help-49

[midnight plank]

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[last slate]

Let $\alpha$ be a root of $2x^2+3x+4=0$. Then find the possible values of $2{\alpha}^5+4{\alpha}^4+7{\alpha}^3+8{\alpha}^2+9\alpha+3$

[grand pond]

deltaG

[last slate]

i want to know if my solution is correct, posting it in a while

[twilit field]

ouch

looks nasty

[last slate]

nah bro trust me after looking at the solution youll kiss it

[west iron]

polynomial division? lol

[last slate]

since $\alpha$ satisfies the quadratic equation, $2{\alpha}^2+3\alpha+4=0$.\
using long division,
$2{\alpha}^5+4{\alpha}^4+7{\alpha}^3+8{\alpha}^2+9\alpha+3 = (2{\alpha}^2+3\alpha+4=0$)({\alpha}^3+{\alpha}^2\frac{1}{2}+{\alpha}\frac{15}{8}) + \frac{3}{8}\alpha -\frac{9}{2}$\
$\frac{3}{8}\alpha -\frac{9}{2}$ is the reduced version of that 5 degree polynomial and $\alpha = \frac{-3+- \iota\sqrt{23}}{4}$
and then i just put this value of alpha in that linear poly

[grand pond]

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

[edgy schooner]

so, then?

[last slate]

then i get the value

but my question is

why is this allowed? considering dividing by zero is illegal

[last slate]

yeah

[west iron]

the actual quotient you're getting looks dubious

Ah I see

[west iron]

Because you're dividing a polynomial by another polynomial

[next current]

Can I ask a question

[west iron]

you're not dividing the numbers you get when you plug in something into that polynomial

[midnight plank]

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[next current]

Ight sorry

[last slate]

ohh

makes sense

[west iron]

It's actually very nice because now the only thing you care about is the remainder (since the divisor evaluates to 0 when you plug in your root)

[last slate]

true, my mind was blown when i saw our math teacher do it

[west iron]

But it's not like dividing by the polynomial 0 which would be not so good

[edgy schooner]

0/0 is not undefined, it is indeterminate

it's not illegal in legal sense as well

[last slate]

but how do we know that the numerator is also zero

alpha isnt a root of the above polynomial

[west iron]

This is a limit thing only

[last slate]

sorry, i didnt understand what you said here

[last slate]

does it now

[west iron]

0/0 is undefined, calling it indeterminate really only makes sense if you're a precalc/calc 1 student

[last slate]

and also it must be approaching to zero, right?

[twilit field]

not really, no

[west iron]

Well there's the polynomial that's always equal to 0, called the zero polynomial, and you can't divide by that

[last slate]

isnt $2{\alpha}^2+3\alpha+4$ always zero?

[grand pond]

deltaG

[west iron]

Not as a polynomial

For example α=1

[last slate]

oh in that manner

i get it now

[west iron]

You need to think of polynomials as objects in their own right

That can be added subtracted multiplied divided without plugging specific values in

[last slate]

thanks :D

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[west iron]

Np

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[manic bison]

How do we find the minimum value for this

[manic bison]

Nvm got it

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[craggy yew]

i got a = 6, d =5, b=7 and c=8

can someone verify my answers?

[outer pilot]

put it in and check if all of the equations are correct

[craggy yew]

ok

...they are so wrong

[outer pilot]

🙃

[craggy yew]

if i subtract 2 from eq 1

i get a+d=1 right?

whats that

[grand pond]

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[craggy yew]

sadly it aint free

and i dont want the solution i wanna solve it myself

ohhhh

[craggy yew]

.

i get a+d=1 right?

nice

[craggy yew]

.

[outer pilot]

maybe try to use a matrix

[craggy yew]

nah i have to use algebra

[outer pilot]

ok thats fine

[craggy yew]

so i get a+d=1?

[outer pilot]

its just would a lot of steps

a-d

[craggy yew]

oh

a-d=1

[craggy yew]

and subtracting eq 3 from eq 4 gives me b-c=1?

@outer pilot (sorry for ping but i need hlep)

[outer pilot]

-1

8-9

it pings me when you react to my message...

I just went to get some food

[craggy yew]

well if a-d=1

is there any way to find a+d?

[outer pilot]

youll always get - if you are subtracting equations

[craggy yew]

lol dumb me forgot

[outer pilot]

if you sum them you can get +

[craggy yew]

ill try

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[dry ore]

Not sure where I did wrongly prob the multiplacion part but I'm not sure how to simplify that

[dry ore]

Ans was 0.1 if I'm not wrong

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@dry ore Has your question been resolved?

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[craggy yew]

can someone provide the step by step solutions?

solve using basic algebra

[sweet wing]

have you had matrices?

