help-0

[noble sinew]

No

[abstract prairie]

Okay

This better

[noble sinew]

Why does 3 come from?

[abstract prairie]

-7i > -21u AHA

Okay then

Both divide bij -7 right

@noble sinew

Or

7i < 21u

Both devide by 7

@noble sinew i <3 u

[lone heart]

@abstract prairie Has your question been resolved?

[abstract prairie]

Need response

[scenic heath]

just do it

nothing is impossible

[alpine sable]

Send a shot of the problem

[stark plover]

me when i invent my own trig indentities

[lone heart]

@abstract prairie Has your question been resolved?

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[median mauve]

question solved

[median mauve]

.close

[lone heart]

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[shrewd sentinel]

What is the line of greatest slope in a 3D plane?

[velvet pelican]

I think it just means line AE

[gritty matrix]

The point E needs to be chosen on CD such that the slope of AE is as large as possible

[lone heart]

@shrewd sentinel Has your question been resolved?

[shrewd sentinel]

Thx

[gritty matrix]

any further question?

[lone heart]

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[raven bolt]

I need help with 14 i

[lone heart]

@raven bolt Has your question been resolved?

[raven bolt]

No it hasn’t

<@&286206848099549185>

[misty hollow]

cry abt it

[raven bolt]

Wdym? Don’t you get help on this server?

[misty hollow]

No

[raven bolt]

I don’t understand

[gray isle]

ignore them

what have you tried?

[devout summit]

Assuming cos(x)≠0 [presence of tan(x)], what do the individual sides of (i) simplify to?

[raven bolt]

I’ve tried to turn cos into sin/tan and tan into sin/cos

[gray isle]

can you show a pic of your work

[lone heart]

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[raven bolt]

Can I turn 1-cos^2x into sin^2x

Would that help?

.reopen

[lone heart]

✅

[raven bolt]

I’ve sent a pic

[gray isle]

Can I turn 1-cos^2x into sin^2x
Would that help?
yes

try expanding the left side

[raven bolt]

Like that?

[gray isle]

note quite

recall your order of operations

$$a(b-c)\neq ab -c$$

[ocean seal]

ℝamonov

[raven bolt]

Oh ok

[raven bolt]

Like this?

[gray isle]

yes

[raven bolt]

This is what I got but the answer at the back said it wasn’t the answer.

[gray isle]

express those cos in terms of sin

[raven bolt]

Like cos^2x = 1-sin^2x?

[gray isle]

yes

[raven bolt]

It still doesn’t equal to eachother

[gray isle]

expand and simplify the numerator

[raven bolt]

Oh yah ok

Lemme see

Got the answer thank you

[devout summit]

For the restrictions, cos(x)≠0

[raven bolt]

Can you help with question 14 iv?

[devout summit]

Notice that you can replace the 1 in the numerator by sin²(theta)+cos²(theta)

[raven bolt]

Yah I did that

[devout summit]

Then, a²+b²+2ab=(a+b)²

[raven bolt]

How did you get that?

[devout summit]

How did the denominator become 1?

[raven bolt]

Or is it just something you need to know

Oh I crossed it out so it didn’t become a fraction

Oh never mind

I can’t do that here

[devout summit]

(a+b)²=(a+b)(a+b)=a²+ab+ba+b²=a²+2ab+b²

[raven bolt]

I get what you are doing but is there anything that can help me come to that conclusion

[devout summit]

Notice that the numerator takes this form.

[raven bolt]

I get how you did it but I wouldn’t have been able to think about it on my own

[raven bolt]

It just seems like this method can only be applied to this question but you have to use different methods for other questions that are equally as hard to come to the conclusion of.

Is it just a matter of practice?

[weak prairie]

Yeah trig identities are mainly just practice, you start to see patterns like turning 1 to sin^2 + cos^2

It’s useful to look at the other side of the equality like if there’s only sins in it then you should change the cos to sines etc

And if there’s numbers you wanna look for factors that cancel to give a number

[devout summit]

Practice does help.

[raven bolt]

@devout summit Did I get this right for question 14 ii?

Wdym?

[devout summit]

Sorry, don't think it is right

You didn't show why sin²(x)-cos²(x)=sin⁴(x)-cos⁴(x). You just wrote it

[raven bolt]

I thought it just equals to eachother

[devout summit]

You need to show steps that take you from sin²(x)-cos²(x) to sin⁴(x)-cos⁴(x)

[raven bolt]

That’s why I asked since I don’t know how to show that

Oh ok

[devout summit]

Notice, $\sin^2(x)-\cos^2(x)=(1)(\sin^2(x)-\cos^2(x))\=(\sin^2(x)+\cos^2(x))(\sin^2(x)-\cos^2(x))$

[ocean seal]

Euclid31415

[raven bolt]

Sorry I don’t understand. Also aren’t you only allowed to change the left side?

[devout summit]

I am changing the left side itself

[raven bolt]

Oh ok

I don’t really understand how it can change to that

[devout summit]

We went from 1-2cos²(x) to sin²(x)-cos²(x)

[raven bolt]

Yah I get that

Did you get that by using identities?

[devout summit]

Yes. The basic identity sin²(x)+cos²(x)=1

[raven bolt]

Yah

I know that

But this has a minus sign

So how would you use that identity?

[devout summit]

Any questions about these steps?

