help-42

[glad sinew]

so this is a contradiction

alpha should be rational then

[upper sparrow]

Yep, that's the idea, fractional parts being equal leads to alpha being rational (but that can't be )

[glad sinew]

wait why would you even care if a is rational or irrational

[upper sparrow]

Because we're trying to show that the set we made with alpha irrational must be dense

[glad sinew]

ohh yea

gotta reread the question again

trying to understand the process that I forgot about what we are even trying to solve

[upper sparrow]

Hehe

The statement needn't be true if alpha were allowed to be rational (e.g. if you could have set alpha = 1/2, that wouldn't get you a dense set, it'd just get you all integer multiples of 1/2 really)

[glad sinew]

Well, off topic and yea I think this is too difficult for me at the moment. Im gonna jump back to smth easier

Im not quite sure what this means

Also, yea I think Im gonna give up on that question for now, to study other topics.. cause I really dont have much time

[upper sparrow]

Awww how long do you have again?

[glad sinew]

all these topics I gotta go through

I have two days

yeaaa.a......

its bad

[glad sinew]

why since 1/n is in C, eps is not a lower bound

All I remember for archimedes is that for eps>0, there exists 1/n<eps

I dont understand why I just remembered the words

[upper sparrow]

You found an element in your set that's smaller than your supposed lower bound

[upper sparrow]

Oh wow shiieetttt that's so soon

[upper sparrow]

Esp to cover those in that time

[glad sinew]

my fault, I spent alot of time before trying to finish a calc assignment. I could've done the assignment sooner and start studying for this sooner

now after this exam, Ive got another calc assignment due in two days time

I have bad time management

and Im lazy so I procrastinate

[upper sparrow]

Relatable, I've done very little these past few days despite having the time to i don't even know with me

[glad sinew]

I just wanna graduate, get a job and not be broke🥲

[upper sparrow]

Awwww

[calm coral]

@glad sinew Has your question been resolved?

[glad sinew]

im still hereee

[leaden thunder]

This thread got off topic and hard to find your actual question and current understanding. Do you mind just opening a new one with your progress

[glad sinew]

sure

Im working on a question rn

but I wanna try it out first

play around with it

this is the question

Actually yea I'll open up new thread

.close

[calm coral]

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[median cloak]

how to solve it

let's discuss

fk

[old falcon]

you might think of a lower case

its easier to draw

[median cloak]

what is a lower case

[old falcon]

how about we do like

[median cloak]

wdym

[old falcon]

find the probability that it takes zach 3 or more throws

I would think of a tree

[median cloak]

how many points he scores within these 3 throws

it seems like he will play this fking game forever, and that's what have been making me confused

[old falcon]

sorry i was drawing

that says throw 1 and throw 2 and throw 3, i just cant write

the R means Ringer

does the layout make sense?

[old falcon]

he might, but it ultimately wont matter

well use the complement to only consider a handful of cases

[median cloak]

it does

[old falcon]

okay

[old falcon]

this is what we want to know

we could alternatively write: 1 - P(zach takes 2 or less throws)

because P(thing happens) = 1 - P(thing doesnt happen)

[median cloak]

the complement is that he cannot score any point in his first 4 rounds.

[old falcon]

in your problem, itd be that he does score a ringer in the first 4 throws

[median cloak]

So (0.6)^4

[old falcon]

no

so, lets say it in words

[median cloak]

ohhh

[old falcon]

Zach makes a ringer in the 5th (or later) round

this is the opposite of Zach makes a ringer in the first, second, third, or fourth round

now you just total these up

P(zach scores in the first round) = 0.4

P(zach scores in the second round) = 0.6 * 0.4

do you see the pattern? do you see why these numbers?

[old falcon]

hmm

[median cloak]

I just ran it over brutally, but i do not think it is a good way to solve it

[old falcon]

so itll be, i think,

[median cloak]

how does the complement works in this case

i didn't see it

[old falcon]

$\sum _{i=0}^3 (0.4)(0.6)^i$

[potent lotus]

jan Niku

[median cloak]

yes

binomial thereom would work perfectly in this one

[old falcon]

we want to know about the cases where it takes him 5 tries, or more

[median cloak]

as it is an "yes or no" situation

[old falcon]

so we look at cases where it takes him 4 times, or less

[median cloak]

to score his first point.

[old falcon]

we argue that these cases together make up everything that can happen

yea

so P(it takes him 5 or more tries) + P(it takes him 4 or less tries) = 1

[median cloak]

I see

how we calculate "P(it takes him 4 or less tries)"

[old falcon]

then P(it takes him 5 or more tries) = 1 - P(it takes him 4 or less tries) just by subtraction

[median cloak]

it seems troublesome

tiredsome

[old falcon]

its not

so say hes on his first throw

tbh im trying to think of

i think they want you to use geometric its bugging me

[median cloak]

it is 0.4 for he scores at first round.

[median cloak]

"geometric" do you means graph it

[old falcon]

no, a geometric series

[median cloak]

ohhh

[median cloak]

I didn't see that it is related to geometric series

[old falcon]

because P(he scores on the ith round) is (0.4)*(0.6)^(i-1)

[old falcon]

right, and if he doesnt, then probability 0.6 that he tries again

[median cloak]

ohhh case is getting systemetric and general

[median cloak]

I suppose that is "P(it takes him 4 or less tries)"?

