help-26

[teal gust]

is one perpendicular to the page

can you visualize that

[tropic zephyr]

Well it says the axis is perpendicular so I'm assuming the radius is vertical distance

[teal gust]

sorta like this ( imagine it coming out of the page )

so the r here is the distance of that mass from the axis

[tropic zephyr]

But that looks like a parallel line

[teal gust]

just by looking at the diagram can you figure out what that r is ?

[teal gust]

yeah just take a page and poke a pencil through it

i cant really describe it to you since

yk

its 3d

[tropic zephyr]

Maybe half of 10

[teal gust]

why half?

[tropic zephyr]

Because it's radius

d/2 =r

[teal gust]

huh

[tropic zephyr]

d is diameter and r is radius

[teal gust]

if you look at the axis from top view

itll be like this

so L here is just 10

like this

so from that Axis through A

the distance of b is just 10 cm

[tropic zephyr]

Do you mean this

[teal gust]

you dont need the arrow

ijust plucked a random image off the internet

ignore the circle

the black dot represents the axis

the orange ball represents the mass B ror mass C

and L represents the distance between the axis ( black dot ) and mass B/mass C

[tropic zephyr]

I think I sort of see what you mean by the top view

[teal gust]

yeahh

so does 10 cm being = r make sense now

[tropic zephyr]

Sort of

[teal gust]

then youve got your answer

just plug in these values into the formula

for B and C

[tropic zephyr]

To me this is what the top view would look like

[teal gust]

i mean

B and C wont be

exactly 180 degrees

to each other

but youre close

[tropic zephyr]

What I've noticed about these two problems is you kind of have to be creative. I didn't know you could split it into a triangle for example

[teal gust]

yeah

the more questions you do

you'll get the hang of it

[tropic zephyr]

I guess that's it for this help channel. Thanks for the help, this homework makes no sense but I try to do most of the problems on my own

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[mossy kraken]

what happens to a line when you multiply a negative inside f(x)? (I've already done letter A I just need help with B inside the picture)

[neon iron]

It reflects across the Y-axis

[mossy kraken]

ohhh okay thank you

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[tropic zephyr]

Don't mean to bother you but how do you get that rainbow border on your avatar

[chilly cedar]

Discord lootbox

You can see that option in your discord settings

(on pc)

[tropic zephyr]

Does it cost money the loot boxes

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[worthy quiver]

The diagram is a plot of the function w=f(x,y) in the plane(instead of in space). How do you find $\Delta s$ in the direction <1,-1>?

[thorny flame]

e=mc^2

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[vocal escarp]

In permutations and combinations when do you add them and when do you multiply them?

[vocal escarp]

For example in some cases arrangements are added and in other they are multiplied

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[neon iron]

how do i prove that its anti-symmetric?

[neon iron]

can i start with saying we have b R a iff b = a^2?

and then do something with a = b^2 and b = a^2 to get a = b?

[woven creek]

let (b^2,b) belongs to R
if (b,b^2) also belongs to R
then b = (b^2)^2
b = b^4
which is true for only b = 0,1
but then the ordered pair is of the form (1,1) or (0,0)

[neon iron]

oh right yeah (a, b) is just a tuple in set R makes sense

that makes sense, thanks

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[neon reef]

Find the probability that jose missed his bus given that he is late for school

E is loses bus and F is late

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[rain sluice]

How do u do B

[noble laurel]

Well if you did part (a)

you know it is its own inverse if (ax+b)/(x+c) = (b-cx)/(x-a)

[rain sluice]

how

[neon iron]

Inverse is just finding out x in terms of f(x) or y , then swapping the x,y

[rain sluice]

yea but how do I show that if a + c = 0

[neon iron]

Let a + c= 0 , it will be proven thereafter

[rain sluice]

then what

[neon iron]

U can substitute a= -c given that

Hope this helps 🫡

I just tried, we get inverse as -cx+b/ x+c

[rain sluice]

what does that do tho

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[boreal mica]

hi

[boreal mica]

nvm

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[finite mortar]

Why are they both not dY

Would it be wrong to write dY instead of dg

[nimble garnet]

you could maybe justify it by saying:
Y = g(i, u)
we do dg/di because we want the change in g over the change in i specifically
but then, we want the change in f over the change in g(i, u) as a whole, which is Y

[finite mortar]

Ohhh i see

Why do you want change in g over change in i specifically though, just because the question asks about functions of f and g?

[nimble garnet]

the question wants dH/di right?