[severe cloud]

they just said basic algebra

[old storm]

you dont even need matrices for this, just solve the system on your own

[craggy yew]

yes

[sweet wing]

oh right, matrices is linear algebra

try subtracting equations from each other

[old storm]

or substituing

[craggy yew]

i tried a lot

[old storm]

did you now

[craggy yew]

cant find a solution

[sweet wing]

\begin{align*}&a+b+c=7\-(&b+c+d=6)\end{align*}

[craggy yew]

whats done here?

[grand pond]

Flappie

[sweet wing]

what happens if you subtract the bottom equation from the top one

[feral sedge]

the easiest way to do this problem is to start by adding all four equations

[craggy yew]

hmm i didnt try that

[craggy yew]

wouldnt that make it long and complicated?

[feral sedge]

no

it absolutely would not

[old storm]

7 - a + d = 6

d = a - 1

c + a -1 + a = 8

c = 10 - 2a

a - 1 + a + b = 8
b = 9 - 2a

a -1 + 9 - 2a + 10 -2a = 6
3a = 12

a = 4
d = 3
c = 2
b = 1

[feral sedge]

messy way of doing it

[old storm]

its super simple

[sweet wing]

substitution is the brute force, but simple way to do it

[old storm]

i literally did it on a chat

the numbers are wayyy too easy not to subsitute this away

[old storm]

this is your solution @craggy yew

[craggy yew]

ohhh

[feral sedge]

this is not a good solution

[craggy yew]

i feel so dumb

[feral sedge]

are you going to try this

[craggy yew]

is there a easier way to do the same?

[feral sedge]

i have been telling you for 5 min

[old storm]

bro the solution i gave him is like 10 lines

[feral sedge]

and it's messy

[old storm]

ah you get 30

cute

[craggy yew]

3a+3b+3c+3d=30

[old storm]

but there is no way someone who only knows basic algebra is going to think of your solution

[feral sedge]

now divide by 3

[old storm]

then a + b + c + d = 10

cute

if a + b + c = 7
and a + b + c + d = 10

find d

ah

that is very clever

[feral sedge]

this is standard way of doing such problem

[old storm]

is it really

[feral sedge]

which is why i say substitution is messy / not good

[old storm]

where was my teacher

[craggy yew]

damn

thanks guys

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[craggy yew]

what am i supposed to do in the question

[twilit field]

add a+b,b+c and c+d first I think

[craggy yew]

no like

what is meant by computing the greatest possible value

am i supposed to find the answer in algebraic form?

or is there a number u get from cancellation?

[twilit field]

of that I'm unsure

[craggy yew]

adding all three equations and cancelling gave me 2b = -c

[twilit field]

great

now solve for a,b,c,d

[craggy yew]

but isnt that wrong?

[sweet wing]

why would it be wrong?

[craggy yew]

because b is supposed to be a positive integer

[twilit field]

yes

[sweet wing]

b is an integer, nothing is said about c

[twilit field]

and c can be negative

[craggy yew]

but like

for c to be negative

doesnt b have to be neg?

[twilit field]

a has to be negative

[craggy yew]

no like

in 2b=-c

for c to be negative

b has to be negative right?

[twilit field]

why?

what is -(-1)?

[craggy yew]

ohhhhhhh

im dumb sorry

[sweet wing]

why does it matter if c is negative or positive?

there is no restriction on c

[craggy yew]

sorry i got confused

[twilit field]

this is a good question, it's normal to be confused, don't insult yourself like that

[dreamy lichen]

arguable

[craggy yew]

how do i use 2b=-c to solve

[twilit field]

you can then express a in terms of c too

right?

and then d in terms of c too

does that make any sense?

@craggy yew

[midnight plank]

@craggy yew Has your question been resolved?

[craggy yew]

hmm

[craggy yew]

how

[twilit field]

2b=-c

right?

just for context, this is the problem

$a+b=-2b$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

so a is ?

[twilit field]

.

[craggy yew]

-3b

[twilit field]

yes

now you know b an c in terms of b

find d in terms of the same

[craggy yew]

-b?

[twilit field]

yes

so what's a+b+c+d?

where everything is in terms of b

[craggy yew]

-5b

[twilit field]

good

and b is a positive integer

I'm taking that to include 0

so then the answer is?

[craggy yew]

?

[twilit field]

do the positive integers include 0?

[craggy yew]

whats that supposed to mean

[twilit field]

I'm asking if you think we include 0

what values can b take

in other words

[craggy yew]

ohhh

[craggy yew]

5b?

i dont know

[twilit field]

the sum is -5b , right?

[craggy yew]

yes

[twilit field]

and b takes values 1,2.....

so what's the maximum value the sum can assume

[twilit field]

?

[craggy yew]

infinity?

(sorry battery died so went afk for a while)

[twilit field]

I think the answer is -5

is that right?

