help-23

[safe radish]

@lean otter Has your question been resolved?

[stable hearth]

consider that it’s a downward facing parabola with vertex at x=.75

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[analog harbor]

hey, i got this quadratic thing and i want to ask that if i should do this:

[analog harbor]

this is the form which is the thing but divided by 2

should i divide it by 2?

as i showed

so i can do this

[trim swan]

Yeah, looks good so far

[analog harbor]

thanks but this website says otherwise

:

[trim swan]

Well, you're not done yet, but it's right so far

[analog harbor]

it doesnt divide it by 2

and does the quadratic equation

[trim swan]

Oh, yeah. You don't have to divide both sides by 2

But you can

and it makes it easier

[analog harbor]

oh

dude but it changes the answer

[trim swan]

No it doesn't

Keep going

[analog harbor]

lemme try

oh yeaaaaahhh

it doesnt

[trim swan]

Yeah

[analog harbor]

thanks man

appreciate it

[trim swan]

Because you didn't just divide by 2, you divided both sides of the equation by 2

The other side was just 0 so it didn't change

[analog harbor]

the thing was this:

i can still do the quadratic equation

and get |an| the answer

i get it now

[trim swan]

Yeah, those two equations have the same solutions

[analog harbor]

aight

thanks dude

[trim swan]

No problem 👍

[analog harbor]

that clears the whole thing

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[lean otter]

$$I_2 = \frac{(R+C)V_b+C\cdot V_a}{(R+C)-C^2}$$

[flat frigate]

broke

broke

[lean otter]

If i divide both sides by Va then i will still have a Va on the RHS and i'm trying to have it only on the LHS

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[late frost]

Bonjour je ne comprends pas la différence entre les 2 questions.

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[late frost]

.reopen

[safe radish]

✅

[late frost]

quelqu'un ? 🙏

[safe radish]

@late frost Has your question been resolved?

[safe radish]

@late frost Has your question been resolved?

[pliant warren]

@late frost Ici, tu dois poser tes questions en anglais.Mais je vais essayer. 🙂

Pour la première question, on ne recherche que les résultats positifs, donc P(le résultat est positif).
Une personne peut être testée positive sans être malade. La deuxième question recherche alors les deux :
être malade à condition que le test soit positif.

[late frost]

Ok will see

[late frost]

for the first question I found 5,7%, and for the second, 94%, is that correct ?

[pliant warren]

first one is correct.

but I get something different for the second, but maybe I miscalculated. Could you show me what you did?

[pliant warren]

no that is not correct, have you talked about bayes theorem in class?

[late frost]

no

[pliant warren]

conditional probabilities?

[late frost]

yes

[pliant warren]

ok you are looking for the probabilty, that a person is sick under the condition that the test is positive. So you have to calculate a conditional probabilty here.

[late frost]

we did not see the difference between conditional and inconditional

[pliant warren]

ever seen this before?

[late frost]

maybe

[pliant warren]

Ok I'll try and explain: A = sick , B = positive test you are looking for $P(A \vert B)$. P(B) is the probability that a person has a positive test. You calculated this in the first part: 0,057. $P(A \cap B)$ is the probability that a person is sick AND has a positive test, that is what you are missing.

[flat frigate]

djuji

[late frost]

so the first question I was right but not the second ?

[pliant warren]

yes

[late frost]

atleast I did something good

[pliant warren]

🙂

So what we have to find is $P(A \cap B)$ the probability that a person is sick AND has a positive test. Any idea how to get that?

[flat frigate]

djuji

[late frost]

0,05 x 0,95 ?

[pliant warren]

yes.

now put that into the formula above.

[late frost]

0,475

47,5%

but this seems strange

irl this would not be like that

[pliant warren]

it seems strange because you are not done yet.

[pliant warren]

you just have the upper part of this fraction.

[late frost]

0,475 / 0,01

[pliant warren]

no for both. you have to devide by the probability that a person has a positive test. So your solution from the first part.

0,475/?

[late frost]

0,057

[pliant warren]

Yes!

[late frost]

wait its 0,0475*

0,05 x 0,95

[pliant warren]

You have to devide 0,0475 with 0,057

[late frost]

0,83333...

[pliant warren]

Yes!

That's your solution

Looks more realistic?

[late frost]

yes!

I now have to redact

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[safe radish]

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[leaden scaffold]

I got 5 for this question, but I need help in understanding some of it.

We’re supposed to recall the formula: f(a+h)-f(a)/h, right?

The tangent line formula??

[lean otter]

dang he just has a blackboard

[elfin yew]

point slope formula

[leaden scaffold]

So I’m not supposed to use the formula I just mentioned?

