help-28

[forest widget]

cut the hexagon into pieces then rearrange

[torn jolt]

I did the wrong assumption that a regular hexagon must have always the same angles

[gilded tapir]

mmm do i just draw a line ae bd

and the midlle left is a recrangle

@forest widget

can you pls help me its 2 am and i have to finish thees problems today

@forest widget

@torn jolt

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[shut patrol]

$x=f(y) | x=f-1(y)$

[glossy valve]

marulk

[shut patrol]

are the both of these ways to represent an inverse function

[pale oriole]

if you mean $x=f^{-1}(y)$ then yes

[glossy valve]

Flappie

[pale oriole]

but f^-1 only exists when f is bijective

so be careful

[shut patrol]

ok thank you

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[astral valley]

can someone help me with this? its about slope

[pale oriole]

!status

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6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[astral valley]

1

[pale oriole]

do you know what they mean by "rise" and "run"?

[astral valley]

i'm supposed to find out the slope of the line

[pale oriole]

in the question they are talking about using the "rise" and "run"

[pale oriole]

i am asking you a question

do you know what they mean by rise and run

its a yes or no question

[astral valley]

sorry, no

[pale oriole]

okay

then i will explain it to you

[astral valley]

thanks

[pale oriole]

imagine you are standing on a hill

it is sloped

you are walking up the hill

until you have moved horizontally, 100 meters

the change in height would be the rise

you rise a certain amount of meters during your walk

[astral valley]

oh

that helps so much

[pale oriole]

i know the drawing is shit

but something like this

then your slope

is $\frac{\text{rise}}{\text{run}}$

[glossy valve]

Flappie

[astral valley]

Thank you so so so much

[pale oriole]

can you find the slope in the graph now?

what would be a rise and what would be a run

[astral valley]

i did it!

[pale oriole]

what did you get as answer?

[astral valley]

close

[pale oriole]

use .close

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@astral valley Has your question been resolved?

[rose parcel]

@pale oriole Sorry for my last question, I had to go.

[pale oriole]

no problem

but start a new thread rq

dont go into this one

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[torn jolt]

\let\1\preccurlyeq\let\2\prec
Determine whether $(3,5)\2(4,8)$ in the poset $(\Z\by \Z, \1)$ where $\1$ is the lexographic ordering

[torn jolt]

is this as simple as saying, that because $3 < 4$ we have $(3,5)\prec(4,8)$

[torn jolt]

yes

an also because 5 < 8

both

ok

thanks

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[ivory moon]

Is sqrt(x) continuous at 0?

[torn jolt]

yes

[ivory moon]

Yeah I thought so. Just wanted to make sure. Thanks

[fair oracle]

Maybe the confusion is approaching from the left of 0

you don’t need to approach as long as it’s not in the domain

[ivory moon]

Yeah got it

Thanks

[torn jolt]

you can only talk about continuity on the domain where the function exists. If you are sceptical because of the non-existence of a left-sided limit then the thing is that we dont even talk about it due to the fact that the function doesnt even exist there

[ivory moon]

👍

Thanks both

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[rose parcel]

Hi! I have this exercice:

[rose parcel]

The first derivate is:

$f'(x)=(14x)/(x^4-2x^2+1)$

[glossy valve]

GANSO

[rose parcel]

The second derivate is:

$f''(x)=(-42x^2-14)/(x^6-3x^4+3x^2-1)$

[glossy valve]

GANSO

[rose parcel]

So

The first objective that i have is to find the poinst that cancels the first derivate of the function

To do this I made f(x)=0. Then I have x=0

Now I'm stuck here, understanding if it is correct, or instead I have to find the numbers that cancels the denominator of the fraction. In this case 1 and -1

If I'm not clear pls tell me

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[thorny prawn]

Help with solving this please (please help with first steps and setting up problem but pls pls dont give it away i want to understand)

Would it simply be adding 22 + 12?

Or are there other steps I haven't taken into account

[warped gulch]

you need to subtract as both vectors are opposite

[thorny prawn]

Are they opposite because of their directions?

Acutally nvm

i can answer that myself

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[warped gulch]

yes

[thorny prawn]

what about this on

e

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[hardy jewel]

Are u aware of the concept of components? @thorny prawn

[thorny prawn]

no

i need to learn

[hardy jewel]

So u need to learn before attempting this question

[thorny prawn]

yes

[warped gulch]

yeah you need to learn basics of vector first

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[finite marlin]

hello

[finite marlin]

I don't know how to compose my function n times, I've been trying to make it for an hour

anyone can help me please

I tried a "for i in range(n)" loop but it seems more complicated

[summer igloo]

well, this isnt a programming server, but here's one possible solution anyways:

in your f_o_f func, you can pass in another parameter "n"
then return f_o_f(f(x, y)[0], f(x, y)[1], n-1)
if n > 1
else return f(x,y)[0], f(x,y)[1]

your original code though, is not really optimized

[finite marlin]

hm a recursive function

I'm going to try it

[summer igloo]

cuz f(x,y) already returns both x and y outputs, but in f_o_f, youre running the f function twice. that'll hog up a lot of space(?) and time

[finite marlin]

Of course, I was trying to see what happenned in my program !