[raven bolt]

Yah I have questions on how you used that identity to get that

[devout summit]

Every number a = 1*a=(1)(a)

I replaced the "1" by sin²(x)+cos²(x)

[raven bolt]

Ok

Ohh

Ok

[devout summit]

Notice the result you get after expanding the last brackets

[raven bolt]

Ok I get the 1= to sin + cod

But

I don’t get how you got two other brackets

Of numbers

[devout summit]

That is just another way to write multiplication

a*b=(a)(b)

[raven bolt]

What I’m saying is that I think sin^2x - cos^2x should equal to (sin^2x+cos^2x)(sin^2x-cos^2x) I don’t know where you got the rest from

[devout summit]

I don't understand. Are you asking why I considered sin²(x)+cos²(x)?

[glass lichen]

those are equal, yes

[raven bolt]

Yah but I don’t understand where he got the rest from and how adding the rest will make it equal.

[devout summit]

Maybe the more intuitive way to go about (ii) would be from RHS to LHS by factorisation

[raven bolt]

I’m confused how you added another sin^2x-cos^2x

[glass lichen]

Yeah RHS to LHS is more obvious for ii imo

[devout summit]

I didn't "add" another sin²(x)-cos²(x). I just rewrote it as (sin²(x)+cos²(x))(sin²(x)-cos²(x))

[raven bolt]

Ohhhhh

I seeee

I didnt see the equal sign

So I thought it just continued

Sorry abt that

And when you simplify it it will equal to the right inside

*side

Thanks

[devout summit]

Hope you understand this

[raven bolt]

I don’t really get what you mean by that. But I get the answer. Also my teacher said never touch the right side.

[glass lichen]

Then your teacher is leading you astray tbh

There's no reason to only work with the LHS, since equality is reflexive

[raven bolt]

Well I think it’s because I could solve it by the right side but then I wouldn’t know how to do the left side.

He says the left side is supposed to be harder for a reason

[devout summit]

There are multiple ways to show equality. You could manipulate RHS and make it look like LHS or in reverse. You could even show RHS and LHS equal the same expression and use transitivity. Only using manipulation on LHS to make it look like RHS is an arbitrary choice.

[glass lichen]

The "harder" side is the one that simply looks harder

powers of 4 are generally harder than squaring

[raven bolt]

Ok thank you.

[lone heart]

@raven bolt Has your question been resolved?

[lone heart]

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[void verge]

So I have this problem here, and ik a simple, "elegant" way of doing it, but how would you do it using the Chinese Remainder Theorem, if applicable

[placid zinc]

So note that x = 1(mod 2) contains strictly less info than x = 3(mod 4)

Knowing the second means you know the first

You have three unique pieces of info here:
x = 2(mod 3)
x = 3(mod 4)
x = 4(mod 5)

By CRT, this is exactly the same info as knowing x (mod 60)

[void verge]

ok.... so then what? sorry i dont really understand CRT that much, i was trying to do substitutions but it got messy and i couldnt figure it out

[placid zinc]

The big thing is to realize the correspondence I put above. Knowing a number in (mod 3), (mod 4), and (mod 5) is exactly the same as knowing it in (mod 60)

To actually get this number, you do this:
https://math.stackexchange.com/questions/2441003/using-the-chinese-remainder-theorem-to-solve-the-systems-of-congruences

[lone heart]

@void verge Has your question been resolved?

[void verge]

ok

after some time

i got 60l + 59 = x

so how do I imply the congruence of x

with this

@placid zinc

<@&286206848099549185> ?

[void verge]

as in

why is it

x = 59 (mod 60)

whats the proof ig

i think i get ig

oh well

my problem is solved so

.close

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[bold tapir]

Not really sure how to solve this integral. What I tried doing was letting x=cos(theta) to get 1/sin(theta) dx but didn't know how to proceed from there. Integral is $\int{\frac{1}{\sqrt{1-x^2}},dx$

[ocean seal]

samarium 62
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[velvet pelican]

Did you remember to change dx into d(theta)

@bold tapir

[bold tapir]

Yeah but I don't know how to do that though

[velvet pelican]

You find dx/d(theta) by differentiating

Suppose dx/d(theta) = f(theta)

Then dx = f(theta)d(theta)

Well not exactly

But you can replace dx by that

[bold tapir]

ok

[velvet pelican]

Should be a standard integral after that

[bold tapir]

alright thanks

[velvet pelican]

Welcome

[bold tapir]

oh wow yeah I was overcomplicating it

I don't remember why but I ended up trying to differentiate arcos(x)

Thanks!

.close

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[sterile otter]

can someone help with this

[sterile otter]

<@&286206848099549185>

im not sure what i have to do

[cloud skiff]

You can work out angle ABC: 47º + angle ABC = 180

Do you know the law of sines?

[sterile otter]

Yea

[cloud skiff]

ok

tell me when you work out all the angles in triangle ABC and I'll help you from there. Let me know if you need help

[sterile otter]

Ok

Yea I'm done

[cloud skiff]

ok

[cloud skiff]

give me a sec

yep you're right

you can now use the law of sines to work out BC

[sterile otter]

yea

ok

1 min

yea done with that

[cloud skiff]

yep

you can now use normal right-angled trigonometry in triangle BDC to work out CD

[sterile otter]

ok thx

[cloud skiff]

no worries

[sterile otter]

i might have a couple more i need help with , ill be back if i need more help

[cloud skiff]

ok

[sterile otter]

is this correct?

[cloud skiff]

I'll check it quickly

yes

[sterile otter]

ok tq

i have no idea what to do here

[cloud skiff]

we'll call the point where the hot air balloon is C

what is angle BAC?

[sterile otter]

already did that

°

[cloud skiff]

good

[sterile otter]

do i use the law of sines again?