[finite mortar]

Yeah

No i understand the reasoning i just don’t know how you’d know to write it

[nimble garnet]

🤷‍♀️

i find you're unlikely to find good justification for "why did they use this strange notation" type questions without asking the professor/teacher/whoever wrote that

all we can really do is guess lol

[finite mortar]

Oh really i just thought it would be formulaic

Hmm

Thanks anyway

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[mossy garnet]

mine

[mossy garnet]

how to find the limit of this sequence

[past cradle]

isolate a_n and take the limit as n goes to infinity on both inequalities

[mossy garnet]

which grows faster n! or n^3

[mossy garnet]

i will, gimme a sec

[past cradle]

n! is basically a polynomial of degree n

so that grows the fastest

[mossy garnet]

damn

didnt knew that

[past cradle]

n! = n(n - 1)(n -2) ... 1

so yeah

behaves like that in our case

[mossy garnet]

what now?

[mossy garnet]

,rotate

[past cradle]

take the limit as n goes to infinity

a bunch of stuff will vanish

[mossy garnet]

RHS is troubling me

look

,rotate

[mossy garnet]

how to do rhs?

this factorial and 3^n is troubling me

[past cradle]

factorial grows faster than exponential

so it goes to 0

[mossy garnet]

limit is -3/2

[past cradle]

yeah

[mossy garnet]

squeeze

[past cradle]

yes

[mossy garnet]

nice.

ty

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[left knoll]

can someone make me a proof of this:

[woeful stone]

Have you search math stackexchange?

[left knoll]

yes

but I didn't find anything

[woeful stone]

You have typed the actual latex?

[left knoll]

ofc

[odd pagoda]

isnt this just markovs inequality?

[left knoll]

i dont know

[odd pagoda]

it is

[left knoll]

this

?

[odd pagoda]

thats chebyshevs inequality

you know, like the line directly above it on wikipedia says

[left knoll]

but that is not the same as my assignment

[odd pagoda]

but it looks very similar

maybe instead of X you can look at a different random variable

[left knoll]

X=X-E[X] ?

[odd pagoda]

well lets not call it the same thing

[woeful stone]

The random variable is |X - E[X]|

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[proud idol]

The proof I found online says to use CR equations
I don't want to use them for this
Why shouldn't I get the same result using standard method?

[proud idol]

I am sure I did something wrong
Please tell me what I did wrong
It's only supposed to be differentiable at z=0 and nowhere else

@lone nest ur well versed with this
If you could spare some time
This human shall be eternally grateful 🙏

Something striked me later on
Please check

<@&286206848099549185>

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[proud idol]

Opening again

[proud idol]

This one

<@&286206848099549185>

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[lone nest]

Do not ping particular users for help when they're not already engaging with you, please.

[proud idol]

ok

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[buoyant wing]

What is the number of combinations of the back line of a chess game which is a 12x12? In normal chess it is 960 if you consider that there needs to be castling and a dark and light square bishop, and that rooks and knights are the same. Assuming these must still be accounted for but there are no additional duplicate pieces, what's the total for a 12x12?

[buoyant wing]

<@&286206848099549185>

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[buoyant wing]

Aw man

I already have to close it

?

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[pine sonnet]

can some one help me with coming up with an related rates equation for this. I was able to solve all the other parts w just algebra except this one

[pine sonnet]

This is all i got but it didnt look right so im confused

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<@&286206848099549185>

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<@&286206848099549185>

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[alpine lion]

Is it possible to create the following Table in Latex?

[torpid sparrow]

$[
\begin{array}{c|ccc}
\multicolumn{4}{c}{G_1 \text{ Table}} \

• & 0 & 1 & 2 \ \hline
  0 & 0 & 1 & 2 \
  1 & 1 & 2 & 0 \
  2 & 2 & 0 & 1 \
  \end{array}
  ]$

[thorny flame]

