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[lone heart]

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[alpine sable]

Is $\sqrt[n]{\lvert a^n \rvert} = \sqrt[n]{\vert a \rvert ^n}=\lvert a \rvert$ ?

[ocean seal]

Bla

[lone heart]

@alpine sable Has your question been resolved?

[rose sigil]

yes

though maybe you should declare what a and n are

[lone heart]

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[rose sigil]

if a is a real number and n an integer then yea

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[meager pendant]

This problem is asking me to answer either "u=" "no solution" or "all real numbers are solutions".

[meager pendant]

I'll send my work can someone confirm if I did it correctly?

also if I didn't mess up I'm guessing it might just be no solution?

[vale wigeon]

handwriting issue + no need to guess

the equation really does have no solution

[meager pendant]

so I did it correctly?

also how would I know if it ends up being all real numbers

what would that even look like @vale wigeon

[alpine sable]

if you subtract 6u from both sides and add 3 you get 0u=-1

so for every real number you multiply it by 0 you get -1

that aint possibl

[vale wigeon]

something like 6u-4=6u-4

[alpine sable]

yeah just 0u=0

[meager pendant]

so whenever the variable is zero it's always no solution by default?

[vale wigeon]

don't confuse "variable" with "coefficient"

a letter isn't the same thing as a number attached to the letter

[meager pendant]

oh yeah I meant the number in front of the variable

[alpine sable]

when you ave the equation 0*x≠0 it has no real solutions

because youre essentially multiplying a number by zero and getting a non zero result which isnt possible in R

surely in R idont know in any other sets lmao

[meager pendant]

ok I see.

also for the next part,

this is just basically an equation with just two variables in each side right? just solve as normal? the first one threw me off a bit

[vale wigeon]

there is only one variable: y

it appears multiple times in your equation, but that does not change the fact that there are no other letters involved

solve this as you normally would

[meager pendant]

yeah I think I got the correct answer

[alpine sable]

What about if a is function

[meager pendant]

.close

[lone heart]

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[wintry sun]

Hey guys. I shall proof, that if (zn) in C is a Cauchy-Series in C, then (cn) converges in C. I tried to proof:
(zn) Cauchy -> Re(zn), Im(zn) Cauchy -> Im(zn) and Re(zn) converge in R, thus cn converges in C. Is it the right approach?

[tacit arch]

At a very high level yes

[wintry sun]

Wdym by that? 😄

[tacit arch]

You have zero details about how you're going to bound the modulus

|cn - w| where w is the limiting complex nunber

[wintry sun]

Can i give you a quick summary on how my proof goes along? Maybe theres a flaw in it, it seems a little short.

[tacit arch]

Sure anyone can check

[wintry sun]

Okay,
We try to proof: (zn) in C is a Cauchy-Series --> Re(zn) and Im(zn) are both Cauchy-Series. I want to proof by contradiction.
Lets assume (without loss of generality) Re(zn) wouldnt be a Cauchy-Series.
This means, that there exists at least one E>0 s.t. for no n_0 in N:
|Re(zn)-Re(zm)|<E for all n,m>n_0.
This however means that |Re(zn)-Re(zm)|>=E for an infinite amount of n.

But: |zn-zm| is equal to |z_(n-m)|>=(Re(zn)-Re(zm))>=E for infinitely many n. (z_(n-m) being some complex number)
Thus |zn-zm|<E for all n,m>some n_1 cant be true and thus |zn-zm| cant be Cauchy. --> Re(zn) is Cauchy. Argument is the same for Im(z)

Any obvious flaws?

[rose sigil]

if you're allowed to use that cauchy implies convergent in R that looks convoluted

[wintry sun]

Yeah, we already showed that any Cauchy series in R converges

[wintry sun]

And is it now enough to add that:
Re(zn) -> a in R, Im(zn) -> b in R --> (zn) -> a+bi? We also already showed that zn -> z0 is equivalent to Re(zn)->Re(z0) and Im(zn)->Im(z0)?

[rose sigil]

for any m,n, we have |Re(z_n) - Re(z_m)| <= |z_n - z_m|

so if {z_n} is cauchy so is {Re(z_n)}

similar for {Im(z_n)}

done?

[wintry sun]

YOOO, why do you have to humiliate my proof like that? D:

Thats waaaay smarter.

[rose sigil]

[lone heart]

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[lone heart]

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[quartz summit]

Okay, I need help with proving a potential theorem (or a counterexample)

[quartz summit]

1. for every function f differentiable, continuous and defined for all real numbers, there exists a real number x so that f(x) ≥ f(-x)

proving this is trivial.

2. for every two such functions f and g, there exists a real number x so that f(x) ≥ f(-x) AND g(x) ≥ g(-x)

1. can be proven in the following way:
   y is a real number
   f(y) ≤ f(-y): x = -y
   f(y) ≥ f(-y): x = y

I see no way 2) could be proven.

[rose sigil]

do you think it's true or false?

[prime badge]

you must mean something else

x = 0 always works, that's how you prove 1) and 2)

[quartz summit]

true

[quartz summit]

hmm, I see

[prime badge]

ok my bad

that's what you meant apparently

[quartz summit]

Is it possible for x ≠ 0?