[cloud skiff]

now you can use law of sines to work out AC and BC

[sterile otter]

ok yea

[cloud skiff]

you only really need to work out one though

[sterile otter]

ok

[cloud skiff]

God needs to go back to heaven now, he has some thoughts and prayers to answer

[sterile otter]

ok

[sterile otter]

when will u be back

[cloud skiff]

I'll check your working quickly but I have to go soon

[sterile otter]

ok sure tq

but just tell me how i find the height

[cloud skiff]

CB is right

drop an altitude from C to AB

then right angle trigonometry

[sterile otter]

oh ok

tq

[cloud skiff]

all good

[sterile otter]

bye god

[cloud skiff]

I will respond to all your prayers if you private message me ❤️

[sterile otter]

ok

[cloud skiff]

(I might not get back to you for 30min though)

[sterile otter]

its fine i have time

[cloud skiff]

cya

[sterile otter]

cya

[lone heart]

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[ivory herald]

a worker has to make 15 products a day to supply the store. He decided to make 21 products a day, so he did his work 4 days early and has 60 leftover products. How much products did he have to do to make his job in time

[vale wigeon]

this problem is confusing.

did you get it from a textbook or homework assignment or similar?

[ivory herald]

well i tried translating it

[vale wigeon]

translating from what language?

[ivory herald]

bulgarian

[vale wigeon]

hm

i don't speak bulgarian but i do speak russian. can you send the original so i can see if it's possible for me to figure it out?

[ivory herald]

uuh idk if u will understand

wait let me try to translate it again i will type here

[vale wigeon]

send the original

i will be the judge of what i do and don't understand

[ivory herald]

okok

wait bruh i said 21 but it was 20 sorry for loosing your time the equations is very easy

sorry again for loosing your time

thanks tho

.close

[lone heart]

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[median dirge]

??

[vale wigeon]

[median dirge]

Haha

If you need help, just a capybara, then you can reopen

[lone heart]

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[sweet parrot]

substruct of 2 upper triangular matrix gives me upper triangular matrix, right?

[sweet parrot]

ping me

[quaint sorrel]

@sweet parrot yes, because matrix addition/substraction happens elementwise

in the lower triangular, where we have the zeroes, we'd always calculate 0-0 = 0

assuming both matrices are quadratic and of the same dimension

[sweet parrot]

thank you, another question if i may

is it correct to say that
det(λI−A)=0 == det(λI−REF(A))=0
where REF is reduced echelon form

[quaint sorrel]

yes , of course

[quaint sorrel]

unfortunately, I am lost on this case, hopefully someone else can answer <@&286206848099549185>

[sweet parrot]

🙏

[lone heart]

@sweet parrot Has your question been resolved?

[sweet parrot]

The answer is no. Question is how do I apply ref to it to simplify it bc my matrix is big?

[noble sinew]

If your matrix is above 3x3 you always use a software to calculate determinant

[sweet parrot]

My problem is that I'm studying software engineering and need to make a code to calculate this

Is there a way to simplify what I said?

[noble sinew]

The fastest is PLU iirc

[sweet parrot]

what is plu

[noble sinew]

https://en.m.wikipedia.org/wiki/LU_decomposition

[sweet parrot]

so my steps will be: (?)
Find ref of A
use LU to solve
det(λI−A)=0

[lone heart]

@sweet parrot Has your question been resolved?

[lone heart]

@sweet parrot Has your question been resolved?

[quaint sorrel]

@sweet parrot reduced echelon form is the form we get after we apply for example gauß-algorithm=

?

In my numerical mathematics course I learned the following

A = LR. I write R instead of U (Germany 🤷 ) for the upper triangular matrix

Let's assume $$ A \in \mathbb{R}^{n\times n} $$ and that we can apply LR onto it.

[ocean seal]

mrbrown

[quaint sorrel]

In A = LR the upper triangular matrix (here R) is the reduced echelon form. We can for example get R if we apply Gauß-alogrithm onto A.

for the eigenvalues we usually start with the equation $$ Ax = \lambda x$$ whereas, $$\lambda$$ is the eigenvalue to the eigenvector x.

[ocean seal]

mrbrown

[quaint sorrel]

Now, let's say we know that A = LR exists then $$ Ax = LRx = \lambda x $$

[ocean seal]

mrbrown

[quaint sorrel]

🤔 🤯

$$ L^{-1}Ax = Rx $$

[ocean seal]

mrbrown

[quaint sorrel]

I only know that det(A) = det(LR) = det(L) det(R) = 1 det(R) = det(R)

xD

[sweet parrot]

i will try to mess with it, thank you

[quaint sorrel]

$$ LRx = \lambda x \iff Rx = \lambda L^{-1}x $$

[ocean seal]

mrbrown

[sweet parrot]

https://rosettacode.org/wiki/Reduced_row_echelon_form

what do i get from this? R? L?

[lone heart]

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[tawdry saffron]

Can someone help me prove this

It's dutch but they want us to prove it using mathematical induction

So proving it for n+1

[lone heart]

@tawdry saffron Has your question been resolved?

