help-0

[tacit arch]

Read the question again. 3060 degrees represents what?

[alpine sable]

The sum of the interior angles

Okay I got it

Thank u

.close

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[pliant cedar]

,rotate

[pseudo ice]

Remember how both cot and cosec are defined

The identity is generally true, provided you don’t do something illegal…

[lone heart]

@smoky condor Has your question been resolved?

[smoky condor]

What's illegal

[crystal thistle]

cot zero

[smoky condor]

♾ I think

[crystal thistle]

also $\theta\neq (2n+1)\frac \pi2$ is unnecessary

[ocean seal]

moriaritie

[smoky condor]

Because this condition will make it ♾

[crystal thistle]

yeah so that's undefined

[smoky condor]

.close

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[smoky condor]

.reopen

[lone heart]

✅

[smoky condor]

@crystal thistle

[crystal thistle]

yes?

[smoky condor]

But they are asking which one is incorrect

And we got that theta not equal to (2n+1)3.14/2

[crystal thistle]

c is incorrect

[crystal thistle]

well

[smoky condor]

.close

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[lament mantle]

How would you solve question a?

[restive flume]

let B = number of blue buttons, and R = numbers of red buttons.
think: if you half R, then add 12, you end up with B. can you write an expression for this?

[lament mantle]

62= 12+0.5R?

[restive flume]

yes, this can be used to find R. sorry, having B was unnecessary

now let Y = number of yellow. knowing a quarter of R equals Y, what expression can you make?

[lament mantle]

IM SORRY IM CONFUSED😭

[restive flume]

it's okay!

[restive flume]

based off this, we can say 0.25R = Y

do you understand how?

[lament mantle]

OHHH

Wait

YEA

OKAY

[restive flume]

okay good!

not sure if the question wants a single equation? based off the wording it does, which is strange

[lament mantle]

That's why i was so confused😭😭

[restive flume]

we'll try smth

given we have
62= 12+0.5R
we have
50 = 0.5R

and given we have
0.25R = Y
we can say
0.5R = 2Y
we did this by multiplying both sides by 2

so now you have
50 = 0.5R and 0.5R = 2Y
so we have
50 = 0.5R = 2Y

that's my best guess for having a singular equation. normally, it's much more satisfactory having a system of multiple eqs.

[lament mantle]

That's what I was thinking😞

I'm pretty sure I get it

Also for question B does red equal 25🥹

[restive flume]

walk through me how to solve
62= 12+0.5R

for R

what is your first step?

[lament mantle]

62-12

50÷0.5

OHHH

[restive flume]

yes, what is 50÷0.5?
note it's not 25. that would be 50 * 0.5

[lament mantle]

I see where I messed up😭😭

Ok so it's 100 right

[restive flume]

yes!

[lament mantle]

Okay I think I get everything now🙆🙆

THANK U

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[neon garden]

My teacher showed me the answer without explaning.I tried my best to calculate it and i still cant get the correct answer.
I not sure whether should i calculate accerelation or have other ways to calculate it

[covert topaz]

wait so speed is on the x axis?

[neon garden]

Speed is on y axis i think

x axis is for time

[covert topaz]

oh i see now yeah that makes sense so we want to know the speed of runner after 6 seconds?

[neon garden]

Yes

[covert topaz]

we know the integral of whatever our velocity equation from 0 to 11 is 100 right because its 100m runner

so you know its constant from 6 to 11 which means the area under the line from 0 to 6 is equal to 100-5s

i think its meant to be plug and chug

so if you plugged in all values when you're solving for the area under the curve you just find the one where the area is 100

[neon garden]

ohh alright thanks a lots!ill try it out now

[covert topaz]

gotchu lemme know if you need any clarifications or questions, just @ me

[neon garden]

Okay thanks

@covert topaz i plugged in all the values but i dont find any of the answer is correct though

Im supposed to find this area right?

[covert topaz]

so you know that (26)+1/2(6(s-2))+5s=180

hmm

[neon garden]

where did the 26 came from?

[covert topaz]

sorry 12 not 26

the rectangle under the triangle

[neon garden]

Ohh yes

[neon garden]

How you get the 180 though?

[covert topaz]

sorry trippin again 100

[neon garden]

No problem,ima calculate rq

[covert topaz]

we can simplify this out:

12 + 1/2(6(s-2)) + 5s = 100
1/2(6(s-2)) + 5s = 88
1/2(6s-12) + 5s = 88
3s - 6 + 5s = 88
8s - 6 = 88
8s = 94
s = 11.75

[neon garden]

Ohh i got it thanks a lot!

[covert topaz]

np anytime!

[neon garden]

.close

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[viscid lake]

Help

This is my trying

But my teacher said this is wrong

[lone heart]

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[vale wigeon]

!msgdel

[lone heart]

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[viscid lake]

This is my try

[covert topaz]

could you just prove that f(x) = 1x + c?

cause thats the only way that summation is true right?

[viscid lake]

Yes

But my teacher said this way of proof is wrong

[covert topaz]

wait no because the sum is to n

[viscid lake]

He said use the rolle's theorem

[vale wigeon]

!msgddel

!msgdel

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[vale wigeon]

the channel will slam shut on you when you least expect it!

go get a new one.

[viscid lake]

Why

[vale wigeon]

Channel closed due to the original message being deleted.

