help-42

[exotic cosmos]

isn't the vector sum there

for kv + lw if k = 1 = l

which is like what i meant with the parallelogram

[ancient thistle]

but we aren't taking their sum

we're taking a vector on the line segment joining them

[exotic cosmos]

oh that's what u mean

okay

[ancient thistle]

d*kv + e*lw for some d + e = 1, 0 <= d,e,k,l <= 1

[exotic cosmos]

yeah makes sense i think

what are we doing here anyway

like the main goal here is to

show what exactly?

like we left off at c + d + e = 1

showing c,d,e >= 0?

[ancient thistle]

well

we didn't construct every point in the triangle

we only found the black edge

and i should be using primes for all these

d*kv' + e*lw' for some d + e = 1, 0 <= d,e,k,l <= 1

[exotic cosmos]

we found the black edge but didn't we work backwards

and get 1 here

[exotic cosmos]

yep i understsand these steps

but like i'm lost what the role of it is

[ancient thistle]

no that was just to tell you that we weren't done

[ancient thistle]

okay so

essentially

[exotic cosmos]

oh wait

we found the black edge

c + d + e = 1

right?

wait huh

[ancient thistle]

yes but it turns out c + d + e = 1 is not really a constraint

that's like the least important one

[exotic cosmos]

ohh

so like okay

we also need something about the signs of c,d, and e

[ancient thistle]

you can get that for free from my argument here without any mention of our line segment stuff

so yeah we need to worry about the signs

[exotic cosmos]

yeah i get that part

[ancient thistle]

but back to d*kv' + e*lw'

[exotic cosmos]

why?

i mean aside from c + d + e = 1

not being a constraint

why does something like

[ancient thistle]

well i can set c = 10000

[exotic cosmos]

c = -100 and then d + e = 101

not work

[ancient thistle]

it's outside the triangle

c + d + e = 1 gives you every single point in the plane

[exotic cosmos]

oh yeah that's the equation of a plane

😭 all these concepts i just haven't learnt yet kinda

[ancient thistle]

for any point in the plane containing u, v, w, you can find c, d, e such that it's cu + dv + ew

which isn't something you need to know

but it's true

[exotic cosmos]

yes that's correct

x + y + z = 1 is a plane where 1 is roughly the distance from th eorigin or smth okay yeah

fine

[ancient thistle]

yeah

[exotic cosmos]

the entire plane can't possibly be the triangle

[ancient thistle]

we've concluded that every point in the triangle can be expressed as d*kv' + e*lw' for some d + e = 1, 0 <= d,e,k,l <= 1 however

more importantly, you can write that expression as
(dk)v' + (el)w'

where 0 <= dk, el <= 1, dk + el <= 1

[exotic cosmos]

yes correct

but like something feels off cuz we're working back in the diagram where u' = 0 😭

[ancient thistle]

so we may as well chuck out the fact that it's dk and el, and just write it with a single symbol each

[exotic cosmos]

when we came back to u = whatever

[ancient thistle]

this is okay, pretend we never went back to u

we do that at the very end

[exotic cosmos]

so you mean dk v' + e lw'

and then sub v' and w' in

[ancient thistle]

yeah

but we can just write it as
dv' + ew' where 0 <= d,e <= 1 and d + e <= 1

[exotic cosmos]

0<= k,l <= 1 and d + e = 1 for black line so yeah sure 0 <= k,d,e,l <= 1

[ancient thistle]

dk -> d, el -> e

[exotic cosmos]

hmmmm

yeah okay

i think

so 0 <= d,e <= 1

[ancient thistle]

ye

[exotic cosmos]

since dk -> d and el -> e

wait

okay fine yeah

[ancient thistle]

and d + e <= 1 now

cuz before d + e = 1

if you multiply by k and l

dk + el <= 1

[exotic cosmos]

yeah

[ancient thistle]

cuz k and l are both <= 1

okay nice

so we are basically done

in our version where u' = 0

the triangle can be described as
dv' + ew', d + e <= 1, d,e >= 0

[exotic cosmos]

indeed

[ancient thistle]

translating back to u, we get
u + dv' + ew' = (1 - d - e)u + dv + ew

the calculation i did earlier

d,e >= 0 we have

but the additional constraint d + e <= 1 means that the coefficient of u is
1 - (d + e) >= 0

[exotic cosmos]

yeah and we already knew that d + e <= 1

what about c 😭

no bounds for that?

[ancient thistle]

c = 1 - d - e

it's the coefficient of u

[exotic cosmos]

wait i'm dumb

[exotic cosmos]

oh yeah

this is what u meant

so c >= 0 and d >= 0 and e >= 0

and that's it

wait

yeah nvm

the sum is indeed 1

which wasn't dependent on c

oh my god that was definitely one of the hardest questions i've done this week

😭

tysmmm, i feel really bad i think for wasting so much time of yours

u could've probably written half a book LOL

[ancient thistle]

if you want to read more context about this problem

you can look up barycentric coordinates

[exotic cosmos]

ohh

isn't it competition math

[ancient thistle]

uh

i wouldn't say it's competition maths per se

like it's a thing that exists which people use

[exotic cosmos]

and briefly how does that relate? like okay i search up barycentric coordinates but what exactly did we do such that it was motivated by barycentric stufffs

[exotic cosmos]

oh it's something about mass thing right

[ancient thistle]

kind of?

you can reframe it in terms of mass

but in the most basic form

barycentric coordinates is a way to give 3 numbers to describe a point inside of a triangle

and the 3 numbers satisfy exactly the constraints we derived

c + d + e = 1, all non-negative

[exotic cosmos]

ohhh

so with barycentric coordinates (i'll research more about this)

would've just solved our question easily cuz it's a tool

that lets us

find the region inside triangles or something?

