help-26

[drifting swift]

no

fuck

draw me a picture of how a and b should look

[neon iron]

alr

[drifting swift]

what is this picture. i don't understand what is where.

where are the labels

i cannot read your mind

[topaz sinew]

Channel closed

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[neon iron]

.reopen

[topaz sinew]

✅

[neon iron]

(12, -5) is the line a

with a magnitude of 13

and the other i

is b, magnitude of 15 but we don't know the direction

[neon iron]

does that help

[drifting swift]

not in the slightest

ok look

draw me a coordinate plane and put the vector a on it with its tail at the origin

[neon iron]

so thats a

(ignore blue I go this pic off the web)

[drifting swift]

would have preferred if you marked the tip of a with an arrow as is typical

but whatever

now i want you to draw a line of length 15

starting at the tip of a

such that the new line's tip

ends up only 2 away from the origin

[drifting swift]

task failed ssuccessfully

now i want you to draw a line of length 15
starting at the tip of a

im out lol

[neon iron]

👍

[neon iron]

oh my I didn't read tip of a

[neon iron]

how would a line even

[drifting swift]

i said i'm out, don't ping me any more.

[neon iron]

alas, the patience games have been lost. thanks for so far

.close

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[celest bough]

Why cant i do the limit by the first way?

[sweet shard]

You have to take the limit of all terms

Not just ones you conveniently choose

[celest bough]

Ahh right thank you

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[covert jetty]

prove that the least value of x+2y+4z is 4root3 where x,y,z are positive real numbers satisfying the condidtion x^2y^3z=8

[covert jetty]

this problem i think could be easily done by weighted am-gm inequality but im tryna understand upon what i would apply the result

[stray wraith]

x²y³z=8
x²/4 × 8y³/27 × 4z = 64/27
x/2 × x/2 × 2y/3 × 2y/3 × 2y/3 × 4z = 64/27

(x/2 + x/2 + 2y/3 + 2y/3 + 2y/3 + 4z)/6 >= (x²/4 × 8y³/27 × 4z )^1/6 = 2/root3
(x+2y+4z)/6>=2/root3
(x+2y+4z) >= 4root3

Motivation behind first few steps is that you wanna break the required sum (x+2y+4z) into required product

I.e. in product we have x²y³z
So we want to break the 1 x we have in the sum into 2 so that after am gm we get a term containing x²
Similarly we wanna break 2y into 3 y's and so on

[covert jetty]

tqsm bro really helped

i hav one more question

if a,b,c,d be all positive real numbers , prove that
(a^2+b^2+c^2+d^2)^3 > (a^3+b^3+c^3+d^3)^2

[topaz sinew]

@covert jetty Has your question been resolved?

[covert jetty]

<@&286206848099549185>

[covert jetty]

no help yet :>

<@&286206848099549185>

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@covert jetty Has your question been resolved?

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[exotic arch]

how do I go on with about simplifying (2n-1)!!/(2n)!!

[exotic arch]

with the aim of not having factorials

[forest bloom]

is that double factorials?

[exotic arch]

yup

[neon iron]

simply just ignore the factorials bcs why is there punctuation in maths 😁👍

[exotic arch]

fr like how does 0!=1!

divide both sides by !

0 (angrily) = 1 (angrily)

[forest bloom]

[exotic arch]

alright well

using D'alambert's criterion

a_n+1/a_n

ugh how do I write it

[knotty ledge]

$\frac{(2n+1)!!}{(2n+2)!!} \cdot \frac{(2n)!!}{(2n-1)!!}$

[thorny flame]

ΣΑC

[exotic arch]

so Ik I can write (2n+2)!! as (2n+2)(2n+1)!!

and that cancels out at the numerator

but how do I go with about the denominator

[knotty ledge]

(2n+2)!! is equal to (2n+2) * (2n)!!

[exotic arch]

oh?

I don't think I have a solid understanding of this

[knotty ledge]

double factorials are the product of every other term lower

until you hit 1 or 2

i.e. 6!! = 6*4*2

7!! = 7*5*3*1

[exotic arch]

right

[exotic arch]

but how do I connect this to (2n+2)!!

is there like a formula to use

[knotty ledge]

think of an example

how would you connect 7!! to 5!!?

[exotic arch]

hmm well 7!!=7*5*3*1 and 5!!=5* 3* 1

so it's -1

fucking discord styling

[knotty ledge]

right but can you relate them in an equation

[exotic arch]

one number is off

[knotty ledge]

similarly to how you do for normal factorials

[exotic arch]

well 7!! tends to have (n)(n-2)...

yeah?

n!!=(n)(n-2)

yeah?

[knotty ledge]

etc etc yes

[exotic arch]

and so on and so forth

[knotty ledge]

so if you have (2n+2)!! = (2n+2)(2n)(2n-2) ...

you can recognise that as (2n+2)(2n)!!

[exotic arch]

Oh?

[exotic arch]

Yes!