[viscid lake]

Okey

[lone heart]

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[smoky condor]

The 11th

[storm ridge]

Are you aware of the sum and product of roots in terms of coefficients?

[smoky condor]

?

[storm ridge]

p/q isn't p tho

Coz p/q is sin+cos

[smoky condor]

✅

[storm ridge]

But p/q is sin+cos
And 1/q is sin*cos

Can you think of any identity that relates sinx+cosx and sinxcosx?

Isn't it p/q

Ok how about

[smoky condor]

Right

[storm ridge]

You square p/q

,rotate

[storm ridge]

1/(cos*sin) = sec cos

And sec cos is q

So it's q(sin+cos)=p

sin+cos = p/q

[storm ridge]

Right, so try this now

[smoky condor]

Thanks

[storm ridge]

Np

[smoky condor]

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[flint acorn]

cant find a single way to actually solve this shit

[vale wigeon]

what is the goal?

actually, isn't there missing data here?

[flint acorn]

circumference is what i need to solve for

[vale wigeon]

perimeter, you mean?

but ok

progress?

[flint acorn]

dont know the correct translation, im swedish mb

just that the opposite sides are 6 and 8

respectively

[edgy merlin]

So all angles are 90 degrees?

[flint acorn]

yes

[edgy merlin]

Ok

Imagine you move the most top line (the short one) down to get a rectangle

[flint acorn]

yeah

[edgy merlin]

you get a rectangle + 2 lines

you can do that because all angles are 90 degrees

[flint acorn]

yeah true

[edgy merlin]

Now the perimeter of the rectangle is?

[flint acorn]

i dont know, we dont know the length of that line

[edgy merlin]

but because you moved that line you get a line which is 8 units long

because its a rectangle

[flint acorn]

oh yeah

[edgy merlin]

so whats the perimeter of the rectangle?

[flint acorn]

oh no wait i mean 28

[edgy merlin]

correct

now you just add 2 lines of length 2

so your answer is 32

[flint acorn]

ahaaa

tyy

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[granite coyote]

Hello! I wanted to ask, if I have a new procedure to find a math formula, where can I publish it?

[tacit arch]

https://vixra.org/

[granite coyote]

Thank you!

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[cinder breach]

Hello, can somebody please tell me if this is right? Thanks

[lone heart]

@cinder breach Has your question been resolved?

[lone heart]

@cinder breach Has your question been resolved?

[crystal thistle]

it seems good

[cinder breach]

thank you

idk how to check by myself if the proofs are right

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[weak ridge]

A particle P of mass 0.5 kg rests on a rough plane inclined at angle α to the horizontal, where
sin α = 0.28. A force of magnitude 0.6 N, acting upwards on P at angle α from a line of greatest slope
of the plane, is just sufficient to prevent P sliding down the plane (see diagram). Find
(i) the normal component of the contact force on P,

[weak ridge]

Please help

[nimble fern]

did you try to draw the forces acting on P?

and then break the forces along and perpendicular to the inclined plane?

[spice crane]

need help

[lone heart]

@weak ridge Has your question been resolved?

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[nova barn]

hello..

i meant i've tried to find the inverse but it went horrible
other is i know that inverse of a function implies that the graph of the inverse is mirror image of original function in y=x ..

and i don't know what to do..

[lofty creek]

what happens when you have f(y)=y^3+3y+1=x and solve for y?

[nova barn]

there is on one root and inflection at x=1..

oh so i can do it like that

oh ok i got it..thanks a lot..

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[shrewd zenith]

Can someone help me with inequality

[lofty creek]

!da2a

[quasi vector]

[lone heart]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[shrewd zenith]

I have x1 times x2>0
and i have a=k-4, c = k-4
so that means x1 times x2=1
and what does thta inequality mean then

the original thing i need to do is find k if both roots are the same signs

so i got the D>=0

But i don't know what to do with x1x2 one

[lone heart]

@shrewd zenith Has your question been resolved?

[shrewd zenith]

<@&286206848099549185>

[stark cosmos]

i think no one wants to answer because it is hard to see

maybe you should make it easier to read

[shrewd zenith]

well my only question is that i have to do inequality x1 times x2 > 0 and x1 *x2 = 0 and that means i get 1 > 0 so what does thta even mean

should i go k-4/k-4 > 0 ?

and then i get good range

and i find intersection

.cclose

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[primal berry]

How do you even solve this I just need the answer

[wispy mulch]

Okay

But idk if thats in the rules

To just show the answer

I think we cant do tests for you

[hidden crest]

4/8

1/2

50%?

[primal berry]

This is not a test

[wispy mulch]

Ye

[hidden crest]

3/12
1/4
25%

[wispy mulch]

3/12 = 1/4

[primal berry]

It's the book and I have 3 semesters of high school left and I am in credit recovery trying to catch up on old math work before I just end my life honestly

Thanks for the help everyone

[finite flax]

!noans

[lone heart]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[primal berry]

Just need to survive

[hidden crest]

Well she said i just need the answer
Idk

[finite flax]

Is this like a joke

[wispy mulch]

Um

[primal berry]

Why would it be a joke at all

[finite flax]

Because this is a workbook for students at a level younger than those who are authorized to be on this server

[primal berry]

I know I'm pathetic or whatever but I y'know

Am very shitty at math

I am 16

Have not done grade 10 or 11 math yet

In credit recovery currently

Trying to do my math.

