help-13

[steel jewel]

alr thanks :)

.close

[cedar kiln]

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[stoic gale]

yes

[tepid sail]

hmm what if you made a basis with e^x and cos(x)

they are linearly independent

and then created a differentiation matrix

[stoic gale]

when you get Ae^(2x)cos(3x) for some constant A

[tepid sail]

nvm that wouldn't work

[stoic gale]

haha im sure one could figure out some interesting linear algebra way to solve this

[tepid sail]

okay so usually when you integrate by parts

you let f'(x) equal to like

[dull oxide]

Tabular?

[tepid sail]

e^x or some cos(x) or sin(x)

that would usually do well

[tepid sail]

haha yeah

[stoic gale]

also known as DI method (shout out to bprp) - basically just IBP

keep doing tabular until you get an expression which includes e^(2x)cos(3x)

[tepid sail]

just write it out tbh

that's where ur messing up

well sin is alterating

and e^2x doesn't cchange

so like 2 times

yes

also don't use this method

just write it out

[stoic gale]

yes you erase the last line and you would be done really quickly

[tepid sail]

pencil don't blame others for trying to help you

I would check your work

once you get a better answer lmk

hmm

just write it out again

pencil

I'm trying to help you

remember the formula for integration by parts

and please write it out!

it's not recommended for you

it can be done with tabular

just write it out properly

and don't use the shortcut method for now

solve it using the standard integration by parts method, and I'll let you know

okay, so do it and ping me

after you solve it normally, I will tell you

Good job!

Now

Expand it out

And bring the last term over to the other side

Okay so almost there

The last line is incorrect

[plucky sphinx]

Bro help me what happens in here im new

[tepid sail]

Idk

Sometimes we do like hs math like here

Sometimes we do like college level stuff

I’m a freshman in university

Aight bro you’re kinda being an ass

[plucky sphinx]

I wanna get into programming ngl

[tepid sail]

When I was a sophomore in hs

I wasn’t talking all this

[plucky sphinx]

Dayum

[tepid sail]

I’m not helping you further but just look at both sides of the equality

On the step before the last one

[plucky sphinx]

Anybody here wanna help

Me

[tepid sail]

Bring the last term on the right hand side over the left hand side

I’m guessing ur like a sophomore or sm

And bro you gotta calm down

You’re taking like basic calc

You’re privileged to get any help

[cedar kiln]

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[lone void]

Simple

so what u asking

[lusty trench]

I'm sure I went through the process correctly to get the answer but it's still saying I got it wrong

would you like an image of my work

[lone void]

yea

[lusty trench]

jimmy a minute then

here @lone void

[lone void]

hmm

[lusty trench]

I forgot the extra part with $f_{xy}(x_0, y_0)$ so I added it at the bottom, but I checked everything and it feels right from what I see

[wraith dagger]

hanzoh

[lone void]

what did u forget

o

gimme a min

[lusty trench]

In the Q(x,y) formula I needed to use another line to finish the formula, when solving I forgot that part on the extra line at first but added it in later

[lone void]

that part is fine

[lusty trench]

-12(x+3)(y-1)

ok

[lone void]

hm

[lusty trench]

good

[lone void]

I mean I think the maths right

that’s all I know

let me get my notes I forgot this

[lusty trench]

that's alright

take ur time

[lusty trench]

I also looked it over like four different times, I don't think there's a mistake in my work, but if there is it wouldn't be easily noticable

[lone void]

idk man

[lusty trench]

that's alright, I'll probably just email my professor

[lone void]

gimme like 5 minutes

I got u

[lusty trench]

aight

[lone void]

I don’t think I did it right

u should email your professor

[lusty trench]

aight

thanks for the help either ways

[lone void]

Yup

[lusty trench]

I appreciate you taking the time to check, have a good day good sir

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[cedar kiln]

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[white isle]

I'm confused, does graph like these have inflection points, I understand that inflection point occur when the graph goes from concave up to concave down, vice versa. to me the graph doesn't swap concavity, it only looks like it is increasing, so would this graph not have an inflection point at (0,0)?

[violet flume]

only the interval included?

[violet flume]

but yea, i think youve got it

you can firm this because f'' = -f

[white isle]

there are no interval given

[violet flume]

i mean we dont know that but clearly

this is sinx

[hollow trail]

the graph does seem to be concave up at the start and concave down at the end

[white isle]

I was taught to read the graph from left to right, to me it look like it's only increasing, is there any better way of understanding it

[violet flume]

well reading increase is only going to get you critical points

what you should look for is if theres a maximum of the rate of change

it does seem like near the left side, the rate of increase starts increasing

then it maximizes around x=0

then it starts to slow down

if we increased the graph in either direction it seems like wed see the function turn around

however we are clearly looking at an interval where the function begins to increase, tops out, then starts to increase much more slowly, nearly stopping

[white isle]

so any small changes of increasing/ decreasing can indicate that graph potentially has an inflection point

[violet flume]

i mean its not really a small change

you just have to know what to look for

[violet flume]

you want to look for places where it seems like the rate of increase of the function tops out and turns around

like here, the function becomes more steeply increasing towards x=0

but after x=0 it looks like its becoming a much more slowly increasing function

this suggests x=0 is an inflection point

[white isle]

would it be bad to visualize it with tangent lines? also are there function in which inflection points don't exist in?

