help-23

[light shoal]

notice that Mv = 0 is the same as Mv = 0v

[patent lodge]

yyup

[light shoal]

so it's actually an eigenvalue-eigenvector equation in disguise

[patent lodge]

WOAH

ok

wow and thats it?

[light shoal]

yea, any nonzero solution to Mv = 0 is an eigenvector corresponding to the eigenvalue 0

[patent lodge]

i feel like i havent used the other parts it asks for then if i just solve it for the eigenvalue of 0

[light shoal]

and if there is no nonzero solution, then 0 is not an eigenvalue

[patent lodge]

yeah but

have i used the other parts in that solution

?

[light shoal]

hmm, well do you have two distinct solutions to the equation in part b?

[patent lodge]

no

i have a vector thats like [7/5+z, 1/5+z, z]

[light shoal]

do you have some formula for all of the solutions? if so you could just pick two of them

yea, try plugging in two values of z to get two specific solutions

[patent lodge]

why?

[light shoal]

then as part (i) suggests, take the difference of the resulting vectors

[patent lodge]

yeah?

[light shoal]

that will give you the v0 and v1 that are referred to in c(i)

[patent lodge]

yeah thats how i solved it

[light shoal]

then you form v = v1 - v0 to get the solution to c(ii)

[patent lodge]

i just let z be z1 and z2

[light shoal]

i mean take two actual numbers for c(ii)

like z=0 and z=1 or something like that

[patent lodge]

yeah

but where does that get me sorry im dumb

[light shoal]

then that gives you two specific vectors v0 and v1

that are not functions of z

[patent lodge]

yes

[light shoal]

then you compute v = v1 - v0

[patent lodge]

yep

[light shoal]

which is a new vector that is not a function of z

[patent lodge]

yes

[light shoal]

and that vector will solve c(ii)

[patent lodge]

so like

ohhh

ok so you are saying

find an eigenvector with like M-eI|0

and then find an explicit one by subbing in random values for v1 and v2?

[light shoal]

well the v1 and v2 have to be solutions to the equation in b

so you sub in random values of z

into your formula from b

[patent lodge]

yeah sorry

but why couldnt i just get my eigenvector of the matrix and then sub in a number if it isnt explicit?

[light shoal]

can you give an example of what you mean?

[patent lodge]

so when i row reduce

with the zero vector as the augmentation

it gives me, [1 0 -1 0; 0 1 -1 0; 0 0 0 0]

so i get like x=z y=z z=z

what do i do now lol

[light shoal]

wait, how did you get all zeros in column 4?

>> A = [1 -2 1 1; 2 1 -3 3; 4 -3 -1 5]
A =
     1    -2     1     1
     2     1    -3     3
     4    -3    -1     5
>> rref(A)
ans =
                         1                         0                        -1                       1.4
                         0                         1                        -1                       0.2
                         0                         0                         0                         0

in short, doing row reduction isn't solving Mv = 0, it's solving Mv = x0

where x0 is the column vector (1, 3, 5)

[patent lodge]

oh you added the 1 3 5

[light shoal]

but yea for sure, you can solve Mv = 0 if you use the augemented matrix with zeros in the 4th column

you don't have to solve it the way they're doing it

but it's what they intend

[patent lodge]

im confused lmfao

so

why do we have the 1; 3; 5 part

[light shoal]

let me walk through it in symbols

[patent lodge]

yes pls

[light shoal]

you found a generic solution to Mv = x0

where M and x0 are as in the problem statement

your generic solution is a formula involving a parameter z

now suppose i have two specific solutions (plug in two different values of z), namely v1 and v2

they are both solutions to Mv = x0

i.e.

Mv1 = x0 and Mv2 = x0

Now subtract those two equations

you get M(v1-v2) = 0

thus v1 - v2 is a solution to Mv = 0

[patent lodge]

yes

[light shoal]

since you've already done all the work necessary to get such a v1 and v2, this is kind of a shortcut

[patent lodge]

ive proved that like another way but yeah ive proved v1 - v2 is a solution to Mv = 0

[light shoal]

you don't need to do a new row reduction to solve Mv=0

[light shoal]

right, so now they just want you to use that fact to find a specific solution

like a numerical vector

[patent lodge]

yes

that satisfies lambda = 0

[light shoal]

well any nonzero solution to Mv = 0 is an eigenvector associated with lambda = 0

because as mentioned earlier,

Mv = 0 is just Mv = 0v in disguise

[patent lodge]

right

[light shoal]

i.e. it's Mv = (lambda)v with lambda = 0

[patent lodge]

sorry ye

[light shoal]

putting it in linear algebra terms, the null space of M is equal to the eigenspace associated with lambda = 0

if you know what those terms mean

[patent lodge]

i can like use context clues to understand lol

[light shoal]

ok cool

anyway you're overthinking this i think

[patent lodge]

but what do i doooo

[light shoal]

you just need to find an example solution

[patent lodge]

hoe

hoe

how

lmao

[light shoal]

just use the info that you already proved

choose a v1 and v2 that solve Mv = x0

you already have a generic solution

plug in two different values for z

[patent lodge]

wait

[light shoal]

that gives you your v1 and v2

[patent lodge]

yeah

ok

wait

so

[light shoal]

show me your answer for that part

[patent lodge]

i really dont understand lol

what do i do

how do i find my v1 and v2

[light shoal]

so what as it, you had v = [7/5+z, 1/5+z, z]

as your generic solution?

