help-0

[stable spire]

very simple qn

u in which std

[alpine sable]

Not sure what to do when 2tan is involved

[alpine sable]

Std?

Like grade?

[stable spire]

grade

yeah

[alpine sable]

Grade 12

[stable spire]

in which country

[alpine sable]

South Africa

[stable spire]

see tan = 3/2 thats ofc

sin = 3x and cos = 2x

[stable spire]

or take sin as just 3

[alpine sable]

What do you mean

[stable spire]

take sin as 3

my answer is

116 ig

oral guess

corect me if i am wrong

[alpine sable]

Why is sin 3

[stable spire]

see tan = 3/2

so sin = 3

cos =2

it is the way to solve these problems

[stable spire]

is my answer coorect ? @alpine sable

[alpine sable]

Wait
Isn’t sin 3/sqr13

[stable spire]

is my answer correc ?

[alpine sable]

How must I know

[stable spire]

sin is 3/root 13 by your calcs

so the answer is 8

ok the answer is 8

u are noob at trigo

[alpine sable]

?

[stable spire]

u know nothin

tan is 3/2

ans sin is 3/root 13

[alpine sable]

I just said that

<@&286206848099549185> can someone help me pls

[bitter vault]

tan theta = 3/2, and since 180 < x < 360, sin theta = -3/sqrt(13).

This gives sin^2(theta) = 9/13, and so we get the required quantity as 8

[alpine sable]

so
do u remember this identity
tan^2(x)+1=sec^2(x)
so now u can find cos^2(x), similarly find sin^2(x)
and tanx=3/2
so plug in the values, because we got sin^2(x) term we dont need to be bothered by sign convention

mb i didnt see it was answered

[alpine sable]

Where does 8 come from

[bitter vault]

$13\sin^2(\theta) - 2/3 \tan \theta = 13 \times 9/13 - 2/3 * 3/2 = 9 - 1 = 8$

[ocean seal]

1345631

[alpine sable]

Oh thank you sm

.close

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[potent cairn]

Hi, I have a deck of 130 cards, and I want each of them to have 1-4 special traits. There are 8 types of traits in two classes, 4 Common types (75% of all traits) and 4 Rare types (25%).

How can I get a randomization WITHOUT repeating the type of trait on the same card?

[potent cairn]

So far I'm using excel, and I gave each card a random number of abilities from 1-4. ✅
Then I can assign up four columns, where each one generates a random number from 1-100 (depending on the number of abilities that card has), which I use to determine which trait type they have. 👀
The problem is that I am not controlling for duplicates of the traits per card. 🚩

How should I go about it?

[lone heart]

@potent cairn Has your question been resolved?

[prime badge]

i don't even know what you;re looking for

you mean the algorithm of some sort, but i don't know what decks are ok

ok i got it

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@potent cairn Has your question been resolved?

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@potent cairn Has your question been resolved?

[potent cairn]

.close

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[swift tangle]

x^(x/3) = 6x

[swift tangle]

How can you solve this problem without guessing? I have an idea of how to understand problems, such as x^(1/2) = 9, but I have no idea how to solve tis equation where the variable is in the exponent

[torn forge]

I mean,

you could cube both side

wait

[swift tangle]

pretty sure it's a bit difficult to cube 6x, yeah

[brave solar]

what do you mean by solving here?

[swift tangle]

solve for x in the intial pinned question

[brave solar]

if you want an algebraic solution that probably doesnt exist

i think its a transcendental equation

[swift tangle]

which is?

[brave solar]

https://en.wikipedia.org/wiki/Transcendental_equation

what is the source of the problem btw?

[torn forge]

i got two roots

x=0 , x=6

[swift tangle]

yeah, it's the right answer

[torn forge]

It’s just

cube it

minus 216x^3

factor out x^3

[swift tangle]

it's from a book that i got in a diffrent language

[torn forge]

you know that x=0 is a root

then extract the other factor

and solve that

and pow

get root x=6

[swift tangle]

I've come here
x^(x) - 216x^3 = 0
How do you factor it out, exactly?

[torn forge]

i mean, the root x=6 was mostly “guessing”

But you could at this point factor out x^3

x^3(x^(x-3) - 216) = 0

x^(x-3) = 216

[brave solar]

there is one more root but its transcendental most probably

[torn forge]

you could graph it if you wanted to ig

[brave solar]

yeah you can know the number of roots by graphing

[swift tangle]

I'm pretty sure the equation became a bit more complicated for me
i guess i'll give @brave solar's article a read

[torn forge]

Ah, alright.

[brave solar]

you cannot "solve" these equations as i said

[swift tangle]

I still appreciate your help, mate

I really do appreciate it

[brave solar]

,w x^(x/3)=6x

[ocean seal]

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[swift tangle]

So the only way of really "solving" these is by basically graphing them out?