[knotty ledge]

just like we could have said that $7!! = 7\cdot 5!!$

[thorny flame]

ΣΑC

[exotic arch]

ohhhhh

I see it!

so then for (2n-1)!!

(2n-1)!!=(2n-1)(2n-3)(2n-5)...

yeah?

[knotty ledge]

things should cancel nicely for you since you have ratios of double factorials of numbers that differ by two:)

ye

eh well

[exotic arch]

so (2n-1)!!=(2n-1)(2n-3)!!

[knotty ledge]

this isnt gonna help you

[exotic arch]

ik ik

[knotty ledge]

really you want to start with the (2n+1)!!

[exotic arch]

I'm just enjoying this information

(2n)!!=(2n)(2n-2)(2n-4)...

[knotty ledge]

since you have a (2n-1)!! floating around

[exotic arch]

(2n)!!=(2n)(2n-2)!!

[exotic arch]

so

I don't work with (2n+1)!! or (2n-1)!!

wait

actually

(2n+1)!!=(2n+1)(2n-1)!!

[knotty ledge]

those two are going to cancel nicely precisely because the numbers inside the double fact are two apart

i.e. 2n-1 = (2n+1) - 2

[exotic arch]

so for (2n+2)!!=(2n+2)(2n)!!

[knotty ledge]

yes yes

[exotic arch]

then I don't have to deal with (2n)!!

because it cancels out already

[knotty ledge]

exactly

same for (2n-1)!!

[exotic arch]

tysm! This was very helpful!

[knotty ledge]

ur welcome!

[exotic arch]

.close

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[tidal grove]

Hey

[tidal grove]

How to get this form

[drifting swift]

image incoming?

[tidal grove]

Yes

Here

From this

They are numerically equivalent

If u sub k1 as 0.5

And k2 as 4

U get 0.093

2dp so im correct

Bit of background i obtained this expression from subbing the following equation for time, into the concentration equation:

[tidal grove]

And I’ve arrived here

Which again is numerically equivalent

But mot algabraically

[tidal grove]

Yes

Even python cant do it

[sweet shard]

Where even is the question

[tidal grove]

Here

[sweet shard]

You've shown like 5 images and none of them have complete instructions

[tidal grove]

Using this

The instruction is to show the crmax/ca0 using t_max

[knotty ledge]

question is why is this true, nightmare

[tidal grove]

I know

Its numerically equivalent

Which is so weird

I subbed k1 as 0.5 and k2 as 4 and got the same for both forms

[knotty ledge]

just trying to clean things up, if you sub a = x/y you get:

$$\frac{a}{1-a}\left(a^{\frac{a}{1-a}} - a^{\frac{1}{1-a}}\right) = a^{\frac{1}{1-a}}$$

[tidal grove]

Wait whattt

Can you explain a lil

[thorny flame]

ΣΑC

[knotty ledge]

i just let a = x/y

[knotty ledge]

why this should be true i dont know immedately, lemme check if it even is

[tidal grove]

Let me know

Ima lil baffled tho but i gotta apply to my situtation

[knotty ledge]

yeah they're equal

this is a lot of effort to get it into a different form so chances are the method you did was "wrong", as in doing it whatever way you did it doesnt get you to their form in any easy way

[tidal grove]

Ah okay okay

So what did u do then

From the get go

What was ur starting point

[knotty ledge]

i just subbed your tmax into your cR(t)/CAO

[tidal grove]

Okay so u did not no my way

Well my expansion

Olay

Okay

And u said immediately

A = x/y

Is x/y = k1/(k2-k1)

[knotty ledge]

regardless of your problem, im trying to see why this is true, why those two things are equal

[tidal grove]

Ohhh

Is it not solved?

[knotty ledge]

well i was trying, with the a = x/y it becomes just one variable which is easier to keep track of

i was trying to see algebraically

[tidal grove]

Oh okay

[topaz sinew]

@tidal grove Has your question been resolved?

[autumn tiger]

hi

@tidal grove

what is your doubt

[tidal grove]

Hey

[tidal grove]

Its how to sub tmax into cb/cao which are here

[tidal grove]

To get this

[tidal grove]

I got to here which is numerically equivalent

[topaz sinew]

@tidal grove Has your question been resolved?

[topaz sinew]

@tidal grove Has your question been resolved?

