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[rapid cosmos]

That makes sense

thanks

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[marsh citrus]

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[solar fulcrum]

how do u find cosx = -1/root2, [-pi, 0]

[solar fulcrum]

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[flint reef]

A sin function with a period of pi/2 is represented as sin4x

[flint reef]

Why is this the case? What is the method of craeting the correct function to fit the period

[late geode]

period of sin(kx) is 2pi/k

[flint reef]

I wondering the method for writing a function with certain periods

How do you determine the vert stretch/comp

For example, a cos function has a period of 3pi/2. Write an equation for it

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[haughty sandal]

can someone pls help me with this?

[haughty sandal]

I got part a but idk where to start for b

[fathom ridge]

if y is decreasing at x = 4 it implies that the derivative at 4 is negative

see if you can use that

[haughty sandal]

can I equate the derivative with 0 and then modify the eq so that it is less than 0??

[cunning jackal]

Just let the derivative <0 and solve the inequality, why equate first

[haughty sandal]

oh okay

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[vocal cloud]

we have to show that g(x) = c *f(x)

so this is what I tried

lets say f(x) = g(x)/f(x)

f'(x) = (g'f - f'g)/f^2

g'f = gf as g' = g

[past frigate]

Does it say f'(x)= f(x) as well? It's cut off for me for some reason

[vocal cloud]

yesss

[past frigate]

Alr alr

[vocal cloud]

and f'g = fg

so f'(x) = 0

hence f(x) = constant

is this right

[past frigate]

Mmm

If f(x) is constant

Then f'(x) is not equal to f(x)

Since f(x) is not 0

You could divide both sides by f(x) and integrate

[crystal lintel]

[past frigate]

Wrt dx

[vocal cloud]

we are not allowed to do that

[past frigate]

Oh dear

[vocal cloud]

if i somehow prove that f'(x) = 0
then f(x) = constnat so
c = g(x)/f(x) this g(x) = cf(x)

[past frigate]

No but then f(x) has to be equal to f'(x)
So it cannot be a constant

[crystal lintel]

you are given f' is not 0

[vocal cloud]

oh shit i actually confused the variable

this is what i actually meant

m(x) = f/g
m' = (f'g - g'f)/g^2

[hushed egret]

😦

[vocal cloud]

lol

idk why that happened

[past frigate]

Lol

[vocal cloud]

so then
f'g = fg and g'f = fg so m' =0 hence m = c

c = f/g

f = gc

[crystal lintel]

the memers have arrived

[past frigate]

Nice

[marsh citrus]

@vocal cloud Has your question been resolved?

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[marsh citrus]

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[vague latch]

can anyone help me with the method of this question

[sullen bronze]

You would have to divide meters per year by seconds per year in order to get meters per second.

That would be (9.45 * 10^15) / (3.15 * 10^7).

From there, you can simply divide the 10's (which would be subtracting its exponents) to get 10^8 in the numerator. Once again, divide 9.45 / 3.15 to get 3.

[vague latch]

tsym bro ur a lifesaver

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[sturdy lagoon]

what am i to do if i get a decimal...

[still temple]

hey guys I'm having some problems in clash royale arena 2... one of my cards costs 100 and another one costs 100 and I can't figure how much would cost if I would upgrade both... please help me

[sturdy lagoon]

200

?

[still temple]

okay

thanks :}

and quick question does that 200 correspond to any factorial... like you have factorial out of 5 which is the closest to 200 only of by 80 but can you have a factorial out of smth to make some random ass numbers?

ik it's a dumb question

but I'm really curious

[marsh citrus]

@sturdy lagoon Has your question been resolved?

[timber hull]

literally what are you on about

@full rover this is a crucial and truly thought provoking question you may all need to band together to answer this one

[sturdy lagoon]

[timber hull]

your entire method is wrong

first factor the 4 out

in the 2nd line

i mean 3rd line

[sturdy lagoon]

alr i was jus doin what they showed

i'll try

[timber hull]

and divide by 4 on both sides

have you dealt with equations of a circle?

