help-27

[devout snow]

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[tough pollen]

find positive intergers x,y,z satisfy:

[devout snow]

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[viral sonnet]

Hello

[viral sonnet]

Merp

3x + 10 = -(1/3)x + 9
3x + (-(1/3)x) + 10 = 9
(10/3)x = 9 - 10
(10/3)x = -1
10x = -1(3)
x = -1(3)/10
x = -3/10
x= -0.3

[viral sonnet]

WHY? to find intersection point between two straight line must use this?

[sweet ridge]

You need to check if ur solution is correct?

[viral sonnet]

well then...

is this correct?

[winter patrol]

sign issue in the first step

[viral sonnet]

so
y = -(1/3)(3/10) + 9
y = -(1/~3~)(-(~3~/10)) + 9
y = 1/10 + 9
y = 0.1 + 9
y = 9.1

[winter patrol]

that - shouldn't be there,
the line that followed ignored it regardless
end result is fine

[viral sonnet]

dun

now

WHY we must use

linear function = linear function

to know where the point of intersection of these two lines?

please ping

must to!

[frigid jetty]

Lol, if they are equal means they have somewhere intersection means it's obvious to do innit

Idk

[viral sonnet]

yes, but WHY?

[heady plinth]

Okay

Let’s say you have y=3x+2

What we really have here is a collection of points (x,y) that satisfy the above equation

So if you have a different equation, let’s say y=2x-1

And you want to find the intersection

The intersection point (a,b) will by definition make both these equations true

So then 3a+2=b=2a-1

And you can solve that for a

Makes sense?

[viral sonnet]

what happen if the lines never crossed?
(of course without it be draw on)

will still can? or what?

[heady plinth]

Are you asking what happened algebraically or what

It only happens when they have the same slope

So let’s say y=3x+1 and y=3x+5

Then we try 3x+1=3x+5

But that leads us to 1=5

Womp womp

[viral sonnet]

so it doesnt make anything to x and y ??

[heady plinth]

There are no such points (x,y) that satisfy both equations simultaneously

I don’t know what you’re asking lol

[viral sonnet]

sorry for my uncapability of writing logic english, because this is not my main language yet im using it to learn the worst lesson yet fun

[viral sonnet]

instruction unclear, yet understanable

[heady plinth]

Русский?

[viral sonnet]

My country have good relationship with Russia tho

[heady plinth]

China?

[viral sonnet]

more south

[heady plinth]

Hmm

Vietnam?

[viral sonnet]

more south

[heady plinth]

Not India right

[viral sonnet]

south not west XD

[heady plinth]

Thailand?

I give up

[viral sonnet]

two more

[restive river]

what was the question?

[heady plinth]

Indonesia

[viral sonnet]

XD

[viral sonnet]

this

[heady plinth]

Do you have any more questions

Does it make sense

[viral sonnet]

🥳

[viral sonnet]

unclear, but understanable
(nope, but i know where it goes)

[restive river]

jujur kgk ngerti soalannya apa

[restive river]

klo udh ngerti close aja kak

[viral sonnet]

🥳

[heady plinth]

How is kgk a word

Crazy

[restive river]

its a simplified word from "kagak". basically informal when writing

[heady plinth]

I see

I’m gonna sleep gl with this

It’s 230 where I’m at

[restive river]

lol 3 30 pm

[viral sonnet]

me never understand the math, that's the fun

[viral sonnet]

opposite

welp

[restive river]

😭 🙏

[viral sonnet]

no sense, but undertanable

See you guys later!

.close

[devout snow]

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[devout snow]

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[restive river]

can i apply x^-1 = ln|x| to term 2 here?

[restive river]

like that?

or does it only work for one term?

[safe knoll]

$$\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| +c $$

[woven radish]

JustToPro

[restive river]

so is that a yesss?

man i sucg at math

[safe knoll]

is derivative of (e^x - 1) present?

f(x) = e^x - 1
f'(x) = e^x

[solar goblet]

you should see what JustToPro sends and determine if your result is correct or not for yourself

[restive river]

no

[restive river]

so like is this the extended version of the formula? the only thing that was taught to us was integral of (1/u)dx = ln|u| + c and to just express everything as u

[safe knoll]

yes cuz derivative of u is 1

[restive river]

oh

[safe knoll]

with respect to u

[restive river]

yeah that actually makes sense 💀

[safe knoll]

also integral of (1/u)du = ln|u|

[restive river]

oh right

yes

god i keep forgetting too many stuff and theres a quiz tomorrow aaa

anyways thank you to you two ill recheck my answerrr 🙏

.close

[devout snow]

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[devout snow]

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[restive river]

Is it right? Have I done any mistake?

[exotic dagger]

cos(x) = 1/2 => pi/3 not pi/6 so should be pi/3 + 2k*pi

[restive river]

Why 1/2 is pi/3?

Isnt it √1/√4?

[exotic dagger]

wdym?

[restive river]

I missunderstood sin with cosine.

[twilit comet]

sqrt 1 / sqrt 4 = 1/2 lmfao

[restive river]

Any other mistake?

X= pi/2 +kpi, pi/3* 2kpi

[exotic dagger]

not that I can see

[restive river]

But the given answear is: 2kpi +- pi/2, 2kpi +- pi/3

Whelp

[exotic dagger]

yeah so cos is an even function, when you get for cos(x) = 0 that x = pi/2 then automatically you know that x = -pi/2 is also a soln. And for cos(x) = 1/2, you get -pi/3 and pi/3. All of these are solns to the eqn between -pi and pi. So you have to add 2k*pi to all of them as cos has a period of 2pi

[restive river]

Any video to understand even or odd functions?