Samuel

$\[
\begin{array}{c|ccc}
\multicolumn{4}{c}{G_1 \text{ Table}} \\
+ & 0 & 1 & 2 \\ \hline
0 & 0 & 1 & 2 \\
1 & 1 & 2 & 0 \\
2 & 2 & 0 & 1 \\
\end{array}
\]$
```Compilation error:```! LaTeX Error: Bad math environment delimiter.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
 ...                                              
                                                  
l.49 $\[
        
Your command was ignored.
Type  I <command> <return>  to replace it with another command,
or  <return>  to continue without it.```

[torpid sparrow]

There must be a weird thing im missing now so i get that error but its something like this

[alpine lion]

Oh wow thank you

Shows up like this in overleaf @torpid sparrow

nvm figured it out

[
\begin{array}{c|ccc}

• & e & a & b \
\hline
e & e & a & b \

a & a & b & e \

b & b & e & a \

\end{array}
]

[thorny flame]

nogo

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[hasty jewel]

i applied green's theorem to get $\int_{0}^{1}\int_{y^{2}}^{\sqrt{y}}\left(-x^{2}\right)dxdy$

[thorny flame]

Jaxx

[hasty jewel]

this is correct but im confused on the first integral because i think im missing a negative somewhere during the process

$\int_{y^{2}}^{\sqrt{y}}\left(-x^{2}\right)dx = -\frac{x^3}{3}$

[thorny flame]

Jaxx

[hasty jewel]

how does this become $-\frac{y^{\frac{3}{2}}-y^{6}}{3}$

[thorny flame]

Jaxx

[hasty jewel]

rather than $-\frac{y^{\frac{3}{2}}+y^{6}}{3}$

[thorny flame]

Jaxx

[hasty jewel]

i.e. why is the negative not distributed

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[scenic osprey]

How the sigma flippity flip flop do they read this steam and leaf plot to get those numbers using this context and in general

[icy sky]

stem refers to the tens digit and leaf refers to the ones

[scenic osprey]

I don’t understand what that means

I’m completely new to this

[icy sky]

tens digit
v
30
^ ones digit

i honestly think this is a dumb way of making a table

[scenic osprey]

Ohhhhh

I get it

Yea this is rly dumb

[icy sky]

but anyway, lets say i spend 57 minute browsing the internet, you would put 5 on the left, 7 on the right

[scenic osprey]

Yesss

[icy sky]

then if another person spends 52 minutes, you just add a 2 on the right

[scenic osprey]

that makes sense

Thank you

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[icy sky]

youre welcome

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[worn grove]

I am stuck to find the solution given the initial value. if $x/=0, then C_1 = C_2 = 0$

[worn grove]

I am stuck to find the solution given the initial value. if $x \neq 0, then C_1 = C_2 = 0$

[thorny flame]

ocean bleu

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[magic grail]

Hello, can you help me to know what is the middle term in a syllogism and how to identify it?

[magic grail]

For example, here:

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<@&286206848099549185>

[jolly fossil]

Yez

@magic grail

[magic grail]

Hi

[jolly fossil]

Hi

[magic grail]

My question comes because I was looking for a way to quickly find out if a syllogism is valid or not and it was mentioned that in a valid categorical syllogism, the middle term must be distributed in at least one premise.

And it was not clear to me what that middle term is and also what it means that it is distributed.

Do you know?

[jolly fossil]

I’m very sorry but I don’t study that maybe u could find others

@dark swan come check this question out

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[static jacinth]

Is it true that if $f_n \to f$ in measure and there exists $f_{n_k} \to f$ in $L^1$ then $f_n \to f$ in $L^1$?

[thorny flame]

Casiel368

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[tall plover]

You know for the identity trigo cos²x + sin²x = 1

[tall plover]

Is it the same thing for :

cos²5x + sin²5x = 5

[noble laurel]

no

[onyx yacht]

Not 5

5x is just some other angle

[tall plover]

so what it is ?

[noble laurel]

cos^2(5x)+sin^2(5x)

you're plugging in the same thing

"5x"

into cos^2

and sin^2

and adding them

so it should be 1

relabel 5x=y

cos^2(y)+sin^2(y)=?

you'd definitely say 1

[tall plover]

this is one ig ?

or idk

so if understand well

it is sin²5x + cos²5x = 1 ?

[noble laurel]

yes

[tall plover]

tysm

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[topaz sinew]

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[paper sable]

or just take "5x" as another variable known as "u" or something, hope that helps 👍

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[thin agate]

hello

[thin agate]

i need help

with precalc

[topaz sinew]

@thin agate Has your question been resolved?

