help-26

[icy sky]

Technically it depends on your domain and range

For real numbers, log is often undefined for negatives

[neon iron]

but putting x = 3 satisfies my eqn and its real

[icy sky]

Yeah i know

If your domain was negative real numbers and range was positive integers and you have an injection, this would be a valid function

Wellt echnically its still a valid function

Also ypur calculator might just be coded to only consider positive values for the base so it took a shortcut

[neon iron]

not my pic lol

[neon iron]

so your saying 3 should be "one" of the answers?

solutions*

[icy sky]

logs are a lot weirder than they seem

So yes, this is a solution

[neon iron]

wodefok

[icy sky]

Often considered the principal solution

ok hpld on lets ask wolfram alpha

,w log_{-2}{-8}

Heh

I guess it isnt the principal solution

[neon iron]

the fuck

[icy sky]

Lemme just boil it down to this: negative logs get weird and often involve complex numbers

[neon iron]

ye

they did so much but where tf is the final solution panel

[icy sky]

Technically 3 works in the way that -2^(3)=-8

But its complicated

[neon iron]

ye

[neon iron]

PROVE ME WRONG MORTAL

dont say its complicated

[icy sky]

Its not wrong

[thorny flame]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by garlicbredfries

[icy sky]

Ok yeah lemme explain

[neon iron]

i dont even see a 3 in there

and you say its (my method) not wrong

im probably the wrongest ive ever been

[icy sky]

Youre not

3 is just not the principal solution

Do you want to know how they got this?

How familiar are you with log properties

[neon iron]

ik

[neon iron]

ik

may be dumb but i think i know

[icy sky]

ok, so wolfram alpha calls "ln" "log"

[neon iron]

is this prob the combined eqn of all the solutions?

[icy sky]

The comboned equation would be $$\frac{log(8)+ni\pi}{log(2)+mi\pi}$$

[thorny flame]

GarlicB

[icy sky]

Ayy first try

Actually i lied

n and m have to be odd integers

But if you take n=3 and m=1, you get 3

Or really any n=3m, for all m odd

[neon iron]

man.

[icy sky]

Ok wait

[neon iron]

3 is the answer.

but it shouldnt be

[icy sky]

3 is an answer

[neon iron]

yea

but if it is an answer

then that means log _ -2 (-8) has a solution

[icy sky]

Yes

[neon iron]

BUT THAT ISNT DEFINED?

log cant have neg shit

[icy sky]

Lets say your function is log_{-2}(x)

What is your domain and range?

[neon iron]

no

that aint defined

[icy sky]

This is important

[neon iron]

i meant no as an answer lol

[icy sky]

Technically it is if you define the right domain and range

If you pick* the right domain and range

[neon iron]

i've never done that

• plus that aint gonnna come in my exam too

its just gonna fuck my brain 😭

[icy sky]

Well log_{-2}(-8) should be coming on your exams either but thats just my opinion

[neon iron]

idk my teacher never thought it plus never saw it in pyqs

[icy sky]

So if you want the simple answer, 3 is technically correct
If you want a more complicated answer, let me know

Otherwise feelf ree to close

[neon iron]

plus idek tf concept we are on with that

are we in real world or complex?

[icy sky]

Im assuming your teacher only wants the real answer

[neon iron]

a

my teacher didnt ask that

its just a diff guys que and i thought he raised a good doubt

and i thought yes 3 should be an answer

and you said it is

so im correct

[icy sky]

It is indeed

But it gets weird

[neon iron]

but idk the concept so im not confident

yea alr

[icy sky]

So i woul say "yes, but..."

[neon iron]

i just thought smth must be wrong in my solving

[neon iron]

k lol

ty

.close

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[icy sky]

Lol np

Feel free to ask me if you want the full explnanation

I kinda sprung it onto you when you didnt want it

[topaz sinew]

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[neon iron]

All endomorphisms of a field are automorphisms

[neon iron]

?

[lyric patio]

help

[neon iron]

Since every homomorphism of a field is injective

Then all endomorphisms of a field are injective

[vital relic]

injective doesnt imply surjective?

you might need choice, but u can have non automorphisms afaik

maybe thats bs

i misremembered, seems ur right

https://math.stackexchange.com/questions/4094392/which-fields-have-the-property-that-every-field-endomorphism-is-an-automorphism

wait no comments/answer say otherwise

https://math.stackexchange.com/questions/201951/find-an-example-of-a-non-surjective-field-endomorphism

yh good idea to try think of examples before seeing

they arent too hard

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[simple orchid]

I'm not sure how v_i which is 18.975 changed into v_i1 which is 12.65

v_i1 and v_f should be the same?

[pastel juniper]

,calc (30 - (0.5)(9.8)(2.25))/(1.5)

[simple orchid]

?

[pastel juniper]

they want up to be negative?

ok

[thorny flame]

Result:

12.65

[pastel juniper]

who's 18.975?

[simple orchid]

Height 1

[pastel juniper]

what they did was calculate the velocity when you saw the melon

which is 12.65

then taking that as final velocity for the motion before you saw the melon and 0 to be intitial (dropped), they get the top small height

and sum the heights

[simple orchid]

Yeah, but why are there two values for v_i1?

