help-10

[spiral void]

what if they're equal?

is it still a valid triangle?

[lapis quartz]

nope

[spiral void]

okieee

thats all

thank uu 🙏

[lapis quartz]

no problem

[spiral void]

.close

[obtuse pebble]

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[obtuse pebble]

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[fluid snow]

What was done in this step?

How does that d appear

1 - z = 1/x^3, sure

Now to incorporate - C, we use d * ?

[grizzled shore]

You took both sides to exp right?

exp(-3ln|x| - C) = exp(-C)(exp(ln|x|))^(-3) = exp(-C)|x|^-3 = exp(-C)1/x³

then d = exp(-C)

Well, not quite, they removed the absolute values so now d can be anything

[obtuse pebble]

@fluid snow Has your question been resolved?

[obtuse pebble]

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[spiral void]

would AAA be enough?

[tawny copper]

Yeah

[spiral void]

angle A and M are corresponding angles

and so are angles B and N

[lapis quartz]

angle c is common

[spiral void]

and then we have c

[tawny copper]

U need to prove AB // MN first though

[spiral void]

ohhh

right

hmm

we cant say that points m and n are midpoints right?

[tawny copper]

Yeah

[spiral void]

can i have a clue?

[tawny copper]

Want hint?

[spiral void]

yes pleasee

[tawny copper]

Not a hint

But u can reverse the property u have from similar triagle to prove they're similar

[spiral void]

huhhh

OHHHH

like by cross multiplication?

something like that?

[spiral void]

aahhhh

[tawny copper]

That property derived if they're similar, so if it's true then they must be similar right

[spiral void]

uhuhhh

[lapis quartz]

its called SAS similarity criteria

[spiral void]

yeahh

got it got it

THANK YOU GUYSS 🙏

i thought it was going to be aaa hehe

.close

[obtuse pebble]

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[dawn osprey]

can someone tell me why they wanted Pi and not -Pi. Ive done a million questions like this and i cant figure it out

[dawn osprey]

It auto simplifies btw

[teal turret]

@timid silo #latex-testing

[dawn osprey]

?? want me to ask there or smth?

[teal turret]

No, not you. i pinged Amelia

[dawn osprey]

oh alr ty

Another example if you want (2nd last answer)

<@&286206848099549185>

[obtuse pebble]

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[distant fossil]

Would there be a shorter way to write : (2x3)(5x6)(8x9)...(3n-1)(3n) or (4x3)(7x6)(10x9)...(3n+1)(3n)

[obtuse pebble]

Please don't occupy multiple help channels.

[hybrid gyro]

for the first case you can write
$\prod_{k=1}^{n} 3k(3k-1)$

[warm shale]

lgkoo

[distant fossil]

My class does not use the notation for repeated muplication if I'm not mistaken, I guess there is no other way to write it?

[red ice]

Yes, if you don't use the product notation, then you have to write out a few terms and use the ...

It's not a bad thing honestly
It can be clearer than the notation

[distant fossil]

well thank you very much

[red ice]

No worries

[red ice]

But how they got to this notaiton is first by rewriting each term as 3 * 2, 6 * 5, 9 * 8 etc

And notice that 3 - 2 = 6 - 5 = 9 - 8 = 1

So 3k where k is a positive integer gives you 3, 6, 9...

3k - 1 gives you 2, 5, 8...

So you have 3k(3k - 1) as the general term

The fancy Pi symbol tells you to multiply all of these terms, and the bounds tell you to do it from k = 1 to k = n

Indeed the last term is (3n - 1)(3n)

[distant fossil]

thank you again 😄 I appreciate the extra explanation

[red ice]

No worries

[obtuse pebble]

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[random olive]

"In which points and for which/which choices of the constant d does the ellipsoid have ...
and the plane 2x + 14y + 9z = d a common tangent point?"

[random olive]

does anyone know how to solve this?

do i start with dividing the rhs with 57?

[obtuse pebble]

@random olive Has your question been resolved?

[random olive]

<@&286206848099549185>

2x^2/57 + 7y^2/57 + 3z^2/57 = 1

is this correct so far?

[obtuse pebble]

@random olive Has your question been resolved?

[obtuse pebble]

@random olive Has your question been resolved?

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[chilly pawn]

how to calculate area of BCD and ABD

[obtuse pebble]

Please don't occupy multiple help channels.

[chilly pawn]

i did it like
BD = root of 164

that's all i know

[timid silo]

bcd is just 8*10/2, abd is bcd - (10-6)*8/2

[chilly pawn]

it`s 12,8

[chilly pawn]

why? can you explain

[timid silo]

the area of a triangle with a 90° is the area of the rectangle divided by two

i hope it makes sense

the area of abd is the rectangle - the bcd, which is is still bcd, and then minus the missing triangle on the left, whichs base is 10-6

does this makes sense?

assume that corner i made is E
BCDE = 10*8 = 80 ADE = (10-6)*8 = 32 BCD = BCDE/2 = 40 ABD = BCDE - BCD - ADE = 80-40-32 = 8

im adding `` to it so the * dont turn into itallic shit

[chilly pawn]

oh

[timid silo]

got it or still doubts?