I had made some mistake fk

[old falcon]

yea you have too many terms

there should only be 4

[median cloak]

The Mickey should be turn into 2 as a coefficient

[median cloak]

I do not think so?

the 0.4 stands for the Prob that he scores in first round

and (0.4)^2 for he scores in first and second

so on (0.4)^3

[old falcon]

this doesnt really make sense

so

if he scored in the first round

itd be probability 0.4

but if he fails, this has 0.6 probability

if he scores in the following round, he must fail in the first

its implied he stops if he scores

so you should only ever have one factor of 0.4

if he scores in the ith round, he will have failed (i-1) times

[median cloak]

ohhh once he score then the game is over?

[old falcon]

yea

[median cloak]

I didn't know that

is that it

[old falcon]

this is the complement, yea

[median cloak]

so 1-mickey = the answer

[old falcon]

i think so

[median cloak]

Well

what if he would continue playing even he had score?

[old falcon]

he could, it wont matter, the problem says first

im gonna sleep

i let other helpers know youre here, if they are around, if you have other questions

[median cloak]

good night

[old falcon]

@median cloak oo i found the sum way

idk if you are interested

[median cloak]

sum way?

[old falcon]

yea, with a series

[median cloak]

Im not sure what it is

show me

[old falcon]

so, would you agree if i said like

$P($zach scores on the ith round$) = 0.4 \cdot 0.6^{i-1}$

[potent lotus]

jan Niku

[old falcon]

0.4, for the 1 time hes successful

then i-1 many failures

by laws of probability,

$\sum _{i=1}^\infty P($zach scores on the ith round$) = 1$

[potent lotus]

jan Niku

[old falcon]

$\sum {i=1}^4 P($zach scores on the ith round$) + \sum{i=5}^\infty P($zach scores on the ith round$) = 1$

[potent lotus]

jan Niku

[old falcon]

we can use this, now

$\sum _{i=1}^4 0.4(0.6)^{i-1} + \sum _{i=5}^\infty 0.4(0.6)^{i-1} = 1$

[potent lotus]

jan Niku

[old falcon]

the second sum is geometric

well, they both are

which formula you remember is the one you can use

i always remember $\sum _{i=0} ^\infty b^i = \frac{1}{1-b}$

[potent lotus]

jan Niku

[median cloak]

I see

[old falcon]

we just need to reindex a little

[median cloak]

but this questions seems a little boring?

we have to change the question, so the situation is more complex

[old falcon]

idk i really do have to sleep

[median cloak]

what if these results of each throw are dependent.

oohhh

good night

[old falcon]

hrm

idk then

good luck

oh, if you need it

i think the answer is 0.1296

[median cloak]

wut

[median cloak]

ohh

never mind, I thought you was able to know the answer, after we making these events dependent, within a mins and some calculation in your mind.

[old falcon]

independence it very convenient dependent is harder

[median cloak]

yea, but it makes the question more enjoyable.

we love complex questions

[calm coral]

@median cloak Has your question been resolved?

[calm coral]

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[proud gull]

Find the number of words with or without meaning which can be made
using all the letters of the word AGAIN. If these words are written as in a dictionary,
what will be the 50th word?

[proud gull]

i got the first part as 5!/2!

but im not sure how to start on the second part of the question

wait nvm i think i got it , its just a bit tiresome

[blazing coyote]

You'll have to work it out by first ordering the words starting with A

[proud gull]

is it naaig

[proud gull]

yeahhh okie , noted

[blazing coyote]

,w rank of naaig

[potent lotus]

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[proud gull]

uhh

[blazing coyote]

,w 5!/2!

[potent lotus]

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[blazing coyote]

oh, just a mo

4! is the number of words starting with A

[proud gull]

yeah

24
then we start with g?

so like 4 !/2 !

[blazing coyote]

Yes

[proud gull]

24+12 = 36
then I also 4!/2!

so like 36+12 = 48

49th letter would be starting with N

i.e Naagi

so 50th would be naaig

??

is that cool or i missed something

[blazing coyote]

yes, I think that's right

[proud gull]

i actually dont have answer to this too , i had this question in my revision question sheet

ah cool thank you!!!

.close

[calm coral]

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[calm coral]

Please don't occupy multiple help channels.

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as this action is irreversible, and this channel may abruptly lock.

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[burnt pecan]

hey guys, I need help to confirm if my calculations are correct and the graph

[willow gyro]

(you might want to translate it, so more people can help)

[burnt pecan]

ok, then gimme a sec

[halcyon rock]

wow I have never seen this words

are you phd student?

[burnt pecan]

ahaha, no, it's another language

[hot creek]

the calculations are correct

[burnt pecan]

thanks,

another thing i need to know is, that they said I should also calculate the intersection of the lines, I calculated the first coordinate to be 3 (on the x-axis), which I can see is correct. But when I calculated for the second coordinate (on the y-axis), maple said it was 1, but when I look at it the point of intersection, it's -1 and not 1.