Thanks !

[summer igloo]

oh, btw, to make your life simpler. when using recursion, you wont even need f_o_f func
you can do the recursion on f(x, y, n) alone

try that out as well

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[dire rampart]

I'm playing around with R-formulae to see if they work when I swap them around. The first formula worked just fine but when I tried to make it in terms of sin, it's not the same. Can someone please tell me why?

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[robust slate]

$\cos x=\sin \left(\frac{\pi}{2}-x \right)$

[glossy valve]

Civil Service Pigeon

[robust slate]

use this ^

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[modern glacier]

Given a collection of centered random variables X1, ..., Xm such that E[Xi] = 0 for all i = 1 to m taking values in [a, b] in R
How would i show that E[Xbar] <= (1/p) * log(E[e^(p*Xbar)])? Note XBar is the maxi Xi and p > 0

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@modern glacier Has your question been resolved?

[wicked aurora]

Seems like you want to use Jensen's inequality.

But I cannot answer how you would use all of the information

To me, it seems like this holds no matter the distrubution of Xbar.

[modern glacier]

how would jensen's inequality apply here?

assuming i don't use all the information

[wicked aurora]

Let me see: Let Y = e^(p*Xbar) and g(Y)=ln(Y)/p.

Wouldn't then g(Y)=Xbar?

[modern glacier]

yh

p*XBar lne / p so = XBar

[wicked aurora]

Okay, and g is concave, so then by Jensens inequality, you directly get the inequality directly.

Well, (checking the internet for criterion), perhaps you need the other criterion to show that E(Y) and E(Xbar) are finite

However to me that sounds excessively formal.

[modern glacier]

ok thanks, got it now

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[glossy valve]

Mycobacterium

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[frozen coral]

Seems like bro forgot to change deg to radians on calculator

[nova basin]

Not even
It's not 0.01 rad

[grim skiff]

Not sure how you got 0.01

Because I got 26 in degrees and 0.4 in rads

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[hexed halo]

Can someone please explain me this?

[frozen coral]

Context?

[hexed halo]

Page number 97

Please delete it

[frozen coral]

These are empirical probabilities that depth of the snow is observed inches

So, for example, there is 2/6 probability that depth of the snow is 6 inches

[hexed halo]

I see. So, is it possible to know why author put 2/6?

[frozen coral]

In the sample space there are 6 observations out of which 2 observations are of 6 inches

[hexed halo]

Sorry, I am not understanding it.

[frozen coral]

There are 6 blocks and depth of snow in each block is
4, 5, 6, 6, 7, 8
respectively

So, to calculate the empirical probability,
We have to divide favourable no. of observations(2 in this case) by total no. of observations(6 in this case)

[hexed halo]

Makes sense!

Thanks mate 🙂

[frozen coral]

My pleasure ☺️

[hexed halo]

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[velvet storm]

Hello, I play a card game, and here's my question on probability.

In a 33 card deck, what is the probability that I get exactly one copy of a card I have two copies of (Card A) or a copy out of two copies of (Card B) in an opening hand of 5 cards?

What is the probability that after my opening hand, in the top nine cards of my deck, is there at least a copy of (Card A), assuming no copies of (card A) in my opening 5 cards?

[fathom saddle]

What's a "copy"?

33 card deck? What's in that?

[velvet storm]

A card. Like, a king, or a jack. This is for Arkham Horror card game, but that's not important to the math

[keen agate]

So you are looking for 2 cards out of 33 total?

[velvet storm]

Basically, if I have two of (Card A) and two of (Card B), what is the probability of only having one of any of those 4 in an opening hand?

After that opening hand, what is the probability of having one or two of card A in the top 9 cards of my deck?

[keen agate]

so each one is a 1 in 33

with 4

so 4 in 33 chance

how many do you draw in an opening hand?

[velvet storm]

5 cards

[keen agate]

so there are 28 cards in the deck remaining after the first draw

2 in 28 per place of 9, so I think it's 9 * 2/28 or 18 in 28 chances that they are in the top 9 cards (assuming you didn't draw one in the opening hand) I think this is corect but I am not 100%, statistics can be a funky from time to time

[velvet storm]

Okay, thank you!

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[modern glacier]

Given that i proved this result

[modern glacier]

how would i pick lambda such that it solves this

XBar is max Xi btw, and E[X_i] = 0 (suppose we have collection of random variables X1,...Xm)

[finite sun]

what math is this?

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[torn jolt]

asked to find range of this

[torn jolt]

I've worked it out to be y>=12

this is how I did it

I know that y=|x|, must be greater than 0

so $|x|-12\leq0$

$|x|\leq12$

[glossy valve]

mj

mj

[wicked aurora]

What if x=-20?

[torn jolt]

this comes after btw

[torn jolt]

hm

|-32|

[wicked aurora]

What is y?