[elfin yew]

oh nvm sorry i said something wrong

its asking find the derivative at the point 2

[merry sleet]

you are supposed to use it, this is the formula of the derivative at point 2

or the definition rather

it also happens to be the slope of the curve at point 2

which is by definition the same as the slope of the line tangent to the curve at point 2

all of these concepts are linked

[leaden scaffold]

I was told I should of added 4 to h in f(2+h)-f(2). My question is why, if I never use the 4?

[merry sleet]

you shouldnt i'm pretty sure

[leaden scaffold]

Got it

[merry sleet]

what was the justification for this 4?

[leaden scaffold]

Okay, so for this problem

Why must I plug 4 in to calculate the slope of the secant line passing through those points but not the tangent line?

[merry sleet]

because those are two different things

to calculate the line passing by two points

you do

(y2-y1)/(x2-x1)

which in this case is

(f(6)-f(2))/(6-2)

so it happens that if you plug in 4

you get the denominator

[leaden scaffold]

So I always plug the denominator back in to find the slope of the secant line?

[merry sleet]

i meant the numerator

the to find the slope of a secant line

you do as you would with any line

[merry sleet]

like this

[leaden scaffold]

Wouldn’t I always be plugging in the value of the denominator back into the equation of the line given if I use the limit definition of the secant line?

[leaden scaffold]

You there?

[merry sleet]

theres a limit definition of a secant line?

[leaden scaffold]

What is f(x)-f)a)/x-a..?

[merry sleet]

its not a limit

its just a slope of a line

for the tangent line or the derivative we take the limit of the value of the slopes

[leaden scaffold]

But we use different formulas for calculating slopes of secant and tangent lines, no?

[merry sleet]

as the two points get closer together

i think you need to understand where these formulas come from

it look slike you are applying them without knowing

[leaden scaffold]

Please inform me I am trying to piece it together.

[merry sleet]

first thing about secant line

then the tangent line is just the limit of the secant lines

as x2 gets closer to x1

thats why we take the limit

[leaden scaffold]

And we always take the limit as h approaches 0?

[merry sleet]

as a+h is closer to a

the approximation gets better and better

[merry sleet]

yes h going to 0 means a+h gets closer to a

[leaden scaffold]

Okay. So back to my question here. Wouldn’t I always be plugging in the value of the denominator back into the equation of the line given if I use the slope of the line formula?

[merry sleet]

wdym?

using the slope to find the formula of the line?

[leaden scaffold]

Okay so for this problem

Do you see how I plug 4 back into the equation given to us?

[leaden scaffold]

Must I always plug the value of the denominator of the equation given to me in a problem when using the formula:f(x)-f(a)/x-a..

[merry sleet]

you plug back 4 because its this formula

it just happens here

that they give you what is f(x+h)-f(x)

and you chose h such that

x+h = x2

[leaden scaffold]

Could you give me some example problems real quick? I’ve ran out on my worksheet.

[safe radish]

@leaden scaffold Has your question been resolved?

[safe radish]

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[ornate oracle]

Hi if we have a vector n (1,1,1) in the x,y,z system and we want a direct orthonormal system how do we do ? The other components are t1 and t2

[ornate oracle]

I would like to know the coordinates of the vectors n, t1 and t,2, pls 🙏

.close

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[alpine ravine]

can someone help me solve this?

[safe radish]

@alpine ravine Has your question been resolved?

[alpine ravine]

<@&286206848099549185>

[alpine ravine]

.clsoe

.close

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[earnest galleon]

you need to post the rest of the question @lean otter

[lean otter]

i need to find g(-3)

[merry pelican]

Find the y-value at x = -3

[lean otter]

I think it's a straight line from -6 to 0

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

i do, but its not at a point and i cant figure out how to estiamte it

[merry pelican]

As srikanth said; it looks like a straight line from -6 to 0 and from that assumption the gradient of this line is pretty clear

I think it's safe to assume that g(-6) = 2, so you can go from there

[lean otter]

so then g(-3) would equal 1?

<@&286206848099549185>

im confused

[merry pelican]

If g is assumed to be a straight line from x = -6 to x = 0 what can you say about the gradient of this line by looking at the graph?

[lean otter]

i calculated the gradient and i found -1/2

[merry pelican]

Why negative?

[lean otter]

oh wait

no wait

[lean otter]

because (5 - 2)/(-6 - 0) = 3/-6 = -1/2

[merry pelican]

You switched up your x-values

It should be (0 - (-6))

[lean otter]

ohhhhhh

[lean otter]

so (5 - 0)/6

so 5/6

[merry pelican]

Still (5 - 2)

[lean otter]

yall good?

[lean otter]

wait now im confused

ohhh wait

so its positive 1/2

[merry pelican]

Yes

And if g(-6) = 2 and it goes in a straight line with gradient 1/2, what is g(-3)?

[lean otter]

1?

[merry pelican]

Why would it be 1?