And I said pi/2+kpi, book said 2kpi+ pi/3...

2kpi or just kpi?

[exotic dagger]

I'm sure there's plenty of videos on yt. But in general, odd func satisfies f(-x) = -f(x), e.g. odd degree polynomial and sin since sin(-x) = -sin(x). And even func satisfies f(-x) = f(x), e.g. even degree polynomial and cos since cos(-x) = cos(x). Even func is symmetric in the y-axis and odd is basically the positive part rotated by 180 degree/pi radian about the origin

[restive river]

Any good video about this?

[exotic dagger]

idk, I've never watched any video to learn odd/even

https://www.youtube.com/watch?v=6pCVGSLTzgM&ab_channel=Mario'sMathTutoring

this one seems short but enough to cover all you need to know

[devout snow]

@restive river Has your question been resolved?

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[worldly sonnet]

Whats the easiest method to find the inverse of a 3×3 matrix and a 2×2 matrix

[tight prawn]

augmenting it by the identity and row reducing to get the identity on LHS

[stone stump]

well for 2x2 its easier to remember the formula tbh

[tight prawn]

thats what i was typing

[worldly sonnet]

Nah but when I use that augmented matrix thing

[tight prawn]

i dont wanna do the tex for a bmatrix rn tho xd

[worldly sonnet]

Idk what happens

The left side like I cannot get it to look like Identity

[tight prawn]

then your matrix is not invertible

[worldly sonnet]

Like is there a specific thing like get the lower triangle first

No like it is cuz the question has a solution but like there must be a like way to first get this element to zero then that and low and behold we have identity on the lhs

Uk what I mean

I'll see some YouTube videos

Thanks

.close

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[devout snow]

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[dusty mesa]

Hi. I think i messed up somewhere. Can someone show me where exactly?

[exotic dagger]

I don't see any mistake so far

[dusty mesa]

Then how do i continue ah

[valid quest]

Here is the answer

[dusty mesa]

I see

[exotic dagger]

you can simplify further by cancelling the $x \sqrt{y}$ in the denominator and cancel a factor of (x-y) in the first fraction, see if that helps

[woven radish]

lgkoo

[dusty mesa]

Okay wait

Is this overall correct?

[exotic dagger]

you tried to cancel a factor of (x-y) in the numerator and denominator but you forgot to do so with -y^2+xy

[dusty mesa]

Wait

what do i do with that

[exotic dagger]

well -y^2+xy = y(x-y) right

if you try to cancel a factor of (x-y) in the top and bottom of fraction, you also need to cancel that too

so $\frac{2\sqrt{x}\sqrt{y}(x-y)+y(x-y)}{\sqrt{x}(x-y)} = \frac{2\sqrt{x}\sqrt{y}+y}{\sqrt{x}}$

[dusty mesa]

That is possible?

[woven radish]

lgkoo

[exotic dagger]

well yeah that's in fact what you have to do

[dusty mesa]

Okay i get it 🤧, thank you for being patient 🙏🏻

[exotic dagger]

np

[dusty mesa]

.close

[devout snow]

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[devout snow]

Available help channel!

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Remember:
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[foggy fossil]

can someone explain me the implication ? i dont get it
If there is an example that implies a sentence, then every example would imply that sentence.

[calm spruce]

hey i am new is there anyone of class 7

i need help

[sterile geyser]

create a new help channel please

[calm spruce]

can u tell me how to create

[fossil locust]

Go to say #help-38

Post your question there

I put help-38 cause it's under this section

[calm spruce]

thanks

[fossil locust]

np

[devout snow]

@foggy fossil Has your question been resolved?

[devout snow]

@foggy fossil Has your question been resolved?

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[umbral cradle]

I'm solving 25b

What i did was

[fossil locust]

You can ofc find the two roots of the equation and solve for x

Yeah so what is the distance of the road inside the circle, in terms of the two roots?

[umbral cradle]

ohhh

I solved it

thanks 🙂

.close

[devout snow]

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[fossil locust]

No worries

Also yeah the square roots cancel out

Npnp

[devout snow]

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[torpid plume]

hi can someone help me with this:

[torpid plume]

i asked about this before and 4got to solve it, i tried again by following the preivous suggestions, but im stuck

what i tried is, since i know it's arithmetic sequence, i can do something like:
2nd term - 1st term = c
3rd term - 2nd term = c
4th term - 3rd term = c
etc...
but im not getting anywhere with it, like i tried the 4th term - 3rd term:
x/y - xy = c
x(1-y^2) = cy

idk what 2 do next with this

[fossil locust]

Can you write an equation

Yeah okay okay you get the idea then

[fossil locust]

Then say xy - (x - y) = (x - y) - (x + y)

xy - x + y = -2y

So xy - x - y = 0

Now the trick is to add 1 on both sides

You get (x - 1)(y - 1) = 1

So then you know that x/y - xy = (x - y) - (x + y) = -2y

x - xy^2 = -2y^2

[fossil locust]

It's messy but solve for x here and then substitute that into the other equation

[torpid plume]

wait ill try

[torpid plume]

i think this should be xy - x -3y = 0 ?