[wispy mulch]

But you’re not trying to understand it

You just need the answer

[finite flax]

But you said you just want the answers

[primal berry]

That's a fair point but I have so little time left so I'm rushing

[wispy mulch]

Why didn’t you do it faster?

[primal berry]

I'm open to learning but idk how well it serves me atp

[primal berry]

Catching up on other things beforehand

Was super behind in terms of courses

I kind of suck at school if it's not the arts

[wispy mulch]

But you will still be behind if you’re not studying

[primal berry]

Yea

I only have g10 and g11 mth left so I just wanna get it out of the way

Then never worry about this again

[wispy mulch]

🤷🏻‍♂️ ups to you

[finite flax]

You're more foolish than you know if you think you will never need to know what half of something is again

[hidden crest]

What's the point when you don't even learn something ?

[stark cosmos]

wtf

[wispy mulch]

Thats true, i would even say thats common sense

[primal berry]

Rather learn it later

[stark cosmos]

this is obviously a kid

[hidden crest]

This looks elementary to me not high school!

[wispy mulch]

You’re 16?

[stark cosmos]

study hard

[primal berry]

I just literally Do not know any shit

About this

[hidden crest]

Omg

[wispy mulch]

And you dont know what half is

Tf you been doing

[finite flax]

@primal berry aside from being lazy and snarky about it, I think you dont understand what you are being asked to do

Did you shade the figure like they told you to yet?

[primal berry]

I know what half is I literally just don't know about all this other specific stuff

Or don't understand it well enough

[wispy mulch]

Um

[hidden crest]

Hmm

[finite flax]

Did you follow the directions yet

[primal berry]

If someone actually took the time to go ahead and it explain it I'd understand but not only is this not my book (it's like whoever's so I can't necessarily share it in) I'm trying not to have to deal with anything like this next semester so

And what after break I have one more month of school before semester two

[finite flax]

Shade the picture.

Like the instructions told you to

[wispy mulch]

^

[primal berry]

Alright

I appreciate the help

[finite flax]

Are you done yet?

[primal berry]

I just explained that it's not my property and multiple after me are gonna use this

[finite flax]

Then copy the picture

[primal berry]

Kk

[primal berry]

Done

[finite flax]

How many pieces are shaded

[primal berry]

Half, 4

[finite flax]

And how many pieces are in the entire pizza

[primal berry]

8

[finite flax]

Look at the write-in spaces

[primal berry]

I see

Makes sense

[finite flax]

The portion of pizza that one person gets is 4 out of 8

Where does the 4 belong

[primal berry]

On top

I understand now

Just kind of gave up off-rip because I just assume I can't really understand any of this

[finite flax]

You also said that the 4 pieces represented half of the pizza

[primal berry]

Or at least understand it fast enough

[primal berry]

Yes

[finite flax]

So then 4 out of 8 pieces is half of a whole pizza

[primal berry]

Yea

[finite flax]

(Per the shading)

So 4 out of 8 pieces is equal to 1 out of __ pieces

[primal berry]

2

[finite flax]

she's ready.

You have become one half with the pizza.

[primal berry]

She/her

[primal berry]

Thank you 😊

[finite flax]

You are welcome

[lone heart]

@primal berry Has your question been resolved?

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[upbeat eagle]

Hia

[upbeat eagle]

Find x values

That 2, makes me unable to do it idk what to do

$2log2 (x/(7x+9))$

[ocean seal]

Equity

[upbeat eagle]

is that .......

Idk...

im very new to logs

[storm ridge]

$m log_a {b} = log_{a} b^m$

[ocean seal]

Lorentz

[upbeat eagle]

log2 (x^2 / 7x+9)

[storm ridge]

Ye ye

[abstract furnace]

express 1 as log_2 2

,, \log_2 \qty(\frac{x^2}{7x+9}) = \log_2 2

[ocean seal]

kanna

[upbeat eagle]

Uhhh bring 7x +9 over?

and then remove the logs through e?

[storm ridge]

e?

[abstract furnace]

[upbeat eagle]

ln e or somthing removes logs

no?

[abstract furnace]

ln is log_e

but this is log_2

[storm ridge]

Ye

[upbeat eagle]

[storm ridge]

$log{a} = log{b} \implies a=b$

[ocean seal]

Lorentz

[upbeat eagle]

oh

2 (7x +9) - x^2

[abstract furnace]

the inverse of log_2 (x) is 2^x

,, \log_2 y= x \implies y = 2^x

[ocean seal]

kanna

[storm ridge]

Bases should be the same tho, which it is in this case

[abstract furnace]

raise both sides by 2

[upbeat eagle]

I thought the log_2 can just cancel out no?

is it because of the base 2?

[abstract furnace]

[upbeat eagle]

[abstract furnace]

this is not log_2 being multiplied to (x^2/(7x+9))

where you divide both sides by log_2

this is the parameter of log_2

,, \cancel{2}^{\cancel{\log_2} \qty(\frac{x^2}{7x+9}) } = \frac{x^2}{7x+9}

[ocean seal]

kanna

[lofty creek]

Please include this in your first message

Then we can easily see in the pins

[upbeat eagle]

Oh i didnt know it does that

Will do next time

Omg omg omg

You can do this

[abstract furnace]

no....

