help-27

[near jolt]

and then you wire the and gate for connecting the two not gate?

[cold crow]

Your making the expression more complicated by that if you group it like that since we're trying to simplifying it it is easier to simplify the expression in the formula I gave

[near jolt]

💀 because it might be anbiguious if you miss some and gates

i see you by saying a not b you might mean a and (not b)

[cold crow]

There's literally no way you could miss it unless your eyes is just dizzy from all these wires but you don't need to add more and gate just follow the expression it is literary the actual direction to wiring

[cold crow]

Dude that just first group

Look how long this is😭😭

We're arguing over one group we have alot more to simplfiy😭😭🙏

[near jolt]

i suggest writing a table for it

[cold crow]

I actually prefer not to

[near jolt]

it just ive never seen people writing "a not b" or stuff like that, afaik its ambiguous

[cold crow]

I'm practicing to simplify it in this long expression

Remember I said I write them all in paper

[near jolt]

ok then, let me write out the expression in latex

[cold crow]

I intend to write it as long as possible to show off lol

But i don't even know if the calculator would work😭😭

[cold crow]

Oh I just write it like that because this keyboard don't have b with dash on top likew this ā and other letters

[near jolt]

$$\overline a b \overline c d \overline e + \overline a b \overline c d e + \overline a b c \overline d \overline e + \overline a b c \overline d e + \overline a b c d \overline e + \overline a b c d e + a \overline b \overline c \overline d \overline e + a \overline b \overline c \overline d e + a \overline b \overline c d \overline e + a \overline b \overline c d e + a \overline b c \overline d \overline e$$

[woven radish]

qwertytrewq

[cold crow]

Holy shit

How do you use that

[cold crow]

But this expression still look long too I don't think it is simplified enough

[near jolt]

ok alr alr i get you but id much rather you just drop all the and in that case, it confused me with half and and half no and

[near jolt]

oh i just copied your text and converted it into boolean

[cold crow]

Could you simplify it

I could simplify three group but this kind of long expression I get lost lol

[near jolt]

well one way is just group similar terms together

first three terms seems to group with the last three term

wait actually

every 2 terms has the first 4 terms being equal

[cold crow]

Can you do the whole thing🙏🙏

[near jolt]

Try grouping every two terms

[near jolt]

this is a help channel we are supposed to guide you rather than sending the whole solution

[cold crow]

I thought yall supposed to give answers

I'm literally begging bro I'm not doing all this😭🙏🙏

[near jolt]

no?

[cold crow]

There's 12 more

Broo

I'm dead

[cold crow]

Bro please

This is not even the longest expression I got😭😭

[near jolt]

we are just here to guide you to the solution, there are better sources to go to if you just want the solution. heck it, gpt can probably do this in a couple second

[cold crow]

This is too long for chatgpt

[devout snow]

@cold crow Has your question been resolved?

[devout snow]

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[devout snow]

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[pine jolt]

question

[devout snow]

Please don't occupy multiple help channels.

[pine jolt]

what would be the most approppriate method to solve this

[supple knot]

!1c

[devout snow]

Please stick to your channel.

[pine jolt]

?

[supple knot]

You have a channel open already

[tough juniper]

Im trying to show that the area of the parallelogram, or v x w, is the determinant after transforming the unit vectors v,w with a general matrix. Where did my math go wrong?

[pine jolt]

huh

[supple knot]

.close

[devout snow]

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[supple knot]

What frost said

[supple knot]

#❓how-to-get-help

[tough juniper]

Ok

[devout snow]

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[cloud osprey]

anyone here?

[devout snow]

@cloud osprey Has your question been resolved?

[devout snow]

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[cloud osprey]

uh no

[wintry flare]

whats the question

[wintry flare]

ixl goes cray

[cloud osprey]

fr

[wintry flare]

whatre you having trouble with

[cloud osprey]

compound figures

mind if I dm u?

[wintry flare]

if you want

[wintry flare]

just seperate it into its respect shapes

you can see 3 here, no?

[cloud osprey]

yeah but I'm js stupid like that

[devout snow]

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[restive river]

i have trouble factorising 😭

[restive river]

final answer

[devout snow]

@restive river Has your question been resolved?

[restive river]

.close

[devout snow]

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[visual hazel]

why did your x² suddenly became -x

[restive river]

uh

[restive river]

could u clarify which one 😭

cuz im confused myself

[visual hazel]

reopen first so it doesnt go poof

[restive river]

oh

.reopen

[devout snow]

✅

[restive river]

go poof haha

[restive river]

oh

oh wait

must have been a mistake

i thought i wanted to factor out 2x

from x^2(-2x
to 2x(-x^2

but i genuinely dk if that works

[visual hazel]

yeah it can work

maybe you should've tried distributing it first then factor

so just make it -2x³

then factor the common 2x out

[restive river]

but honeslty i tried a different approach 😭

[restive river]

uh oh

[visual hazel]

ok we're cookin

small mistake though

it should be x² in the second term

because you have x³ and you factored only one x

so you are left x²

[restive river]

ohhhh no wonder

why i was confused how is the answer 1-3x^2

YESS

I FINALY

GOT IT

THANK YOU SM 😭

i didnt notice my mistake

.close

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[errant wyvern]

i have no idea how to do this problem

domain cant be negative

and i somehow got the minutes right

someone please help

[knotty sage]

got your steps written ?