[violet flume]

no, as long as you know what you are looking for

[white isle]

ok, I think I understand, in this case my graph will have an inflection point at (0,0) because the rate of change decreases after passing that point

[violet flume]

yes

[white isle]

also are there function in which inflection points don't exist in?

[violet flume]

yes

not to be crass but you should invent one

can you think of a function where the rate of change only increases?

hint: ||what if it was just a positive constant?||

[white isle]

ohh

[violet flume]

and loosely you should be able to visualize what these look like

i guess really loosely

[white isle]

yup an function like f(x) = x

[violet flume]

yea thats sort of a uhh

i mean thatd be like a dummy example

heres a function with no curvature

y=2 works in the same way

you might consider a less trivial example like x^2

(the rate of change only increases)

and like wise maybe x^2+4, x^4 + x^2,

[white isle]

got, may I ask another question separate to the one above

[violet flume]

yea although its encouraged to start a new channel

i will hang out

[white isle]

I was given a function f(x) = (x+1)^5-5x-2 and it asked whether it was concave up bewteen interval (-1, infinity), I got x = -1, but I confused on how I would draw an line of interval and understand the right hand side of the interval

[violet flume]

understand?

What do you mean in this context

what do you mean understand the right side?

[white isle]

I know x = -1

[violet flume]

can you add more context here, what do you mean you got x=-1

[white isle]

was given a function f(x) = (x+1)^5-5x-2 and it want me to find whether concave up bewteen interval (-1, infinity), to find concavity at the point I learned that we must differentiate the function twice, which I did

[violet flume]

okay, so you differentiated f(x) to get f''(x)

[white isle]

yea, I got x from simplifying f"(x)

[violet flume]

so you mean that you started from f(x), you obtained f''(x), and you found that x=-1 is a solution to f''(x)=0

can you classify the point x = -1 in the context of this discovery? what kind of point is it?

[white isle]

I think it, that this point is potentially a point of inflection, but the question wants us to determine whether it is concave up

[violet flume]

so you found an inflection point

[white isle]

I lost on how to do that

[violet flume]

inflection points are places where the concavity of the function changes

potentially

lemme see theres an image that might be helpful

im gonna toss a few

whats important to know is that the concavity is given by your second derivative

so you can create some equivalent statements, like

if f'' > 0 on an interval, the function is concave up there

the function has positive concavity i guess you could say

just like how we might say, if f' > 0 on some interval, the function is increasing there

all of this to mean, you found the place where the concavity changes, right

you solved f'' = 0 and got x = -1

does f'' change at x=-1 to become positive?

aka, does the function have an inflection point at x=-1, and then become concave up?

@white isle doin okay?

[white isle]

I'm confused

how would we interept the left side and right side of the inflection point

[violet flume]

how do you feel about critical points?

i know this is not what you asked

[white isle]

f'(x)?

[violet flume]

yea like

if i said oh heres some function

find where its increasing

give me all the maximum and minimums using the derivative

find where its decreasing

conceptually do you feel okay with that process? how the derivative helps you to find these intervals and points?

[white isle]

yea

[violet flume]

so, one way to think about all this inflection point stuff, is to just kind of use blinders, and apply all of this intuition again to a new function (the derivative of f)

you feel good with being given f, and using f' to find all that stuff

so now pretend f' is just some g, and we want to look at g'

[violet flume]

so we interpret it just like with f'. if f' was 0 somewhere, and to the left it was negative, and to the right it was positive, youd have found a critical point

and to the left f would be decreasing, and to the right f would be increasing

you just change the words around

if f'' was 0 somewhere, and to the left it was negative, and to the right it was positive, youd have found an inflection point

and to the left f would be concave down, and to the right f would be concave up

idk if this is helpful i promise im trying lol

[white isle]

I understand what you are saying, but what confuses me is mainly the definitions, I guess, for example f'' > 0 on an interval, the function is concave up what am I suppose to do with my x value?, once I get f'(x) or f''(x) and the values of x, I don't really understand what i suppose to do with the values of x that I get from solving these functions

[violet flume]

well its like this

you have f'', so you can answer questions about concavity

they ask you if f is concave up somewhere

so all you need to answer is if f'' > 0 there

these are equivalent

presumably you found all the places where f'' = 0

its on the edge of the interval

so all that remains to check is that f'' > 0 somewhere on our interval

[white isle]

so we plug in any number into our f"(x) function to see whether positive or not

[violet flume]

yea, which works here because there are no 0s on our interval, right?

you have this nice, continuous f''

it has no 0's on our interval

so if its positive somewhere, it has to be positive everywhere

because otherwise itd have to cross through 0 to become negative

it doesnt matter what theorem it is

intermediate

intermediate value theorem

[white isle]

oh got it, if postive then concave up, if negative concave down. for critcal numbers they just local minimum and maximums, but what happens when f'(x) = DNE, what does that tell use about f"(x)?