Shouldn't v_i1 be 18.75m/s?

[pastel juniper]

i havent watched the video serrafim

i just told you how the problem is solved

[simple orchid]

Then is the answer 17.58 correct?

https://youtu.be/wYN-qjaVqXE?t=416

This is an embedded link with timestamp

If you wanna take a look

[pastel juniper]

lemme watch it

,calc 30 - (0.5)(9.8)(2.25)

[thorny flame]

Result:

18.975

[pastel juniper]

this is

tbh wrong notation

[simple orchid]

?

[pastel juniper]

like really messed up

[simple orchid]

What is?

[pastel juniper]

they just performed half calculation

and then did other half

[simple orchid]

I got my height for the second one 17.58

Used g = 10

Not sure if you can see my work process

[pastel juniper]

i can

[simple orchid]

So the total height is d + 30m

[pastel juniper]

yeah nothing wrong

[simple orchid]

That woman got 38 m

for her answer though

tf?

She's wrong, right?

[pastel juniper]

umm

idk lemme watch

[simple orchid]

What kind of bs is this

[pastel juniper]

,calc (30 - 11.25)/(1.5)

[thorny flame]

Result:

12.5

[pastel juniper]

your calculation is wrong

._.

[simple orchid]

Which part?

Why?

[pastel juniper]

you shouldnt get 18.75

[simple orchid]

Why?

[pastel juniper]

you forgot to divide by 1.5

[simple orchid]

Oh

That makes sense

Hold up

Oh

I get 7.8 m for the second height

That's much similar ig

Ohhh

[pastel juniper]

[simple orchid]

That parentheses were multiplication

I thought it was some kind of parameter

2:44 am moment

Thanks though

Love ya

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[soft escarp]

isn't this wrong?

i think i^100 should be -1

sqrt(-1) to the power of 2 aka squared should be -1

[thorny flame]

Dyssrupt

[simple orchid]

Imaginary numbers after the 4th power are constantly repeating

[thorny flame]

Dyssrupt

[soft escarp]

oh wait

i thought i^2 was equal to i^50

i just made a stupid assumption

nvm guys lol

thanks for correcting me

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[soft escarp]

wow

it got me again

in the second example lol

i made the assumption that i^51 would be just i because

i^100 is even

and i^100 is 1

so i just multiplied by i

but that assumption turned out to be wrong

i exponent property is very trappy

[neon iron]

heh

if you remeber the value of i i2 i3 i4

you probably wont have much prob with theses type of ques

[thorny flame]

Dyssrupt

[jade thunder]

If you know what an argand diagram is the power it rotates it 90 degrees

[neon iron]

just dividing the power by 4 and looking at the remainder will be the power of i

[pastel juniper]

discord's lagging

[neon iron]

im bad at typing ngl

[jade thunder]

i^0 is “1, 0” as in (1 + 0i)
i^1 is “0, 1” as in (0 + 1i)
i^2 is “-1, 0” as in (-1 + 0i)
i^3 is “0, -1” as in (0 + (-1)i)
Then it repeats

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[tardy dagger]

$P(r+s,n) = \sum_{k = 0}^{n} \binom{n}{k} P(r,k)P(s,n-k)$

[thorny flame]

Ziggy

[tardy dagger]

$\frac{(r+s)!}{(n-r-s)!} = \sum_{k=0}^n \left(\frac{n!}{k!(n-k)!}\right) \left(\frac{r!}{(r-k)!}\right)\left(\frac{s!}{(s-n+k)!}\right)$

[thorny flame]

Ziggy

[tardy dagger]

$\frac{(r+s)!}{(n-r-s)!} = 2^n \left(\frac{r!}{(r-k)!}\right)\left(\frac{s!}{(s-n+k)!}\right)$

[thorny flame]

Ziggy

[tardy dagger]

can i do that?

[silk shard]

What is k in that expression?

[tardy dagger]

but thats not true?

$$\sum_{k=0}^n \Big(\dfrac{n!}{k!(n-k)!}\Big) = 2^n$$

[thorny flame]

Ziggy

[supple sun]

Ye sorry

[tardy dagger]

the size of the subset

[supple sun]

But what I wanted to say is that $\Big(\dfrac{n!}{k!(n-k)!}\Big)\neq 2^n$

[thorny flame]

Joseph.P

[tardy dagger]

yes but the sum does

[supple sun]

Oh thought all the expression was in the sum

[tardy dagger]

oh i suppose it is

but you can isolate the one

cant you?

[supple sun]

No you can’t actually because if so you would have $$(a+b)^n=\sum_{k=0}^n\binom{n}{k}a^kb^{n-k}=2^na^kb^{n-k}$$ and in the last expression you don’t even know what is $k$.

[thorny flame]

Joseph.P

[tardy dagger]

hn

hnm

[supple sun]

This expression is wrong

[tardy dagger]

okay

this one isnt though

$P(r+s,n) = \sum_{k = 0}^{n} \binom{n}{k} P(r,k)P(s,n-k)$

[thorny flame]

Ziggy

[supple sun]

First what’s P ?

