help-10

[wise talon]

from now on

[silver plover]

why 69

[wise talon]

well

[silver plover]

derpz ur acc hella dirty minded

ive noticed

[wise talon]

you'll wait for a sequence

that converges to 69

but the terms of the sequence are

[silver plover]

[wise talon]

irrational numbers

your help will arrive in 1min

1/2 min

1/4 min

1/8 min

...

thats what will happen if u do not take analysis 2

[silver plover]

...

im doing all stats modules next year

plus differential equations and groups and rings

[wise talon]

measure theory? :O

take measure theory

which will require analysis 2

[silver plover]

❌

not doing complex either

[wise talon]

why would u not do complex

complex is nice

at least take complex analysis

[silver plover]

is it like real?

[wise talon]

ehh its nicer

i heard

[silver plover]

if its like real i aint taking it

[unreal musk]

Depends on the place I think, for me analysis 2 was nicer than complex

Well at least in terms of learning it anyway lol

[wise talon]

what does analysis 2 cover at your school anyway @silver plover

[silver plover]

lets see

[wise talon]

oh why would you not take this

these arent too bad

[silver plover]

cuz groups and rings is easier

[unreal musk]

Beautiful

[wise talon]

[silver plover]

its between this or group and rings

[wise talon]

later:

omg I hate abstract algebra

[silver plover]

cuz im doing maths and stats

so i have to do stats modules + differential equations

[wise talon]

we better not see you complaining about abstract algebra here

if not

your helps will be answered with

[silver plover]

lmao

[wise talon]

"trivial"

[silver plover]

...

[wise talon]

that is worth taking tho

[unreal musk]

"exercise for the reader"

[wise talon]

anal2

[silver plover]

u wanna see

group and rings?

[wise talon]

like differentiation and integration is worth learning

dont you have a soul

dont you wanna go find some calc kids

and then go

proof:

for 2h

[silver plover]

heres groups and rings

[wise talon]

yep

[unreal musk]

Wonder if they cover PIDs, UFDs and the like

[silver plover]

easier than real 2?

[unreal musk]

Euclidean domains

[wise talon]

err

honestly real analysis 1 is just hard because thats ur first exposure to analysis

once u get thru it

analysis 2 becomes MUCH easier to deal with because of your increased mathematical maturity

so seriously

dont be scared

it's genuinely genuinely worth taking it

the big struggle of analysis 1 is that it's your first proof based course

[silver plover]

true tbh

[wise talon]

so you have to learn to write proofs AND analysis at the same time

by now your mathematical maturity is quite sufficient and analysis 2 won't be so bad

it really builds on analysis 1

for example

we can use the algebraic limit theorem for sequences

to prove the algebraic limit theorem for functional limits

by characterizing how functional limits work as sequences

so don't waste your struggles in anal1

go forth and take it

seriously

you worked very hard to prove a bunch of things in analysis 1

and in 2 you get to taste the juicy fruits bore from the effort

you won't regret it

take analysis 2

ok if you take analysis 2

you will get help in

well as fast as possible

[unreal musk]

[wise talon]

if you dont take analysis 2 we have

[unreal musk]

You won't be alone cause we love lurking

[wise talon]

seriously tho

[unreal musk]

[wise talon]

i live for your analysis questions being opened in these channels

otherwise ill have to go look at a stats problem

now back to the list of angery emotes

if you dont take analysis 2

and of course

[unreal musk]

All posts you make will be instantly sullied

[wise talon]

^

the entire helper team will sully every post you make if you do not take analysis 2

fr tho

we'll help you

plus we do analysis bookclub in the secret vc which u can use now

plzzzzz

stats problems always get ignored

but ur analysis problems will always get jumped on to help

PIDs might be in algebra 2 for sa2017

what they described seems to be algebra 1

[silver plover]

ok

i will consider it

.

[wise talon]

do it

also

if u do analysis 2 u can do measure theory which can

really enlighten stats

fr

[silver plover]

dang

.close

[obtuse pebble]

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[plain jungle]

What value should we take as 'x'?

[tardy epoch]

it tells you in the solution

[plain jungle]

I did not learn sets

[tardy epoch]

that's an interval

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:functions/x2f8bb11595b61c86:introduction-to-the-domain-and-range-of-a-function/v/introduction-to-interval-notation

[plain jungle]

I don't have time for this

Can I get to know what value we should take for'x'

That's it

[tardy epoch]

watch it on 2x

[tardy epoch]

x isn't a single number

[tardy epoch]

you'd understand if you'd watch this

[obtuse pebble]

@plain jungle Has your question been resolved?

[obtuse pebble]

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[near pivot]

Hey guys could someone help me with this pls?

[near pivot]

So I've tried doing it normally... however, I have to do 1/2 CHOOSE 2

which my calculator gives as error

I've also tried using the formula

n!/r!(n-r)!