[patent lodge]

yes

[light shoal]

ok

say i plug in z=0

i get v1 = [7/5, 1/5, 0]

now plug in another z, say z=1

i get v2 = [12/5, 6/5, 1]

[patent lodge]

yep got them

[light shoal]

those are two solutions

[patent lodge]

no its one solution when you minus them

?

[light shoal]

those are two solutions to Mv = x0

now you subtract them to get a solution to Mv = 0

namely:

v2 - v1 = [1, 1, 1]

[patent lodge]

oh fuck yeah sorry lmfao

yes

[light shoal]

so if we didn't make any arithmetic mistakes, [1, 1, 1] should satisfy M[1, 1, 1] = [0, 0, 0]

[patent lodge]

ok im following so far

[light shoal]

(column vectors of course, i'm too lazy to typeset)

[patent lodge]

yes

ok

i am following

[light shoal]

that's all there is to it

that's your answer

[1, 1, 1]

[patent lodge]

right

and then any multiple of [1, 1, 1] is an explicit eigenvector?

[light shoal]

yep

[patent lodge]

sorry the lectures never taught the term excplicit

opk jesus thank you so much

[light shoal]

ahh

yea explicit just means like "an actual vector of numbers" in this case

like an example solution

[patent lodge]

thanks for going thru it stepo by step lol

yeah i see

[light shoal]

sure

[patent lodge]

ok thank you so much :))))

.close

[safe radish]

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[uneven veldt]

well do you know what is sin(-x) = ?

[light shoal]

hint: the only way this can happen if both terms on the LHS are 1

(assuming we're dealing with real values of x)

[uneven veldt]

use basic trig properties like sin(-x) = -sin(x)

[sonic geyser]

Do this

[uneven veldt]

yes, this implies that there is so such value of x which will satisfy this equation

[light shoal]

didn't the equation use to say sin(x) - sin(-x) = 2 ?

double check that you typed it right

ahh

then yea, 0=2 means there's no solution

fraid not

sin(-z) = -sin(z) is still true for complex z

is there anything else stated in the problem?

can you show a screenshot if you have one?

yea, alas there's no solution, you can also see that from the red text

there's no x for which both of those red equations are true

what does he write next in the video? "april fool" hahah

he is messing with your mind

or he made a typo on the whiteboard

probably same as "no solution"

[dull sequoia]

$\frac{e^{ix}-e^{-ix}}{2i} + \frac{e^{i(-x)}-e^{-i(-x)}}{2i} = 2$

does just cancels doesn't it

oh oops

[flat frigate]

Frosst

[dull sequoia]

yeah minus for sine

misremembered

this just solves for 0 = 2

[light shoal]

nah, the LHS cancels to 0

[dull sequoia]

complex i doesn't help either

or i mean complex version of sine doesn't help either

[light shoal]

nope, there no complex solutions either

[dull sequoia]

i mean im sure you can define some solution for it

[light shoal]

"let x be a value such that 0=2"

[dull sequoia]

let x be the solution to the equation sin(x) + sin(-x) - 2 = 0

[light shoal]

but sin(-x) = -sin(x)

so that reduces to -2 = 0

[dull sequoia]

not for this particular x

[light shoal]

ohhhh

a special x

[dull sequoia]

well i mean that's how you get imaginaries too no?

you define that i is the solution to √-1

[light shoal]

well when you introduce the imaginaries, they don't contradict any existing math...

[dull sequoia]

well it's actually i^2 + 1 = 0

you just define i such that i is the solution to the equation i^2 + 1 = 0

[light shoal]

sure

[dull sequoia]

here we can define let's say j such that j is the solution to the equation sin(j) + sin(-j) - 2 = 0

i feel like that's doable

[light shoal]

well you'd have to define what sin(j) is

[dull sequoia]

you would

it doesn't, but j is like lexically closer to i

[light shoal]

it does have a dot on top

[dull sequoia]

which is kinda the vibe im going for with this j

[uneven veldt]

If you have no further doubts, then sure

[safe radish]

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[grizzled skiff]

Stuck on part b

How do we know it >= 6 when completing the square?

[wise blaze]

every real number will not be negative if squared

[grizzled skiff]

Sry bro I don’t understand

I got the completing square part

But how is it >=6?

[wise blaze]

(a+2)^2 must be larger or equal to 0

right?