[chilly pawn]

I'm chewing on the information you gave me

[timid silo]

i dont really know how else i could explain it, sowwy :(.

[chilly pawn]

but i need area

like cm^2

[timid silo]

oh ok wait a minute

BCDE = 10cm*8cm = 80cm² ADE = (10cm-6cm)*8cm = 32cm² BCD = BCDE/2u = 80cm²/2u = 40cm² ABD = BCDE - BCD - ADE = 80cm²-40cm²-32cm² = 8cm²

yes its the same thing, i just added the units and fixed bcd formula

btw you can type ² with alt_gr + 2, assuming you're on keyboard

[chilly pawn]

but answers says:
ABD = 24cm^2
ABCD = 64cm^2

[timid silo]

uhm i really dont know how ABD would be 24cm², and you did not ask for ABCDcm²

ABCD would be ABD + BCD

[chilly pawn]

my bad, i need area in cm^2 of ABD and ABCD

[timid silo]

i really dont know how ABD is supposed to be 24cm²

you can also get ABCD with ABDE - ADE

oh nooo i get it

wait

[chilly pawn]

and also i need AD side

[timid silo]

BCDE = 10cm*8cm = 80cm² ADE = (10cm-6cm)*8cm / 2u = 16cm² ABCD = BCDE-ADE = 80cm² - 16cm² = 64cm² BCD = BCDE / 2u = 40cm² ABD = ABCD - BCD = 64cm² - 40cm² = 24cm² AD = ²/(AE²cm²+DE²cm²) = ²/(4²cm²+8²cm²) = cm*²/(16cm²+64cm²) = cm²*²/80 = 4cm*²/5

i forgot to divide ADE by 2 the first time

sowwy

[chilly pawn]

huh?

what is AD

[timid silo]

uhmm you asked for AD didnt you?

[timid silo]

here

[chilly pawn]

yes

so what is AD?

[timid silo]

4 times the square root of 5

oh i forgot units

wait

AD = ²/(AE²cm²+DE²cm²) = ²/(4²cm²+8²cm²) = cm*²/(16cm²+64cm²) = cm²*²/80 = 4cm*²/5

[obtuse pebble]

@chilly pawn Has your question been resolved?

[haughty forge]

BDC = 10 * 8 / 2 =40 cm^2
ABD = 6 * 8 / 2=24 cm^2

[obtuse pebble]

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[ember oar]

Question: solve te following system of equations using elementary operations. Interpret the results geometrically

[ember oar]

1. 2x + y - z =-3
2. x - y +2z = 0
3. 3x + 2y - z + = -5

My work

The answer is "the point at which the tree planes meet is (-1, -1, 0)

I'm trying to figure out where I got wrong

<@&286206848099549185>

[stiff glade]

Yes, how can I help you ?

[ember oar]

Question is above

[stiff glade]

I can't understand your handwriting

[ember oar]

My cam is shit

Hold on

[stiff glade]

Where is the equation

[ember oar]

The relevant information here is the numbers I just wrote down the steps so you can ignore that writing

[ember oar]

I put it here

And it's on the page as well

[stiff glade]

And where is it on the page

[ember oar]

First page, top left corner?

[stiff glade]

Oay

You already solved the equation, what do you need help with ?

[ember oar]

Well, it's not correct

The real answer is (-1, -1, 0)

My answer is nowhere close

[stiff glade]

You have a point

You know

I have the same problem with the Barycentric Coordinates

[ember oar]

I'm literally in pre calc idk what that is

But are you able to help me?

[stiff glade]

Yeah

[ember oar]

ok thank you

I'm just scratching my head bc it don't work lmao

[obtuse pebble]

@ember oar Has your question been resolved?

[obtuse pebble]

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[low fossil]

I found part a) by brute force which is 15 but I dont know if brtue force will help me with the other parts and dont know if theres an actual methods to systematiccally work through these

[low fossil]

i also managed to do d by doing 45-10 = 35 but have no idea how to compute b or c

[obtuse pebble]

@low fossil Has your question been resolved?

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[meager palm]

why is this. wrong

[meager palm]

for 3)

is it actually 10 choose 10

<@&286206848099549185>

[timid silo]

Which question cinna

[meager palm]

for 3)

[timid silo]

Do u know binomial theorem

[meager palm]

yeah

[meager palm]

did I use it right

[timid silo]

It's 1

10C10

U r rt

[meager palm]

yeah why is it 10 then

i thought it was n - k

[timid silo]

Wait a sec , lemme check again

It is 1 damn

[meager palm]

whyy

[timid silo]

There is only 1 coeff of z¹⁰

Is the answer 10

[meager palm]

idk what the answer si

[timid silo]

Then the answer is obviously 1

It's 10!/10! 0!

[meager palm]

so 10 choose 10 is the same as 10 choose 1?