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[shadow sequoia]

Here is my problem, I'm completely lost.

[copper dome]

What grade is it

[shadow sequoia]

12th

[coarse mica]

you got the -5 and 6 right but check the areas where numbers will work

[shadow sequoia]

oh, was it C

[coarse mica]

for example, plug in -6, 3, and 7 to see what happens

[shadow sequoia]

yeah it was

i was sure i got the right answer but it wasn't working, ty

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[copper dome]

help

Can anyone help me with math

[copper dome]

Can u help me

[coarse mica]

yeah

what?

[copper dome]

With transformations

[coarse mica]

where?

also wrong place

[copper dome]

Huh wdym

Like this

[vernal matrix]

Get an unoccupied help channel, this one belongs to someone else and as they've closed it, it might lock

One from this section

[copper dome]

Oh

But I didn’t. Lock

It

[vernal matrix]

The OP did

[topaz sinew]

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[sonic harness]

how can i do this 2b

[merry hill]

what

[fickle pelican]

its not right angled

[merry hill]

thats only for right triangle

[paper sable]

OOPS

WTF

[merry hill]

yeah

[paper sable]

oh nah i need to sleep

5;30 am

wtf id wrong with me

[fickle pelican]

2b, try using sine law first to find the side length opposite to 101 degrees

then cosine law but use the third angle

180-101-50

[versed cairn]

id just find 3rd angle and use sine law once

[fickle pelican]

real

[versed cairn]

but i doubt they know sine law cuz of the hint

[sonic harness]

i dont really know the law that much

[fickle pelican]

oh

[sonic harness]

but i use it to find the third length

i try to find this one?

[fickle pelican]

u dont have to

[sonic harness]

what about this

[fickle pelican]

whats the third angle?

[sonic harness]

29deg

[fickle pelican]

real

okay so are u aware of sine law

[sonic harness]

i have notes let me check it out

oh

[sonic harness]

so sinA/a=sinB/b=sinC=c

[merry hill]

yes

[sonic harness]

and for a triangle with ABC

when do you use this "sinA/a=sinB/b=sinC=c" and "a/sinA=b/sinB=c/sinC "

whats the difference

[versed cairn]

the first one is wrong

sinC is not equal to c

neither is sinA/a equal to c or sinB/b equal to c

oh wait u meant sinC/c , both of them are same , u get same results so dont worry which one u use

[sonic harness]

yeah. i meant the difference when the lowercase abc (side) is at the top and captial ABC (angle) at the bottom

and same thing when captial ABC is at the top and abc at the bottom

[versed cairn]

same thing

sinA/a = sinB/b
take reciprocal
a/sinA = b /sinB

[sonic harness]

sothere is no difference?

[topaz sinew]

@sonic harness Has your question been resolved?

[topaz sinew]

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[left sage]

im stuck on the first question here

would it just be (x-c,y+s)??

[topaz sinew]

@left sage Has your question been resolved?

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@left sage Has your question been resolved?

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[true crow]

What should the domain of this piecewise function be?

[restive inlet]

can you think of any valules of x where the function isn't defined?

[true crow]

i do think that the domain is all real numbers

but mathway tells me that it’s

[restive inlet]

mathway is wrong

[radiant sparrow]

[restive inlet]

or you didn't type in the question properly

[neon iron]

yes

[true crow]

i wrote it correctly tho lol so its probably its fault

[slate terrace]

why does this part suddenly negative? im kind of lost

[true crow]

.close

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[grand marsh]

i think the first derivative is 1/ln(x) and the second is -1/2ln(x), how do I solve for the relative extrema and point of inflection?

[grand marsh]

nevermind i go tit

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[topaz sinew]

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[uneven pollen]

I’m supposed to find r, the growth rate, but I’m stuck on how to continue from here

[uneven pollen]

anyone got any suggestions?

or find any mistakes in my math so far

[neon iron]

ur writing is majestic

[uneven pollen]

<3 <3

[ornate magnet]

Where's the 100 in denominator in 2nd step?

[uneven pollen]

When I add the value for K? or when I separate variables?

[ornate magnet]

1-n/100 = (100-n)/100

@uneven pollen

[uneven pollen]

uhm

sorry I'm a little confused

just trying to understand where

[ornate magnet]

See 2nd line on left

[uneven pollen]

oh

since you get (1 - N/100) I just multiplied by 100 to get rid of the fraction