[brave solar]

yeah pretty much

or numerically

[torn forge]

or just “guessing”

[brave solar]

,w plot x^(x/3)=6x

[torn forge]

lol

[swift tangle]

Geogebra doesn't seem to enjoy graphing out such problems, either

[ocean seal]

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[brave solar]

there's a root at ~0.15

[torn forge]

oh wiat right 0 is undefined

[swift tangle]

Calculators also have strokes while solving such problem

[brave solar]

you need to solve these numerically

using newton's method or something

[swift tangle]

Wolframalpha is essential to solve this

[swift tangle]

I haven't reached such point, so I guess guessing is my easiest path

or wolframalpha

[brave solar]

you can graph these always

[swift tangle]

'

[vale sapphire]

that x/3 in the power messes it all up

otherwise you can get a closed form using lambert's W function

[brave solar]

you dont have to graph like this

[vale sapphire]

(but since it's a special function, it's not the closest of closed forms either)

[brave solar]

plot them individually

[swift tangle]

having x as an exponent to itself is indeed quite confusing

[swift tangle]

could you clarify?

[vale sapphire]

That's the lesser problem

[torn forge]

you just plot em individually to see where they intersect

[swift tangle]

Oh, fair enough

OK - so the solutions are basically wolframalpha and graphing by plotting the sides individually and find their interesct point

(and guessing)

[vale sapphire]

x^x = a has closed form solutions for just about every real number a

x^(x+1) = a doesn't

the particular form of this equation is such that it doesn't seem to have a closed form

[swift tangle]

Yeah,
All right, much appreciated for helping me out, guys

.close

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[silver bear]

Can anyone explain the algebraic simplification happened inside the bracket in 2nd row. The only way I can think of is factor the 1-e^(-2t) to (1+e^-t)(1-e^-t). That still give the same the answer as those step

[brave solar]

L hospital

[silver bear]

The l hospital yielded (-e^(-t+1) +1) which then simplify to pi/2

So idk what wrong

[tacit arch]

Show your attempt

[silver bear]

@tacit arch

[brave solar]

this is not how l hospital works

you need to differentiate the numerator and denominator separately

[silver bear]

Ok

Thanks

I get it now

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@silver bear Has your question been resolved?

[silver bear]

Yes

[lone heart]

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[solemn cipher]

.open

ah dang it

[solemn cipher]

there we go, that works, okay so all i need is for somebody to teach me how domains and range works, if anybody could do that just tell me how and like give me a few practice problems that'd be great

[viral oracle]

idk how translate english so i give site

https://www.mathsisfun.com/sets/domain-range-codomain.html

[tacit arch]

that's a good link aol posted

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@solemn cipher Has your question been resolved?

[solemn cipher]

aight, i'll heck it out

[lone heart]

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[queen vigil]

How did they do that? I'm so confused on how they turned sec into cos and tan into sin

That's what they essentially did, but I have no idea how they got there

[limpid spade]

my simplifiying

u know sec is 1/cos and tan is sinx/cosx

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[bitter grail]

hi, so i have a project and i have no clue if my display of maths is good an i wonder if anyone could help me and show anything i could fix

[wary stream]

The trig part is wrong

If sin = o/h, what is o, and h?

[bitter grail]

opp adjacent hypotenuse

[wary stream]

I know

I mean the values

[bitter grail]

ohh

[wary stream]

If sin = o/h, what are the values for opp and hyp?

[bitter grail]

opp is 34 adj is 15 and hyp is 37

its on dark mode so you cant see it

[limpid spade]

34/15

[wary stream]

Which side is 34 and which is 15?

[limpid spade]

that's opp/adj

[bitter grail]

so would it be cos

[limpid spade]

no

[wary stream]

You're not helping, I'm trying to show the OP what they did wrong

[bitter grail]

acc ill take a pic

[wary stream]

So, as I said sin = o/h

What is the value for opp and hyp?

[bitter grail]

value for opp is 34 m

hyp is 37

[wary stream]

So in your work, said sin = 34/15

[bitter grail]

lemme fix that and plug it into my calculator

[wary stream]

Meaning that you said opp = 34 and hyp = 15

[bitter grail]

aight i fixed it

[wary stream]

Same thing applies with cos

You did that one wrong too

[bitter grail]

is it

soa coa toh?

[wary stream]

No

SOH CAH TOA

[bitter grail]

ty

?

what is this about

[wary stream]

<@&268886789983436800>

[bitter grail]

@wary stream im stuck

[wary stream]

Where?

[dreamy cedar]

dealt w/

[bitter grail]

i cant get angle c

x

keeps turning back as 66 degrees

i used the wrong sin formula

[wary stream]

For angle x, what is opp, and hyp?

[bitter grail]

opp is 34

hyp is 37

[wary stream]

No

[bitter grail]

how

[wary stream]

At angle x, what is opp from x?

[bitter grail]

hyp

[wary stream]

What is the value that is opp from x?

[bitter grail]

15

[wary stream]

Then why did you say 34?

[bitter grail]

i got mixed up

is there anything else i could fix?

[wary stream]

Not that I see of

[bitter grail]

what about angle y

[wary stream]

Where is y?

[bitter grail]

nvm got it

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[ocean seal]

epiphonically

[crude rose]

how do i find what this converges to?

ive been trying the telescoping series but im not sure how to work through it

[alpine sable]

ok

what i would do is turn it into each into a geometric series

for 1/(x-1), factor a negative out

so it becomes - (-1)/(1-x)

+1/(1-x)

do you see what im saying?

i don't think it telescopes

[crude rose]

ohhh

[alpine sable]

you also want the index to start at 0

do you're gonna have to do a variable transformation

[crude rose]

and you would divide by 2

for the left

[alpine sable]

yes

but first make n=0

so say m=n-5

then replace it with m

[crude rose]

ya

[real solar]

Is the question to find the exact value it converges to or to find if it converges or diverges

[crude rose]

eaxt value

[alpine sable]

yeah no doubt in my mind geometric series

[real solar]

That definitely makes it bit harder

I dont see how geometric could work

[alpine sable]

wdym

[crude rose]

you write it as two series right?