[timid silo]

https://tenor.com/view/dr-strange-one-just-one-gif-22683317

[timid silo]

10 C 10 = 10 C 0

[meager palm]

okay wait im so confused can we start from the beginning

i thought it was n choose r

[timid silo]

10 C 1 = 10 C 9

[timid silo]

Ok

Do u know this formula

[meager palm]

yeah and r here is 10 - r = 10 ?

which means r = 0?

[timid silo]

Yeah

[meager palm]

so then its 10 choose 0

[timid silo]

I mean both r equivalent to each other if u see

There is only one way of selecting 10 things out of 10 things

And there is only one way of selecting 0 things out of 10 things

[meager palm]

okay cool so either is valid

[timid silo]

Yup

It is one of the combinatorial identity

[meager palm]

so this is valid

[meager palm]

or is it x^0 and z^10

[timid silo]

U want to find coefficient of z¹⁰ right

[meager palm]

yes

[timid silo]

Which 10 C 0 or 10 C 10

[meager palm]

yes

[timid silo]

Z have a negative sign here , but as u want to find coeff of z¹⁰ , even power , the minus sign converts to positive, and your final answer is 1

I hope u understood now 🙏🏻

[meager palm]

can i just leave it as like

(10 10). x^10 z0?

im more so confused because my friend did (10 0) x^0 and z^10

[timid silo]

U can use ... , I mean it's binomial expansion, and the need is to find coeff of z¹⁰ , or u can either start counting from x powers or z powers 😄

[meager palm]

so like it doesnt matter

[timid silo]

Not at all

[meager palm]

cool tysm for helping me

i need to review this when my brain is not fried

.close

[obtuse pebble]

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[meager palm]

.reopen

[obtuse pebble]

✅

[meager palm]

can i ask one more q

n choose k is the same as n choose n - k right

[timid silo]

Yup

Ur brain not accepting this or what?

[meager palm]

yeah im really tired

i pulled an allnighter

should not be doing work rn

ty for helping me

[timid silo]

If u select something, u also reject something too keep that in mind 😁

[meager palm]

.close

[obtuse pebble]

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[obtuse pebble]

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

Hi

[mild laurel]

<@&286206848099549185>

[versed stratus]

have a question?

[mild laurel]

Do u guys help with ixl

[versed stratus]

ixl?

[wary badger]

it’s an online

math

like middle school thing

[versed stratus]

if it's math related/physcis related yes

[wary badger]

it’s probably algebra 1 level

[slow prawn]

Wtf is this channel name lol

[mild laurel]

It’s shapes but it’s that there are so many and I get confused

[wary badger]

right lol

[mild laurel]

And they’re not like normal shapes

[slow prawn]

show us then

[wary badger]

what’s confusing

which do you think it is

[mild laurel]

I can pick multiple

[wary badger]

hence why i said which

[fossil sedge]

Knief can you go to my channel I need help

[mild laurel]

Parallelogram

[broken tendon]

We cannot give you answers, only hints

[mild laurel]

Oh

[broken tendon]

You see that the 4 sides are grouped in 2 groups of 2 sides that are equal

[mild laurel]

Yes

[broken tendon]

Which shape has this property?

[mild laurel]

Kite?

[broken tendon]

That's the only question you should ask yourself

[broken tendon]

First, let's remove the choices that don't apply and give me reasons

[mild laurel]

Oh

I already answered I got it right tho

Though*

But now Ik the drill

[fossil sedge]

Rectangle, parellagram and quad

[mild laurel]

Thank u!!

[fossil sedge]

What grade math is this

[mild laurel]

I wish I were as smart as u guys in math

8th

Im good at every subject except math

[wary badger]

maybe change your name

could be offensive

[mild laurel]

lol

Ok

[wary badger]

bit immature

[fossil sedge]

How is that 8th grade math

[wary badger]

it is lol

america

[slow prawn]

yo is it bad that i dont know any of what this is

[wary badger]

yes

[slow prawn]

like

wtf lol

[mild laurel]

Idk that’s what it says

[fossil sedge]

Bruh I’m not that smart but I can do it

[wary badger]

i’d hope so

this is middle school math

[fossil sedge]

Egzactly

[mild laurel]

So can I get help pls

[fossil sedge]

Fine

[mild laurel]

Thx

[fossil sedge]

Parrelgram rhombus

[mild laurel]

I got it wrong u see

It’s so hard

[broken tendon]

Don't give answer

[mild laurel]

Can I get help pls

[broken tendon]

Ok, what defines a square

[mild laurel]

Wait no het got it wrong

I have a new question now

[broken tendon]

What?