[grizzled skiff]

Yes

[wise blaze]

then (a+2)^2 +6 must be larger or equal to 6 right?

[grizzled skiff]

Ohhh right yeh

I’m with u

Could we not also say

0

Say > 0 ?

[wise blaze]

what if a= -2

[grizzled skiff]

It’s still > 0 no?

0 + 6 = 6

6 > 0

?

Or am i being stupid again lmao?

[wise blaze]

alright you mean that

its right

you can say its >0

[grizzled skiff]

Oh okay cool

Thanks man

U explained it very well !

Rlly appreciate it 🙂

.close

[safe radish]

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[wise blaze]

No problem!

[safe radish]

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[lime grove]

$505x-673y=1$

[flat frigate]

RulzerFly.

[lime grove]

$and they gaved me this \left(505\cdot4=2020\right);\left(673\cdot3=2019\right)$

[flat frigate]

RulzerFly.

[little sorrel]

what do u need to solve?

[lime grove]

u thought that x = 4 and y = 3 but it's wrong

[lime grove]

for x, y

[wise blaze]

in terms of?

[lime grove]

integers

[wise blaze]

How

[lime grove]

what ...

[fast gazelle]

are we supposed to find all of the solutions, or just one of them?

[lime grove]

all

pls don't give me the answer directly

[dense prism]

A single linear equation in two variables has infinite pairs of (x,y) satisfying it.

You can't manually find each solution lol they're infinite

There has to be another equation I believe which makes it a pair

[lime grove]

yes

how to find that pair..

[dense prism]

There are infinite pairs, for example
x=1/505 and y=0.
x=0 and y=-1/673 and so on...

[wise blaze]

interger

te

[dense prism]

Oh does it only demand integer solutions

[lime grove]

yes Z

[grizzled shoal]

You used x=4 and y=3

How did that not work?

[dense prism]

True

[lime grove]

it works but it's not just one solution

[high mica]

does x and y have to be positive?

[dense prism]

Even for integer solutions there are infinite solutions, of the form
x=673n+4 and y= 505n+3

[lime grove]

no

[high mica]

cuz if they dont then i think theres unlimited sols

just get lcm of 505 and 673 amirit

[dense prism]

Keep changing the value of n and you'll get infinite solutions

Even for integers

(673n+4,505n+3)

So the question is incomplete

[lime grove]

how did you got that

[high mica]

wait

[high mica]

no

it isnt

[high mica]

this is the solution

no one said that cant be a solution

[grizzled shoal]

Suppose we have one solution for
505x+673y=1
If we add 505 onto our y and 673 onto our x, we will also get a solution

[high mica]

this where n is integerful

[dense prism]

I don't think that is what @lime grove desires

[high mica]

crap i mean is an integer

sorry i keep adding ful to words that dont need a ful

[dense prism]

He meant he wants a pair

[high mica]

but then the question is

wait

oh no

[dense prism]

I said there are infinite

Can you send the full problem please?

[high mica]

well time to blame it on the problem

[dense prism]

@lime grove

[lime grove]

.

the problem is complete

[dense prism]

Do you have screenshot of some sort

Of the problem, because I want to clarify what it is asking

[lime grove]

you will understand nothing

wait wait i saw your bio

[high mica]

i refuse to believe

[lime grove]

you speak arabic?

[dense prism]

Not really it's from the Qur'an

A verse

[lime grove]

oo okay i know it's talking about heaven and actions

[dense prism]

Yes

[lime grove]

where are you from btw 🙂

[dense prism]

Kashmir

[lime grove]

kashmir?? it's a country ..??

[dense prism]

It's in India

[lime grove]

ooo okay indians are intelligent

[wise blaze]

Agree

[lime grove]

about what ?

[wise blaze]

indians are intelligent

Indeed

[lime grove]

oo okay

now can we solve the equation pls

[dense prism]

I'll need a screenshot of the problem for that 😄

[lime grove]

and americans the dumbest

[wise blaze]

Bruh

though im not american

[lime grove]

you're an asian ??

[wise blaze]

yes

[lime grove]

south of korea?

[wise blaze]

what th

[lime grove]

china?

[lean otter]

india

[wise blaze]

Em

technically yes

but different

[earnest wagon]

I noticed an abrupt pattern in the question btw, on putting x=4n and y=3n at same time, the value of LHS=n

[lime grove]

you eat dogs right??

[wise blaze]

Never

Its illegal

in hong kong

[lime grove]

it won't work

[lime grove]

in hong kong only?

[wise blaze]

idk

but many of them dont eat

[lime grove]

but you eat snakes

[earnest wagon]

True but it was interesting to find this pattern

[lime grove]

yes

[wise blaze]

hk is not china

[lime grove]

....... okok

[wise blaze]

good

[lime grove]

i have to admit it china has some great green tea

[wise blaze]

japan better i think

[lime grove]

i'll try it then judge which one is better

[wise blaze]

ok

[lime grove]

<@&286206848099549185> who's good at math no one ...