[ancient thistle]

not exactly

i'd rather say that what we did was prove that barycentric coordinates exist

c, d, e in your problem are the barycentric coordinates of a point in the triangle uvw

[exotic cosmos]

oh

so the coefficients

kinda

that sounds like something weird

where do you learn about this

if you follow a general learning trend/path

or is this just extra learning and doesn't fall into any specific "subject"?

[ancient thistle]

idk it probably comes up in linalg at some point

[exotic cosmos]

have u used this book before?

the gilbert strang one

or is this axler stuff

[ancient thistle]

i haven't used either book to learn linalg

[exotic cosmos]

oh yeah you're the genius

wait omg

you're right

[ancient thistle]

i've only read axler because other people use it here and i end up helping them

[exotic cosmos]

it kinda looks like what we did

[exotic cosmos]

do u just use lecture notes or smth?

*did

[ancient thistle]

uh

i guess?

if you're doing computer graphics programming, you'll have to use what we found today to test if points lie within triangles

i've had to program this at least a couple of times

[exotic cosmos]

oh i'm not that good at it

but that's barycentric

in a nutshell right?

[ancient thistle]

yeah

[exotic cosmos]

like doing this

and that also happened

to be what we did in this problem

[ancient thistle]

yes

[exotic cosmos]

okkkkkk so wait

it shows up in geometry?

like where is this introduced even

[ancient thistle]

linalg is kinda the backbone for coordinate geometry

[exotic cosmos]

oh so this does

show up here

after all

here = linear algebra

okay i was curious cuz

[ancient thistle]

yeah

[exotic cosmos]

i can watch a video sure but idk where to find questions pertinent to it

cuz this topic

is definitely not in my book

the gilbert one i chose

so that's that

[ancient thistle]

it gets used in plane geometry sometimes so there will be material on that

[exotic cosmos]

but also i heard barycentric is advanced comp math topic so i don't want to do hard questions which idk how to do either

[exotic cosmos]

oh

[ancient thistle]

yeah that's what i mean by plane geometry

[exotic cosmos]

i see

i'll look into it

but yeah looks very related

exercise 2:

okay tysmm once again

sorry for taking alot of your time

[ancient thistle]

lol

the notation is scarier than what it's really trying to say

very typical of maths

[exotic cosmos]

😭 yeah looks complicated but also approacable cuz like

it's what we did

and this just looks like

[ancient thistle]

ye

[exotic cosmos]

a copied version of that

hilarious but

everything after page 1

is out of my level already

[ancient thistle]

lol

[exotic cosmos]

so only that half page i think

[ancient thistle]

dw about it then

[exotic cosmos]

yeah i'll just watch a video and hopefully if i ever see something similar

i can watch it again

okay bye

[ancient thistle]

[exotic cosmos]

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[livid mauve]

Hi, I need help in linearisation

[livid mauve]

I have a downwards facing parabola and idk how to linearise it, I want help in its linearisation

<@&286206848099549185>

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@livid mauve Has your question been resolved?

[livid mauve]

<@&286206848099549185>

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@livid mauve Has your question been resolved?

[livid mauve]

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[toxic elm]

guys, can anyone teach me what this probability generation function is doing? why can you plug in i and -1 to it??? And how did the last step work in the first picture? (the one arriving at (i-1)/6+(-i-1)/6+1)) why can we just use the same formula for the experiment. What's the difference between specific formula of the quesiton and the definition of pgf (since the information allowed us to construct a function solving for P(xmod4=0), I believe there are information differences in the two expressions)

[leaden thunder]

Your notes are quite confusing

do you have the original problem and definition of probability generating function

[toxic elm]

definition

the problem is to find the probability that the sum of 10 dice rolls (each independent) is divisible by 4

sorry our TA's lecture style is all over the place...

[calm coral]

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[foggy salmon]

I need help with the 15th question

[foggy salmon]

I don't know what else I'm missing..

I've only got 2 equations

[trim jungle]

find p'(0)

ur missinf cx from ur equation

[trim jungle]

u have written it wrong

[foggy salmon]

P'(0) = c, and they said x is the only real factor so c must be zero.. atleast that's what I thought

[trim jungle]

ok

also

[foggy salmon]

Sorry I seem to have skipped a few steps

[trim jungle]

yea

p(x) is greater than 0

and p'(x) is less than 0

[foggy salmon]

Why so?

I wasn't sure abt the sign of d so I couldn't say anything abt P(x)

Even for x>0, it would still depend on d right?

[foggy salmon]

I don't understand

[trim jungle]

because it is given that p(-1) is less than p(1) which means a<0 and b<0 which mean p'(x) is greater than 0

[foggy salmon]

Oh you're right P'(x) won't depend on d so.. for x>0, P'(x)>0 right?