[alpine sable]

yes

btw you meant to write $\sum_{n=5}^{\infty}\frac{2}{2n-5}-\frac{1}{n-1}$

[ocean seal]

kaynsu

[crude rose]

yae

not x

cause that would make no sense

[alpine sable]

yessir

if you need help just ping me im a god at doing these problems

[crude rose]

kk

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[cunning trout]

For the graph of this function,

[cunning trout]

I don’t understand why it looks like how it’s graphed

[void niche]

what about it do you not understand

[cunning trout]

The graph

[void niche]

what about the graph do you not understand

[cunning trout]

The way it’s graphed

[void niche]

k good luck

[last ether]

Yeah

[cunning trout]

?

[last ether]

So let's take a look at $|x| + 3x$

[ocean seal]

Umbraleviathan

[cunning trout]

Ok

[last ether]

which has two possible outcomes

[cunning trout]

Yeah

[last ether]

$-x + 3x$ or $x + 3x$

[ocean seal]

Umbraleviathan

[cunning trout]

Right

[last ether]

So

When x is negative, it'll have the slope of -x + 3x

When x is positive, it'll have the slope of x + 3x

[cunning trout]

True

[last ether]

So look back at this graph

When x is negative, what slope does it have

[cunning trout]

Sure

So basically it’s translated 4 units down

I took |x| and 3x-4 as seperate graphs

[last ether]

Yeah

Well don't take them as different graphs

[cunning trout]

Oh ok

[last ether]

Right here

When x is negative is has a slope of 2x but when x is positive it'll have a slope of 4x

Because |x| is {x, -x}

[cunning trout]

Yh

Should I not take graphs separately in these types of questions?

[last ether]

Well

I mean you can but I would just break it up in conceptual terms

Not physical terms

You know that for |x|, it will be -x for x<0

And x for x>0

So you can think of $|x| + 3x$ as a piecewise

[ocean seal]

Umbraleviathan

[cunning trout]

Yes

[last ether]

Where it is $-x + 3x, x<0$

[ocean seal]

Umbraleviathan

[last ether]

$0, x = 0$

[ocean seal]

Umbraleviathan

[last ether]

$x + 3x, x>0$

[ocean seal]

Umbraleviathan

[cunning trout]

Right

[last ether]

Translate it 4 units down, and it shows why the graph looks the way it is

[cunning trout]

Ok I get it now

Thanks

.close

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[small swift]

Why is this bayes theorm?

and not independent

cuz i did law of total probability

[alpine sable]

because it

you have to write a probabilty tree

[lone heart]

@small swift Has your question been resolved?

[small swift]

for the first part

i did 40 choose 3 and then 3 choose 3

so first i pick 3 numbers from 40

and then out of those three they must match

but that doesn't seem to be correct

how can I solve this?

[sudden hinge]

There are (40 choose 3) possibilities you could pick the numbers. But only one of those combinations gets all 3 numbers correct. So the probability would be 1 / (40 choose 3)

[small swift]

could you reword by chance?

3 choose 40 is us selecting 3 numnbers

so we can pick 40 choose 3 numbers

but how do i say that all 3 must match

and don't in eed to first pick the 3 numbers

so 40 choose 3 then 3 choose 3 ?

[sudden hinge]

(40 choose 3) gives us the number of all combinations {a, b, c} you could make out of 40 numbers

[small swift]

yes

[sudden hinge]

They all differ by at least 1 number, though

as we count combinations, not permutations

[wary stream]

You mean 40 choose 3

[small swift]

like abc vs cba?

[wary stream]

40 C 3

[sudden hinge]

And there's only one winning combination, where all 3 numbers are correct

[small swift]

yes

[sudden hinge]

All combinations are equally likely to be picked

so the probability of picking that specific one that gets all 3 right is 1 / (40 C 3)

[small swift]

ok

and for the second part

i have 2/3 that match

so that's 3 choose 2

and then to get my third number i do 37 choose 1

[turbid magnet]

hey is anyone able to help me

[sudden hinge]

Yes

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@small swift Has your question been resolved?

[lone heart]

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[unique echo]

How do I do this question?

I understsand I need to integrate

but I have no idea how to integrate this thing

actually i think i found the answer

.close

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[opaque inlet]

Just a question. How would you differentiate equations like y=2^(2x-3)

[shell widget]

Log on both sides

@opaque inlet

[plucky lynx]

chain rule and the fact derivative of a^x is ln(a)a^x.

[opaque inlet]

Hm ok

log_2(y) = 2x-3

Wait but what do I do after the log

[shell widget]

actually you don't need to take log on both sides, you can just do what plegasus said

[opaque inlet]

I don’t get what he means tho

[shell widget]

You must've learned that derivative of a^x is ln(a) * a^x

No?

a is some constant > 0

[opaque inlet]

Nope

[wary stream]

$\frac{d}{dx}a^x = a^x ln(a)$
That's a common derivative identity

[ocean seal]

dldh06

[opaque inlet]

Ok

So it’s just

2^(2x-3) * In(2)

?