[topaz sinew]

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[polar blaze]

is this statement true?
if 6+n = 0 mod 30, then n^5 = 0 mod 30

[polar blaze]

sorry reverse the statement

wait nvm

.close

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[empty fossil]

I’m trying to find all numbers divisible by 2,3,5,7 between 1 and 100 and I’m getting something wrong , = 50 + 33 + 20 + 14 - (16+6+ 10+ 4+7+2) + (3+2+1) = 78

[empty fossil]

the anwser is apparently 75

[trail scarab]

Find the number of multiples of 2,3,5,7 under 100

becareful of double counting

[gleaming thunder]

They're already implementing PIE
I think that advice isn't quite relevant

[wooden moon]

Multiples are 2,3,5,7 are non-primes between 1-100

Except 1

[empty fossil]

Yeah so there is 25 primes so the anwser should be 75 but I keep getting 78

[wooden moon]

So, your answer is correct

There are 78 multiples if we count 2,3,5,7+74 non-primes

[empty fossil]

oh yeah so my anwser is correct

[wooden moon]

1 is excluded

[trail scarab]

you're right. sorry

[empty fossil]

Right ok so the textbook is being silly thanks

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[ivory sorrel]

In the solution to this problem, they have assumed it;s of the form $(x-1)(x^4+ax^3+bx^2+c)$I;m not sure I understand why. Shouldn't it be $(x-1)P(x)$?

[thorny flame]

Why am. I here

[ivory sorrel]

source:- Jee mains 2022

[grim jacinth]

f has to be at most of degree 5

Because of the RHS

f: degree 5

f’: degree 4

[ivory sorrel]

ah, OK.

[grim jacinth]

Etc

[ivory sorrel]

thanks

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[grim sable]

sum them?

[vestal hatch]

,calc 27.23 + 6.9 + 4.268

[thorny flame]

Result:

38.398

[vestal hatch]

5 sig fig

[grim sable]

so 38.4

[vestal hatch]

yeah

[grim sable]

oh wait no

38

2 sig figs

[vestal hatch]

ah

fiuc

i already answered it

.clkose

.close

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[neon iron]

does this work only because a is a constant

or why does it work

[cursive thorn]

Yes, since cos(a) and sin(a) are constants with respect to x, you can pull them out

[proud sand]

it works because a is a constant

[neon iron]

thanks!

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[neon iron]

.reopen

[topaz sinew]

✅

[neon iron]

how does this happen

or rather why

[proud sand]

so you can divide nicely

[neon iron]

i mean sure, but how could one accomplish that

[proud sand]

think of it like (sin x+cos x) + (cos x - sin x)

and then use u sub

for denominator

[neon iron]

im not asking that

[proud sand]

u= cos + sin and du=cos-sin

[neon iron]

how did he turn the first one into that

[proud sand]

cos x = that long term

[neon iron]

yes

[proud sand]

multiplied by 1/2

[neon iron]

yes

how

[proud sand]

add like terms

[neon iron]

am i allowed to do that?

[proud sand]

cos x + sin x + cos x - sin x = 2 cos x

[neon iron]

in an integral

[proud sand]

Yes bruh

[neon iron]

okay then, thanks:)

.close

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[neon iron]

Find the domain in which $-3\pm\sqrt{3x+3}$ would be a function

[thorny flame]

mj

[marble gate]

!status

[topaz sinew]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I 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

[neon iron]

I had some thoughts going but I'm pretty sure the answer can be anything over -1

so what I thought firstly was

$x\geq-1$

[thorny flame]

mj

[neon iron]

because, sqrt can't be under 0

[marble gate]

that sounds fine to me

[neon iron]

so then when it came to plotting both instances

,w plot -3+sqrt(3x+3), -3-sqrt(3x+3)

[thorny flame]

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[neon iron]

this is why it isn't a function

because it fails the vertical line test

[marble gate]

oh, I glossed over the +/-

[neon iron]

so using this as the domain

but flipping it

where it is

$x\leq-1$

is close

[thorny flame]

mj

[neon iron]

but we're including -1

which would also make it fail the vertical line test

so wouldn't the domain to make this a function, have to be

$x<-1$

[thorny flame]

mj

[neon iron]

@marble gate let me know if you agree with this because I don't understand how the answer key got what it got

which makes no sense tbf

only thing that worked when I plotted with desmos

was the first one

$x\leq-3$

[thorny flame]

mj

[neon iron]

but how come it isn't what I have?

[trail scarab]

not unles 3x+3=0. otherwise f would be multivalued

[neon iron]

so my working is correct?

even in terms of inverse domain

the answer key seems to contradict itself

[trail scarab]

this is obviously wrong

[neon iron]

this is the inverse domain of the answer key for 7b

so having it be

x>= -1

would make it

a non-function

so reversing it in that sense

would make it

a function

x<-1

[trail scarab]

im unsure: what is f(x) ?

[neon iron]

sorry for not providing context

this is 7b, the question you are required to find the inverse for

upon doing this

you will get this as your inverse function

through this, you can determine domains and ranges for both the normal and inverse

The question afterwards asks which of the following is not a function, and what can you do to the domain to make it a function

the answer key says 7b is the non function

which I am ready to accept

but it claims that

making the domain either $x\leq-3$ or $x\geq-3$ would make it a function

[thorny flame]

mj

[neon iron]

^ this I don't understand

<@&286206848099549185> anyone care to chip in?

[sonic escarp]

are you talking about f(x) or its inverse? i mean the inverse is not defined for x<-1, how should x<-3 make this a function?