just to simplify

<@&286206848099549185> sorry for ping

Was hoping someone could help me find r

If it helps, the answer is supposed to be r = ln (9/4)

[final jungle]

You can take the exponential on both sides to get rid of the natural log

[obtuse garden]

I just wanna say your handwriting is phenomenal

[final jungle]

True

[uneven pollen]

Would you multiply with 10o on both sides first?

[final jungle]

Yep

[uneven pollen]

so you get e^(100rt)*C ?

okay

[final jungle]

yes

It equals to ...?

[uneven pollen]

N/N-100

[final jungle]

Now solve for N in terms of t

[uneven pollen]

working on it, tysm!

so if I insert one of the initial values into this, I guess I can solve for r

or wait

I need to find C first

[final jungle]

You need both initial values since there are 2 constants to be determined. One is r and the other is constant of integration

[uneven pollen]

yeah

if i insert t=0 first i can find c

then use t=1 to find r

Almost got it!

Just made a mistake with the 100 somewhere

struggling to find it I'm not very good with logarithms 🥲

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[ember ledge]

I have found the 2 equations, how can I find the restrictions?

I don’t understand how restrictions work, can I get an example if possible?

[regal charm]

x is in [-2,0] for the left parabola

there's nothing more to it lol

since i'm guessing you had something like y = -2x^2- 4x which models the entire parabola

[split cloak]

When there's no dot means it's restricted value idk

[regal charm]

but you need only the portion where x is in [-2,0] for the logo

[ember ledge]

?

[regal charm]

what?

[split cloak]

Look at the graph

[ember ledge]

Is that the restriction? so sorry I’m really confused at what its asking

[regal charm]

This is the "graph"

[ember ledge]

Yes

[split cloak]

we don't need other parts, just these from x -2;2

[regal charm]

you want to impose restrictions such that the graph is only this

which is done by restricting x on [-2,0] for the left side of the logo

do the same thing for the right side of the logo

?

can you just say it in words?

[ember ledge]

Parabola 2: X cannot be greater than 2, x cannot be less than 0

[regal charm]

okay but what about when x = 2

[ember ledge]

Parabola 1: x cannot be less than -2, x cannot be greater than 0

[regal charm]

and when x = 0

[ember ledge]

Parabola 2: x can be greater or equal to 0, x can be less or equal to 2?

[regal charm]

okay better

x can be greater or equal to 0,but x has to less or equal to 2?

now put this in mathematical terms

[ember ledge]

Parabola 2: x>0 , x<2

Make that “>” “<“or equal too

Cuz my keyboard can’t type

[ember ledge]

^

[regal charm]

uhhhh i guess that works yeah

but why not just

[ember ledge]

Thank you so much!!!

[regal charm]

,, -2 \leq x \leq 0

[thorny flame]

nyxie9151

nyxie9151

[ember ledge]

thank you!!

[regal charm]

seems more tacit to me

sure np

[ember ledge]

so sorry, I probably made you rlly rlly pissed.. I just started this unit of work in year 8 and I’m rlly confused hahah

Thank you!

.close

[topaz sinew]

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[regal charm]

not really i was just doing something else

anyway np

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[ember ledge]

HELLO I AM BACK. 😭😭😭

[ember ledge]

would these restrictions now just be x must equal to -3 or x must be equal to 3?

X=3, x=-3?

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[topaz sinew]

@ember ledge Has your question been resolved?

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[low bramble]

This question got the a's value but i could not get it as the determent is not getting as same as the options available

[radiant torrent]

what are you getting for your determinant?

[low bramble]

just a sec i will send the solved

[radiant crag]

Why you need to solve the exact a out

use $\delta = b^2 - 4ac$ will ok

[thorny flame]

e_waste

[radiant crag]

oh wait no

Ok I know how to do it

wait a sec

[low bramble]

ok

[winter egret]

You probably want to do three cases separately, a<0 a=0 and a>0

[radiant crag]

So this is a problem requires you to transform the equation to quadritic functions

[winter egret]

Note that you don't want just two roots, you want the roots to have opposite signs

[low bramble]

ok i will try this one

[radiant torrent]

you know that if the roots are real, they are guaranteed to have opposite signs

due to vietas

so just make sure the roots are real using discriminant

[radiant crag]

So this is a function with no roots of opposite signs

[low bramble]

yes but the option availabe give answer A