[torn jolt]

wait sorry

-12

wait so |x| must be greater than 20?

but what if

x is less than 20

[wicked aurora]

You claimed y>=12, but I believe I proved you wrong since y=-12 is not greater than 12

[torn jolt]

so

I'm sorry I don't understand

we know that in all instances 5|x+20| must be positive

[wicked aurora]

Yes!

[torn jolt]

it can be 0

so no not always

and 0

[wicked aurora]

Yes!

[torn jolt]

but the trick here is that

y cannot be -12 or less

oh I see

okay

y can be -12

yeah

just not any lower

in the instance

it is 0

due to the abs

so y>=-12

Yes

okay, I'm devoting like 10% brain capacity rn so I didn't catch on earlier lol

[wicked aurora]

Welcome to math

[torn jolt]

😂

thanks

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[red siren]

ax + by + c = 0
if its > 0 then its x,y above the line
if its < 0 then x,y below the line

[red siren]

ive gone through examples to kinda get an intuition for how this works but is it actually that obvious?

like is there a formal proof for these conclusions somewhere

[eager obsidian]

I think you can formally prove it by taking the (shortest) distance of the point from the line and checking its sign

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[storm dragon]

I'm studying this integral for different values of alpha. I've reached the conclusion that it diverges when alpha > 0 and converges to -1/alpha when alpha < 0. I'm unsure of what happens when alpha = 0 since it seems to run into division by 0.

[warm pulsar]

:/ no, just plug in 0 and see
u end up with this

$\int_0^\infty dx$

[versed drift]

you can find the value of the indefinite integral then check what happens in the limit of zero and infinity

[storm dragon]

Oh I see what I did wrong, I worked out the limit first and only then tried to plug in the 0

Thanks for the help

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[rustic shoal]

The top question

[austere cove]

!status

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

[rustic shoal]

1

[austere cove]

it looks like there's a typo in g(x)

and it's intended to be 3x^2 + 1

[rustic shoal]

?

[austere cove]

,rotate

[austere cove]

So you have g(x) = 3x2 + 1, I'm saying that g(x) should be written as 3x^2 + 1

probably

[rustic shoal]

I got this from math counts

[austere cove]

especially because it's asking for the sum of all of the roots

[rustic shoal]

So is it solvable?

[austere cove]

yes.

[rustic shoal]

Ok

[austere cove]

assuming that the part cut off says "what is the sum of the x-values for which h(x) = 0?"

[rustic shoal]

My bad here’s a better one

[austere cove]

,rotate

[austere cove]

ok yeah

So we need to find the roots of h(x), so first we need to find h(x) itself, what is h(x)?

[rustic shoal]

Im not sure

[austere cove]

well, h(x) is defined as f(x) + g(x), what is that value?

[rustic shoal]

2x + f(g)?

[austere cove]

huh?

[rustic shoal]

Idk

I added the x together

[austere cove]

I see. Well, that's not an allowable transformation. What you do is replace f(x) with "5x - 3" because f(x) = 5x - 3.

and similarly you replace g(x) with "3x^2 + 1" because g(x) = 3x^2 + 1

then you add these two expressions together.

[rustic shoal]

Ok

So -2x

[austere cove]

h(x) = -2x?

[rustic shoal]

Yea

[austere cove]

I could just tell you you're incorrect, but it would be more helpful to you if you showed me how you arrived at that conclusion so I can point out where you made your mistake

[rustic shoal]

Yes please

[austere cove]

well, I need you to show your work before I can do that

[rustic shoal]

I subtracted 5 from 3 since they are like terms

And that how I got -2

[austere cove]

There's a lot going on here that isn't correct.

So first, it isn't 5, it's 5x, so 5x isn't like terms with -3. Next, you're subtracting 5 from 3, when you'd be normally subtracting 3 from 5 if they were like terms (which they're not). Next, you haven't used g(x) at all. Finally, you reported -2x but you claimed to have calculated -2, so where did the x come from?

[rustic shoal]

I read it wrong I thought 3 was 3x

[austere cove]

let's take this step by step

h(x) = f(x) + g(x)

What I want you to do is take that equation, and replace f(x) with the definition of f(x), g(x) with the definition of g(x), and then report just that, and do not attempt to simplify

[rustic shoal]

What do you mean by defection of f(x)

[austere cove]

definition

[rustic shoal]

Detention

[austere cove]

you're doing this intentionally

[rustic shoal]

No I swear I’m not

[austere cove]

bullshit, find someone else to help you.

[rustic shoal]

It’s auto correct

My bad

Have a good day

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[torn jolt]

quick question regarding K-maps

[torn jolt]

for the following K-map

<karnaugh-map>
\begin{karnaugh-map}[4][2][1][$yz$][$x$]
    \minterms{0,1,2,5}
    \implicant{1}{5}
    \implicantedge{0}{0}{2}{2}
\end{karnaugh-map}

[torn jolt]

can we not make both the red and green implicants be one?

like do they all need to be like in a rectangular-like formation to be considered an implicant

[vast fossil]

Yeah, from what I know, you could have just one term in the simplification if those 1's formed a square

[torn jolt]

oh yeah im dumb

yeah its because its a grey code

you want one variable to change between each row and column

[vast fossil]

Yeah

[torn jolt]

fair

thanks bean

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[torn jolt]

.reopen

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✅

[torn jolt]

oh actually

i guess a funny question regarding grey zero

im looking to prove the algorithm i described to you using induction

like,
"if there are odd amount of 1's, flip the bit one position above of the last 1
if there are even amount of 1's, flip the last bit
"

you can like recover
0000
0001
0011
0010
0110
0111
1111
and so on

using this

but i wonder if it is proveable that it holds for any bit length n using induction

ok we can talk about it elsewhere

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[misty fractal]

Hi, dumb question but how would you combine equivalence relations?