[tardy dagger]

permutation

it makes me want to use the binomial theorem somehow

[supple sun]

You sure it’s not arrangements ? Because if it’s not you got the wrong formula

[tardy dagger]

$(x+y)^n = \sum_{k = 0}^{n} \binom{n}{k} x^ky^{k-n}$

[thorny flame]

Ziggy

[tardy dagger]

what i've written is correct

[topaz sinew]

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@tardy dagger Has your question been resolved?

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[simple pawn]

hi so ive made some progress

[simple pawn]

this is what ive done so far but i am getting a wrong answer

maybe it should be y(t) = (t^3+1) (e^(t^2))

?

[acoustic pecan]

this seems like a good shout

your working seems fine to me

[simple pawn]

lol i have one more attempt left

alright ima take a shot

LETS GO BABY

thanks a lot for your help yesterday too man

[acoustic pecan]

no worries

[simple pawn]

luv u. have a good wknd

[acoustic pecan]

glad to see youre doing well

you too

[simple pawn]

:3

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[outer steppe]

How to find pyahtogeream therom

[outer steppe]

for iscoelse triangle

[acoustic pecan]

could you be more specific, do you have a picture of a question you want this for?

[outer steppe]

the bottom side is 4

[acoustic pecan]

the pythagorean theorem applies to right angle triangles
this isosceles triangle can be split into two right angle triangles (see the dotted line)

the only info youre missing is the base length of one of those two right angle triangles, what would that be?

[outer steppe]

4

alr i got it thanks

.close

[acoustic pecan]

thats the base length of the isosceles

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[outer steppe]

thanks

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[pseudo locust]

Learning some real analysis, I want to check a proof I did that $a_n=n^3$ for $n\in\mathbb{N}$ is unbounded from above, so we are working towards the statement $\forall C\in\mathbb{R},\exists n\in\mathbb{N}, a_n>C$

[thorny flame]

bostock

[pseudo locust]

so for all $C\in\mathbb{R}$, there exists $n$ such that $n^3=n\cdot n^2>C$ which is true as $n\in\mathbb{N}$ and $n^2, C\in\mathbb{R}$ so the Archimedean property applies

[thorny flame]

bostock

[pseudo locust]

Therefore the statement we are working towards is true so $n^3$ is unbounded from above

[thorny flame]

bostock

[pseudo locust]

<@&286206848099549185>

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@pseudo locust Has your question been resolved?

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[boreal mulch]

how would i do b? i tried by parts, and got stuck at integrating sec^3 theta, thanks

[boreal mulch]

,rotate

[keen venture]

Trig sub,
x = sec(t)

[boreal mulch]

yes i did that

[keen venture]

You'll get part a back

[boreal mulch]

Is this the correct approach

[keen venture]

OH sorry I thought part b was the other one

[split inlet]

more integration by parts ig

[boreal mulch]

oh

lemme try that

[split inlet]

sec^2x and sec x

[keen venture]

I wouldn't start that way. Start with a trig sub x = 2sec(t)

[split inlet]

oh wait not x

i meant theta

same thing though

[boreal mulch]

ye im trying this rn

do i by parts tan^2t*sect?

[split inlet]

yes

[keen venture]

Turn it into sin and cos
No wait, that won't be it

[split inlet]

tan^2=sec^2-1

[boreal mulch]

wait im getting a bit closer

[split inlet]

update pic pls

[split inlet]

no wait

its sec(sec^2-1)

so sec^3-sec

[boreal mulch]

which step?

[split inlet]

at tan^2*sec right

[boreal mulch]

ohh

[split inlet]

so since you have 2 terms

you can seperate them into different integrals

[boreal mulch]

yep i wrote 4integrate(sec^2u-1)secu du

[keen venture]

Integral of tan²(t) is not tan(t) - t

[boreal mulch]

it is

,w integrate tan^2x

[thorny flame]

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[keen venture]

Huh, I stand corrected

[boreal mulch]

Consider the tan^2x as sec^2x-1 and integrate that

[keen venture]

Smart

[boreal mulch]

so anyways now im stuck at integrate sec^3u - secu du

[keen venture]

sec³ is dreaded and I was hoping we didn't have to go there

[split inlet]

werent we doing sec^3 earlier

u=sec

dv=sec^2x or something

[boreal mulch]

Wait

we were at 4 integrate tan^2usecu right?

[split inlet]

yeah

[boreal mulch]

then turn the tan^2 into (sec^2u-1)

[split inlet]

yeah

[keen venture]

Here's a clever way
https://math.stackexchange.com/questions/179935/how-to-integrate-sec3-x-dx

[split inlet]

uhh

[boreal mulch]

yo waittt

[split inlet]

what

[boreal mulch]

i am getting something here hold on

[split inlet]

so what's the integral again

[split inlet]

bingo

wait what was the original question

[boreal mulch]

integrate sqrt(x^2-4)

[split inlet]

so we plug that into the top pic?

this thingy?