[tawdry saffron]

Ping me when answering

[analog olive]

@tawdry saffron

[tawdry saffron]

Yes

[analog olive]

so you checked for n=1

[tawdry saffron]

Yes

[analog olive]

I assume you are getting stucked in the third case where n=k+1

[tawdry saffron]

Yea

1 sec

Hope it's readable

So that's where I'm stuck

Don't mind me checking for n=4

[analog olive]

ok so you have (1+2 + ... + n) ^2 + (n+1)^3 and if we prove that
(1+2 + 3 + .... + n + n+1)^2 is equal to this, we will be done

[tawdry saffron]

I don't know how to work out the term ^3

[analog olive]

you don't need to simplify that anymore.. the term I wrote below will get simplified to that

so you could assume (1+2 + 3 + .... + n) = m and (n+1) = p
so (1+2 + 3 + .... + n + n+1)^2 = (m+p)^2 = m^2 + p^2 + 2pm
= (1+2 + 3 + .... + n)^2 + (n+1)^2 + 2(1+2 + 3 + .... + n)(n+1)

tell if you understand till here

[tawdry saffron]

I understand

[analog olive]

yeah and (1+2 + 3 + .... + n) = n(n+1)/2 so replace that in last term. you will get n(n+1)^2 and then take (n+1)^2 common from last two terms

[tawdry saffron]

1 sec

Writing it all down

Yes

So $(n+1)^2(\frac{n^2}{4} + n + 1)$

[ocean seal]

Ruffy🎄

[analog olive]

huh

$$(1+2 + 3 + .... + n)^2 + (n+1)^2 + 2(1+2 + 3 + .... + n)(n+1) $$
$$(1+2 + 3 + .... + n)^2 + (n+1)^2 + 2 \frac{n(n+1)}{2}(n+1)$$
$$(1+2 + 3 + .... + n)^2 + (n+1)^2 + n(n+1)^2$$

[ocean seal]

[redacted]

[tawdry saffron]

Oh

I also replaced the first terl

Term

With $\frac{n(n+1)}{2}$

[ocean seal]

Ruffy🎄

[analog olive]

ohk

Now take (n+1)^2 common from last two terms

[analog olive]

We weren't replacing that because we had to obtain (1+2 + ... + n) ^2 + (n+1)^3 and we already got (1+2 + ... + n) ^2 , so we aren't disturbing it

[tawdry saffron]

Okay thanks

1 sec more writing down

I am a slow writer

Got it

[analog olive]

so you are done right with proving the 3rd part

[tawdry saffron]

Yep

@analog olive have another 1

[analog olive]

yeah

[tawdry saffron]

Have to prove this

The same way

Is where I am atm

My n looks like a 4 but it is an n

[analog olive]

so we need to prove that (1+x)^n (1+x) >= 1 + (n+1)x

and we have (1+x)^n >= 1+nx

[tawdry saffron]

Yes

[analog olive]

so
$$(1+x)^n (1+x) \geq (1+nx)(1+x)$$
$$(1+x)^n (1+x) \geq 1+nx + nx^2 + x$$
$$(1+x)^n (1+x) \geq 1+ (n+1) x + nx^2$$
So ,

$$(1+x)^n (1+x) \geq 1+ (n+1) x $$

[ocean seal]

[redacted]

[tawdry saffron]

Wait how is it (1+nx)(1+x)

On the first line

[noble sinew]

\begin{align*}
(1+x)^n (1+x) &\geq (1+nx)
(1+x) \ &= 1+nx + nx^2 + x \
&= 1+ (n+1) x + nx^2 \
&\geq 1+ (n+1) x
\end{align*}

Much better

[ocean seal]

ScapeProf

[analog olive]

$$ (1+x)^n \geq 1+nx$$
Multiplying (1+x) both sides :
$$(1+x)^n (1+x) \geq (1+nx)(1+x)$$

[ocean seal]

[redacted]

[tawdry saffron]

Oh

Right

I did 1+(n+1)x

Okay thanks!

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[tawdry saffron]

$\lim_{x \to +\infty} (\sqrt[3]{(x-a)(x-b)(x-c)}-x)$

[ocean seal]

Ruffy🎄

[tawdry saffron]

How to calculate this limit

With L'Hopital

Should I do *x/x?

[lone heart]

@tawdry saffron Has your question been resolved?

[devout summit]

Try replacing "x" with variable "1/k" where k approaches 0+. Then, you can apply L'hôpital's rule once you bring the k to the denominator. @tawdry saffron

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@tawdry saffron Has your question been resolved?

[rugged lily]

if you end up solving it can you send the solution

im kind of confused as to how to solve that as well 😅

[coral pagoda]

I think difference of cubes will come in handy.

$a^3-b^3=(a-b)(a^2+ab+b^2)$.

[ocean seal]

dackid

[coral pagoda]

Yea, this'll work. Multiply the top and bottom by a^2+ab+b^2, where a is the cube root term and b is x

I'm gonna simplify things. Let $u=(x-a)(x-b)(x-c)$.
Then rationalize the fraction like so: [ (u^{\frac{1}{3}}-x) \cdot \frac{u^{\frac{2}{3}}+u^{\frac{1}{3}}x+x^2}{u^{\frac{2}{3}}+u^{\frac{1}{3}}x+x^2} = \frac{u-x^3}{u^{\frac{2}{3}}+u^{\frac{1}{3}}x+x^2}]

[ocean seal]

dackid

[coral pagoda]

Hope this helps @tawdry saffron

[tawdry saffron]

Thank you!

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[alpine sable]

Why would D be the right answer?

[alpine sable]

If CB=BA, then aren't triangles CBD and CAE similar, and thus won't Angle CBD = Angle BAE?