[trim jungle]

exactly

[foggy salmon]

Okay

[trim jungle]

after diffrentiation it won't matter

[foggy salmon]

Yea

[trim jungle]

but u got what i said right

about increasing and decreasing

[foggy salmon]

Yea, abt P'(x) being greater than 0.. for x>0

[trim jungle]

ok

so u know the answer now?

[foggy salmon]

No I still don't know what to do with that information

I'll cook a bit with whatever you said for now

[trim jungle]

see we can find that p(1) is max....you got that right?

[foggy salmon]

Wait how?

[trim jungle]

increasing function....

[foggy salmon]

Hm in the interval [0,1], I think so

[trim jungle]

yess

[foggy salmon]

But what abt [-1,0)?

[trim jungle]

now what is the condition for decreasing function?

[foggy salmon]

f'(x)<0

[trim jungle]

what is your answer for decreasing coming?

[foggy salmon]

I didn't get one-

Where do I apply that condition?

[trim jungle]

it will come greater than 0

[foggy salmon]

what will

[trim jungle]

4x^2 + 3ax + 2bx > 0

right?

[foggy salmon]

Yess

In the int [0,1]

[trim jungle]

yes

so it is not less than o

[trim jungle]

.

you said the condition yourself

[foggy salmon]

Yes..

So it's increasing right

[trim jungle]

that for it to be decreasing

it should be less than zero

[trim jungle]

yes

now tell me is it option 1 or 2

you got it right?

[foggy salmon]

How did you decide P(1) is now the max

[trim jungle]

cause the condition-

f'(x) > 0 - increasing....

in [0,1]

[foggy salmon]

Say there's a value, in the interval [-1,0], that's larger than P(1)

[foggy salmon]

Yes, it's the max in the interval [0,1]

[trim jungle]

so u tell me now is p(1) max of p?

or not

[foggy salmon]

Idk, cause I haven't figured out the nature of P(x) in [-1,0)

Gimme a sec to process

[trim jungle]

bro i think you should ask a teacher....

[trim jungle]

okk

[foggy salmon]

Ok I'm sorry my bad, P'(x) is increasing in the whole interval [-1,1], meaning P(1) is the maximum in that interval

[trim jungle]

yesss

[foggy salmon]

Wouldn't that mean P(-1) is min?

[trim jungle]

no

because

f'(x) > 0

[foggy salmon]

I thought that's why it's the min-

[trim jungle]

but a > 0 and b > 0

so p, (x) can never be less than 0

[foggy salmon]

What about d?.. and wouldn't the x³ term change the sign?

[trim jungle]

d is not at all relevant here

[foggy salmon]

I think I'm missing something, is there some condition to be applied for when a polynomial power 4 is always positive?

[foggy salmon]

I don't get why

[trim jungle]

i really think you should ask your teacher tom

oe watch a video on increasing and decreasing functions

cause you are getting more confsed by me

sorry

[foggy salmon]

Yea I feel stupid ill do that, thanks

[trim jungle]

or ping helpers again

[trim jungle]

nah i am also stupid as i cannot help u bro

it is ok

[foggy salmon]

Nah thanks for sticking with me

[trim jungle]

no one is stupid just because they can't solve a few questions

don't underestimate yourself

you can do it with a little help from teachers and practice

[foggy salmon]

[foggy salmon]

Okkk I finally get it it's only inc in [0,1] ;-; it's decreasing in [-1,0] and I could have found that the same way I said it's increasing as long as x>0.. so when x<0 P'(x) is negative, i.e., decreasing.. sooo P(0)<P(-1)

THANK YOU I FINALLY GOT IT

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[lean aspen]

how should I factor x^3 - 6x^2 + 11x?

I pull out the x to get x(x^2-6x+11) but I'm stuck trying to factor it from here.

[lean aspen]

this is in reference to a partial fractions problem

[glad parrot]

You can start by x

And factor the 2nd degree polynomial that will show up

[lean aspen]

i said i already did that

[glad parrot]

Indeed

[upper sparrow]

You could split the partial fractions you get into the form
[
\frac{A}x + \frac{Bx + C}{x^2 - 6x + 11}
]

[potent lotus]

@upper sparrow

[lean aspen]

hmm thats not a bad idea]

ight thanks I'll work with that

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[karmic willow]

Super confused rn my homework is asking to simplify log_7(49)

[karmic willow]

I made it exponential and solved for x and got 7

[velvet osprey]

well that's quite far off

can you show what happened after you made it exponential?

[karmic willow]

Nevermind I got it mixed up it's 2

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[frank isle]

please how dow i find Alpha????

[ancient grotto]

Set up a system of equations using trigonometric functions.

Note that the foot of P and Q are at the same point.

Thus, what trig function should you use?

[pale blade]

Force equilibrium on the horizontal/x direction.

[frank isle]

this one?

[ancient grotto]

Yep.

[frank isle]

aight

but how do i get alpha

[ancient grotto]

Ah, I misread.

Sorry.

I mean, it might work.

[ancient grotto]

Use cosine on P.

[frank isle]

wht u mean

[calm coral]

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[stiff dragon]

Ok so I know it happens at the x value 2

bc y' = 6x - 7
5 = 6x - 7
12 = 6x
2 = x

but now, i thought you plug in 2 back in the original equation of y which gives you 3 so its (2,3)

and then you use the x and y to solve for y = mx + b

but thats wrong

LOL

because i got y = 5x - 12 but the answer is y = 5x - 7

[velvet osprey]

how did you get y=5x-12

[manic vault]

,calc 3(2)^2-7(2)+5

[potent lotus]

Result:

3

[velvet osprey]

sounds like an execution error to me

[manic vault]

Show your work.