[boreal mulch]

wait im rewriting everything to make it clear

[split inlet]

kk

[boreal mulch]

,rotate

[thorny flame]

You must be a bot manager to use this command!

Sep 16, 2023, 1:55 AM

[iron harness]

!status

[topaz sinew]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[boreal mulch]

alright i solved it

,rotate

[split inlet]

nice

[boreal mulch]

thanks guys

.close

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[split inlet]

ok

do you know where to start

[undone dawn]

Make the slope perpendicular?

[split inlet]

yes

if it's perpendicular, then the it's the negative reciprocal of the original slope

[undone dawn]

Y=10x+6

Not sure how to implement the (-1,-8)

[empty sail]

You know the slope of the perpendicular line but that line passes through (-1, -8)

So the equation of a line is y = mx + b. You know m

You use that point to find b

[undone dawn]

Yeah that’s ab where I got

U sub something

-8 with y or something like that?

[empty sail]

Yeah

And you know the x value too

[undone dawn]

Y=10x+2

[empty sail]

Yeah that's right

[undone dawn]

Ok thanks!

Also i got this one. Ik i gotta isolate x

By dividing y

Dividing x

Yeah idk bro

[restive inlet]

how would you solve for $x$ if you were given something like
$$5 = 7x - 9$$

[thorny flame]

ℝam()n()v

[undone dawn]

Add 9 to 5

Then divide

[restive inlet]

by?

[undone dawn]

14/7

X = 2

[restive inlet]

same idea applies here

[undone dawn]

First wouldn’t I have to move y to the other side and remove the x

[restive inlet]

same idea applies here

the equation is in the exact same form, except i replaced y,6m,2b with numerical values
the process for isolating x is exactly the same

separate terms with x from terms without

divide by coefficient of x

[undone dawn]

So i subtract y with 2b

[restive inlet]

no

reword that

[undone dawn]

Uhh

[restive inlet]

recall what you did in the first step

of
5 = 7x - 9

[undone dawn]

I added

[restive inlet]

added what exactly
and apply the exact same idea to what you have here

[undone dawn]

So add -2b

[restive inlet]

no

did you add -9 in my example?

[undone dawn]

Yup

[restive inlet]

you did not

[undone dawn]

9

I added 9

[restive inlet]

yes

[undone dawn]

So i add just 2b

[restive inlet]

9 is not -9

yes

add 2b to both sides

that'll give you
$$y + 2b = 6mx$$

[thorny flame]

ℝam()n()v

[restive inlet]

and recall what you did when you had
$$14 = 7x$$

[thorny flame]

ℝam()n()v

[undone dawn]

Divide 6m both sides ?

[restive inlet]

yes

[undone dawn]

Y+2b/6m=x

[restive inlet]

() are needed

to clearly indicate numerators and denominators

and they will be necessary for the system to interpret the desired result

(y + 2b)/(6m) = x

[restive inlet]

oh, if the system allows you to input fractions like that then those () aren't needed then

they would be for single line text input

[undone dawn]

But its fine if they are kept right

[restive inlet]

yes

[undone dawn]

Ok

Is this (3,0)?

[restive inlet]

no

[undone dawn]

Not sure what y1=y2 us

Is

[restive inlet]

y_1 and y_2 are lines

[undone dawn]

Yeah

But them being = and what x is

[restive inlet]

that's the x-coordinate of the intersection point

which is just 3 only

[undone dawn]

Wouldn’t that be (3,0)

The coordinate

[restive inlet]

no

[undone dawn]

Ah ok

[restive inlet]

the vertical coordinate of the intersection isn't even 0

(3,0) would refer to that green point

[undone dawn]

Ohh

And this one I’m not sure how this pulled together into a slope intercept form

[restive inlet]

you are essentially given two points

use a method of determining a straight line equation using two points

like two-point formula
or first determining the slope then applying point slope formula
(which is pretty much the same thing)

[undone dawn]

I got like y=5/6x+.33

[restive inlet]

you messed up somewhere

b value is wrong

[undone dawn]

To get b value u choose one of the ordered pairs right

So I’d have to put 4 in for x

So 5/4 x 4

[restive inlet]

1 sec,

the slope was also wrong

can you show your work

[undone dawn]

Yeah

[restive inlet]

firstly, those indexes should be subscripts not superscripts
$$x_1,y_1,x_2,y_2$$
NOT
$$x^1,x^2,y^1,y^2$$
secondly you've misidentified their values

[thorny flame]

ℝam()n()v

[restive inlet]

f(3) = 4
indicates that when x=3, y=4

[undone dawn]

Oh shoot

6/5 @restive inlet ?

[restive inlet]

for the slope, yes

[undone dawn]

-6/5(-2)=2.4

2.4 - (-2) = 4.4

@restive inlet

F(x)=6/5x+4.4

[restive inlet]

where's

-6/5(-2)=2.4
coming from

[undone dawn]

F(-2)=-2

[restive inlet]

says nothing about why you're doing

-6/5(-2)=2.4

what's the purpose of that

[undone dawn]

To find b?