[craggy yew]

the book doesnt have solutions

[twilit field]

let me look it up

[craggy yew]

and i cant find the arml 1990 solutions

[craggy yew]

ok

[twilit field]

yup

think about why this is the answer

@craggy yew

[craggy yew]

why should b be as small as possible?

[twilit field]

b is a positive number

and we already know the sum is negative

[twilit field]

does that make any sense

[craggy yew]

no

[craggy yew]

which sum?

[twilit field]

a+b+c+d=-5b, yes?

[craggy yew]

yes

[twilit field]

and b >1?

[craggy yew]

why cant it be 1?

[twilit field]

I meant $b\geq 1$

[craggy yew]

the latex is not working

for me

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

now?

[craggy yew]

yes

[craggy yew]

yes

[twilit field]

so now does it make sense

[craggy yew]

no

we have to find the greatest possible answer so why do we have to minimize

[twilit field]

is -5 bigger than -10?

[craggy yew]

ys

[twilit field]

okay, so how would I MAXIMISE -5b

by minimsing 5b, right?

[craggy yew]

wait what is meant by maximising?

[twilit field]

here just finding its maximum value

[craggy yew]

well since b can be any value

[twilit field]

greater than or equal to 1

[craggy yew]

shouldnt it be multiplied by -infinity?

[twilit field]

b is more than 1

always

[craggy yew]

negative infinity?

[craggy yew]

im confused so can u reveal the ans?

[twilit field]

b=1

[craggy yew]

why

[twilit field]

we want to get a value for -5b that's as big as possible right

and $b\in [1, \infty) \in \Z$

[grand pond]

ƒ(Why am. I here)=I don't know

[craggy yew]

yes

[twilit field]

so what value of b would you chose

[craggy yew]

infinity

[twilit field]

why

that would give you $-\infty$ right?

[grand pond]

ƒ(Why am. I here)=I don't know

[craggy yew]

because its a big number

[craggy yew]

ohhh

[twilit field]

ok, is $-5 \cdot 1$ bigger than $-\infty$?

[grand pond]

ƒ(Why am. I here)=I don't know

[craggy yew]

yes

ohh

[twilit field]

does the answer make sense now?

[craggy yew]

i see why its 1 now

[twilit field]

cool

[craggy yew]

ah

i finally get the whole thing

[twilit field]

[craggy yew]

thanks for helping me so much!

[twilit field]

no problem, this was a fun question

[craggy yew]

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[craggy yew]

.reopen

[midnight plank]

✅

[craggy yew]

(sorry for ping) @twilit field one last question

[twilit field]

sure

[craggy yew]

how did u get the answer of -5 by -5b?

[twilit field]

b=1

[craggy yew]

that was a dumb question

sorry for wasting ur time

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[last slate]

https://tinyurl.com/4a2449b6

[last slate]

please help with assignemnt

@here

[midnight plank]

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[midnight plank]

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[eternal pine]

how do i solve this integral

[eternal pine]

this would be my approach dunno if it is correct

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[slow moon]

can someone help me with this

[slow moon]

i need help with g

and h and i

what i have so far

it seems i’m wrong for concave up and down

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[slow moon]

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[sweet wing]

!status

[midnight plank]

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

[young spire]

2

/1

[nova yoke]

show where you are stuck?

[young spire]

I’m not sure where to move what

[nova yoke]

well start by clearing the denominator on the RHS

[young spire]

Like I tried moving the 3n-2 over to the left but didn’t know what to do next

[nova yoke]

then rearrange so you have all the terms containing N on one side

and everything else on the other

[young spire]

Lemme c

Ok Ty

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[young spire]

I have got the wrong answer

Is it possible to not expand the brackets

[sweet wing]

!original

[midnight plank]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[young spire]

Solve each literal equation for x

After expanding

[sweet wing]

!original

[midnight plank]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[sweet wing]

thank you

why cant you expand the brackets?

[young spire]

Ye but my teacher did it without expanding and when I did I still got the wrong answer but was close

[sweet wing]

you can also divide

[young spire]

Divide pi r

[sweet wing]

so then what do you get

[young spire]

A/pi r= r-x

[sweet wing]

now, isolate x

[young spire]

Ye-r

(A/pi r) -r= -x

This I where I got stuck and made the wrong choice

[sweet wing]

what did you do here

that was wrong

[young spire]

Was it

[sweet wing]

you said you made the wrong choice

no?

what was that

[young spire]

After this step I mean

[sweet wing]

yes

so what did you do

after this

how can i find out what you did wrong

[young spire]

Don’t u divide -1

[sweet wing]

if you dont tell me what you did

[sweet wing]

then what do you get?