[misty fractal]

suppose s = 2 (mod 3) and s = 2 (mod 5) and s = 2 (mod 7)

i actually know a way

i combine s = 2 (mod 3) and 2 (mod 5) to get s = 2 (mod 15)

and then s = 2 (mod 15) and s = 2 (mod 7) to get s = 2 (mod 105)

but is there a "trick' that you can do if all the remainders are the same?

maybe can you straight away say lcm(r_1, r_2, r_3,...) + remainder

[warm kite]

https://www.khanacademy.org/computing/computer-science/cryptography/modarithmetic/a/what-is-modular-arithmetic

[misty fractal]

what?

[warm kite]

give it a read

[misty fractal]

but why? I know it

[warm kite]

ik you know it

just read addition etc on modular

[twilit leaf]

Search up "chinese remainder theorem"

[misty fractal]

yes that's what i used

it's "sieving" under chinese remainder theorem

but is there a "trick" i can exploit

i think most of my questions are actually

with same remainders

i know how to generally do it

but i'm wondering if i can just say lcm(a_1, a_2,a_3) + r

[twilit leaf]

If theyre all the same number, say C, then you know its also C

[misty fractal]

ah so my thing is valid right?

my "hypothesis"

[twilit leaf]

Well yeah

[misty fractal]

okay thanks~

[twilit leaf]

Youre welcome

[misty fractal]

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[torn jolt]

In planar embedding, G divides the Euclidean plane exclusively into squares (with 4 corners) and triangles (with 3 corners) (including the surrounding area).  The number of squares is exactly 18.

 Using the Euleric Polyhedron Formula, determine the number of triangles.```



F - |E| + |V| = 2
S is the amount of squares (18)
T is the amount of triangles.


Handshaking Lemma:
sum of all degrees = 2|E|
4 regular so 4|V| = 2|E|
|E| = 2|V|


Formulas I made up:
• 4S + 3T = the sum of all degrees = 2|E| = 4|V|
Solve by |V|:
S + ¾T = |V|
• S + T = F

now put them in the euler:
F - |E| + |V| = 2
(S+T) - 2|V| + |V| = 2
(S+T) - |V| = 2
(S+T) - (S+¾T) = 2
¼T = 2
T=8
Whats my error? Why does T stay the same regardless of S? I'd assume it's the 4S+3T = 2E formula, but I can't think of a better one...

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[torn jolt]

<@&286206848099549185>

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[wise seal]

what do I do here

[wise seal]

The bullets are what it is given

Shall I replace f(3) with f(f(1))?

Will that allow me to remove the fs after?

[torn jolt]

look at the second point

[sharp flame]

replacing x with 2/3 is probably better

[misty lodge]

Just put x=1 and see what u get

[vast fossil]

Plug in x = 1 and x = 2/3

[misty lodge]

Im not even sure how f^{-1} is relevant here

[vast fossil]

You will get a system of equations with f(1) and f(3) as unknowns

Solve for f(3)

[wise seal]

I need it later for another question

[misty lodge]

Why would you plug 2/3. He just needs to find f(3),no?

[vast fossil]

It's given that f(2/3) = 1

[misty lodge]

Dont see how its useful

[vast fossil]

And everything else given is not sufficient

[vast fossil]

Can you solve it without using f(2/3) = 1?

[misty lodge]

Yes🤷‍♂️

[vast fossil]

Ah you are right

[misty lodge]

Plug 1 instead of x. Easy peasy

[vast fossil]

I had a mistake in the notes

[wise seal]

Oh yes I get it

[misty lodge]

@wise seal look its not that complicated. We are given that f(1)=3 and we need to find f(3). We have f(f(x)) so its clear that if we plug 1 instead of x we will get f(f(1)) which is f(3), from there you calculate f(3)

[wise seal]

Yes I got it thanks a lot!

[misty lodge]

Glad to help

[wise seal]

Now I have to solve f^-1(x)=3 which gives me f(x)+x-11=0

am I supposed to do anything else from here?

[vast fossil]

f^-1(x) = 3 is means x = f(3)

And you have solved for f(3)

[wise seal]

wait what

Ohhhhh oh

so x=2

[vast fossil]

Yeah

I just noticed that f^-1(1) = 2/3 implies f(1) = 3

Perhaps the latter (the statement f(1) = 3) was not supposed to be given

[wise seal]

Wait I didn’t get that

How is f(1)=3?