If the gradient is 1/2, you go up 1/2 on the y-axis every time you go 1 to the right on the x-axis

[lean otter]

wait

so it would be 3/2?

[merry pelican]

You go up 3/2 from x = -6 to x = -3 yes

So if you add the 2 you were already at when you where at x = -6, you get?

[lean otter]

7/2

[merry pelican]

Si

[lean otter]

yessss

wait why would we add the 2?

[merry pelican]

But I still think the point of the question was just to read it directly on the graph

[lean otter]

there was no way to really estimate, so i got kinda confused

i have another question as well

for the same graph

[merry pelican]

You could safely assume it was directly between 3 and 4, so 4,5 (which is the same as 7/2) would have been a sufficient guess

[lean otter]

okay

[merry pelican]

What's the question

[lean otter]

my next question is find g(5x) if g(x) = 4

and its the same graph

i have no clue what im supposed to do

[merry pelican]

At what x is g(x) = 4?

[lean otter]

-2

[merry pelican]

Yes

So 5x would be?

[lean otter]

-10?

[merry pelican]

Yep

So if 5x = -10, what is g(5x)?

[lean otter]

uhhhhhhhh

ohhhhhhhhh

0

[merry pelican]

Si

[lean otter]

gracias

[lean otter]

.close

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[brittle onyx]

I need help figuring out if i got this right
write the quadratic function in vertex form. then identify the vertex

g(x)=0²+7x+2
g(x)=x²+7x+...+2+...
the dots equal to 49/2
so g(x)=x²+7x+49/2+2+49/2
(x+7/4)²-45/2
x=(7/4,45/2)

[stray socket]

What's the original function?

[brittle onyx]

the top one

g(x)=x²+7x+2

[stray socket]

Are you forced to use complete the square

Because like

[brittle onyx]

i accidentally put 0 instead of x

[brittle onyx]

no

[stray socket]

Then use this

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[lofty jetty]

what is the question

[lean otter]

find area of traingle

[lofty jetty]

do you want to see the solution or do you want to get some hints

and try to solve it yourself

[lean otter]

can give some hints

[split ether]

I think you can solve for the lengths

[vestal sedge]

try drawing the tangent to the circle perpendicular to the bottom side

[lofty jetty]

lines of the same color have the same length

[lean otter]

hm

[lofty jetty]

we know that blue line is 2 cm long right?

[lean otter]

oh yea

[lofty jetty]

then we can calculate green line's length

[lean otter]

oh

is it 5

[lofty jetty]

yes

so what do you think we should do next

[lean otter]

uh

[lofty jetty]

we need to calculate red line's length

you can try to come up with solution or ask for hint

[lean otter]

umm

(2+r)^2 + 7^2 = (5+r)^2?

[lofty jetty]

exactly

now you just need to solve this equation

[lean otter]

ok so thats like

(2+r)(2+r)
4 + 2r + 2r + r^2 + 49 = 25 + 10r + r^2

[lofty jetty]

yes

[lean otter]

and then

r^2 + 4r + 53 = r^2 + 10r + 25

[lofty jetty]

yes

[lean otter]

so 28 = 6r

[lofty jetty]

yep

[lean otter]

and then r = 28/6

😄

[lofty jetty]

yep congrats

[lean otter]

wait how did u know that those lengths were equal

[lofty jetty]

whenever you have a circle in a figure and this circle is touching all of its sides, the lines between anchors and where circle touches that figure are the same length

sorry for my english, ill try to draw it

its always true for any figure. the only condition is that circle must be touching ALL of its sides

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[acoustic violet]

please show us your work. if you've forgotten the concepts, you may search online for "function transformation".

[safe radish]

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[lofty holly]

I understand that sin is the relationship between the opposite side and the hypotenuse in a right angle triangle. What I don’t understand is sin in terms of 90°

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@lofty holly Has your question been resolved?

[buoyant shadow]

at 90° you're at the top of the circle, where the height is 1

at 180° it's 0, at 270° it's −1

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[lofty holly]

.reopen

[safe radish]

✅

[lofty holly]

Ok

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[charred glacier]

i'm struggling with this

[tiny wraith]

Just watch the video about it

[charred glacier]

i didn't know there was one

can u send the link

i saw it on his shorts channel and i tried solving it

i didn't know he solved it in a video

[tiny wraith]

Oh

https://youtu.be/wyORoB5Q2Hg

There is a second derivative here though

[foggy salmon]

OMG THE EXACT SAME?

wait nvm 2nd derivative

[lean otter]

Lmao the fact that u have to compute an integral for where the limit is going to

[tiny wraith]

The second derivative made it a lot easier I think

[foggy salmon]

o lol

[tiny wraith]

It made stuff cancel in the denominator

[lean otter]

The summation is the worst looking thing there I think

That power series is yikes

[uncut cloak]

Wait you just differentiate then the things cancel no?