I’ll prob just search it up

Thanks anyways

[wary stream]

,w derivative of 2^(2x-3)

[ocean seal]

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[wary stream]

Wolfram simplifies

[opaque inlet]

.close

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[leaden aspen]

Hi this may be kind of simplistic but I'm confused on how the coefficient of the natural log gets decided

[leaden aspen]

I'm able to get the answer - just kind of missing this piece of the algorithm

[merry depot]

at some point in rewriting it you should get $\int \frac8u \ du$ which is equal to $8\int\frac{du}{u} = 8\ln(u)$

[ocean seal]

Zybikron

[leaden aspen]

Ahh thank you! that makes sense it looked like a substitution problem, just wasn't in the substitution section so I wasn't thinking of that.

[merry depot]

yeah, it's a u-sub. It's a little non-standard though which throws people off

[leaden aspen]

right - nothing in brackets. the quotient reminded me of a few sub problems. much appreciated again!

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[stable spire]

in how many ways can we send six urgent letters if we can use three messengers and each letter can be given to any of them ? so that each has at least one.

[placid zinc]

There's not a lot here, we can count these manually

Another way to do it:

• Find the number of ways to distribute the letters with no restrictions
• Subtract the number of ways to distribute with somebody getting 0 letters

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[quasi swift]

i know i use induction but idk what the base case should be

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@quasi swift Has your question been resolved?

[naive valley]

what are these, fibonacci numbers?

[quasi swift]

Yes

[tacit arch]

Just do n=2. But in the future, just do n=1,2,3 and pick one that makes sense to you. The work isn't that hard

And it often leads to work appearing in the proof

[naive valley]

n=1 also works as a base case; on the LHS you have F_1 and on the RHS you have F_2, and you know these are equal

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[misty bobcat]

we have a grid 2xn

[misty bobcat]

we want to write letters A,B, C into cells

in how many ways can we do it ?

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[merry sundial]

I’m confused on how to do this with elimination method

[alpine sable]

what did you try?

u have to factor the first formula actually

[merry sundial]

Ohhh ok I just found out I was confused on something my bad

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[tawny fable]

Hi there

[tawny fable]

Question In calculus

I was wondering:
Can I say that T_n = sum ( 1/n) - ln(N) ?

Or should it stay as it is?

[naive valley]

yes that is equivalent

[tawny fable]

I see

[naive valley]

did you try using the hint?

[tawny fable]

I mean

I tried to understand it basically

But I don't really understand where do they want me to go

It is a part of the proof we did in class of the Integral Test

[naive valley]

probably the most direct way to apply it is to rewrite the integral as $\int_1^N \frac{dx}{x} = \sum_{n=1}^{N-1} \int_n^{n+1} \frac{dx}{x}$

[ocean seal]

OurBelovedBungo

[tawny fable]

oh really?

[naive valley]

hmm or maybe putting it another way...

[tawny fable]

So we take a look at each part of the integral and sum it all up together

[naive valley]

when you increase N to N+1, what is T_(N+1) - T_N

if you can show that it's always negative then it means that the sequence {T_n} is decreasing

so can you find an expression for T_(N+1) - T_N ?

[tawny fable]

Let me see

Ok

I got this

[tawny fable]

I am struggling to prove that it is negative

[naive valley]

your answer looks right

i think to show that it's negative it's better to switch back to the integral:

$\frac{1}{N+1} - \ln\left(\frac{N+1}{N}\right) = \frac{1}{N+1} - \int_N^{N+1}\frac{dx}{x}$

[ocean seal]

OurBelovedBungo

[naive valley]

and just notice that the integrand 1/x is greater than or equal to 1/(N+1) in that interval of integration

and therefore -1/x is less than or equal to -1/(N+1)

hence the integral of -1/x over that interval is less than or equal to -1/(N+1)

and so the overall result is <= 0

as desired!

in retrospect it was probably better to just stick with the integral throughout instead of rewriting in terms of logs

[stable spire]

How to do it without restrictions

that formula for combinations of at least 1

??

[tawny fable]

I don't think I get that

[naive valley]

which part did you not follow?

[tawny fable]

How do you get that it is greater than or euqal to 1/N+1

[naive valley]

ah because 1/x is monotonically decreasing

so in any interval [a,b] its smallest value happens at x=b

in this case at x=N+1

[tawny fable]

Ohhh

That is the actual hint then

[naive valley]

yeah we got to use it after all

[tawny fable]

k

let me write it down and update you 🙂

[naive valley]

sure

[lone heart]

@tawny fable Has your question been resolved?