[smoky adder]

i don't think so

[rose sigil]

take f(x) = x and g(x) = -x

[smoky adder]

counterexample would be like y = x and y = -x

[rose sigil]

lol

[quartz summit]

yeah, thought of that just now, hmmm

[lone heart]

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[hasty basin]

does anyone know the explanation of why we use these formulas to find the lower and upper limits?

[stuck fractal]

Are you aware of the box-plot diagram which represents this?

[hasty basin]

the box and whisker plot?

[stuck fractal]

Precisely

[hasty basin]

yeah i'm aware of it

[stuck fractal]

So

In the figure, the difference between min and max provides the range of the dataset,

The median is the centre-point of the ordered data
Q1 is the first quartile (25%) and Q3 is the third quartile (75%)

IQR is the difference between Q3 and Q1

Which part of the equation are you confused about?

the 1.5? part?

[hasty basin]

the whole thing, i somewhat understand the 1.5 part but the rest i don't completely understand

im doing a math project and we have to explain the math we use in it

[stuck fractal]

We're measuring dispersion here, and values beyond Q_+1.5IQR are considered as outliers, I'm assuming you're aware of that.

Are you aware of Gaussian distribution?

[hasty basin]

i've read about it a little

thats how they got the 1.5 i think

[stuck fractal]

That's correct.

Can you describe the function of the lower and upper bounds

[hasty basin]

uhhh, show the range of the data?

[stuck fractal]

If you have a set of data. Ordered data

And you wanted to provide, a description of the data statistically

You can gain valuable data like the IQR, Q1, Q2, Q3, range so on so forth

IQR is a good description of the 'middle' 50% of the data-set.

The IQR can be calculated as follows:

IQR = Q3-Q1

[stuck fractal]

You can re-arrange this and define accordingly to set the boundaries of the whiskers using the 1.5 IQR value.

---

The whiskers must end at an observed data point but can be defined in various ways, not needing to be 1.5*IQR

[hasty basin]

gotcha, im just a little unsure of why we use the IQR in that formula

[stuck fractal]

A measure of spread. How spread out 50% of your data-set is.

Then your whiskers (from your equation you're querying) is a function of that spread

The formulas are essentially saying...

If a datapoint is below $\text{Q1}-1.5\times \text{IQR}$ then it is a potential outlier.

[ocean seal]

@stuck fractal

[stuck fractal]

The factor of 1.5 is a common multiplier (close to 3\sigma) to determine how far below Q1 data must be before it is considered outlier.

This is generally a rule of thumb though, and other multipliers like 3 etc can adjust the sensitivity of what is and isn't an outlier.

The rationale behind it is that the equation provides a threshold where data points beyond a certain min and max are used to identify potential outliers.

[lone heart]

@hasty basin Has your question been resolved?

[lone heart]

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[proper tinsel]

I'm not quite sure how to write this proof

[proper tinsel]

i guess more specifically, i have issues formalizing a definition of "f is green"

[lone heart]

@proper tinsel Has your question been resolved?

[acoustic sundial]

think about when P and Q are really close

Q being greater than Q' means essentially that in between P and Q the derivative was increasing

and that all derivatives in between P and Q are greater than the derivative at P

[placid zinc]

I'd personally find the y-value of Q'. Then you know P is > that

[lone heart]

@proper tinsel Has your question been resolved?

[proper tinsel]

what if Q < P though?

[proper tinsel]

How can i draw any conclusion about P and Q'?

[placid zinc]

You'll have to assume some things. Let's say the x-value of P is a, and the x-value of Q is b.

Totally random variables I just came up with, a and b.

[proper tinsel]

okay

[placid zinc]

And of course you have the function f and f'

[proper tinsel]

yes

[placid zinc]

I'd personally construct that tangent line first

[proper tinsel]

the tangent line at P?

then the tangent line would be something like g(x) = f'(a)(x-a)+f(a)?

assuming a is the x coord of P

i see that the Q' would be g(b) = f'(a)(b-a)+f(a)

and then g(b) < f(b) by definition

so then f'(a) < [f(b)-f(a)]/(b-a)

but im not sure how to get the other inequality

[placid zinc]

Notice that if we took the tangent line at Q instead, then P' < P

[proper tinsel]

er sorry could you guide thorugh a bit of the process

[placid zinc]

You've done all the hard work already. Just, same problem except the tangent line is at x = b now

We're just putting the tangent line at the other point.

[proper tinsel]

ah sorry i was doing this but i didnt flip the inequality so i was left with f'(b) < f(a)-f(b)/a-b

i got it now

thank you so much

.close

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[olive ice]

I need to solve the diff equation in terms of y

what I did is divide by x^2

y' + 2y/x = cosx/x^2

the coefficient function of the y term is 2/x, so

let some value k = e^(integral of (2/x) with respect to x)

therefore k = e^(-2/x^2)

i left the constant out because it will cancel anyway after this step here:

multiply the equation by this k value

i.e. ky' + 2ky/x = cosx*k/x^2

here that constant would cancel if i had included it

so i'm expanding k:

y'e^-2/x^2 + 2(e^(-2/x^2))y/x = (e^-2/x^2)cosx/x^2

oh phuck. i see the issue

.close

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[vale wharf]

I’m very confused on conditionals and converses

[summer dirge]

what specifically is the problem?

which aspect do you not understand?