[lean otter]

U can do that?

[uncut cloak]

You can differentiate inside the sum

Im p sure

[lean otter]

Damn

In my mind that would change the whole sum because you are kind of replacing one thing with another

But no clue 😵‍💫

[uncut cloak]

It’s like how you can differentiate in integral

So probably with sum as well

[lean otter]

Yeah I will research about the eventually

[safe radish]

@charred glacier Has your question been resolved?

[uncut cloak]

You know I think it works because it’s just like differentiating term by term

It’s just a sum and differentiation is a linear operator

[charred glacier]

ty am @tiny wraith

tysm

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[true estuary]

@lean otter soz to ping

is it ok if we can go over what u said

as i have tried some of it

[lean otter]

Oh yeah sure

Go ahead

[true estuary]

so

take the limit of f(x) to infinity and minus infinity and find possible horizontal asymptotes.

take the limit of f(x) to where it is not defined on its domain. Thus finding any possible vertical asymptotes.

in terms of this

as x tends to - inf and inf

there is an asymptote at y = 0

do u agree?

[lean otter]

Uhh

Yep seems right I guess since e would always be negative and grow larger than x

[true estuary]

cuz as x gets larger, xe^-x^2/8 gets smaller

and tends to

0

[lean otter]

Yes

[true estuary]

cool

niw

quate f(x) to 0 to find x intercept, equate x to 0 to find find y intercept

for this

there is an intercept at (0,0)?

*intercept

[lean otter]

Are we talking about the vertical asymptotes?

Oh

[true estuary]

intercepts sry

[lean otter]

Yes there would be at (0,0)

[true estuary]

cool

[true estuary]

now this take the derivative and equate that to 0 and find the critical points.

right

how do i differentiate this

i thought product rule but

how u different e^-x^2/8

[lean otter]

Uhh either quotient or product rule doesn't matter

Well let's just use Quotient

[true estuary]

im just thinking tho btw

there must be a quicker ethod

cuz

this is from a past exam paper for real analysis which lasts 50 mins

and im assuming they dont want us to spend long on this

[lean otter]

You can uh

Oh nvm actually

[lean otter]

Sheesh real analysis? I mean I am just applying what I know from at most calculus 2. If you guys have some more refined "tricks" Then I don't know of that

Well quotient rule isn't that bad

Let's just do it

[true estuary]

ok

so product rule then

u = x

[lean otter]

Yeah sure

[true estuary]

du = 1

v = e^-x^2/8

dv = ...

[lean otter]

Okay so use chain rule

u = -x^2/8

Or is this a double composition hmm

[true estuary]

du = -1/4x

[lean otter]

Yeah okay

Yeah just do that

Result is (-xe^-x^(2/8) )/4

That's lowkey unreadable but I hope u know what I meant lmao

[lean otter]

Yep seems good

Now equate that to 0

[true estuary]

x = 0

[lean otter]

Uhh seems like it's only at 0 itself to me,

[true estuary]

e cancels out

[lean otter]

Yes

[true estuary]

so -1/4 x^2

= 0

bruh

its 0

bruh

[lean otter]

Nope nope wait

It's not at x = 0

That's just going to be 1

[true estuary]

have we differntiated wrong then

[lean otter]

I don't think so

It's correct

[true estuary]

how u get 1

when u make the diff = 0

u get x = 0

[lean otter]

Well plug in x = 0 to the derivative

[true estuary]

yh 1

[lean otter]

Left hand side is gone and right hand side is e^0 more or less which is 1

[true estuary]

but u intially said make it equal to 0

[lean otter]

Yeah we need to find what makes it be 0

[true estuary]

yh thats 0 tho

when x = 0

f'(x) = 0

[lean otter]

But it is 1 though? At x = 0?

[true estuary]

oh yh

im confused tho cuz u said this take the derivative and equate that to 0 and find the critical points.

[lean otter]

Yes this is how you find critical points to my knowledge I'm not inventing stuff

[true estuary]

then why did u say now to make x = 0

[lean otter]

I made a mistake at the beginning then corrected myself by saying that is in fact 1 when x = 0

[true estuary]

ok ok

sry to ask alot

just wanna clarify

[lean otter]

I don't think that derivative has any zeros

[true estuary]

but ok x = 0 f'(x) = 1

now what?

[lean otter]

So let's look back at our derivative and make sure it's correct

If there are just no critical points, then the function is just monotonic lmao

[true estuary]

ok

yh

theyre isnt any crictical points

and our differntial is correct

[lean otter]

Yes this function is either forever decreasing or forever increasing

Let's plug in some random value for x and see what is it

[true estuary]

ok

[lean otter]

Hmmm

But I'm doubtful

This should not be monotonic

[true estuary]

yh icl pal

think weve done it wrong

ight

imma just show the desmos pic

[lean otter]

Yes I think so too

[true estuary]

great

[lean otter]

Well seems like the critical point was at

-2 and 2

[true estuary]

our asymptoes were correct

y = 0

but what did we do wrong with the critical points

[lean otter]

It's pretty hard to find the zeros for that so not sure..