Still didn't work...

Any help would be great!

[obtuse pebble]

@near pivot Has your question been resolved?

[glossy basalt]

Why would I immediately think of taylor expansion about 0 instead of binomial expansion

[obtuse pebble]

@near pivot Has your question been resolved?

[near pivot]

sry wdym?

[timid silo]

$\begin{pmatrix}1/2\2\end{pmatrix} = \frac{\frac{1}{2} \cdot(\frac12-1)}{2!} = -\frac{1}{8}$

[warm shale]

Landau08

[timid silo]

in analogy to, for instance, $\begin{pmatrix}4\2\end{pmatrix} = \frac{4\cdot 3}{2!}$

[warm shale]

Landau08

[timid silo]

see for example: https://en.wikipedia.org/wiki/Binomial_theorem#Generalizations

Newton was, I believe, the first to generalize the binomial theorem to fractional powers thereby anticipating Taylor's expansion in that special case

[obtuse pebble]

@near pivot Has your question been resolved?

[obtuse pebble]

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[spark blaze]

There is a sign in the elevator that reads "6 people or 500 kg". What is the probability that the weight of 6 people will exceed 500 kg if the weight of a single person has a normal distributon N(75 kg; 10 kg)?

[rigid lintel]

10kg standard deviation or 10kg variance?

[spark blaze]

variance

[sweet edge]

U need to *6 and then get the z score for 500 and do 1 - P(Z < z) since u want P(X>500)

[rigid lintel]

the sum of all 6 is normally distributed

[spark blaze]

*6 arguemnts?

[sweet edge]

the distribution given is for 1 person

Ur curious about 6

So u gotta scale it appropriately

[spark blaze]

How do I scale it

[rigid lintel]

just the normal rules for expectations and variances

theyre iid after all

independant identically distributed

[spark blaze]

Sorry heavytf2 I don't know the "normal rules"

[rigid lintel]

alright well

suppose all X_i independant

[spark blaze]

ok this is the case becouse every person in the lift is different

right?

[rigid lintel]

then $var(\sum_{i = 1}^{n} X_i) = \sum_{i = 1}^{n} var(X_i)$

[rigid lintel]

yes

[warm shale]

heavy

[rigid lintel]

var meaning variance

[spark blaze]

ok but we need to scale distribution

We don't need var

[rigid lintel]

and the expectation stays the same

well

it gets multiplied by 6

so you have total weight is a random variable thats N(75*6, 60) distributed

[spark blaze]

Why we got the var if we just created distribution with *6 arguments

I don't get what's going on

[sweet edge]

The distribution is for a single person

But since u want to care about 6, u need to multiply the mean and var by 6

this will make the distribution suited for 6 people now

[spark blaze]

So why did we mess with var?

[rigid lintel]

because you need to know the variance of the total weight of 6 people

[spark blaze]

Don't I need to just get the area under the curve for this new distribution that we created?

Area to certain point ofc

[rigid lintel]

yeah thats it

[spark blaze]

So I normalize the distribution and get the anwser

Without use of var

[rigid lintel]

you cant normalize a distribution without thinking of var

[spark blaze]

ok what if I got other distribution? This "multiply arguments" methods works for them as well?

[rigid lintel]

not for all

suppose the exponential distribution

[spark blaze]

I always use this formula

[rigid lintel]

the sum of exponentially distributed random variables is not exponentially distributed

[spark blaze]

The First one

[rigid lintel]

but it is gamma distributed

for normally distributed random variables however

it does work

provided the random variables are iid

[rigid lintel]

yes that works here

[spark blaze]

One more thing

Earlier, when I summed distributions, I needed to use an integral

"The random variable X has an exponential distribution with parameter λ = 2, a
the random variable Y has an exponential distribution with parameter μ = 10. It is also known that X and Y are independent. Determine the distribution density of the random variable Z = X + Y ."

Integral method always works and the scale args is just for normal distribution

Or how is that

[rigid lintel]

yes exactly how you said it

normal and iid

[spark blaze]

iid?

[rigid lintel]

independant identically distributed

[spark blaze]

ok, so when not sure - use integral method

[rigid lintel]

exactly

[spark blaze]

Thank you heavytf2

❤️

[rigid lintel]

np

[spark blaze]

ok, I'm closeing this ticket

have a good night ❤️

.close

[obtuse pebble]

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[old halo]

ACD=CAD=31.5
from triangle ABC angle ABC=ACB
the sum of angles in a triangle is 180 degrees, so ABC+BAC+ACB=180
ACB+(31.5*2)+ACB=180
ACB=58.5
BCD=ACB-ACD=58.5-31.5=27

[earnest oriole]

woah?