[sturdy lagoon]

yea

[timber hull]

ok

complete the square

on x^2-4x

and y^2-+7y

[sturdy lagoon]

alright

x(x^2-4x) and y(y^2-+7y)

[timber hull]

??

like how have you just multiplied by x and y

[sturdy lagoon]

🐶 💔

idk man

[timber hull]

do you know how to complete the square

[sturdy lagoon]

nah i forgot

[timber hull]

unlucky 😂

[sturdy lagoon]

anyway they wanted it done like this

I had the answer alr i jus wanna do it this way

[timber hull]

yeah if you read it legit says complete the squares on x and y and you cant do that so gl

[sturdy lagoon]

ok i'll be fine thanks

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[sterile rock]

What have i done wrong?

[sterile rock]

trying to find this

i think ive misunderstood one of the rules...

[worthy notch]

I can't tell what you were doing after this part, but the part I circled is the correct answer

[sterile rock]

i tried to just factor out (2x-3)^6(x)(4-x)^4

what did i do wrong?

nvm

got it

[jovial lance]

remember to .close if youre done with help

[sterile rock]

nvm

ok

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[slim moon]

How do I use the characteristic polynomial function from Linear Algebra to find the eigenvalues (and then the eigenvectors) of a non-diagonal matrix?

I have the function in the bottom right of my page

and the matrix [T]b near the top left

I tried doing some elementary row operations to turn it into a diagonal matrix without modifying the determinant, but I don't think I can use this since the eigenvalues for that one are not the same as my original matrix

I am trying to find the eigenvectors to then compare them to the bases from my linear transformation (which I have already computed into [T]b)

this is the problem from my textbook

[slim moon]

main question

[marsh citrus]

@slim moon Has your question been resolved?

[weak crow]

You don’t need to have a diagonal matrix to compute the determinant of it

So u can still compute the characteristic polynomial even though it’s not diagonal because all you really need to do is compute the determinant

[marsh citrus]

@slim moon Has your question been resolved?

[slim moon]

and then I can just get the eigenvalues and use them to get my eigenvectors, and then compare with the set b?

[weak crow]

Yea

It’s pre much the exact same process whether it’s diagonal or not

[slim moon]

i get the feeling they wont match the bases then

(not that they have to)

Ok, I'll try to figure smth out

thanks

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[still temple]

how do i solve this question?

.close

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[honest cedar]

can someone tell me if im putting in the critical values right or if i should just put the x value?

[honest cedar]

also for c should i be using U or ,

[hollow crescent]

looks fine to me

[honest cedar]

okay i wasn't sure cause my hw is really touchy with U or , and when i think its one it switches up on me and wants the opposite

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[dawn socket]

y=2f(x)

[dawn socket]

whats transforming or changing

the y or the x

[proven jacinth]

y

y = 2 f(x)

read through the equation word by word

y is equal to 2 times f(x)

if y is equal to 2 times f(x) then it is two times as much as f(x)

so if your graph of f(x) has some height y then your new graph is twice a high

as high*

[dawn socket]

i think i got it

ok

so the y is changing

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[still temple]

Determine the exact value of the trig ratio

Cos15

[still temple]

I'm getting this wrong

my steps are turning it into cos(30-15)

which translates into (pi/6-pi/12)

which translates into

$\sin\frac{\pi}{6}\cos\frac{-\pi}{12}+\sin\frac{-\pi}{12}\cos\frac{\pi}{6}$

[elfin berry]

mj

[still temple]

which gives me root6-root2/4, which is wrong

not sure what I'm doing wrong

[near oar]

i think $\cos(\alpha - \beta) = cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)$

[elfin berry]

Macacofonico

[lone heart]

This isn't the right identity to use

[still temple]

I remember it being opposite sign

oh wait it is

[still temple]

I have cos(30-15), which would mean it's a +

in middle

[lone heart]

Yes but you still used the wrong identity

[still temple]

how?