[vast fossil]

Let's say only the first and third conditions are given

[wise seal]

yes

[vast fossil]

\begin{align*}
f\left(\frac23\right) + f\left(f\left(\frac23\right)\right) &= 3 \cdot \frac23 + 2 \
1 + f(1) &= 2 + 2 \
f(1) &= 4 - 1 \
f(1) &= 3 \
\end{align*}

[glossy valve]

A Lonely Bean

[vast fossil]

The second condition is redundant

[vast fossil]

Hence my guess

[wise seal]

ohhhhhh

The denominator equals 3x based on what I’ve been given

So it’s

Cos0 is 1 and sin0 is zero so it’s 3/3x ?

So 1/x ?

[vast fossil]

Is f continuous?

Ah wait that's unnecessary

[vast fossil]

Yeah, the numerator goes to 3 and the denominator goes to 0

Meaning the limit does not exist

[wise seal]

if x was 0+ for example would it be infinity ?

because 3>0 and 0 is + ?

[vast fossil]

Yeah

[wise seal]

okay I see I see

thank you thank you

It is given that f is twice differentiable and that f(x^2+x+1)=x^3 and I have to show that the tangent of the f is parallel to the x axis at x=1

So I have to solve f(x)-f(1)=f’(1)(x-1) and find that f(x)=0 right?

if I solve what is given for x=0 I get that f(1)=0 is that right?

but how do I find the derivative of f

Oh do I differentiate the f(x^2+x+1)-x^3?

And then solve for x=0 so it gives me f’(1)?

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[wise seal]

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[broken vault]

need help on this equation

[silver lintel]

try to rewrite it in the form $f(x) dx = g(y) dy$

[glossy valve]

LF

[broken vault]

would that be (1/x)dx = (1/(y**3+2y))dy ?

[silver lintel]

y^3-2y

[void nova]

Yep sorry didn't see it

[broken vault]

I feel theres is something wrong and I can t find it

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[willow sedge]

oops mbv

[blazing dust]

Hi beard

[sick berry]

this is fractinal part

[blazing dust]

How to tackle it?

[sick berry]

the numerator

its discontinuos when its an int

so break the integral

0 to 2

then 2 to 4

[elfin stream]

[x] is integral part right?

[sick berry]

yes

[blazing dust]

Yes

[elfin stream]

well simply notice that x/2 - [x/2] can be either 0 or 1/2

depending on if x is odd or even

if x is odd

then

this becomes 1/2

if its even

its 0

[elfin stream]

therefore, this is unnecessary

[blazing dust]

I understand it

[blazing dust]

Pretty simple

So for all even numbers it will be 0

e^0 dx

=x right?

For odd e^1/2 dx
e^(1/2)/2

[elfin stream]

well u have to evaluate it too

[elfin stream]

the odd numbers are 1,3,5,7,...,1999

right?

so

[blazing dust]

yes

[elfin stream]

well how many of them are there in total

@blazing dust

[blazing dust]

999

[elfin stream]

right I think

so now

how many evens are there?

[blazing dust]

1001?

[elfin stream]

no

[blazing dust]

0,2,4....2000
,0 is considered or not?

[elfin stream]

0 is

wait then me is stupid

cus it is 1001 ig

[blazing dust]

2000= a+(n-1)d

[elfin stream]

let me get to the point

u can write the odds as 2n+1 correct

and the evens as 2n correct

so ur integral becomes

[sick berry]

do you have the answer?

[elfin stream]

wait even that doesnt matter

[elfin stream]

i am being stupid

[sick berry]

is the asnwer 2000(e-1)

?

[elfin stream]

I cant for some reason find how many odds and evens are there between 0 and 2000

[blazing dust]

I don't know the answer actually

[sick berry]

2000(e-1)

should be the answer

because

{x} is a periodic function right

[sick berry]

this is the question basically

[elfin stream]

wut..

[sick berry]

yes

{x} is a periodic functionm

[elfin stream]

answer should be x evaluated at 1000 - evaluated at 0 + sqrt(e)x evaluated at 1000 - evaluated at 0

so 1000 + 1000sqrt(e)

so 1000(sqrt(e)+1))

[sick berry]

no..

[elfin stream]

how did u get this

[sick berry]

period of {x} is 1 right?

[elfin stream]

e^7l means?

[elfin stream]

what r u even integrating w.r.t to

[sick berry]

x obvio

here period is 1

so I took 2000 outside

[elfin stream]

hmm

write

so

according to u

it should be

$2000 \int_{0}^{1} e^{\frac{x}{2} - \left[\frac{x}{2}\right]} dx$

right?