[upbeat eagle]

2nd line, log_2 makes the 1 go to 1^2

YES

[abstract furnace]

it's the other way around

[upbeat eagle]

I SWEAR wait

look

[hidden crest]

Slow Down

[hidden crest]

2^1

[abstract furnace]

like i said it's the "other way around"

[hidden crest]

Not 1^2

[upbeat eagle]

oh

lets pretend

ok

Ty kanna and M4

[hidden crest]

Ur wlc

[upbeat eagle]

.close

[lone heart]

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[limpid shore]

I'm looking for the radius of convergence of the following sum $$\sum _{k=1}^{\infty }\frac{\left(3k+1\right)!}{\left(k!\right)^4}x^k.$$ I guess there's some trick which I do not spot. Any hints are appreciated.

[ocean seal]

Philip

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[limpid shore]

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[alpine sable]

In the following figure BD=4cm, DC=25 cm. The line AD is perpendicular to BC and AD=10 cm. Find angle BAC.

[rich quiver]

!status

[lone heart]

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

[modest osprey]

Arctan(4/10)+arctan(25/10)

[alpine sable]

90° or 60° or 115° or 30° or 100°?

which one is the answer

i need the formula

to solve it

[modest osprey]

Well you would have to solve the tan inverse

I just took tan of bad and dac and just inverse it and then add

Btw if you were asking how to solve addition of 2 arctan then there is a formula

Arctan(a)+arctan(b)=arctan(a+b/1-ab)

Oh yeah there was a condition too

[alpine sable]

tan of bad is ad/bd?

[modest osprey]

ab>1 for above

[modest osprey]

Bd/ad

Wait you don't know how to take tan of an angle in a right angle triangle?

Then you should use Pythagorean theorem

To find ab and ac

[alpine sable]

it comes in decimals

[modest osprey]

And then use Pythagoras on whole abc to see if it's 90° or not

[alpine sable]

but the answers are one of this

[whole quiver]

ur calculator is in radians

[modest osprey]

But if you use the formula for tan inverse addition it gets equal to 90

[modest osprey]

Lol that could be the case too

Try multiplying your answer by 180/π

[alpine sable]

inverse tan of bad plus inverse tan of dac?

[modest osprey]

Yes

[modest osprey]

Now use this formula

You get arctan(infinity)=90°

[alpine sable]

I don't understand that formula sorry

[modest osprey]

Have you studied inverse trigonometry?

[alpine sable]

inversing and adding those two tans gets me 89.1....

[alpine sable]

a little bit. I forget most of the stuffs

[modest osprey]

Well try to see if the values you input are correct

Cuz i got exactly 90°

[alpine sable]

what do u get for inverse tan of bad?

[modest osprey]

Try using arctan(2/5)+arctan(5/2)

21.8°

[modest osprey]

Of bad

[alpine sable]

yeah that's what i got too

[modest osprey]

And 68.2 of the other angle

[alpine sable]

oh i took the 68.19

thanks i see it now

[modest osprey]

Doesn't matter as you got the value of arctan(bad) as something more than 21.8 but you rounded it off

So if you add it still gets to 90°

[alpine sable]

yeah

thanks a lot man

[modest osprey]

Welcome

[alpine sable]

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[alpine sable]

In 1985 the population of town A and town B were the same. From 1985 to 1995 the population of town A
increased by 60% while the population of town B decreased by 60. What percentage of the population of
town A was the population of town B in 1995?

[alpine sable]

oh what's the process of this maths?

[modest osprey]

Consider population at 1985 for any one as x

Then for A after 10 yrs it's x+.6x and for B its x-.6x

So you get (1.6x)y=0.4x [y is the percentage requires]

[alpine sable]

could you explain it?

[modest osprey]

Well the percentage is 100y

[modest osprey]

Uhh 60% of x is (60/100)x=0.6x

So if it increases by 60% then total become 1.6x

If it decrease it becomes 1-0.6 =0.4x

[alpine sable]

the answer is supposed to be 60% or 75% or 25% or 45% or 30

[modest osprey]

Yeah so you calculate y and multiply it by 100

[alpine sable]

uh why?

[modest osprey]

Because y is percentage of A that B has

Lets say you have 100 units and i have 20 units so that means I have 20% of your units

So basically you did 100y=20

[alpine sable]

could you explain this line too?

[modest osprey]

Then found out y and multiplied it by 100 to get the percentage

Well the pop that A has is 1.6 and B has 0.4

So what would be the percent pop that B has of A?

1.6y=0.4

Calculate y

×100

[alpine sable]

25 percent

[modest osprey]

Yeah you got it

[alpine sable]

thanks a lot again

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[hoary onyx]

Hello, I need help with a calculus problem.
Where gamma is the circumference with center (0, 1) and radius 3, clockwise.
Could somebody help me please? Thank you.

[tacit arch]

Did you learn any theorems

About line integrals

[hoary onyx]

Yes, I know Green's Theorem

Tried applying it without success

Tried to find the potential function too

[tacit arch]

Show this

[hoary onyx]

I did the double integral of partial Q / partial x - partial P / partial y dxdy

I got zero from that

But I don't know how to continue

Sorry, could you give me some guidance please?

[tacit arch]

Show the work

[tacit arch]

Yes if you show your work

[tacit arch]

Why aren't you done then?

[hoary onyx]

So I got that
partialQ/partialx = (1 * (x²+2y²)-x * 2x)/(x²+2y²)²
= (2y²-x²)/(x²+2y²)²
and partialP/partialy has the same value.