[restive inlet]

how's that supposed to help find b

[undone dawn]

Add a value to x

[restive inlet]

that just looks like you're multiplying slope to an x coord

[undone dawn]

What else would u do

[restive inlet]

show full equations to clearly show what's going on

so you don't misplace values, screw up signs / other operations

if you want to find b that way, start with something like
y = mx + b
do the substitution and do clear manipulation to get b

[restive inlet]

yes

continue

[undone dawn]

^

[restive inlet]

yes

ideally you should write the 0 before the .

[undone dawn]

So y = 6/5x + .4

[restive inlet]

ideally you should write the 0 before the .

[undone dawn]

Yea

Is this correct?

[topaz sinew]

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[grim swift]

im low intelect

idk how to go about this

oh its 10y

.close

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[vital moth]

its the first

[acoustic tangle]

Looks like g to me

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[zinc glen]

The question is : determine whether f and g r inverse functions

[lapis raft]

,rotate

[lapis raft]

are you doing d?

[zinc glen]

Yes

[lapis raft]

so, what's wrong?

[zinc glen]

And I want to make sure if I answered correctly

[lapis raft]

you did

[zinc glen]

Ty!

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[jovial current]

6
Σ (1/n+1)
n=1

[jovial current]

help me solve this please

[neon venture]

hmm

do 1-n+n = n+1 - n on the numerator
and then telescoping

[jovial current]

telescoping?

[neon venture]

wait a sec

$\sum_{n=1}^6 \frac{1-n+n}{n+1} = \sum_{n=1}^6 \left(1 - \frac{n}{n+1} \right)$

[thorny flame]

Herels

[neon venture]

hmm it wont work i guess

[jovial current]

😭

i dont understand

[neon venture]

squaring what

[jovial current]

no

wrong send

[jovial current]

HOW DOES IT WORKK?

[neon venture]

no need to yell

Like i said, I just add n-n on the numerator

[jovial current]

oops my bad i wasntt T-T sorryy

[jovial current]

so its 1?

[neon venture]

no, do you know how to manipulate fractions ?

$\frac{1+n-n}{n+1} = \frac{n+1-n}{n+1} = \frac{n+1}{n+1} - \frac{n}{n+1} = 1-\frac{n}{n+1}$

[thorny flame]

Herels

[neon venture]

thats basically what I did

but it wont work anyway

[jovial current]

oh 😭

i dont understand it in that way

[neon venture]

wdym

[pastel juniper]

this is a HP

Harmonic Progression

[neon venture]

yes

should give something like this :

[jovial current]

what does harmonic progression even mean

[neon venture]

ln(n+1) + γ + O(n)
I believe

[neon venture]

well

Its just a sum

$\sum_{k=1}^n \frac{1}{k}$ is the harmonic progression

[thorny flame]

Herels

[jovial current]

dang what does k means

i havent learned it yet

[neon venture]

its just a variable... nothing deep

[jovial current]

so the answer is 1-n/n+1 ??

[neon venture]

Ive never said that...

[jovial current]

I was asking.............

because i literally dont understand any of this T-T

[neon venture]

Basically, your sum is not possible to calculate with regular ways

And we said your sum is like the harmonic progression

$\sum_{k=1}^n \frac{1}{k} = \ln(n) + γ + O(1)$

[thorny flame]

Herels

[neon venture]

with γ the Euler mascheroni constant

I don't know the result for your sum

[jovial current]

OMG WHAT R THOSE SYMBOLS

[neon venture]

let ignore that

anyway, your sum is not possible to calculate

with regular ways

[jovial current]

😔😔😔

idk how to start calculating it

[neon venture]

Me neither

thats why we talked about the harmonic progression

,w \sum_{n=1}^6 \frac{1}{n+1}

at least wolfram alpha gaves the result

[thorny flame]

Results provided by WolframAlpha

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

Requested by herels

[jovial current]

aaaaahhh ill jst deal w this later

anyway

look

am i doing this correctly

[neon venture]

yes

[jovial current]

omg im finally doing smthing rifght for the 1st time in a while

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[shrewd prairie]

anyone help with this ? just answers is enough

[drifting swift]

we don't give out answers.

[shrewd prairie]

oh ok my bad

[drifting swift]

!status

[topaz sinew]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[shrewd prairie]

1

equation perhaps

[drifting swift]

do you know basic algebra

[shrewd prairie]

yes

[drifting swift]

ok

[shrewd prairie]

sorta

[drifting swift]

let x be the amount of money invested at 4%

can you tell me an expression for how much money was invested at 7%?

both in dollars obviously.

[shrewd prairie]

let y maybe

should there be 2 unkowns ?

[drifting swift]

you can do it with one

[shrewd prairie]

oh ok

[drifting swift]

since you know how much was invested in total

(76000)

phyllis has 76000 dollars of which he invests x into a 4% account, and the rest into a 7% account.

how much did he invest into the 7% account?

[shrewd prairie]

76000-x

[drifting swift]

right

what's the interest he earned from the 4% investment?

[shrewd prairie]

76000 times 4% ?

[drifting swift]

no

how much did he invest into the 4% account?