[old halo]

Something wrong?

[earnest oriole]

thats right

but woah

.close

[obtuse pebble]

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[river ember]

I've tried to use the slope formula to solve this, like y2 - y1 over the time x2-x1. Then I set up a proportion for 1 hour, but when I substituted the numbers from the chart, it didn't really work out. I think the reason for this is because it's exponential functions, but I'm not too sure. If anyone could help, I'd be grateful!!

[dark stirrup]

You are right. It's because it's an exponential decay

[river ember]

What do you think I should do first? I think I really do need some advice

[gusty halo]

it looks pretty linear to me

[dark stirrup]

Look again. Rate of dropoff decreases over time.

[gusty halo]

barely

[dark stirrup]

That's exponential decay

enough

First few drop-offs are 17

last are 11

That's a 35% difference

Quite significant

Also, the drop-off rate is always decreasing

If it were increasing and decreasing randomly, then I would argue linear

[gusty halo]

yeah i guess ur right

[dark stirrup]

How were you modelling it linearly? Were you doing a best fit method?

[dark stirrup]

I will admit though, the problem asking for "hourly rate" does kinda suggest linear tho

I wish it would "hourly rate of decay"

[gusty halo]

thats what i thought at first

@river ember what math class is this?

[river ember]

Algebra 1

[dark stirrup]

ah

[gusty halo]

probably linear then

[dark stirrup]

Maybe it is linear then..

yea

[gusty halo]

otherwise you need calculus

[dark stirrup]

But you know about exponential functions

It's a weird problem

What's the context of your other problems. Have they all been linear or exponential?

[gusty halo]

have you seen exponential functions in class?

[river ember]

Kind of

[dark stirrup]

I'll leave this to you ☹️☹️

I gotta take a shower

[gusty halo]

im just gonna assume its linear

@river ember what were you using for y1 and y2

and x1 and x2

i would recommend using 0 hours and 4 hours to get the most accurate measurement

or if you know how to do linear regression that works as well

hopefully that’s enough to help bc i gotta walk my dog now hes barking at me

gl

[obtuse pebble]

@river ember Has your question been resolved?

[obtuse pebble]

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[fathom flicker]

I need help with this integral

[fathom flicker]

,, \int{e^{-t}sin(3t)dt}

[warm shale]

AustinU

[fathom flicker]

I am sending my work now

I need help finding my mistake

[ivory trout]

I have no paper w me but this integral screams IBP

[fathom flicker]

I want to solve using euler's method, not by parts.

I can do it by parts

and at the final step, here is the graph

Showing the two do not differ only by a constant

but such a small difference makes me feel like I was close and only had a small mistake

so if anyone could point it out to me I would appreciate it

if my solution (the one on the right) had arctan of positive 3 instead of negative three then the only difference between the graphs is that one is a sine function and the other cosine

this could come from where I represented 3i-1 as e^(itheta)

possibly there is a mistake there?

@tardy epoch

<@&286206848099549185>

[stone bane]

Honestly this looks like it screams laplace transform

[fathom flicker]

well

I can do the integral

using by parts

I am just trying to use this method to solve it, and I know there is a way

but also I don't know how to do laplace transforms yet

either

[stone bane]

Do you know some complex analysis or the leibniz rule?

Every easy way to solve this seems out of reach lol

Ohhhh

I see now

[fathom flicker]

Integration by parts is an easy way to solve

just do by parts until you get the original

and then it will be /2

I know the solution

[stone bane]

I’m not versed enough in the method you’re using to help too much, sorry

[fathom flicker]

dang

thats okay

putting these back at the top

<@&286206848099549185>

TLDR: I am getting a slight phase shift in my solution, I need help in identifying my error. I know this can be solved using integration by parts, I am trying to use this method for practice. Any help is appreciated. Thanks

for the integral

,, \int{e^{-t}sin(3t)dt}

[warm shale]

AustinU

[wary vigil]

i'm not entirely sure why you didn't continue from here

you almost have it

@fathom flicker

[fathom flicker]

I did

did I go in the wrong direction?

I started getting a substitute for the fraction in terms of

e^(itheta)

what should I have done instead?

[wary vigil]

you should've multiplied with the complex conjugate

top and bottom

and seperated the exponential into the real number and the complex number

[fathom flicker]

Okay, you're right that would work. I do believe that this method also works, do you see my error in doing it this way though?

[wary vigil]

you can then apply euler's formula again

[fathom flicker]

because a+bi can be rewritten as re^(itheta)

so if I did it correctly it still should've worked out

[upbeat plinth]

@fathom flicker this is one reason why we discourage asking for help outside your channel

[fathom flicker]

why?