Can someone tell me why?

[tender raptor]

what is 3+3 please help me 😭

[hot creek]

!occupied

[calm coral]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[hot creek]

also stop trolling

[calm coral]

@burnt pecan Has your question been resolved?

[burnt pecan]

thanks for the help

.close

[calm coral]

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[outer stump]

how do i factorise this

[hollow totem]

well ordinarily you should combine them and then factor the result

it happens to work in this instance to factor each part separately as well but in general that's not reliable

[outer stump]

ok

well normaly we learnt that x^2 - y^2 = (x-y)*(x+y) but idk what is quare root of (18x^2 - 8)

[hollow totem]

it isn't a square

[outer stump]

or the quare root of (9x^2 + 12x + 4)

ok

[hollow totem]

this one does happen to be a square but it's not used the way you're thinking

you've seen (x+y)^2?

[outer stump]

yeah

this is what this one's about

[hollow totem]

do you know how to expand (x+y)^2

[outer stump]

yeah

x^2 + 2xy + y^2

[hollow totem]

right

that's what 9x^2+12x+4 is

it's in that form right now

[outer stump]

how come?

[hollow totem]

because the person who wrote the problem made it that way

[outer stump]

no, like what is x and y in this one

[hollow totem]

hint: the plus signs stay the same

so look at the first term, what squared equals that

[outer stump]

9x^2 = (3x)^2

12x idk

[hollow totem]

12x is the "2xy" term

[outer stump]

4 obviously (2)^2

[hollow totem]

so it's not a square

[outer stump]

OH

OHHHHHHH

[hollow totem]

let's say (a+b)^2 to make it easier

[outer stump]

ok

[hollow totem]

so this is a^2 then

[hollow totem]

this should be 2ab

[hollow totem]

and this b^2

[outer stump]

yea

thank u very much

what abt the second term

[hollow totem]

after you pull out a common factor it becomes a difference of squares

[outer stump]

idk what do u mean by that

[hollow totem]

18 and 8 are both divisible by 2

factor out the 2

[outer stump]

i half understand

[hot creek]

it's basically dividing and multiplying 2

do that

[outer stump]

9*2x

-4*2?

[hot creek]

pull the 2 out now

[outer stump]

2(9x-4)

OH

OHHH

[hot creek]

9x?

[outer stump]

oh ye

[hot creek]

you mean $9x^2$

[potent lotus]

斯韋裡

[outer stump]

yea

well now everything is falling into place

[hot creek]

it's correct then

[outer stump]

= (3x + 2)^2 - 2 (9x^2-4)

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

wait now im confused again

[hot creek]

wait

$2(9x^2-4) can be written as 2[(3x)^2 - (2)^2]$

[potent lotus]

斯韋裡

[outer stump]

yea?

[hot creek]

now it is in the form of $a^2-b^2$

[potent lotus]

斯韋裡

[hot creek]

do you know this identity?

[outer stump]

yeah

[hot creek]

apply it

[outer stump]

idk where to go from now

[hot creek]

where are you stuck now

[outer stump]

im here atm $(3x+2)^2-2[(3x)^2-2^2]$

[potent lotus]

chillmystery

[hot creek]

the identity is $a^2-b^2= (a+b)(a-b)$

[potent lotus]

斯韋裡

[hot creek]

use this

[hot creek]

just keep it there

okay just wait

$2(3x+2)(3x-2)$

[potent lotus]

斯韋裡

[hot creek]

it can be written like this

[outer stump]

how??

[hot creek]

i just applied this identity

[outer stump]

idk how u did it

[hot creek]

you can just write it that way

[outer stump]

im confused

[outer stump]

is this the second term only or both of them

wait x^2 + y^2 = (xy)^2???

[hot creek]

only the second term

[hot creek]

nope

[hot creek]

$x^2 * y^2 = (xy)^2$

[potent lotus]

斯韋裡

[outer stump]

oh ye

[hot creek]

the - was from the start

[outer stump]

oh i just realized

a^2 - b^2

i understand now how u can write it that way

[hot creek]

now take (3x+2) common

[outer stump]

so $(3x+2) * [(3x+2) * (3x-2) *2]$?

[potent lotus]

chillmystery

[hot creek]

absolutely wrong

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

[potent lotus]

斯韋裡

[hot creek]

now dividing (3x+2)

you get

$(3x+2) - 2(3x-2)$

[potent lotus]

斯韋裡

[outer stump]

im confused again

[hot creek]

just divide by this

[outer stump]

wdym devide

[hot creek]

okay since you can see (3x+2) in both the terms you can divide and get rid of it

[outer stump]

wdym devide

i dont see any fraction

man imma go i spent almost 1 hour on this problem and i have many others

!close

.close

[calm coral]

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[calm coral]

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[orchid summit]

Could anyone help me out with this chem question please?