Oh I see

They cancel at x = 2

Well, not sure how you could find that but I'm pretty dead brained right now so have at it ig

Well anyways x = 2 is our critical point @true estuary

[true estuary]

ight cool

thanks for ur help anyway mate

appreciate it

[lean otter]

Yeah I mean , pretty tough one

[true estuary]

yh it is

[safe radish]

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[surreal echo]

did i draw this diagram correctly and how would i solve this question?

[safe radish]

@surreal echo Has your question been resolved?

[safe radish]

@surreal echo Has your question been resolved?

[surreal echo]

pe;las

https://tenor.com/view/travis-scott-astroworld-travis-head-travis-scott-head-travis-scott-headache-gif-23896240

[viscid grail]

you have an isosceles triangle

[surreal echo]

i have an isosceles triangle?

[viscid grail]

yeah 120 + 30 + 30 = 180

[surreal echo]

OHOKAYYYYY

now what

[viscid grail]

what's a depression?

[surreal echo]

this

[viscid grail]

I see

I think distance means a perpendicular line

to another line

but dunno about this problem

[surreal echo]

darn okay

[viscid grail]

sorry that I can't help much

no sleep and I'm tired as hell

[surreal echo]

no worries man thank u for taking the time anyways 🫡

hope u get a good rest

anyone else

[lean otter]

Uhh @surreal echo

I believe I solved this exact same problem like several weeks ago funnily enough lmao

[surreal echo]

fr?

[lean otter]

But anyway s

[surreal echo]

https://tenor.com/view/mae-brucelee100-poland-gif-19906479

[lean otter]

Eh actually nvm it's not the same. Mine was something like calculating the height of a building with those angles

[surreal echo]

ooohh

similar i guess

[lean otter]

Also, don't really understand why you assumed the angle 97 was supplementary to the 120

[surreal echo]

oh yeha

idk why i did that

lol

the diagram is wrong

[lean otter]

Shouldn't it be like, two triangles inside a big triangle if that makes sense?

Because ig the dude is looking in the same direction or whatever

[surreal echo]

yeah that makes sense

cuz the angles wouldnt work out for a regular triangle

[lean otter]

I was thinking something like this

[surreal echo]

OHHHHH

YEAH THAT MAKES WAY MORE SENSE

[lean otter]

Don't mind my amazing cropping there because I couldn't be asked to find an image of the actual triangle

Uhmm

[surreal echo]

but it wouldnt be a right triangle

[lean otter]

I think that's fine, let's just get the image right first and foremost

[surreal echo]

okay that makes sense

[lean otter]

Okay so like draw the tree thingy up 10 metres first

[surreal echo]

okay

[lean otter]

Then draw the first triangle of the coyote

The 97 degree is bottom right angle and it's right angle

[surreal echo]

okay

[lean otter]

Okay second, the rabbit is same story and make it a 120 angle there

But also do write the angle of depression top left too

Of that triangle

Show me ur drawing once u r done so we make sure

[surreal echo]

wait so

the rabbit is the angle furthest to the right>

?

[lean otter]

Yee exactly

Yeah it is

It's the one with the 120

[surreal echo]

ohhhhh

[lean otter]

And then u have to find x which Is the little bottom side of the triangle the rabbit makes

[surreal echo]

so like this

[lean otter]

Yes I believe so

[surreal echo]

so the distance between the reabbit and the coyote

[lean otter]

Wait no

Nvm actually but yeah

Anyways

[lean otter]

Yes

Okay

[lean otter]

Here is what I'm thinking

You take tan(20) = y/10

y being the other length

That the coyote makes with the angle 97

Which would mean that y = tan(20) * 10

[surreal echo]

okay

[lean otter]

So we know that side now whatever that is

Next, we find the length of the general big triangle with angle 50 degrees

[surreal echo]

okay

y= 3.64

[lean otter]

Let's define the whole length of the bottom side as z

[surreal echo]

okay

[lean otter]

tan(50) = z/10 which is z = 10(tan(50))

Which means that since y + x = z then x = z - y

But hm.

This method seems sketchy

[surreal echo]

why?