[young spire]

R-(A/pi r)=x

[sweet wing]

thats correct

i dont see the issue

[young spire]

Maybe the textbook got it wrong

[sweet wing]

what does the textbook say

[young spire]

They got

I

I

i

[hard umbra]

it's the same

[sweet wing]

yeah its the same thing

....

[young spire]

Ok Ty

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[steep jasper]

<@&286206848099549185>

[steep jasper]

Please help with the 7th one

[forest jackal]

(1-x)(1+x)=1+x-x-x^2

i just multiplied the terms

[steep jasper]

Yes n then

[forest jackal]

+x-x cancel out

[steep jasper]

Okay

[forest jackal]

(1-x)(1+x)=1-x^2

[steep jasper]

Okay

[forest jackal]

did you get it now?

[steep jasper]

Yesss

[forest jackal]

so now we have (1-x^2)(1+x^2) in numerator

which gives 1-x^4

[steep jasper]

How

In numerator?

[forest jackal]

we just got 1-x^2 from multiplying 1-x and 1+x

[steep jasper]

Okay

[forest jackal]

and 1+x^2 is already there in numerator

[steep jasper]

Okay

[forest jackal]

so now multiply them

[steep jasper]

Thi

This*?

[forest jackal]

yes

good

[steep jasper]

Yayayayya

[forest jackal]

now just multiply 1-x^4 with 1+x^4

because its already there in numerator

[steep jasper]

Okay

Okay

[forest jackal]

which gives 1-x^8

[steep jasper]

Yessss

Nthen

Do

Same?

Again?

[forest jackal]

(1-x^8)(1+x^8)=1-x^16

[steep jasper]

Which will give 1-x^16

[forest jackal]

(1-x^16)(1+x^16)=1-x^32

[steep jasper]

Yess

N now what abt the denominator part?

[forest jackal]

we already had 1-x in denominator

so its proved

[steep jasper]

Okay

THANK YOU SO MUCH

[forest jackal]

np

[steep jasper]

I GOT IT

Ya smart

[forest jackal]

now close the channel fr

[steep jasper]

Lmao alr

.close

[midnight plank]

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[slate ferry]

(Sorry for interrupting but, what grade are you in? Or are you in collage?)

[steep jasper]

9th

Just started

Why tho?

Oh ik that's easy but my head is messed up so I wasn't able to do

[slate ferry]

(We cant talk here after you closed the channel, dms if u wish)

[steep jasper]

Nah it's okay

[steep jasper]

Alr

[midnight plank]

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[young spire]

Question d is c2=a-b2 corrrect

Squared btw

Does b squared plus c squared equal (b+c) squared

[gritty holly]

you want to make a the subject

[young spire]

Yes

[gritty holly]

you need help with d)?

[young spire]

Yes

[gritty holly]

ok

[young spire]

It says square first

[sturdy glacier]

@storm09 can you sovle i plase 🛳️

[young spire]

?

[gritty holly]

get your own channel

[forest jackal]

not this guy again 💀

[forest jackal]

he has 2 of his own

[gritty holly]

on the right track but next subtract b^2 from both sides

[young spire]

Subtract?

[gritty holly]

💀

my bad add

[young spire]

Ye

[young spire]

.

[forest jackal]

no no no

[sinful trout]

no

it doesn't

[young spire]

Ye that’s what I thought

[sinful trout]

$(b+c)^2 = b^2 + c^2 + 2bc \neq b^2 + c^2$

this is GIVEN that 2bc is not equal to 0

[grand pond]

ren

[young spire]

So I get a equals b^2+c^2

[sinful trout]

fr

[young spire]

But it says (b+c)^2

Book Error maybe idk

[forest jackal]

the question here is different from what you wrote

c=sqrt(a)-b

add b on both sides

b+c=sqrt(a)

[young spire]

The book wants me to square first

[forest jackal]

oh

then that gives

c^2=a+b^2-2bsqrt(a)

[young spire]

So the book telling me one thing but gets the answer differently

[forest jackal]

yes lmao

[young spire]

Where u get 2broota

Root a

Just to c

O u have to square the whole side

Like (sqrt a-b)^2

[forest jackal]

yes

[young spire]

Ok Ty I understand now

.close

[forest jackal]

np

[midnight plank]

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[acoustic sparrow]

find d to maximise theta

[acoustic sparrow]

the calc solution is horrendous algebra,

a geometric sol would be greatly appreciated

[midnight plank]

@acoustic sparrow Has your question been resolved?

[chrome vessel]

here is one way: try and look at which theta works

hint: if theta works, what would the circumcircle passing through the triangle with angle theta and length 5.6 look like?

[acoustic sparrow]

pardon?

theta works?