[velvet osprey]

cause your planning sounds correct the way you described it

[stiff dragon]

oh

OMG

LOL

i realized for x and y i did both 3

[manic vault]

You might have mixed up the x and y-coordinates, or something along those lines.

[stiff dragon]

i forgot that x is 2

[manic vault]

Ah.

[stiff dragon]

thank you so much lol

i was worried that my steps are wrong

but that means i just need to be more careful on the small details

thank youu

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[still terrace]

how do you check if a vector is coplanar?

[restive elk]

U cant

[mortal orbit]

coplanar to what

[still terrace]

uhhh

this exercise says

show that the vectors arent coplanar

[mortal orbit]

so vector is coplanar TO another vector

[still terrace]

yeah i guess

but how do i do this

help?

<@&286206848099549185>

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[calm coral]

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[upper sparrow]

Did you have any idea how to begin this one?

[manic vault]

Try to draw it.

Always helps out.

[viral abyss]

i did and i got 82.1 but idk if its rifht

[manic vault]

Show your work.

[viral abyss]

i cant ss on pc but the hypotenuse of teh triangle i drew has hypoteuse 210, long leg is 5 and 23 degeree is in between them

[manic vault]

Nevrmind

[upper sparrow]

I don't think that's how I interpret what they mean in the question

[manic vault]

^

[upper sparrow]

Of course, "distance to the drone" is a bit ambiguous, but I assume that's meant to mean the horizontal distance

[manic vault]

The drone is not necessarily parallel to you

[viral abyss]

then where is the5

[manic vault]

Well that is how I originally interpreted it

[manic vault]

Below your eyes

You basically construct a rectangle below the triangle

[viral abyss]

what

[keen grail]

Pretty sure by distance to the drone they are referring to the hypotenuse

[manic vault]

Let me draw what I have on paper

[viral abyss]

ok now what

[manic vault]

Let me mark it

[keen grail]

I think you’re making it more complicated/wrong lol

You find the opposite side of the angle using sin rules and just add 5 at the end

[manic vault]

It still does same thing

[keen grail]

From the way I interpret the question, the triangle starts at eye level

[manic vault]

You want black+green

[viral abyss]

yes

[manic vault]

So use sin(23) to get black

[keen grail]

But you have the hypotenuse labelled as the entire thing from the ground

[manic vault]

No

You misinterpreted it

[keen grail]

K

[viral abyss]

i think i got this.. if not ill ask teacher tmrw

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[opaque yoke]

Could someone help please

I'm not sure what to do

[high sentinel]

Look back at what your notes tell you or watch a video from the organic chem

Im not a helper but i put this through gemin and this is what it says (might be wrong) but look back at notes

1. Recognize Similar Triangles

Since angles A and B are congruent, and we can assume the triangles share the same angle at the unlabeled vertex, the two triangles are similar by Angle-Angle Similarity. This means their corresponding sides are proportional.

2. Set Up a Proportion

Match the corresponding sides of the triangles. The side of length 4 in triangle A corresponds to the side of length 2 in triangle B. The side of length 6 in triangle A corresponds to the side of length x in triangle B. We can set up the following proportion:

4/2 = 6/x
3. Solve for x

Simplify the proportion:

2 = 6/x
Multiply both sides by x:

2x = 6
Divide both sides by 2:

x = 3
4. Check the Answer Choices

Since 3 is not one of the answer choices, the correct answer is E. NOTA (None of the Above).

its probably wrong but idk

[opaque yoke]

yeah its wrong

angle-angle cant be used i think

idk

[upper sparrow]

!nogpt, please don't do that

[calm coral]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[opaque yoke]

.close

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[pearl iron]

How would I rigorously prove that F(s) + G(t) ≥ st

Not geometrically

[calm coral]

@pearl iron Has your question been resolved?

[calm coral]

@pearl iron Has your question been resolved?

[ionic venture]

is this calc

[potent smelt]

Because g = f^(-1) we have that f(x) = y implies g(y) = x. Therefore, if we have a point on the curve (x, f(x)) = (g(y), y) then consider a second point: (x + Δx, f(x + Δx)) = (g(f(x + Δx)), f(x + Δx)) we can construct the following two triangles: the first with base Δx and height f(x) and the second with base Δy = f(x + Δx) - f(x), and height g(y). Notice that Δy can be approximated as df/dx Δx, and this approximation becomes exact as we take the limit as Δx -> 0.

Then from here it's a matter of limit manipulation, and using the definitions. Assuming I haven't made a mistake in the setup, which is possible. Give it a go.

@pearl iron

[calm coral]

@pearl iron Has your question been resolved?

[calm coral]

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[calm coral]

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[scarlet lichen]

can someone help me out with the concept of projecting vectors

[scarlet lichen]

i don’t rlly get it

the whole shadow thing

[winter elbow]

!show

[calm coral]

Show your work, and if possible, explain where you are stuck.

[scarlet lichen]

there’s no work it’s the concepts

i understand the formula

but the concept

[winter elbow]

like you don't know why the formula looks like that?