[high mica]

im not good at math but i can try

[calm zenith]

do you need to find x and y?

[wise blaze]

the question is solved

[lime grove]

it's not solved

[lime grove]

yes

[wise blaze]

x=673n+4, y=505n+3

[calm zenith]

i got a feeling the question has infinite solns or its incomplete

[lime grove]

but how

[wise blaze]

you can sub n to be any integer

[calm zenith]

hit and trail

[lime grove]

if i write this like that i'll get 0/5

[wise blaze]

if it worth 5 marks

i dont think the question is like this

[calm zenith]

yeah

[wise blaze]

any photo

[lime grove]

it is it's part of an exercice worth 5 points so if i miss the first one it's done..

and it's the first one..

[calm zenith]

welp there can be infinite solutions

[wise blaze]

exactly

[calm zenith]

is there any specific answer given for the question?

[lime grove]

wait i'll check the answer

[calm zenith]

sure

[wise blaze]

add oil i gtg

[lime grove]

x= 673n+4 y=505n+3

[calm zenith]

^

[lime grove]

yeah he's right but how to make it appear from 0

like building up the answer

[safe radish]

@lime grove Has your question been resolved?

[lime grove]

imean just forget aboutit

.close

[safe radish]

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[minor quail]

Does it work for negative numbers?

[minor quail]

No

.close

[safe radish]

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[stiff sequoia]

Algorithms and calculation methods, I almost completed the task but I can't do it completely. The first picture is my notebook and the second is an example

[stiff sequoia]

<@&286206848099549185>

[safe radish]

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

[stiff sequoia]

okay

[high mica]

what;s the question

[stiff sequoia]

Finish the task

[safe radish]

@stiff sequoia Has your question been resolved?

[stiff sequoia]

<@&286206848099549185>

[safe radish]

@stiff sequoia Has your question been resolved?

[raven cosmos]

i have a question

[safe radish]

@stiff sequoia Has your question been resolved?

[safe radish]

@stiff sequoia Has your question been resolved?

[safe radish]

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[hearty dagger]

Isnt it supposed to be
V enemy ship- 60 km/hr=20 km/hr
Then V enemy ship= 80 km/hr ?

[safe radish]

@hearty dagger Has your question been resolved?

[lean otter]

no because they relative to the destroyer the ship is getting closer so its -20, thus 60 - 20 = 40

[hearty dagger]

So its gonna be
V enemy+ 60= 20
Bec its getting closer?

[lean otter]

sure depending on the direction, but its gonna be 40

[hearty dagger]

Ok ty

.close

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[west bough]

$\int_1^\infty{\frac{1}{x}}dx\leq\sum^\infty_{n=1}{\frac{1}{n}}$

[flat frigate]

dark

[west bough]

how do you understand this analyticially

there is an analytical proof that i need to understand after getting this

[obtuse jackal]

for all x in [n, n+1[, 1/x <= 1/n
Integrate over [n, n+1[ then sum over all n

[west bough]

bruh

[obtuse jackal]

Works for all decreasing functions btw

[west bough]

that is written word for word in the solution to the question but i only just got it for some reason

tbf the question wrote it in a messy way

thanks

.close

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[west bough]

closed bracket since x can equal n+1.

[obtuse jackal]

It's just to make it a neat disjoint union. In fact, it doesn't matter at all

[safe radish]

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[west bough]

i mean since you wrote 1/x<=1/n. But since the inequality is used for a direct comparison test for convergence it wont matter

[west bough]

or what do you mean with the disjoint union?

[obtuse jackal]

Forget about it

[obtuse jackal]

It doesn't matter because points don't matter to integrals

[west bough]

is there something that explains how you can have open brackets but still use smaller instead of smaller equal

[obtuse jackal]

Wdym

[west bough]

nevermind

.close

[safe radish]

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[safe radish]

@trail anchor Has your question been resolved?

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[fickle abyss]

i know the value of \ $a_1 = 2$\$a_2 = -4$\$a_3 = -1$\$a_4 = 2$

[lean otter]

is a_r defined for r > 4?

[flat frigate]

yomiko

[fickle abyss]

no

[lean otter]

then it can be anyhting

post the full problem

[fickle abyss]

why does r have to be >4?

[still anvil]

This was very relevant

[fickle abyss]

its pretty obvious what i sent is a periodic order of 3

anyways

[lean otter]

no?

[still anvil]

lol

[fickle abyss]

?

[lean otter]

thats just the first 4 values of the sequence

anyways

[plucky elk]

it wasn't

[fickle abyss]

it doesn't matter now anyways, you guys know that now

[plucky elk]

it matters that you realize you keep withholding relevant information from your questions

[fickle abyss]

the answer it gave is this

[plucky elk]

do you know what this means?