[fossil locust]

Ah shit, xy - x + 3y = 0

Then yeah (x + 3)(y - 1) = -3

[torpid plume]

ok so u added -3 to both sides 2 reach here right

[fossil locust]

Yes

[devout snow]

@torpid plume Has your question been resolved?

[torpid plume]

im getting cubic equation

[torpid plume]

is this correct

[devout snow]

@torpid plume Has your question been resolved?

[devout snow]

@torpid plume Has your question been resolved?

[devout snow]

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[steady lintel]

how do i factor and find all the zeros for f(x)=x^3+9x^2+23x+15; x+5

[steady lintel]

they gave me one factor

which is x+5

[sterile geyser]

synthetic division

[steady lintel]

o

how do i write the answer again

i got 1,4,3,0

[sterile geyser]

thats impossible

the highest order is 3

there must be 3 roots

one root is -5

have u learnt factor theorem?

[steady lintel]

yeah

[sterile geyser]

so u know why -5 is a root right?

f(-5) = 0

[steady lintel]

isnt this synthetic division

[sterile geyser]

yeah

[sterile geyser]

ohh u got this

for ur synthetic division?

[steady lintel]

mhm

[sterile geyser]

okay write like this

x^2+4x+3

u just reduce by a power

now find the roots of the quadratic

[steady lintel]

how do i do that

[sterile geyser]

$x^3+9x^2+23x+15=(x+5)(x^2+4x+3)$

[woven radish]

axxy

[sterile geyser]

right?

based on the synthetic division

[steady lintel]

mhm

[sterile geyser]

now u want f(x) to be 0

u can either do that when x+5=0, which gives x=-5

or when

$x^2+4x+3$

[woven radish]

axxy

[sterile geyser]

is equal to 0

do u know how to find the roots of that quadratic?

[steady lintel]

not really

[sterile geyser]

have u learnt the quadratic formula

[steady lintel]

ax^2+bx+c=0

[sterile geyser]

and the solution to it?

[steady lintel]

im not sure waht abc are in the equation tho

[sterile geyser]

they are the coefficients

[sterile geyser]

a is 1, b is 4, and c is 3

[steady lintel]

yep

[sterile geyser]

for example

[steady lintel]

so 1x^2+4x+3=0

[sterile geyser]

yess

[steady lintel]

so is x -1

bc 1^2+3= 4

so 4x needs to be -4

[sterile geyser]

why are you doing synthetic division before learning quadratics

have u at least learnt factoring?

its okay if u havent

[steady lintel]

tbh i dont know what im doing 😭

[sterile geyser]

https://www.youtube.com/watch?v=IlNAJl36-10

[steady lintel]

:0

thx

i learnt all this i just have a hard time memorizing it

[sterile geyser]

its fine

did u find the roots of the quadratic?

[steady lintel]

one sec

@sterile geyser

oh wait theres more

[sterile geyser]

that is correct

so x=-3 and x=-1

[steady lintel]

yeah

[sterile geyser]

finally, u can write it like this:
(x+5)(x+1)(x+3)

this is because

say x+1=0
x=-1

which is one of the roots

x+3=0
x=-3

so in general, a third degree polynomial can be written like this:
f(x)=a(x-x1)(x-x2)(x-x3)

where x1, x2, x3 are roots

[steady lintel]

so if it was positive 3 would it be (x-3)

[sterile geyser]

yupss

[steady lintel]

is the 5 from the given factor

[sterile geyser]

yes

[steady lintel]

oke :)

tysm u rly helped a lot

[sterile geyser]

np

type .close once u r done

[steady lintel]

.close

[devout snow]

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[devout snow]

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[stable lantern]

how can I show that this is true?

[dense jay]

you can probably apply the addition formulae for sin

[devout snow]

@stable lantern Has your question been resolved?

[devout snow]

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[lament schooner]

lets say x is infinity

[lament schooner]

and y=(-x)^2

is infinity positive?

im saying in this context

[heavy current]

you can do a change of variables yes... but what does that have to do with infinity here

[lament schooner]

oh forgot aout that

ty

.close

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[high cypress]

is #8 a "many solution" inequality? or no solution?

[restive river]

what do you think

[high cypress]

i think many

but i was sick when my teacher taught this so im just guessing

[restive river]

so what is the usual method u use to solve those inequalities?

[high cypress]

i dont usually do no solution or many solution problems

[restive river]

right but ok you can think of it intuitively anyways i guess

so like

this is saying 2x + 7 < 2x + 7, both of those are representing the same thing right?

[high cypress]

yea

[restive river]

so like

does it make sense that a number is bigger than itself?

like 1 < 1

[high cypress]

noo

[restive river]

righttt

so can this inequality ever be true

[high cypress]

ooh okay so its no solution?

[restive river]

yeaaah

[high cypress]

tysm!

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[lament schooner]

cosx just go between -1 and 1

what can x ---> infinity do here

[uncut crow]

what is that 1 doing there

[lament schooner]

jsut chilling

[uncut crow]

lol

possible it's a typo

what were the instructions?

it's kinda uncommon for something like this to come up in calc hw problems

if there are no typos

[lament schooner]

well

it is ig

teacher on crack

[uncut crow]

ok maybe it's fine

[topaz beacon]

are you familiar with how to work with improper integrals?

[lament schooner]

id o know the basic yes

[topaz beacon]

so what happens if you set up the limit?

[lament schooner]

lim t ---> infinity

[topaz beacon]

ok

[lament schooner]

and idk what -cos(t) is

cuz it just goes up and down between -1 and 1

[topaz beacon]

so does the limit exist?