I am butchering everything you said sorry about that

[violet flume]

hmm can you think of a f which is defined at some x=a but f'(a) is not defined but f'' is defined

[violet flume]

mostly f' tells you about f'' in the same way that f informs you about f'

that is, maybe, its more helpful to think of them in the other way around, if youre hunting for intuition

higher derivatives give you information about lower derivatives, not the other way around

idk if that makes sense, but thats how id think about it

[violet flume]

critical points of the derivative of the derivative are inflection points

critical points are to f' as inflection points are to f''

and if f' being DNE somewhere means f'' is DNE there i think it could be an inflection point too

but just by definition

see i think like

1/x is maybe an example

f'(0) is not defined

neither is f''(0)

and in fact the curvature of this function never changes

youd call x=0 a critical point because f' does not exist there

if i remember rightly

but the curvature doesnt change, so its not an inflection point

and actually i think that a point where f'' is not defined is not an inflection point even if the concavity changes sign there

these are pathological cases though

or maybe better say they arent interesting because they are either defined to be or not be

sorry im babbling

[white isle]

ok, i got it

[violet flume]

the curvature can change over a point where f or f' or f'' is not defined

or it can not change

its not enough information

f(x) = 1/x^2 has a change in curvature at x=0

f(x) = 1/x doesnt

just depends

[white isle]

thanks for your help, it has been a pleasure

[violet flume]

hope it was helpful

have a good one

[white isle]

you too

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[cedar kiln]

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[lunar radish]

Practicing for extracurricular math test, these are the questions i'm completely stuck on and made no progress

1. Let a be a number such that a^4 + a^3 + a^2 + a + 1 = 0. find a^2000 + a^2010 + 1
2. Polynomial functions satisfy f(y)=3y^2 - y + 1 and F(G(y))= 12y^4 - 62y^2 + 81

what are all possible values for the sum of the coefficients of G(y)

3. Cuz i don't understand the symbols. Define f : {x|x<=0, xER} => R with f(x)=|x| and find inverse

4. Prove or disprove that the graphs y=f(x) and x=f(y) must intersect

5. For what values of a is the function f(x) = x/(x-a) its own inverse

[violet flume]

on 3, the problem is saying: Define a function f, which maps non-positive real numbers to the reals, by f(x) = |x|

[lunar radish]

thx

anything else?

[violet flume]

you should be able to figure out 4 by just guessing

[lunar radish]

how

[violet flume]

by guessing, guess an example that disproves it

what functions have you tried

[lunar radish]

well none really

hm

i can't figure out a counter example

[violet flume]

you havent tried enough

lets start with a constant

x=2 y=2

whats the problem

[lunar radish]

any value would work?

wouldn't it just become two equations of 2=f(2)

[violet flume]

no im asking

like

lets design one

im telling you its false

they say x=f(y) and y=f(x) always have to intersect

its not true

whats the problem with f=2

[lunar radish]

i know

[violet flume]

so x=2 and y=2

this is just a preliminary guess, its the easiest guess

if it doesnt work, maybe we can qualify the reason why it doesnt work, and make a better guess

[lunar radish]

why doesn't it work

[violet flume]

something slightly less easy but better informed

right, why doesnt it work

what does the graph of these two functions look like

what quality of the graphs make it not work

[lunar radish]

wouldn't x=2 and y=2 just give y=f(2) and x=f(2)

so y=x

[violet flume]

x=2 is just a line

[lunar radish]

i'm kinda confused

[violet flume]

its all the places where x is 2

[lunar radish]

right

[violet flume]

so any y value, but x is 2

so, x=2 is a vertical line, its straight

y=2 is the other way, all x, but anywhere y =2

[lunar radish]

ad y=2 is horizontal

[violet flume]

so its horizontal

yea, exactly

[lunar radish]

but they intersect

[violet flume]

whats the problem here

yea

at (2,2)

but i mean

what do you expect, theyre straight, right

[lunar radish]

yeah

[violet flume]

if they curved

maybe we could get lucky

[lunar radish]

yeah

[violet flume]

so, really, we want a function that kind of bends up and away

can you think of a function like that

[lunar radish]

sqrt

[violet flume]

something that maybe sort of bottoms out, and goes up towards infinity on each side

okay, we can try that

[lunar radish]

oh

x^2

[violet flume]

sure

its better, but its still not perfect

,w plot y=x^2 and x=y^2

[wraith dagger]

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[lunar radish]

+5

[violet flume]

,w plot y= x^2 + 5 and x=y^2 + 5

[wraith dagger]

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[violet flume]

[lunar radish]

yay

OH

WAIT

i'm kinda dumb

they're basically two things

that are

like

perpendicular

i just can't visualize it

[violet flume]

i get what you mean

this x=f(y) and y=f(x) thing flips the functions over the line y=x

[lunar radish]

yeah

[violet flume]

so, if you can find a function that doesnt cross that line

the functions wont intersect, either

[lunar radish]

can yo uwork me through the polynomial one

Let a be a number such that a^4 + a^3 + a^2 + a + 1 = 0. find a^2000 + a^2010 + 1

i feel like i just missed a concept

not supposed to be a hard question

[violet flume]

idk this is like challenge math

not my area of expertise

[lunar radish]

k it's fine

[violet flume]

for 5, im sure theres some way to guess

but, you could just use the definition of inverse

you want f is its own inverse

so f(f(x)) = x

its not fun, but this will get you the answer

[lunar radish]

i got x^2-ax^2+a^2x=x from that 💀

[violet flume]

close

wait

yea thats fine

[lunar radish]

wait i didn't do it right i think

[violet flume]

nah, this seems right

assume x != 0

so x - ax + a^2 - 1 = 0

then uhh

[lunar radish]

yeah

i'm screwed

[violet flume]

we want this to be true regardless of x value right

so i think we can just match up coefficients

write it x(1-a) = 0 and a^2 -1 = 0

require both equations be true

this gives 1-a=0 and a^2 = 1

(a = 1)

if we did good, the function should be symmetric about y=x

(because f(y) must equal f(x))

[lunar radish]

yup

desmos confirms

[violet flume]

you can solve 3

i think thats all i can help with at the moment

[lunar radish]

wait

3 is just

[violet flume]

3 says, you have a function that takes in a non-positive number and returns its magnitude

[lunar radish]

f-1(x)=-x where x>=0 right?