[lean otter]

I am not sure. Because the angles they provided would be redundant

Unless the question is meant to be that

But anyways

Can u check if the calculations would be consistent? @surreal echo

[surreal echo]

consistent?

wdym

like

[lean otter]

What do u get for z and x

[surreal echo]

gimme a sec

okay so rounded

i got y= 3.64

z = 11.918

[lean otter]

Yep

[surreal echo]

x = 8.278

[lean otter]

Hmmmmmm

I still feel off about something. And I don't want to fuck u over by something that I thought of at 2 am lmfao

[surreal echo]

ahh okay LOL

the calculations seem to be correct though

like

i dont see why it would be incorrect

or how

[lean otter]

Just the wording of the question keeps throwing me off

[surreal echo]

oh yeah that's what i had a problem with too

it's weird

like tricky

[lean otter]

Ahh I guess the angles were added there just to show that the coyote is lesser than the rabbit

Honestly, I don't really see what else it could be at this time. But I'd double check if I were u anyways

[lean otter]

Yeahh I hate word problems lmao

[surreal echo]

okay, that makes sense

i really appreciate what you're doing by the way

i think ur the kindest stranger i've ever met

[lean otter]

Oh it's fine I'm just chilling going around here lmfao. But thank u

[lean otter]

Aww thank u so much. You are also very kind and I appreciate the patience lmaoo

[surreal echo]

yeahhh all you man

anyways ill let u be on ur way lol

thanks again sm for the help :D

[lean otter]

It's guccii, I hope u have an amazing day/night

Good luckk with ur studies

[surreal echo]

thank u!!

.close

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[vague coral]

?????? Where do I even begin?

[sand minnow]

take the derivative of the proposed solution and perform the matrix multiplication with the original function and see if those two are equal

[vague coral]

How do I take the derivative of a matrix though

[sand minnow]

it would just be the derivative of each part of the expressions, since the vectors of $\vec{y}(t)$ are constants then you can treat them as constants

[flat frigate]

Duh Hello

[sand minnow]

here is the derivative of the first part of $\vec{y}(t)$ and maybe you will understand how to do the 2nd one

[flat frigate]

Duh Hello

[vague coral]

2nd one is the same thing

[sand minnow]

notice that you can just treat $c_1$ and the vector as constants, as they have no time dependence

[flat frigate]

Duh Hello

[vague coral]

It just has a -2

[sand minnow]

yea

[vague coral]

Ok

Buur like it makes me think

Now what

I got the derivatives but now what

[sand minnow]

that is the left side of this equation

so now you need to find the right side and see if its the same

and thats just matrix multiplication

[vague coral]

Ok?

[sand minnow]

if they are the same then that is a solution to the differential equation

which is part of what the question is asking

[vague coral]

Okay….

[sand minnow]

you dont seem convinced

[vague coral]

So I multiply by y

[vague coral]

It’s not that I’m convinced it’s just that I,m extremely not confident

[sand minnow]

for differential equations when they give you a solution you just need to find that the equality in the original diff eq holds at all times. so if you just plug everything in you should get complete cancellation

if you just really look at the equation it will tell you what to do, so yes in this case you perform the matrix multiplication between that matrix and y

[vague coral]

That would be y, 3y,3y,y

Right?

But then what lol

[safe radish]

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[dense shuttle]

is my diagram correct?
if so, for c), trying to figure out why 2, 4, 8 is the maximal line
couldnt you do 1 -> 6 -> 2 -> 4 -> 8 ?
also i dont understand b) either, is a vertex which isnt related maximal & minimal? surely ive messed up somewhere..?

[safe radish]

@dense shuttle Has your question been resolved?

[safe radish]

@dense shuttle Has your question been resolved?

[lean otter]

hm

idk what poset is

[safe radish]

@dense shuttle Has your question been resolved?

[dense shuttle]

<@&286206848099549185> 😉 any help would be great
my exam is tomorrow and if i pass this course (🙏) then it will be my last math course so u wont be bothered by me anymore (probably)

[safe radish]

@dense shuttle Has your question been resolved?

[dense shuttle]

.close

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[vital smelt]

How to know why the Lim changes here?

[split ether]

lnx approaches -infinity as x approaches 0 from above

[vital smelt]

Oh so x↓0 is always x→-Inf
And x↑0 is always x→Inf
?

Is that correct @split ether ?

[split ether]

No?

[vital smelt]

Hmmm then I still don't get why in this case you change the limit to > -Inf
Sorry

[split ether]

x -> 0
lnx -> -infinity

It doesn't say x -> infinity

[vital smelt]

But is x > 0 is not the same as x ^ 0 right?

[split ether]

x ^ 0 means that it approaches 0 from below

[vital smelt]

Yeah from the left right? So the x value keeps increasing and it goes to --> Positive infinity

[split ether]

Yes, but when x approaches 0 from the left, lnx doesn't exist

[safe radish]

@vital smelt Has your question been resolved?