[echo socket]

Yes that should be true

But I don't understand what's the question

[alpine sable]

Oh, it's asking whether:
(A) Quantity A > Quantity B
(B) Quantity B > Quantity A
(C) Quantity A = Quantity B
(D) Not Enough Information

[echo socket]

The triangles would be similar if BD || AE, which doesn't seem to be given

[alpine sable]

I guess that maybe the answer is D because we don't know enough for it to be similar? Just Googled the similarity rules again (haven't done geometry in a while), and it seems that 2 angles need to be equal, a side-angle-and a side, and side-side-side. Only know 1 side is equal

[alpine sable]

Oh yeah, ok, got it

Damn I was very quick to naturally assume that they were parallel looking at the picture

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[amber marsh]

How do i do a

i subbed in c as 0

and gave me an error

so now im confused

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[jade prairie]

I need to know how can i find the value of [H]^2 in details

[glass lichen]

,rotate

[glass lichen]

isolate for the ln(1/H^2)

then isolate for 1/H^2

then isolate for H

[jade prairie]

Thats it?

[glass lichen]

yep

You just do it step by step

[jade prairie]

Okay thank u so much

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[faint lark]

I was reading something and couldnt find the exact definition of something;

A typical human fingerprint pattern has a period of around 500 μm, with a width of the ridge ranging from 100 to 300 μm, so the structural parameters were determined to exhibit a ridge width of 100 μm, a period of 200 μm, and a groove depth of 50 μm

what is 'the period' referring to? I assumed statistics but I cant google it properly.

[glass lichen]

I assume you've done trig?

[faint lark]

yes but not in english 😄

[glass lichen]

@faint lark it's analogous to period of trig functions

it's just how long until it repeats the pattern

(long distance, not long time)

[faint lark]

okay thank you!

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[stone current]

Hi, so I've been trying to solve this problem: the sum of three numbers in arithmetic progression is 12. The sum of their squares is 66. Find said numbers. My first thought was to somehow translate that into a system of equations and solve it, but I am not sure how to translate it into algebra. Or whether I should use one of the known arithmetic progression formulas and plug whatever is given in

[glass lichen]

let the arithmetic sequence start at a and have a common difference of d

[glass lichen]

that's correct

x=a, y=a+d, z=a+2d

so in fact you know one of the numbers has to be 4

[stone current]

So then I replace one of the variables by 4 and solve?

[glass lichen]

well it makes it somewhat easier to solve..

since you know y=4.

[oak chasm]

That won't be enough to solve.

You can expand:

(a + d)² = 4²
a² + 2ad + d² = 16

Now, expand your whole equation:

a² + (a + d)² + (a + 2d)² = 66

Then you'll be able to get a multiple of a² + 2ad + d² separated from the rest, and you know that that's 16. Substituting in the 16 will leave you with one variable, which makes that variable easy to solve for.

Just remember that x² = .... has two solutions.

[stone current]

Thank you Mosh and Chai T. Rex

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[crude sonnet]

2-x/3+x = 1/2

[crude sonnet]

basicly what i did was make it into a polynominal with the formula a/b = c/d

by cross multiplying

i got ad-bc = 0

which then i got

2(2-x) = 1(3+x)
4-2x = 3+x
4-3 = 2x + x
1/3 = x

[glass lichen]

oh it's (2-x)/(3+x)

not 2-x/3+x

[crude sonnet]

wait what

[glass lichen]

2-x/3+x is read as $2-\frac{x}{3}+x$

[ocean seal]

Mosh

[crude sonnet]

oh

that i did not know

so how would i then solve this?

[glass lichen]

well is it (2-x)/(3+x)?

[crude sonnet]

hmm

its

let me draw it out

[glass lichen]

or take a picture of it.

$\frac{2-x}{3+x}=\frac{1}{2}$?

[ocean seal]

Mosh

[crude sonnet]

yes

that is right

[glass lichen]

then everything you did is correct

[crude sonnet]

okay now next question

sorry not next but a follow up

so when i tried to verify

i plugged in 1/3 for x

and im having a bit of troubles solving..

2 - 1/3 / 3+1/3 = 1/2

[glass lichen]

what's 2-1/3?

[crude sonnet]

i think i multiply by common denominators

so that would be 6/3 - 1/3

[glass lichen]

yes.

[crude sonnet]

which is 5/3

[glass lichen]

yeah

[crude sonnet]

for 3 + 1/3 i got 10/3

[glass lichen]

yes, so you have $\frac{5/3}{10/3}=\frac{5}{10}$

[ocean seal]

Mosh

[crude sonnet]

hmm

u got half of it right

the last part is 1/2

[glass lichen]

yes

[crude sonnet]

yea

so how do we solve from here

[glass lichen]

what.......?

[crude sonnet]

thats the part im a bit lost

5/3/10/3

[glass lichen]

You already solved a while ago

x=1/3 is the solution

[crude sonnet]

yes

but i need to verify

this is the verification

[glass lichen]

Yes, and we just did that.....

so you're done

nothing else to do.

[crude sonnet]

okay but im not sure im understanding

how would we do 5/3 / 10/3

[glass lichen]

Didn't sound that way

??

[glass lichen]

.

[crude sonnet]

isint there a fraction property

to flip this into multipication

[glass lichen]

yes.. division of fractions is multiplication by the reciprocal

however it's easier to just multiply through by 3

[crude sonnet]

oh yeah

[glass lichen]

then simplify 5/10 into 1/2

[crude sonnet]

ohhhh

so theres 2 ways

reciporical thingy

and multiply by common denominator

now what if we had 5/4 / 9/ 7 what would we do here?

[glass lichen]

multiply by the reciprocal..