[radiant crag]

And there's also one down below

you work these two out and the answer is A

[radiant crag]

Oh this works for $ax^2+bx+c$

[thorny flame]

e_waste

[winter egret]

oh yeah that's probably the best approach

[low bramble]

ok

[radiant crag]

Oh yes I forgot Vieta's

yes

[low bramble]

just a sec

[radiant crag]

My approach is too complex

[low bramble]

i forgot what is vieta

[radiant crag]

$x_1 + x_2 = -\frac{2a}{b},x_1 \cdot x_2 = \frac{a}{c}$

[thorny flame]

e_waste

[radiant crag]

where x is the root of function $y=ax^2+bx+c$

[thorny flame]

e_waste

[radiant crag]

@low bramble

[low bramble]

okok i get it

thx

.close

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@low bramble Has your question been resolved?

[half edge]

.close

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[shadow idol]

how do i solve this?

anyone?

[nimble garnet]

which one

[shadow idol]

(a)

i tried subbing in (1,8) into the equation

[pseudo quest]

the max y value would be k, so 8 in your case,
the x value at the highest point is -h, so for you its -1
to find the y intercept you have to take x=0

[restive inlet]

you are given the max point / vertex,
the values of h and k can be identified directly from that

[pseudo quest]

so its 8-3(1)^2 = 5, so m = 5

[nimble garnet]

!nosols ?

[topaz sinew]

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.

[pseudo quest]

okay apologies

[topaz sinew]

@shadow idol Has your question been resolved?

[topaz sinew]

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[jade scarab]

"Calculate the volume of the mental body bounded by the cylinder z = 7y
2 − 10y + 74 and the
hyperbolic paraboloid z = −x
2 + 6y
2 + 14x + 36."

[jade scarab]

can i start but putting them equal to each other?

[topaz sinew]

@jade scarab Has your question been resolved?

[golden blade]

I think yes, to find the region first

[jade scarab]

Yes okay

[golden blade]

i'll be back later, at work now

[jade scarab]

okay thanks

[topaz sinew]

@jade scarab Has your question been resolved?

[golden blade]

I think you simply do this

[thorny flame]

𝔸dωn𝓲²s

𝔸dωn𝓲²s

[jade scarab]

Hmm okay I am just eating now but I will try that

[jade scarab]

I simplified this and got x = -7 +- sqrt(11)

[golden blade]

How, where did y disappear?

[jade scarab]

Oh your right I don’t know

[golden blade]

It should be a circle

[golden blade]

yea do that 🍲

[jade scarab]

okay i silplified it to y(y-10) + x(x-14) + 38 = 0?

i think

[golden blade]

you silplified it nice

lemme check

[jade scarab]

is f(x,y) = 7y^2 -10y +74?

[golden blade]

Well technically we can use absolute value in case if we get negative volume

f-g or g-f would give the same basically

brb

[jade scarab]

okay so we can do |(7y^2-10y+74)-(-x^2+6y^2+14x+36)|?

[golden blade]

ok

[golden blade]

correct

[golden blade]

correct

To see what I mean

Say x and y are any numbers

[jade scarab]

yes i get the same thing as before but with the absolut value

[golden blade]

x - y
and
y - x = -(-y + x) = -(x - y)

So you can see both yield the same just one negative

[jade scarab]

yes

[golden blade]

yes

,, \abs{ \iint_R (7y^2-10y+74)-(-x^2+6y^2+14x+36) \dd A}

[thorny flame]

𝔸dωn𝓲²s

[golden blade]

We would need now to determine the region

Like bounds-wise

I would first silplify the integrand

chances are, we are better off with polar coordinates since we integrate over a circle

Does that make sense?

dont cry

you got this

[jade scarab]

haha yes i think so

so we have |(y(y-10)+ x(x-14) + 38)|= 0

[golden blade]

really?

lol ok

gimme a sec

Ok

,w plot to 7 Power[y,2]-10 y+74=-Power[x,2]+6 Power[y,2]+14 x+36

[jade scarab]

wait do we not haha?

[golden blade]

yes gimme some ti

[jade scarab]

okay

[golden blade]

Let's complete the square

y² - 10y + x² - 14x + 38 = 0

We would need a +(10/2)² = 25 and +(14/2)² = 49

(y-5)² + (x-7)² = 25+49-38 = 36 = 6²

So now we know the radius and center

[jade scarab]

hmm okay

[golden blade]

somethings is off

i dont understand this

[jade scarab]

yeah no i am lost

[golden blade]

WTF

That is the region

We simplified it to (y-5)² + (x-7)² = 6²

radius 6 and center (7,5)

now we good

[jade scarab]

yeah okay

[golden blade]

,, \abs{ \iint_R y^2-10y + x^2 - 14x +38 \dd A}

[thorny flame]

𝔸dωn𝓲²s

[golden blade]

Ok now we would use polar coordinates

x = rcos(θ)
y = rsin(θ)

what is r = ?

from our region R

[jade scarab]

does r not variate?

or can we say that is it 0?

[golden blade]

haha no

yes

[jade scarab]

okay

[golden blade]

sorry mb

yes

[jade scarab]

its okay

[golden blade]

haha

ok

can you tell the bounds

[jade scarab]