[errant wyvern]

I just did inverse tan on the calculator

But then got negative so I added 360

But there’s supposed to be 4 answers

[knotty sage]

well you can add or subtract 180 bcz of tan(x) period is 180

[errant wyvern]

ohh ok thanks

thanks for the help i totally forgot

cya

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[stark zealot]

geometry - need help with this problem

[fossil locust]

do you know the angle in the centre theorem?

this is the specific case you need

the key is that the two angles need to be on opposite sides

[stark zealot]

So the center (d) is 2x an angle? Like w

[fossil locust]

yes 2x is an angle

x corresponds to angle TWP here

[stark zealot]

Pt is 280?

As 140x being x multiplied by 2

[fossil locust]

that's the reflex angle correct

[stark zealot]

Is tw just half that?

[fossil locust]

no, we're not on part b yet

also the triangle isn't isosceles clearly (3 different angels)

[stark zealot]

What do i do to find PT then

[fossil locust]

so by definition of measure that's angle POT

you already have POT but it's the reflex angle (on the other side)

[stark zealot]

So 32?

[stark zealot]

Yeah

[fossil locust]

no

[fossil locust]

this is the angle you need in red

[stark zealot]

Wait no

48

Im usnurr

[stark zealot]

24 x 2 so 48

[fossil locust]

literally just 360 - 280 = 80

you've been using the wrong angle PTW this entire time

[fossil locust]

if you backtrack to here

reflex angle POT = 280

that's all you needed

[stark zealot]

Oh I did that initially I thought it was wrong

Okay I understand it now Loll

[fossil locust]

couldn't you just check it on the software

[stark zealot]

For what my instructer assigned no

[fossil locust]

ah

okay now for part b

we need to use the middle image here

if you draw line segments OT and OW you'll see how it matches

so actually the first step is to find the third angle in the triangle, TPW

[stark zealot]

Yeah im lost

[fossil locust]

what's TPW then?

the third angle in the triangle

[stark zealot]

40

[fossil locust]

24 + 140 + x = 180.......

[stark zealot]

16

[fossil locust]

okay yes

[stark zealot]

so

[stark zealot]

arc pt is not 40?

[fossil locust]

so basically this is what happens

[fossil locust]

you need to learn the process first

don't jump straight to the answer

that's some advice for learning all of maths

oh wait let me fix the diagram

[stark zealot]

i have the answer written on paper here

i dont remmerb how i got it in the first place

was in class

and i think tw was 48 or so

Ah

[fossil locust]

ok fixed

[fossil locust]

so y = 16 don't forget

[stark zealot]

and that angle is 32

[fossil locust]

yep and that's the answer!

so yeah like we move the angle first to make it simpler, cause you're allowed to do that here (if you really want to, try understanding the proof from isosceles triangles)

and then bingo

[stark zealot]

.close

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[devout snow]

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[lime wedge]

There are 10 people, 7 wear red, 3 wear blue, 2 wear green. All people are unique. They first sit down in a row and then sit down in a round table. For both cases, find the probability that no person wearing blue sits next to a person wearing green.

[lime wedge]

Not sure if what I’m doing is right

I let there be 8 spaces between red and used combination for all the cases where blue and green aren’t together

[gray summit]

I think you're missing their arrangements

Like, this first case, it's (8 5) but they green and blue can arrange themselves so it should also have 5!/(3!2!) multiplied to it

[lime wedge]

Oh, should it be permutation then?

I just replace all the combinations with permutations and I won’t have to add anything?

[gray summit]

Yes

U did do it for the pairs though, not for the individual people

Just using permutations solved the issue ig

[lime wedge]

👍 For round table if I just keep the same things but instead of 8 spaces I use 7 and the total ways is 9! Will it work?

[gray summit]

Mostly, it would

[lime wedge]

I’ll try that then for the round table problem

Thanks for help

.close

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[devout snow]

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[hushed void]

İs there a way without trigo?

[hushed void]

Like with only basic circle infos

İt asks for |CD|

[pure flower]

uh

full translation?

[hushed void]

uh it basiacly saying r=sqrt(21)

Ask for the cd

[faint zinc]

Sum of the arcs is 120°

[hushed void]

Yes

[hushed void]

Yw

Ye

İs there a problem?

Guys what happened😑

[hushed void]

Somebidy help

<@&286206848099549185>

[hushed void]

Helpp😨

[hushed void]

😑😑

[faint zinc]

@hushed void sorry for dipping, something came up.

[hushed void]

Np

[faint zinc]

Maybe there's a trick where you can make a quadrilateral

[hushed void]

hmm🤔

Whats that

Move the beams? You mean

[devout snow]

@hushed void Has your question been resolved?

[hushed void]

Omg someone help

[hushed void]

<@&286206848099549185>

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Please do not ping individual helpers unprompted.

@hushed void Has your question been resolved?

[hushed void]

.close

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[devout snow]

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[sick valley]

can someone explain what they did in the "another method" ?

[olive snow]

Surely

[sick valley]

i did the first method

but i dont understand why in the other one they use a and b

[olive snow]

Let z = your quotient

With z = a+bi

And then multiply by -1+i

[sick valley]

so z = (3-2i)/(-1+i)

[olive snow]

Find z

[olive snow]

.

[sick valley]

ok z(-1+i) = 3-2i

[olive snow]

And after developping with cartesian form they take real part and imaginary part and solve

For a and b

Which is a very long way

[sick valley]

@olive snow are you sure i had to multiply by -1+i ?

not -1-i ?