[broken tendon]

Ok, so this is really similar to one of the problem from earlier. Look at your previous questions in the chat and you will find your answer

[mild laurel]

Yes

I got it right

[broken tendon]

Good job

[mild laurel]

Rectangle

Is what I think

But sometimes it’s more

[limpid frost]

hmm

its arectangle

[mild laurel]

I got it wrong

[wary badger]

how many sides

does it have

and how many sides do quadrilaterals have

[mild laurel]

I only need like 3 more right answers

And the I’m finally done

Then*

I’ve been working on this for 40 mins

[humble cobalt]

dang

I'll try to take you past the home stretch

[wary badger]

real philanthropic

[mild laurel]

Wdym

[humble cobalt]

lol it's came from a inside joke with friends

[humble cobalt]

I'll help you get this done

[mild laurel]

Ty sm

[humble cobalt]

so from the shape shown we see that:

there's 4 sides

the sides are equal in length of each other

there are not any right angles present

so the 4 sides shows that it's a quadrilateral

and the sides being equal length would be a square, but since there are not any right angles, it has to be a rhombus, not a square

and then with the quadrilateral hierarchy, that would mean that it would also fit the criteria of a parallelogram

[mild laurel]

I don’t thinks it’s a square

[humble cobalt]

thats right

[mild laurel]

NVM I FINISHED IN INE QUESTION

TYSMMMM

[humble cobalt]

so ur good?

[mild laurel]

I mean I could do another ixl with u if u want

[humble cobalt]

nah that should be good

nice job!

[mild laurel]

Ok thx

[humble cobalt]

alr cya

[obtuse pebble]

@mild laurel Has your question been resolved?

[mild laurel]

Yes

[obtuse pebble]

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[light spindle]

Help

[obtuse pebble]

Please don't occupy multiple help channels.

@light spindle Has your question been resolved?

[light spindle]

No

[obtuse pebble]

@light spindle Has your question been resolved?

[obtuse pebble]

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[remote pivot]

I'm like so lost here

What are we supposed to be doing here for A

I know a derivative is supposed to be used for finding the slope of a tangent line, but I don't know whats happening here

[old isle]

so are you familiar with

say you have a function

f(x) = 10x + 5

[remote pivot]

yes

[old isle]

do you know how to do f(2x)

[remote pivot]

yes

f(x)=10(2x)+5

[old isle]

you just do that here

[remote pivot]

20x+5

[old isle]

f(x + h)

- f(x)

[remote pivot]

what am i supposed to passin hto

[old isle]

f(x)

instead of 2x ur doing (x + h)

[remote pivot]

so I know the function that I will be passing into is $f(x)=5x^2+6x+5)$

[warm shale]

guy

[remote pivot]

right?

[old isle]

correct

[remote pivot]

so im supposed to do $f(x+h)=5x^2+6x+5$

[warm shale]

guy

[remote pivot]

?

[timid silo]

It is not equal

[old isle]

$f(x + h) = 5(x + h)^2 + 6(x + h) + 5$

[warm shale]

aurelianus

[remote pivot]

okay ye

yes

but how did you know to pass in x+h

[old isle]

because that's what its asking

f(x + h) - f(x)

[remote pivot]

ohhh

i see

[timid silo]

How did you know to pass 2x in place of x for f(2x) lol

[old isle]

^

[remote pivot]

okay thank you

i appreciate it, i get it now

💙

.close

[obtuse pebble]

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[tight nexus]

hi, doing some geometry work and wanted some help- don't exactly know where to start besides drawing the venn diagram images, I know areas of both circles are pi and diameter is 2 if radius is 1, but other than that I'm stuck

[timid silo]

So the problem pretty much needs you to find the Area of one of the circles minus the common area between them

[tight nexus]

do u know how u would be able to find the common area?

[timid silo]

ok so u should draw 2 triangles in that intersection

formed by the radii of both circles

[tight nexus]

like this?

[timid silo]

yeah

so those in blue are all equal to 1

[tight nexus]

Yup

[timid silo]

ans the middle thing is entirely made of 1

so now

draw a bisecting line right in the middle of that

[tight nexus]

ok

[timid silo]

so you split it into two equal halves

[tight nexus]

Ok

[timid silo]

also can you label like the corners just so its easy to reference the angles im going to mention xd

[tight nexus]

yeah sure

[timid silo]

ok perfect

so the middle ones are right angled obviously

[tight nexus]

yup

[timid silo]

so firstly

you know DH = 1/2 and DA = 1

can you find AH?

[tight nexus]

ya one sec

wait I got square root 0.75

[timid silo]

yuh thats correct

but uh

lets keep it in fractions

[tight nexus]

ok 3/4

[timid silo]

,,\3{0.75} = \3{\434}= \4{\33}2

[timid silo]

so i wanted u to calculate that because this is a famous triangle

do you know what triangle has like a sqrt(3) in it?

[tight nexus]

yes like the measured angles inside are like 60 90 30

[timid silo]

yeaa

can you label what thr angles are then

[tight nexus]

ya

[timid silo]

oh god a bit crowded now LOL

but anyways

[timid silo]

So now try calculating the area of the sector ADC

[tight nexus]

is it (square root 3)/8?

oh wait that's ADB

[timid silo]

hmmm are u sure

it's gonna be $\ds \412\21^2\2\4{2\pi}3$

[timid silo]

@tight nexus

[tight nexus]

wait I'm confused lmao

[timid silo]

like

[tight nexus]

I thought I was finding area of the ADC triangle

[timid silo]

the area of a sector is given by $\ds A = \412 r^2 \theta$ with $\theta$ given in radians. You have $\theta_{ADC} = 120\degrees$ which corresponds to $\theta_{ADC} = \4{2\pi}3$

[timid silo]

no i said area of sector

[tight nexus]

oh

[timid silo]

the sector is all of this part

[tight nexus]

the entire middle thing?