[shrewd prairie]

oh

x

[drifting swift]

yes, so how much interest did that give him

[shrewd prairie]

?

whats the equation of simple interest?

ive only learnt compound interest

[drifting swift]

you don't need to know either one

it's a 1 year deposit

it makes no difference at all lmao

[shrewd prairie]

is it x times 4%

[drifting swift]

of course it is

0.04x

[shrewd prairie]

ooh ok

[drifting swift]

and how much did he earn from the 7% investment?

[shrewd prairie]

(76000-x) times 7%

[drifting swift]

0.07(76000-x)

ok

so now can you write the equation that says the guy's total earnings are 4240?

[shrewd prairie]

added both are equal to 4240?

thats the equation?

[drifting swift]

yes but write it out

[shrewd prairie]

0.04x + 0.07(76000-x) = 4240

ooh ok i got the ans ty

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[mint violet]

is this euclidean algorithm correct i think its wrong at the step 2=(1)(2)+0 please verify

[mint violet]

should it be a=a.1+0 instead

<@&286206848099549185>

?

shouldnt every reminder go down 3 times

[flat kindle]

what

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[crisp stag]

Hi can I have help with this math problem. I have gotten the answer 2/4

[crisp stag]

Hi can I have help with this math problem. I have gotten the answer 2/4

and can I have a explanation

[acoustic tangle]

First of all, don't you think 9 should be included in the range?

Since there is a point on the graph with y-coord equal to 9

I.e., (1, 9)

[crisp stag]

i don't really understand what you mean by that

9 is in the range

[acoustic tangle]

Yes, but you put ) instead of ] next to the 9 in your answer

[crisp stag]

okay

so the top part is correct

domain

[acoustic tangle]

The domain that you have given is correct, yes

[crisp stag]

okay

[acoustic tangle]

Another issue with your answer for the range is that you are including some negative numbers in it

It would mean the graph somewhere goes below the y-axis

[crisp stag]

oh

[acoustic tangle]

That, as you can see, does not happen

[crisp stag]

yes

[acoustic tangle]

It's always above the y-axis except when the graph intersects it

[crisp stag]

okay

[acoustic tangle]

So you instead have [0, 9] as the range

[crisp stag]

okay

so for the range it would be (1,9)

[acoustic tangle]

With [] instead of ()

Which parenthesis you put matters

[restive inlet]

and 0 not 1

[crisp stag]

oh okay

[acoustic tangle]

Oh and that too yeah

[crisp stag]

okay could i have help on this problem as well

I got this incorrect too

[onyx shard]

what is the definition of the amplitude?

[crisp stag]

to be honest i am not 100 percent sure

[onyx shard]

look it up then in your notes

[crisp stag]

it is the the amount by which the graph of the function travels above and below its midline.

[onyx shard]

yeah

where is the midline here?

[crisp stag]

i dont know how to find the midline or what it is

[onyx shard]

mid line

[crisp stag]

middle \

[onyx shard]

well, its what you think it is 😄

[crisp stag]

okay

[onyx shard]

the axis in this case

so whats the amplitude?

[crisp stag]

2?

[onyx shard]

yeah

[crisp stag]

so i was correct

when answering the problem

[onyx shard]

and what is the definition of period?

[crisp stag]

but it says i got a 0/2

so 2 was wrong

.close

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[supple sun]

Let $z\in\overline{\mathbb{Z}}$ an algebraic integer of conjugates $z_1,...,z_n$. Let $P\in\mathbb{Q}[X]$ also be a polynomial of degree at most $n−1$, and $p\in\mathbb{N}$ a prime number. Which of the following statements are always true?

All $P(z_1),...,P(z_n)$ are conjugates of $P(z)$. \

The conjugates of $P(z)$ are among $P(z_1),...,P(z_n)$. \

If $P(z)$ is an algebraic integer, then $P\in\mathbb{Z}[X]$ \

The sequence $z,z^2,z^3,...$ of the powers of z contains at least one term worth 0 or 1 modulo p. \

$\overline{\mathbb{Z}}$ modulo $p$, i.e. $\overline{\mathbb{Z}}/p\overline{\mathbb{Z }}$ is a body. \

There are an infinity of algebraic integers two to two distinct modulo p.

[thorny flame]

Joseph.P

[supple sun]

I think the third one and the fourth are wrong

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@supple sun Has your question been resolved?

[supple sun]

<@&286206848099549185>

<@&268886789983436800>

Oh wow

[urban grove]

#groups-rings-fields might be a better place to ask

but also, instead of body, do you mean "field"?

is this translated from german/french?

[supple sun]

Translated from the french

[supple sun]

In French it’s "corps" so maybe

[supple sun]

Thanks

Do I close this channel ?

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@supple sun Has your question been resolved?

[supple sun]

.close

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[stuck bronze]

How do I solve this

[stuck bronze]

nvm

.close

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[stuck bronze]

am i correct?

[stuck bronze]

nvm

I figured it out

[acoustic pecan]

Where did 2 come from in b

[stuck bronze]

ty tho 😭💖

[north canyon]

(c) should be 7

[stuck bronze]

7.0

[north canyon]

Yeah

[stuck bronze]

yea

I do have another qs tho

how to I solve this

[north canyon]

Do you know the basic limit laws?

