]

right

thanks ann