So one minus the other gave me zero

But I don't know how to calculate the integral from that

[sour verge]

What do you get when plugging that into Green's Theorem?

[hoary onyx]

I get that the curl is 0

[sour verge]

The theorem tells you that your path integral is really just the integral of the curl over the domain

And the curl is 0

So...

[hoary onyx]

So the path integral is zero? My friend got the answer -2pi

[sour verge]

If the curl is 0 then the path integral is 0, yes.

[lone heart]

@hoary onyx Has your question been resolved?

[tacit arch]

Then check your friend's work

[lone heart]

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[brave lagoon]

From this https://www.nature.com/articles/s41598-020-64839-y#Sec12. I have to convert the image model to a bayesian model. So far I havent bene able to do much but It is to compare the frequentist and bayesian models. This is what I have so far. Here is the dataset. https://github.com/lauraleemoore/Growth-curve-fitting/blob/master/growth_data.csv

[brave lagoon]

hello, my assignment to convert a frequentist model to a bayesian one

in R

can I get help with thi?

the R part can be done later but rn its to convert the model to a bayesian one

with priors and stuff

this is what I have rn {r}
currentmodel <- ulam(
alist(
mass_obs ~ dnorm(mu,sigma),
mu <- massexp(k(t-time)),
t ~ dnorm(3, 2),
k ~ dnorm(0, 0.5),
sigma ~ dexp(1)
), data = list_data, chains = 4
)

<@&286206848099549185>

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@brave lagoon Has your question been resolved?

[lone heart]

@brave lagoon Has your question been resolved?

[lone heart]

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[shrewd zenith]

Hello, does anyone have like 2 mins to check my answer, ill send the ss thanks in advance. ( I need to get K so that function is positive for each x -> a>0 , D<0 and i get 2 ranges that doesnt have intersection so i dont know what the answer should be)

[shrewd zenith]

The thing that confuses me here is that i dont have intersection for those 2 k ranges, so idk if i should type that K like that doesnt exist or should i type k(4,+inf) UNION ( - inf, 3 )

I would really appreciate help, thanks in advance

[tall topaz]

Looks right to me and yeah you need to find the intersection, so in this case that would be the empty set.

[shrewd zenith]

So k does not have answer?

[tall topaz]

Union wouldn’t make sense here because we need those two things to be true at the same time.

[tall topaz]

Yeah

[shrewd zenith]

Yes thats right

thank u so much man

So basically

[tall topaz]

[shrewd zenith]

When i have something like

Both roots positive, both roots negative, both roots same sign, both roots different sign

For all those things i need to find ranges for all the conditions, and then find intersections right?

[tall topaz]

I mean those can’t all be true at the same time

How can both roots be positive and negative at the same time

[shrewd zenith]

Nono i mean like seperately

Like all seperate cases

For all those cases i find intersections?

[tall topaz]

Sure

[shrewd zenith]

Okay thank u so much

Have a great day!

.close

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[humble wolf]

Is the limit of this negative infinity?

[humble wolf]

I used L'hopitals rule and end up with -2/0

[lofty creek]

is it (-cos) or -cos?

[humble wolf]

-cos i would believe

[lofty creek]

okay

[humble wolf]

Sending the work in a sec

[lofty creek]

are you saying here that [cos(x)^2]'=sin(2x)??

[humble wolf]

isnt the derivative of cos(x)^2 2cos(x)sin(x)?

or am i high

[lofty creek]

yes it is

ah i see what you did

[sour verge]

-2/0 does not go to -infinity.

[humble wolf]

does it not?

[sour verge]

I think you'd have to check left and right sided limits.

[gray isle]

derivative of cos^2(x) is **-**2sin(x)cos(x)

[lofty creek]

but its -cos^2(x)

[gray isle]

yes. which means you shouldn't still have a - sign there

[sour verge]

Still -2sin(x)cos(x) = -sin(2x)

[humble wolf]

ah yeah thats right

[sour verge]

Just off by a sign I guess

[lofty creek]

ah good point

[humble wolf]

yeah but it doesnt matter rly tho since its x->0 no?

[sour verge]

Also

Isn't the derivative of e^(-x^2) meant to be -2x e^(-x^2)?

[humble wolf]

good point

then its 0/0 again

[sour verge]

Indeed.

[humble wolf]

gosh this is a nasty limit

[sour verge]

I thought it was weird because the graph doesn't actually blow up at 0.

[humble wolf]

Am i supposed to do Taylor or something for this?

[lofty creek]

perhaps squeezing?

[humble wolf]

nah what the hell is this 💀

wild lhospital

[sour verge]

You actually still have 0/0 since the sign on 2cos(2x) isn't correct.

[humble wolf]

yiiiikes

[sour verge]

HMm

[humble wolf]

this is a wild limit

imma try my odds with taylor i think

[sour verge]

I just did some computation, and l'Hopital technically works after FOUR ITERATIONS

[humble wolf]

hahah yikes

[sour verge]

It's only after four times that the numerator's derivative is nonzero

[humble wolf]

This is calc 1 for reference so im assuming im missing something here

[sour verge]

Yeah seems odd

[humble wolf]

whats funnys is that the other limits are a joke compared to it

someone cooked on the 3rd haha

mods

<@&268886789983436800>

Geeeeez my ears

[sour verge]

What happened?