[chrome vessel]

analyse for which theta does there exists a right triangle looking like the graph

[acoustic sparrow]

???

oh so your telling me to find max alpha

[chrome vessel]

no, it more like asking the question: does theta=80 degrees work? how about 70 degrees

[acoustic sparrow]

??

i dont see why it wouldnt

[chrome vessel]

if you can distinguish which theta works and which theta doesn't, then you can just find the maximum of theta

[acoustic sparrow]

right

i'll try

[chrome vessel]

and the hint is drawing the circumcircle of the triangle with angle theta

note that the angle theta uniquely defines a circle passing through BC

[acoustic sparrow]

why does it?

[chrome vessel]

all the points that forms angle of value theta with BC lies on the circle

[acoustic sparrow]

?????

[chrome vessel]

given B,C the set of points {D: angle BDC=theta} will look like a circle

this is because if D, E are on the same circle as B,C then angle BDC=angle BEC

and if angle BDC=theta, then the circumcircle of BDC will have center O such that angle BOC=2theta

you can prove that there is only 2 possible O that satisfies this

[chrome vessel]

may I ask what is your background in geometry?

[acoustic sparrow]

i get you can define a circle with 3 points

but i cant understand what youre writting

[chrome vessel]

Can you prove the following:
given two points B,C on the plane, show that the set of points {D: angle BDC=theta} is the union of two circular arcs?

[acoustic sparrow]

no

[chrome vessel]

try to prove it, it will help getting an understanding of the relation between circle and angle

do you understand the statement though?

[acoustic sparrow]

no

[chrome vessel]

take 32 degree for example

I am saying all the points that forms a 32 degree angle lies on two circular arcs:

as outlined by the red line

i am saying all the points that forms 32.019922285598 degrees with BC lies on those arcs

[acoustic sparrow]

yes

[chrome vessel]

now do you understand the statment?

if so can you attempt to prove it?

[acoustic sparrow]

look its almost 12

i just wanted a solution to the other one

[chrome vessel]

I can't give a solution until you understand those things (as these things are needed in the solution)

[acoustic sparrow]

oh my god

ill just hit the hay

[chrome vessel]

but you can show that the circumcircle must be tangent to the d there

[acoustic sparrow]

.close

[midnight plank]

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[midnight plank]

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[deft pier]

so im doing limits

and i can't go any further in it

i did p=f(x)-mx

but can't go further

[midnight plank]

@deft pier Has your question been resolved?

[midnight plank]

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[heavy falcon]

the answer takes the integral from 0 to 2

shouldnt it be from -inf to 2

[midnight plank]

@heavy falcon Has your question been resolved?

[heavy falcon]

<@&286206848099549185>

[zenith cradle]

In this kind of questions they ask for an enclosed area like shown here, the keyword is that the area is bounded by C, the line d and x=2. Origin could be said to be a boundary to the area, that is to say, the cyan and yellow areas are separate.

[heavy falcon]

but it wasnt specified that the area is taken to be bounded by x=0

what matters is that it is bounded by (C),(d) and x=2

also how do i know that 2 is the upper bound not the lower one for for example

shouldnt the curve be draw from x=0 to indicate that at least

[heavy falcon]

to indicate this i mean

[zenith cradle]

the thing is I see yellow and cyan areas as different bounded regions, and they ask area of a specific bounded region. By the topological definition of boundary, it being the points of closure (those that limit the region) and are not inside the region itself, the yellow and cyan areas would be strictly speaking different regions.

[heavy falcon]

although what i am considering to be one region still satisfies all the named conditions no

because the area from -inf to 2 is bounded by (C) and (d)

i dont see how it isnt

[zenith cradle]

but the cyan area is not also bounded by x = 2

[heavy falcon]

but the area of cyan + yellow

is

i am finding area of yellow+cyan

which is bounded by x=2

and also bounded by (C) and (d)

the yellow region alone is also bounded by (C),(d) and x=2

but one cant determine which is the one to take if it is not specified am i wrong

and since the only boundary for x is 2

then we will have to consider it unbounded by any line of equation x=a from the other side right?

[zenith cradle]

yes, apparently you are right, here's one example where a single point is not enough to limit a region. Need to check again what a region is I guess.

[heavy falcon]

yes here the region is specified to be between 0 and pi

and pi isnt an intersection point that lets the are be enclosed by a "closed" shape

same thing for 0

and this is the same as the one that you split into 2

so what to do now

[zenith cradle]

What's remarkable at least is that they make a distinction between the regions R1 and R2 and also they say R is a region not a bounded region. In any case, I don't think I can help much more, since I am not that knowledgeable about these concepts.

[heavy falcon]

yes they split them at the intersection point

because one of them is below the other before the intersection

and this changes after the intersection

you are finding the area so you will have to integrate the abs value of the difference

thats why

[heavy falcon]

and it isnt necessary to mention if it is bounded or not the boundaries are already clear

tysm

and sorry if i wasted your time

[heavy falcon]

<@&286206848099549185>

[midnight plank]

@heavy falcon Has your question been resolved?