[tawny fable]

Ok yeah I got it

Thanks

On the second part I need to show that it converges

T_N

Using the fact that it is decreasing

So in my intuition it will converge only if -ln(N) will take the sum down faster than it get's up

if you understand what I am trying to say

[naive valley]

looks like T_N is always positive?

because $\frac{1}{n} \geq \int_n^{n+1}\frac{dx}{x}$

[ocean seal]

OurBelovedBungo

[tawny fable]

Yeah

We are not talking about series here though

So I don't have any tools

except taking it to inf which is unknown

[naive valley]

and therefore $\sum_{n=1}^{N} \frac{1}{n} = \frac{1}{N} + \sum_{n=1}^{N-1}\frac{1}{n} \geq \frac{1}{N} + \sum_{n=1}^{N-1}\int_{n}^{n+1}\frac{dx}{x} = \frac{1}{N} + \int_1^N \frac{dx}{x} \geq \int_1^N \frac{dx}{x}$

[ocean seal]

OurBelovedBungo

[naive valley]

which means that $T_N = \sum_{n=1}^{N}\frac{1}{n} - \int_1^N\frac{dx}{x} \geq 0$

[ocean seal]

OurBelovedBungo

[naive valley]

so now you have that T_N is decreasing and bounded below (by zero), hence it converges

[tawny fable]

ohhh

Using that thing

decreasing + bounded = converges

yup

[naive valley]

exactly

so that shows that the euler constant exists

[tawny fable]

Why did you take out the 1/N here

?

[naive valley]

finding the value is another story, do you get to do that?

[tawny fable]

Nope. That's on my second year in statistics lol

[naive valley]

ah because when you break the integration interval [1,N] into subintervals of the form [n,n+1] there are only N-1 of them

[tawny fable]

hmmm

[naive valley]

namely [n,n+1] for n=1 through N-1

so when we want to compare with $\sum_{n=1}^{N}\frac{1}{n}$, the latter sum has an extra term, so we just set it aside and work with the part from 1 to N-1

[ocean seal]

OurBelovedBungo

[naive valley]

no big deal, the extra 1/N makes the sum even bigger, which makes T_N more positive

amazing that it's not even known if it's rational or irrational

i would be astonished if it turns out to be rational though

[tawny fable]

Well

I guess one day we will find out

Anyways

@naive valley

Thank you for your help!

[naive valley]

pleasure

[tawny fable]

Do you mind if I ask some more questions about the connection between integrals and series?

[naive valley]

sure

[tawny fable]

Ok so my next question is some prove/disprove

And what I am really looking for, more than the answer, is the intuition.

Its these

rn I am in (a)

struggling to really understand what should the intuition be

[naive valley]

(a) is surely false

i mean you can take an integrable f(x)

and give it any new values you like at integer values of x

that will not affect the integral at all

but it will completely control the sum since the sum only looks at integer values of x

[tawny fable]

Oh I see

[naive valley]

(c) is also false if negative values of a_n are allowed

[tawny fable]

That is because the integral does not change if it is different than an integrable one by finite number of points

thats the thing?\

[naive valley]

take $a_n = (-1)^n \frac{1}{\sqrt{n}}$

[ocean seal]

OurBelovedBungo

[naive valley]

right, so the integral over any bounded interval is unchanged if you change the function value just at integer values of x, and therefore the improper integral over all of [k,infinity) is also unchanged

so you can take any f(x) that has a convergent integral, and choose your favorite divergent series and set f(n) equal to the n'th term of that series

while leaving f(x) unchanged for non-integer values of x

(d) is in fact true

if sum(a_n) is absolutely convergent, then |a_n| < 1 for all sufficiently large N, and therefore a_n^2 < |a_n| for all sufficiently large N

therefore (for large enough n) sum(a_n^2) is sandwiched between the convergent series 0 and sum(|a_n|)

[tawny fable]

I see, I need to go over it. It will take me some time.

About (b), what do you think?

[naive valley]

well probably you should look at the integral of 1/(1+x^2)

that's a monotonically decreasing function for x >= 1

so, similar to 1/x, you should be able to compare the integral to the series

and the integral has the advantage that you can calculate a numerical value

since the integral of 1/(1+x^2) is arctan

[tawny fable]

Yeah

So I know that the integral of f(x) is pi/2

[tawny fable]

I've got to the point where I have

But I can not take say it is smaller than 3pi/4

@naive valley Please let me know what you think.

[lone heart]

@tawny fable Has your question been resolved?

[tawny fable]

Still stuck at that. @naive valley

<@&286206848099549185> Anyone else can help?

[lone heart]

@tawny fable Has your question been resolved?

[tawny fable]

.close

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[dawn quail]

Is it normal to feel incredibly lost the first time you are introduced to imaginary numbers?

https://tenor.com/view/desmond-abc-lost-lost-abc-desmond-hume-hume-gif-21758830

[alpine sable]

Atleast I didn't got lost ig

Are you including complex numbers btw?

[dawn quail]

Just this screen so far

Maybe I need to take a break and come back fresh to it

[alpine sable]

Uh ok

it's pretty straightforward I mean

[dawn quail]

For √-x right?

[alpine sable]

ya

[dawn quail]

As it’s impossible to do so with real numbers

[alpine sable]

Do what?

Calculate sqrt of negative numbers?

[dawn quail]

Find the square root of a negative number

Yeah

[alpine sable]

Ya u can say so

almost ig coz idk what happened to me one day

[dawn quail]

Hence “imaginary” numbers

[alpine sable]

I calculated a Taylor approx for sqrt

Evaluates at -1

Got -30

Don't take this seriously tho

[dawn quail]

I feel like some mathematician is just making up rules here

[alpine sable]

Yes

[dawn quail]

Out of thin air

[alpine sable]

$i^2=-1$

[ocean seal]

Pi Creature

[alpine sable]

Is by definition actually

[dawn quail]

https://tenor.com/view/trigger-gif-6113599

This bastard

[alpine sable]

Coz u can also say

$\sqrt{-1}^2=\sqrt{-1}\sqrt{-1}=\sqrt{-1-1}=\sqrt{1}=1$

[ocean seal]

Pi Creature

[alpine sable]

which is not true

[alpine sable]

Wdym I love complex numbers

Even more than real numbers

They are super cool

When u learn more about them

[dawn quail]

I feel like imaginary numbers could lead to problems, when it comes to positive and negative for real numbers, but I will keep learning.