[lament schooner]

dont think so?

but teacher didnt give a DNE option

[topaz beacon]

so what does that say about the integral

[lament schooner]

its improper?

[sinful pier]

D

[lament schooner]

oh

wait

yeah

i

lol i just forgot

[topaz beacon]

lemme clarify these terms for you
convergent - the integral gets arbitrarily close to a certain value
divergent - not convergent

[lament schooner]

okay basically its just infinite

yeah

[topaz beacon]

no its not infinite

but its also not convergent

[lament schooner]

so the area is not infinite?

[topaz beacon]

no, but it does not approach a value

[lament schooner]

wait why not

cosx just goes on infinitely

so if x approach infinity why isnt area under curve infinite

[topaz beacon]

because the area alternates from negative to positive

[lament schooner]

ah right

[topaz beacon]

take the sequence (-1)^n, it is divergent because the terms never approach a value

[lament schooner]

but since it mirrors the x axis and it just cancels out...couldnt we just calculate the area? or not cuz x is inifnite

[topaz beacon]

they just alternate

[lament schooner]

ahh kk

tysm!

.close

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[lament schooner]

i realized my thought was kinda dumb LOL

[topaz beacon]

youre welcone

it happens

[devout snow]

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[livid carbon]

Is it even possible to do the inductive step (i.e showing (P(2) ∧ P(3) ∧ ... ∧ P(n)) => P(n+1)) just using algebra? I'm stuck on this and getting nowhere with just algebra

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@livid carbon Has your question been resolved?

[devout snow]

@livid carbon Has your question been resolved?

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@livid carbon Has your question been resolved?

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@livid carbon Has your question been resolved?

[eternal tapir]

you get two cases, floor((k+1)/4) = floor(k/4) <= floor(k/4) + 1 and floor((k+1)/4) = floor(k/4) + 1

and when floor((k+1)/4) = floor(k/4) + 1, we have that k must be of the form $4j + 3$, using that, and evaluating, $T(k+1) = 2T(j + 1)+ \sqrt{k + 1} \leq 2\cdot 3\sqrt{j + 1}\log_2 (j + 1) + \sqrt{k + 1}$

[woven radish]

M8 of 48

[eternal tapir]

and from there i assume you can do some clever manipulation to get back 2sqrt(k+1)log(k+1)

can try exploring that

[eternal tapir]

best i get is $2\cdot 3\sqrt{j + 1}\log_2 (j + 1) + \sqrt{k + 1} = 2\cdot 3\sqrt{j + 1}\log_2 (j + 1) + \sqrt{4j + 4} = 2\cdot 3\sqrt{j + 1}\log_2 (j + 1) + 2\sqrt{j + 1} \leq$ not sure if this inequality is best approach $ 2\cdot 3\sqrt{j + 1}\log_2 (j + 1) + 2\sqrt{j + 1}\log_2(j + 1) = 2 \cdot 4\sqrt{j + 1}\log_2 (j + 1) = 2 \cdot 2 \sqrt{4j + 4}\log_2(j + 1) = 2\cdot 2\sqrt{k + 1}\log_2(j + 1)$

[woven radish]

M8 of 48

[lavish prawn]

You can form a nice induction with the hypothesis "Assume it holds for all numbers $< 4n$ for some $n \geq 1$. Then we show it also holds for $4n + k$ where $k \in {0, 1,2,3}$.

[woven radish]

Lartomato

[lavish prawn]

step 1: using definition
step 2: applying induction hypothesis
step 3: pulling 2 into square root
step 4: using monotonicity of square root to replace 4n with 4n+k and adding a factor of 3 into the second summand
step 5: pulling factor sqrt(4n+k) out of both summands
step 6: replacing 1 with log(2) and using log(a) + log(b) = log(ab)
step 7: using monotonicity of log to replace 2n with 4n + k

@livid carbon it can be done!

\begin{align}T(4n + k) &= 2 T(n) + \sqrt{4n + k}
\ &\leq 6 \sqrt{n} \log_2(n) + \sqrt{4n + k}
\ &= 3 \sqrt{4n} \log_2(n) + \sqrt{4n + k}
\ &\leq 3 \sqrt{4n + k} \log_2(n) + 3\sqrt{4n + k}
\ &= 3 \sqrt{4n + k} (\log_2 n + \underbrace{1}_{= \log_2 2})
\ &= 3 \sqrt{4n + k} \log_2(2n)
\ &\leq 3 \sqrt{4n + k} \log_2(4n + k)\end{align}

FUCK YEAH we did latex

[woven radish]

Lartomato

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[jaunty jackal]

help me pweasee

[radiant atlas]

In what?

[rustic jetty]

!question

dang

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[cerulean void]

hi i just have a rly quick trig question if anyone would be willing to answer, can't find anything online and it's honestly p dumb so 💀

the pythagorean identity sin^2x+cos^2x=1 can also be expressed as cos^2x-1=-sin^2x right?

[cerulean void]

idk if the sin would be positive but it makes no sense to me to make the sin positive

which is why im asking

[trail eagle]

That seems ok, what's the problem with it?

You can just subtract sin^2(x) on both sides and subtract 1 on both sides.

[cerulean void]

okay awesome

yeah that's all i wanted confirmation on HAHAH

thank you!

.close

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[shell salmon]

not to sure how to work this one out

[elder shadow]

do you remember the requirements for using l'hopital's rule?