[fickle abyss]

yes

its repeats every 3 sequence

im guessing they did 80/3

[plucky elk]

they used a1 = a4 = a7 = ... = a (3 * k + 1), and similarly for a2 = a5 = ..., a3 = a6 = ....

you have to solve for k

[fickle abyss]

I've got k = -2

[plucky elk]

k is a positive integer

[fickle abyss]

no its not

k is a constant

[plucky elk]

a1 = a4 = a7 = ... = a(3 * m + 1), and similarly for a2 = a5 = ..., a3 = a6 = ....
solve for m

[fickle abyss]

wait im not sure how you got from a1=a4=a7 into an equation

[plucky elk]

m is the greatest positive integer that 3m + 1 is still part of your sum

3m + 1 is the last index in the sum where a1 = a(3m+1)

1, 4, 7, 10, ..., 3m + 1

solve for the greatest m such that 3m + 1 <= 80

[fickle abyss]

where did you get 3m+1 from?

[plucky elk]

1,4,7,10, has a pattern

find that pattern

[fickle abyss]

right okay

yh im confused

a1 = 2
a2 = -4
a3 = -1
a4=2

so im finding how many times does 2 repeat in 80 terms

• how much -4 repeats in 80 terms and so onn

[plucky elk]

you're just adding (a1+a2+a3) m-many times

• leftover couple terms since 80 isn't divisible by 3

[fickle abyss]

im gonna watch some guy do the question cuz i kinda get it but not rlly

thanks tho

.close

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[hollow coyote]

quick question

[hollow coyote]

does this look right

7b

[fathom fable]

7b is good

8 isn't quite right, since your series is actually a little less than 1/n for all terms, so a straight comparison test with 1/n tells us nothing

You'd probably want a limit comparison test

[hollow coyote]

why not?

[fathom fable]

So, 1/n diverges

but a comparison test with a divergent series only works if your series is always bigger than the divergent series

Since your series is always a little less than the corresponding terms in 1/n, you can't just do a direct comparison with 1/n

[hollow coyote]

ah

so why does the limit version work?

[fathom fable]

Intuitively, the limit comparison works here because while your series is always a little less than the terms of 1/n, it's not by much - they really are pretty much the same series in terms of how fast they go to zero, just one is shifted a little bit if that makes sense

[hollow coyote]

ah alright, so I take the limit of the series?

[fathom fable]

You take the limit of one series's terms divided by the other series's terms

if it comes out to be a finite constant, then they either both diverge or both converge

[hollow coyote]

can the other series be k^2 / k^3?

[fathom fable]

Absolutely

That's what I'd recommend even

[hollow coyote]

alright, great

but the limit test doesn't tell me much right

[fathom fable]

Since the ratio of the terms of one series to the other's is constant, then each series is basically a multiple of the other as we get really far into the sequence

So if 1/n diverges, and you get a finite constant, whatever series you also compared it to will diverge

Which limit test

[hollow coyote]

limit comparison

[fathom fable]

Limit comparison can tell you a lot

[hollow coyote]

so I do lim of (a_k / b_k) right

but u said

it's either both diverge or converge

how do I know exactly?

[fathom fable]

Well, does k^2/k^3 diverge or converge?

[hollow coyote]

it diverges?

wait no

[fathom fable]

Right

[hollow coyote]

huh?

how

[fathom fable]

The sum of k^2/k^3 = 1/k diverges

[vast cloud]

i could be makib it up but might you be able to compare instead to like. 1/2k instead of 1/k to solve this

[fathom fable]

I think so, but you'd need to show that 1/2k is less than the series we're evaluating for all the terms - or not "all" the terms, but for some infinite tail sequence

I think limit comparison is a little more direct

Basically, what the limit comparison test allows you to do is take a series you don't know if it converges or diverges, and compare it to one you do know, like 1/k

and if you get a finite nonzero constant, you know they behave the same

[hollow coyote]

how do I know k^2 / k^3 diverges?

ahhh

that makes much more sense

[fathom fable]

so, your unknown sequence behaves the same as your known sequence

Right, there we go

and that's a consequence of the fact you're basically showing one sequence ends up being just a multiple of the other

[hollow coyote]

ohhhh

alright

so if this sequence / (k^2 / k^3) = a number, then it's divergent?

@fathom fable

[fathom fable]

Yup

Make sure you take a limit though

[hollow coyote]

got it

thanks

[fathom fable]

Any time

[hollow coyote]

i evaluated the limit to be 1

so it diverges

sweet

how does k^2 / k^3 diverge though

[fathom fable]

k^2 / k^3 diverges because when you simplify it, it becomes 1/k

which diverges

[hollow coyote]

OH

i am dumb

thanks!!

.close

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[ornate jungle]

can i get an explanation of where i went wrong ?

[dull sequoia]

f(x) ≠ x³/3

What is f(x)?

What’s the function you’re integrating

[ornate jungle]

-x^2 + 4 ?

[dull sequoia]

So then what’s the integral of that

[ornate jungle]

-x³/3 + 4x

?