[olive snow]

For their "another method" ja

[sick valley]

ah ok

[olive snow]

Thats for the usual method

Which is clearly better

Cuz this conjugate technique not only work with complex but also with sqrt for example

Really useful

Thats why train the conjugate instead of the "another method"

[sick valley]

im here

-a + ai - bi - b = 3 - 2i

now i have to group i ?

ok i got this thanks

.close

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[devout snow]

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[timber pebble]

what do you know about a sum of logs

[harsh lodge]

This is my working

ive been stuck on it for 30 mins

its killing me

[wind mason]

||Telescopign series?||

[tranquil current]

Write out just the first 2 elements of the series in full, and see if you notice anything

[wind mason]

Try to write out the first few terms of the sum first.

[harsh lodge]

ima be hoenst idk im A level

[harsh lodge]

i did

[timber pebble]

it might be easier to use rule 1

[wind mason]

Try it again. You should notice something.

[timber pebble]

than rule n

[harsh lodge]

but they dividing

[timber pebble]

sure but try rule one

[harsh lodge]

alr ill try give me a few

[wind mason]

Do not try to split up each series: evaluate it as a whole.

[timber pebble]

log(3\2) + log(4\3) = ?

[harsh lodge]

omds

[rare kernel]

after this step write the series

[harsh lodge]

log and then those multiplied

[timber pebble]

and what cancels

[harsh lodge]

idk who to follow

[harsh lodge]

3

[timber pebble]

yea, sorry

[timber pebble]

sure

[harsh lodge]

wait what should i do

[timber pebble]

how about log(3\2) + log(4\3) + log(5/4) = ?

[harsh lodge]

i mean i get all this

its jus

when i acc go solve it

wait its so hard to explain

can anyone od vc

[timber pebble]

theres no vc help here

[harsh lodge]

dang

[timber pebble]

theres nothing special to do except see the pattern imho

[harsh lodge]

ok so

[timber pebble]

can you see where this goes

[harsh lodge]

no we need to solve it to get 2

[timber pebble]

yea

so if the answer is 2

what should the argument of the log be

$\log_5 x=2$

[woven radish]

jan Niku

[timber pebble]

whats x?

we can work backwards, if its helpful

[harsh lodge]

25

[timber pebble]

yea

[harsh lodge]

wait would u be comforatbel doing vc if i add u

[timber pebble]

and you sum of logs looks like this

im on a phone lol

at a park

[harsh lodge]

i think we are both approaching it in different ways

ohh

[timber pebble]

[harsh lodge]

so nice uk is so dull

[timber pebble]

so i cant but mb someone else

[harsh lodge]

ill show u the mark scheme

[timber pebble]

do you agree that the sum, with rule 1* is thxs:

$\log \qty( \frac 3 2 \cdot \frac 4 3 \cdots \frac{50}{49} )$

[harsh lodge]

yeah so i got that

and trhen the steop after

[woven radish]

jan Niku

[harsh lodge]

should be add right

[timber pebble]

theres no sum left

we destroyed it by applying rule one

and turning a sum of logs into a single log of a product

[harsh lodge]

oh right yeah i was thiking as seperate logs

[timber pebble]

yea, theres only one log left

[timber pebble]

what has to happen here to make the argument 25

we kind of already know

[timber pebble]

you saw the pattern here

3s must cancel...

[harsh lodge]

wait im confusded

[timber pebble]

4s...

[harsh lodge]

is this for the top fraction or bottom fraction

[timber pebble]

its the entire left hand side

[harsh lodge]

OH YEAH

i see

[timber pebble]

we had $\log \frac 32 + \log \frac 43 + \dots + \log \frac{50}{49}$

[woven radish]

jan Niku

[harsh lodge]

now how do we get the correct proof leading to 2

[timber pebble]

use the pattern

$\frac 32 \cdot \frac 43 \cdot \frac 54 \cdots$

[woven radish]

jan Niku

[timber pebble]

which factors die here

[harsh lodge]

wdym pattern

rhe 3

and 2 and 4

idk

[timber pebble]

start cancelling stuff

we have a 3 on top and bottom

so 3s go away

[harsh lodge]

yeah

[timber pebble]

so does 4

[harsh lodge]

yeah

[timber pebble]

the next term excluded will kill that 5

and the next 6

so whats left, once we stop

of the top of the previous term kills the bottom of the next

[harsh lodge]

why dou have 3 terms

i only did the first 2 terms and last term

[timber pebble]

we can continue as long as we want

[timber pebble]

dont kill the 2 with the 4

kill the 4 with the 4

[harsh lodge]

alr

whats left then

is

[timber pebble]

thats the million dollar question

[harsh lodge]

10

[timber pebble]

which numbers dont pair up?

[harsh lodge]

i feel like a child im so stupid

[harsh lodge]

5 and 2

[timber pebble]

you just dont see it yet

its no biggie

[timber pebble]

yea, in the one i wrote

now what if we include another term

6\5

[harsh lodge]

illl never get into med school man i have my finals in 50 days

[timber pebble]

now the 5 dies

and you have 6\2 left

if we include 7\6

now 6 dies, we have 7\2

you see?