[timid silo]

nop

[tight nexus]

Just the thing until it reaches the other blue lines on left

[timid silo]

yesh it includes the entire middle thing except the white parts i left out on the left

[tight nexus]

Ok

[timid silo]

so like

that's thr area of the sector

but likr

[timid silo]

we wanna calculate this area now

because it'd be one half of the intersection's area

[tight nexus]

ok

[timid silo]

so we need to subtract off thr area of the trisngle ADC

[tight nexus]

ok so area of ADC is square root 3 / 2?

or /4

[timid silo]

should be /4

so sqrt3/4

anyways

[timid silo]

so subtract that off from this

@tight nexus

[tight nexus]

ok

How would I simplify 2pi/6 - square root 3 /4

[timid silo]

u cant its irrational completely i judt wanted u to write it out

i mean 2pi/6 = pi/3

[tight nexus]

ok

[timid silo]

so there is that

okay so

so thats one half

so uh

multiply by 2 the entire expression

and thats ur full area

[tight nexus]

ok

wait for the sector thing

is that just an equation

[tight nexus]

ohhh ok

thank u!

[timid silo]

ye

so like now uh

ur intersection's area is 2pi/3 - sqrt3/2

so now u want to find the visible area

so just calculate the full area of the circle and subtract it off

[tight nexus]

Isn't pi the area

Of the circle

[timid silo]

ye

so ur answer is just

[tight nexus]

So pi - (2pi/3 - sqrt3/2)

[timid silo]

,, \pi - \4{2\pi}3 + \4{\33}2

[timid silo]

ye

[tight nexus]

Wait why would it be plus

[timid silo]

,w evaluate pi - 2pi/3 + sqrt3/2

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by qylooo

[timid silo]

yeah that sounds good enough

[timid silo]

i distributed the minus

[tight nexus]

Oh

Ok I see

[timid silo]

it simplifies to $\ds \4{\33}2 + \4{\pi}3$ if you add up the two terms of $\pi$

[timid silo]

ok i guess we are done

[tight nexus]

Ok ty

[obtuse pebble]

@tight nexus Has your question been resolved?

[obtuse pebble]

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[keen rune]

Hi

[keen rune]

How many 4 letter combinations can you make from the word OUTPUT?

Can you please help me solve it using the box method/permutations? I do not quite get the concept and find it counterintuitive

I know that if it was the first case, you'd take only the letters that don't repeat, which is 4!, 24

[buoyant igloo]

Isn't it just 6 choose 4

[keen rune]

Yes

That's true

[buoyant igloo]

So an Intuitive way to see 6C4 would be

Total combinations are 6!

[keen rune]

But the answers supposed to be 90

I don't quite get it

[narrow vault]

6c4 also results in dupes unless you treat all letters as unique
consider choosing OTPU (not choosing first U) vs OUTP (not choosing second U)

[keen rune]

True

This is just giving me a headache

[narrow vault]

casework probably works fine

[keen rune]

Wdym case work

[narrow vault]

no repeats: 4!
one repeat (eg TTUO):
two repeats (eg TTUU):

ngl it feels like there should be some more direct way (and there are but complicated) but this feels fine ish

[obtuse pebble]

@keen rune Has your question been resolved?

[obtuse pebble]

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[ashen crest]

Hello, can you please help me with this :

[grizzled shore]

If P(A’) = 0.2 how does P(A) = 0.6

[ashen crest]

Its p(a|b) (our french way to write it with the b' under the a')

Mb p(a'|b') whatever

[obtuse pebble]

@ashen crest Has your question been resolved?

[ashen crest]

So p(a'|b')=0.2

P(a)=p(b')=0.6

[astral silo]

so what does p(a') =?

then u know the P of a' and b' and u can tell the overlap of both ' which is nothing

[obtuse pebble]

@ashen crest Has your question been resolved?

[crimson basin]

P(C|D)P(D) = P(C∩D)
P(C) = P(C∩D) + P(C∩D')

these tools are all you need

[obtuse pebble]

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[ashen crest]

Thank you guys !

[ashen crest]

.close

[obtuse pebble]

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[twin mesa]

I need help

[timid silo]

You can post the question

[obtuse pebble]

@twin mesa Has your question been resolved?