[mint violet]

just substitute all the values

[topaz sinew]

@stuck bronze Has your question been resolved?

[stuck bronze]

oui

but

for question e

do you think my answer will get accepted if I write DNE

or should I write D.N.E

I have no idea fr

oop-

nvm

.close

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[vital moth]

helllo everyone E_1 and every E_alpha are subspaces of E

[vital moth]

i dont understand the $\pi_{\alpha}\circ i_\beta$ part

[thorny flame]

calculus is fun

[topaz sinew]

@vital moth Has your question been resolved?

[vital moth]

<@&286206848099549185>

[opal vault]

Write every x as sum(x_gamma) if you need a third index

If x is in E_alpha, then you can write x = x_alpha

[vital moth]

are you talking about the piecewise map pi_alpha circ i_beta

[opal vault]

So first of all, if alpha=beta, then $\pi_{\alpha}\circ i_\beta(x) = \pi_{\alpha}\circ i_\alpha(x_\alpha) = \pi_\alpha(x_\alpha) = x_\alpha = x$

[thorny flame]

rafilou2003

[opal vault]

If $\alpha \neq \beta$, then $\pi_{\alpha}\circ i_\beta(x) = \pi_{\alpha}\circ i_\beta(x_\beta) = \pi_\alpha(x_\beta) = 0$

[thorny flame]

rafilou2003

[opal vault]

This is because x is in E_alpha so we can write x = x_alpha

[opal vault]

And this is because x is in E_beta so we can write x = x_beta so the coefficient in x_alpha is null

[vital moth]

isnt it i the output of the function at beta=alpha?

i didnt understand what was the letter written

[opal vault]

"i" is for identity here

[vital moth]

oh ok

[opal vault]

So here alpha=beta, then $\pi_{\alpha}\circ i_\beta(x) = x = i(x)$

[thorny flame]

rafilou2003

[opal vault]

So we can write in this case the equality between linear maps $\pi_{\alpha}\circ i_\beta = \iota$

[vital moth]

yea sure

[opal vault]

I prefer the notation Id tho

[thorny flame]

rafilou2003

[opal vault]

Oh yeah it's a iota

[vital moth]

but how is $\pi_{\alpha}\circ x_{\beta}=0$?

[thorny flame]

calculus is fun

[opal vault]

When alpha neq beta

[vital moth]

for alpha neq beta

yea i wanted to continue the question

[opal vault]

Ok

So for $x \in E_\beta$, as mentionned, we can write $x = \sum_{\alpha}x_\alpha = x_\beta$

[thorny flame]

rafilou2003

[vital moth]

is it bc $x_{\beta} \notin E_{\alpha}$?

[thorny flame]

calculus is fun

[opal vault]

Nono, it's because E_alpha and E_beta are in direct sum

So if x is in E_beta, its projection in E_alpha is 0

[vital moth]

how do you know that they are

[opal vault]

I can read "direct sum of subspaces"

[vital moth]

oh it says $E=\sum_{\alpha}{E_{\alpha}}$ is a direct sum of subspaces

[thorny flame]

calculus is fun

[opal vault]

Yep

So E_alpha and E_beta are in direct sum

So for any vector of E_beta, its projection on E_alpha is the 0 vector

[vital moth]

so for $\beta \neq \alpha$ we have $E_{\alpha}\cap E_{\beta}=0$

[thorny flame]

calculus is fun

[finite cipher]

Guys

I have 5 questions and need help in them

[opal vault]

#❓how-to-get-help

[vital moth]

so if x_beta is in E_alpha then it must be 0 otherwise it isnt in E_alpha and its projection will also be 0?

[opal vault]

Yes

[vital moth]

why do we consider it 0 if x_beta notin E_alpha

[opal vault]

We dont consider it, it is

[vital moth]

why dont we say that pi_alpha circ x_beta doesnt exist

[opal vault]

Because it does exist and is 0

[vital moth]

why is it 0

[opal vault]

Since E_alpha and E_beta are in direct sum, any element x of E_beta can be written IN A UNIQUE WAY AS x = x_alpha + x_beta + ...
(Where x_alpha in E_alpha, etc...)

Since we also can write x = 0 + x + ...

Then this IS the unique way and its coefficient x_alpha = 0

[vital moth]

do you mean x_beta

forgive me if i am being stupid but i am heading to bed after this and you can say i am starting to get into bot mode

[opal vault]

If x is in E_beta, then x_beta = x and x_alpha = 0

Because x = x_alpha + x_beta + ... = 0 + x + ...

[vital moth]

yea i got what you mean

by coef in x_alpha

the term ax_alpha in the sum that represents x_beta is just 0

[opal vault]

Exactly

[vital moth]

so its projection onto E_alpha is 0

[opal vault]

Yep

[vital moth]

tysm for your time i showed maximum stupidity in this and the other channel about continuity

[opal vault]

Nah its alr

[vital moth]

i reached absolute max 😂

i gtg sleep now cyaaa gn

.close

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[soft lily]

I don't understand how this fits

this is a 3x4 matrix

how can there be a vector x_4 with 4 entries?