[timber pebble]

youre basically there

youre just having trouble generalizing the pattern

[harsh lodge]

oh yeah

but how can i word that

bcuz in the solution what they did was

They got this

[timber pebble]

yea

the first is by rule 2

[harsh lodge]

And I was confused on how they got to the log 5 50 - log 2

[timber pebble]

the second is by rule 1

rule 2 is easier imho

[timber pebble]

because when the denominator gets split intoa log, it is minus

[harsh lodge]

we got to the 2nd step together but how did trhey et to the 3rd step

[timber pebble]

and when its on the top its plus

its the same logic were using our way

this is called telescoping

like a telescope is long but it closes all small

a lot of terms are paired up and disappear

[harsh lodge]

likse converging

oh nvm

[timber pebble]

like in our big fraction

[harsh lodge]

im still confused

how did they jus not account all the other terms and get log 50 - log 2

[timber pebble]

what is (3\2)(4\3)(5\4)(6\5)(7\6)(8\7)

[timber pebble]

because they telescope

only 50 and 2 are not BOTH on the top and bottom

[harsh lodge]

4

[timber pebble]

2 isnt paired up to anything

neithe039

neither is 50

they only appear once

[timber pebble]

8/2

now what if we go out to 50?

can you see it

what is (3\2)(4\3)(5\4)(6\5)(7\6)(8\7)...(46\45)(47\46)(48\47)(49\48)(50\49)

[harsh lodge]

oh so if they both appear at top and bottom they cancel out so hence they disapear btu because 50 is the final term it has nothing to pair up with so its only gonna apeear at the top likewise for 2

[timber pebble]

yea

[harsh lodge]

idk

[timber pebble]

its 50/2

[harsh lodge]

25

[timber pebble]

im just extrapolating your answer here

if we go up to 8/7 and get 8\2

[harsh lodge]

wiat

[timber pebble]

assume the pattern continues

[harsh lodge]

ive sene this before somewhere

its similar to ther proof of

this reight

u list them out and then divide one by another

[timber pebble]

oh that i dont knowoff the top of my head

isnt this done by induction

[timber pebble]

anyways

ah

[harsh lodge]

anyways thanks

[timber pebble]

it comes up ALL the time

like uhh

[harsh lodge]

uff my brain is fried

sorry for taking up so much fo ur time

[timber pebble]

i can do better when im home

nah its good

im waiting out the sun

[harsh lodge]

ur crazy good at maths

[timber pebble]

ifonly

[harsh lodge]

drop the tips man

[timber pebble]

take a break n come back

youll see it

[harsh lodge]

i keep forgeting concepts even after so much practice

[timber pebble]

youre on the last step

itll make sense w a fresh head i think

[harsh lodge]

i get the top and btoom so they dont cancel but ima be hoenst i dont get the pattern

[timber pebble]

you can do it forward too it can make more sense

but its crazy to start over now

if youre tired

[timber pebble]

you can try a vid on telescoping

[harsh lodge]

idk if u have heard of A levels

i think telescoping is more advanced than a level maths

[timber pebble]

https://www.youtube.com/watch?v=H8HVYiEqzBo

its usually covered in a calculus class here

first or second semester of college for engineers/math

[harsh lodge]

ill chekc ity out

[timber pebble]

good luck

[harsh lodge]

thanks man

how do i close this

.close

[devout snow]

Channel closed

Closed by @harsh lodge

Use .reopen if this was a mistake.

[devout snow]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[unkempt smelt]

hi

[tender sandal]

?

[wind mason]

Got a question?

[unkempt smelt]

I need to show that A has infinty indexes and I dont know how

without functions

[wind mason]

“Infinity indexes”?

[unkempt smelt]

numbers

[finite obsidian]

What do you mean infinity indexes

If you mean has infinitely many elements, can’t you just like

n =/= m means 2^n =/= 2^m

[timber pebble]

assume there are finitely many as a first step?

[unkempt smelt]

ok

[timber pebble]

then ||try to use the finite collection to construct an element which cant be in that collection||

[devout snow]

@unkempt smelt Has your question been resolved?

[devout snow]

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[vestal egret]

Hi, I'd like help for this determinant equation:

| 2 x x x |
| x 2 x x |
| x x 2 x | = 0
| x x x 2 |

I tried doing zeros below the main diagonal, I know it's obviously wrong as it gets ridiculously wild (check the images 💀), but I don't see how else it could be solved with the determinant properties I know, which are:

"1. The determinant of a matrix always coincides with the determinant of the transpose of that matrix.
2. |A · B| = |A| · |B|
3. If a determinant has an equal row or column (two or more equal rows or two or more equal columns), the determinant equals zero.
4. If a determinant has a row or column consisting of zeros, the determinant equals zero.
5. If a row or column in a determinant is multiplied by a scalar, the determinant is multiplied by that scalar.
6. If a determinant comes from an upper triangular matrix, a lower triangular matrix, or a diagonal matrix, the value of the determinant is the product of the elements of the main diagonal.
7. If in a determinant, one row is exchanged for another row or one column for another column, the determinant changes sign.
8. If in a determinant, we have a row or column that is a linear combination of one or more other rows or columns, the determinant equals zero.
9. The value of a determinant does not change if a row or column is represented as a linear combination of other rows or columns (respectively).
10. If a row or column of a determinant is made up of terms that are sums (or subtractions), the determinant can be separated into a sum (or subtraction, respectively) of two determinants."