[violet flume]

so given the magnitude of a number, what function maps the magnitude back to the corresponding nonpositive number?

[lunar radish]

negative

[violet flume]

yes

its just negative

this is actually part of the definition of |x|

[lunar radish]

but the domain has to be a certain thing right?

[violet flume]

well, you can see it there

[violet flume]

right

well, you can see this too

you know for an inverse to exist, what youre really looking for is that like

say you write f(x) = y

you want that f(y) = x is also a function

[lunar radish]

range is domain

domain is range

[violet flume]

so if you look at abs value

,w graph |x|

[wraith dagger]

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[lunar radish]

yeah okay

[violet flume]

if you rotate this thing 90 degrees

[lunar radish]

i gtg

[violet flume]

it fails the vertical line test

[lunar radish]

sorry

[violet flume]

not a function

[lunar radish]

yeah

[violet flume]

youre good

[lunar radish]

but the range

[violet flume]

have a good one

[violet flume]

if we just take half the pointy function

its just a line

it does fine

[cedar kiln]

@lunar radish Has your question been resolved?

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[somber flax]

For this problem, I use the tan sum identity but i am confused because tan(pi/2) is undefined. I am not sure how to work out (tan(theta)+tan(pi/2)) / ( 1 - tan(theta)tan(pi/2). This is the tan sum identity after everything was plugged in. can i have some direction?

[rain drift]

I would do it in terms of sin and cos

$tan \left(\theta + \frac{\pi}{2}\right) = \frac{sin(\theta + \frac{\pi}{2} \right)}{cos(\theta + \frac{\pi}{2}\right)}$

[wraith dagger]