[upbeat plinth]

you have someone helping you here while someone is already helping you in another channel

[fathom flicker]

this person came from the other channel

where I asked for help

and 2 brains are better than 1

[upbeat plinth]

we'd rather not have someone waste time on a problem thats already being helped with

[wary vigil]

have you tried using the angle addition formula

[fathom flicker]

I have not, but currently the two are not equal so applying an identity wont make them equal

[upbeat plinth]

so pls continue in only 1 channel

[fathom flicker]

okay, how do I tell howl to help me in this channel without "advertising it in another channe;"

[wary vigil]

isn't austin just here now?

[fathom flicker]

I am being helped in calculus aswell now

I asked in there again before you started helping me in here

[wary vigil]

ah ok

[fathom flicker]

except howl actually hasn't said anything yet

just that he would work on it

[wary vigil]

tell them to come to here

then we're all good

use the angle addition formula for sin

then use the relationship between the inverse trig functions and the trig functions

see what happens

[fathom flicker]

at what step of my problem

should I use the angle addition formula

the very end?

[wary vigil]

in the sin you have at the end of the first page

[fathom flicker]

okay

[wary vigil]

i'm looking at the identities and it's looking good

like you're getting the same result as the answer good

[fathom flicker]

should I also do that with the other side then

with the cosine

[upbeat plinth]

i mean its too late to worry about that. just keep it in mind for future reference, and i advise to get to know the rules in #❓how-to-get-help

[wary vigil]

only with this

[fathom flicker]

okay so ignoring the denominator

I get

,, sin(-arctan(-3))cos(3t)+cos(-arctan(-3))sin(3t)

[warm shale]

AustinU

[fathom flicker]

^ thank you

[fathom flicker]

I was about to ask about that

lol

when was I supposed to learn those identities?

just curious

[wary vigil]

you weren't

[fathom flicker]

ah

[wary vigil]

but you weren't supposed to do it this way either 😄

[fathom flicker]

wait

I have negative arctangent

do I just use that cos(-x)=cos(x) and sin(-x)=-sin(x)

[wary vigil]

arctan is

[fathom flicker]

is that true

[wary vigil]

you could

you could also do

[fathom flicker]

gotcha

where are you pulling these identities from

you are quick with them

[wary vigil]

wikipedia

i have both pages open 😛

[fathom flicker]

lol

how did you become familiar with them?

[wary vigil]

doing these kinds of problems

means you quickly become familiar with the wiki pages for trigonometric identities

[fathom flicker]

ah so you just went and browsed for the right ones? or did you know these would be there already

[wary vigil]

i knew there is a formula for angle addition

[fathom flicker]

well yah

[wary vigil]

don't remember what it is precisely

[fathom flicker]

okay so I am getting

[wary vigil]

i know that arctan is antisymmetric

[fathom flicker]

,, \frac{3}{sqrt(10)}cos(3t)+\frac{1}{sqrt(10)}sin(3t)

[warm shale]

AustinU

[wary vigil]

and i knew there had to be some formulae for inverse trig in trig

yup

[fathom flicker]

and I should now do the same thing to the other side?

[wary vigil]

other side?

you almost have it

[fathom flicker]

because I don't yet see how that is equivalent to -cos(3t-arctan(1/3))

where are you trying to take me to?

I have been trying to show the equivalency

[fathom flicker]

what should I be doing

[wary vigil]

i thought you wanted to show that what you got through the re^iphi

was correct

[fathom flicker]

yes that is what I want to do

[wary vigil]

yeah so you had the imaginary part

which should be equivalent to this

[wary vigil]

this

[fathom flicker]

yes

[wary vigil]

and we're almost there

😄

[fathom flicker]

one second let me gather my thoughts

[wary vigil]

ok

[fathom flicker]

okay yes so the thing at the bottom of the page

got it

I was looking at the bottom of the other page

[wary vigil]

this

[fathom flicker]

and they are similar

[wary vigil]

=

[fathom flicker]

so it worked out

[wary vigil]

this

[fathom flicker]

yah

[wary vigil]

by using those trig identities

you found that yes you got the right answer

that indeed the method you used was correct

[fathom flicker]

hold on okay

then how come when I graph it

[wary vigil]

if a bit heavy on the trig identity sides

side

[fathom flicker]

they aren't the same?

[wary vigil]

computer doesn't always understand these functions perfectly

[fathom flicker]

I am actually missing a negative sign

I probably just lost it

but

yes I did

found it

okay I see the equivalence now

so I was correct the whole time?