for r its from 0 to 6?

[golden blade]

yes

that

and theta

[jade scarab]

okay and then do we have theta

oh yes

[golden blade]

it's a circle

so

[jade scarab]

from 0 to 2pi?

[golden blade]

yummy, yes

[jade scarab]

haha

[golden blade]

we integrating the radius and then rotation is 360°

[thorny flame]

𝔸dωn𝓲²s

[golden blade]

That doesn't look bad

We can sort some terms

,, \abs{ \int_0^{2\pi} \int_0^6 \left ( r^2(\sin^2\theta + \cos^2\theta) -10(r\sin\theta) - 14(r\cos\theta) + 38 \right ) r : \dd r \dd \theta }

[thorny flame]

𝔸dωn𝓲²s

[golden blade]

can you see something special

it's ok if you don't

[jade scarab]

its symmetrical?

[golden blade]

can you spot an identity

i'll let you cook

[jade scarab]

sin^2(theta) + cos^2(theta) =1?

or no?

[golden blade]

you cooked 😎

yes

[thorny flame]

𝔸dωn𝓲²s

[golden blade]

We can integrate this wtr to r

it's technically like polynomials

yesn't it?

[jade scarab]

yes i think so haha

[golden blade]

Can you do this?

starting with r³

[jade scarab]

hmm maybe hold on

[golden blade]

i will hold on

[jade scarab]

r^4/4 is the first one

[golden blade]

oha yes

continue

you treat sine and cosine as constants, keep that in mind

[jade scarab]

okay

sin theta (10/3 r^3)?

[golden blade]

yes

[jade scarab]

okay

then

cos theta (14/3 r^3)

[golden blade]

ja

[jade scarab]

haha

and then 38r^2/2

[golden blade]

jaa

[thorny flame]

𝔸dωn𝓲²s

[jade scarab]

jaa

[golden blade]

my dc is ass on pc

3 business days for a meSage

[jade scarab]

haha

now we plug in?

[golden blade]

yes haha

[thorny flame]

𝔸dωn𝓲²s

[golden blade]

lower bound disappears

[jade scarab]

yes

[thorny flame]

𝔸dωn𝓲²s

[jade scarab]

yes!

[golden blade]

oha nice

Do you think you can integrate it?

[jade scarab]

yes hold on

[golden blade]

I hold on and tight

[jade scarab]

324 is gone

[golden blade]

how

Did you eat it

[jade scarab]

i mean does it not dissepear

haha

when integrating

no wait

[golden blade]

We integrate a constant

[jade scarab]

i was kidding

[golden blade]

yes!

as always😏

[jade scarab]

haha

wait are we intagrating first wrt theta?

[golden blade]

yes

[jade scarab]

dont we get 324 theta for the first one?

[golden blade]

yes

easy

[jade scarab]

okay then 720 cos theta?

[golden blade]

almost

look at the minus

oh wait i am dumb

[jade scarab]

noo

[golden blade]

continue haha

[jade scarab]

then 1008 sin theta

[golden blade]

ja but with minus

ok

[jade scarab]

wait why was it not minus before

[golden blade]

ok

[jade scarab]

haha

[golden blade]

Do you know this

sin(x)
cos(x)
-sin(x)
-cos(x)

and then repeat

[jade scarab]

oh your right sorry

[golden blade]

no sorry!

haha

[jade scarab]

then 684 theta

[thorny flame]

𝔸dωn𝓲²s

[golden blade]

ja we can actually add 324 and 684

[jade scarab]

yes and then put in the bound

haha

1008 theta?

[golden blade]

yes wait on me

[jade scarab]

okay

[golden blade]

slow down you too fast haah

[jade scarab]

haha

[thorny flame]

𝔸dωn𝓲²s

[golden blade]

ok now plug in

[thorny flame]

𝔸dωn𝓲²s

[golden blade]

Now simplify this

[jade scarab]

Okay so we get

[golden blade]

yes

dare it

[jade scarab]

720 + 1008 * 2pi -720?

witch becomes 1008 * 2pi?

[golden blade]

yes valbourg

[jade scarab]

hahaha

[thorny flame]

𝔸dωn𝓲²s

[jade scarab]

witch is 2016pi

yes

[golden blade]

oui

sweet number

prolly wrong

[jade scarab]

noo dont say that

[golden blade]

ok prolly right

[jade scarab]

yes

[golden blade]

😂

[jade scarab]

so that is it?

[golden blade]

uhmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

[jade scarab]

[golden blade]

jo

[jade scarab]

okay

[golden blade]

99%

right

and done

[jade scarab]

i dont know if i will compleat this weeks questions

they jus become harder and harder each week

[golden blade]

compleat

damn

that reminds me of some

[jade scarab]

complete

[golden blade]

no compleat is right to me :D

[jade scarab]

haha okay

[golden blade]

just try what you got

and give your best

[jade scarab]

this just looks complicated

haha yes i will

[golden blade]

yea (only because it's Swedish)

[jade scarab]

haha true

okay well enough for tonight

[golden blade]

yes (time for spotify)

[jade scarab]

thanks for the help

haha yes

[golden blade]

thank you too

[jade scarab]

no

[golden blade]

yes

[jade scarab]

nej

[golden blade]

jo

[jade scarab]

no bye

[golden blade]

yes bye

[jade scarab]

.close

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[rapid wedge]

so I know u gotta do the first derivative test

but when I find the derivative of f(x)

and set it to 0

I get -1 = 0

which isnt true