I'm completely sure this equation can be solved extremely more easily with different properties, I just don't know if it can be solved with these ten properties I've been taught and I'm just blind and I can't see it, or if you need a different property from these to solve it easily. There's just definitely no way it takes this much algebra to solve it. Thanks!

[vestal egret]

Please try to consider (for the resolution) just the properties I wrote if possible, if not then ok

[unkempt crane]

what is the question asking for

[vestal egret]

solve the equation

[unkempt crane]

solve for x?

[vestal egret]

yeah

i guess

[unkempt crane]

ah

[vestal egret]

there aren't any other unknowns so

ig it's x

[unkempt crane]

cant we just look at the section graphs

so basically x should be mod 2

-2 does satisfy

so its -2 ig

[vestal egret]

huh

[vestal egret]

its 2

[unkempt crane]

?

2 and -2 both satisfy oh

I am saying x = -2

hello??

alr imma go now

[stone stump]

if you knew about eigenvalues it wouldnt be too hard to see that the determinant is (2-x)^3(3x+2) as thats the product of the eigenvalues

[vestal egret]

no, they don't

only x=2 does

[vestal egret]

then determinant is -256

[stone stump]

you could have made your life a bit easier if you simplified some more fractions along the way

[vestal egret]

i in fact do not know about eigenvalues

[stone stump]

eg (x^3-3x^2+4)/(2-x^2/2)

but well in the end you get a quartic with three obvious roots and then -2/3 as the last root

[vestal egret]

yeah a bit

tho it wouldn't have helped a lot

and that would have still required to factorise it

so overall idk if i would have saved much more time

[vestal egret]

basically it's an exercise that's supposedly meant to be solved with the ten properties i mentioned

i just don't see how

[stone stump]

wdym

you did it

[vestal egret]

yeah but

isn't there another method to do it

i mean

the determinant just has 2 on the main diagonal and all the rest of the terms are x

idk it feels like sth that has a way easier way to be solved, by properties

[stone stump]

I cant think of something easier with just those properties

but frankly you also have to get used to not everything being super nice anymore. sometimes you do just have to fight your way through some algebra

all things considered this really isnt that bad

[vestal egret]

no i agree

it's just that we only do simple stuff for these things

[unkempt crane]

then whats the problem u solved it

[vestal egret]

it's the last grade before university so it isn't usually that complicated

[vestal egret]

what i mentioned earlier

i thought a different property could be apply

[stone stump]

x=2 is not the only solution btw

[vestal egret]

turns out not

[vestal egret]

yeah i just realized

idk why i didn't get -2/3

[unkempt crane]

ye

[vestal egret]

ig i miscalculated at some point

[stone stump]

happens ¯_(ツ)_/¯

[unkempt crane]

uh -2 satisfies I think

[vestal egret]

ok thanks for the help 🙂

[vestal egret]

it doesn't

[stone stump]

no

you could say that "obviously" if you choose x=-2/3 then the sum of the columns is 0, so therefore the columns are lin dependent

but frankly I wouldnt expect anyone to just see that

[unkempt crane]

it kinda does

idk im dumb

[stone stump]

on the other hand thats basically the eigenvalue argument I had in mind earlier

[vestal egret]

no xd

if x=-2, the determinant is -256

a bit far from 0

[unkempt crane]

yk the mid column has a negative remainder

[vestal egret]

OHHHHH

wait that makes a lot of sense

[vestal egret]

what mid column

[stone stump]

but without the eigenvalue angle I'm not sure how to argue that 2 and -2/3 are the only roots

without actually computing the whole quartic

[devout snow]

@vestal egret Has your question been resolved?

[vestal egret]

no wait a moment

ok nvm close it

[vestal egret]

yes

[quick seal]

.close

[quick seal]

u gotta write .close

[vestal egret]

actually wait

[vestal egret]

i checked all the algebra closely

and x=-2 is indeed a solution

of the equation

like when you're factorising and all, the way I did it, you get x=2, x=-2/3 and x=-2

but for some reason

-2 just doesn't work

when you sub 2 and -2/3 into the determinant, it equals 0

but not with -2

ahhh ok it's cuz it makes the denominator zero

so does 2, but it still works for some reason

.close

[devout snow]

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[lean hamlet]

Huhu, one could not call logic logic because you call it irrational, subjective.... Can describe logic? And my idea is that as soon as you can describe it, it is a valid logic. Or do I overlook something? Thanks for the tips

[pseudo basin]

https://tenor.com/view/what-huh-whut-confused-grin-gif-17587850

[misty crest]

word

[lean hamlet]

Is it because of my English or my idea that causes confusion?

[pseudo basin]

maybe both?

i can say that your question makes absolutely no sense to me

[rare kernel]

dude this is a math server

[lunar eagle]

sir this is math server

[lean hamlet]

After thinking about logic, and I got a question about that. And on the subject of logic. Is there any unlogication at all and if so, how can you verify/falseify it? And do I understand logic as a tool correctly? Logic is there to be able to communicate via the same. And yes, logic belongs to math

It dosent metter thx 😄

.close

[devout snow]

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[tall aspen]

fasr

[tall aspen]

just give me the answer

nothing else

[rare kernel]

;/

i dont think we are supposed to do that

[tall aspen]

why?

bullshit, i just wanna know the answer

[rare kernel]

A = 1x2
B=2x2
AB would be possible if
columns in first matrix = rows in second matrix
2 = 2

[acoustic leaf]

!noans

[devout snow]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[tall aspen]

cuz i guess its given wring in the book

[rare kernel]

nbo wait

[tall aspen]

wat?