[wary vigil]

you were

[timid silo]

🤣

[fathom flicker]

well I'm really glad I spent an hour and 30 minutes to confirm that

[wary vigil]

lol

[fathom flicker]

seriously thank you for the help though

[wary vigil]

next time try the way i proposed at the start 😄

or IBP

[fathom flicker]

the conjugate?

what is IBP

[wary vigil]

ye

[fathom flicker]

oh

by part

yeah

[wary vigil]

yes

[fathom flicker]

I can do that in like 2 sec

lol

howl did you solve it

[timid silo]

Nope don't know anything about integrating using Euler's identity :catthink: something new to learn

[fathom flicker]

it's something I learned in diffeq actually

or in a online diffeq lecture

I haven't really taken diffeq yet

well thank you guys for your time

.close

[obtuse pebble]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[karmic hedge]

https://i.gyazo.com/cb3dd07f52ed0704bd11b24bd83c7f09.png

[fathom flicker]

that looks tedious

[karmic hedge]

I got:
g(x) = x^3 - x +9
g'(x) = 3x^2 - 1
h(x) = x*sqrt(x) + ln(x)
h'(x) = sqrt(x) + x/2sqrt(x)

[karmic hedge]

It was

[fathom flicker]

stop

I see an issue

[unreal musk]

h'(x) is wrong

[fathom flicker]

h'(x)

you forgot the ln(x)

to derive that

[karmic hedge]

omg

yes you are right

[unreal musk]

• convert $x\sqrt{x} = x^{\frac{3}{2}}$

[warm shale]

chartbit

[karmic hedge]

Oh yeah didnt spot that either 🤦‍♂️

[karmic hedge]

Is it incorrect if I dont do this

and just use product rule

[unreal musk]

You could use product rule of course, it will get you the same thing!

[karmic hedge]

Ah ok nice

Cool, thanks guys!!

❤️

.close

[obtuse pebble]

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[idle spade]

can't figure out the blanks

[idle spade]

i understand that the derivative of u is 6x-pi

but i'm not sure how to incorporate that into the substitution to get f(u)

[lone jetty]

So understand that the derivative of u is with respect to x

So you have du/dx = 6x - pi

So you can isolate dx to get $dx = \frac{1}{6x - \pi} \cdot du$

[warm shale]

Alfie

[idle spade]

correct yes

[lone jetty]

You can replace $3x^2 - \pi x + \pi$ with the definition of u

[warm shale]

Alfie

[lone jetty]

and then replace dx with this definition

and the fraction will cancel out xs from the top

so you are left with the integral only in terms of u

[idle spade]

okay you lost me there

[lone jetty]

So we have $\frac{12x - 2 \pi}{3x^2 - \pi x + \pi} \cdot dx$

[warm shale]

Alfie

[lone jetty]

We can first replace the denominator with u

$\frac{12x - 2 \pi}{u} \cdot dx$

[warm shale]

Alfie

[lone jetty]

Right?

[idle spade]

yes

[lone jetty]

And we know $dx = \frac{1}{6x - \pi} \cdot du$

[warm shale]

Alfie

[lone jetty]

So we can now write this as $\frac{12x - 2 \pi}{u} \cdot \frac{1}{6x - \pi} \cdot du$

[idle spade]

right

[warm shale]

Alfie

[idle spade]

ahhhhh thats where the cross kicks in

[lone jetty]

Yeah, cancels out

remember to rewrite your bounds in terms of u as well

you're integrating between 0 and 1 in terms of x. you're integrating between something else in terms of u.

[idle spade]

yes between

pi and 3

[lone jetty]

yep looks correct 😄

[idle spade]

ahhh okay

the cross is where it gets me

like getting to the cancellation

the step just before that

[lone jetty]

you were almost there

just write it all out in long form and it should click

[idle spade]

but i see that now its just combining the two

yes i recall that now

fully lost it until you helped me out there though

thank you very much

[lone jetty]

np 😄

[idle spade]

ah wait

sorry so from here, do i just plug in my f(u), a and b into the integral and solve that?

then go back in terms of x based on the output

wait no

i just solve in terms of u and it should answer it

okok got it all good

thank you!!

[obtuse pebble]

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[timber osprey]

in this system of equation, the rank of the augmented matrix is 2 right? therefore it has infinitely many solutions?

[timber osprey]

0x+0y=0 (last row of the row enchlon matrix)
infinitely many solutions right?

[obtuse pebble]

@timber osprey Has your question been resolved?

[gilded needle]

can you show your RREF ?

according to matlab, the last row is 0 0 0 1, which would be inconsistent

[timber osprey]

yes one second

@gilded needle

[gilded needle]

wait hold on

shouldn't your augmented matrix be 3x4?

like this: ```
1 1 -3 2
2 1 1 1
3 2 -2 0

[timber osprey]

There's only 2 variables though no?