[orchid summit]

I don't really know where to start because everything I do is different from the answer of 42.7 KJ/mol

nevermind I'm pretty sure the textbook is wrong

.close

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[jade mantle]

I need some help justifying this formula shows the the area of this cone and cylinder (without the top)
The volume of the cone is 40 dm3

[jade mantle]

i've come to far as to bind s(r) = 12,7324/(r^2 -0,6666667r)

and come to this result

But i don't see what i'm missing

[calm coral]

@jade mantle Has your question been resolved?

[dull wagon]

whats s

[jade mantle]

s is the height of the cylinder

[dull wagon]

are you given that info?

[jade mantle]

I'm not given the value of s
s is shown to be the height on the image i sent

[dull wagon]

because those formula you gave don't have s

[jade mantle]

I bound s to r, so i can find the area O(r)
As the total volume needs to be 40, i bound s to r with this function

[dull wagon]

oh, total is 40, not just the cone as you initially wrote?

[jade mantle]

My bad,
volume of cone and cylinder is 40 in total.

[dull wagon]

can you show the work for your result

[jade mantle]

Formula to find volume of the entire thing, and since V=40 the equation looks like this

To find the area of both, the formula is, where t is the side length of the cone

I bound t to r as such,
since height of cone is 2r and radius is just r

So to find surface area cylinder+cone i've reached this formula now

And s i have already bound to r, so i concluded this must be the function to find surface area of cylinder+cone

[dull wagon]

ideally you should work with exact values

and fractions

you shouldn't be using a calculator for this at any point

leave pi as pi, sqrt(sutff) as sqrt(stuff) etc

apart from un-ideal rounding,
your issue here is that you didn't apply pythag properly

you need (2r)^2, not 2r^2

[jade mantle]

lol
the fact that it's that, that i forget, out of this

There must be something i'm still missing. Even when i correct for pythagoras, the answer i'm getting is still off

[dull wagon]

missing () around the s when subbing

pain to find with all your rounding

[jade mantle]

oh, thanks

Sorry^^

.close

[calm coral]

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[tight compass]

What is the graph of function a^x if a > 1?

[next herald]

Can't you just use Desmos to figure that out

[calm coral]

@tight compass Has your question been resolved?

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[eager harness]

can someone please help me with number 2 😭

[eager harness]

<@&286206848099549185>

[calm coral]

@eager harness Has your question been resolved?

[calm coral]

@eager harness Has your question been resolved?

[calm coral]

@eager harness Has your question been resolved?

[calm coral]

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[calm coral]

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[graceful pendant]

How can i add this radical expression?

5√3　+　3√2　-　2√3

[remote mural]

,, a\3c+b\3c = (a+b)\3c

[remote mural]

you can use this here

for example, you can start by associating what your a,b, and c's are in your expression

[graceful pendant]

Can i leave the √2√5 together?

Or i need to multiply

[remote mural]

where did √2√5 come from tho

you have [
5\33+3\32+2\33
]
right?

[graceful pendant]

Ill show you picture instead

[graceful pendant]

Yess

[remote mural]

sadly that does not work

[graceful pendant]

Its taking long to send sorry

[remote mural]

so ill keep it easy for you: you can only do this if the square roots have the same number inside

like

[graceful pendant]

Herr

E

[remote mural]

Yes

i see it

[graceful pendant]

Do i have to multiply the two radicals

[remote mural]

,align
2\33 + 4\33 &= 6\33\
2\33+4\34 &\c r \ne 6\33\34

[remote mural]

no

you just evaluate $5\33 - 2\33$

[graceful pendant]

What happens to 3√2

[remote mural]

leave it as is

you cant do anything to it really

[graceful pendant]

Okay

Thank you

[graceful pendant]

So it is just cancelled out

[remote mural]

no

it isnt

it just stays there

[graceful pendant]

Do I still have a single radical with it

[remote mural]

wdym

[graceful pendant]

It says i have to combine into single radical

Sorry I have english bad

[remote mural]

i dont think thats possible

are you sure u copied the problem right?

[graceful pendant]

Yes

5 ✓3 + 3✓2 - 2✓3

[remote mural]

there really is no simpler term with only a singular radical

that i can find

[graceful pendant]

I had tried cancelling ✓3 and I got 6✓2 from it

[calm coral]

@graceful pendant Has your question been resolved?

[calm coral]

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[teal radish]

hi

[calm coral]

Please don't occupy multiple help channels.

[teal radish]

kk

[lunar meteor]

please don't ask puzzels here

help channels are to solve math issues or problomes

[calm coral]

@teal radish Has your question been resolved?

[calm coral]

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[lusty merlin]

Couldn't understand the right part of it

[calm coral]

@lusty merlin Has your question been resolved?

[potent lotus]

spq_64_t

[lusty merlin]

@potent lotus

I understand

But I have a little doubt

[lusty merlin]

Why is it mentioned here j€J

[olive fox]

Because {I_j} is indexed by J?

[lusty merlin]

Wait

There is a small j just the end of the equation in the original image

Is that mis print?

[olive fox]

I don't see it, unless you mean the semicolon

[lusty merlin]

Shit

I was considering it j

[olive fox]

RIght, I mean it's weird to have a semicolon at the end of the formula

Guess the book is old enough

[lusty merlin]

Ya this halmos's set theory

@olive fox thankyou

[olive fox]

np

[calm coral]

@lusty merlin Has your question been resolved?