[crude sonnet]

i see

you're a very smart person

5/4 x 7/9 correct?

wait yes i understand this

[glass lichen]

*7/9, yes

[crude sonnet]

gotchaaa

thanks bro

i really appreicate you spending the time to help me

cheers!

Also i used to also play lunar client on minecraft

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[frozen matrix]

can someone help

[glass lichen]

what have you tried?

[frozen matrix]

x^2018=u

[glass lichen]

Do I need to tell you that won't help?

[frozen matrix]

yeah it didnt

tried integration by parts too but idk

[glass lichen]

is the integrand even or odd?

[frozen matrix]

god damn it

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[frozen matrix]

oh wait

.reopen

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✅

[glass lichen]

Hint: If the integrand is ugly to integrate, think smart not hard

[frozen matrix]

its multiplication how do we know its even

[glass lichen]

it's odd

[frozen matrix]

I mean after you take integral

[glass lichen]

??

The integrand is an odd function

[frozen matrix]

is it odd right now in integral?

[glass lichen]

hence the integral is 0

yes, x*exp(x^2018) is odd

[frozen matrix]

oh

yeah

[glass lichen]

and recall the odd integrand property and you're done

odd functions on intervals symmetric about 0 will integrate to 0

(assuming bounded interval obviously)

[frozen matrix]

btw can we solve it

like the function

[glass lichen]

no

god no

you'd have to be able to integrate $e^{u^{1009}}$ or something like that

[ocean seal]

Mosh

[glass lichen]

and given $e^{u^2}$ isn't possible

[ocean seal]

Mosh

[glass lichen]

isn't possible within elementary functions*

[frozen matrix]

oh okay

thanks

[glass lichen]

if the 2018 was a 2, then you could

however you'd still get 0, cause it's still an odd function

[frozen matrix]

wait what

why does it matter

oh because x^2=u

okay

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[glass lichen]

u-sub works for xexp(x^2) yeah

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[alpine sable]

Hello

[alpine sable]

Will the answer of 11th question be 0?

After doing algebra, I am getting x(6/x + 1/x^2)

As x approaches to infinity.

[plucky lynx]

"No algebra" require you can intuitively see 1/x -> 0 as x -> \infty so you left with 3^2 = 9.

from there you see the limit should be 0, if your not convince you can prove it.

[alpine sable]

Okay. Thank you 🙂

[plucky lynx]

np.

[alpine sable]

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[ebon bronze]

Does anybody know whether the partial sum of hyperharmonic numbers has closed solution for not natural orders?

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@ebon bronze Has your question been resolved?

[charred flint]

do partial sums of harmonic numbers for natural orders have a closed solution?

[ebon bronze]

Hmm, they do have a solution that gets rid of sum and uses just the hyperharmonic numbers themselves, so yeah

I mean, harmonic numbers are still sums, but that's something

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[echo socket]

What have you tried so far

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[chilly tide]

these are to be done by constructing simultaneous linear equations. anyone know how to begin contructing the equations before moving on to solving them???

[alpine sable]

Not yet

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@chilly tide Has your question been resolved?

[stable sinew]

in the first one,
let the number of outback models sold be O, and the number of bush walkers sold be B --> you want to find one of them to find the other, so you need to write equations that link the two. from the first piece of info you have, O = (100% + 20%) * B. from the second piece of info, the profit from selling O number of outbacks is price of one unit * O. do the same for your bush walkers and you have your equations. apply the same process to q16 and u should get the answers

[chrome salmon]

Alright

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[severe sluice]

hi sam

[alpine sable]

https://tenor.com/view/spider-man-we-one-gif-18212100

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[wispy olive]

Wait a minute-

WTF.

[plucky umbra]

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[median dirge]

@chilly tide hi

[chilly tide]

hi

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[flint mango]

Can someone help please

[echo socket]

Are you allowed to use L'Hopital's rule?

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[loud sentinel]

i have a question regarding ii)

im not sure on how to continue

the two attachments are my working up to that point

[severe sluice]

hint: 0<=A<=2pi
draw a horizontal line y=k, and see how many times it intersects the sine wave in the interval [0,2pi]

[loud sentinel]

y=k where ?

sorry im unclear

@severe sluice

[severe sluice]

y=k is... the line?

[loud sentinel]

which line

im so sorry im rlly dumb sometimes

[severe sluice]

when sin(2A) = k, we should have the functions y=sin(2A) and y=k intersect, right?

[loud sentinel]

yeah i guess

is it smtg like this

like the green line or the blue one

[severe sluice]

both works

because those are all lines with -1<=k<=1

[loud sentinel]

ok so lets say the green line is at 1 it intersects the graph 2 times right

but if its in the range it intersetcs it 4 times right

same as -1 it intersects twice

so are my answers 2 and 4?

for the number of solutions i mean

[devout summit]

If k=0 then 5 times since you had interval [0,2π]

[loud sentinel]

oh right

so i have three answers then is it ? 2,4 and 5

[severe sluice]

that's right, but you should also point out how it's usually 4 and there are only special cases where it's 2 or 5

[loud sentinel]

oh okay i got it

thanks @severe sluice and @devout summit

[severe sluice]

np!

i'm just sam....

[loud sentinel]

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[tepid zodiac]

Hello guys

[tepid zodiac]

If i want to find a normal vector to a curve f, do i do grad f?

[severe sluice]

how do you even get the grad of a curve?

[tepid zodiac]

I think I might i get it wrong guess

Here is the problem I'm solving

It's number 2

grad of that curve?