[dull sequoia]

Yes

[ornate jungle]

oh was i only missing the 4x ?

[dull sequoia]

Now do your F(b) - F(a) again

[ornate jungle]

ohhhh okok

[dull sequoia]

Yes

[ornate jungle]

another question

how would i find total area

on same interval

[dull sequoia]

Your teacher is right you should write out what is f(x)

[ornate jungle]

okok

[dull sequoia]

Counting under the x axis as well?

[ornate jungle]

uh the question on the paper was "Compute the total area under y = f(x) on the interval [0, 3]

[dull sequoia]

It says net on here

[ornate jungle]

oh i meant theres a part b to this question

and it says to compute the total

[dull sequoia]

Ok show part b then

[dull sequoia]

That’s a weird question

[ornate jungle]

😭

[dull sequoia]

I’d say it diverges to inf? I’m not too sure

[ornate jungle]

ahh ok

thanks for the help

[dull sequoia]

Is that something you’ve seen before?

[ornate jungle]

ummm

[dull sequoia]

It’s weird to make the distinction between net area and total area in this sense

Since if you integrate anything without bound it nearly always diverges

[ornate jungle]

idk what that means ngl

just a guess but would i do F(b) + F(a)

?

[dull sequoia]

No!??

Why would you add them

[ornate jungle]

idk i just saw total and thought maybe add? just a random guess

[dull sequoia]

Guesses still have to be reasonable, adding them together makes no sense

[ornate jungle]

o

[dull sequoia]

Ok

I figured it out with the help of some people

Your diagram is bad

The parabola does not intersect the x axis at (0,3)

[ornate jungle]

gotcha

[dull sequoia]

Net area takes area under the x axis to be negative

total area separates them and adds them together

So you need to separate your integral, 1 for the area above, 1 for the area below then add them

[solar pike]

True, for total area you need to integrate the absolute value of the function. In the case of this function, you only need to split it into two sub-intervals.

[ornate jungle]

do i change anything for the net area i calculated earlier?

[dull sequoia]

Yes, read what I said

Also draw a better diagram

[ornate jungle]

okay

x intercept at 2 ?

[solar pike]

f(2) = 0

For x > 2, f(x) is negative

In [0,2) f(x) is positive

[ornate jungle]

i think i did something wrong ?

[dull sequoia]

Why have you done from 0 to 2 then from -2 to 0

Where did -2 to 0 come from

[safe radish]

@ornate jungle Has your question been resolved?

[ornate jungle]

what

oh wrong pic

[ornate jungle]

.

[dull sequoia]

So why have you chosen from 0 to 3

[dull sequoia]

Did you read these messages

[solar pike]

He's calculating the net area

[dull sequoia]

Oh that’s for part a

[solar pike]

Yes

I think it's correct unless I'm missing something

[ornate jungle]

that correct for net right ?

[solar pike]

Yes

[ornate jungle]

total area is the integral from 0 to 2 then 2 to 3?

[solar pike]

In [0,2] you will get a positive area and in [2,3] you will get a negative area

You need to add both absolute values (with a positive sign) together

To get the total area

in other words, you should find $\integral_{0}^{3} |f(x)| dx $

[ornate jungle]

is the answer 19 ?

WAIt

[solar pike]

Tell me what did you get on each interval

[ornate jungle]

not that

[solar pike]

Integral_0^2 f(x) dx=

Integral_2^3 -f(x) dx = ?

[ornate jungle]

i got 3

interval 0 to 2 was 8/3 + 8

interval 2 to 3 was -5 - 8/3

[solar pike]

You are missing a minus

It's -x^3/3 + 4x

So -8/3 + 8

for [0,2]

[ornate jungle]

ahhhh

so -16/3 + 3 ?

[solar pike]

Nope

The second interval is also false

[ornate jungle]

oh

[solar pike]

Note that it must be POSITIVE

F(3) =3 , you already calculated it in (a) right?

F(2) = -8/3+8, we just did it

F(3) - F(2) = 3 +8/3 - 8 = -5 + 8/3 = -7/3

This is negative area, because the curve is under the x axis

But we need to take the absolute value

[ornate jungle]

total area is 7/3

[solar pike]

Nope

7/3 is only for [2,3]

You need to add [0,2]

[ornate jungle]

7/3 + 3

[solar pike]

Nope

We said that for [0,2] it's 16/3

[ornate jungle]

OH YEA

[solar pike]

F(2) = -8/3 + 8 = 16/3

[ornate jungle]

add 16/3 + 7/3

[solar pike]

= 23/3

[ornate jungle]

😭

thank you guys so much

[solar pike]

You are welcome

[ornate jungle]

❤️

.close

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[tawny mirage]

the first one was p straightforward, but now what? lol

[tawny mirage]

i think it has something to do with logs but im unsure..