X and Y

[timber osprey]

And 3 equations so 3 rows

[gilded needle]

what are "-3z" and "z" and "-2z"

[timber osprey]

Oh my

I'm sorry I sent the wrong question lol

[gilded needle]

oh haha

[timber osprey]

They're very similar

[gilded needle]

ok, pls send the one you intended

[gilded needle]

ok

[timber osprey]

Also I only calculated REF not Reduced REF
I don't think it's needed for calculating the rank right

[gilded needle]

matlab agrees that the last row is 0 0 0

yeah you don't need to go all the way to rref

    1.0000         0    0.6000
         0    1.0000    0.2000
         0         0         0

[timber osprey]

so any x and y value will work

because 0x+0y=0 right

[gilded needle]

no, there are three equations and 2 unknowns

getting 0 0 0 for the last row just says that the equations are not inconsistent

so you can ignore the last row and focus on the first two: ```
1.0000 0 0.6000
0 1.0000 0.2000

this is the RREF so you can read off the solution directly

x = 0.6 and y = 0.2

unique solution

[timber osprey]

confused here
did you simply ignore the last equation? thats allowed?

[gilded needle]

if it's 0 0 0 then that is equivalent to 0x + 0y = 0

[timber osprey]

yes

[gilded needle]

a true equation but a useless one as it doesn't tell you anything

so in that case you can ignore it

if it had been instead:

0 0 1

that would mean 0x + 0y = 1

i.e. 0 = 1

which is impossible, hence that would mean there is no solution

[timber osprey]

so when a zero row exists in a augmented matrix of a system of equations, i cant say anything about how many solutions it has?

[gilded needle]

right, a zero row , say 0 0 0 in this case, just means
0x + 0y = 0
which is true for all x,y

it's equivalent to 0 = 0

so it doesn't tell you anything at all about x and y

[timber osprey]

yes
also Rank = number of unknowns so there exists exactly one unique solution

i think i got it thanks

[gilded needle]

rank is at most the number of unknowns

it can be less

[timber osprey]

when it is less, no solutions right?

[gilded needle]

no, it depends

if the equations are consistent, then rank < number of unknowns means there are infinitely many solutions

if the equations are inconsistent then there is no solution, regardless of the rank

[timber osprey]

ah i see

thanks

[gilded needle]

sure

[obtuse pebble]

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[violet ingot]

I do not understand how the answer could be 1.5 and not -2.5

[obtuse pebble]

@violet ingot Has your question been resolved?

[obtuse pebble]

@violet ingot Has your question been resolved?

[sweet edge]

hmm

oh oops

,w 4^2-7*4+14

[warm shale]

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[sweet edge]

,w 4.5^2-7*4.5+14

[warm shale]

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[sweet edge]

,w (2.75-2 )/ .5

[warm shale]

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[sweet edge]

yep./

@violet ingot so you messed up the S(4) part

that's all

[violet ingot]

Yes

[sweet edge]

yep!

[violet ingot]

Thx

[sweet edge]

np

[obtuse pebble]

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[glossy basalt]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

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[twilit pelican]

would appreciate any help with this

my current work

[obtuse pebble]

@twilit pelican Has your question been resolved?

[twilit pelican]

<@&286206848099549185>

[twilit pelican]

<@&286206848099549185>

[glad valley]

<@&286206848099549185> come on helpers blanket has helped answer alot of people's questions, it's time we give back

[obtuse pebble]

@twilit pelican Has your question been resolved?

[twilit pelican]

<@&286206848099549185>

[fathom flicker]

blanket

I'd love to help you, but it is very late at night for me, if you DM me this I will get you help / a solution in the morning

or keep waiting for others

[twilit pelican]

hmm alright

ill dm before i sleep if i dont get anything from others or myself by then

ty!

[obtuse pebble]

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[jade perch]

Im paying a loan $260000 at 2.36% interest which would make the A=468765.3033 monthly repayments should be 1562.551 (im not sure) i then want to pay an extra $250 a month from the get go i need to work out how much interest that cuts off and how much time

[jade perch]

I need a formula aswell

[mellow glacier]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

@jade perch Has your question been resolved?

[jade perch]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[obtuse pebble]

@jade perch Has your question been resolved?

[jade perch]

<@&286206848099549185>

[lost island]

<@&286206848099549185>

[jade perch]

I love getting ignored ngl

[jade perch]

<@&286206848099549185>

[sweet jacinth]

?

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

If a and b are well formed formulas, prove

1. a and b are tautologically equivalent
2. (a <-> b) is a tautology

[obtuse pebble]

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@timid silo Has your question been resolved?

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[obtuse pebble]

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[prisma flax]

i have 32 students in my class, what is the possibility that 10% of them has the same birthday?

[drifting wraith]

0?

it's just impossible, the number is not divisible by 10, 3.2 students would never have the same birthday

[prisma flax]

just so what is the possibilty that 3 students has the same birthday?