and doesnt help me with this question

I use the quotient rule to get to this point

then I get -1 = 0

[rapid wedge]

I put this photo into photomath and it also gets -1 = 0 from this statement

[limber ridge]

yeah

this just means the derivative never equals 0

[rapid wedge]

so how am I supposed to find the critical points?

[limber ridge]

there are no critical points

[rapid wedge]

how am I supposed to know when f(x) is increasing or decreasing then

the course teaches us to use the first derivative test

[limber ridge]

firstly look at the derivative and check if it has any discontinuties

if it doesn't, then not having a zero means that the derivative is either always positive or always negative

makes sense so far?

if it does have a discontinuity, then check the derivative on either side of the discontinuity and the sign you will get will not change on either side since the derivative never crosses the x-axis

@rapid wedge are you follwoing me?

[rapid wedge]

kind of, so first find any discontinuity

this function has a discontinuity at 4/3

[limber ridge]

ok cool

so check whether the derivative is positive or negative at each side of the discontinuity

[rapid wedge]

got it

answer is A

thanks

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

How do i find BX?

[little talon]

I’ve spent like an hour on this and I haven’t gotten anywhere, that’s beside the point though.

[molten harness]

What have you tried?

[neon iron]

,rotate

[little talon]

Nvm guys sorry

.close

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[topaz tulip]

how do i solve this for lambda?

[limber ridge]

ill show you a trick for cubics that usually works when the polynomial has integer roots

you know how cubic can be written in factored form in terms of its roots $P(x)=(x-p)(x-q)(x-r)$

[thorny flame]

The د

[limber ridge]

this means when you expand it to standard form, the constant term will be -pqr

or just the product of all the roots

so usually if you factor the constant term you can check the factors to see if they are roots

and check by plugging them into the function

because by the polynomial remainder theorem a is a root of P if $P(a)=0$

[thorny flame]

The د

[limber ridge]

@topaz tulip you following me?

[topaz tulip]

is this syntesis division

synthetic*

[limber ridge]

yeah after you find the first root you can use synthetic division to find the quadratic which you can solve easily

[topaz tulip]

oh so you wanted to give explain synthetic division or is there another easier way?

[limber ridge]

ill show you

here your constant term is 4

what are the factors of 4

[topaz tulip]

1 5 8 4

oh

1 2 4

[limber ridge]

yeah

and the negatives of those as well

now try doing plugging some of these factors in the polynomial to see if it gives you 0

so $P(1)=\ ?$

[thorny flame]

The د

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@topaz tulip Has your question been resolved?

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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

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[static venture]

Hello quick question for part b

Why at the end we add 9

I understand it has a periodic order of 3 and there's 40 periods happening

With remainder of one more term left

If the period is patterned
2,-1.5,9 why is the next term to be adding 9

[neon iron]

a1 is 9

[static venture]

How do we know that :0

[neon iron]

note

[static venture]

If starting from a1,
9, 2, -1.5, 9, 2
Why not it be 2

[restive inlet]

its all in the first section of work

considered both cases for k, rejected the second one since a_1 = a_2 which they don't want

[static venture]

Do we find a1 by subbing in a3 terms into the equation
As a1 should be periodically be equal to a4

[restive inlet]

yes

[static venture]

Ok I know where a1 comes from now

So the period is order 3 with
9, 2, -1.5

It says from 1 to 121
So from 1 to 120 there's
3x40

40(9+ -1.5 +2)

Ohhh and then

After -1.5 is 9

Hence add extra 9

Thanks ^^

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[topaz tulip]

with factors of 5+- and 1+- i will never get 0 when trying synthetic division

is it bcs the roots are complex?

[limber ridge]

did you try -1?

1 and -1 you should always try

[topaz tulip]

oh..

yea i got 0

thx

[limber ridge]

yeah yw

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[deft wing]

I need to prove that the following sequence is convergent and to find the limit.
$(a_n)$ is defined as $a_1 \coloneq 2$ and $ a_{n+1}\coloneq \sqrt{1+a_n}$ for $n\leq 1$

[deft wing]

how would I go about this?

[thorny flame]

Lukas

[neon iron]

limit at n = infinity ?