[ebon goblet]

can someone explain why the quartic functions with no roots are like wonky parabolas? the functions im looking at are 6x^4+8x^3+5x^2+3x+2 and 5x^4+4x^3+5x^2+4x+5

[obtuse pebble]

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[ebon goblet]

.reopen

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[ebon goblet]

^

[twin mesa]

wait

what

my question wasnt resolved

[dim temple]

open a new channel

ill help you there

[ebon goblet]

can you help me too lol

[twin mesa]

help 44

[dim temple]

im not exactly sure how quartics create some wierdly shaped parabolas tho

[ebon goblet]

ok

[buoyant igloo]

they don't always

create weird parabolas

[timid silo]

Please try with other software also

[buoyant igloo]

they may still have turning points

[ebon goblet]

no but the ones that i did

[buoyant igloo]

just all above the x-axis

[ebon goblet]

they just look odd

[obtuse pebble]

@ebon goblet Has your question been resolved?

[ebon goblet]

<@&286206848099549185> how do i do b?

these are hints if it helps

[obtuse pebble]

@ebon goblet Has your question been resolved?

[obtuse pebble]

@ebon goblet Has your question been resolved?

[obtuse pebble]

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[stone moon]

help pls

[inland matrix]

! status

[obtuse pebble]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[stone moon]

1

[buoyant igloo]

use sigma notation for the inside

[torpid pasture]

try geometric series for the inside

[inland matrix]

The terms inside the brackets can be simplified, they're in a GP

[stone moon]

i used but i got stuck in 1-{1/(n^2-1)}

now idk wat to do

cuz it needs to be multiplied

[inland matrix]

$\frac{1}{1-(1/n^2)}$

[warm shale]

Sulphur

[buoyant igloo]

multiply top and bottom by n^2

[inland matrix]

This is what the sum would be

[stone moon]

😮

[buoyant igloo]

and then you get n^2/(n^2-1)

now for the product

write a few terms out to see what happens

[stone moon]

ok

[buoyant igloo]

hm

you get

[stone moon]

WOW

THX A LOT MAN

BRO U R THE GOAT THX A LOT

[buoyant igloo]

did it work?

[stone moon]

YE

2

is ans

ig

[buoyant igloo]

I haven't verified it

but do your terms cancel out or smth

[stone moon]

😮 Ye

THX A LOT MAN

.close

[obtuse pebble]

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[modern bison]

This is of binomial theorem and idk how to start this one

[torpid pasture]

write 9^(n+1) as (8+1)^(n+1)

[modern bison]

Okay

[torpid pasture]

everything would be divisible by 64 except the last 2 terms in the binomial expansion

take them out and do some algebra with the -8n-9

[modern bison]

How? How do we know (1+8)^n+1 is divisible by 64 then

[torpid pasture]

the binomial would become
8^(n+1) + (n+1) 8^n + ((n+1)n/2) 8^(n-1) + ... + (n+1) 8 + 1

yeah its messy

basically the idea is that

all the terms except the last 2 involve 8 to some power >=2

[modern bison]

.close

[obtuse pebble]

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[obtuse pebble]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[tiny temple]

wait a second bro
i am not done posting yet

the last two equations I got for B0 and B1 seem wrong
I am not getting whatever wolfram said
and even my book has a different answer than both of this

,w a(n)= 4a(n-1) -3a(n-2)+(2^n)+n+3

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by the_path_finder

[tiny temple]

the coefficients are not matching
help me find the calculation error

also there shouldn't be a constant term
fuck i just realized that

[obtuse pebble]

@tiny temple Has your question been resolved?

[obtuse pebble]

@tiny temple Has your question been resolved?

[tiny temple]

<@&286206848099549185>

forgot this existed

[obtuse pebble]

@tiny temple Has your question been resolved?

[tiny temple]

<@&286206848099549185>

[tiny temple]

Somebody please
Help

Now 4 hours from now I will be asleep and this thread will close..
I just can't get anything right

It'll be 3 am and I have to sleep for the quiz tomorrow anyhow
I can't afford to not sleep

[obtuse pebble]

@tiny temple Has your question been resolved?

[tiny temple]

Sigh
Lol
Well I leveled up and did it on my own

.close

[obtuse pebble]

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[hybrid violet]

`

https://cdn.discordapp.com/attachments/903430332324405288/1228124573275263037/image.png?ex=662ae723&is=66187223&hm=6638d2892d834f8b194c3c2283273f8f5c17e5295061b4af8f2856cc795aa1ba&

no idea where to go from here

[obtuse pebble]

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[lilac path]

Why does Los's theorem imply A_T?

[obtuse pebble]

@lilac path Has your question been resolved?

[lilac path]

<@&286206848099549185>

[lilac path]

<@&286206848099549185>

[zealous glen]

yoyoyoyo

how can i help

helping like 4 ppl rn so sorry for any delay

js ping me with every msg

[lilac path]

@zealous glen

[zealous glen]

Sir if you dont get that, I am absolutely useless

Chatgpt might be able to help

But for extra motivation, if ur smart enough to get into a course like that then ur miles smarter than i am

[obtuse pebble]

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[tardy peak]

hi, does anyone maybe understand how they got to the 1/2pi and -1/2 pi?

for the θ

i understand that r^2 = x^2+y^2

and dydx -> r drdθ

so the formula is r^2

and that the sqrt(1) and sqrt(9) give the limits for "dr"

but i dont understand how the θ has been calculated

<@&286206848099549185>

;-;

<@&286206848099549185>

i feel so annoying, sorry

.close

[obtuse pebble]

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[zealous glen]

.reopen

@tardy peak

ay bro

n o problem

[tardy peak]

oh

[zealous glen]

lemme see if i can help

[tardy peak]

.reopen

[obtuse pebble]

✅

[zealous glen]

alright so what donmt u get

[tardy peak]

i dont understand where they get the θ limits from

the 1/2 pi and -1/2 pi

the upper and bottom integral

[zealous glen]

been a fat minute since i studied this lemme make some sense of this and ill ping u when i got it

[tardy peak]

ahh okay great thanks!