MellowDramaLlama

$tan \left(\theta + \frac{\pi}{2}\right) = \frac{sin(\theta + \frac{\pi}{2} \right)}{cos(\theta + \frac{\pi}{2}\right)}$
```Compilation error:```! Missing \endgroup inserted.
<inserted text> 
                \endgroup 
l.57 ...right)}{cos(\theta + \frac{\pi}{2}\right)}
                                                  $
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.```

[somber flax]

how would tan(theta + pi/2) be equivalent to sin(theta + pi/2), can you explain that, i dont understand?

[rain drift]

its not hold on I'm getting syntax errors

[somber flax]

oh i see

[rain drift]

it's this

x = theta, alpha = pi/2

[somber flax]

oh okay i will try that

thank you

[rain drift]

lemme know if you get stuck

[somber flax]

no that worked out perfectly, i didnt ever think to use the quotient identity in this problem. thanks a lot!

.close

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

i need help with factorising using common factors

[fair mortar]

please post a specific postition you need help with

[crimson sedge]

@crimson sedge ^

.close

[cedar kiln]

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[tired inlet]

Could someone help me out?

[tired inlet]

How do I write using the reciprocal identities?

I know sec(x) = 1/cos(x)

[livid hound]

apply that

[tired inlet]

Although does the 3 go to the numerator or denominator?

[livid hound]

in combo with exponent laws

stuff gets multiplied to numerator

$a\times \frac bc = \frac{ab}{c}$

[wraith dagger]

ℝαμΩℕωⅤ

[tired inlet]

Here is the full question

Thought I could do it, but really confused on how to integrate it.

[median holly]

Sec²x

Is direct integration

[tired inlet]

Oh true

my bad I forgot

[livid hound]

this did not require you to express sec in terms of cos

[tired inlet]

Yeah, I was just overthinking it

[livid hound]

its recommended that you ALWAYS post the original at the very start

[tired inlet]

Will do 👍

Still getting a different answer when I enter it on my calculator

Is this the correct way to write it?

<@&286206848099549185>

[livid hound]

poor notation, that may be interpreted differently be certain calculoatrs

if the calc doesn't recognise cos^2(x),
use more (): (cos(x))^2

is this purely for checking an answer you got?

[tired inlet]

Yeah

[livid hound]

show what you're getting

[tired inlet]

I'm getting 0.87189

[livid hound]

and a pic of what the calc is pumping out

[tepid sail]

,w integrate from π/6 to π/4 \frac{3}{cos(x)^2}

[wraith dagger]

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[tired inlet]

When manually integrating I get this

[tired inlet]

But when I check my calculator I get this

[tepid sail]

what are you entering into your calculator

[tired inlet]

^

[tepid sail]

no like

exactly

[livid hound]

poor notation, that may be interpreted differently be certain calculators

[tepid sail]

yeah that's fair

I kinda assumed they were inputting (cos(x))^2

[livid hound]

seems like it might be computing
$$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{3}{\cos(x^2)} \dd{x}$$

[wraith dagger]

ℝαμΩℕωⅤ

[livid hound]

,w integerate 3/cos(x^2) from pi/6 to pi/4

[wraith dagger]

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[tepid sail]

yeah

[tired inlet]

Isn't that correct?

[tepid sail]

no

[livid hound]

no

[tepid sail]

they're very different

[tired inlet]

What is wrong with it?

[livid hound]

square in the wrong place

[tepid sail]

you are squaring cos(x)

not x

[livid hound]

$\cos^2(x) = \cos(x)\cdot \cos(x) = (\cos(x))^2 \redneq \cos(x \cdot x)$

[wraith dagger]

ℝαμΩℕωⅤ

[tired inlet]

Oh yeah I see now

[cedar kiln]

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[cedar kiln]

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[cursive thicket]

i dont get it

im trying to calculate for question 16

but its telling me that justin has to lower his mark to be in a higher percentile ???

like what

please help

[gusty forum]

you haven’t given any information that lets us help you other than something’s clearly gone wrong

[cursive thicket]

im good now

[tepid sail]

use .close pls

[vital reef]

.close

[cedar kiln]

@cursive thicket Has your question been resolved?

[cursive thicket]

.close

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[cursive valve]

if this grid represents that fraction (4 column x 3 rows) what number equals to that diagonal line ?

[cursive valve]

Also how would one define math ? What is it ?

[royal loom]

that diagonal line divides the grid in half, maybe you'd call it 2/3

however this doesn't really have an answer

[cursive valve]

It does

[royal loom]

and neither does your second question

[cursive valve]

I may be missing the context

[royal loom]

you are

[cursive valve]

Okay i give you a time stamp

https://www.youtube.com/watch?v=ZQElzjCsl9o 8:02

Help me understand what he said.

[royal loom]

he is asking about the length of the line

it is a triangle

and a special type of triangle

[cursive valve]

Equilateral ?

[royal loom]

where you know the base lengths are 3 and 4

so the length of the hypotenuse must be 5

[cursive valve]

Oh thank you.

I confused it with matrix.

[cursive valve]

Why ?

There is no definition for math ?

[royal loom]

it is just a word

use a dictionary if you want I guess

,w math

[wraith dagger]

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[cursive valve]

Then what does mathematics even mean ?

[royal loom]

the study of math

the subject of math

[cursive valve]

Study of a word ? -.-

[royal loom]

this is not a specific math question, if you have a problem you need help with, post it, otherwise close this channel as this is not the point of them

[cursive valve]

Where should i ask that question ?

In what channel ?

Before i close this.

[royal loom]

#discussion or #math-discussion or #chill

[cursive valve]

Sure.

Thanks for the help.

!close

.close

[cedar kiln]

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[weak haven]

What’s the difference between these two rules

And when do we use each?

One of them

[stone flame]

they are the same

tge second is just the first but square rooting both sides