[rare kernel]

both a,r are true

and r is correct reason

[tall aspen]

okay how?

[rare kernel]

^

[tall aspen]

,close

.close

[devout snow]

Channel closed

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

[devout snow]

@static gorge Has your question been resolved?

[indigo flicker]

try connecting a line from the center of the circle (triangle) to one of the vertices of the triangle

[indigo flicker]

yeah, then solve using your sine cosine laws

wait have you learnt that yet

[static gorge]

I already solved the question using sin law

but I'm looking for other ways

[indigo flicker]

i mean u can use circumcircle radius of an equil triangle = side length / √3

[static gorge]

huh?
explain.

or just derive it

[indigo flicker]

its just the law of sine proof

where you use s / sin 120° = r / sin 30°

(s is the side length, r is circumcircle radius)

then r = s/√3

[static gorge]

$\frac{s}{\frac{\sqrt{3}}{2}} = 2r$

[woven radish]

Isaac

[static gorge]

oh ok

i see it

[indigo flicker]

yeah

[static gorge]

$r=\frac{s}{\sqrt{3}} \ \
w\sqrt{3} =\frac{s}{\sqrt{3}} \ \
w\sqrt{3} = \frac{292}{\sqrt{3}}$

ooo

[indigo flicker]

no its r = s/√3

not s√3

[woven radish]

Isaac

[indigo flicker]

ye there we go

[static gorge]

292/3

nice

[indigo flicker]

yup thats the answer

[static gorge]

while you're still here look at this

.close

[devout snow]

Channel closed

Closed by @static gorge

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[static gorge]

b and n are integers

$f(x) = ab^\frac{x}{n} \ \ \
ab^\frac{4}{n} = 5 \ \
ab^\frac{7}{n} = 135 \ \
ab^\frac{9}{n} = ??$

[sand dove]

can you find some equation that only involves b and n from those first two?

[finite obsidian]

What have you tried

So true

[sand dove]

xdd

[sand dove]

then try this

[woven radish]

Isaac

[finite obsidian]

But what gave you tried to pin down what f(9) should be

Yes

[static gorge]

you want me to solve for b in terms of a, x, and n?

[finite obsidian]

Look for it, what do you know you can do with equations like the values at 4 and 9

What they suggested is to get rid of a by something using those

[static gorge]

[indigo flicker]

i just solved the question
tip: try getting rid of a and find b^(3/n) first

[finite obsidian]

I mean, what operations for a pair of equations involving ax and ay could let you remove a

[finite obsidian]

Yes that is what has already been suggested

[indigo flicker]

well i just came back and didnt check chat

[static gorge]

getting rid of a and n?

[indigo flicker]

only a

[finite obsidian]

Just a

You can’t get rid of n yet

[static gorge]

because n affects x?

[finite obsidian]

(And you don’t need to ever)

[finite obsidian]

No

Ignore n for the moment

It’s constant, so what if we just said c= b^(1/n)

[static gorge]

$b^\frac{x}{n}$

[woven radish]

Isaac

[finite obsidian]

Then we know $$ac^4 = 5$$ and $$ac^7 = 135$$

[woven radish]

Sharp, paragon of destruction

[finite obsidian]

What can you do to get c alone

[static gorge]

divide...

[indigo flicker]

exactly

[static gorge]

where have you been

come solve it already

[indigo flicker]

f(7)/f(4)=(ab^(7/n))/(ab^(4/n))=135/5

[static gorge]

$f(7)/f(4)=\frac{ab^\frac{7}{n}}{ab^\frac{4}{n}}= \frac{135}{5}$

[indigo flicker]

yes

[woven radish]

Isaac

[indigo flicker]

yeah

[static gorge]

$b^\frac{3}{n}$

[woven radish]

Isaac

[indigo flicker]

now find b^(1/n) and sub in the original functions

[finite obsidian]

People can not solve your problems for you

[indigo flicker]

dont rely on us

[static gorge]

no

you can solve it I'll then understand

[finite obsidian]

Tf u mean no

[finite obsidian]

No

That’s called cheating dawg

[static gorge]

it's like you're teasing me

[finite obsidian]

Good

Only way forward is to figure it out

So you know c^3 = 135/5

What’s c

[static gorge]

cubic root 135/5

[finite obsidian]

Yeah, what is that

[static gorge]

b^(1/n)

[finite obsidian]

Yeah what number is it

Anyhow, you now know b^1/n, so you can plug that in to a b^4/n = 5 to find a

[static gorge]

k

$\frac{ab^\frac{7}{n}}{ab^\frac{4}{n}} = \frac{135}{5} \Rightarrow b^{\frac{7}{n}-\frac{4}{n}} = 27 \Rightarrow b^\frac{3}{n} = 27 \Rightarrow b^\frac{1}{n} = 3$

[woven radish]

Isaac

[indigo flicker]

yes

[static gorge]

wow

couldn't you explain like this

[indigo flicker]

i mean at first i was trying to solve this but the other guy jumped in so i let him explain

[static gorge]

oooh ok

no problem then

[indigo flicker]

eventually i thought the c = somthing stuff was too overcomplicating so i decided to jump in

[static gorge]

$f(x) = a \times 3^x \
f(4) = a \times 3^4 = 5 \
a = \frac{5}{3^4} \ \ \
a = \frac{5}{81}$