Wasn't looking haha

[humble wolf]

Some new account linked some earrape clip

idk if virus aswell

[sour verge]

Oh rip

Doubt it lots of trolls try their luck around

But yeah someone decided c) was ending you all

[humble wolf]

haha yeah its a crazy one

Im moving on from that one, thanks for all the help and ty for the quick response by the moderators

.close

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[waxen turtle]

Uhh is it G composed of S

[sour verge]

Yeah.

[waxen turtle]

Answer C

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[desert needle]

Can someone explain why for case 1 it's x<5/2, and for case 2, it's x>3? (I get this bit now_

I'm also kind of confused where the 3-x>0 comes from

Would I hvae to go through the same process of negative and positive possibilities for any inequality question like this?

How can I figure out if the answer is AND or OR?

[alpine sable]

when multiply negative numbers in both sides of the inequality you must change the direction of the inequality symbol (this is actually for your second question)

this also happen when negative exponents

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@desert needle Has your question been resolved?

[desert needle]

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[vivid flame]

Hello, i am learning college linear algebra asa self learner from zero using axler book. I need help with the proof of the additive inverse in complex arithmetics. Could you tell me if my proof is valid, i feel somehow it is wrong. But i do not know exacly why. Here is my proof on the following property : for all number "alpha" in C, there exists a unique "beta" such that alpha+ beta=0

[vivid flame]

I feel that the last implications are too quickly made

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[rocky cloak]

whats the name for the group defined by rotations where you always rotate the system to a reference rotation frame first.

I.e, you always rotate the vector so its pointing directly up first, then apply a second rotation to reach the new target position.

The reason this is being done is because the vector is the normal of a mesh and I want the tangent and binormal to be uniquely defined for any normal given the tangent and binormal are provided in the reference position

[lone heart]

@rocky cloak Has your question been resolved?

[limpid turret]

What's a binormal? Also, this seems like extra work. What are you trying to do fully?

!xy

[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.

[rocky cloak]

the tangent and binormal are the two vectors orthogonal to a normal vector

in 3d the definition of the tangent and binormal is arbitrary as long as they are perpendicular to eachother and the normal.

if you rotate the co-ordiante system with random rotation matricies then rotate it so the normal is back in the origianl position again, its possible for the coordiante system to be "twisted" and for the tangent and binormal to be in difeferent places to where they started

I'm trying to see if there is a sub group of SO(3) / rotation where the tangent and binormal are uniquely defined for all oritentations of the normal given you have a reference frame with all 3 vectors defined.

[lone heart]

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@rocky cloak Has your question been resolved?

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[hybrid ravine]

Hello, idk how to do this

[misty mauve]

since you are dividing a polynomial of degree four with degree one, it must result in a cubic. The question mentions that the remainder is -4 which means the constant term must be -4 for the cubic.

assume the roots of the polynomial are alpha, beta, gamma and delta

let alpha = beta since this thing is supposed to have a double root at x = 1

so you get an equation for sum of roots as 2alpha + gamma + delta = -b/a

lets keep these as known facts and apply factor theorem on the polynomial. Since -4 was our remainder f(x) + 4 must be divisible by x+1. Which means f(-1) + 4 = 0

f(-1) = a - b + c + e

alpha^2 * gamma * delta = e/a using product of roots

alpha^2 + 2alphagamma + 2alphadelta + gamma*delta = -c/a

f(-1)/a = 1 - b/a + c/a + e/a

ehhhhh, sorry i got lost in between, let me think through that again

[hybrid ravine]

yep alg take ur time

[tidal lichen]

Have you tried doing the polynomial division?

using a,b,c,and e

[misty mauve]

yeah right f(-1) + 4 = 0 sp f(-1) = -4. We also already know f(1) = 0 since x = 1 is a root

so a+b+c+e=0

and a-b+c+e =-4

subtracting both equations we get 2b = 4

so b = 2

which means a + c + e = -2

[hybrid ravine]

mhm i see that

[misty mauve]

we still need to solve for a, c and e, which is three variables for which we need 3 equations to solve and we already have one in a+c+e=-2

now let me see how to get the other 2

pretty sure it has to do with alpha, gamma and delta i will try it on paper first before typing

alright i got a lead

so there are four roots alpha, beta, gamma and delta

x = 1 is a double root so take alpha = beta = 1

product of roots = e/a

so gamma*delta = e/a

sum of roots = -b/a

i know that b = 2

so 1 + 1 + gamma + delta = -2/a

and then sum of couples of roots should be c/a

so there are six couples since there are 4 roots and 4 choose 2 is 6

you will get 1 + gamma + gamma + delta + gamma*delta = c/a

which can be rewritten as 1 + -2/a - 2 + e/a + gamma = c/a

the above equation came from using the sum and product formulae

now only gamma has to be eliminated

[ashen chasm]

Can you help me find the perimetre of this shape

[misty mauve]

we know that gamma*delta = e/a and gamma + delta = -2/a - 2

[hybrid ravine]

wrong channel

[ashen chasm]

sorry which channel do i have to go

?