[calm coral]

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[jagged lichen]

.help

[calm coral]

Commands:

• clopen: .close, .reopen
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• factoids: .tag
• help: .help
• version: .version

Type .help <command name> for more info on a command.

[jagged lichen]

This is wack

and I don't know if I should put 110 for it because 112 + 70 = >180

[dull wagon]

diagrams aren't drawn to scale

pay close attention to what the question says and which sides are actually parallel

[calm coral]

@jagged lichen Has your question been resolved?

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[inner zealot]

Set up an equation(relationship) between each person.

[thin yew]

n = 3s

n = a+7

holy shit can someone actually help me

[inner zealot]

No need to use profanity.

[amber bolt]

3s + s + (3s−7) = 70

[thin yew]

finally

thank u

[thin yew]

wait why 3s-7

[inner zealot]

And this is why you don't just give people the answer. Help them figure it out. 🤦‍♂️

[thin yew]

at least he gave me something

💀

what did you give me

a little lecture?

not helping very much

.close due to horrible helpers

[calm coral]

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[ancient plank]

how do i solve this :

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

y'(4) = 24

*its not the letter 'u' *

[calm coral]

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[pine schooner]

.close

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[uncut lava]

I need help with combinatorics.
You have a set of 5 numbers {2,3,5,7,8}
a) How many numbers of 4 digits without repeating are you able to form?
b) With the result of a. How many of them are even and how many are odd numbers?

I'm stuck at B.

[wet path]

separate into two cases: case 1 where the last digit must be either 2 or 8 and case 2 last digit must vs either of 3, 5, or 7

to verify, these two cases must total into (a)

btw just to check whats youe answer in (a)?

[uncut lava]

5P4 = 120

[wet path]

yeah thats correct

there's another way to think about it

_ _ _ _
you have 4 digits

there are 5 possibilities for the first digit, and after filling that in there are only 4 remaining for the second digit, etc...

the way we get 120 is from 5*4*3*2

[uncut lava]

ohhh so when i put the 2 at the end, there's 4 digits instead of 5

[uncut lava]

i think i get it

i will try it, thank you

so it would be 4P3 + 4P3

right?

[wet path]

the key here is to simply start with the ones digit

[wet path]

nope that only adds up to 48

wait

[uncut lava]

oh that's right you told me that it has to sum up 120

[wet path]

yeah yeah

[wet path]

try this way of thinking, i think it makes it easier

[uncut lava]

yes but i have to do it with my calculator

it's a requirement of my professor

i think it would be... 5+4+3 for both cases

but that won't sum up to 120

oh i get it

i don't have to think with 4 digits

i have to think with 5 because 2 can still be a part of the combination

so

5P3+5P3

and it gave me 60

that would be it, right?

no that can't be

because 2 shouldn't be repeating

ok i understood now

the sum of ALL combinations (even or odd) has to sum up to 120

[uncut lava]

so this would be ok but i have to do it with all 5 (2 even, 3 odd)

24x5=120

is that it?

48 even combinations, 72 odd combinations

[wet path]

that is technically correct but im not sure your prof will accept that solution 100%

[uncut lava]

why?

i have to use the formula?

[wet path]

well i dont know about your prof's standards

it reasons out well intuitively

but it's not very rigorous

[uncut lava]

i think that he isn't being rigorous because we are just starting with combinatorics

[wet path]

ah then you can just make it clear how you arrived it at that i guess

[uncut lava]

but, what is the solution that you think i should arrive at?

i would like to understand this perfectly if possible

[wet path]

well i would've done it intuitively too using the argument i proposed above

[wet path]

.

here instead of starting to fill in with the 1st digit, we should start with the last digit

because that will decide evenness/oddness of the number

so in the case of even theres 2 choices for the ones digit

and then the rest of the digits is 4 3 2

that'll be 48 even

same with odd except you start with 3 choices in the last digit, which gives 72 odd

[uncut lava]

factorial?

[wet path]

doing 2*4P3 and 3*4P3 might work but again idk what your profs standards are

[wet path]

yeah that's basically how you arrive at the P formula

[uncut lava]

so, you are saying that instead of 4P3x2 i should be doing 4x3x2 + 4x3x2

?

i have another part to the problem that i can't solve either

c) how many of those numbers are minor than 5000?

oh i guess i would have to use 2 and 3 as the first digits because 5 7 and 8 would give a number major than 5k

ok, i think that my question was resolved. thank you very much hskthca

.close

[calm coral]

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[calm coral]

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[pure berry]

Hello. I don't understand why if the diagonals of a quadrilateral are equal, it must be no particular shape and could be an isosceles trapezoid, rectangle, or a square.

In addition, if a quadrilateral has one right angle, it must be no particular shape and could be a quadrilateral. Why is the answer "could be a quadrilateral" when it says that the shape is a quadrilateral in the question?

Thanks for your help.

[calm coral]

@pure berry Has your question been resolved?

[pure berry]

<@&286206848099549185>

[leaden thunder]

Yes all those shapes you mentioned have equal length diagonals

But there are other quadrilaterals without equal diagonals

[pure berry]

Why isn't the answer must be an isoceles trapezoid and could be a rectangle or square?