[weak haven]

Ohhh

Alright alright

[stone flame]

the second is more useful if you are looking for one specific side, while the first is better for finding the angle A or other sides if thats what you meant

[weak haven]

But they both could be used

Right

[stone flame]

but they are essentially the same

[weak haven]

Alright thank you love!!!

.close

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[burnt hamlet]

can some help me solve this

.close

[cedar kiln]

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[modest lily]

Hi I have a question it was e^(2ln(y)) = y +12 and I got two answer y=4 and -3 but I am confused whether -3 is a correct solution as in this form its a math error but can't it be rewritten as (-3)^2 because e and ln cancel

[tepid sail]

so y^2 - y - 12 = 0

(y-4)(y+3)

y = 4 and y = -3

but then

recognize

ln(-3) IS NOT DEFINED

so y = -3 is not a solution

[modest lily]

yeah i was just wondering about whether the form would make it (-3)^2 but it doesn't, thanks for the clarification

[tepid sail]

yep

.close when ur done

[modest lily]

.close

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[rugged tusk]

I have two pouches, each with five seemingly identical coins, but we know that in the first pouch, there is one fake coin, and in the second pouch, there are two fake coins. We randomly drew one coin from each pouch, and then from this pair, we randomly selected one about which we confirmed that it is fake. What is the probability that the second selected coin is also fake?
Solution: 4/15

[rugged tusk]

can somebody help me with that?

i dont get how they got this solution

[cedar kiln]

@rugged tusk Has your question been resolved?

[rugged tusk]

Let F1 denotes that 1 coin is false and F2 denotes that two coins are false.
We want to find out $P(F_2 \vert F_1)$.

From the Bayes theorem we can get $P(F_2 \vert F_1) = \frac{P(F_1 \vert F_2)P(F_2)}{P(F_1)}$

[fair geyser]

it's not "one coin is false"

we also randomly pick it out from two

[rugged tusk]

then i dont know how to move forward

[wraith dagger]

Michal

[fair geyser]

in this case and in majority of cases i've seen bayes theorem doesn't help you

you have enough information to do this

bayes rule pushes you back one step

there are three ways you end up with a false coin picked:
it comes from the first pouch: (1/5)(3/5)(1/2)
it comes from the second pouch: (4/5)(2/5)(1/2)
or they are both false: (1/5)(2/5)
added together, they give P(B)

[rugged tusk]

if P(A) means that i picked 1 false coin, what is P(B)?

and how to calculate the intersection? if A and B are independent we can calculate it as a product of P(A) and P(B), otherwise bayes...

[fair geyser]

yeah i can see that, I'm implicitly multiplying (1/5)(2/5) by 1 to get P(A & B)

thanks

[rugged tusk]

$P(A\vert B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)P(B)}{P(B)} = P(A)$ ?

[wraith dagger]

Michal

[rugged tusk]

then why we calculated P(B)

[fair geyser]

maybe i'm not doing that

[rugged tusk]

Let F1 denotes that 1 coin is false and F2 denotes that two coins are false.
We want to find out $P(F_2 \vert F_1)$.

From the Bayes theorem we can get $P(F_2 \vert F_1) = \frac{P(F_1 \vert F_2)P(F_2)}{P(F_1)} = \frac{1 \cdot \frac{2}{25}}{P(F_1)} $

[wraith dagger]

Michal

[rugged tusk]

i think that this was pretty good start, now i need to calculate only the denominator

$P(F_2 \vert F_1) = \frac{P(F_1 \vert F_2)P(F_2)}{P(F_1)} = \frac{1 \cdot \frac{2}{25} }{ 1-\frac{2}{25} - \frac{12}{25} } = \frac{2}{11}$

[wraith dagger]

Michal

[rugged tusk]

hmm, its still incorrect

[somber lantern]

First, your $P(F_1)$ is incorrect. It should simply be 1 - 12/25 = 13/25.
Second, $P(F_2 | F_1)$ is not the probability you are looking for. You need to account for the probability that you pick a false coin. ie. you are twice as likely to pick a false coin from two false coins than from a false and true coin.

[wraith dagger]

chencking

[rugged tusk]

So I don't get it how to calculate that

[somber lantern]

Try making $F_1$ the outcome that you pick a false coin from your two coins and $F_2$ the outcome that both coins were false. You want to find $P(F_2 | F_1)$.

[wraith dagger]

chencking

[fair geyser]

what

how is that different

that's what it was from the start

[somber lantern]

Yes. Those are the outcomes the problem gave in the start.
He mistakenly used incorrect outcomes. That is why he got the wrong answer.

He was calculating the wrong probability.

[fair geyser]

i copypasted (1/5)(3/5)(1/2) twice in my solution

sorry

(1/5)(2/5) + (1/5)(3/5)(1/2) + (4/5)(2/5)(1/2) is the denominator, 1 − 3/50 − 4/25− 12/25 if you want to subtract instead

[cedar kiln]

@rugged tusk Has your question been resolved?

[cedar kiln]

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[small cloak]

how do i find the critical point of this

its gonna be -(t-5) and t-5

i find the derivative of both

gonna be -1 and 1

-1 & 1 = 0 is underfined

now what

[crystal raptor]

What are the instructions of the question

[small cloak]

find the critical point

[crystal raptor]

What is your definition of crticial point

[small cloak]

idk the definition

[spice kraken]

critical point c means f'(c)=0 or f'(c) is not defined

[small cloak]

i just know thats the value of x

after finiding the derivative

[small cloak]

so 0 is the critical point?

the answer key says 5 tho

[spice kraken]

does f'(0)=0?

also 0 is not even in the domain

we want to find f'(x)=0

[steep umbra]

we need to find a point where the f'(x) is not defined

means it is not differentiable

[small cloak]

oh so the critical point is what value we substitue is f`(x) to get 0 or underifned?

[steep umbra]

yea

if you see the normal |x| function

it is not differentiable at x = 0

so we need to make t - 5 = 0

do u get it?

[small cloak]

i think so , thanks

another question , please

[steep umbra]

sure if i can help

[small cloak]

48

i got this

so only critical point is 0, right?

because if we substitute 0 , we get undefined

[steep umbra]

i dont think this is correct

[small cloak]

what is it

[steep umbra]