[lone heart]

You have sincos + sincos

[still temple]

oh your right

give me a minute

[still temple]

I get sqrt2/2

[lone heart]

Show your work

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

sorry about that

I changed up my thing so that I used special angles

aka 45-30

instead of 30-15

I still seem to not get it though, maybe I'm putting it into my calc wrong

$\cos\frac{\pi}{4}\cos\frac{-\pi}{6}+\sin\frac{\pi}{4}\sin\frac{-\pi}{6}$

[elfin berry]

mj

[still temple]

^^ is what I'm putting in

$\frac{\sqroot{6}-\sqroot{2}}{4}$

[elfin berry]

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

[lone heart]

Why do you have -pi/6?

[still temple]

because my thingi s

cos(45-30)

so x=45, y=-30

in cosxcosy+sinxsiny

or is it not supposed to be -ve

[lone heart]

The negative doesn't follow

Theta = 45

[still temple]

righttt

[lone heart]

Phi = 30

[still temple]

that would probably give me the answer I'm looking for then

and it does

thanks for your help

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[marsh citrus]

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[still temple]

@marsh citrus

[marsh citrus]

Please don't occupy multiple help channels.

[main idol]

.close

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[azure bone]

Are there more integers than natural numbers

[native frigate]

no

there is a 1 to 1 correlation, and therefore they are the same cardinality @azure bone

[azure bone]

but likee

integers go off into the negatives fr

[native frigate]

they sure do

[azure bone]

and naturals are just positive while integers is liek both

Fr

[native frigate]

natural numbers is a strict subset of integers, but they are the same cardinality

[azure bone]

why

[native frigate]

there is a 1 to 1 mapping

0 -> 0
1 -> 1
2 -> -1
3 -> 2
...

how to map natural numbers to integers 1 to 1

each natural number clearly maps to a unique integer, and all integers are clearly expressed

@azure bone

[azure bone]

idk wat that menas

Means

[native frigate]

https://www.youtube.com/watch?v=OxGsU8oIWjY

this video gives pretty good intuition

[azure bone]

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[empty sierra]

What did i do wrong

[hollow sparrow]

your antiderivative is incorrect.

integral of 1288t = (1288/2)t^2 = 644t^2
integral of -10576 = -10576t

[empty sierra]

So like that

Wait

Im confused

[hollow sparrow]

well then you need to adjust your evaluation

also you are multiplying instead of subtracting in you integral

$\int_{12}^{17} (1288t - 10576)dt \\= [644t^2 - 10576t]_{12}^{17} \\= [644(17)^2 - 10576(17)] - [644(12)^2 - 10576(12)] \\= [186116 - 179792] - [92736 - 126912] \\= 6324 - (-34176) \\= 6324 + 34176 \\ = 40500$

[empty sierra]

Ohhh

Wait im dumb as hell

Thanks

[hollow sparrow]

nah you're good

well actually

somethin in my math is wrong lol

the answer should be 40500

but same procedure

[elfin berry]

MellowDramaLlama

[hollow sparrow]

there we go lol

[empty sierra]

Yea that looks right i copied the procedure and got 40,500 and im like wait what did i do wrong😭

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[mellow prawn]

Sample question that im using to help me figure it out

[final skiff]

This is suppose to be in the form e^(at) where a is a number

[final skiff]

This looks good

Find what ln(1/2)/1620 is and that will be your a

To 5dp

[mellow prawn]

Ohh whoops

[mellow prawn]

When I try the sample question I still get a different answer than they get

[final skiff]

What r u doing and getting

[mellow prawn]

1.8626 😀

Wait no

I did it again

Waaaiittt

Yayyy

I’m good now

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[lost stratus]

I would like my proof checked 🙂

[brave spire]

Eh

Its correct

But i would also show that b^2/4 can't be of the form ak-1/2

Where a is an integer

Not divisible by 4

[lost stratus]

Ak?

[brave spire]

a*k I mean

[marsh citrus]

@lost stratus Has your question been resolved?

[marsh citrus]

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[frosty hatch]

,rotate

[marsh citrus]

@sage schooner Has your question been resolved?