<@&286206848099549185>

please hewlp me

[loud tangle]

do you have their specific ages because then we could count leap years as well

[prisma flax]

i mean

the same year

not year

like data

for example

1.1

@loud tangle

[loud tangle]

if it was in say 2020 the chance would be slightly lower due to the fact that there is an extra day

[prisma flax]

i mean

it is called

birthday paradox

i just need to calc it with 32

[prisma flax]

ignore the year

just the date

[loud tangle]

are we allowed to assume they are all the same age?

[prisma flax]

yes

[loud tangle]

in the sense that theyre born in the same year

[prisma flax]

same age and month, day is known

[loud tangle]

well you have to do case work for when the month is 30 or 31 days long

or 28/29 for february

[prisma flax]

we can just say

365 dys

[loud tangle]

you said same age and month so that means they all share a month of bdays

[prisma flax]

the chance of a student having the same date as the other is 1/ 365

[loud tangle]

unless im misinterpreting

[prisma flax]

not same month

just same year

[loud tangle]

well wouldnt it be 1/365² due to the fact that 1/365 is just having a specific date as a bday

[prisma flax]

there is 365 days so the possibility for 1 having the same birthday as other one is 1 in 365

so i need to calculate it for 32 students

[loud tangle]

that 10% of them share their birthday?

or that all do

[prisma flax]

so what is the possibility that 3 of them has the same birthday

yes

[loud tangle]

1/365³

because for 1 student to have a speciric bday it would be a 1 in 365 chance

[prisma flax]

yes

[loud tangle]

if it were 2 it would be 1/365*1/365=1/365²

and same goes for 3

[prisma flax]

but theres a problem

u only checked for 1 student

u need to check for the entire class

each student with another student

[loud tangle]

it applies to any 3 students though

[prisma flax]

yes

[loud tangle]

because youre not specifying what birthday

ur just saying they have it in common

[prisma flax]

do u want me to explain again?

[loud tangle]

what is it that you dont understand

/genq

[prisma flax]

I hve 32 students in my class, what is the chance that 3 of the students in my class share the same birthday

ignore the year

lets assume they are all born in the same year

https://en.wikipedia.org/wiki/Birthday_problem

[proper forum]

A silly question but why doesn't the probability depend upon the total number of students in the class? I've probably confused "possibility" and "probability".

[proper forum]

I think OP was trying to ask the same thing.

[loud tangle]

i think i might have had the wrong approach

ah i see

i didnt consider the number of possible cases propperly

[obtuse pebble]

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[obtuse pebble]

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[bold grotto]

Is the answer -root3/3<f(x)<root 3/3

[bold grotto]

I differentiated it at 0

And got in cos form
Solved for x value

I got pi/3

And input that into the formula to got output

And answer of Is the answer -root3/3<f(x)<root 3/3

[unreal musk]

For one, you can tell the range will have something like “0<=f(x) <= …” just by looking at the picture oh they only did it for a part, nvm me

,w sin(x)/(2-cos(x))

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[unreal musk]

= as well but basically

[bold grotto]

Oh

Damm

Sorry gove me a sec to look at that

Is -1/root 3 tge same as -root3/3

[unreal musk]

Tl;dr here

[bold grotto]

Oh okay still it's intresting to see it visually

Very cool!!!

Thank you!!!

.close

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[exotic creek]

why is it minus 1/2(x "minus" 9) -3

why is it not plus?

$x+9$

[warm shale]

yomiko

[trim portal]

Because absolute value is lowest when the expression inside is 0

And you know it is lowest at x=9

So the inside expression must be 0

and x-9=0 at point x=9

[exotic creek]

hmm

[exotic creek]

wouldn't $(x/2 + 9/2)-3$ be the AB line so it should work by subbing y =0 on the equation?

[warm shale]

yomiko

[exotic creek]

because there is one positive grad and one negative grad so its just 2 straight line equation

[trim portal]

That should work I think

lemme check

[exotic creek]

it does work i calculated it wrong

my bad

[trim portal]

Oh wait doesnt

it should be -9/2

[exotic creek]

i got the answer tho

1 min

nvm it doesnt work

[trim portal]

https://www.desmos.com/calculator/fsuvxulwnf Try this, observe what happens with the absolute value when you change the coefficients

[exotic creek]

but why doesn't $(x/2 + 9/2)-3$

[warm shale]

yomiko

[exotic creek]

why doesn't it work

[trim portal]

Because there needs to be -9/2

The blue line is the one you proposed, and the red is the actual absolute value

Absolute value has minimum when the expression inside=0

in this case the expression inside is x+b, and you know that it equals 0 when x=9. (By point A) So 9+b=0, solve for b and b=-9

[exotic creek]

right ok

thanks

.close

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[obtuse pebble]

Please don't occupy multiple help channels.