Because don't both rectangles and squares fit isoceles trapezoid's definition

If the diagonals of a quadrilateral are equal, doesn't it have to be an isoceles trapezoid at the very least then?

[leaden thunder]

No

Do you consider squares and rectangles isosceles trapezoids?

Oh I guess it's not universal
https://en.m.wikipedia.org/wiki/Isosceles_trapezoid

[leaden thunder]

Depends on your teacher's convention then

[pure berry]

My teacher does

I think

Because according the textbook we use, quadrilaterals → trapezoids → (I believe isoceles trapezoid should be here) → parallelogram

[pure berry]

So, according to the convention I am using, it should be:

Must be: isoceles trapezoid
Could be: squares + rectangles

[calm coral]

@pure berry Has your question been resolved?

[leaden thunder]

I don't understand what you're trying to convey with these arrows

[leaden thunder]

Think?

You should clarify with them

[pure berry]

I meant that I am positive

Sorry

[pure berry]

Let's use circles

So quadrilaterials is the big group

And the trapezoids fall into the category of quadrilaterals

And then isoceles trapezoids fall in the category of trapezoids

And then parallelograms fall into the category of isoceles trapezoids

And etc.

<@&286206848099549185>

[calm coral]

@pure berry Has your question been resolved?

[calm coral]

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[vernal bramble]

Hello, I'm trying to factorise a quadratic question and I'm unsure of what to do next. The example question is confusing me. I'll show the question I'm doing and the example question in just a moment

[pure kayak]

okie dokie

[vernal bramble]

Example questions

Question I'm on

[pure kayak]

,rccw

[bright lagoon]

[pure kayak]

you split it into -8 and 2

[vernal bramble]

Ye

[pure kayak]

but theyre supposed to multiply to 16, not -16

oops nvm, im blind

ignore that lol

so whats up

[bright lagoon]

?

-16, -6

[pure kayak]

i was looking at the top right

[bright lagoon]

Oh ok

[vernal bramble]

I'm stuck on the last step if you see the example questions

[pure kayak]

do you know what it is theyre doing in the last steps?

[bright lagoon]

I don't see any problem

[vernal bramble]

I believe their factorising but I'm not sure...

[pure kayak]

youre completely right

[vernal bramble]

Their isn't a problem

[bright lagoon]

Moreover, if you factored out the 2 at the start, this would just be the "Big X Method", and not the "AC Method"

[vernal bramble]

I'm stuck on the last step and don't know how to continue doing the question

[vernal bramble]

I'm not sure what u mean

[pure kayak]

imagine the x-4, for visual purposes is just P
then you have 2xP+2P

does that look easier to factorise

[vernal bramble]

One moment

So 2x + 2p + 2?

[pure kayak]

im not sure what happened there

[vernal bramble]

Yeah that doesn't make sense nvm

[pure kayak]

maybe P was a bit of an odd choice
2xy+2y, is probably more standard looking

[vernal bramble]

Is x times or x?

[pure kayak]

the variable

i use * for multiplication

[vernal bramble]

Kk

Wait

Oh so I just literally times 2x and 2 by the brackets in front of them?

Wait no

[pure kayak]

more that youre just factoring out the entire bracket

[vernal bramble]

OOOOOOOh

So, 2x + 2 would be answer

[pure kayak]

not quite

the x-4 doesnt disappear

its (2x+2)(x-4)

or 2(x+1)(x-4)

[vernal bramble]

I must be thinking of rational expressions

Oh so I put the things outside in a bracket and the x - 4 in another bracket

[pure kayak]

yeah, in the same sense that 2xy+2y=(2x+2)y

[vernal bramble]

Well okay. I think I understand how to do now. Thanks for the help!

I'll close it down

.close

[calm coral]

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[calm coral]

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[potent igloo]

an infinite number if loops are allowed

[lunar token]

no

i got 50

but idk

[pure kayak]

what are the rules exactly? once you leave a room you cant go back to it?

[potent igloo]

!show

[calm coral]

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

[lunar token]

pretty vague

i did 2 x 3 x 1 x 4 x 2

then added 2 for the bottome square

so i used addition and producte rule

[potent igloo]

lgtm

[lunar token]

what

[calm coral]

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[cobalt folio]

i need to find the sequence,100,62,31,15,8,4,2,1,0

[brazen elbow]

what do you mean by "find the sequence"?

do you mean find the pattern or?

[remote mural]

are you sure u copied this correctly?

0, 1, 2, 4, 8, 15 make sense as they seem to describe the sum of the tribonacci numbers (having looked at OEIS), but starting from 31 this stops being anything that is consistent

[calm coral]

@cobalt folio Has your question been resolved?

[cobalt folio]

i promise, i am confused by the pattern as well

[cobalt folio]

yes

[calm coral]

@cobalt folio Has your question been resolved?

[calm coral]

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[remote mural]

Given that $\tan(\theta)=-\frac{1}{2}$ and $\sec(\theta) < 0$ what quadrant is the function in?