Actually I think I got the mathematician wrong 😑 it was not Descartes.. he actually coined the term as fictitious and useless. It was Geroalamo Cardona that came up with the unit “i”

[alpine sable]

Originally coined in the 17th century by René Descartes as a derogatory term and regarded as fictitious or useless, the concept gained wide acceptance following the work of Leonhard Euler (in the 18th century) and Augustin-Louis Cauchy and Carl Friedrich Gauss (in the early 19th century).

This is what is written on wiki

[dawn quail]

https://tenor.com/view/math-problems-hilarious-funny-drinking-draw-the-actual-thing-gif-17025729

[dawn quail]

Would this be true sometimes or I haven’t learned that part yet?

[alpine sable]

Well generally the way complex numbers are defined they won't contradict anything in real numbers

[dawn quail]

Ah OK

So imaginary numbers are kinda like variables in the way they interact with real numbers and math rules except they cannot change into real numbers?

[alpine sable]

change into real numbers

what does that mean?

i^2 does change into -1

[dawn quail]

Like x can be 2 if you calculate the expression

But i cannot become 2

[alpine sable]

$i^x$ ok

[ocean seal]

Pi Creature

[alpine sable]

Ofc $-2i^{2}=2$

[ocean seal]

Pi Creature

[alpine sable]

Wdym by change?

Exactly

[dawn quail]

I’m not sure lol

https://tenor.com/view/huh-gif-19545825

I think I will just need to keep learning to find out

[alpine sable]

Probably

[dawn quail]

Ty!

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[dawn quail]

.reopen

[lone heart]

✅

[dawn quail]

Just thought of something

You know how 3² = 9

To get there it’s 1 * 3 * 3

Could imaginary number represent the 1?

Hence giving i² = -9

Something like that

[trail flicker]

3^2*i=-9

[dawn quail]

Got it

[trail flicker]

i on its self is sqrt(-1)

[dawn quail]

Oh

[dawn quail]

You mean that’s the mathematical definition if i in math?

[trail flicker]

yes

[rare gale]

i^2 = -1 not -9 🙂

[dawn quail]

And the universal acceptance for what i represents

[trail flicker]

yes

[dawn quail]

Ahh OK I thought it just meant any non real number made up

[trail flicker]

i is the answer to sqrt(-1)

i mean imaginary

[dawn quail]

OK

Ty

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[viral oracle]

$\frac{sqrt{3+sqrt[3]{x}}}{sqrt{3-sqrt[3]{x}}}=3$

[ocean seal]

aol

[viral oracle]

dang

[gray isle]

\

$\frac{\sqrt{3+\sqrt[3]{x}}}{\sqrt{3-\sqrt[3]{x}}}=3$

[ocean seal]

ℝamonov

[viral oracle]

yes that

i dont understand how to solve this

i got to

$\frac{3+\sqrt[3]{x}}{3-\sqrt[3]{x}}=9$

[ocean seal]

aol

[harsh latch]

t=x^⅓
sqrt(3+t)/sqrt(3-t)=3
sqrt(3+t)=3sqrt(3-t)
3+t=9(3-t)
3+t=27-9t
10t=24
t=12/5
x=(12/5)³

[viral oracle]

what did you do beetween 3rd and 4th line

Oh

^2

x =1728/125

[harsh latch]

actually youre almost done

[viral oracle]

Yes i understood

$3+\sqrt[3]{x}=9*(3-\sqrt[3]{x})\
3+\sqrt[3]{x}=27-9\sqrt[3]{x}\
\sqrt[3]{x}+9\sqrt[3]{x}=27-3\
10\sqrt[3]{x}=24\
\sqrt[3]{x}=24/10\
\sqrt[3]{x}=12/5 |^3\
x=1728/125$

[ocean seal]

aol

[viral oracle]

.close

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[misty radish]

need probability of these two

[misty radish]

(dersom = if)

got these formulas and table, but doesn't help much

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[winter lotus]

can someone please post a step by step on how to solve this? tryna find the area

[supple tundra]

Sure i will solve your problem for you

give me a sec

[winter lotus]

thanks

Find the area of:

[pine kettle]

@winter lotus area of a triangle is 1/2 b * h

[winter lotus]

yeah, i know

but, how do you solve it lmao

[tall hearth]

b = h = 5√6 - 2√3

[oblique pier]

@winter lotus still need help?

[winter lotus]

yeah

[oblique pier]

alr

[ocean seal]

! बाशुदेव

[oblique pier]

construct the height of the triangle call it AH, AH cut BC at H

apply the formula

1/AH^2=1/AB^2+1/AC^2

to find AH

after that apply the Pythagorean theorem to solve for BC

there we go

we got our number for 1/2 x b x h

[winter lotus]

so, what would the final answer be?