[indigo flicker]

thats the answer

[woven radish]

Isaac

[static gorge]

he was just trying to help

ok let's get this done with

I already have answer so idk what he was talking about

but I needed to understand

[indigo flicker]

aight now you do

[static gorge]

$\frac{5}{81} \times 3^x \ \
\frac{5}{81} \times 3^9 \ \
\frac{5}{81} \times 3^4 \times 3^5 \ \
5 \times 3^5 \ \ \ = 1215$

[woven radish]

Isaac

[static gorge]

https://tenor.com/view/emoji-smile-thumbs-up-okay-good-job-gif-15442359769564278345

ty @indigo flicker

[indigo flicker]

no problem

[static gorge]

.close

[devout snow]

Channel closed

Closed by @static gorge

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[lost crag]

whats "xV"?
x \in complex numbers,
V = vector space \subset of complex numbers

[sullen island]

V = vector space \subset of complex numbers
subset?

[finite obsidian]

Saying that xV = {xv| v in V} is a subset of V

I.e. applying “multiply by x” does not throw you out of V

Incidentally, it has to be equal for nonzero x, since multiply by 1/x

[lost crag]

ty mate

.close

[devout snow]

Channel closed

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[visual stone]

!help

[devout snow]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[visual stone]

how do i find the sign for -3x^2 + 7x - 2

[knotty sage]

the signe ?

[visual stone]

yeah like for a graph

[knotty sage]

like direction ?

[visual stone]

nah like f(x) >/ 0,k

or f(x) > 0,k

im confused tbf

[knotty sage]

have a picture of the question ?

[visual stone]

the picture wont help cause it doesnt provide more info

but basically theres a graph right

i gotta find the domain

image

zeros

variations

sign

.close

[devout snow]

Channel closed

Closed by @visual stone

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[odd geyser]

Open multiple views in geogebra 6 desktop

[odd geyser]

Hi I would like to have a 2d and 3d view side by side in geogebra 6 but cant find how to do that

I think I managed to do it before

Sorry if this isnt very math related didnt know where to ask

[tame palm]

In the same window?

Or did you want to separate the windows?

[odd geyser]

Same window

btw I called it geogebra 6 but now im not even sure if thats the version

I saw that theres one called "Classic" and you can select the views you want but cant find that here

[tame palm]

Click on the 3 dots to open another window.

You can only float windows in Geogebra 5.

[odd geyser]

should i change verisions?

I dont even know the difference

[tame palm]

Are you trying to do this?

[odd geyser]

yes

[tame palm]

Yeah, just click on the 3 dots on the top-right and click on "3D Graphics".

That should open another window.

If you want to float windows, you can only do that in Geogebra 5. That's useful if you have dual monitors.

[odd geyser]

the thing is I dont see the three dots

or im dumb idk

[tame palm]

Are you using the web app?

[odd geyser]

no

desktop

[tame palm]

On Windows?

[odd geyser]

yes

Windows 10, just reinstalled the app

[tame palm]

Hmm, that Window looks different from mine.

[tame palm]

It looks like you are opening the wrong program. There is a Geogebra.exe file which is the wrapper for the entire suite. It looks like you are opening geogebracalculator.exe which opens the same window for me.

C:\Users<user_name>\AppData\Local\GeoGebra_6\app-6.0.8812

The directory may be different depending on which version you have.

[odd geyser]

The folder is Geogebra Calculator instead of GeoGebra 6

https://www.geogebra.org/download?lang=en

[tame palm]

It looks like you didn't installed the entire suite.

[odd geyser]

thats weird

yeah theres something weird now

[tame palm]

https://geogebra.github.io/docs/reference/en/GeoGebra_Installation/

[odd geyser]

Like I cant even find it in programs

[tame palm]

If you scroll down, you will see a section called Geogebra 6 Classic. That is what you want to download.

[odd geyser]

thanks!

finally

thank you so much

[tame palm]

If you have dual monitors, you may want Classic 5 which will allows you to float windows.

[odd geyser]

I do not but maybe in the future

ill close this

thanks again

[tame palm]

yw

[odd geyser]

.close

[devout snow]

Channel closed

Closed by @odd geyser

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[devout snow]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[leaden swift]

Among the 32 participants of the literary meeting with the author of popular historical books, 2 people will be selected through a draw, with each receiving a different book by the author. How many different possible outcomes are there for this drawing?

[polar chasm]

how many books did the author write (or how many kinds of book does he give away)

[leaden swift]

bro are u srs

[leaden swift]

he give away 2 books

[polar chasm]

okay, lets assume they are the same 2 books all of the time

[leaden swift]

it doesbr matter bro

[polar chasm]

so first the "winner" of the first book is drawn, how many ppl are there to choose from?

[polar chasm]

it kinda does, if he was choosing from e.g. 4 different books and always chose 2 to give away, then there would be many more different outcomes

[leaden swift]

31 ig

[polar chasm]

32

there are 32 participants to choose from

after the first book is given away to the chosen one, there will be just 31 remaining participants to choose from (for the 2nd book)

[leaden swift]

no

because there cant be a winner for 2 books

2 people are chosen

not 2 rewards are chosen

[leaden swift]

yes

[polar chasm]

Yeah, so to choose the first winner, you have 32 options and for the second one you have 31 options

[shell plover]

2 people are chosen from 32 and then they are given 1 book each (different books)
is that the question?