[neon iron]

it's inverse

we are trying to find what restrictions we can apply to the domain

that would make the inverse a function

[sonic escarp]

again: the inverse is not defined for x < -1. how should x< -3 help anything? can you provide the original question, and not only some parts of it?

[sonic escarp]

OK, xour question is obviously how to restrict the domain of f(x) (!) in such a way that the inverse is a function.

[neon iron]

yes

I get 7b being the question in which the inverse is not a function

[sonic escarp]

i asked you "are you talking about f(x) or its inverse?" you answered: "its inverse" that is something completly different.

[neon iron]

I am trying to answer question 8

in which it clearly asks about how to restrict the domain so that the Inverse would be a function

but I get what you might be trying to say

[sonic escarp]

every parabola is symmetric, you have to restrict it to the left (or right) of the symmetry axis to get the inverse a funtion.

[neon iron]

so question 8 is asking what changes can you apply to the original function

to make the inverse

a function

that would make a lot more sense, it is poorly worded then

[sonic escarp]

this.

[neon iron]

yeah fair enough

so restrict the domain making it not symmetrical about the y axis

,w plot 1/3(x)^2+2x+2

[thorny flame]

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[neon iron]

as we can see the vertex is at -3

ah now I see

restricting it either way

would make it

a function

thanks a bunch

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[vestal hatch]

Is this question as simple as it seems or is it a trick?

I'm assuming since it slows and doesn't change course of direction it would still be due east? So no change in "direction"

[whole geode]

well, it's asking for the net change in velocity.

the change in velocity is 15 m/s

[vestal hatch]

no its asking the direction

[whole geode]

but what direction is that 15m/s such that when it gets added to a east 20m/s it produces an east 5m/s?

[sonic escarp]

as the velocity reduces the direction of the reduction must be the other direction.

[vestal hatch]

wdym other direction

r u sayihng

west

?????

what

welp

I already entered east.

hm

yeah that makes sense

[sonic escarp]

think about it. how do you reduce velocity? something is working against you.

[vestal hatch]

ok

object is stationary

[vague ether]

someone help with surds

[vestal hatch]

The object maintains a constant velocity for approximately 6 seconds then the velocity drops to zero.

The total displacement of said object is zero.

yeah

i did just notice that

I see.

Yeah I didn't realize

that

the entire object

was above 0

Should have paid attention to the y axis

slope in velocity time graph is uh

acceleration

object is maintaining a constant acceleration

ugh

now they want me to find Vy

yes

Vy

see

no

yes

The formula is :

vertical velocity at time

idk how to find that

-160

17.1

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[slender stone]

This is what I have so far

DP= 1 I made a lil mistake

[tribal torrent]

It’s wrong

[slender stone]

Oh shit

What is

[tribal torrent]

Idk I got no idea

[slender stone]

whut

does anyone know this?

<@&286206848099549185>

[topaz sinew]

@slender stone Has your question been resolved?

[slender stone]

No

[charred hedge]

?

[slender stone]

What?

[charred hedge]

do you want help?

[slender stone]

Yes please

[charred hedge]

ok

[slender stone]

My question is above

[topaz sinew]

@slender stone Has your question been resolved?

[slender stone]

<@&286206848099549185>

[sage patio]

what

[slender stone]

Euhm my question is above

This is what I have so far idk if this is right and euhm yea I need help

I made a lil mistake DP=1 not 0

No one has been able to help so far

I will close this no one has been able to help me for the past days with this if you have any ideas feel free to send them to me!

.close

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[coral sun]

Can’t post in the forum sadly, I just asked a question there maybe that’s why, but is this true ?

[coral sun]

I suppose I'll get no answers 💀

But who knows

[worthy storm]

which part are you asking about?

the inverse images?

[stuck ether]

yes, you are correct @coral sun

[coral sun]

yeah, but i got told i was correct ^^ thank you all

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[thorny flame]

Iusgnol

[midnight sinew]

Thiis is the proof for surjectivity, I wasn't able to do it the usual way, would this work?

And is there a more direct way to do this, meaning starting from let y be an element of the codomain and then finding an explicit x in the domain st f(x)=y

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[green gulch]

Hi

[green gulch]

i need some guidance

what can i do now

cuz of the x idk how to get it to In-1

i n-1

and i dont want to look at the solution yet

just realised i forgot to make it x^2

GET OUT MY CHANNEL YOU STUPID

<@&286206848099549185>

[fair lintel]

what u need

[neon iron]

!occupied

[topaz sinew]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[green gulch]

read what i wrote

idk how to proceed

ban this rude guy

your question doesnt even make sense

solve $x^3+y^3+z^3=k$ solve what?!?!?!