[midnight plank]

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[rocky mountain]

im stuck on i, i thought that i needed a domain for the range?

[slow walrus]

range would be all real numbers for both

a and b

part

[rocky mountain]

so just any number works?

[slow walrus]

when x belongs to R, range of F(x) is R

for a and b

part

[rocky mountain]

solve for x/

?

[slow walrus]

for i part of the q

[rocky mountain]

okay ive got it now sorry🤣 tysm

.close

[midnight plank]

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[slow walrus]

u got ii,iii,iv?

too

[rocky mountain]

yh should do

ty

[midnight plank]

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[craggy yew]

how did we get the underlined equation

[twilit field]

Ok, so person A is stationary wrt to the belt

right

so what's their speed

[craggy yew]

whos person A

[twilit field]

Al

[craggy yew]

oh

6 ft/s

[twilit field]

ok, what about Bob

[craggy yew]

10 ft/s

[twilit field]

yes

and Cy?

[craggy yew]

8 ft/s

[twilit field]

shouldn't it be 16?

[craggy yew]

i might be wrong

[craggy yew]

why tho

[twilit field]

the walkway is moving at 8m/s

and the person is walking at 8

[craggy yew]

cy is not walking on the walkway

hes walking beside it

[twilit field]

ah

then 8

yes

ok, so now what

[craggy yew]

after some time bob will be ahead and al will be in the middle

[twilit field]

good

[craggy yew]

and we have to find the distance between al at that point and the beginning of the walkway

i think

[twilit field]

you want to find the distance between Al and cya

or al and bob

[craggy yew]

yes

my main question is that how did this equation come

[twilit field]

the time is measured from the time the first person steps on the belt

so one person's coordinates are $8(s-4)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[craggy yew]

whys it 8(s-4)

[twilit field]

they step on the belt 4s later

[craggy yew]

isnt that the distance?

[twilit field]

yes

[craggy yew]

but like

bob's overtaking and al going in the middle will take a lot of time right?

why is it only 4 seconds

[craggy yew]

oh wait is this supposed to be its original position when cy stands on the walkway?

[twilit field]

that's cy's position at the instant we're observing the sysetm

*system

[craggy yew]

so why is the time s-4

[twilit field]

cy starts walking only after 4 s

[craggy yew]

wait what time is s

[twilit field]

when Al is half way between cy and bob

[craggy yew]

at what position are the people in s-4

[twilit field]

I don't follow

[craggy yew]

why is 4 subtracted from s

[twilit field]

because cy starts moving 4 seconds later

[craggy yew]

yes but why do u need to do that

[twilit field]

because you're measuring time with respect to the person who started first

[craggy yew]

what happens during s-4

[twilit field]

that's when Al is inbetween

[craggy yew]

what happens during s

[twilit field]

read the solution again please

it explains it better than I can

[midnight plank]

@craggy yew Has your question been resolved?

[midnight plank]

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[twilit field]

I don't follow the second step here

[twilit field]

from let k be the largest element ...

like how are we sure dividing by a common factor will give is $v_k$

[grand pond]

ƒ(Why am. I here)=I don't Know

[slow walrus]

which book?

mind sharing?

[twilit field]

axler

search for axler springer open access

[slow walrus]

kk thanks

[prime hornet]

what do you mean?

a_1v_1 + ... + a_kv_k + ... + a_mv_m = 0, so we rearrange for v_k

such a k must exist cause v1,...,vm is LD

so Axler is just choosing the largest k for this

[twilit field]

ah

[sage olive]

all they're doing is dividing both sides by a nonzero coefficient

totally valid

[twilit field]

I thought we couldn't assume this statement to be true

sorry

thanks

[prime hornet]

[twilit field]

.close

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[oblique torrent]

what's the value?

of I

.close

[midnight plank]

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[oblique torrent]

.reopen

[midnight plank]

✅

[oblique torrent]

wait

[sage olive]

bro didn't even give us a chance to help 😭

[oblique torrent]

now u can tho

[sage olive]

do you know the value of the gaussian integral?

e^(-x^2)?

if you need to derive that from scratch it's doable but it's a pretty famous result

you can just make a u-substitution with u = x / sqrt2 to turn this into half of the gaussian integral

[oblique torrent]

i dont think so

[sage olive]

is there a specific technique they wanted you to use?