[zealous glen]

its asking for it en theta

[tardy peak]

i understand that the radius of the cone is like 1 to 3

but the θ of the cone

like on the circle, i dont understand why the upper and bottom integral are 1/2pi and -1/2pi

like that, how was i supposed to know from the question that those are the upper and bottom integral

[zealous glen]

im lost as for the 1/2 pi and the -1/2 pi but if i'm reading this even partially correctly, then its because of the diameter being pi

so then obviously the readius will be 1/2pi

upper and lower bounds fit accordingly

again, im no genius, so definetly get a second opinion but thats what i think

[tardy peak]

ahh okay, thank you!!

.close

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[slow prawn]

bit confused here

like

is it

pi(1-x^3)

[cinder basalt]

if you meant pi(1-x^3)^2

then no

this was the volume you calculated

you can find the red volume and substract the blue volume

outer - inner

[slow prawn]

so can i

(1-(1-x^3)^2)

[cinder basalt]

red volume --> r=1 do you see why?

blue volume --> r=(1-x^3)

[slow prawn]

ah

so i can

just to make sure my brain is dead asf rn

[cinder basalt]

sorry for the vague notation

[red ice]

The way you should think about this is imagine revolving f(x) first, then you take out the volume that is formed by revolving g(x)

[cinder basalt]

i think y=x^3 has an axis of rotation at y=0 not y=1

[red ice]

Ah it would just be easier to move everything 1 unit down so that y = 1 becomes y = 0 and y = x^3 becomes y = x^3 - 1

So yeah it's just $\int \pi (x^3 - 1)^2 \ dx = \int \pi (1 - x^3)^2 \ dx$

[warm shale]

south

[obtuse pebble]

@slow prawn Has your question been resolved?

[slow prawn]

ahh tyty

[obtuse pebble]

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[remote rapids]

In this why is r not 170 but r^2 is?

[remote rapids]

In this why is r not 170 but r^2 is?

[atomic parcel]

Wdym?

[brazen viper]

The equation of the circle has r^2 there, so r^2 = 170.

If the equation has r there it would be r instead

The reason why it's r^2 instead of r is because of the Pythagorean theorem.

A point (x, y) is a distance √(x^2 + y^2) from the origin

If we let that distance be r, then we see that x^2 + y^2 = r^2

[red ice]

Also it's worth checking your answers to see if they make sense
You're saying that the radius is 170
But then the x-distance is 11 - (-2) = 13 and the y-distance is 5 - 4 = 1

So that doesn't line up

[obtuse pebble]

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[clear root]

can someno please gelp me

[obtuse pebble]

Please don't occupy multiple help channels.

[hexed gull]

duplicate

.close

[obtuse pebble]

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[charred carbon]

I just need help with the bottom 3 not the top

[obtuse pebble]

Please don't occupy multiple help channels.

[green epoch]

Hlo

U can factorize the denominator then split it

For (d)

I think it'll be (3k+1)(3k+4)

[charred carbon]

Wait what would be the factor

[green epoch]

This

[charred carbon]

No like what 2 numbers would you use to factor that out

[brazen viper]

@charred carbon would you prefer this thread or the other closed as a duplicate?

[charred carbon]

This one is fine my bad for using 2

[green epoch]

12 and 3

[charred carbon]

oh okay but how does that change in finding the value of r

My main issue was finding r because the ratio was different depending on the value of k

[green epoch]

Yeah I don't see it becoming a geometric series, I was using the manual method

Basically after factoring the denominator you can write the series as summation of 1/3[1/(3k+1) - 1/(3k+4)]

Just write first few and last few terms, everything will get subtracted

[charred carbon]

wait how did you get that

isnt the formula a/1-r

[green epoch]

Partial fraction

Yeah that's when there's a common ratio

[charred carbon]

oh okay

What about the other 2 is it the same thing?

[green epoch]

part f looks like a normal GP

(e) can again be split

[charred carbon]

what do you mean by GP

[obtuse pebble]

@charred carbon Has your question been resolved?

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[formal basin]

How can this be simplified further?

[crude coral]

Factor out cotx

And multiply cos and sin together , u will get something like this

$\cot x (\sin^2 x + \cos^2 x)$

[warm shale]

JustToPro

[formal basin]

oh

and sin^2(x) + cos^2(x) are just 1 right

.close

[obtuse pebble]

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[warm shale]

alee

[open mountain]

why is this limit "e"?