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[coral parcel]

how do you change from the equation above to the one below

[lusty widget]

same denom

cross multiply

move over term without y’ to RHS

and factor y’ on LHS

and divide

[coral parcel]

is this correct

[lusty widget]

where did

the cos go

also why is there xy in the denom

[coral parcel]

i factorised out y'

and moved 3y/xy over

to the other side

and i multiplied 2x by xy so they have the same denominator

[lusty widget]

how’d u move it over?

also where is the cos

[coral parcel]

wow your handwriting is really neat

i mean the cos y is still there

i just factorised out the y'

yeah i think i just gotta take things step by step

[safe radish]

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[teal skiff]

Hey, all, there's a certain theorem similiar to the factor theorem, but the factor theorem is only when the leading term has a coefficient of one. what's the theorem called to find a factor for a polynomial where the polynomial is non-monic?

[safe radish]

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[full kettle]

Does the given number series converge?

[full kettle]

I'm trying to understand what my teacher did. So if we have 0<=(a)n<=(b)n and b(n) is convergent

then (a)n is also covergent

[lean otter]

compare it with 1/sqrt(n)?

[full kettle]

but he did something like this

and I totally don't get it

[winged cloak]

i think he starts from bottom to top

[full kettle]

If I understand it right, for every n b(n) should be bigger than (a)n but it isn't for 1/2sqrt(n)

[fiery merlin]

Compare it to the harmonic series.

[flat frigate]

Chai T. Rex

[fiery merlin]

The terms of that are less than the terms of yours after n gets to three, I think.

[full kettle]

but I can also do this, right?

and 1/sqrt(n) isn't convergent

so also 1/(sqrt(n)+1) isn't

am I right?

[fiery merlin]

Well, you need it to be ≥ a series that diverges to infinity.

I mean 0 ≤ 1/sqrt(n) as well, but the sum of a bunch of 0s converges.

[lean otter]

maybe limit comparison with 1/n might be more intuitive for you

[fiery merlin]

The 1/(2 sqrt(n)) works, though.

And that's the same as the series of 1/sqrt(n) with a 1/2 factored out of all the terms.

[lean otter]

yes, did not deny that, but they said they did not get that method

so just giving another solution

[flat frigate]

Chai T. Rex

[full kettle]

but this means that every 1/(2sqrt(n)) is smaller than 1/(1+sqrt(n)), right?

[fiery merlin]

Yes, that's right.

And so the thing you started with is greater than that series with 1/(2 sqrt(n)).

All the terms after a small n are greater.

[full kettle]

OOOOOOOOOOOOH

wait now I see

and since (a)n isn't convergent

then (b)n isn't either

wait

nvmd

[flat frigate]

Chai T. Rex

[fiery merlin]

So, 1/sqrt(n) diverging means 1/(2 sqrt(n)) diverging, which is less than 1/(1 + sqrt(n)), so that diverges as well.

[full kettle]

okay now I get it

thank you very much

[fiery merlin]

No problem.

[full kettle]

.close

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[keen idol]

Please can I have some help with question 7

I thought the tension in each string from the 20N would be the weights, but that's wrong when I tried

So I'm not sure what to do

<@&286206848099549185>

[fiery merlin]

No, they have tension pulling the weight up vertically and fighting each other horizontally.

They balance out horizontally and lift up 20N vertically.

So, how would you get the horizontal components of the tensions?

[keen idol]

t1 x sin50 and t2 x sin65

[fiery merlin]

Right, and those balance out or their meeting point would be moving horizontally.

[keen idol]

yeah

[fiery merlin]

So, W₁ sin(50°) = W₂ sin(65°)

And then the vertical tensions sum to 20N upwards, so what are their vertical components?

[keen idol]

t1 cos 50 and t2 cos65

so they are equal to 20?

when added

[fiery merlin]

OK, so W₁ cos(50°) + W₂ cos(65°) = 20 N.

Now you have two simultaneous equations.

[keen idol]

ahh i forgot the horizonal bit

thanks mate

[fiery merlin]

You're welcome.

[keen idol]

i found that second ewquation butn i completely forgot about the horzional component

im not used to that sort of pulley diagram lol

[fiery merlin]

Oh, OK.

[keen idol]

thanks again

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[fiery merlin]

No problem.

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[foggy salmon]

whats

det(A)

[mighty phoenix]

where

[foggy salmon]

um

determinant of a matrix

[mighty phoenix]

in my book there is a formula to solve this using elementary row opreations

[lean otter]

solve with identity matrix

[mighty phoenix]

yes yes

[lean otter]

and do normal row reduction operations

just do the same on the other side with the identity matrix

[mighty phoenix]

A = I2A

[foggy salmon]

eh

isnt it goood to check if inverse exists first

[mighty phoenix]

It exists i think

[foggy salmon]

think-

[lean otter]

they usually teache inverses and this stuff before they get into determinants i believe

[mighty phoenix]

How to solve from now on

[lean otter]

in most curricula

[mighty phoenix]

....

[foggy salmon]

o

sry im

2 low cultivation

[lean otter]

its okay, if it has pivots and shit everywhere then it has an inverse

[lean otter]

you want help with

row reduction?