[thorny flame]

tim

[green gulch]

what the hell are you on about

you cant even speak proper english you absolute buffoon

now get the hell out of my channel so i can get help

[fair lintel]

stop guys

[coral fable]

<@&268886789983436800>

[sour sandal]

<@&268886789983436800>

lol

[fair lintel]

make sure to read the hole conversation to understand who started lol

[green gulch]

start ur own channel then

bro that question doesnt even make sense

my guy

you are so braindead

[fair lintel]

man stop HAHAHHAA

[craggy haven]

both of you need to learn to behave

and be nice to each other

[fair lintel]

facts

[craggy haven]

.close

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[mellow venture]

Question: We want to compare the behavior of physics and chemistry students by recording their attendance at the first mathematics lecture. 88 of the 101 physics students are present, compared to 73 of the 115 chemistry students. Can we say, with a confidence level of 95 percent, that aspiring physicists are more motivated than chemists?
Solution:
Null Hypothesis: $p_1 = p_2$ (where $p_1$ population proportion of physics and $p_2$)
Alternative Hypothesis: $p_1 > p_2$
We will use Z socre because the population samples are both larger than 30.
$$Z = \frac{p_1 - p_2}{\sqrt{p*(1-p)*(\frac{1}{n_1} + \frac{1}{n_2})}}$$
We get a value of 1.71. The critical z-value for 95 percent with n> 30 is 1.645.
So we have $Z > 1.645$, so we reject the Null Hypothesis. We cannot say with a confidence level of 95 percent, that the aspiring physicists are more motivated than chemists 😦
Am I correct?

[mellow venture]

where $p= \frac{n_1 * p_1 + n_2 * p_2}{n_1 + n_2}$

[thorny flame]

Fractalogist

Fractalogist

[mellow venture]

<@&286206848099549185>

[reef fjord]

conclusion is wrong

if you reject null hypothesis, that means you have evidence that proportions aren't the same, yes?

[mellow venture]

yes

[reef fjord]

so what should your conclusion be? because what you have written seems to convey the opposite

[mellow venture]

omg i see thxx

I used the right formula for Z-score right?

[topaz sinew]

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@mellow venture Has your question been resolved?

[mellow venture]

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[junior stratus]

I have a basic question. Suppose I have a series of the form $\sum_{k\in\mathbb Z}f_k(x)$. Can I apply the Weierstrass M-test to this series although, in almost every text I've come across, it is only stated for the index set being $\mathbb N$?

[thorny flame]

Philip

[junior stratus]

Do we require the sequence of M's to run from -infinity to +infinity then?

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[fallen sun]

okay so i am wondering about the meaning of notation f²(x), does it mean f(x)² or f(f(x))

[craggy haven]

unfortunately this is somewhat ambiguous

[fallen sun]

ah okay

[craggy haven]

like, sin²(x) always means sin(x)²

but for other things it seems more common for it to mean f(f(x))

if it has any meaning at all

[fallen sun]

so i'll just have to guess ig...

thanks though

[craggy haven]

well... guess or rely on context. what are you seeing this in?

[fallen sun]

v is a polynomial and f is defined as f(v) = X * v' - v'' and its in a chapter about linear transformations and their matrices

where v' is the derivative of v

okay now that i think about it here it probably means the function composition meaning

so that is that

👍 thanks 👍

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[mellow venture]

Yesterday, January 19, 2012, the following wind speeds (in km/h) were measured at 12 noon:
Uccle = 24
Chevre =32
Florennes = 43
Kleine Brogel = 32
Melle = 32
Middelkerke = 14
St-Hubert = 22
Ukkel = 24
The wind speed in Florennes is immediately noticeable. What are the chances that you measure such a high wind speed, especially knowing that all places here are relatively close together?

[mellow venture]

So I used t-test, because samples and standard deviation is not given. And because the mean is also not given I used the following formula to find the standard deviation
$$\sigma = \sqrt{\frac{N-1}{N^2} \sum (x_i - \overline{x})^2}$$

[thorny flame]

Fractalogist

[mellow venture]

with mean x = 27, i found sigma to be 7.2

then i used the t-test score formula $$t = \frac{X - \overline{x}}{\sigma}$$ where X=43 temparure of Florennes

[thorny flame]

Fractalogist

[mellow venture]

i got t=2.22, so I went to look at the t-table for two sided and n=8 and i found it to be near and bit larger than 0.95%

so, there will be a high chance

is this the correct way to solve it?

coz i find it to be a weird question

<@&286206848099549185>

[topaz sinew]

@mellow venture Has your question been resolved?

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@mellow venture Has your question been resolved?

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@mellow venture Has your question been resolved?

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@mellow venture Has your question been resolved?

[strong oak]

This is a weird question to me too, I haven't done one exactly like this. I didn't verify your numbers, but your approach makes sense to me. If you didn't already, I think it makes sense to not include the 43 when calculating your mean, standard deviation, and sample size (pretend you have sample size 7 and only use the 43 at the very end when calculating your answer for t).

Regardless though, I think you'll get a very high t-score that's off the table. Your interpretation of the answer also seems off. I interpret question as what is the chance of getting a wind speed of 43 or greater? The "or greater" part means you'd subtract 1 from 0.995 and get that the chance is less than 0.005 or 0.5%.