[scarlet lichen]

idc abt the formula right now question is asking me to just draw the scalar projections of one vector on the other

like what’s the goal

are we trying to make a right angle or what

idk

[eternal shard]

https://www.youtube.com/watch?v=PLcniG5HmRk

[scarlet lichen]

doesn’t rlly make sense to me

how to find projection of b onto A

becuase it’s a straight line down so there’s no shadow

[swift laurel]

,tikz
{[rotate around={30:(0,0)}]
\draw[dashed] (-3,0) -- (5,0);
{[thick, -Latex]
\draw[\red] (0,0) coordinate(O) -- node[midway, left]{$\vb a$} (2,2) coordinate(A);
\draw[\red] (O) --node[midway, below right]{$\mathcolor{black}{\proj}{\mblue{\vb b}} \mred{\vb a}$} (O -| A) coordinate(PA);
\draw[\blue] (PA) -- node[midway, below]{$\vb b$} ++ (1,0);
\draw[\green] (O) -- node[midway, above right]{$\vb c$} (-2, 1) coordinate(C);
\draw[\green] (O) --node[midway, below right]{$\mathcolor{black}{\proj}{\mblue{\vb b}} \mgreen{\vb c}$} (O -| C) coordinate(PC);
}
\draw[dashed] (A) -- (PA) (C) -- (PC);
\path pic[draw, angle radius = 7]{right angle =A--PA--O} pic[draw, angle radius = 7]{right angle =C--PC--O};
}

[potent lotus]

cloud

[swift laurel]

as you can see the projection can point in either the same direction or the opposite direction as the vector being projected onto

[calm coral]

@scarlet lichen Has your question been resolved?

[calm coral]

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[scenic atlas]

I got till here, not sure what to do nest

Oh nvm got it you have to use hopital

.close

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[hidden ice]

Im confused about part b, the only think important from part a is that when you remove the discontinuity at (2, 3/10) you have f(x) = 3x/(2x+1)(x+2)

This question should be really easy, its literally just algebra, but I got the wrong answer 🤔

[velvet osprey]

did you get lost in the algebraic sauce

[hidden ice]

I guess I did 😭

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

3x/(2x+1)(x+2) = ((4x+2) - x - 2))/(x+2)(2x+1)

[velvet osprey]

did you try to rawdog it starting from f

[hidden ice]

But then they wouldnt be equal right?

Cause of the discontinuity?

[velvet osprey]

this doesn't answer my question and i do not understand yours either

[hidden ice]

I thought that we needed to remove the discontinuity in order for f(x) and g(x) to be equal

Since it says 'Except for the perforation'

[velvet osprey]

yeah so your working will pre-assume x != 2

that's all

[hidden ice]

But if thats the case what difference would it make?

I will still get the same answer right?

Or no?

[velvet osprey]

idk what you mean by "the same answer"

i actually have little idea what you are talking about anymore?

[hidden ice]

Ok I will just leave it

[hidden ice]

I will try this

(3x^2 - 6x)/(2x + 1)(x^2 - 4) = ((2)(2x+1) - (x+2))/(x+2)(2x+1)

(3x^2 - 6x)/(2x + 1)(x^2 - 4) = (4x + 2 - x - 2)/(x+2)(2x+1)

(3x^2 - 6x)/(2x + 1)(x^2 - 4) = (3x)/(x+2)(2x+1)

Wait

That was literally our original equation with the discontinuity removed 👀

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

(3x)/(x+2)(2x+1) = (3x)/(x+2)(2x+1)

Ok ya that was easy 💀

Thanks though

.close

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[hidden ice]

This is the g(x) theyre referring to

The algebra is loooooooong but for h'(x) = 0 I got x = +-1 and g'(x) = 0 I got x = 1

[verbal finch]

What did you try?

[hidden ice]

Yes wait im telling you

[verbal finch]

Alr alr

[hidden ice]

The answer sheet confirmed these solutions, thats now what I was confused about

Im confused why I got 2 solutions for h'(x) = 0 but only 1 solution for g'(x) = 0

Does that mean its NOT true?

Or does that mean its only true for x = 1?

Or what does it mean im confused

[verbal finch]

Ok so u said u got 2 values ie ±1

But he mentioned that interval is from 0 to 3

So just ignore -1

[hidden ice]

Ohhhhhhhhhhhhhh

I forgot about that

[verbal finch]

Lol

[hidden ice]

So what do I actually say as my answer

Something like 'As you can see from the algebra above, they are both on the same interval (1, y- coordinate)'?

[verbal finch]

Yes basically now that u have x=1 for both equations

I proved that they attain their maxima at the same point

If u want to go super clear then just prove that f"(x)<0 for x =1

[hidden ice]

Ahhh yeyeye

Ok I get it now

Thanks for your help!

❤️

.close

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[wispy aurora]

Im stuck at how to open these

[wispy aurora]

does anyone have an idea to solve this

[random adder]

An idea would be to factor and just factor out that common -1

[wispy aurora]

what do you mean?

im sorry im not a native english

[random adder]

$$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$

[potent lotus]

casework

[wispy aurora]

wait ill try

[random adder]

You should know
$$\sum_{a = 1}^n a^2 = \frac{n(n + 1)(2n + 1)}{6}$$

[potent lotus]

casework

[wispy aurora]

What is this

[random adder]

A formula for the sum of first n squares

[wispy aurora]

I think im too dumb to understand this

.close

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[glossy haven]

whats this answer

[glossy haven]

i got 8 and 17.5

but they are wrong

[calm coral]

@glossy haven Has your question been resolved?