[obtuse pebble]

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[celest musk]

It should be without the negative sign before x at the end

What's wrong?

[obtuse pebble]

@celest musk Has your question been resolved?

[celest musk]

Nvm im dumb

.close

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[open mountain]

Can anyone help me see if the series is convergent or divergent

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[charred carbon]

Only need help with question h

Tried to find r but there is no consistent ratio for the series

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[agile knoll]

for 15

why is the answer “e” and not infinity?

the answer keys says its e but i keep working it out to be infinity

[fringe rampart]

because n an x^n has the same convergence as an x^n

[hazy peak]

The answer is $\limsup \frac{n}{(n!)^{\frac{1}{n}}} = e$

[warm shale]

themadchessplayer

[agile knoll]

what

how do you even get to that?

x doesn’t cancel out anywhere

[hazy peak]

Umm... formula for radius of convergence.

[hazy peak]

The given series is a power series in x

Why would x feature in the radius of convergence of the series?

[agile knoll]

wait im confused

the way we’ve always solved for radius is through the squeeze theorem

by first taking the limit of the series through the root or ratio test

when i did that it came out to be 0

[hazy peak]

https://en.wikipedia.org/wiki/Cauchy–Hadamard_theorem

[agile knoll]

wth

is that really calc 2?

[swift stream]

You usually learn it in a complex analysis class

[hazy peak]

Maybe? Our educational system does not work that way

[agile knoll]

im cooked

my teacher didn’t mention anything like that at all

[hazy peak]

It's there in baby Rudin pretty early on

[fringe rampart]

you don't really need it to the exercice i think tho

[hazy peak]

Because it does not matter whether we have real or complex x

[agile knoll]

i can’t find a similar problem in my notes either

[fringe rampart]

but it's efficient 😋

[hazy peak]

Also it is pretty easy to prove, so...

[agile knoll]

i haven’t ever learned it

can you give me a quick run down please

[warm shale]

themadchessplayer

[agile knoll]

can it be solved using the ratio test at all?

[next wind]

yes i just solved it this way, want my proof ?

[agile knoll]

cuz we haven’t learned that at all and i genuinely don’t know what the sup

[agile knoll]

yes please

[next wind]

sure, ill write it again with decent handwriting and will send a photo in a minute

the way you applied the ratio theorem is not correct as well, the ratio should be a_(n+1) / a_n, the x shouldnt appear @agile knoll

this is where your error comes from imo

[agile knoll]

do you not plug in n+1 for every n

that would give you x^(n+1)

[hazy peak]

limsup of a sequence {a_n} is the maximum of the limit points of the sequence. Alternatively, it is the minimum number L such that all but finitely many a_ns are larger than L

When the sequence a_n converges, limsup coincides with limit

[next wind]

@agile knoll

[agile knoll]

why do you disregard the x^n

[next wind]

yes, the ratio theorem is for the a_n x^n series

there shouldnt be any x in the a_n

so in your series, the a_n is as i wrote it

that is where your error comes from

[agile knoll]

in my notes

this is one of the examples my teacher worked

[next wind]

ok I see what your teacher did

one second

I used D'Alembert's rule, which states that if you have a series in the form of a_n x^n, the limit of | a_(n+1) / a_n | dictates the convergence radius

but your teacher's rule also works, you just have to do an extra step

[agile knoll]

yeah im following my teacher

[next wind]

try to look at my proof, where most of the reasoning is done, and adapt it to your teacher's method

[agile knoll]

but i keep getting infinity

[next wind]

if you want I can rewrite my proof with your teacher's method

[next wind]

show me your reasoning for getting infinity

we'll go from there

[agile knoll]

alr i gotta drive to bank rq

give me like 10

[next wind]

sure

my reasoning and your teachers is the same, you just multiply every step of the way by x and you should have your teacher's method

anyway, just send me a pic of what you got once back

ping me as well just in case

[agile knoll]

@next wind this is what im getting

[next wind]

the last = is wrong

the limit of what you got isnt 0

[agile knoll]

why wouldn’t it be 0?

oh is it because of x

[next wind]

no, the limit of what you have (if we remove x) isnt 0 either

why do you think it is ?

forget about the x for now, as it is a constant when we look at the limit in regard to n

[agile knoll]

i got told that an arbitrary number over a larger arbitrary number equals infinity

(n+1)^n is a larger function than n^n

so wouldn’t the lim be 0?

[next wind]

no, and this is a prime counter example

to persuade you of this result, you can simply substitute n with a big number

this isnt a proof of course, but it will persuade you for now that the "rule" you used is false

[agile knoll]

yeah i just graphed it

and it isn’t an exponential function

so is that rule only true in some cases?

like take for example an exponential function over a linear function

[next wind]

yes, the rule works sometimes, it just depends on "how big" one number is compared to the other

you will have a more formal feel for it when doing equivalences

do you have any idea how to solve the limit ? now that we have established it is not 0