[sick berry]

2000*

its 2000 in the question

[elfin stream]

right sorry

[sick berry]

yeah

[glossy valve]

ItzKraken

[sick berry]

yes

[elfin stream]

how do u plan on evaluating this

[sick berry]

this part is x

[elfin stream]

hmm true

[sick berry]

done

[elfin stream]

but u dont get 1-e then

[sick berry]

0 to 1

not 1 to 0

[elfin stream]

[sick berry]

think again

[blazing dust]

Ye odd wala samjh na aa rha h

[elfin stream]

the integral of e^x from 0 to 1 is e^x evaluated at 1 - evaluated at 0

[sick berry]

arre nahi aisa mat karo

ye property use karo

{x} ka period 1 hota hai

[sick berry]

exactly

e - 1

[elfin stream]

u said

1-e

[blazing dust]

$\int_{0}^{1} e^{\frac{x}{2} - \left[\frac{x}{2}\right]} dx$

[glossy valve]

Harry

[blazing dust]

at 0 it will be 1

at 1
e^1/2

[sick berry]

ha

[sick berry]

toh ye basically question ye hai ab

[blazing dust]

only x kaise

[sick berry]

kyunki 0 to 1

hai na

ab

x = 0.5 ke liye

{x} = 0.5 hi hai na

[blazing dust]

no 0 hoga

[sick berry]

kya

GIF 0 hoga

[blazing dust]

e^(0.5/2-0)

[sick berry]

OH WAIT

x/2 hao

FFF

MY BAD

question galat

sorry

[blazing dust]

e^1/2-1 hoga na integration

then 2000(e^1/2-1)

[sick berry]

meri baat suno

samajhne ke liye

ye karo

x/2 = t manlo

dx = 2dt

[blazing dust]

good idea

[sick berry]

ha

ab solve karo

1000(e-1)

hoga answer

[blazing dust]

yar ni ho rha

e^1/2 aa rha h mera toh

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[fluid prawn]

This right?

And this

[sharp lagoon]

Expand the factored form.

Or try those xs in the original equation.

[tired sedge]

That's right

[sharp lagoon]

No, it isn't.

[tired sedge]

Lemme check

[twilit leaf]

Bro didnt even check 💀

Simplify t in the first one

[sharp lagoon]

(-2)^2 + 28 = 12(-2)
4 + 28 = -24

[twilit leaf]

The second one is not right yeah

[sharp lagoon]

14 isn't right either.

[tired sedge]

Damn what a strange roots, you should use D

This shit i mean

I forgot the name

[sharp lagoon]

14^2 is larger than 12(14) already.

[fluid prawn]

So only my first was correct?

[sharp lagoon]

You should simplify 11/55.

[tired sedge]

I guess you missed -

[fluid prawn]

Yeah I just realized

How about my second how was I suppose to solve it

[tired sedge]

Everything is correct,

[sharp lagoon]

You can't do the easy ways, you need the quadratic formula or something like that.

[dull seal]

Complete the square

[sharp lagoon]

The problem with your factoring is that -14 times 2 isn't 28.

[fluid prawn]

Yeah but it was correct factoring right

[sharp lagoon]

No, expand it to see why not.

[tired sedge]

x1×x2=c x1+x2=-b

[sharp lagoon]

The constant term will be -14 times 2, but you started with 28, not -28.

[tired sedge]

Hard to get it with vieta

So use formula

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[torn jolt]

I don't understand what an "image" is

and any idea I do have doesn't really correlate with specific points

Whayet a Minhute

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[urban sedge]

are there two series that satisfy this but their limits are not equal?

[undone vector]

If a_n converges to a and b_n converges to b, then a_n+b_n converges to a+b. Same idea for subtraction.

[urban sedge]

i did this and conculed that there are none that satisfy that but their limits are not equal

did i do anything wrong?

[undone vector]

I think if you want to be more rigorous you should assign a value to each limit, then show that the limits values are equal implying the limits are equal.

[urban sedge]

ok ty

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[lunar bramble]

A 13 ft ladder is leaning against a wall and sliding towards the floor. The top of the ladder is
sliding down the wall at a rate of 7 ft/sec. How fast is the base of the ladder sliding away from
the wall when the base of the ladder is 12 ft from the wall?

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[lunar bramble]

im doing this cus i cant find my other help channel it disappeaed after i clicked off it

y^2=13^2-x^2

solve for y

and then i differentiatied

dy/dx=(-x(dx/dt))/(sqrt(169-x^2))

i plugged everythgin in

solve for dx/dt

and got 35/12

the answer is 91/12

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[sleek quest]

how would i do this?

[sleek quest]

i drew it out

[hollow sable]

I assume it would just be angle chase

But there’s a kill with Japanese theorem (which is shown by angle chase so I think this I would be similar)

[sleek quest]

i dont know japanese theorem

[sleek quest]

i should probably use this theorem right?

[hollow sable]

Yeah

[sleek quest]

im not sure what to do with what i get

ill send a picture of mypaper

you there?