[potent lotus]

BlazeStorm81

[remote mural]

I'm leaning towards either 2 or 4 because cosine is negative there but I don't know how to decide between them

[upbeat venture]

where is tangent negative?

[upbeat venture]

also this is incorrect

[remote mural]

either 2nd or 4th

[upbeat venture]

good yes

where is cos negative?

[remote mural]

2nd or 3rd quadrant

oh so it has to be in the second quadrant because thats the only place where tan and cosine can be negative together right?

[upbeat venture]

yep :D

[remote mural]

okay thank you!!!

.close

[calm coral]

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[fossil heath]

Could you explain how they got the final answer ? I did it myself and am getting (cos(3p))^-2

[red wind]

yea its the same

[fossil heath]

ohhh

[swift laurel]

[ \frac 1{\cos x} = \sec x] by definition

[potent lotus]

cloud

[fossil heath]

i/cos(3p)^2 = sec^2(3p)

[red wind]

yess

[fossil heath]

got it thanks

.close

[calm coral]

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[half raven]

can someone explain how this is

i understand where ln(1 + y^2) comes from

but not 1/2

[leaden thunder]

use u substituion

u = 1 + y^2, then du = ?

[half raven]

y dy

2y dy

[leaden thunder]

yes. so what number should go here
y dy = ? * du

[half raven]

i got it

ig i just forgot how to integrate for a sec

[leaden thunder]

missing + C but yea

[half raven]

:P

ty

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[calm coral]

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[remote mural]

,, a^{n-m} = \4{a^n}{a^m}

[remote mural]

us this

[remote mural]

what is it if u apply this

[calm coral]

@remote mural Has your question been resolved?

[leaden thunder]

factor out 2^n from both terms now

yes

simplify (1 - 0.5)

then multiply

[calm coral]

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[cobalt cave]

this is the second time i see this type of question and second time it appear as a wrong answer. i calculated probability of (odd + even) numbers and found B yet book says correct answer is D. what do you think?

[hollow totem]

You also have to calculate the probability of having even + odd

[cobalt cave]

oh i get it

yeah lol

would it be the same if it was odd x even

[hollow totem]

Well there's 3 ways to get an even product

[cobalt cave]

no i didn't mean that

if the question was asking me to find probability of it being an odd number when i do not sum but multiplication would i still need to do odd x odd twice

or maybe remove squares but if it's {1,3} do i still need to take probability of {3,1} too?

[hollow totem]

nah just odd*odd once

like example if we choose from 1 to 4

1+2, 1+4, 3+2, 3+4, 2+1, 2+3, 4+1, 4+3 are all odd

Every odd+even and every even+odd

[cobalt cave]

i remember it being something like for example C(5,1) x C(4,1) and multiply it with 2! as it represent numbers arrangement

[hollow totem]

But for products

The only combos are 1*1, 1*3, 3*1, 3*3

[cobalt cave]

alright so 3x1 is still taken despite there is 1x3

[hollow totem]

yeah

the first time you choose any odd and the second time you choose any odd though

It's not a duplicate calculation that's just one of the 4 ways to do that

but yes you take them both

[cobalt cave]

how about same numbers

1 x 1 and 1 + 1 taken only once?

[hollow totem]

Yes

1+1 is the same sum as itself

[cobalt cave]

yea makes sense when you think it as a list

[hollow totem]

but 3+4 is not the same sum as 4+3

The reason why not is that every sum needs to be equally likely

Like if you flip two coins, you're more likely to get 1 head and 1 tails then 2 heads

[cobalt cave]

like when thinking as lists from function as i remember logic goes as parent of number 1 is 2 and parent of number 2 is 1 but when it's the same number one is parent and it can't be listed twice

right logic here?

[hollow totem]

Maybe I can't really understand

[cobalt cave]

wait

i think it's called venn scheme?

anyway i better move away for this before i further confuse myself

i'll keep this in mind

thank you for helping

.close

[calm coral]

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[slow plume]

Given:

• Circle C1 of radius R and its center X1, Y1.
• Set of N points P and each of their positions, where all points P are contained in circle C1 ("contained" includes the points lying on the circle's edge).

Need to prove:

• There must exist a circle C2 of the same radius R such that such that at least two points from the set of points P lie on the circle's edge, and the circle contains all other points.

I can visualize this by imagining myself "tug" on the circle, where the points are fixed, and balancing it on two other points. However, others I've discussed this with are not fully convinced and I am unsure how to prove that this is true for any set of points contained within any circle of radius R.

[frozen ibex]

at least two points or exactly two points ?

[slow plume]

at least, I'll edit it

[frozen ibex]

Idk what you mean by tug, tell me if it's the following :
You make a "copy" of C1 (Let's say C2) You position C2 such that all points are in C2, but one point is on its edge (Let's say P1), you then rotate with rotate C2 on the axis of P1 such that another point gets on the edge ?

If so it seems like the proof will use the IVT

[slow plume]

Yes, I think that gets the same idea across.