[stray socket]

All of these have to do with logs

[tawny mirage]

i did use log on the first

[stray socket]

First of all, 125 = 5^3 and 25 = 5^2 so rewrite everything with base 5 exponents

[tawny mirage]

okay i see u

[quartz edge]

can anyone help, i got 4/15 when i added but and i dont know how to simplify it further

[tawny mirage]

so 5^3x and 5^3x-21

no wait..

5^3x and 5^2x-14

there

aaand since they have the same base is it okay to trash the bases and solve for 3x=2x-14 ?

or no? lol

[tacit venture]

yes

[tawny mirage]

sick!

thank ya

alright yea i got this, thank y'all

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[maiden sphinx]

Question regarding ring theory:
let Rad(0) be the radical of 0, is the 0 still in the set Rad(0) or not?
With the definition of Rad being $$Rad(I) = {x \in R | \exists n \in \N: x^n \I}$$

[flat frigate]

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

[maiden sphinx]

because R/Rad(0) is a reduced Ring but via definition of Rad(0) 0 shouldnt be a part anymore

[drowsy moss]

why not?

0^n = 0 is in I?

[maiden sphinx]

yeah but if Rad(0) contains 0 how come R/Rad(0) is a reduced Ring

meaning it contains 0 itself

but no other nilpotent element

[drowsy moss]

Because R/I is still a ring, so it has a 0 element

[maiden sphinx]

elaborate further

cause by def. I contains the 0 element or not?

[drowsy moss]

R/I is a factor ring. It consists of the elements r+I
Basically, it's the ring version of factor groups.

the additive identity in R/I is 0+I.

[maiden sphinx]

we havent had factor rings yet

but R/I means all elements of R who are not in I or am I wrong here

[drowsy moss]

you're wrong

do you know what factor groups are?

[maiden sphinx]

no

[drowsy moss]

What is your definition of R/I then?

[maiden sphinx]

well the script we use is shit

its just {a+I | a in R}

[light shoal]

that's the set difference of R and I, usually written R \ I (note backslash, not forward slash)

[maiden sphinx]

uff you right on that as well

thanks guys

.cloe

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[safe radish]

@dusty plank Has your question been resolved?

[safe radish]

@dusty plank Has your question been resolved?

[safe radish]

@dusty plank Has your question been resolved?

[safe radish]

@dusty plank Has your question been resolved?

[safe radish]

@dusty plank Has your question been resolved?

[safe radish]

@dusty plank Has your question been resolved?

[novel breach]

Hallo

Ich hab eine Frage

[lean otter]

The answer is actually 1/2

[quartz edge]

Ty, I got it!

[oak jetty]

kid definitely under 13 if that the kind of problem they doing

[safe radish]

@dusty plank Has your question been resolved?

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[fresh birch]

I’m confused on #2 bc the answer key says the area’s 400(pi)-300sqrt3, but is it 300 and not 400?

Plz help

[merry sleet]

did you find the radius?

[safe radish]

@fresh birch Has your question been resolved?

[fresh birch]

Yes

I’ve already found everything

[merry sleet]

ok so show what you did

[fresh birch]

Ok

Hold on

Sorry w8

It’s this one ^^^

So I already know that the perimeter’s right

But the second boxed answer, which is the area, is not

Bc on the answer key it says that it’s … -300sqrt3

Not 200

Which is why I’m confused

@merry sleet u there?

[merry sleet]

why do u have 40 pi?

[fresh birch]

That one’s right

The perimeter’s correct

It’s just the area that’s apparently wrong

[merry sleet]

ok so you have a problem with the area of the triangle right?

it should be height = 30

base/2 = sqrt(300)=10 sqrt(3)

and then the area of the triangle is the product so 300sqrt(3)

maybe you took the height equal to 20

but its 30?

[fresh birch]

How’d you find the height? @merry sleet

Don’t u just do 10 * 2?

[merry sleet]

you alreayd found it

here the 20 is what lacks on top of the 10

by symmetry of the triangle

its the same on top

[fresh birch]

Nvm, I got it

Makes sense, thanks

Also

One other question

It’s question #2 here

[merry sleet]

1?

find the inner radius and outer radius

using trig and pythagoras

[fresh birch]

So I don’t have an answer key for any of these, but here’s what I did

No, #2

Here’s what I did @merry sleet

#2 here

[merry sleet]

i dont have the same for the third side

of the right angle triangle

,w sqrt(812 - (812/4))

[flat frigate]

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[merry sleet]

ok my bad

then its just

,w 6 * (9sqrt(2) * 9 sqrt(6)/2 /2 )

[flat frigate]

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[merry sleet]

is it what you got?

[fresh birch]

I believe I just got 243

I’m not sure where u got the sqrt3 from

[merry sleet]

idk i just found the area of the small triangle

and multiplied by 6

because there are 6 of them

[fresh birch]

Right so

[merry sleet]

sqrt(6) doesnt disappear

[fresh birch]

Then how come I didn’t get that

[merry sleet]

sqrt(2)*sqrt(6) = 2sqrt(3)

[fresh birch]

Hold on, lemme try again

Ok so

What did u get for the length of one side length? @merry sleet

[merry sleet]

i got the same as you for the length

9 sqrt(6)

the height is just then 9sqrt(2) + 9 sqrt(2) /2

[fresh birch]

Wait so isn’t the area

(486 swrt3)/4 @merry sleet

[merry sleet]

which is?