[vale wharf]

Let’s take for example this one I don’t know how to write the converse for “if you are 18 years old, then you can vote “

And then once it starts adding letters and numbers it gets more difficult for me

[summer dirge]

alright

so "if I am 18, then I can vote" is a statement

more specifically, it is an implication

meaning if you know that I am 18, then you can automatically assume that I can vote without asking me anything else

we say that me being 18 implies that I can vote

[vale wharf]

So then I would say “if I can vote, I am 18?”

[summer dirge]

you got it!

[vale wharf]

Okay the actual difficult one I need help with 2,3

[summer dirge]

in general, if a statement is in the form P -> Q, then the converse of the statement is Q -> P

[vale wharf]

Hm

[summer dirge]

okay, what's the issue?

[vale wharf]

Okay

[gaunt grotto]

Initially I think that there all false except the last

[vale wharf]

“If x is a number divisible by 4, then x is an even number”

[summer dirge]

!nosols

[lone heart]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[gaunt grotto]

hang on

[summer dirge]

okay, what is the converse statement?

[vale wharf]

Thanks but I would like to learn it even tho that’s the easiest way to get it

[gaunt grotto]

x may be even but not divisble by 4 e.g. 2/4 (2 is even) = 1/2 (not an integer)

[summer dirge]

note that whatever comes after if is always P, and whatever comes after then is always Q @vale wharf

[vale wharf]

If x is an even number, x is divisible by 4?

[summer dirge]

so "if I have 18" = P, and "then I can vote" = Q

[summer dirge]

👍

[summer dirge]

the converse just means to flip the if, then order basically

[summer dirge]

do you think this is true or not?

[vale wharf]

So 3 would be “ if AC = CB, then c is on the bisector of AB”

[vale wharf]

Uhhh

[summer dirge]

yes, that's the converse

[vale wharf]

And I need to find out if it’s true

[summer dirge]

yes

[vale wharf]

No

[summer dirge]

why not?

[vale wharf]

I may be wrong though

Actually

[summer dirge]

you're right, but why is it not true?

[vale wharf]

Idk 🤷🏽‍♂️

I’m not good at math

[summer dirge]

no worries

[vale wharf]

I don’t get it

[summer dirge]

think about this way

to show that a statement is not true, you just have to find one counterexample to it

for example, "All sheep are white" is easily disprovable

i just have to find a sheep which is not white

the same applies to your converse statement

[vale wharf]

So I just have to find an even number which isn’t able to be divided by 4?

[summer dirge]

correct.

any even number that isnt divisible by 4 will do

just pick one

[vale wharf]

2

[summer dirge]

👍

that works

[vale wharf]

So that would be false

[summer dirge]

correct

in general, just because an implication is true, doesnt mean the converse is true

[vale wharf]

Ok

“ if AC = CB, then c is on the bisector of AB” would be true

That’s the converse

For 3

[summer dirge]

why do you think it is true?

[gaunt grotto]

I'm not sure that is true

[vale wharf]

because they are all connected

[summer dirge]

elaborate a bit?

[vale wharf]

Or wait since it would have to ACB

ACB since it not in the direct line of it

C wouldn’t be in the bisector ?

Or am I just saying nonsense

[gaunt grotto]

Yes!

[summer dirge]

on the left is the original statement, on the right i have a counterexample to the converse

oh wait i mislabelled

hold on

[vale wharf]

K

[summer dirge]

here

[vale wharf]

sooooooooo

False

[summer dirge]

oh wait

i just realized, C is on the bisector here

[vale wharf]

Ok

[summer dirge]

my bad lol

[vale wharf]

Yeah

All good

[summer dirge]

but the statement is still false

[vale wharf]

Okay thanks 🙏 ima try the rest by myself

[vale wharf]

If I have trouble can I dm you?

[summer dirge]

here, B and C are the same point, so i dont think the converse is true here

[vale wharf]

Hmm okay

[summer dirge]

uh, I'd prefer you not to be honest

[vale wharf]

All good

Thanks anyways

[summer dirge]

if you need more help, just open another channel

odds are someone will help you

[vale wharf]

👍🏼

Your a life saver

I have a game tomorrow and if I don’t pass i can’t play 🥲

You too @gaunt grotto

Thanks

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@vale wharf Has your question been resolved?

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[proper radish]

Hi

[proper radish]

Why is (c) a partia, order but not total I’m failing to comprehend

[limpid turret]

For problem 1 I'm guessing?

@proper radish Consider -1 and +1. Are they comparable?

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@proper radish Has your question been resolved?

[lone heart]

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[jade kayak]

how do i do b and c

pls help lol, gotta sleep soon

[stark crater]

Have u tried taking the 2nd derivative of g

For c just apply IVT

[jade kayak]

Yeah I got g''(x) = [f''(x^2-x) * (2x -1)] * [2(f'(x^2-x)]

but i have no idea what to do after

[jade kayak]

ohh ok

[stark crater]

Solve for the f’’ part. There are two values where x^2-x=0 but one of them is more useful because of what u r given

[jade kayak]

ohhh

[worthy horizon]

Plug in 0 and see if + or -

[jade kayak]

So with the double derivative equation of g''(x). I isolate f''(x) and plug in to solve for it

[worthy horizon]

Yea

[jade kayak]

got it thanks

.close

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[brazen temple]

Can someone explain the third step?