[misty mauve]

remove the delta in the sum formula using the gamma from the product formula

gamma = e/(a*delta)

[hybrid ravine]

u go to the section called math help (available)

[misty mauve]

so gamma + e/(a*gamma) = -2/a - 2

oof now we got a quadratic for gamma, there has to be a smarter way

[hybrid ravine]

;-;

[slate vortex]

I think I have a semi-non-rigorous solution

[misty mauve]

how far did you get with this problem

[slate vortex]

If you use the fact that x=1 is a double root,

[hybrid ravine]

only up to b=2

[slate vortex]

We have that $P(x) = (x-1)^2 Q(x)$ for some polynomial $Q$.

[ocean seal]

Ari

[misty mauve]

Q(x) must also be quadratic right?

[slate vortex]

So then we just have to interpolate $Q(x)$ so that multiplying with $(x^2-2x+1)$ gives you $P(x)$

[ocean seal]

Ari

[slate vortex]

yes

Since $P(x)$ has a $ax^4$ term, the leading term of $Q(x)$ must be $ax^2$.

[ocean seal]

Ari

[slate vortex]

Then $ax^2(x^2-2x+1)=ax^4-2ax^3+ax^2$.

[ocean seal]

Ari

[misty mauve]

now we can compare the coefficients

[slate vortex]

$P(x)$ also has a $2x^3$ term, so the next term in $Q(x)$ needs to make the cubic term's coefficient $2$.

[ocean seal]

Ari

[slate vortex]

that's essentially the idea yeah

So then next term in $Q(x)$ is $2(a+1)x$.

[ocean seal]

Ari

[slate vortex]

which makes the cubic term 2x^3 when it cancels out

Ok

So now, notice the constant term of $P(x)$ is $e$, which means the constant term of $Q(x)$ is also $e$.

[ocean seal]

Ari

[hybrid ravine]

ye?

[slate vortex]

So $Q(x) = ax^2 + 2(a+1)x + e$.

[ocean seal]

Ari

[slate vortex]

Multiplying that by $(x-1)^2$, you get the $x^2$ and $x$ terms in the resulting product to be $(-3a-4+e)x^2 + (2a+2-2e)x$

[ocean seal]

Ari

[slate vortex]

And based off $P(x)$, we now have that $-3a-4+e=c$ and $2a+2-2e=0$.

[ocean seal]

Ari

[slate vortex]

2nd equation tells us that $e=a+1$. Substituting that into the 1st equation yields $2a+c=-3$ as desired.

[ocean seal]

Ari

[hybrid ravine]

ohhh i get it

thx 🙂

.close

[lone heart]

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[misty mauve]

ah that was nice, thanks

[slate vortex]

yep ❤️

[lone heart]

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[slate vortex]

Similar triangles probably is the key

[alpine sable]

can u help

idk how to approach this

[violet urchin]

similar triangles

like he just said

[alpine sable]

ill think abo t it

i dont see any

i only see 2 congruent triangles

but idk how that helps

[slate vortex]

I'm guessing you mean the 2 big right triangles. Can you see any smaller triangles within those triangles?

[lone heart]

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[alpine sable]

the unshaded ones are similar?

i would have never seen that

so k=2

k squared is 4

[slate vortex]

Yes.

They're similar to each other but also to the big right triangles

[slate vortex]

Key question is do you see why they are similar

They look similar yeah but you should be able to explain why mathematically

[alpine sable]

because if you rotated the top one left

the line that is drawn is the exact same

so it creates the same angles

u dont assume that the line is the same u know that cuz it goes from 1/4 of the line to corresponding corners

[slate vortex]

uh I guess lmao

It's pretty easy to show that the 2 slanted intersecting lines are perpendicular (using either vectors or coordinates), so then the triangles have 2 angles the same, which implies similarity

[alpine sable]

is it this

[slate vortex]

Anyways it's more important to know how to use similiarity to get the answer

[slate vortex]

That's not quite right

[alpine sable]

wait

i meant 26/128

idk why i put 10a there

i meant to put 11

[slate vortex]

You don't have a 1:15 ratio in the left triangle

[alpine sable]

isnt it 1/16

[slate vortex]

The sides aren't corresponding. The top unshaded triangle has its hypotenuse equal to 1. The big right triangle has its long leg equal to 4.

So their area ratio isn't 1:16

[alpine sable]

oh yeah

so what do u do now

[slate vortex]

You can compare hypotenuses instead

The top triangle has hypotenuse 1

The big triangle has hypotenuse sqrt(17) by pythag

so the area ratio is 1:17, so it'd be a and 16a

[alpine sable]

yeh that much i get

i see now

[slate vortex]

ok you should be able to finish the problem on your own then 👍

[lone heart]

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[calm tulip]

is there a specific rule i could use or do i just use long if i get a fractional remainder to check

[violet urchin]

on the synthetic division side, you divided by x+3/2

but we wanted to divide by 2(x+3/2)

[storm ridge]

You could use the remainder theorem if it's only the remainder you're after

[calm tulip]

huh, idk if im blind but i cant see it (these are also notes my teacher gave me)

[violet urchin]

well you divided out x-(-3/2)

right

but thats (1/2)(2x+3), which was the thing we originally wanted to divide out

thats why it didnt work, you forgot to divide out that extra factor of 2

[calm tulip]

ohh

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[ripe topaz]

can someone simply understand this basic stats?

[ripe topaz]

i'm tyring to figure out the chance player 2 hits based on the fact player 1 hit

can this be deduced from the correlation score or something?

[lone heart]

@ripe topaz Has your question been resolved?