[glossy haven]

<@&286206848099549185>

[ionic sail]

is that all the information given to you

[glossy haven]

there is a graph

all i need is A now

im just gonna get the graph now

[glossy haven]

here

[ionic sail]

from the graph you can tell that there are 40 penguins
median is the mass of the 50%th percentile that is the mass of the 20th penguin
to find the interquartile range you need the mass of the 75th% percentile and 25th% percentil

and then subtract those masses

[glossy haven]

so the interquartile range would be 30?

tried to shorten interquartile range to the beginning letters and it didnt send

[ionic sail]

nope
what is the 75%th and 25%th percentile of 40

[glossy haven]

75th is 30 and 25th is 10

[ionic sail]

yes

what are the masses corresponding to these valus

values*

from the graph

[glossy haven]

19 and 23

[ionic sail]

just subtract these to get the interquartile range

[glossy haven]

4

[ionic sail]

yep

[glossy haven]

Tysm

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[glossy haven]

.close

[calm coral]

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[regal charm]

I have no idea where to start on this, and my teacher never covered this in class 😭

[regal charm]

i have NO idea why it's on our test review

[ancient grotto]

Do you know the dimensions for work?

[leaden thunder]

example 6.5.6
https://math.libretexts.org/Courses/Monroe_Community_College/MTH_211_Calculus_II/Chapter_6%3A_Applications_of_Integration/6.5%3A_Using_Integration_to_Determine_Work

[regal charm]

doesn't really ring a bell?

[balmy grove]

Triangles are half a quadratic shape?

[regal charm]

wait imma check if one of the other professors at my uni went over it in his videos

work's the integral kxdx between two limits

[ancient grotto]

Well, it’s a Newton-meter (N*m).

[regal charm]

oh that

[ancient grotto]

Try to conclude from there.

[regal charm]

wait isn't work either N*m or joules?

depending on if youre using imperial or metric or something like that

understood

[ancient grotto]

Yeah

I personally think in this context that N*m is better used.

[regal charm]

wait a second

how have i never heard of libretexts up until now

this site is awesome

gotcha

[ancient grotto]

Yeah, has it for all science and mathematics.

[regal charm]

I need this

gotcha

.close

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[scenic atlas]

Where did the series expansion fail?

Is missed a minus at the x^2 part but cant find any other error

[near seal]

take the limit to numerator and denominator

then multiply and divide by 4x

its done

[scenic atlas]

No i solved it

Just wanna know why series expansion failed

[leaden thunder]

doesn't look like series expansion enough terms in denominator

[scenic atlas]

Hmmm

.close

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[deep violet]

x takes values from between 0 and pi.
Maximize the following:
2cos(x/2) + sin(x)

[glad parrot]

What have you tried ?

[deep violet]

Calculus is not allowed.

We need to MAXIMIZE ++++++ the value of that expression.

[velvet osprey]

one non-calculus insight you can make is this:

[tardy chasm]

[velvet osprey]

letting f(x) be the expression you aim to maximize, you can immediately say f(x) > f(pi-x) for any x ∈ [0, π/2)

[glad parrot]

Rewrite sinx = 2cos(x/2)sin(x/2)

[remote mural]

hello, i'm 16 and i haven't been on school since my 10, i lost alot of practice in almost everything and my parents won't do anything, don't bully me pls 🙏

[velvet osprey]

since sin(x) = sin(pi-x) but cos((pi-x)/2) < cos(x/2)

[velvet osprey]

hi, this channel is occupied.

[remote mural]

ty

[velvet osprey]

i mean this in the gentlest way possible but you should move to #discussion

these help channels are for when somebody needs help with a specific question

[remote mural]

i need help

[velvet osprey]

#❓how-to-get-help has instructions for how to claim your own help channel

so that it has your name on it

mq do not encourage channel hogging please.

[velvet osprey]

did you think we would not see, read, or understand the word "maximize" the first time around?

[calm coral]

@deep violet Has your question been resolved?

[calm coral]

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[eternal shard]

What did you try?

[noble narwhal]

wait, i found the answer but now i need help for the second question: show that if z = exp(itheta) or z = exp(-itheta); the series is divergent

if z = exp(itheta)

@eternal shard and then?

[glad parrot]

Z=|z|exp(itheta)

You have to let the |z|

Since its going to be a condition for the convergence

[noble narwhal]

and then

i dont understand

[eternal shard]

can you post the original

[noble narwhal]

soooo

[vast flare]

Give me 5min

I’ll do it for u

[eternal shard]

I guess this one fails the divergence test if you were to take the limit

or no

[noble narwhal]

wha-

[eternal shard]

i have no idea leave me alone

[velvet osprey]

!nosols

[calm coral]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[noble narwhal]

wdym

[velvet osprey]

!nosols

[calm coral]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[glad parrot]

I thought you was asking for the first question

[manic vault]

A complex number z is composed of the modulous (|z|) and the argument (theta). In polar form, this is $z=re^{i\theta}$, where $r=|z|$.