[shell salmon]

yes

subbing in infinity there must be a limit and 0 there must be a limit and f'(a) when subbing in a it must give a limit

but not sure how to get 1/10

[elder shadow]

subbing in infinity?

in this case, the requirement in an indeterminate form when subbing in the value we are approaching

sub in 25 and check if you have 0/0 or inf/inf

[shell salmon]

oh yea ok

but how to get 1/10 ??

[elder shadow]

to use lhopitals you replace the numerator and denominator with their derivatives, and you will have a limit you can evaluate by substitution

[shell salmon]

ah got it

1/(2*sqrt(x))/1

[elder shadow]

yes

[shell salmon]

1/10

eq

ty

[elder shadow]

👍

[shell salmon]

.close

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[bronze willow]

I need explanations on this, how does the exact thing I find for the question is not even the answer?

[bronze willow]

Pinging helpers for every 15 mins

<@&286206848099549185>

[devout snow]

@bronze willow Has your question been resolved?

[bronze willow]

<@&286206848099549185>

[devout snow]

@bronze willow Has your question been resolved?

[bronze willow]

<@&286206848099549185>

If anyone's up to answering these questions I'll post them here

Question 1 (b) [RESOLVED]
Got a hint on reddit but I don't understand it so I need help

Question 5, Finding P(D) [RESOLVED]
Got no hints yet, still needs an explanation

Question 6(b), finding P(Z) [RESOLVED]
I don't understand how I can find the probability when the tree constructed is going to be that huge

Question 8(c) finding the probability that one object is yellow, and two objects are balls [RESOLVED]
I get very confused about this question as the answer I got was not the answer

I'm sorry this feels like a lot but my main weakness in math is probability because of the word problems in it

If needed, I'll ping helpers every 8 minutes just to get someone's attention

Answers:
1(b) 5/9
5. P(D)= 0.437 (rounded to 3 d.p.)
6(b) 2/35
8(c) 0.0877 (rounded to 4 d.p.)

<@&286206848099549185> 4 questions can you do it?

<@&286206848099549185>

[bronze willow]

<@&286206848099549185>

[bronze willow]

<@&286206848099549185>

[bronze willow]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

1 hour

Im not getting anywhere with this am I

[primal yacht]

@bronze willow hi...inbox

[bronze willow]

just do it here

hhhagshagdagg

[solid osprey]

chill, idk why but today theres kinda a lack of helpers

most of the time there are normally alot more helpers

[wooden pasture]

you have been harased too to be spamed by @ helper 😂😂😂😂😂

[devout snow]

@bronze willow Has your question been resolved?

[bronze willow]

<@&286206848099549185>

[bronze willow]

Well thanks to reddit I got the explanations to everything

.close

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[rigid sigil]

french?

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@rigid sigil Has your question been resolved?

[fossil locust]

.close

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[fossil locust]

Please don't take up more than one help channel

[devout snow]

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[flat raft]

If we have a 3x3 system of equations, when do we say one of the equations is a linear combination of the other two? And does this mean the equation has no solution?

[wicked turtle]

well if you have eq1 and eq2

and then you have eq3 = a(eq1) + b(eq2)

that's what it means for eq3 to be a linear combination of the other two

a and b are scalars

as long as the eq1,eq2 system has a solution, then the eq1,eq2,eq3 system will also have a solution, but it won't necessarily be a unique solution

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[flat raft]

Thanks a lot

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[south tapir]

I need help to Calculate a Null Hypothesis

[south tapir]

I have forgotten how to calculate it and i need help on how to procede

English is not my first Language

i have exampels of earlier work but i still can't figure i out

I would prefer if i could talk to the one helping me because texting in English is relativly hard for me

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[prisma frost]

for B why do i differenitate x to find the maximum

[prisma frost]

them make it 0

[flat raft]

dy/dx =0 means the function is flat, at the maximum/minimum the function is flat(neither increasing nor decreasing)

you have to solve for x to find the maximum point

[prisma frost]

oh ok

ty

so if it ask me to prove if something is max or mini just make dy/dx = 0

[flat raft]

Not all time

but it works in most problems

functions can be flat for other reasons apart from max/min

and sometimes dy/dx is not defined at the max/min

[prisma frost]

oh ook

would i to do the second derivative to prove that is a max or no?

[flat raft]

no

you can sketch a picture of the graph

check:

1. Where dy/dx =0
2. dy/dx is not defined
3. Endpoints as x tends to infinty

These are candidates for a max/min

[prisma frost]

oh ok

ty for the help

[flat raft]

goodluck on exam

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[hasty palm]

I don’t understand these questions

[opaque haven]

lets look at the first one

what types of energy does the coaster have on top of the mountain

[hasty palm]

KE at top of mountain is 360J

Wait r u talking abt ski or roller coaster

[opaque haven]

the roller coaster

[hasty palm]

Kk

[opaque haven]

lets not look at the values at first, lets focus on the idea

[hasty palm]

Kk

[opaque haven]

so on top of the mountain we know the roller coaster is moving right? so it must have some kinetic energy

but since its so high up there what other kind of energy will it have?

[hasty palm]

At the top of mountain yes kenetic

Potential

Mechanical

[opaque haven]

gravitational potential, right

[hasty palm]

Ya

[opaque haven]

at the bottom of the hill now, lets check what goes on there

does it have kinetic energy?