[brazen temple]

-S = ...

Rotated image:

[quasi vector]

they subtracted 2S from S

[brazen temple]

RHS?

[quasi vector]

it's the first step - the second step

[brazen temple]

1 . 2 + 1 . 2^2 + 1 . 2^3 + .... + 1 . 2^100 - 100 . 2^101

this^

[quasi vector]

yes

[brazen temple]

how is it - 100 . 2^101? I didn't get that.

[quasi vector]

all of the terms line up except for the 100* 2^101 term, so when you subtract it comes down with a - sign

[brazen temple]

can you please elaborate?

I didn't get that.

[brazen temple]

^

[quasi vector]

do you understand WHY they are multiplying the S by 2 and then subtracting?

[brazen temple]

no

[quasi vector]

i assume you are being taught AGP's right now?

[brazen temple]

yes?

[quasi vector]

well

the point is to transform the AGP into a GP

since we know what the sum of a GP is

the second step basically "shifts" all the terms to the right, so that when you subtract, you are left with a GP

[crude kindle]

Guys help

[quasi vector]

the -100 * 2^100 shows up because it is an excess term

[crude kindle]

Please help me

[lone heart]

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[brazen temple]

oh ic

[brazen temple]

nvm got it.

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@brazen temple Has your question been resolved?

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[vagrant agate]

in a simulatenous equation y = 4x-1 and 2x-y+5 = 0

[vagrant agate]

if i sub y = 4x-1 in it would become 2x -4x -1 +5 = 0

then -2x = -4 so x = 2

[stark crater]

When you sub y in you get 2x-(4x-1)+5. You need to distribute the negative in front of (4x-1) before proceeding

[vagrant agate]

oh so 2x - 4x + 1 +5 = 0

[modern skiff]

yep that's right

[vagrant agate]

then -2x = -6

x = -3?

3*

oh ok ive got it thank you

.close

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[meager gull]

Is there a better solution than to apply l´hoptial again?

[solemn juniper]

you're supposed to differentiate both parts of the fraction, not differentiate the fraction

[meager gull]

omg

true

😂

[alpine sable]

yes, use cos(x+π/2) = sinx identity
factorise the denominator
then use the result
x tends to 0 sinx/x = 1

[vale wigeon]

i'd begin by substituting t := x-3 lol

[deep moon]

Why differentiation

You can just use cos(pi/2+x)=-sinx

[lone heart]

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[marble ibex]

Would my work be right

i think YT is 17.6

its 17.57

but rounded up its 17.6

[indigo vine]

wtf is thus persons pfp it looks like a naked women

[marble ibex]

probably a photo shoot

[tough pilot]

seems like an average discord pfp

ive seen worse

[indigo vine]

💀

like

[marble ibex]

ay yall we answering the wrong questions frfr

[tough pilot]

i got sqrt 112

[marble ibex]

how tf

lmao

the square root of 309 is 17.57

[marble ibex]

where did we get 23 from

[tough pilot]

16+7

thats xz

[marble ibex]

16^2 + 7^2 = yx^2

[tough pilot]

why did u do 16^2 + 7^2

[tough pilot]

yx is not the hypotenuse

xz is the hyp.

[marble ibex]

the hypotenuse is usually the opposite direction from the right triangle ?

[tough pilot]

wait ima write it on paper

[marble ibex]

there are three triangles

X Y Z and X Y T and Y T Z

[tough pilot]

,rotate

[tough pilot]

@marble ibex

[marble ibex]

give me a minute just looking over

I do not think that is right.

please explain your work

@tough pilot

[tough pilot]

xyt is a right triangle

where xy is the hypotenuse

so xy^2 = 16^2 + x^2

(x is YT)

now, triangle ytz is also a right triangle with YZ as the hypotenuse

so YZ^2 = x^2 + 7^2

now we take the big triangle, here XZ is the hypotenuse

XZ^2 = YZ^2 + XY^2

XZ is equal to 23

so putting in the values, we have

23^2 = (16^2 + x^2) + (7^2 + x^2)

@marble ibex do u get it till here?

[marble ibex]

yes

[tough pilot]

now just solve the equation
2*x^2 = 23^2 - 16^2 - 7^2

2*x^2 = 224

x^2 = 112

[marble ibex]

so 10.6

damn

you got the same answer as my friend

but

you used a whole different way

the hillbilly way

[lone heart]

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[pliant palm]

How do I get the area of the shaded region

[vale wigeon]

,rccw

[vale wigeon]

the two sectors minus the kite APBQ, surely

[pliant palm]

Ok thanks

Is there any other way if I don't get the area of the kite?

[vale wigeon]

could also split it along PQ and calculate it as two circular segments

[pliant palm]

Ok got it. I did by just getting the area of the triangle and doubling it. Thanks

.close

[lone heart]

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[steep jackal]

how do u know what test to use in sequence and series

[steep jackal]

when given a random question

anyone got a cheat sheet of tricks or algorithms

i dont have time to solve qquestions

[lone heart]

@steep jackal Has your question been resolved?