[potent lotus]

mathisfun

[eternal shard]

modulos is just 1 here bruh

[manic vault]

Do you know what power series are?

[manic vault]

Not for question 1.

[eternal shard]

y'all went all the way back just to skip the part where q1 is done

[noble narwhal]

i dont understand this

[manic vault]

Do you see the condition?

[vast flare]

What didn’t you understand

[noble narwhal]

yes

[noble narwhal]

for convergence?

[manic vault]

This is overcomplicated. All you have to do is just apply the result from part 1.

[manic vault]

Yes.

[elder pawn]

@vast flare dont give full sols. i deleted it

[leaden thunder]

it was also wrong

[vast flare]

Okay sorry

[manic vault]

What is the modulous of e^(i*theta) and e^(-i*theta)? (adwin gave it away)?

[noble narwhal]

1

[eternal shard]

roketto should delete my solution too

[manic vault]

Good. And does that fall within the convergence?

[noble narwhal]

why u giving wrong answers

[eternal shard]

how ungentleman to give wrong solutions

[noble narwhal]

no

[manic vault]

So what can you conclude?

[noble narwhal]

that if z is that, the series does not converge

[leaden thunder]

i don't think the solution to a) that hhh showed proved the radius of convergence was equal to 1, just that it was less than 1

[manic vault]

Do you know how to calculate radius of convergence?

[noble narwhal]

youre right

[eternal shard]

just checking the equal case

prob

[leaden thunder]

yea i don't think you can use part a) to prove b)

[manic vault]

For which question in part b)? Both? Or just the radius of convergence?

[noble narwhal]

the first

i think..

[leaden thunder]

the first part of 2

[manic vault]

Hmm

[manic vault]

Can we use $\lim_{n\to\infty}\bigg\vert\frac{a_{n+1}}{a_n}\bigg\vert$?

[potent lotus]

mathisfun

[noble narwhal]

to find the radius?

[manic vault]

That is what I think we should use.

And also, for convergence/divergence.

[noble narwhal]

but the limit is hard to find

i dont think we can directly use this

[manic vault]

How so?

[eternal shard]

I think comparison test is a lot easier

you can just bound cosine

[noble narwhal]

because of theta, also i dont know why they would ask us to deduce the radius if we can just immediately apply the formula

[noble narwhal]

where

[eternal shard]

nvm i just forgot the partial terms have to be non-negative for that one

[manic vault]

Hmm

[leaden thunder]

<@&268886789983436800> scam

hehe con-mmunity. these bots find identify themselves

[eternal shard]

Dirichlet's test

[noble narwhal]

ok

Xn decreasing sequence would be

Xn = 1/nexp(-itheta)?

wait

[eternal shard]

it has to be a real decreasing sequence

according to wiki

[noble narwhal]

1/n then?

[eternal shard]

yea

so we have to find a contradiction that shows the complex sequence cos(nθ)e^(±iθ) is unbounded

[noble narwhal]

so we take a partial sum?

[eternal shard]

wait

[noble narwhal]

no srry

so we just take exp(itheta)?

because cos(ntheta) is bounded

?

uh

noo

what am i saying

the series cos(ntheta)exp(itheta)

[eternal shard]

this was what you had

[noble narwhal]

yes

[eternal shard]

this is like

[noble narwhal]

yes

[noble narwhal]

where do we separate

i mean

[eternal shard]

mid fraction

[noble narwhal]

the two sequences

[eternal shard]

in my notes i noticed the taylor series of ln(1-x)

\sum x^n/n

[noble narwhal]

uhm ok

but why tho

cant we just look at any partial sum of exp(i2ntheta)+1?

[leaden thunder]

she's asking how splitting the sum is justified

[noble narwhal]

yes like

[eternal shard]

it isnt 😦

[noble narwhal]

where is the taylor series of ln(1-x) in this

[eternal shard]

but i am unsure rn

this is what i had in mind

[noble narwhal]

but then it converges ??? or...

[eternal shard]

no

what about the other series

1/n

but i still dont know if that's a good justification

[noble narwhal]

uhm yeah it doesnt

it looks good

[eternal shard]

in my notes i only had that \sum x^n/n converges iff -1 < x \le 1

so i am unsure

[noble narwhal]

so we didnt even use dirichlet

[eternal shard]

ye

on wikipedia it says that what i used only converges iff |x| < 1 so i think that's not the right way to do it

[leaden thunder]

can you just use the taylor series for log on the first term

log(1-exp-1) or something

[noble narwhal]

is it the second part now?

to find the radius?

[eternal shard]

,, \frac{1}{2}\sum_{n\ge 1} \frac{e^{2in\theta}+1}{n} = \frac{1}{2}\sum_{n\ge 1} \frac{1}{n} \cdot (e^{2in\theta}+1)

[leaden thunder]

profound observation adonis

,tex .Maclaurin log

[eternal shard]

forgot the name

[leaden thunder]

rage quit

,tex .maclaurin

[potent lotus]

riemann

[eternal shard]

y

[leaden thunder]

bottom one with x = -numerator or something

[eternal shard]

i am just trying how one could use dirichlet but i am getting confused cause e^2intheta seems to be bounded 😕

[leaden thunder]

@noble narwhal did you even learn dirichlet

[noble narwhal]

uhm... yes but it was last year but now i remeber

so what can we say

to conclude

in conclusion, the radius is 1?