[oblique torrent]

yes

uhm

wait

[oblique torrent]

polar coordonate system

[sage olive]

https://youtu.be/mrcRODAx-Vk?si=8I-BA7oIQKvwBur9

[oblique torrent]

ohhh

thanxx

.close

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[rocky mountain]

what rules are used here? im tryna mark my hw but the answers skip steps or im struglling to follow along with whats happening😭

[long fox]

Formula : sinAcosB - CosAsinB = sin(A-B)

And 2sinAcosA = sin2A

[rocky mountain]

.close

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[rocky mountain]

ty

[midnight plank]

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[tidal turret]

find all $x \in \mathbb{R}$ such that the series converges, $\$indicate for which values the convergence is absolute $\$and for which values the convergence is conditional $\\ \sum_{n = 1}^{\infty} \frac{x^n}{n^3}$

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[midnight plank]

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

[tidal turret]

1

[night egret]

do you know what the interval of convergence is?

[tidal turret]

the interval for which the series converges

[night egret]

do you know how to find the interval of convergence using the ratio test?

[tidal turret]

,, \lim_{n \to \infty} \left| \frac{\frac{x^{n+1}}{(n+1)^3}}{\frac{x^n}{n^3}} \right| < 1

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

this?

[night egret]

yes
now let’s try to simplify this

[tidal turret]

sure

,, \lim_{n \to \infty} \left| \frac{n^3(x^{n+1})}{x^n(n+1)^3}\right| < 1 \ \lim_{n \to \infty} \left| \frac{n^3 \cdot x^{n} \cdot x}{x^n(n+1)^3}\right| < 1

I did nothing fancy

let me simplify

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

okay

,, \lim_{n \to \infty} \left| \frac{n^3(x^{n+1})}{x^n(n+1)^3}\right| < 1 \ \lim_{n \to \infty} \left| \frac{n^3 \cdot x^n \cdot x}{x^n(n+1)^3}\right| < 1 \ |x| \cdot \lim_{n \to \infty} \left| \frac{n^3}{(n+1)^3}\right| < 1 \ |x| \cdot \lim_{n \to \infty} \left|\ \frac{n}{n+1}\right|^3 < 1

lmao

we can pull x out of limit

[night egret]

yep

[tidal turret]

we need to expand (n+1)³

[night egret]

or

[radiant trout]

n^3/(n+1)^3=(n/n+1)^3

[tidal turret]

true

ty

is this euler limit or something

,w lim n to infinity (n/(n+1))^3

[night egret]

well what does n\n+1 approach

[tidal turret]

if we factor by n and cancel both n from numerator and denominator the limit is approaching 1/1

[night egret]

you can’t factor by n

[sweet wing]

dont forget that |x|

[night egret]

but you can do lhopital

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

so -1 < x < 1

[sweet wing]

yep

[tidal turret]

$x \in (-1, 1)$

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

but about the convergence being conditinal convergence or absolute convergence?

which values of x absolutely converges which values conditionally converge

[sweet wing]

you used the ratio test, right?

[tidal turret]

yes we used ratio test

[sweet wing]

what does the ratio test say when its conclusive

[tidal turret]

> 1 for divergence

< 1 for convergence

[sweet wing]

< 1 means its absolutely convergent

[tidal turret]

okay

and for which values is it conditional convergence doe?

[sweet wing]

absolute convergence is stronger than conditional convergence

if $\sum |a_n|$ converges, then $\sum a_n$ also converges

[grand pond]

Flappie

[sweet wing]

if $\sum |a_n|$ doesn't converge, then you cannot say anything about $\sum a_n$

[grand pond]

Flappie

[sweet wing]

with the ratio test, you conclude that $\sum |a_n|$ converges

[grand pond]

Flappie

[sweet wing]

thus, it is absolutely convergent

[tidal turret]

yes there is no x in R such that it conditionally converges

because |a_n| converges

anyways thank you

.close

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[severe glade]

My teacher gave us problem (b) and said that if we wanted to, we could use the “difference of squares formula” to solve it. I used that formula in the second picture because it’s much quicker.

However I wanted to try figuring it out without the formula in the first picture, did I do it correctly?

[midnight plank]

@severe glade Has your question been resolved?

[severe glade]

<@&286206848099549185>

[sterile latch]

looks right to me

[dusky flint]

yeah

looks right to me too

[sterile latch]

in general though, i agree with your teacher

[severe glade]

Okay thank you, is the difference of squares considered a formula?

[sterile latch]

stick with the difference of squares formula, it will be much quicker

[severe glade]

Or a property

[sterile latch]

formula

properties are things objects have (ie integers are either even or odd)

the same way we consider the quadratic formula a formula and not a property

[severe glade]

Also is there a list of formulas I should I know? Cause tbh I feel like every other class my teacher mentions a “formula shortcut” that we can use to figure out but I always feel like I don’t rememebr it when she brings it up

[sterile latch]

what class are you taking?

[severe glade]

Like for example when she mentioned this one in class I didn’t remember it at all

[sterile latch]

algebra 2?