[mellow venture]

Thanks!
So when something asks for "greater" I have to subtract it from 1? If "smaller" I let it so?

[strong oak]

Normally yes, although you should pay attention to how the table defines things. This one seems to calculate for "less than" so you need to subtract 1 from it for "greater than". I saw ones online that did the opposite so you wouldn't do that

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[thin jackal]

I just wanna clarify a few things before taking my test in a few hours since I've been out of school for 4 years

(6x^2+3y)(2x^3-y)
12x^5-3y^2

Can I combine different exponents of x like this

[neon iron]

no

[thin jackal]

I had a feeling I couldn't

[neon iron]

(6x^2+3y)(2x^3-y) = 6x^2 * 2x^3 + 6x^2 * (-y) + 3y* 2x^3 +3y* (-y)

[thin jackal]

Alright all I needed to know forgot to do the Y part I was just curious if the x squared and cubed combined or not thx

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[molten crystal]

i have this

[topaz sinew]

Please don't occupy multiple help channels.

[molten crystal]

i want to find B

so i am doing
-sqrt(b^2-4) = sqrt(b^2-4)

it's more of an algebra question really

if i want to solve for b here, i could square both sides but that also makes both sides positive. and i can't to much with it. but i feel i'm missing something.

if i move everything on one side then yes i can find b.

but why i'm stuck if i square both sides?

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@molten crystal Has your question been resolved?

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@molten crystal Has your question been resolved?

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@molten crystal Has your question been resolved?

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@molten crystal Has your question been resolved?

[whole geode]

@molten crystal it might help you to realize that this means that sqrt(b^2 - 4) = 0 (because -a = a implies a = 0)

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[shy tiger]

Hey! I'm having some trouble to start solving this limit. Can you give me a hint please?

[fading condor]

and where the hell is the limit

[shy tiger]

Oh yeah sorry, it's the limit when n -> infinite

I figured I could factorize the inside of the ln with exp(n)

well that works sorry

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[kindred patrol]

Problem 30

[kindred patrol]

Where do I go from here?

[marble gate]

well 3^1 = ????

or rather, what is 2*3^1

[kindred patrol]

yea

whag do I do next

[marble gate]

yes, can you simplify 3^1

[kindred patrol]

ok

then

[marble gate]

and what do you get

[kindred patrol]

then it goes like that

[marble gate]

...no, 3^1 is not 1

[kindred patrol]

so

I thought that’s what u wanted me to do

[marble gate]

can you write 3^1 as just a regular number without an exponent

[kindred patrol]

yeah

2 * 3 * 3^x

[marble gate]

yep, and 2*3 = ...?

[kindred patrol]

6

[marble gate]

and you're done

[kindred patrol]

that doesn’t make sense

that goes back to the original equation

I thought we were trying to get the same base

[marble gate]

you needed to show that 2*3^(x+1) = 6*3^x

you started with 2*3^(x+1) and then made it into 6*3^x, which is what the question asked

[kindred patrol]

ohhhh

I get the question now

we were trying to solve that it equals

[marble gate]

prove that blahblahblah = something means "take blahblahblah, change it around without changing the value by simplifying things, and end up with something"

[kindred patrol]

yea ok I was confused on what it was asking but I get it now thx

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[white grotto]

isn't this wrong ??

[white grotto]

i think that the correction is wrong

[vale oriole]

Yeah

For example 5/2

[white grotto]

then why it's written there

[white grotto]

yeah yeah i'm tottaly okay with you it's wrong but they used it to prove

[vale oriole]

There is special thing

If {x}<1/2 then its true

And i see it in b

@white grotto

[white grotto]

but they didn't mention anything about this

[vale oriole]

It equivivalent to x<[x]+1/2

[white grotto]

yeah yeah cuz x - [x] = {x}

yeah now i get the idea

[white grotto]

thanks for help !

[vale oriole]

@white grotto What you prove?

What goal?

[white grotto]

this one

[vale oriole]

But why?

[white grotto]

it's a question

[vale oriole]

Okay competition?

[white grotto]

which competition ??

[vale oriole]

Is it from competition?

[white grotto]

from an exam

[vale oriole]

Okay thank

[white grotto]

btw here's the full correction

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[cinder owl]

can someone go over the definition of recursion theorem with me, it really confuses me

[knotty ledge]

Would help for you to post the definition

[odd pagoda]

I mean this really just says in the most confusing way that recursively defining a function like f(0)=17 and f(n+1)=n^2+7f(n) or whatever, works

so a=17 and h(x,y)=x^2+7y

[cinder owl]

yeah this is weird

this just sounds like how induction is defined

[odd pagoda]

well it basically is

well, induction you also had to prove that it works

[cinder owl]

i mean mostly the h(n, f(n) confuses me

[odd pagoda]

what about it

thats just normal function notation

"some expression involving n and f(n)"

[cinder owl]

why are you mapping the natural numbers to an element belonging to X

and then outputting X

thats what im wondering

[odd pagoda]

I am mapping a pair (n, x) to an element in X

where n is a natural number and x is in X

whatever X is

so for example X could be the set of matrices

and then h(n,x)=x^n

or X could just be modulo 10 and h(n,x)=x+n mod 10

or whatever

or n doesnt even actually have to "appear"

[cinder owl]

is there a better definition for this theorem

slightly less weird

like i think i get what youre saying but

[odd pagoda]

this isnt a weird definition

[cinder owl]

thanks!