[limpid turret]

Depends on the question

[lone heart]

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[solid sluice]

why is this not possible?

[echo socket]

When solving for unknowns, you can't just differentiate both sides because that's not an equality of functions

x is some constant, how do you go about differentiating with respect to it?

[solid sluice]

ah i understand

thank you

.close

[lone heart]

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[limpid turret]

You can solve this with the W lambert function I think

[lone heart]

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[old gust]

How do I solve this

[old gust]

from what I understand

of the bigger square (piece of land) for example can be

(Length of the shed + 10)

and width can be (width of the shed + 40)

[lone heart]

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[old gust]

.close

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[loud crater]

Hello I am french and i need help for my exercice

[vale wigeon]

post both the fr and eng versions

french first

[loud crater]

How i traduct ?

[vale wigeon]

maybe you won't need to *translate

what does the word "parcelle" mean?

ok it means land plot apparently

[loud crater]

Groud

Plot

[vale wigeon]

predictable given the context

ok right

[loud crater]

Ok

In first I have to show that the angles are similar

[vale wigeon]

here is the translation of the problem...

Using the info in documents 1 and 2,

1) Show that triangles ABC and BCD are similar.
2) Help Sarah calculate:
(a) the cost to enclose land plot #2 along the sides CD and BD
(b) the amount of time needed to mow plot #2.

that the triangles are similar.

[loud crater]

Yes but why ?

[vale wigeon]

why what

[loud crater]

Why is it similar

The question is that

[vale wigeon]

do you know in general how to prove similarity for triangles?

[loud crater]

Yes de need two similar angles

We need

[vale wigeon]

two pairs of equal angles.

ok, and do we have that?

[loud crater]

We have one

[vale wigeon]

one and only one pair?

[loud crater]

After is hide but i don’t find it

[vale wigeon]

ok, can you name the two equal angles you see?

[loud crater]

bDc and aCb

[vale wigeon]

uppercase

BDC and ACB, ok

angle ABC is equal to some other angle in the picture

which one?

[loud crater]

BDC ?

[vale wigeon]

no

[loud crater]

Euh fail

Bcd

[vale wigeon]

ok right

so we have two pairs of equal angles

we win

[loud crater]

But nothing proof that

[vale wigeon]

what?

what are you talking about here

are you saying that these angle equalities are unproven?

@loud crater ?

[loud crater]

Yes

Sry i was Trying

[vale wigeon]

so you see two angles indicated with the same mark

and you claim that this is not evidence enough that the angles are equal

do i understand you correctly?

well you are wrong

[loud crater]

Yes

[vale wigeon]

or deliberately ignorant of the conventions of geometric notation

when two angles are marked with the same angle mark in a diagram, it means they are known to be equal

[loud crater]

I Know but have mark only on bDc

And aCb

[vale wigeon]

incorrect

there are two sets of marks

[loud crater]

Ok so the angle b is 90

[vale wigeon]

angle CBD is 90° yes

[loud crater]

Ok

And How calcul cd and bd ?

[vale wigeon]

use the similarity that we just proved

or maybe trigonometry if you have access to that. it might make things a bit easier

[loud crater]

I Didnt know How do that

[loud crater]

Can you give me just one mesure

[vale wigeon]

DC/CB = CB/BA

use this to find DC

[loud crater]

Ok i Gonna ttt

Try

I find 1,25 @vale wigeon

[vale wigeon]

1.25 what?

1.25 kilometers?

[loud crater]

Cb/ ab is 1.25

Oh so is 62.5

?

@vale wigeon

[vale wigeon]

DC is 62.5 meters, yes

find BD by any method you wish

[loud crater]

37,5

With pythagore

[vale wigeon]

ok whatever

[loud crater]

How find the time she cut Now ?

[vale wigeon]

it's proportional to area

[loud crater]

?

[vale wigeon]

cutting time is proportional to area

find the area of plot 1 and the time taken to cut it (from document 2, you are given the start and end times)

find the area of plot 2, or recognize that it's (5/4)^2 times the area of plot 1 (by similarity!!!)

either way set up a proportion or whatever.

[lone heart]

@loud crater Has your question been resolved?

[lone heart]

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[alpine sable]

Q5

I might be completely wrong

Could someone direct me?

[lone heart]

@alpine sable Has your question been resolved?

[deep moon]

Hello

@alpine sable

[alpine sable]

hi

[deep moon]

You can find the poi of 2 graphs

Point of intersection

Then

[alpine sable]

I tried by making the integrals equal to eachother

but it seems to not be working

[deep moon]

Our given area is area covered by 8e^(-x) - area covered by e^(2x)

Both in 1st quadrant

[deep moon]

Explain why you would do that

[alpine sable]

]to find the area shared by both of them?

[deep moon]

You're putting integral ydx as equal for both

That means you're implying that they cover the same area

Which doesn't make much sense, right?

[alpine sable]

yh i guess

so i need to find the poi? Is rthat different to taking away the area?

.close

[lone heart]

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[cosmic acorn]

hi

[cosmic acorn]

ist that correct

is there a difference between : start in 0,0 and beginn in 1,1 ?

I began at 0,0 in this task for example

[lone heart]

@cosmic acorn Has your question been resolved?