[glass lichen]

iirc gradient is orthogonal to the level curve

[tepid zodiac]

So is that a normal vector?

[glass lichen]

yes..

since that's how you define the tangent hyperplane

[tepid zodiac]

I was gonna find the tangent vector and derivate it

But seems a bit complicated to me hahah

Seems like do grad is the best way

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[molten tapir]

I know this is very basic question but what operation do I need to know to achieve this?
I have a maxValue and a minValue lets say min=5 max= 50 and i have a third value (x) which represents the point between min and max....
So when x=0 result=5 and whenn x=100 result = 50

[molten tapir]

its about progression or something?

the third value its always between 0 and 1

[median dirge]

What is result as?

[noble sinew]

“Third value (x) between 0 and 1”
Proceeds to give example of x=100 (third value)

[median dirge]

Thats what is confusing in your question

Explain a bit

[molten tapir]

sorry i got lost with so many roms

ill explain the context :

[median dirge]

Yeah please.

Its quite non-understandable

[molten tapir]

i have a a current height , this height has a maximum and a minimum, with the current height and those maximum and minimums i get a heightPercentage.

Then i have a minVolume=X and maxVolume=Y

I want the volume go from X to Y according to the heightPercentage

like... progressively

proportionally

sorry my math vocabulary sucks

its more understandable?

maybe f(heightPercentage) with max=X and min=Y ?

im sorry my math level is so low

[noble sinew]

Why do you call it percentage when it isn’t a percentage?

If it was max value it would be 100%

Not equal to max value

[molten tapir]

mmmm, the height has a min and a max... so taking those and the current height i calculate the height percentage.... i thought i needed this step in order to do the next step

like... lets say minHeight is 2... and height is 2... then heightPercentage==0%

then i would need the 0% progression between minVolume and maxVolume which would be equal to minVolume obviously

makes sense?

[noble sinew]

If x<=min
f(x)=min
Else if x>=max
f(x)=max
Else
f(x)=x

Is that not what you are saying?

Else it makes no sense

[molten tapir]

im so sorry... let me explain differently. From scratch

[bleak ridge]

I think he means more like (max-min)*percentage/100 + min

[noble sinew]

But he said output was max

If x>=max

[molten tapir]

LEts start from scratch:

I have this:
Height values: currentHeight, minHeight and maxHeight
Volume values: minVolume, maxVolume

when current height moves between min and maxHeight i need to assign a proportional volume considering minVolume and maxVolume

does it make sense now ?

[bleak ridge]

c=(max-min)*proportion + max would be the general formula you can use to solve for that

Im not going to use percentages just for simplicity sake

c is current

[molten tapir]

whats proportion there?

[bleak ridge]

Itd be the thing you want to be equal from volume to height

So youll have one of those fancy equations for height and volume

Lemme solve for proportion so its more useful

c - min=(max-min)*proportion

[molten tapir]

👀

[bleak ridge]

$\frac{c - min}{max-min}=proportion$

[molten tapir]

ah thats what i was calling percentage!! sorry im not native

[ocean seal]

PapaBread

[bleak ridge]

Oh

A proportion is just a percentage thats from 0 to 1

Instead of 0 to 100

[molten tapir]

exactly yes

thats it

so i would need to get the proportion of the height....

[bleak ridge]

You normally would but in this case you can kinda skip that step

Youll have two of those equations

One for volume and the other for height

With all of the max and mins and currents and stuff

So

$\frac{cHeight - minHeight}{maxHeight-minHeight}=proportionHeight$

[ocean seal]

PapaBread

[bleak ridge]

$\frac{cVolume - minVolume}{maxVolume-minVolume}=proportionVolume$

[ocean seal]

PapaBread

[molten tapir]

aha....

[bleak ridge]

Then you want the proportions to be equal right

[molten tapir]

yeah this is what i was missing

[bleak ridge]

So you can just set them equal and solve for what you dont know

[molten tapir]

how?

[bleak ridge]

Uhh

Youd do like

$\frac{cHeight - minHeight}{maxHeight-minHeight}=\frac{cVolume - minVolume}{maxVolume-minVolume}$

[ocean seal]

PapaBread

[bleak ridge]

That

You know all of the mins and maxes

And you know the cHeight or current height

So the one thing you dont know is the current volume

Which you want to be proportional

[molten tapir]

yes

[bleak ridge]

It has to be proportional cause we set them equal

So then the next step would just be solving for cVolume and simplifying

Im just going to abbreviate using h and v

[molten tapir]

ok

[bleak ridge]

$\frac{cH- minH}{maxH-minH}=\frac{cV- minV}{maxV-minV}$

[ocean seal]

PapaBread

[bleak ridge]

So the first step would just be to multiply both sides by (maxV - minV) right

Then add on minV

$\frac{(cH- minH)(maxV-minV)}{maxH-minH}+minV=cV$

[ocean seal]

PapaBread

[bleak ridge]

Which is pretty nasty

It probably simplifies quite a bit

[molten tapir]

omg 😅

okay

this last equation would do everything without the need of using the proportionns

[bleak ridge]

Ye

But its a mouthful

[molten tapir]

whats that

[bleak ridge]

Its just a lot to write

So Ill try to simplify it

[molten tapir]

okay

my stomach hurting because of my ignorance 😅

👀

[bleak ridge]

Actually it doesnt look like it simplifies much

[molten tapir]

then its fine as it is?

since i have to code it its ok

@bleak ridge how do you call this? its a simple equation or...?

does it have a name in math?

like... progression.... equation....whatever?