[lone heart]

@ripe topaz Has your question been resolved?

[lone heart]

@ripe topaz Has your question been resolved?

[lone heart]

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[cosmic acorn]

Hi

[cosmic acorn]

where is the mistake

[alpine sable]

what's the question

[cosmic acorn]

solve the equation

find z

[alpine sable]

So it's 10^z + 10^z+1 = 5?

[cosmic acorn]

yes

the untouched equation

[cosmic acorn]

do you have any idea

[cosmic acorn]

hm. 11?

[alpine sable]

wtf?m

idek where u pulled 20 from

[cosmic acorn]

you have 11(10^z). Where comes ghe 11 from?

[alpine sable]

Ur mistake was step 3

Idek where the 1/2 on RHS came from

[cosmic acorn]

i thought I could sum them up

[alpine sable]

Bro u need to know Ur algebra before u do exp

[upper vector]

10^z + 10*10^z = 10^z * (1+10)

[alpine sable]

^^

[cosmic acorn]

I know algebra, but sometimes idk how to solve some question first. I'm improving :(. I yesterday started with this on my own..

[cosmic acorn]

Understandable, ty

[alpine sable]

nah ur basic algebra is 100% not there

you gotta practice that more

then exp should be a breeze

[cosmic acorn]

Idk where, what. I think, I do improve my alg skills with such questions

[alpine sable]

huh

wdym where what

as in Ur algebra is not good u gotta improve it

[cosmic acorn]

nvm

[cosmic acorn]

i do my best

ty

[alpine sable]

youll get there eventually

the more you do the more you understand

[cosmic acorn]

truly

.close

[lone heart]

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[cinder sundial]

There are two questions

I would like to check whether these two answers are correct

For I don’t have a correct answer with me

Forgot that I can put it on Desmond

[cinder sundial]

No I don’t know how to do it

[wheat isle]

Desmos*

[cinder sundial]

thank you

[lone heart]

@cinder sundial Has your question been resolved?

[alpine sable]

Correct but btw if ab= 0 just solve for a= 0 or b= 0

Because even if let's say a= 0 b can be any value

And the eqn would still equal to 0

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[grizzled zephyr]

can soemone please explain whats this

what rule

how do i even begin

i just started integrals and this looks hard asf

[gray isle]

consider the max value of sqrt(1 + cos(x))

consider riemann sums / rectangles
do NOT attempt to find the antiderivative of sqrt(1 + cos(x)) / explicitly evaluate the integral

[alpine sable]

Hi

[lone heart]

@grizzled zephyr Has your question been resolved?

[grizzled zephyr]

man thats so hard

@gray isle I am sorry for ping but here, what did they do?

whats this 2pi

dont we just plug in 2 and -2 into the function

f(b) - f(a)

to find the value of the integral

thats what I learned

[gray isle]

they divided area by length of interval

[grizzled zephyr]

why they did that

[gray isle]

that's how you get the average

[grizzled zephyr]

i learned to evaulate the integral

[grizzled zephyr]

i do this

[gray isle]

evaluating the integral gives the area

[grizzled zephyr]

ok when u evalute it

it becomes 0

[gray isle]

how are you getting 0

[grizzled zephyr]

√4-(2)^2

and √4-(-)^2

both are equal to 0

[gray isle]

you haven't integrated

[grizzled zephyr]

0- 0 = 0

hm?

they gave me f(x)

[gray isle]

yes.

[grizzled zephyr]

ok so i just plug it in the integral

i feel there is something i am missing

[gray isle]

you're mixing up average value with average rate of change or something

[grizzled zephyr]

i just substitute f(x)

[gray isle]

that's not how you evaluate integrals

[grizzled zephyr]

hmm

[gray isle]

$\int_a^b f(x) \dd{x} = F(x)\eval_a^b = F(b) - F(a)$ \ \
where $F(x)$ is the antiderivative of $f(x)$

[grizzled zephyr]

wuttt 💀 but i literally learned thats how its done

[ocean seal]

ℝαμΩℕωⅤ

[grizzled zephyr]

oh

so if i find the antiderivative

of f(x)

it will work?

anti derivative of this

[gray isle]

for the area yes

the antiderivative of this is slightly tedious

they opted for the simpler approach of what this integral represents

[grizzled zephyr]

so i just learn the geometric approach they used

[gray isle]

(which they explained)

ideally you'd want to learn the long approach as well
as the geometric approach only works if the bounds are from stuff like
-r to r
-r to 0
0 to r
which will give you a nice fraction of a circle

[grizzled zephyr]

mhm , thanks

[crystal thistle]

you start with a substitution, like cos(x)=t

[lone heart]

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[grizzled zephyr]

I do not understand at all

[grizzled zephyr]

additivity rule says +

how did they substract

also shouldnt it be -4 + 6

we go from 1 to 2 first then 2 to 5

[vale crag]

integral from 1 to 2 + integral from 2 to 5 = integral from 1 to 5, that's the addition rule

they just substracted the integral 1->2 from both sides

[grizzled zephyr]

yes

[vale crag]

so, integral from 2 to 5 = integral from 1 to 5 - integral from 1 to 2

sure it's not straight up the addition rule, but it's a pretty easy consequence of it

[grizzled zephyr]

oh thats because we dont have 2 to 5

can you please help me with e here

we dont have 1-7

[vale crag]

well same thing here