[mighty phoenix]

yes

[lean otter]

just do the same thing i told u about last time

just

write it in this way

[mighty phoenix]

i did it 2 steps

what oprations i do now

i can't think of solutions

[lean otter]

Write it like this

,rotate

[mighty phoenix]

yep i did that

i just didn't send here ss

[lean otter]

okay first row scale that by multiplying 1/2

you get 1 2 3

[mighty phoenix]

yaa

[lean otter]

then

multiply by -3 and add with second row

send me ur matrix after u r done with that

[mighty phoenix]

see this

[lean otter]

oh okay sorry

[mighty phoenix]

now

[lean otter]

where did u get the -5 and -7 in second row?

ah okay

matrices are confusing

got it dw

[mighty phoenix]

row 2 change to R2 - 3R1

[lean otter]

although im confused with ur -16 and -1 in ur identity matrix

its gonna be 0 + 0 anyways for that no?

[mighty phoenix]

ohh yess

sorry

[lean otter]

yeah anyways

whats ur new matrix?

[mighty phoenix]

so my L.H.S =
1 2 3
0 -5 -7
0 1 -1

[lean otter]

sounds fine

[mighty phoenix]

R.H.S
1/2 0 0
-3/2 1 0
0 1 1

[lean otter]

yep sounds gucci

next step

scale second row by -1/5

[mighty phoenix]

ok then

[lean otter]

then multiply row 2 with -1 and add with row 3

[mighty phoenix]

you are saying we have to change row 2 or row 3

[lean otter]

just saying scale row 2 first by multiplying with -1/5 so u have a 1 in the pivot point

and then afterwards multiply row 2 with -1 and add that to row 3

[mighty phoenix]

ok

[lean otter]

i am trusting you with the calculations, because i cannot keep track

show me your resultant matrix after u r done

[mighty phoenix]

ok

ok i can slove now

Thanks for help

.close

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[ornate oracle]

Hi, could someone explain to me how to solve this equation please? I'm having trouble isolating x

[split ether]

I'd move (x + 15)/2 to the other side and square both sides of the equation

(Note that 20 - (x + 15)/2 >= 0)

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[safe radish]

@lean otter Has your question been resolved?

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@lean otter Has your question been resolved?

[lean otter]

hmm

basically

(3x+2)(x+2) = (x-1)(x+2), then (3x+2)(x+2)-(x-1)(x+2) = 0

so factor out (x+2),

( (3x+2)-(x-1) ) * (x+2) = 0

so

(2x+3)(x+2) = 0

[lean otter]

kind of the same thing as 3*4 - 2*4

so

3*4 - 2*4 is (3-2)*4

oh

thats just

(3x+2) - (x-1)

oh lol

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[inner nest]

Just a quick question

[inner nest]

What are these points called?

The very edge points respect to an axis

[halcyon carbon]

Is there even a name for it?

[inner nest]

I believe yes

I just forgot that fucking name

[lean otter]

Circle's edge points?

Honestly no clue what you would be talking about, unless it is something about parametric curves that is just flying right over my head

Oh wait @inner nest

Are u talking about vertices or something lmao

And co-vertices

[inner nest]

bruh

I am talking about a circle's equation

not an ellipse

[lean otter]

I mean a circle is an ellipse so

Other than vertices I don't have a single clue

[inner nest]

😑

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[inner nest]

bs

I am closing this channel

.close

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[low flicker]

Given the know quantities of r1, d, and h I am trying to find r2.
I know it has something to do with similar triangles but am not really getting anywhere with that. I'm assuming I'll use Pythagorean theorem but not sure.

[split ether]

r2/d = r1/h

[low flicker]

is that because the ratios of the sides have to be the same?

[tiny wraith]

yes, the triangles are similiar

[low flicker]

.close

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[proud sedge]

I have to solve X can someone help me

And explain to me how to solve it

[stray socket]

Multiply by the conjugate of the LHS

Well actually

Identify what x cannot be

Do that first

[proud sedge]

I have to solve the equation

[stray socket]

That'll help with winnowing out extraneous solutions

Yeah I know

Start with knowing what x cannot be

#❓how-to-get-help this ain't your channel

[delicate nova]

Ok

[stray socket]

Delete the image

[delicate nova]

How

[stray socket]

Hold down in it, press delete message

[delicate nova]

Ok i did it

Sorry for bothering

I suck at algebra

[stray socket]

@proud sedge start by multiplying by the conjugate, or start by identifying what x cannot be

Either way you're gonna have to do both

So you choose what you wanna do first

[proud sedge]

I have zero clue how to multiply them

[stray socket]

I guess we'll start by multiplying by the conjugate

So what is the conjugate of the left hand side

[proud sedge]

3x+1?