@hollow sable

<@&286206848099549185>

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[torn jolt]

whats some top tips of how to study everything for a mock exam or even for the final? I find it really hard to study for math exams that cram all the topics into 5 pages and im never really good at them. i live in england doing gcse level maths and im doing the exam board edexcel on the higher paper

[torn jolt]

please reply to the msg or else i wont see the reply in time

if there are any

[gritty rose]

Don't use help channels for studying tips

#discussion or 2

[torn jolt]

I DIDNT KNOW

I WENT THERE THEY SAID GO TO HELP

[keen agate]

Generally math can be hard to study for, cause of it's vastness in ability to have a question. I would heavily on formulas. Run through some examples. See what you can and cannot do. What you can do, don't focus so much on obvs. In your studying phase it is ok to look up the answer, or even get questions and answers and do the work to show why that answer is correct. Since this is a big test there might be like 'study, practice tests' online that or a list of the types of math on it if not known already

[torn jolt]

thank you so much but its just i just dont know where to start. Do you recommend that i only do a few questions on each topic and move on if im getting it right, or purely on what requires a formula. Here we don't really do a lot of formulas but more equations and formulas

[keen agate]

Try out a little bit of everything just to see what's already good and what needs some work, if you struggle with a question find others like and focus more on those. This doesn't mean don't focus on the ones your good on as there might be a trick way or two on making them complicated, but I would personally just like try a little of everything and focus on the struggle points and retouch on others if there is time

[torn jolt]

Thank you i really appreciate the help and I'll probably try this in the morning because its really late here. Byee and have a goodnight/day

[keen agate]

Goodbye, if you have no further questions use .close

[torn jolt]

thank you

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[sharp fable]

I fixed it.

[sharp fable]

I know the first part is 5 seconds.

But what about the second part?

"What does this tell you about the drone and the toy rocket"?

That they overlap?

[plush egret]

huh what a dumb question

[sharp fable]

Can't even ask questions. 🤦‍♂️

[plush egret]

maybe im an idiot but to me its not really clear how youre supposed to answer that

[plush egret]

no, not your question is dumb

the one theyve given you

[sharp fable]

ohhh

My bad.

[plush egret]

ur good

to me it doesnt seem like a meaningful question, sorry

maybe i should have said nothing

[sharp fable]

Sorry sorry.

[plush egret]

idk maybe you say like

theyre capable of reaching the same height? idk

or the rocket goes higher than the drone?

[sharp lagoon]

The first part isn't 5 seconds. At 5 seconds, R(t) is 0 and D(t) is 20, so you don't have R(t) = D(t).

[sharp fable]

Sorry about that.

I meant 4.5

[sharp lagoon]

There is an answer to the second part.

[sharp fable]

It's pretty ambigious though, no?

[sharp lagoon]

What do R and D represent?

[sharp fable]

The rocket and the drone.

[sharp lagoon]

No, look at what they say.

What about the rocket and drone?

[sharp fable]

the height.

[sharp lagoon]

Right, so at 4.5 seconds, what is true?

[sharp fable]

The height of the rocket and the drone are the same.

[sharp lagoon]

Right, so that's the answer.

[sharp fable]

ahh.

[sharp lagoon]

They're at the same height at 4.5 seconds.

[sharp fable]

Thanks.

[sharp lagoon]

No problem.

[sharp fable]

.close

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[harsh elk]

why does "A" index 4 sqrtroot A^5 = a^9/4?

[torn jolt]

this?

u can rewrite that as

A^1 * A^(5/4)

And combining the powers is 1+5/4

which is 9/4

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[velvet sedge]

!status

[full forum]

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

[velvet sedge]

!showwork

[full forum]

Show your work, and if possible, explain where you are stuck.

@late copper Has your question been resolved?

[full forum]

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[broken dragon]

I need help

[broken dragon]

I did something

But I don't think that's right answer

[shrewd hamlet]

u use lagrange?

[broken dragon]

Nope?

[shrewd hamlet]

whatd u do

[broken dragon]

Second

@shrewd hamlet

here

[shrewd hamlet]

before i check ur answer, lemme make sure ik how to do it myself lol

[broken dragon]

https://tenor.com/view/cat-gangam-style-gangnam-style-kitty-gif-6136640624539375945

[shrewd hamlet]

maybe i just wasted all this time trying it this method

lemme just do it the easy way lol

i cant rlly tell what ur answer is

what exactly did u get?

[broken dragon]

-2.2

a = -2.2

[shrewd hamlet]

i got something a little different

hmmm

[broken dragon]

Damn

Could you send your solution?

So i could compare

[shrewd hamlet]

nah we cant do that here

i can guide u thru it tho

so lets start again, i think u started off on the right track

@broken dragon

start with the first equation

square both sides

u already did that, and got it right

now look at that 2xy term

anything in equation 2 thats similar?

try figuring out what i mean

then lmk what u get

[full forum]

@broken dragon Has your question been resolved?

[broken dragon]

I tried

Nothing

Nah man imma dead

I have finals in 3 days

I don't know what to do

[floral frost]

yoo

whats the question??

[shrewd hamlet]

ok rewrite out the first equation squared

then write the second equation next to it

[broken dragon]

I did it

But I multiplied first one by negative 1

So they could cancel each other

[shrewd hamlet]

nah dont do that

keep it as i said

[broken dragon]

ok

[shrewd hamlet]

see the xy in equation 2

we wanna get that to be just like the 2xy in equation 1

so what should we do to equation 2

[broken dragon]

Btw

Should I use formula in equation 1?