[marsh citrus]

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[jovial cobalt]

I want to ask about 2 b)

[jovial cobalt]

it it just referring to the integer half of the bottom

because then could we place 2/2 and 3/2 in the same category

i dont know im sorta horrible at quotients on sets so my understanding my be incorrect

[copper raven]

yeah they screwed up their example

floor(2/2) = floor(3/2) you're right

if you replace floor with ceiling it works though

@jovial cobalt

[jovial cobalt]

wait so floor isnt good here?

[quaint elm]

floor rounds down to the nearest integer

so like floor(3.14159) = 3

[jovial cobalt]

yeah so it should be ceiling

[quaint elm]

yea

[jovial cobalt]

okay thanks sm @quaint elm @copper raven

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[simple quiver]

Need help on 3

[simple quiver]

How do you get x=30

Do you do 90/3 or sum?

<@&286206848099549185>

[violet glacier]

set the time ran as t

the distance ran would be 25t

and the distance between the second base and the player would be sqrt((90-25t)²+90²)

you can get t with the two thirds info

and you take the derivative and put in the value

[simple quiver]

?

I’m confused

Could you explain the problem from the begining

[marsh citrus]

@simple quiver Has your question been resolved?

[marsh citrus]

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[broken rover]

help needed

[broken rover]

progress made since,but need confirmation for 2nd sequence answers

<@&286206848099549185>

[swift sail]

hi

[broken rover]

hi

[swift sail]

what u need help with

[broken rover]

2nd sequence

[swift sail]

whats the 2nd sequnce

[broken rover]

i mean sec 2 sequence 1

for 5th would it be 33

and 6th 46

@swift sail

[swift sail]

i cant see the photo proply

[broken rover]

section b

first sequence

???

[marsh citrus]

@broken rover Has your question been resolved?

[marsh citrus]

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[sly notch]

Can someone help me with this topology exercise.

It asks to determine the compact subsets of the Kolmogorov line.

That is, over the reals, the topology where open sets are null, the whole, and every (a,infinity) with a in the reals.

[sly notch]

I know about the finite covering but I got kinda stuck

[crystal lintel]

take an open cover of a subset of R. can you find a finite sub over?

think about what must be in the cover

how about you try some different examples of subsets of R

like (0,1), (-inf, 100), (0, inf)

[odd juniper]

Help

Somebody help

@everyone

<@&286206848099549185>

[crystal lintel]

[worthy basin]

You can start by multiplying the whole number "1" by the denominator of the fraction 9

1 x 9 = 9

then add the result to the numerator

9+7

= 16

So 1 7/9 as an improper fraction is 16/9

You can also turn the whole number into a fraction

1 = 1/1

and then find the common denominator

you do that by prolonging the fraction by 9

1/1 prolonged by 9

9/9

and 9/9 + 7/9

[crystal lintel]

not the right channel for this

[worthy basin]

wdym?

[crystal lintel]

this is piqueiras’ channel

[worthy basin]

ohh these have diferent themed names!

[marsh citrus]

@sly notch Has your question been resolved?

[sly notch]

lmao I had to go for a sec

I for example proved every finite set would be compact

I am guessing compacts of usual topology should as well be

(a,inf) should NOT be compact for the same reason as in usual as well

finite join of compacts is a compact right

[crystal lintel]

why?

can you write or describe a cover of (0, inf)

just one

[sly notch]

(1/n, inf)

[crystal lintel]

does that include 0?

[sly notch]

no

[crystal lintel]

oh

[sly notch]

and (0,inf) neither

xD

[crystal lintel]

sorry i had [0,inf) in mind 🤦🏻‍♀️

[sly notch]

yea

[crystal lintel]

so

[sly notch]

any covering of [0,inf) should include some (0-d,inf)

thats a covering of a single element

[crystal lintel]

i think an important thing is whether the set contains it’s greatest lower bound or not

[sly notch]

oh

good point

[crystal lintel]

from thinking about (0,inf) and [0,inf)

[sly notch]

yea

cause if you do some other random set

joint of intervals

Just consider lowest

[crystal lintel]

if the set contains it’s glb, it needs something of the form (glb - epsilon, infinity) to cover it

[crystal lintel]

but otherwise you can do something like that

with the left endpoints converging to glb

[sly notch]

exercise is just characterizing compacts

so cool

[crystal lintel]

yea i think that’s the characterization then

[sly notch]

I also have another if its possible

checking if some sets are compact in the

sogenfrey topology

[crystal lintel]

ok

[sly notch]

R, [0,1), (0,1], and

points of a strictly growing succession

[crystal lintel]

ok any ideas so far?