[hasty palm]

Yes because it’s still moving

Potential energy no bc height is 0

[opaque haven]

of course, its speeding up as well so the kinetic energy is higher

and thats right, potential energy will be 0

[hasty palm]

And the mechanical energy stays the same

[opaque haven]

well we cant quite say that

[hasty palm]

Bc we loose to friction

[opaque haven]

the problem itself says that we actually lose 80kj of energy

[hasty palm]

mechanical energy - 80 ?

[opaque haven]

lets form an equation

[hasty palm]

Bc I get confused with friction

Kk

[opaque haven]

so at the start, on top of the hill, we had K(start)+U(start) energy

right

kinetic and potential

[hasty palm]

$ke_i + pe_i = ke_f + pe_f$

[woven radish]

ella

[hasty palm]

Would we use that

[opaque haven]

yes with a small twist

that equation implies that whatever energy we had at the top makes its way down to the bottom wholly

but thats not true, we lose 80kj

so how will we interpret that in the equation

[hasty palm]

Hmmm

I have no clue

[opaque haven]

$ke_i + pe_i$ is the energy we have at the top

[woven radish]

bill

[opaque haven]

$ke_f + pe_f$ is the energy we have at the bottom

[woven radish]

bill

[hasty palm]

Yes

Oooooh

So we would do this

$ke_i + pE_i = ke_f + pe_f -80$

[opaque haven]

youre on the right track

well actually

which ones are the f

and which ones are the i

[hasty palm]

Ah sorry i accidentally put I twice

Here

[woven radish]

ella

[opaque haven]

well lets look at whether that 80 should get a plus sign or a minus sign

[hasty palm]

Isn’t it loosing 80J

[opaque haven]

so the final energy on the right hand side is 80 less than the initial energy. so to get to the initial energy again and make it so the equality is true, we actually have to add 80 instead of subtracting it

does that make sense?

[hasty palm]

Wait I need a sec to process that info

So yes it’s 80 less that initial energy

And we need to get to initial energy again but why

[opaque haven]

well from the moment we are making an equation we demand an equality right

we want the two sides to be equal

[hasty palm]

Oooooooooooooooooooh

[opaque haven]

otherwise we will just be starting off from a wrong statement

[hasty palm]

So both sides need to have the same value so it’s sort of like 300 = 250 + 50

[opaque haven]

exactly

you cant start off with 300=280 and expect to end up with a correct result, right?

so we need to make sure both sides are equal in order to actually have an equation

[hasty palm]

So they each sides need to have the same value

I get it so that’s for every equation I do ?

[opaque haven]

well yeah, for an equation to hold true both sides need to have the same value

even when we solve equations like x+3=6

what we're basically doing is: what value does x need to have in order for the equality to actually hold?

[hasty palm]

Omg now it makes sense bc x ends up equaling to 3

And 3+3=6

[opaque haven]

exactly

[hasty palm]

That makes so much more sense

So now the equation would be

$kE_i + pE_i = kE_f + pE_f + 80$

[woven radish]

ella

[opaque haven]

exactly

now you get to replace the energy formulas and solve

[hasty palm]

Tysm now I understand bc I was doing -80 instead of +80

The concept behind it now makes sense

[opaque haven]

great

[hasty palm]

Ima solve and let u know

[quaint citrus]

yea its literally the same thing, but whenever i think about this stuff, i do Kei + Pei + W = Kef + Pef, and work would be the negative quantity

[hasty palm]

What’s W

[opaque haven]

its just another way to put it, basically its how much energy we gained during the whole phenomenon

since here we didnt really gain anything and we actually lost 80kj, w is -80

its another way of doing the same thing

when i would do questions like these back in the day we'd use Q, as in the energy lost to the enviroment in the form of heat

you'll see many ways of phrasing the same thing, but the important thing is u get the idea

[hasty palm]

At the bottom of the hill would the height be 0?

I got 21.93 but it says the answer is 16.4m/s

[opaque haven]

well lets do it together then

so for kinetic energy the formula is 1/2mv^2

for potential energy the formula is mgh

[hasty palm]

Yes for ke_i it would be 1/2 (500) (1.20^2)

I got 360J

So now it’s $360 + pE_i = kE_f + pE_f + 80$

[woven radish]

ella

[opaque haven]

we got:
(1/2)(500)(1.2^2) + (500)(10)(30) = (1/2)(500)(V^2) + 80

[hasty palm]

$360 + pE_i = kE_f + pE_f + 80$

[woven radish]

ella

[opaque haven]

so (250)(1.44)+(150000) = (250V^2)+80

[hasty palm]

Huh

[opaque haven]

i did 1/2*500 which is 250

[hasty palm]

I got 147000 for PEi

[opaque haven]

how so?

oh what do you use for g?

10?

or 9.8

[hasty palm]

500(9.8)(30)

[opaque haven]

okay if you use 9.8 then yeah its 147000

[hasty palm]

Ah kk

[opaque haven]

so all in all we have V^2 = 589.12 right

[hasty palm]

I have 446.55 = v2

[opaque haven]

its 147000+360 = 147360

[hasty palm]

Yes

[opaque haven]

and then devide that by 250

[hasty palm]

147360 = 330v2

[opaque haven]

330?

[hasty palm]

What abt the 80

• 80

[opaque haven]

well we have 250v^2 + 80

[hasty palm]

Yes

[opaque haven]

we cant just add the 80 on the 250, the 250 has a v^2 on it

thats not a like term

[hasty palm]

Ah true

[opaque haven]

so instead we take the 147360 we had and we subtract the 80 from that and we get 147280

[hasty palm]

So how 147360 / 250

[opaque haven]

lets start again

right now we have 147360 = 250v^2 + 80 right?