[oblique pier]

i won't do that for you

try it out : D

math is only fun when ur the one who do it

[winter lotus]

i personally like it to have the answer on standby and work towards it

isnt very practical when doing tests tho

coz they wont give u the answer lol

[oblique pier]

yea im not recommend doing that....But

[oblique pier]

i usually use this method to solve inequality

i am allow to use calculator in my country

[winter lotus]

wait, could u please draw the ah, bc stuff

a little confused

i just would like the step by step solving on how

or the working out

[pine kettle]

whats the point of solving it like this

[oblique pier]

why not

[pine kettle]

because we already have 1/2 b * h

[oblique pier]

oh well..

oops

[pine kettle]

$\frac{1}{2}(5\sqrt{6}-2\sqrt{3})^2$

[ocean seal]

sills

[pine kettle]

@winter lotus can you foil that

[winter lotus]

wdym by foil

[viral oracle]

$\frac{1}{2}(5\sqrt{6}-2\sqrt{3})^2

oh u already drew that

[oblique pier]

aa my method was too long

[winter lotus]

oh, that foil

[pine kettle]

yep

so simplify $(5\sqrt{6}-2\sqrt{3})(5\sqrt{6}-2\sqrt{3})$

[ocean seal]

sills

[viral oracle]

$\frac{1}{2}(256-25\sqrt{6}2\sqrt{3}+23)$

[ocean seal]

aol

[viral oracle]

oh ahhh

i wanna write Xs not stars

[winter lotus]

the star means the x

star is just the other way of writing times

[winter lotus]

i gotchu

[viral oracle]

Fuck latex returning to line writing

[pine kettle]

use $\cdot$

[ocean seal]

sills

[viral oracle]

Ok thanks

Anyways

$1/2(150-20\sqrt{18}+6)$

[ocean seal]

aol

[pine kettle]

why you doing the work for t hem

thats not helping

[viral oracle]

Idk

Because i dont know like 80% of stuff in this server

[winter lotus]

i got $162-60\sqrt{2}$

[ocean seal]

Leaky

[viral oracle]

what is that

[winter lotus]

the answer to

[viral oracle]

shouldnt you use the (x-y) ^2=x^2-2xy+y^2

[winter lotus]

@pine kettle what would i do next?

any further steps?

[pine kettle]

this is correct

[winter lotus]

took me way too fucking long

[pine kettle]

now you just need to multiply by the $\frac{1}{2}$

[winter lotus]

jesus christ

[ocean seal]

sills

[winter lotus]

$\frac{1}{2}\cdot162-60\sqrt{2}$

[ocean seal]

Leaky

[winter lotus]

do this?

@pine kettle

[pine kettle]

nooo

dont forget

$\frac{1}{2}(162-60\sqrt{2})$

[winter lotus]

oh shit, yeah

[ocean seal]

sills

[viral oracle]

I got 81-20 * square root of 3

[winter lotus]

i would apply the distributive law first, right @pine kettle ?

[viral oracle]

Yes

[winter lotus]

i see

$81-30\sqrt{2}$

[ocean seal]

Leaky

[winter lotus]

@pine kettle yes?

[winter lotus]

20??

[winter lotus]

is that root over 18?

[viral oracle]

Yes

[winter lotus]

why?

[viral oracle]

Whats wrong

root of 6 * root of 3 = root of 18

[winter lotus]

My internet died @viral oracle

[viral oracle]

where does the 162 come from

[winter lotus]

oh im a fucking idiot

its supposed to be 160

[viral oracle]

actually ot is 162

it

1/2(-60)=-30

[winter lotus]

bro im so tired jesus christ

its 1:40am

i keep messing the damn numbers up

alright ill do it later tomorrow

.clsoe

.close

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[hexed arrow]

Given the following: T1 is a transaction that begins first and is currently the oldest transaction. T2 started second and is the youngest among the two transactions. If T1 is aborted, does that make T2 the oldest?

[hexed arrow]

Also, what is the difference between waiting and aborting?

[viral oracle]

Yes it does

[hexed arrow]

I searched online. Does the only difference between aborting and waiting mean that the locks are released for the former and not for the latter?

[lone heart]

@hexed arrow Has your question been resolved?

[hexed arrow]

<@&286206848099549185>

[lone heart]

@hexed arrow Has your question been resolved?

[lone heart]

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[alpine sable]

Is this right?

[lone heart]

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[upbeat peak]

yes it is

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[molten talon]

when you’re finding how many terminating zeroes there are in x! why are questions dividing by 5 constantly?

[molten talon]

i get that 2 and a 5 make a 10 where there appears a 0, but why 5 out of the 2

is it always gonna be dividing 5 each time. or is it gonna vary between questions and why

[lone heart]

@molten talon Has your question been resolved?

[vale wigeon]

there are more factors of 2 than there are of 5 in the prime factorization of a factorial

[lone heart]

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[shut sleet]

Could someone pls help me convert this limit into an integral

[shut sleet]

disregard the choices

all i know is that the change in x is 3, so the upper bound of the integral - the lower bound is 3

[last ether]

That is quite the integral symbol

Recall your formula

[placid zinc]

I usually set n = 3, and see where the rectangles land

[last ether]

$\lim_{x\to\infty}\sum_{k=1}^{n} \left(a + \frac{k(b-a)}{n}\right)\left(\frac{b-a}{n}\right)$

[shut sleet]

i still dont really know what to do

say what

[ocean seal]

Umbraleviathan

[shut sleet]

and then how do i convert that into an integral

[last ether]

Would be n to infinity

But no

[last ether]

You have a and b

So match coefficients

[shut sleet]

i dont have a and b in the problem, just a-b

[last ether]

You do have a

[shut sleet]

oh is it 1 because k=1?