[haughty sun]

!original

[lone heart]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

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[cosmic acorn]

.reopen

[lone heart]

✅

[cosmic acorn]

what do you need?

Is that correct, yes or no?

[haughty sun]

i need context, what class is that from math, physic, other. what the question. i need details

[cosmic acorn]

math, you can see above the vectors in 2d

I have to draw it in the coordinate system

as vectors

[worn fox]

What you've done is correct

You should start at 0,0 really yes

[cosmic acorn]

what if I start at 1,1

[haughty sun]

oh, it's just a vector problem okay

[cosmic acorn]

is there something different

[haughty sun]

everything would be one point up and one point right

[worn fox]

We like to relate a vector to the point that it's coordinates represent. If you start at 1,1 you lose this correspondence

[cosmic acorn]

I have heard, that if i have a coefficient for example 4*(1,1(vector)) then i have to draw it at first normally, so just 1,1 but the second vector has to be 4 times bigger than the first one. Is that right?

[cosmic acorn]

did you understand my questuon? Xd

my english is not that good

[worn fox]

Yes it would become (4,4)

[cosmic acorn]

so that is exactly then valid if I do not begin at 0,0 right?

[worn fox]

It's slightly complicated

Really vectors don't care where you draw them, you only care about length and direction

So sure you could draw (4,4) starting at the point 1,1 and ending at 5,5

But in isolation there isn't really any point in doing this and you lose the intuitive picture of "the arrow finishes at (4,4)"

[cosmic acorn]

I see

.close

[lone heart]

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[undone saddle]

why is the left area not calculated by (-g(x)--f(x))dx?

[undone saddle]

this part, the intersections are x=-1 and x=0

the graph for f(x)=x^3 +6x

the graph for g(x)=-x^3 +4x^2+12x

ig my question is how do you calculate that area

I know how it works in a positive area of the graph, not a negative one

<@&286206848099549185>

[severe cloak]

Vbhj

Negetives

[haughty sun]

then you can just move the axis

[undone saddle]

if you could why would you need to change the formula into negative if all you do is change the y axis

[haughty sun]

f(x)=x^3 +6x + 10
g(x)=x^3 +4x^2+12x+10

[undone saddle]

hmmm

ig thats maybe true but

I dont think my examiner would appreciate that

[severe cloak]

Yes

[haughty sun]

there are many ways to calculate the area

[undone saddle]

like what?

how do you do it the conventional way thats what im supposed to know

[haughty sun]

$\left| \int_{?}^0{g(x)}dx -\int_{?}^0{f(x)}dx\right|+\left| \int_{0}^?{g(x)}dx -\int_{0}^?{f(x)}dx\right|$
where ? are intersection point

[ocean seal]

nico

[haughty sun]

like this one

[undone saddle]

wait I thought you were asking a question

ok lemme try and fill it out

[haughty sun]

$\left| \int_{?}^0{g(x)-f(x)}dx \right|+\left| \int_{0}^?{g(x)-f(x)}dx\right|$
same thing but wrote differently

[ocean seal]

nico

[undone saddle]

.close

[lone heart]

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[worldly gulch]

Is it right?

Ill move forward

[fallen verge]

No

[worldly gulch]

It should be

+2?

[fallen verge]

Do not distribute the 2

[worldly gulch]

Wdym

[fallen verge]

2(2x-1)³ =/= (4x-2)³

[worldly gulch]

Ok

Thats what i wrote

But ight

[fallen verge]

Just because you wrote it doesnt mean its right

[worldly gulch]

(-)*(-)=+

[feral hemlock]

You can't multiply something with something with another power

[worldly gulch]

So its (4x-2)³

[fallen verge]

NO

[feral hemlock]

Just don't touch

[worldly gulch]

Oh

I open it right?

[fallen verge]

Expand it, yes

[worldly gulch]

I opne the power

Yes

Ok

[feral hemlock]

Firsttly expand (2x-1)³

[worldly gulch]

?

Like that

Why

@cinder tundra

[cinder tundra]

You didn’t multiply 2 with -1^3

I stopped reading there but the rest also looks wrong

[worldly gulch]

(a-b)³

[cinder tundra]

Oh you are just doing the first

Sorry

[worldly gulch]

🤬😡😡😡👍🏽

Jp

It's ok

[cinder tundra]

Forget everything I confused with your mistake in the second image

[worldly gulch]

Forget the - -

Its -

[cinder tundra]

And how much is 2 times 8 times x to the cube?

[worldly gulch]

2*8x³?

[cinder tundra]

Y

[worldly gulch]

Its 16x³

[cinder tundra]

Ok so u got 16x^3-12x^2+6x+1

And u can’t forget that was multiplied by 2 in the beginning

[worldly gulch]

U cant multiply power

[cinder tundra]

? Where

[worldly gulch]

.

[cinder tundra]

I already expanded

Now u multiply individually

[worldly gulch]

I multiplied by 2 the (2x)³

[cinder tundra]

Oh u did in between

Your writing is confusing but ok

It’s better u expand first

[worldly gulch]

yh thats my bad notebook

[cinder tundra]

Put it between bracket and then show how u multiply by 2

[worldly gulch]

no look

It was

2*(2x-1)³

I expand rhe (2x-1)³

It becomes 2*(2x)³............