[hasty palm]

Yes

[opaque haven]

lets move the 80 over to the other side

[hasty palm]

I divide 147360 by 250

Ah ok

[opaque haven]

so it turns into a minus

we cant do that yet

and then 250v^2 = 147280

so now we can devide by 250

[hasty palm]

Yes

[opaque haven]

and get v^2 = 589.12

[hasty palm]

Yes

Root of 589.12 is

[opaque haven]

around 24.27

which isnt the answer that u said before, which makes me think that might be wrong

[hasty palm]

Yes ur exact

Hmmm ya

It has to be bc it says the answer is 16.4

So it must be wrong

[opaque haven]

ah i see the error

so we lost 80kj right?

thats not just 80, thats 80.000 joules

we forgot to convert the units

[hasty palm]

Oooooooh bc it’s kilo joules

[opaque haven]

exactly

so actually we have 147360 = 250v^2+80.000

so we get 250v^2 = 67.360

so v^2 = 269.44

and when we take the sqrt we get 16.41

which is about 16.4

[hasty palm]

I’m so sorry my iPad died and I couldn’t find a Charger I’m back

@opaque haven

I’m confused bc when u subtract 147360 by 80.000 it dosent equal 67.360

[opaque haven]

doesnt it?

[hasty palm]

Oooooh I got confused bc u put a . Not a comma

So I thought u were saying 67.360

As in like decimal

[opaque haven]

ah no

[hasty palm]

Yesssss I now got the answer

Ima try the next one and let u know how it goes

For this is the bottom of the ramp h=0?

[devout snow]

@hasty palm Has your question been resolved?

[heady tulip]

prob better if not

h=0 would make the height of the ramp = 0 from using that frame of reference

[hasty palm]

True so the height in this scenario is unknown

[heady tulip]

im not sure if P should be positive or negative

anyways

same idea

[hasty palm]

P?

Is P potential energy

[heady tulip]

oops

i use p for lost energy

[hasty palm]

Oooh

Kk

[hasty palm]

Yes I’m confused on if Lost Energy should be + or - bc the equation has to be equal but like idk I’m not sure I’m pretty sure + but still idk

[heady tulip]

i would think P would be positive on the RHS

like initial energy = final energy

Ki + Ui = Kf + Uf + lost energy

[devout snow]

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[stuck musk]

Could I get some help with 19.

[stuck musk]

Here is an example:

As far as I got before thinking I’m not doing it right

[devout snow]

@stuck musk Has your question been resolved?

[heady tulip]

lol

<@&286206848099549185>

[devout snow]

@stuck musk Has your question been resolved?

[stuck musk]

<@&286206848099549185>

[devout snow]

@stuck musk Has your question been resolved?

[devout snow]

@stuck musk Has your question been resolved?

[midnight echo]

What’s that lmao

Oh wait, part b is variation of parameters

[stuck musk]

Yep

[midnight echo]

Your work is a bit confusing (maybe cause it’s been a while since I’ve done ODEs), but what did you get for the Wronskian of your homogenous solution?

[devout snow]

@stuck musk Has your question been resolved?

[stuck musk]

Not sure what a wronskian is

C1cosx/2+ C2sinx/2=0 probably

[stuck musk]

Original question is up here too, I pretty much followed the example for my work

[subtle delta]

If you still need help soon, I think I can help you, your complimentary solution is correct, but you’re forgetting a principle of the undetermined coefficients method when it comes to your particular solution

[stuck musk]

yep still need help, what principle?

I'm curious what I'm missing, but also need help mostly with variation of parameters

[faint zinc]

@stuck musk I always seem to need to look this stuff up whenever I have the occasion to solve an ODE. https://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

For 19, because your g(x) = cos x, your y_p(x) guess will be A cos(x) + B sin(x), where A and B are your undetermined coeffs.

These are linearly independent from your general solution set because those are cos(x/2) and sin(x/2)

which you can tell from your Wronskian if you needed to, but you don't seem aware of the Wronskian?

You can find info on the Wronskian here: https://tutorial.math.lamar.edu/Classes/DE/FundamentalSetsofSolutions.aspx

[devout snow]

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[restive river]

g(a) = lim h->0- f(a+h) - f(a)/h
is lim a-> n+ g(a) still f'(n-)

[red turret]

<@&268886789983436800>

[devout snow]

@restive river Has your question been resolved?

[restive river]

Could you write that out, please

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[empty barn]

R = { (a, b) ∈ Z x Z | a - b = 3k for some integer k} how can i find if this relation is equivalence Relation ?? i know how to solve problem in case if there is no k here, for example a - b = 3 but how can i solve problem when there is that variable k?

[empty barn]

do it need to be reflexive for every k or just one?

[restive river]

Ok

For reflexive what do you need to check?

[empty barn]

to chack if relation is reflexive just that

and then if it is transitive and symmetric

but do relation need to be reflexive, transitive and symmetric for all values of k or just for one value of k?

[restive river]

What?

K is some fixed integer, so all values

Idk how it's relavant

[empty barn]

i don't know how to prove if this relation is equivalence relation

how to show that this relation is reflexive for every value of k for example?

[restive river]

If it's a reflexive relation,
$(a,a) \in R , \forall a \in Z$

[woven radish]

fredthebread69

[restive river]

Agreed?