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[slender sonnet]

There is a trapezoid the trapezoids legs continue until they make a triangle the bases are 5x and 3x the area of the whole triangle is 50 need the area of the trapezoid only

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@slender sonnet Has your question been resolved?

[slender sonnet]

<@&286206848099549185>

<@&286206848099549185>

[dense lily]

stop

[split cloak]

You can say that they are congruent

I guess...

[slender sonnet]

so area of the trapezoid would be 4x *h

@split cloak do you know how to do it?

[split cloak]

actually no idea...

i was studying this kinda math 1 year ago

Geometry*

[slender sonnet]

Ok

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[pseudo karma]

Is this right?

Havent finished yet

[storm narwhal]

looks good to me

[pseudo karma]

Alr

[storm narwhal]

its just a simple notable products expression

just develop your equation it will turn out fine.

[pseudo karma]

Multiplied with -6

[storm narwhal]

yeah

[pseudo karma]

Its -12x

Not a😭

[pseudo karma]

Its correct

?

[storm narwhal]

yeah looks about right to me

except you forgot a on that 12

except that

all good

[pseudo karma]

Do i expand this?

[storm narwhal]

isnt a different problem ?also seems like u messed up

[storm narwhal]

u NEED to expand it

[pseudo karma]

Its a diff.problem yh

[storm narwhal]

its notable products

[pseudo karma]

I asked bc Its after the =

[pseudo karma]

Imka do it again

Just in case

[storm narwhal]

just use the table

[pseudo karma]

Yhh

[storm narwhal]

youll be fine

[pseudo karma]

Thx buddy

[pseudo karma]

Do i multiply with a³-9

Or only -9

.close

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[thorny flame]

Mycobacterium

[neon iron]

are A and B same by RxC measures?

AB = I implies that either A = B^-1 or B = A^-1

in that case AB != BA will not be true

[gritty trout]

it turns out that, for square matrices, if there is a matrix B such that AB = I, then there is a matrix C such that CA = I and, furthermore, B = C. but this is kind of unexpected or cool

yeah. if A and B are not square then it can fail

maybe i should think of an example

since you asked for that

oh yeah true. but in that case you'd phrase it as there are matrices A and B such that AB = I but BA \neq I

(also note that the I's are different dimensions in the non-square case)

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[gritty trout]

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[misty anvil]

So in the problem above I am supposed to convert it to cylindrical coordinates and I have the answer and for some reason the interval for theta in the answer is on [0,2π] and I don't understand why because y is positive in this integral and y would only be positive on (0,π)

[misty anvil]

This is the answer btw

[topaz sinew]

@misty anvil Has your question been resolved?

[misty anvil]

oh wait someones here

[strong oak]

I think there's a typo in the question, the bounds for y don't make sense with the bounds for x (plugging in y=3 gives you a negative square root). If you give wolframalpha the integral, it says it doesn't converge

[misty anvil]

Ok thank you very much

.close

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[versed widget]

yo

[versed widget]

i dont understand the fourth step

how does she get that

[versed widget]

i have the step

but i dont have the step after that

i can differentiate it

[vernal matrix]

Multiply both numerator and denominator of circled by 1 - v^2/c^2

[versed widget]

but the simple algebra is kinda failing

[vernal matrix]

That makes the denominator then become to the power 3/2 by rules of indices

[versed widget]

but ought you not to multiply by the denominatgor

like this is what i did

(1-v^2/c^2)/3/2 on one and (1-v^2/c^2)^1/2 on the othr

[vernal matrix]

Do you know how to common denominator something like 1/2 + 1/4?

Similar idea here

[versed widget]

this is what i've done but for those numbers

[vernal matrix]

I mean… that’s correct I guess

But you only need to “make up” the 1/2 to 2/4 to be able to add them, because the denominators have a common factor

So 1/2 + 1/4 = 2/4 + 1/4 = 3/4

[versed widget]

any number has a common factor

their common factor will be a*b

given any real number

48*871 has the common factgor of 41808

[versed widget]

what

[vernal matrix]

Mistyped it originally-

[versed widget]

why only c

oh

[vernal matrix]

But anyways the idea is similarly as here you only need to multiply 1/2 by 2/2 to get 2/4

[versed widget]

yes x^1/2*x^1=x3/2

but x^3/2*x^1 = 5/2?