So 2*8x³

16x³

[cinder tundra]

2(8x^3 - 12x^2 + 6x -1)

Now u do distributive here

And u get 16 -24 12 and -2

With correspondent x

[worldly gulch]

Ohhh

[cinder tundra]

Yes

[worldly gulch]

Thanks

I expaned the (4x+1)³

@cinder tundra

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[worldly gulch]

.close

[lone heart]

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[final badge]

How do i solve this?

[lone heart]

@final badge Has your question been resolved?

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[wooden jewel]

Help how do i do c)

[lone heart]

@wooden jewel Has your question been resolved?

[icy crane]

find a counterexample

[summer dirge]

do you think that c) is true or false?

[lone heart]

@wooden jewel Has your question been resolved?

[lone heart]

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[rocky summit]

$$\lim_{x\to 0^+} \frac{3 \ln x}{x}$$

Can someone please give a hint?

[ocean seal]

Sweet Tea 🧋

[rocky summit]

as $x\to+\infty$ it obviously converges to $0$, but when $x \to 0$ we have $\left[\frac{-\infty}{\infty}\right]$, so idk what to do

[ocean seal]

Sweet Tea 🧋

[tacit arch]

If you can't do l'hoptial then try using log properties

And swap limit with log

[opal ingot]

isn't this just -infinity divided by 0

(leaving for a bit, but I think you got this one)

[rocky summit]

,close

.close

[lone heart]

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[alpine sable]

how do i do this question

[nocturne vapor]

nice pfp

[alpine sable]

thanks can u help?

[nocturne vapor]

i think

so you're given two points

[alpine sable]

yh

[nocturne vapor]

it's asking you to form an equation where n is years after 1995

so you have (191, 0) and (169.68, 13)

so then you can find your slope with those two points

which should be enough to make the equation

[nocturne vapor]

just realized the numbers should be reversed sorry

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[lone heart]

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[stark crater]

the function is negative when the graph is below the x axis

[lone heart]

@rustic jolt Has your question been resolved?

[lone heart]

@rustic jolt Has your question been resolved?

[eternal fable]

By observing the function is negative between 1 and -1

So

-1<x<1

[rustic jolt]

thank you!

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[rapid sundial]

Is the complex contour integral the same idea as the "curtain" integral from calc 3?

[placid zinc]

"curtain" integral?

The only integration I do is horseshoe integration

[rapid sundial]

idk like the integral was like a curtain

and you found the area of the curtain

[placid zinc]

Complex integrals are pretty similar to line integrals

[rapid sundial]

ah ok thanks

that's what they're called

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[candid radish]

One interior angle of a convex polygon is $160$ degrees. The rest of the interior angles of the polygon are each $112$ degrees. How many sides does the polygon have?

[ocean seal]

!ｓａｔｏｒｏ !

[candid radish]

uhm hello I need help with this question

I needa go soon, so I'm kinda in a bit of a hurry

nvm I got it

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[candid radish]

.reopen

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✅

[candid radish]

An equilateral triangle and a square are inscribed in a circle as shown. $ABC$ is isosceles. The triangle and square share a common vertex. What is the number of degrees in the measure of the angle indicated by the question mark?

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!ｓａｔｏｒｏ !

[candid radish]

uhh I need help with this question

nvm I got it

.close

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[kind violet]

Hi

[kind violet]

Why is that not correct?

Or this

Its 3am and my brain is fckong (yes fckong) tired

[severe bridge]

could you explain how you get from s-r=7 to 0,3s-r/3=3?

0,2*

[kind violet]

Look at the first pic

I did + r to have s=7+r and put it in s from the second equation

[severe bridge]

Ok let's start from the beginning

You have s-r=7 and u have to find r?

[kind violet]

you re right

[severe bridge]

Can u send the original question?

Because you can't find r the value of r without the value of s, unless you express r in function of s

[kind violet]

2 Equations

Its not from a function

[severe bridge]

oh okey

sry i missread the problem

[kind violet]

No problem 🙂

There is the linear system of equations and it has four methods. I have used the substitution method.

So I do + r

And then I have s= 7+r

And then I take the s= 7+r and put it in the second equation

[severe bridge]

okey so everything up to the 5th line seems ok

the problem is

0,2r+1/3 is not 1,2r

[kind violet]

TELL ME

Wait

I did * 3

So there is 0.2r + r= 1.6 * 3

[severe bridge]

You have to multiply 0.2r by 3 too

[kind violet]

Why is that?

Actually

If I take this to the other side

I never noticed that

[severe bridge]

So when you are working with an equation you need to make sure that everything remains the same in value

If you multiply one side by 3, you have to multiply the whole other side by 3 too

So 3(0,2s+r/3)=1.6*3

[kind violet]

If I have for example:
3a + 7 = 6 |-7
3a = -1 |:3
a=-1/3

Lets do the division :
3a + 7 = 6 | :3
a+7 = 2 |-7
a= -5

Lets do the way you said:
3a + 7= 6 |:3
a + 2.33 = 2 -2.33
a= -1/3

I see now ig

I never noticed that fr

uff

Thank you!!

[severe bridge]

You're welcome!