@empty barn

[empty barn]

ok yes

[restive river]

To check (a,a) $\in R:
\
$a-a=0$
\
Therefore it's true as it's a multiple of K (0K)

[woven radish]

fredthebread69
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[empty barn]

and a -b = 3k is reflexive when k = 0? but for k = 1 is not

[restive river]

Maybe some integer mean atleast one, I might be wrong

Oh yeah

It is like that

[empty barn]

it can mean that

Have you solved a similar problem before?

[restive river]

Yeah I have

I reckon it's exactly that.

I mean I can't find 2 variables a and b that satisfy
a-b=3k for all values of k
Lol

[empty barn]

So, there is at least one value of k for which the relation is reflexive, and another value for which the relation is symmetric...

ok thank you

[restive river]

For reflexive

$(a,b) \in R
\ \rightarrow a-b=3k
\ b-a=-3k
\ \therefore (b,a) \in R
\ hence \ proved$

[woven radish]

fredthebread69

[restive river]

Makes sense?@empty barn

[empty barn]

yea

[restive river]

Try for transitive

[empty barn]

in this case if k = 1 i can see that relation is symmetric for some a and b and that makes sense to me

[restive river]

Don't think much about k

[empty barn]

so i need to think about k only for reflexive case

id there is k where (a,a) can exist

[restive river]

No need think of k at all

Afterall you start with that assumption

[empty barn]

a - b = 3k and b - c = 3m, a - c = 3k + b - (b - 3m) = 3k + 3m = 3 (k + m), (k + m) ∈ Z

[restive river]

(a,c) in R

[empty barn]

ok thank you again

[devout snow]

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[fast fossil]

help

[midnight grove]

leme see

[fast fossil]

Only C and D

Then E and F is after

[midnight grove]

alright

so you know the general equation of a line?

the slope intercept form

,,y=mx+b

[woven radish]

Renz

[fast fossil]

yes

[midnight grove]

aigt lets start with l1

whats the y intercept of l1

[fast fossil]

4

[midnight grove]

correct

since the y-intercept of the line is b

in y=mx+b

we can substitute 4 in there

,,y=mx+4

[woven radish]

Renz

[midnight grove]

now, all we need to find is m

do you know what m is

[fast fossil]

Uhhh

1/2?

[midnight grove]

youre almost there

i was asking what m stands for

its the slope right?

it can be represented as

[fast fossil]

Yes

[midnight grove]

$m=\frac{rise}{run}$

[woven radish]

Renz

[fast fossil]

1/-2

?

hmm

[midnight grove]

correct

[fast fossil]

Oh!

[midnight grove]

lets simplify that

,,-\frac{1}{2}

[woven radish]

Renz

[midnight grove]

the - goes to the fraction

so our slope is

,,m=-\frac{1}{2}

[woven radish]

Renz

[fast fossil]

ohh

so is it

[midnight grove]

now substitute that value of m here

[fast fossil]

y = -1/2x + 4?

[midnight grove]

yep!

thats the equation of the first line

you do the same process for line 2

[fast fossil]

Ok so...

[midnight grove]

find the y-intercept, then find the slope, which is just rise/run

[fast fossil]

y = x + 2

?

[midnight grove]

correct!

[fast fossil]

Great!

So is that the first answer?

[midnight grove]

yep

[fast fossil]

Okay

[midnight grove]

the system of linear equations will be

,,y=-\frac{1}{2}x+4\y=x+2

[woven radish]

Renz

[fast fossil]

Ok

[midnight grove]

for letter d, just approximate it

[fast fossil]

Ok

One solution?

[midnight grove]

the intersection of the lines will be the solution right?

[fast fossil]

Yes

[midnight grove]

so just estimate the coordinate of the green dot

[fast fossil]

1.2, 3.25?

[midnight grove]

yea go for that

[fast fossil]

Ok

[midnight grove]

i think the answer does not necessarily need to be exact

since its just an estimate

[fast fossil]

Ok

[midnight grove]

however for e

you gotta solve for it algebraically

,,y=-\frac{1}{2}x+4\y=x+2

[woven radish]

Renz

[midnight grove]

what do you notice about the equation

[fast fossil]

uhhh

[midnight grove]

they both equate to what?

[fast fossil]

?

y?

[midnight grove]

ya

[fast fossil]

Ok

[midnight grove]

they both equate to y

[midnight grove]

meaning, you can equate these to each other

you will have

,,-\frac{1}{2}x+4=x+2

[woven radish]

Renz

[fast fossil]

oh and then - 2

?

[midnight grove]

what -2

[fast fossil]

4

[midnight grove]

you just need to solve for x

make the x terms on one side, and make the constant terms on the other side

[fast fossil]

Uhhh

I forgot how, this is the end term assignment and that was at the start of unit

[midnight grove]

ah

alright

lets take things one step at a time

alright on the left

we have a +4 right

how can we undo that

[fast fossil]

-4?

[midnight grove]

correct

we subtract 4 from both sides

,,-\frac{1}{2}x+4-4=x+2-4

[woven radish]

Renz

[midnight grove]

+4-4 cancel each other

whats 2-4

,,-\frac{1}{2}x=x+2-4

[woven radish]

Renz

[fast fossil]

-2

[midnight grove]

ye

,,-\frac{1}{2}x=x-2

[woven radish]

Renz

[midnight grove]

now, theres an x on the right side

we want to remove that

how can we remove an x on the right

[fast fossil]

-x

[midnight grove]

ya

,,-\frac{1}{2}x-x=x-2-x