wait let me get a pen and paper and solve on my own

[small cloak]

the derivative is correct am sure

no need

[steep umbra]

ok

so we need to make g'(x) = 0

so the numerator must equal 0

1 - x = 0

[small cloak]

wait fr?

so in rational functions

like we dont just set the whole equation to 0

actually ill get 1-x=0

[steep umbra]

yea

[small cloak]

but still isnt 0 a critical point

[steep umbra]

you just take the denominator to the other side

[steep umbra]

not always

but here it is

since if the denomiator is 0 g'(x) becomes undefined

[small cloak]

yes

what are domain endpoints?

idk whats that

[steep umbra]

can you find the domain of thr given function?

[small cloak]

i forgot how to get domain 😅

[steep umbra]

its a sqaure root func

so whatever is inside of it

cant be negative

2x - x^2 > 0

whatever x satisfy this

are in the domain of the func

can you solve this now?

[small cloak]

x= 2

[steep umbra]

huh?

That just gives you 0

[small cloak]

how come

2x-x^2 = 0

-x^2 = -2x

divide by x

x= 2

[steep umbra]

yea you set it equal to 0

we need to make it 0 ot greater than 0

when will it be greater than 0?

[small cloak]

when its = to 3?

wait no i mean 2 and less

because x< =2

we change the direction when negating

look

[steep umbra]

yea

wait

[small cloak]

so endpoints is finding the domain of the denominator

[steep umbra]

no the whole function i think

[small cloak]

but according to the solution

its only denominator?

[steep umbra]

nope

Idk how they got only 1 as the critical point tho

0 should also be a critical point i think

[small cloak]

i didnt understand this

[steep umbra]

i just solved the inequation

the domain is [0,2]

so 0 and 2 are included in the domain

since root0 is defined

[steep umbra]

i think this is called 'the wavy curves' method of solving inequalities

[small cloak]

i have more questions can i send 😅

[steep umbra]

umm how many?

[small cloak]

2

[steep umbra]

okay

[small cloak]

i read this 1000x times

[steep umbra]

plenty of people here to help you here even if i go

[small cloak]

i didnt understand how it only have 1 real solution

[steep umbra]

Do you get the part where it says "it crosses the x-axis"

yes i get it i can explain

But you need to ask specific ques

see if it even had 2 real solutions

if would mean that f(x) is zero at 2 points

and if in a continuous function there exists 2 points a and b where the function gives out the same val

there must exists a c between a and b such that f'(c) is zero

[cedar kiln]

@small cloak Has your question been resolved?

[cedar kiln]

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[chrome turtle]

i have done this with arg properties

It took some calculations

but in the solution they did something with geometry i dont get the solution

How do they know z is on the arc

I dont know what arg(z1/z2) actually means graphically

wait actually i know what arg(z1/z2) means its just the difference of angles but in here something different happening i think

[neat dune]

(z + beta) - beta = z, as well as (z + alpha) - alpha = z. In other words, those triangle side lengths represent z + beta and z + alpha

[chrome turtle]

uh lemme get this hold on

i dont understand how we can conclude its the side length

[neat dune]

It's like vector addition yknow

[chrome turtle]

Which side we taking about

other than the hypotenuse?

The other two

[neat dune]

The ones that have the angle pi/4 in between

[chrome turtle]

oh

ok if its like vector addition

Shouldn't it be z + (the side length) = alpha

[neat dune]

There it'd be z - (z + alpha) = -alpha

[chrome turtle]

oh

I am messing up the directions

how do you know its that direct

like the way they point out

i understand how you pointed them

But i dont know why that

[neat dune]

So if you have a vector PQ then that represents a vector in the direction from point P to Q right? so if z represents point P, and Q represents point -alpha, then PQ = (z - (-alpha)) = z + alpha represents the vector pointing from z to negative alpha

[chrome turtle]

bro im ashamed of myself

just when you wrote i noticed

[neat dune]

it's aight

[chrome turtle]

thanks

[cedar kiln]

@chrome turtle Has your question been resolved?

[cedar kiln]

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[crimson sedge]

Pls give reasoning for q1 part (a)?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[chrome turtle]

Sorry for this, it just hit me now u meant here the qp vector is z + alpha right , cause the pq vector is -alpha - z

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[summer pebble]

Umbrella Corporation hired 87 employees, which increased its stff by 40%. What was the original number of employees? How many do they have now?

[winter harness]

Set up an equation

One side is 87 employees

Other side is = to a relationship and 140%

[summer pebble]

wdym relationship?

[north harbor]

it is 87 = 40% from original number or smth

[summer pebble]

so 1.4x=87?

and then x=62.142 (63)

[north harbor]

Do you round the numbers?

[summer pebble]

you cant have 61.14 people

you cant have a fraction of a person

[north harbor]

ye, the number is odd

[summer pebble]

's what makes me think im wrong with this answer

but quora says the same equation so 🤷‍♂️ thank you!

.close

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[wicked badge]

how do I convert this into cosine or sin?

I know how to work with the sum difference identities but I've only done it with sin and cos values

[upper abyss]

You don't have a sum/difference identity for tan? There's a popular one

[winter harness]

Tan is sin/cos

[wicked badge]

yes so is it sin 19pi12/cos 19pi/12?

[upper abyss]

But indeed if you don't want to use it, it's enough to get sin(19π/12) and cos(19π/12), then divide them.

[wicked badge]

ah okay

maybe I'll look at the tan one

I think that'll be easier

.close

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[polar lintel]

anyone know why its wrong?

[polar lintel]

ive tried 20k, 19k, and none worked

<@&286206848099549185>

[safe ermine]

you can apply the limit t -> infinity

on a calculator