,w 486/4

[flat frigate]

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[merry sleet]

almost

[fresh birch]

But I didn’t say 486/4

[merry sleet]

i know

[fresh birch]

I said 486 sqrt3/4

[merry sleet]

but the answer is 243*sqrt(3)

which is twice as much as yours

[fresh birch]

How tho

The formula for an equilateral triangle’s s^2 sqrt3/4

[flat frigate]

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[merry sleet]

f

oh yeah its good

i made a mistake earlier

[fresh birch]

Yk how the formula’s (s^2 sqrt3)/4?

[merry sleet]

idk

[fresh birch]

What’s good?

[merry sleet]

i never use it

[fresh birch]

Is my answer correct?

[merry sleet]

your answer

yes

[fresh birch]

Got it, thank you

[merry sleet]

,w 6 *( (9sqrt(2)/2 * 9 sqrt(6)/2)/2)

this is my way

[fresh birch]

But I thought u said my answer was right

[flat frigate]

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[merry sleet]

its the same answer

[fresh birch]

Oh, right

[merry sleet]

486/2 =243

[fresh birch]

I just hadn’t simplified, yh

Alr thx sm

Do u have time to help me w/ some more?

[merry sleet]

why not

[fresh birch]

Thx sm

So

#3 here

This was my process

Is my answer correct?

[merry sleet]

,w 3600 sin(5°)cos(5°)

[flat frigate]

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[merry sleet]

i think my answer is good

because its close to a circle

of radisu 10

(a 32 gon is almost a circle)

36 gon*

you got x right

but y im not sure

i would do cos(5 ) = y/10

then the area of the triangle is 100 cos(5)sin(5)/2

then double it to get one of the 36 slices

then *36

[fresh birch]

R u sure?

@merry sleet

[merry sleet]

22 000 is too much that i'm sure

because this 36 gon is inside a circle with radius 10

and the area is 100 * pi

[fresh birch]

Lemme try again

[merry sleet]

of the circle

[fresh birch]

One sec

I got the same answer again

@merry sleet

[merry sleet]

where does the 62 come from?

x isnt 62

x must be smaller than 10

since its the hypothenuse

10 is the hypothenuse i mean

[fresh birch]

Ik but

I’m finding the perimeter

To plug into the formula 1/2Pa

[merry sleet]

1/2Pa?

what is 62?

[fresh birch]

It’s the perimeter

It’s not exactly 62 tho

It’s somewhere around there

[merry sleet]

the perimeter of what? the triangle?

[fresh birch]

The entire 62-gon

[merry sleet]

its a 36 gon

oh ok

so its x*72?

[fresh birch]

Yeah

[merry sleet]

but why do you need that?

you have a formula for the area of a 36 gon?

from the perimeter?

[fresh birch]

Just for any polygon

Yeah

[merry sleet]

ok which is?

[fresh birch]

1/2Pa

[merry sleet]

what is a?

i assume P is perimeter

[fresh birch]

The apothem

Yes

[merry sleet]

which is?

[fresh birch]

I believe abt 717

[merry sleet]

no but the definition

[fresh birch]

Oh

[merry sleet]

idk what that is

[fresh birch]

Just a line that forms a right angle by connecting to any one of the sides of a regular polygon

[merry sleet]

so y?

[fresh birch]

Yeah

[merry sleet]

y =717?

[fresh birch]

Abt, yeah

[merry sleet]

in a right triangle where 10 is the hypothenuse?

[fresh birch]

Yeah

[merry sleet]

you know the hjypothenuse is the longest side?

by definition

[fresh birch]

Yeah

Oh

Right

Lol

[merry sleet]

lol indeed

[fresh birch]

So then

How’d u go about the problem?

[merry sleet]

just correct your y

and you should be good

i'm not going to force a method on you

yours seems good

i think the mistake is that when calculating y you took

the perimeter

instead of x

as the third angle

third side*

[fresh birch]

Ohh

Maybe

Lemme try again

One sec

[fresh birch]

Wait

I’m so confused

I got abt 24.4 this time

That can’t be right

I did (62…)(0.8…)/2

@merry sleet

[merry sleet]

0.8 is y?

[fresh birch]

Yeah, isn’t it?

[merry sleet]

no

y should be around 9 something

its almost as big as 10

[fresh birch]

I did tan5=0.8/y

How tho

[merry sleet]

,w 0.8/tan(5°)

[flat frigate]

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Requested by Benjamin#3570

[merry sleet]

yeah

abouyt 9

thats it

so whats the problemù

you forgot to divide by tan(5)