[frozen ibex]

Anyhow, if that's your the idea, you might want to look at the wobbly table theorem, as i think the situations are quite similar

[slow plume]

Also it must contain all points P

[slow plume]

Thank you

[frozen ibex]

If a point exits the circle then it means at some point it was on the edge of the circle

then just stop at this point

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[static stag]

1001101sub2 to decimal is 77 right?

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[eternal bluff]

question 21 please

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[royal raft]

how would I factor out a constant in order to get (1 + x/c) in the square root like I did in the first equation

[royal raft]

sorry the first is wrong

hold on

could I factor out a 1/3 and divide x^2 by 3

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[umbral lark]

can someone explain how that function was simplified into the one on right

[flat laurel]

distributivity property of real numbers

then some exponents laws

[umbral lark]

yea well i get why 6x2/3 but no clue where -8x is coming from

[flat laurel]

a(b+c)=ab+ac right?

[umbral lark]

ohhh right

yea i guess that makes sense now😭

silly question

thanks tho

[flat laurel]

^^

it's particularly called as the distributivity of real numbers

[trail ingot]

destructivity?

[flat laurel]

mb

[trail ingot]

lol

[flat laurel]

@umbral lark gl gl

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[simple loom]

I need help for question 2

You want the function to equal 0

so $9{cos}^2(3x)-6cos(3x)-2(3x-2)(3x+1)=0$

[potent lotus]

lime

[simple loom]

<@&286206848099549185>

[upbeat venture]

well the only way that makes sense to me is to find an x value that makes them both 0 separately

otherwise it's not really doable i don't think

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[gusty lark]

I tried solving this but I'm not sure I got it right I also don't know how to find it's domain

First I defined V = lwh

Then I made l = 3 -2x

and w = 2 - 2x

and finally h = x

I multiplied them out to get 4x3-10x2+6x

Is this the right thing to do for this problem?

And also how would you find domain after for this odd function

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[warm warren]

Stochastic processes:

[warm warren]

im struggling to make sense of this little o

[rancid monolith]

hi

[warm warren]

i understand that lambda is measuring event occurences per some time, so multiplying by the distance of "t + δt" and "t" makes sense

but what about the o(δt)

[spare beacon]

What exactly is your question here

Are you asking what little o notation is

[warm warren]

i've got the definition of the little o but im not sure why it appears

so in this case we're using the 2nd one and g is just the identity function

but idk where this f(δt) = o(δt) is coming from, where is this f from

[spare beacon]

The point is that o(δt) represents a function that behaves in a certain way under a limit, namely that o(δt)/δt = 0 as δt tends to zero.

The specific form is not important

[warm warren]

but why does it appear in the middle line here?

[spare beacon]

You need some arbitrary function that represents the probability of an event occurring more than once for large intervals that describes the nature of how it goes to zero as the interval gets small

lambda dt represents the linear part, o(dt) is the part that decays faster than linear

I'm going to leave now. But if you are still not understanding, if you take a very large interval, you are more likely to see more than 1 event.

As the length of time you are looking for an event goes down, you want the probability of observing one event decreases linearly with the size of the time window dt and the probability of observing more than one to decrease more rapidly.

You need to define your poisson process or whatever it is so that that arbitrarily small observation windows has a nonzero probability of observing 1 event and the probability of observing 2 events is essentially zero.

[warm warren]

i see, it makes sense for the last section, i'll read up a bit more on the poisson process but i get the motivation for it now, thanks

.close

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[calm coral]

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[torpid dune]

,rotate

[torpid dune]

i found the areas of AED and CDF but I cant find the area of EDF

[upper nova]

what is length of CE?

do you know?

[torpid dune]

its 2

[upper nova]

then do you know length of CD?

[torpid dune]

ohhh

root 2

[upper nova]

so length DE is available

[torpid dune]

is [EDF] = pi-pi*sqrt(2)+pi/2

?

@upper nova

can someone confirm that the answer is 3pi - pi*sqrt(2) - 1?

<@&286206848099549185>

[upper nova]

is this assignment?

[torpid dune]

wdym

Is 3pi - pi*sqrt(2) - 1 your answer? @upper nova

[upper nova]

I mean if this is some kind of assignment I think I shouldnt tell you the ans

[torpid dune]

it's not a test or exam

so getting help is allowed

;-;

[upper nova]

what have you got for area of AED and CDF?

[torpid dune]

(pi-2)/4

i hope its not wrong

[upper nova]

I can't do calc rn

dont have anything to write down

[torpid dune]

ok ill wrtie down my working

[AED] =[CFD] = [AEC] - [COD] - [AOD] = 45/60(pi)(2^2) - 1*1/2 = 1/4(pi)(1^2_

which is pi/2 - 1/2 - pi/4

[upper nova]

first
what is $\frac{45}{60}\pi 2^2$

[potent lotus]

Dri111

[upper nova]

AEC?

[torpid dune]

uh

i meant 45/360

🤦‍♂️

cuz like the angle is 45 out of 360

[upper nova]

and 1*1/2 is COD?

[torpid dune]

yeah

the triangle

[upper nova]

where's AOD then?

[torpid dune]

AOD is 1/4(pi)(1^1)

so its pi/4

AOD is the quadrant

COD is the triangle

shouldve probably said that