[sly notch]

its weird, Google has way more info on sogenfrey than kolmogorov 💀

[crystal lintel]

lol i never learned those names i had to google them

we called them like

right ray topology

and left hand topology

[sly notch]

my teacher is sometimes kinda obsessed with names

says R cannot be compact as topology is coarser than the usual one

[crystal lintel]

sogenfrey coarser than usual topology?

[sly notch]

XD

idk but reasoning should be the same

[crystal lintel]

you could just give an explicit cover

[-n,n) for integers n

[sly notch]

growing succ. should not be compact either for same reason

[crystal lintel]

clearly no finite subcover

[sly notch]

if u take natural numbers

[crystal lintel]

if it’s unbounded yea

but does it have to be?

[sly notch]

indeed

no

Just growing

[crystal lintel]

what does it ask exactly?

[sly notch]

to check if they are compact

[crystal lintel]

what is the set?

[sly notch]

reals with the sogenfrey

[crystal lintel]

the growing in succession one

[sly notch]

use the naturals?

[crystal lintel]

if there’s no unboundedness req on it and the set is just the image of some increasing sequence then it could be compact or not compact

[sly notch]

hm

are the naturals compact there

[crystal lintel]

what do you think

[sly notch]

no?

no finite covering at all

[crystal lintel]

yes if you e.g. make a cover by covering each natural number with one set, there is no finite subcover

[sly notch]

yea

hm

[crystal lintel]

i’m gonna go take a nap

[marsh citrus]

@sly notch Has your question been resolved?

[marsh citrus]

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[still wyvern]

Ok so

[still wyvern]

Qs asking for support reaction

No

Oops

Vertical reaction

Answer in book says 142 i keep getting 226.18

What am i doing wrong?

Thats the ws

<@&286206848099549185>

Bois i need to shower

[marsh citrus]

@still wyvern Has your question been resolved?

[marsh citrus]

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[still temple]

Suppose that the distribution function of a discrete random variable $Y$ is given by [
\m{F_Y}y =\env{dcases*}{
0 &if $y < \ff18$ \ 0.4 &if $\ff18 \le y < \ff14$ \ 0.7 &if $\ff14 \le y < \ff38$ \ 0.9 &if $\ff38 \le y <\ff58$ \ 1 &if $\ff58 \le y$
}
]
Calculate the third central moment of $X = \m g Y = Y^2$.

[still temple]

Just to verify, this is just [
\sum_{i=1}^4 \p{X_i - \bm\E X}^3
] right

[still temple]

.close

[marsh citrus]

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[marsh citrus]

@still temple Has your question been resolved?

[still temple]

<@&286206848099549185> anyone online who can help me?

[sacred kestrel]

augmented matrix reduction of [ B | B' ], when you reduce B to I, you get the change of base on the right.

same goes for Q, reducing [ B' | B ]

[still temple]

Sounds simple

my thought proces was:
B = P*B'

Then find the inverse of B'

And say B x B' inverse

[sacred kestrel]

the notation for the matrix P being a change of basis would look like $[P]_{B\leftarrow B'}$

[elfin berry]

b0ngl0rd

[sacred kestrel]

the way i like to remember to set up the augmented matrix is to match the subtext. we find P by reducing [ B | B' ]

[still temple]

yes, but to reduce it

or rather what does the comma inbetween mean

[sacred kestrel]

and actually you only need to reduce once, i remember now the inverse of the change of basis is the opposite change

its defining B and B' by a set of vectors

you can then treat them like matrices using vectors as columns, or column vectors.

so B =

[ 1 2 ]
[ 2 1 ]

[still temple]

so i can just use normal RREF on this?