[last ether]

No

Lemme show you in a sec

I gotta like

Do funny shit

Gimmie a sec

[shut sleet]

lol

id like to see funny shit XD

[last ether]

$\lim_{n\to\infty}\sum_{k=1}^{n} \left(a + \frac{k(b-a)}{n}\right)\left(\frac{b-a}{n}\right)$

$\lim_{n\to\infty}\sum_{k=1}^{n} \left(2 + \frac{k(3)}{n}\right)^2\left(\frac{3}{n}\right)$

[ocean seal]

Umbraleviathan

[last ether]

So now match coefficients

[shut sleet]

oh a is 2

[last ether]

Yeah

[shut sleet]

so then b is 5

[last ether]

Yup

And then you have the (1+3x/2)^2 correct so

[shut sleet]

and then the function is something squared

[last ether]

Yeah

[shut sleet]

but how do i find f(x)

[last ether]

Wait lemme think

[shut sleet]

is it just (2+3x)^2

[last ether]

I'm pretty sure it's just x^2

This I forgot lol

[shut sleet]

actualyl thats right

but how

[last ether]

I forgot lol it's just something I remember

[shut sleet]

i know you can also do the integral with upper bound 3, lower bound 0, of (2+x)^2. How do the bounds get picked in this case?

[placid zinc]

So if you set n = 3, we get:
(2 + 1)² + (2 + 2)² + (2 + 3)²
= (3)² + (4)² + (5)²
We can see the function that these rectangles are finding the area under

There's many ways to interpret these. For any function, you can translate the function horizontally and change the bounds to match

[shut sleet]

wait but do you how to get the bounds if your doing the integral of (2+x)^2?

i know the bounds are 3 and 0, but i just dont know why

i mean i have the answer sheet lol

just trying to understand this

[placid zinc]

So if you set n = 3, we get:
(2 + 1)² + (2 + 2)² + (2 + 3)²

That looks a lot like a right riemann sum of (2 + x²) between (0,3)

Actually, the left interpretation is making me sus. Go with the right sum always lol

[shut sleet]

wait so if you set the function equal to (2+x)^2 for the integral (as opposed to x^2) how would you get the bounds?

[placid zinc]

The 1,2,3

[shut sleet]

so you set n equal to something that would make the fraction 1?

[placid zinc]

That makes it easy to me to look at it, personally

[shut sleet]

but then once you do that, i still dont know how to get the bounds

sorry that its not clicking for me

im just confused

[placid zinc]

So we are kind of doing a "reverse" process here. Normally you'd write an integral as a sum to define it, but in this case we're writing a sum as an integral

With that in mind, you want to be comfortable with the forward process

[shut sleet]

im not lol

this whole part of this for the calc curriculum confused me

well really just converting limits into integrals

[placid zinc]

Here, I'll give an easier question to tackle.
Can you approximate the area under (2 + x)² between (0,3) using a right riemann sum, and 3 rectangles?

[shut sleet]

wait let me try and rewrite that as a summation

so its the summation starting with x=1, terminating at x=3 of (2+3x)^2, right?

[placid zinc]

So find the rectangle length, which is 3/3 = 1

And if f(x) = (2 + x)²
Then the area under the curve is approximately
1f(1) + 1f(2) + 1f(3)

[shut sleet]

yes

[placid zinc]

Or,
(2 + 1)² + (2 + 2)² + (2 + 3)²

[shut sleet]

wait so was my summation correct?

[placid zinc]

No haha

[shut sleet]

okay let me try again then

so its the summation starting with x=1, terminating at x=3 of (2+x)^2, right?

is that it?

[placid zinc]

Let's change the bounds. What about between (6,9)?

[shut sleet]

okay i think now i know how to write then summation

hang on

still 3 rectangles?

and a right reimann sum still?

[placid zinc]

Yeye

[shut sleet]

if thats the case, it would be the summation starting with x=1 terminating at x=3 of (2+x+6)^2, right?

or just (8+x)^2

[placid zinc]

Yes. But without the sigma notation:
(2 + 7)² + (2 + 8)² + (2 + 9)²

We can see the original bounds are in the answer

So if we want to do this backwards and we already have the answer, we look at the values x takes for the original bounds

[shut sleet]

okay thank you so much, i think i understand it!

sorry this took so long lol

[lone heart]

@shut sleet Has your question been resolved?

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[errant zinc]

Need help with part b

[errant zinc]

<@&286206848099549185>

[winter tangle]

show me what you have done so far

[errant zinc]

i have got

i have got to this bit idk what to do after

<@&286206848099549185>

[lone heart]

@errant zinc Has your question been resolved?

[lone heart]

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[frozen wren]

https://puu.sh/IZeTX/9a8cad2a63.png

[frozen wren]

can someone check my work

[lone heart]

@frozen wren Has your question been resolved?

[frozen wren]

<@&286206848099549185>