[short spruce]

what have you tried

[royal basin]

posted in #help-7｜zen1thxyz

.close

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[timid silo]

I can do a taylor series of one function

And I know what it is for cos

But I don't know how to do this

[plain stag]

For multivariable functions, it's the same, you're just differentiating wrt each variable

First order taylor uses both first order partial derivatives, second order uses all second order partial derivatives, etc

[timid silo]

Okay so, it's going to be 1 + jacobin + hessian?

[plain stag]

Pretty much

[timid silo]

That's um

Easy

Lol thank you 🙂

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[clear owl]

please explain the whole problem (11 and 10)

[timid silo]

for (11) there are 3 right angled triangles. Apply Pythagoras theorem formula on the smaller triangle to the left to get the unknown side in terms of 'x'.
Once you get that, apply Pythagoras theorem formula on the triangle to the right to get the other unknown side.
Then apply Pythagoras formula on the big outer triangle. You would get a quadratic in x and can solve for x

[obtuse pebble]

@clear owl Has your question been resolved?

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[carmine token]

i’m getting 7/6

but answer is supposed to be 5/6

do you guys know why?

i did right- left

it’s problem 3in the worksheet

[warm canopy]

it should be the region to the left of the one you shaded

oh wait

nvm

[carmine token]

yeah

the triangle thing right ?

[warm canopy]

the one you shaded is right

you need to find that point of intersection

and do 2 separate integrals

[carmine token]

(0,0) and (1,1)

[carmine token]

why two?

[timid silo]

is that right though

[warm canopy]

because for the first half of the region you want the area under x^2, and for the second half of the region you want the area under x+y=2

[timid silo]

,w x^2 +x -2 = 0

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[timid silo]

i see

[carmine token]

wait should it be from -2 to 1 in my integral?

[warm canopy]

your integral should be from 0 to 2

split at 1

[obtuse pebble]

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[obtuse pebble]

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[languid sky]

How would I calculate the complexity of the following algorithm? (in python)

    if x == 0:
        return y+1
    elif y == 0:
        return f(x-1, 1)
    else:
        return f(x-1, f(x, y-1))

[languid sky]

i forgot how to do it

[tardy epoch]

#old-network for cs server

or find the python one

[languid sky]

"cs"?

[tardy epoch]

computer science

[languid sky]

ah

[tardy epoch]

don't know which link actually

top or bottom

oh it's the top one:
https://discord.gg/eF3Wjsd

[obtuse pebble]

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[native latch]

I'm having trouble understanding this logic. This is my interpretation:
If p is true, q is true = p is false or q is true

[native latch]

p: I press the power button
q: The computer turns off
p->q if i press button, computer turns off
is the same as
i dont press button or computer turns off

it makes no sense

unless the V means "either or" and not just "or"

so it becomes either i dont press button or computer turns off

can someone confirm this?

[timid silo]

erm take it by each case

also i dont like your q because not q is the computer turning on which is something that the power button does

take if its raining outside then i have an umbrella

where p is if its raining

and q is if u have an umbrellla

not p is if its not raining

so the statement: its not raining outside or i have an umbrella can only be false if its raining and you dont have an umbrella

u feel me

[native latch]

Yeah, that makes much more sense. Thanks for explaining!

[obtuse pebble]

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[obtuse locust]

this question has a range of {1.3 ≤ x ≤ 1.75}

[tranquil sonnet]

What I would do is just set an inequality up

Hmm but might be really messy

Actually just solve for x when the volume is 300 more, and 900 more

You’ll get a quadratic

[obtuse locust]

i had to figure out the range values which is x cant be under 1.3 or over 1.75 if it is you get over 900 or under 300

but i wasnt sure if thats all it wanted

[obtuse locust]

what do u mean for this?

[obtuse pebble]

@obtuse locust Has your question been resolved?

[obtuse pebble]

@obtuse locust Has your question been resolved?

[obtuse pebble]

@obtuse locust Has your question been resolved?

[obtuse locust]

[lucid flume]

That's fking easy

See this

You need to compute for two different cubic equations

(x+3)(x+5)(x+11)=900
(x+3)(x+5)(x+11)=300

Solve them and find the value of 'x' from both of them

then set the lowest value of 'x as the lower boundary

and the highest as the higher boundary

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[tired sierra]

what am I doing wrong here?

[sweet edge]

What have u got so far

[tired sierra]

i found |v| which is 5 through sqrt 3^2+4^2

then subbed it into |u||v|cos(0)

so its |u| x 5 x cos60=0

that gives me 5/2 move to other side 2/5

but the answers 4

[sweet edge]

wait a sec

it's not = 0

the dot product is 10

not 0

so u dot v = 10 = |v||u|cos(60)

|v|=5 so
2 = |u|cos(60)

[tired sierra]

oh my god I overlooked the scalar product being 10

thats my fault