OR

[supple flume]

Im guessing its something like 19999 💀

[polar lintel]

how can i do that? sorry not too good with this calculator

its ti-84+

[safe ermine]

i don’t have that model

[polar lintel]

hmm

mine doesnt include an infinity symbol

[safe ermine]

you can just do a really big number

if you don’t have infinity

[supple flume]

If you make t go to infinity the equation become 9450/(0.5 + 0(because it gets smaller)) if im correct

[polar lintel]

yea i did x at 499,999 and it came out to 18900 or so

[supple flume]

Just think of what happens to the number as it gets bigger

The number you add to the bottom gets way smaller

[polar lintel]

ye

[supple flume]

Cus it 1/26.5e^t

[polar lintel]

so would it approach 0

that wouldnt make sense though

[supple flume]

Which if t = infinity 1/infinity is 0

Well not true

But u get my point

As it approaches

[polar lintel]

i feel like there is a method to finding this algebraically no?

by solving for it

[supple flume]

I did it in my head by just assuming t is an infinitly big number

[polar lintel]

okay but t is the denominator

not the numerator rn

[supple flume]

Ye

So it gets smaller

To 0

Then u just solve 9450/0.5

[polar lintel]

so the answer would be 18900?

since you canceled out anything connected to t as 0

[polar lintel]

im braindead

he sent the answer here

[safe ermine]

yeah 💀

[polar lintel]

thank u 🙏

[safe ermine]

:3

[polar lintel]

.close

[cedar kiln]

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[heady sierra]

how do you find the closest approach of two particles A and B?
my notes say that rB|A(t) * vB|A = 0 but we don't have a velocity for the two vectors im trying to find
"The position vector of particles A and B, t hours after 12 noon, are r = 12i + 3j + t(3i + 4j) and r = -3i - 5j + t(2i +6j) metres respectively. c) FInd when A is closest to B and find this distance."
Already found the distance in terms of t and when A and B are 18 m apart
it says in the answers that |AB| is minimised when t = (-2)/2(5)
but i have no idea how that works
i tried solving for the vB|A using the parts that are t(3i + 4j) and the other one in the position vectors and solve that against the vB|A(t) but that didnt work either

[cedar kiln]

@heady sierra Has your question been resolved?

[heady sierra]

<@&286206848099549185>

[crimson sedge]

..?

[crimson sedge]

this is the question?

[heady sierra]

yes

i want to find the distance/time when they are the closest

[crimson sedge]

u used vector subtraction yet?

[novel topaz]

u can simply use the distance formula here

then u get the distance as f(t)

from there u could get the answer

like [12+3t-(-3+2t)]^2+[3+4t-(-5+6t)]^2=distance^2

d^2=5t^2 - 62t +289

???

@heady sierra

[heady sierra]

yes

just a sec

so i did that

[heady sierra]

t = -2/2(5)

which i dont get

[novel topaz]

t=-1/5?

[heady sierra]

yes

should i send a photo of the answeers

[novel topaz]

how did u get distance 18

[novel topaz]

yes

[heady sierra]

they show full working

ok 2 secs

[novel topaz]

my distance is like 22/sqrt5

[heady sierra]

part c

[novel topaz]

ohh

do u know how to get the minimum value of a quadratic function

[heady sierra]

I know using differential calc with methods but I don’t think we’re expected to know dif calc for this

[novel topaz]

right

[heady sierra]

Bc this vectors I learnt before dif calc

Is there any other way

[novel topaz]

well first we would right the equation as a square of something

[heady sierra]

Would dif calc work too just to know for later

[novel topaz]

yes

[heady sierra]

Ok nice

How’d they do it though

[novel topaz]

ok 5t^2-2t + 289 can be written as 5(t-1/5)^2 + 289 - 1/5

[heady sierra]

Yep

They just seemed to pull t out of thin air bc they usually show all equations

[novel topaz]

they skipped the steps

or they directly used t=b/2a

[heady sierra]

I can do that?!

What formula is that

It’s familiar but I need the name

[novel topaz]

i dont know its name but to get the minimum value of a quadratic equation , u just have to do x=b/2a where a is positive

[heady sierra]

I’m so stupid

Bro I’m gonna fail this exam

Thank u so much though

.close

[cedar kiln]

Channel closed

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[cedar kiln]

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[versed fulcrum]

Hello

[versed fulcrum]

Not sure how to solve this

[dusty hazel]

I'll just get an image lol.

[tepid sail]

well use the fundamental thm of calculus

[dusty hazel]

yeah.

[tepid sail]

mhm

[cedar kiln]

@versed fulcrum Has your question been resolved?

[crimson sedge]

you can use leibnitz rule

[wraith dagger]

notamy

[crimson sedge]

ah sorry, didn't see that you've already sent it

[cedar kiln]

Channel closed

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[versed fulcrum]

.reopen

[cedar kiln]

✅

[versed fulcrum]

Ohh I seee

thanks

.close

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
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[minor crystal]

https://cdn.discordapp.com/attachments/1081837589780234241/1173166722891653210/image.png?ex=6562f7a7&is=655082a7&hm=335b46731bb644ac148388708b891b5f24d0868f0e21c01082605009c7aedb2e&

[minor crystal]

1/2 < 1/x + 1/y < 1/3?

do i just have to use process fo elimination on this?

[lunar lynx]

Yeah. Just checking for each option is the simplest way.

Doing analysis for the question is probably an overkill.

[minor crystal]

but my inequality is wrong right

since when was 1/2 < 1/3 lol

[lunar lynx]

Yeah. Lol

[minor crystal]

the teacher wrote that

no the other way doesn't conform the question's specification?

lol

i write it as two i guess?

[lunar lynx]

Actually it's two separate inequalities.

[minor crystal]

yeah so i think there's something i can exploit

[lunar lynx]

1/x + 1/y < 1/3

1/x + 1/y > 1/2

[minor crystal]

wait that doesn't make sense again

lmao

1/x + 1/y is less than 1/3

but is more than 1/2???

[lunar lynx]

If it satisfies anyone, it is in solution set again.

[lunar lynx]

There is an "or" in the question.

It needs to satisfy any one of them.

Since it can't satisfy both.