[polar chasm]

essentially

[shell plover]

ohkk

[polar chasm]

now if you just multiply those two together, youll get ur answer

[devout snow]

@leaden swift Has your question been resolved?

[shell plover]

so we just need to selecct 2 people from 32 and then give them books

I think the answer should be || 32C2 × 2 = 32×31 ||

[devout snow]

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[kindred mauve]

can someone check my answer please

[ripe flax]

the question a says find the angle not the side

[wind mason]

You don’t know angle ABC and thus cannot apply the Law of Cosines.

[ripe flax]

so u still have to do another step

personally i think u should do sine rule here

cuz ur just finding angle, u got 2 sides and an angle given

[kindred mauve]

so

don't do cosine atall

[ripe flax]

mhm

[kindred mauve]

Ok

[wind mason]

And the way you applied it was incorrect.

[kindred mauve]

oh

let me try again

[kindred mauve]

is this it?

[wind mason]

Yeah.

[kindred mauve]

And part b

Idk wat to do for part b

.close

[devout snow]

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[devout snow]

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[thick quartz]

Hi i am confused as to why in a unit circle, why 5pi/6 - 2pi, doesn't land me on the same place as 5pi/6. Because isn't 2pi just equal to 360? so why when i minus 2pi from that, i don't just go in a circle back to the original place (5pi/6)?

[shell plover]

It lands on the same spot

[pseudo basin]

what two different spots do you find yourself landing on?

[thick quartz]

like if i minus 2pi from 5pi/6 dont i land on -7pi/6?

[shell plover]

Actually -7pi/6 ≠ +7pi/6

Negative sign indicates that the angle is measured in clockwise sense

[iron kindle]

,tex .unit circle

[woven radish]

Bonk

[thick quartz]

ohhh ok i understand that now then but could i ask for this question,
solve the equation for -2cos^2x + sinx+ 3= 2 for -2pi is smaller than or equal to x is smaller than or equal to 2pi

i got sin = 1/2 and sin = -1, why is -7pi/6 not part of the answer?

[pseudo basin]

okay right so uh

two things right there

• if you can't type the symbols ≤ and ≥, you should replace them with <= and >= rather than full on words like you did

[shell plover]

it is

[pseudo basin]

• if you don't want to do that, -2pi <= x <= 2pi is better replaced with x between -2pi and 2pi or x from -2pi to 2pi

[pseudo basin]

anyway who exactly is saying -7pi/6 does not belong to the solution set...?

[thick quartz]

my textbook lol

not sure why

[shell plover]

can we get a pic to see?

[thick quartz]

yeah althought it doesnt really explain why

its b)

the question is: solve the equation -2cos^2x + sinx + 3 = 2 for -2pi is smaller than or equal to x is smaller than or equal too 2pi.

[shell plover]

how is -pi/6 part of the solution set

sin(-pi/6) = -1/2

i think the answer is wrong

[thick quartz]

ohh ok that would make a lot of sense

[pseudo basin]

yeah looks like they bungled it

[thick quartz]

would you guys be able to tell me the correct answers?

i just want to make sure i know how to get the correct answers

[shell plover]

-11pi/6, -7pi/6, -pi/2, pi/6, 5pi/6, 3pi/2

[pseudo basin]

{-11pi/6, -7pi/6, -pi/2, pi/6, 5pi/6, 3pi/2} seems to be the correct answer for this one ie what the book should have said

[thick quartz]

ok thanks so much!

.close

[devout snow]

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[zenith spoke]

let $G,G'$ be groups, $H$ be a subgroup of $G$. define a function $f:G\to G'$ such that $\ker f=H$ and let $(G:H)$ denote the number of distinct left cosets of $H$ in $G$. Prove that $(G:1)=(\Im f:1)(H:1)$

[woven radish]

pirateking0723

pirateking0723

pirateking0723

[zenith spoke]

is this correct

[zenith spoke]

nvm this i was editting it but took a while so i cant delete it anymore

[toxic grove]

I don't love the typesetting but the actual math is right

[zenith spoke]

how do i improve it

[toxic grove]

It's just hard to read, and equations are being split over multiple lines

Sorry that was a bit mean

Also you're missing some asterisks

[zenith spoke]

no i didnt get offended by this i was asking genuinely because i too saw that it may be bad for others to read this

no need to say sorry

[toxic grove]

Tex tries very hard to not split equations over multiple lines

But if you don't give it enough leeway then it is forced to

So if you have long math equations, or paragraphs with lots of math and not very many words, you should use display mode

[zenith spoke]

where

[zenith spoke]

do i type \displaystyle right after every time i start with a $?

[toxic grove]

Actually I might be wrong on that, it's just not clear what the difference is between f(xH) and f*(xH)

No you can use double dollar sign

like $$1+1=2\qquad\hbox{this}$$

[woven radish]

depression

[zenith spoke]

well f:G→G' so it cant take xH no ?

because xH is not an element of G

[toxic grove]

You wrote $f^(xHyH)=f(xyH)=f(xyH)=f(xy)=f(x)f(y)=f(xH)f(yH)=f^(xH)f^(yH)$

[woven radish]

depression

[toxic grove]

so either you're missing asterisks or it means something in particular

[toxic grove]

I agree though

[zenith spoke]

ah wait i think that i should define $f^$ as $f^(xH)=f(x)$

[woven radish]

pirateking0723

[zenith spoke]

yea

mb