[severe glade]

Precalcukus

I haven’t taken algebra 2 in a couple years

[sterile latch]

i would say you should know the following

• difference of squares

• sum/difference of cubes

• double angle identity for sine and cosine

[dusky flint]

yes true

[sterile latch]

• half angle identity for sine and cosine

and also the quadratic formula

[dusky flint]

Sridhar Acharya Formula

[sterile latch]

besides those i dont think there are very many formulas i needed to use during my pre calc class

[sterile latch]

same thing ¯_(ツ)_/¯

[severe glade]

Thanks I appreciate it.

Is there not a list of formulas similar to how there’s a list of “law of exponent” and “radicals rules”?

[dusky flint]

yup

[severe glade]

For exponents my teacher gave me a list of exponent rules to remember

And radicals

But I was curious if there was a list for formulas I should know

[dusky flint]

there's also completeing the squares method

[severe glade]

So I can just review them

[sterile latch]

im not sure if theres a generalized one

you're probably better off going to ask your teacher

to see which formulas they would want you to know for their class in specific

[dusky flint]

yeah ig there's a good ig page for generalised formulas

[sterile latch]

b/c obv my advice is personalized to the course i took

[dusky flint]

its called math vibes or smtg

it gives good charts with tricks sometimes to remember those formulas

[severe glade]

Okay gotcha.

Jw in the first picture I sent

Is that method I used called factoring?

Where I broke it down

[dusky flint]

no

its called expansion

[severe glade]

Ohhhh

[dusky flint]

factoring is when u make the stuff smaller

[severe glade]

Didn’t I do that?

Well expansion and Then factoting

Factoring

[dusky flint]

u did at last by applying identity of (a-b)^2

this complete method is called simplification

[severe glade]

Got it thank you both for ur help I appreciate it

[dusky flint]

Always

[severe glade]

.complete

.solved

[midnight plank]

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[midnight plank]

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[white loom]

why isnt it C/(x^2+1)

why Cx + D

[sharp coral]

the numerator should be a polynomial of 1 degree less than the denominator

[midnight plank]

@white loom Has your question been resolved?

[white loom]

so if it was x+1 in the denominator it would just be C?

or C + D

[sharp coral]

just C, if you have the sum of two unknown constants you can combine them into a single unknown constant

[white loom]

.close

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[obtuse wing]

I have to calculate the partial derivative of a function at a point, using the definition

[sweet wing]

!original

[midnight plank]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[obtuse wing]

This is the original function

The task is to: Using the definition, calculate the partial derivatives of the function ( f ) at the point ( P_0 ).

[grand pond]

john3515

[sweet wing]

do you know the definition?

if so, what is it

[obtuse wing]

[ \frac{\partial f}{\partial x}(x_0, y_0, \ldots) = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x, y_0, \ldots) - f(x_0, y_0, \ldots)}{\Delta x} ]

[grand pond]

john3515

[obtuse wing]

this is with respect to x

[sweet wing]

have you tried plugging in f and seeing what happens?

[obtuse wing]

didn't I do it in the picture I sent before?

[sweet wing]

yes you did

so what is your question?

[obtuse wing]

how do I proceed

[sweet wing]

you made a mistake here

its h^2-2h+1

[obtuse wing]

oh right

such a stupid mistake

yeah it works

.close

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[severe glade]

For question D (blue problem) are both my answers technically correct?

[midnight plank]

@severe glade Has your question been resolved?

[severe glade]

<@&286206848099549185>

[midnight plank]

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[last slate]

Say we have a simple circuit with a battery,bulb and a resistor

[twilit field]

yes?

[last slate]

Now we attach a voltmeter

What would the reading be?

Oh wait

[twilit field]

the internal resistance of the battery is 0?

[last slate]

Yes

[twilit field]

equal to the potential across the battery

no?

[main current]

Normally batteries are defined by that V

So if it's a 3V battery, that will read 3V

[last slate]

But didn't we already split the voltage across the bulb and the resistor?

[twilit field]

use KVL

[last slate]

Kirchoffs what?

[twilit field]

Voltage law

[last slate]

That exists?

[twilit field]

yes

[last slate]

I know only junction law and loop law

[twilit field]

learn it

it's very very useful

[last slate]

Oh okay

Thanks

[main current]

It's probably loop law

"Voltage drops = Voltage gains around any loop"

It's commonly referred to as KVL

If that battery adds 3 volts, then the lamp and resistor are going to burn those 3 volts

[last slate]

Oh

Say buld uses up 2 volt and the resistor uses 1

V3 reading would be 3 or 0?

[last slate]

It's 3

But then wouldn't the potential of the whole circuit be equal to 6?

[hard shard]

remember the battery supplies voltage and the bulb and resistor use that voltage