[kind violet]

.close

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[heady vapor]

Jeff finds some bugs. He finds 10 fewer grasshoppers than crickets. He finds 5 fewer crickets than ladybugs. If Jeff finds 5 grasshoppers, how many ladybugs does Jeff find? How many crickets does he find? Write two equations to solve the problem.Jeff finds some bugs. He finds 10 fewer grasshoppers than crickets. He finds 5 fewer crickets than ladybugs. If Jeff finds 5 grasshoppers, how many ladybugs does Jeff find? How many crickets does he find? Write two equations to solve the problem.

[faint monolith]

so let Grasshoppers=g

[heady vapor]

I did the variables of G for Grasshopper/ C for Crickets / L for Ladybugs

[faint monolith]

Crickets=C

oh

ok

so

10 fewers grasshoppers than crickets

can you form a equation with that?

[heady vapor]

I have listed as G = C- 10

[faint monolith]

yes that works

then do that again for "5 fewer crickets than ladybugs"

[heady vapor]

so equation 2 is

C = 5 + L

[faint monolith]

no

you did the first equation correct

this is the same thing

there are 5 FEWER crickets than ladybugs

so if you add another 5 to the amount of ladybugs then it would be 10 fewer

you want them to be equal

[heady vapor]

im stuck then

so do I solve for the first equation?

[faint monolith]

so how did you do this

what was your thinking

when interpretating those words into this equation

[heady vapor]

I was thinking G would mean how many grasshopper Jeff has, so i put it as 5

so i was going to write 5 = C - 10

[faint monolith]

We are not up to that part of the problem yet

first write down your two equations

work step by step

you already got g+10=c

"He finds 5 fewer crickets than ladybugs"

[heady vapor]

so its l-5?

[faint monolith]

How do you make the amount of crickets equal to the amount of ladybugs?

[heady vapor]

L-5?

[faint monolith]

yes but do you understand why

currently you have 5 MORE ladybugs than crickets

[heady vapor]

because he has 5 FEWER crickets

[faint monolith]

so you need to minus 5 to make it even

yes

so now you have

g+10=c

c+5=L

then it tells you g=5 right

[heady vapor]

yea, jeff has 5 grasshoppers

[faint monolith]

so we plug that into our first equation

[heady vapor]

so 5+10 = c so 15 = c

[faint monolith]

correct

then we plug that into second euation

[heady vapor]

so he has 15 crickets

[faint monolith]

to get L

but the question asks for how many ladybugs not crickets

so we're not done

[heady vapor]

5+15 = 20

[faint monolith]

yes

that is the answer

[heady vapor]

oh man, i got confused with the second equation

[faint monolith]

all goods

[heady vapor]

ty for that help really informative

[faint monolith]

your welcome

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@heady vapor Has your question been resolved?

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[frosty prawn]

i am trying to derive the limit form of integral by hand

i am just thinking about it (in terms of areas of rectangles) and got this far

i ran it using code and it seems to work when setting r to a high number like 1000, 10000 etc but idk if this is a generally accepted notation of writing the limit of an integral

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@frosty prawn Has your question been resolved?

[tacit arch]

Key idea 8 here
https://math.libretexts.org/Bookshelves/Calculus/Calculus_3e_(Apex)/05%3A_Integration/5.03%3A_Riemann_Sums

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@frosty prawn Has your question been resolved?

[frosty prawn]

Yess

this is what I needed

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[frosty prawn]

thanks @tacit arch

your name is literally riemann

xD

i only know riemann hypothesis

but not this ty

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[olive oar]

If x is the matrix with ones just on top of the diagonal and and 1 at the bottom left, and y = diag(0, 1, 2, .., p-1) over F with char F = p and L is the Lie algerba spanned by them (we know it's 2 dimensional since [x,y] = x), how do I show that $M = L + F^p$ is solvable?

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Zander

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[pulsar delta]

hi

[pulsar delta]

i have a query

[rapid sundial]

[pulsar delta]

is 3 months sufficient time to cover these areas?

given i have basic understanding of maths

[torpid dirge]

this seems like a basic understanding of maths

[rapid sundial]

"basic" is very vague. Also, I have no idea what your capabilities are nor your time commitment

[pulsar delta]

high school maths

[hidden fable]

it’s nothing too demanding with high school maths and some study, but no idea what your level is

[pulsar delta]

can i post pictures of questions ?

[rapid sundial]

Unit 5 is General English. No idea what that is

[pulsar delta]

unit-1 is enough i can manage all other areas

[hidden fable]

in entrance exams they sometimes have english questions, they’re usually pretty basic stuff

[pulsar delta]

am i allowed to uplaod pdf here?

[rapid sundial]

Again, everything seems pretty standard. I have no idea what your time commitment is nor do I know how fast you learn. If you spend enough time, I'm sure it's possible

[pulsar delta]

can anyone take a quick look at questions? the last section, each question is for 2 marks

part 2 page 12

[hidden fable]

sat vibes

[pulsar delta]

comp sci questions are easy

[hidden fable]

what maths topics did you cover in high school?

[pulsar delta]

actually i took a 2 semester discrete maths course

i have forgotten all of it since its been 3 years

i didnt score well enough also

also i just need to score around 70/100

in this exam

[hidden fable]

you’re gonna make it

[rapid sundial]

You took discrete math in high school, but you're doing this kind of math...you're sort of out of order