Like

x2 + y2

formula

(x + y)2 - 2xy

[shrewd hamlet]

not rlly necessary

yea dont do that

[broken dragon]

okay

[shrewd hamlet]

$x^2 + y^2 + 2xy = a^2 - 2a + 1 \newline \newline xy = a^2 - 7a + 12$

[glossy valve]

Stephen

[broken dragon]

Yeah

[shrewd hamlet]

whats ur answer to this

[broken dragon]

AAAAAAAAAAAAAAA

[glossy valve]

Stephen

[shrewd hamlet]

hmm not sure thats quite right

[broken dragon]

$x^2 + y^2 + 3xy = 2a^2 - 9a + 13$

[glossy valve]

Metrocop

[broken dragon]

So we get dis

[shrewd hamlet]

no no no

[broken dragon]

Okay this gets us nowhere

[broken dragon]

?

why

[shrewd hamlet]

thats not it

ok ill tell u

multiply equation 2 by 2

what do u get

[broken dragon]

bros is in epic mode

[shrewd hamlet]

u like john mayer?

[broken dragon]

Nah i just saw it in your

activity

heheheh

[shrewd hamlet]

lol

[broken dragon]

why

Okay

Ill do it

So whats the point of it

[shrewd hamlet]

tell me what u get then ill tell u

[broken dragon]

$x^2 + y^2 + 2xy = a^2 - 2a + 1 \newline \newline xy = 2a^2 - 14a + 24$

[glossy valve]

Metrocop

[broken dragon]

We get this

[shrewd hamlet]

u forgot the 2xy

[broken dragon]

wait

no

$x^2 + y^2 + 2xy = a^2 - 2a + 1 \newline \newline 2xy = 2a^2 - 14a + 24$

ye

[glossy valve]

Metrocop

[broken dragon]

Sooooooooooooooooooooooo

[shrewd hamlet]

good

[broken dragon]

hmmmm

[shrewd hamlet]

see that 2xy

in equation 2

plug what its equal to INTO equation 1

into the 2xy in equation 1

[broken dragon]

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

Makes sense

$x^2 + y^2 + 2a^2 - 14a + 24$ = a^2 - 2a + 1$

[glossy valve]

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

[shrewd hamlet]

ok good

now get all the terms with an "a" on the right side

tell me what u get

ok im leaving, just find the vertex of the resulting quadratic

pce

[broken dragon]

okay thankssss

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[next basalt]

Utilize inc-excl principle to determine how many numbers are divisible by 3,5 or 7 between 1 <= x <= 138

[onyx glen]

!status

[full forum]

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

[next basalt]

I’m typing

[onyx glen]

k

[next basalt]

I assumed three amounts, A, B and C where A = total amount of numbers divisible by 3 between the interval 1 .. 138. Same with B but divisible with 5

This gives me A=46, B=27 and C=19

How do I calculate using the incl excl principle

[onyx glen]

do you know how the incl-excl principle works in general?

maybe you should not have been so hasty to replace sets with their cardinalities.

[next basalt]

|A| + |B| + |C| + |AnB| + |AnC| + |BnC| - |AnbnC|

This would be my guess by how incl exl works

[hallow walrus]

no, it should be |A|+|B|+|C|-|AnB|-|AnC|-|BnC|+|AnBnC|

[next basalt]

Like that?

[hallow walrus]

yes indeed

[next basalt]

I don’t get it. Why not A + B + C minus all AnC, AnB, BnC

And AnBnC

Why do we add AnBnC

But subtract the others

[hallow walrus]

in the first three terms, you are adding that region 3 times

in the 4th, 5th and 6th term, you are subtracting that region 3 times

[lilac parcel]

Why don't you set it up as the following:

Let A, B and C represent the sets of integers between 1 and 138 that are divisible by 3, 5 or 7. If you know this, and set it up accordingly to the InclExcl principle, what is your equation?

Edit:

If you calculated A, B, and C, then what is A ∩ B? And so on...

A ∩ B ∩ C has to be added back in, because that region is 'removed' when you subtract A ∩ B, A ∩ C, and so on...

[hallow walrus]

you need to count that region in the final count, so we add it back in

[next basalt]

How do I determine what AnB, AnC, BnC and AnBnC is?

Their cardinality respectively

[lilac parcel]

If A represents 46, so 138/3, and your B represents 26, so 138/5 then A ∩ B must be 138/3*5 no? That expression tells us: 138 can be divided by both 3 and 5 simultaneously.

[next basalt]

Makes sense, will try that, one moment

I got it to be 75

However the correct answer (according to the test) is 63

Not sure what he is doing tbh

[lilac parcel]

Jag kan visa dig i privata meddelanden om du vill (:

[next basalt]

Ja tack 👏

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[covert ermine]

Answer the following questions using the sequence in the next 3

11, 17, 23,

[covert ermine]

The bot on my homework isnt working right, i have the answer im just doulble checking

.close

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[torn jolt]

how do i simplify $\frac{34}{\frac{4}{10}\pi}$ without making it a decimal

[glossy valve]

Matt

[torn jolt]

my initial thought was just to flip the 4/10 but that changes the value

.close

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[minor needle]

you