[bleak ridge]

Uhh

[molten tapir]

hahhaha sorry, i suck at this 🤣

[bleak ridge]

I mean the base equation is linear interpolation

[molten tapir]

ahh!

yeah

[bleak ridge]

Id just call it a proportion equation though

Whatever makes sense to you though

[molten tapir]

linear interpolation!! thats the name i wanted to google but i didnt even remember

thats what they taught me inn class

(not very well)

[bleak ridge]

Oh

[molten tapir]

whats this? spam?

[orchid bane]

no

[molten tapir]

@bleak ridge thanks a lot 🙏

[bleak ridge]

I dont think youre allowed to advertise here

[bleak ridge]

Np

[molten tapir]

@bleak ridge

the operation between the numbers on top its a multiplication?

[bleak ridge]

Ye

[molten tapir]

hehe thanks 🙏

[bleak ridge]

Its multiplying the two sets of subtraction

[orchid bane]

?

[bleak ridge]

Perhaps

[orchid bane]

i am dead

[molten tapir]

how do i freethischannel?

like, how to say it is solved?

[lone heart]

@molten tapir Has your question been resolved?

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[mental hearth]

how would i approach this

[remote heron]

just a guess but howd ratio work out

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@mental hearth Has your question been resolved?

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[feral steppe]

Hello, is there anyone that can send me proof of the abc formula without completing the square? Feel free to dm or to ping

[placid zinc]

Why do you want such a thing? Haha.

[echo socket]

What abc formula are you talking about?

[feral steppe]

I'm curious

[echo socket]

The quadratic formula?

[placid zinc]

You can also sub the quadratic formula directly into the quadratic, and see it gives 0

[feral steppe]

the formula to solve ax^2+Bx+c for x

[placid zinc]

Which is a proof that it works, but doesn't give the quadratic formula

[feral steppe]

no i mena like

How the quadratic formula works

Like why is it like it is

[thorny patio]

Its a pretty hefty one to prove

Thats why they dont show it when you first learn it

[placid zinc]

Have you used the quadratic formula? Do you understand how to apply it?

[feral steppe]

Yes

I even know how to find the derivative

[echo socket]

The only formal proof I've seen of that formula is using completing the square, but I think there's a 3b1b video about the quadratic formula which might help you

[feral steppe]

It's Christmas break and I've got nothing to do xD

[thorny patio]

Not mine i took it from google

[feral steppe]

It's with completing the square

I need one without

[thorny patio]

Oh without

[feral steppe]

Yes

[placid zinc]

Well, you can still always construct the vertex using calculus

I don't know if that's cheating

[feral steppe]

-b/2a?

[placid zinc]

Yeye. The vertex is at x = -b/2a

Can get a vertex form, without completing the square to get that form

[feral steppe]

hmm

okay

thx

.close

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[pearl loom]

I was watching a youtube video and the instructor did this but I don't understand why is he keeping +c when we know the original function, after taking the anti-derivative of the derivative of a function, aren't we supposed to get exactly that same function?

[glass lichen]

the integral of the derivative is the function up to a constant

derivative of the integral is the function itself

[pearl loom]

why

I know exactly which function I did it

from

why does it matter

[glass lichen]

you dont

d/dx [f(x)+c]=f'(x)

[pearl loom]

suppose we know f(x) is 3x + 4

then I know f'(x) is 3

[glass lichen]

yes

[pearl loom]

so shouldnt integral of that be 3x + 4

[glass lichen]

no

[thorny patio]

When you integrate something, you end up with infinite possibilities that fit the answer

[glass lichen]

$\int\dv{x}(3x+4)\dd{x}=\int 3\dd{x}=3x+c$

[ocean seal]

Mosh

[thorny patio]

It doesnt matter what the original f was

If i integrate something, theres a plus c

Because after you take derivative and integrate you have manipulated the object

[glass lichen]

• indefinite integrals always have unknown constants.

[pearl loom]

yeah I'm asking why

[glass lichen]

cause vertical translation doesn't affect derivatives

so integrating introduces an unknown aspect: Where the graph is vertically

[thorny patio]

Start from f(x) = 3x + 4

Derivative of that is

f'(x) = 3

When i say integrate f'(x) I'm not asking you what was the original f

Im asking for integral of 3

[pearl loom]

yeah ok that makes sense

[thorny patio]

Integral of 3 is not necessarily the original f

It's only one of infinitely many options so you have to plus c there

[urban pine]

the derivatives of 3x + 4, 3x - 2, 3x + 1000, etc are all the same

[thorny patio]

It depends how the question is asked to an extent

[pearl loom]

aight I get it

thanks

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[tacit kernel]

How is a + (b - a) * t also expressed as a(1 - t) + bt ?

[glass lichen]

a-at+bt

[tacit kernel]

.close

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[late parcel]

36 is also an answer choice

I honestly just don’t know where to start

Do you add all the derivatives up and then integrate that?

<@&286206848099549185>

[remote heron]

Use ftc?

[late parcel]

ftc?

[remote heron]

$\int _a ^b f(x) \dd x = F(b) -F(a)$

[ocean seal]

jan Niku

[remote heron]

for appropriate functions

[late parcel]

Ok let me try it

@remote heron is that 24

Is that just -22-2

[remote heron]

$\int _0 ^6 f'(x) \dd x = f(6) - f(0) = 2 - (-22)$

[ocean seal]

jan Niku

[late parcel]

Wait so it should be negative 24

[remote heron]

no

the upper bound goes first

[late parcel]

Wait

Oh ok

So it’s C @remote heron