[opal vault]

3×4 matrix multiplied by 4×1 vector will output a 3×1 vector

Matrix multiplication

[soft lily]

how is that?

[opal vault]

Remember that an (m×n) matrix and a (k×l) matrix can only be multiplied if n =k, and will then output an (m×l) matrix

[soft lily]

not sure what that means

oh I see

I mean I'm starting to

no

I don't get that

[opal vault]

The number of columns of the first matrix has to be equal to the number of rows of the second

[soft lily]

m = k

n = l

[opal vault]

How many columns does an (m×n) matrix have?

[soft lily]

n

[opal vault]

Yes

[soft lily]

why would the columns and rows switch places

[opal vault]

I suggest looking at how matrix multiplication works again

[soft lily]

again implies I ran into it

which I have not

oh the next 3blue1brown video says matrix multiplication

[opal vault]

Where does this come from? Because if you haven't seen matrix multiplication I don’t think a teacher would let you do those kinds of exercises on your own

[soft lily]

my homework?

[opal vault]

Or at least have you seen how multiplying a matrix with a vector works?

[soft lily]

nope

I saw it once and it made 0 sense

[opal vault]

So, how are you supposed to do this exercise if you can't compute Ax ?

[soft lily]

oh sorry

I misread

I just figured out matrix, vector multiplication

[opal vault]

Ok, so can you explain how does that work?

[soft lily]

row entries of a vertical vector X are paired with the columns of a matrix A

for whatever reason

[opal vault]

Yes

Basically, from an existing vector x, we create a new vector, Ax, by taking the weighted sum of its coefficients

The "weights" are given by the matrix A

[soft lily]

how can you call a matrix bigger than 1 row by x columns or 1 column by x rows a vector?

[opal vault]

Idk if that is your question, but a vector here is just an element of R^n, but n can be different

x can be a vector of R⁴ while Ax can be a vector of R³

[soft lily]

does R^n describe a vector space?

[soft lily]

that is difficult to comprehend

[soft lily]

does that mean a vector can be literally anything?

like a collection of vectors?

[opal vault]

A vector, in general, is an element of a vector space

[soft lily]

what do you mean by element?

[opal vault]

Ie anything that is inside a vector space

For example, (1,2,4) is an element of R³

We write $(1,2,4)\in \bR^3$

[thorny flame]

rafilou2003

[opal vault]

R³ is a vector space, so by definition, (1,2,4) is a vector

[soft lily]

is a 2x3 matrix a vector in R^3?

[opal vault]

No, but it's a vector in $M_{2,3}(\bR)$

[thorny flame]

rafilou2003

[soft lily]

woah wtf is that

[opal vault]

The set of 2×3 matrices

[soft lily]

okay, so you weren't calling a matrix a vector

I was just forgetting how vector multiplication worked

my b

[opal vault]

Here I was trying to keep "vector" specifically for objects in R^n

[topaz sinew]

@soft lily Has your question been resolved?

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[topaz sinew]

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[neon iron]

I need help with this I'm doing part a right now

[neon iron]

Sorry it's blurry here's a better image

I asked chatgpt and it messed me up

[radiant tapir]

I can't help you because I'm heading out but just as a word of advice, don't use ChatGPT to do math. It's often inaccurate or wrong (for the time being)

[opal vault]

Do you know how to find k?

[neon iron]

Not really

Also am I using the right formula

[opal vault]

This is the correct formula

[neon iron]

Because chat gpt gave me 2 different formulas one with e and one with a

[opal vault]

You've already used the starting price (in 2010) as information. There's still one piece of information you haven't used

[neon iron]

Wait I'm sorry could you please define what p0 and e and k mean I want to write them down

And t

I just forgot a lot of this stuff

[neon iron]

Is it the 7 years

[opal vault]

P0 is the starting price

e is just short for exponential function

So P(t) = P0 exp(kt)

Finally, k is called the growth factor

[opal vault]

And yes you haven't used this yet

[neon iron]

But just in general tho

[opal vault]

The starting value

[neon iron]

Thank you

[opal vault]

At t=0

[neon iron]

Ok thank you

[neon iron]

Wait so what does e do?

[opal vault]

e is for the "natural exponentiation"

because the derivative of f(t) = e^t remains e^t

and so we write e = exp(1)

[neon iron]

So p0×e^kt is called an exponential formula?

What's it's name again sorry

[opal vault]

yes, it's a type of exponential function

[neon iron]

If can be used for both growth and decay right?

[opal vault]

yes

when k > 0, it's growth. When k < 0, it's decay

If you use the formula P0a^t, then growth is for a>1 and decay is for a < 1

The link between the two is :
P0a^t = P0e^(ln(a)t)

[neon iron]

Sorry if I'm annoying you with my questions I just wanna write it down

[neon iron]

Should I plug it in for t

[opal vault]

yep

if you plug in t=7

by definition, P(7) = P0e^(7k)