[leaden thunder]

up to you to decide which path. follow the log to justify it converges for the range of x in taylor

[eternal shard]

oh wait

[eternal shard]

this surely isn't bounded

|e^(2iθ)| = |e^(iθ)|^2 = 1

so the geometric series wouldnt converge and the sum of 1s anyway

[noble narwhal]

ok

[eternal shard]

i give up i am going to sleep

[noble narwhal]

waiitt

for the conclusion

if z = ewp(-,+iheta) it diverges

uhm no actually

yeah ok thanks good night

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[azure kite]

The activity is about Anti differentiation by substitution.

I don't know how to properly apply the rules and I'm really really bad at fractions.

Please let me know if I got #1 right, and how to continue numbers 2,3,4,5, and 6

[dull wagon]

is incorrect because your
du is wrong

[azure kite]

What I understand so far:
You use U on the one that is harder to solve, du is the easier one. Then you substitute the values and add +1 on the end

[dull wagon]

derivative of 5x isn't 1

[azure kite]

Please elaborate

Ohhhhh

Lik this?

Mb it's hard to use notes app with fingers lmao

[azure kite]

On #2, What if it's just dx on the numerator? What is its value?

[dull wagon]

du = 5 dx

why are you adding 5 to the denominator and the power

and having 5du at the end is also incorrect

du = 5 dx
dx = ?

[azure kite]

5

?

Or du/5?

[dull wagon]

dx =

du/5?

[azure kite]

?

[dull wagon]

yes

use that when chaning your variable

[azure kite]

Should I do it like this?

[dull wagon]

du shouldn't be there anymore after integrating

[azure kite]

So like it's just 5?🤔

Sorry I'm dumb😭

[dull wagon]

* 1/5

[azure kite]

Ohhhhh ok ok

Where do I apply the dx?

Do I distribute it with the U?
(5x-3) (1/5)

And (101)(1/5)

?

I'll go look for yt vids in the meantime. I'll be back

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[foggy tiger]

yo question whats a function that would go over this perfectly

[ionic moon]

elaborate?

what is "this"

and wdym by "go over"

[vagrant dune]

what

[foggy tiger]

like

hi

so

uhm

[tacit moat]

can you draw out what the function you want looks like?

[foggy tiger]

yeah ofc

[tacit moat]

it's not really possible to have a single function be all of those lines
you can make a function for each line

also, what is this for?

[foggy tiger]

sorry this is for a desmos projecty

if you want i can share a link so you can check it out

actually can you go over it and see if you can help change some lines around to make them look better

im having some trouble on the shoes

[tacit moat]

sure, share the link

[foggy tiger]

https://www.desmos.com/calculator/q9ddh1smd4

if you change anything can you send me the equation and tell me what one you changed

[tacit moat]

ah I don't think I can help much
it already looks pretty good imo, maybe the line weights needs tweaking

if you give me a specific line you don't like, then maybe I can find a better function

[foggy tiger]

Is there anything you can do for the feet? they look so out of place and bad

[foggy tiger]

thank you btw

heres my rubric let me know if i hit the points at like a teacher stand point

thank you sm 😭

[calm coral]

@foggy tiger Has your question been resolved?

[ionic moon]

this is good! id give you a perfect score

[calm coral]

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[pallid apex]

<@&286206848099549185>

[potent smelt]

!15min

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

[pallid apex]

sorry

[low river]

Use the second equation given

P = 80
r = rate, or 0.055
n = how many times per year it is compounded, so 12
t is the amount of time you have from now (age 34) to age 65

plug things in

[calm coral]

@pallid apex Has your question been resolved?

[calm coral]

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[stark hatch]

Someone please teach me how to do PNC questions

[stark hatch]

I promise I know PNC and that I did it really well before my midterms (september) but I've forgotten EVERYTHING

[marsh valley]

You can divide it into cases and use symmetry to avoid repeating calculations

Really this is asking how many ways you can add two numbers to 7 where each is less than 5

[stark hatch]

Yeah ik that but how do I show these calculations

[hushed knot]

Let me give you just one part of the answer and you would understand everything

[marsh valley]

Choose 2 questions from the first group
Choose 5 questions from the second
That’s one way

[hushed knot]

6C2×6C5
This is just one case
There are two groups of 6 questions and total questions are 7

[stark hatch]

Ohh ok

Okay okay got it ty

[hushed knot]

I think you understand

[marsh valley]

So you can just see which pairs add up to 7 and you’ll be golden

[stark hatch]

👍👍👍👍👍

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[heavy nimbus]

I'm so lost here😭

[blazing coyote]

You're trying to prove equality?

[heavy nimbus]

Yep

[blazing coyote]

okay

so there;s a simpler way

find $\frac{2}{2} \log(2) - \frac{1}{2} \log(2)$

[velvet osprey]

\log

[potent lotus]

What a wonderful world !

[heavy nimbus]

I'm definitely doing this wrong💀

[blazing coyote]

uh, yeah

isn't it just $\frac{1 \log(2)}{2}$

[potent lotus]

What a wonderful world !

[ocean hornet]

Do you know about addition property of logarithms when base is same?

|| log(a) with base c + log(b) with base c = log(ab) with base c ||

[heavy nimbus]

Yep, you multiply the arguments?