[vernal matrix]

You can multiply $\frac1{(1 - v^2/c^2)^{1/2}}$ by $\frac{1 - v^2/c^2}{1 - v^2/c^2}$ to make it up to $\frac{1 - v^2/c^2}{(1 - v^2/c^2)^{3/2}}$

[thorny flame]

@vernal matrix

[versed widget]

then u have x^1/x/5/2 = x^-3/2

[vernal matrix]

While true, you shouldn't have this (well unless you want to do it the long way, which I guess if you really wanted to... just means you have to work more later)

[vernal matrix]

You only need this to "make up" the denominator of that first term to have the same denominator as the second

[versed widget]

this will take ages to solve

now i have ((1-v^2)+v^2/c^2)/(1-v^2)^3/2

[vernal matrix]

what did you do with the /c^2?

[versed widget]

this doesnt rly do anything

that the c^2 is still here

actuallyt i can just remove the c^2

this is the longest way

and if i had to all of this by hand

it'll be a page

[topaz sinew]

@versed widget Has your question been resolved?

[topaz sinew]

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[pseudo karma]

|7-x|=x+3

[pseudo karma]

Is it -7+3?

Can someone help me on this one

|7-x|=x+3

[keen matrix]

you're trying to solve for x here is that your solution for x?

[pseudo karma]

Im trying to solve for X i guess

[keen matrix]

seems like it

with absolute value, do you know what it does?

the | | things

[pseudo karma]

Absolute

[keen matrix]

do you know what it does?

[pseudo karma]

Its

|7-x|=x+3
7-x=x+3 or 7-x=-(x+3)

Its ik

Ok

I found it

[keen matrix]

yep that looks right :) now just do you solving as per usual

[torpid sparrow]

But be careful

[pseudo karma]

Oh no

[torpid sparrow]

This one got only one solution

[pseudo karma]

wow

U made it so dramatic

[keen matrix]

make sure you always check

[marble ember]

Yeah i saw this too. The second equation isn't a function, is it?

It isn't a valid linear equation that i can tell.

[neon iron]

becomes 0x=10

[pseudo karma]

.close

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[real finch]

I already got as answer B, however, I'm not sure if it's the correct answer since it kind of came by a deduction method, and not the "formal" way to do so

[proven grove]

prolly need a question translation

[marble ember]

Sorry can u please translate the question?

[real finch]

It technically says: calculate the equation of the straight line L

being OA=AB

[marble ember]

Ah, here are the steps,

Step 1) figure out the equation of the line OA.

Step 2) figure out the distance of OA. That'll also be the distance of AB

Step 3) shift the equation of OA to the distance of AB. There's a formula for that

Lemme, elaborate now.

The equation of OA is y=6x

[real finch]

what

Isnt it 6x-y-35?

[marble ember]

Nope. U won't get that 35 thingy there.

Here's how u understand it intuitively.

The line OA went through the point 0,0

So, the format of the equation will look like y=mx, where m is the slope

Here, the slope is m=6

Here's another way to get OA.

(Y-0) = m (X - 0), here m= (6-0)/(1-0)

[real finch]

yes I know that, I just didn't take 0

I forgot

[marble ember]

yeah. That's step one completed. Now find out the distance OA

[real finch]

yeah, so I found that c must be a multiple of 6

[formal spade]

c sol. b is wrong

[real finch]

C being Ax+By+C=0

[real finch]

why though

root of 37?

[marble ember]

Here, distance OA= √(1²+6²) = √37

And AB will also be = √37

[marble ember]

Well let's figure out!

Now, the lines OA and BL here are parallel. Can you understand this part?

[real finch]

Ok I get all of it, I just messed up on the equation of AO

root of 37 is equal to c/root of 37

so c is equal to 37

in Ax+By+C being the equation of L1

And A and B being 6 and -1

due to the slope being 6

[marble ember]

Yep. Good job

[real finch]

damn I messed up on something simple as AO

[marble ember]

Answer is C, indeed

[real finch]

lmao

thanks though

[marble ember]

It happens dw

[real finch]

.close

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[trail lake]

I want to know which books have these types of topics because in my high school we didn't hear about it
Convex set,concave set
Concavity by differentiation

[pure estuary]

I think any analysis text should cover this

[topaz sinew]

@trail lake Has your question been resolved?

[trail lake]

have you read about it in any analysis book yet?

[pure estuary]

yeah in Rudin's principles of mathematical analysis text

although I don't remember much of it

[trail lake]

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[forest bloom]

hi, this might be a super dumb question, but i wanted to clarify something

[forest bloom]

diagonals divide the triangle into equal portions?

Well i know for triangles, you have the median that does that but hmm this is curious

iirc don't the diagonals have to be like orthogonal to each other?

[shut belfry]

7

[forest bloom]

yes i know

i literally wrote it down lol

[shut belfry]

so what's the doubt?

[forest bloom]

.

for the rectangle don't the diagonals have to be orthogonal for them to split into equal portions?

[shut belfry]

no

diagonals are not orthogonal

[forest bloom]

okay then

a diagonal should split the rectangles into two pairs of congruent triangles