[sacred kestrel]

RREF

[still temple]

can i set it up like this?

[sacred kestrel]

no

1 2  not 1 2
2 1      1 2

[still temple]

like this then

[sacred kestrel]

i suppose you can write it like that, but just remember any row operation done on the left must be done on the right

reduce the left matrix to the identity and youll have the change of basis on the right

then when you invert the resulting change of basis, you get the reverse change

so find P, then P^-1 = Q

[still temple]

so like this:

[sacred kestrel]

yes

that is P

now find the inverse for Q

1/detP * adj P

[still temple]

can i jsut reduce this now

[sacred kestrel]

you could but why do that when you can invert P

[still temple]

alright, ill do that

[still temple]

do you perhaps recall the name of that rule?

[sacred kestrel]

its the formula for the invert 2x2 matrix, otherwise you'll have to do the long way.

[still temple]

i knew that one

[sacred kestrel]

looks good to me

[still temple]

or i hope so at least since that iw what i've done here

[sacred kestrel]

that should be Q now

so we found P and Q.

you can check your work by computing PQ and seeing if = I

[still temple]

sounds right to me

or i suppose logically

since the books ask you to show that Q is the inverse of P

Can i tag you once again in about 10 min @sacred kestrel? I have another quesiton, but i've a short errand to run

[sacred kestrel]

just open a channel. there's quite a few people here knowledgable in linear algebra

[still temple]

yeah, but then i can skip the wait time

[sacred kestrel]

if im here ill probably help. dont sweat it

[still temple]

alright

So

find the base for the kernels and image of linear maps dedfined in terms of multiplication by following matrices

[6,9, 4,6]

this would be my answer

[sacred kestrel]

your answer for the kernel is off

[still temple]

oh

[sacred kestrel]

reduce the matrix

less confusion

now show me the solution for Ax = 0

from this we can find a vector

we need to define an x1 and x2 for our parametric vector solution

try that, i cant do it all

[still temple]

oh

i've done the same

one sec

[sacred kestrel]

you did, but got the wrong answer. im hoping that seeing it in my writing will help you see the mistake

[still temple]

i still get the same x_2 tho

xd

[sacred kestrel]

what is x2

[still temple]

x2

[sacred kestrel]

the reason why we use matrix reduction here is because we can see where the pivots are

x1 is our pivot, not x2 like you worked it out

[still temple]

now i just insert

right

[sacred kestrel]

x2 must be our free var because thats the column where we dont have a pivot

this means x2 = x2

[still temple]

yes

that is what i mean by this

here

[sacred kestrel]

you see how i found a vector though?

[still temple]

i did not do that

the vector part

[sacred kestrel]

the real solution to the kernel is v(-3/2, 1)

the kernel is a line along (-3/2, 1)

i found it by factoring the resulting vector of x

x = (x1, x2), so x is a vector. by finding the dependence relation we find the vector(s) of the kernel space.

[still temple]

so this is the solution

[sacred kestrel]

yes, x2 can actually be labelled anything, but i would accept that as correct if it were me teaching

your answer for the column space is correct. we have a pivot only in the first column so the image is the span of the first column

[still temple]

is there also a constant that can satisfsfy an amount of numbers

here it was 3/2 that would make it correct

so it was clear to anyone

what if it was [812, 190, 8, 10001]

[sacred kestrel]

?

[sacred kestrel]

idk what you mean

[still temple]

in the span quesiton

The first column is a scalar multiple of the second column

This time it was 3/2

with the matrix A = [6,9,4,6]:

So i am asking would that also be the case for [812, 190, 8, 10001], where it here isnt a trivial

[sacred kestrel]

you'll notice when columns are multiples of eachother in the reduction process

as long as you follow the procedure the same, it doesn't matter what entries are in the matrix, if there is a dependence relation it will be revealed through reduction

no need to worry about multiples if you just reduce. you'll find them quickly.

[still temple]

So in this one

they all just equal 0?

but the image image of the matrix is the entire space of R^3, correctg?

[sacred kestrel]

if the work is correct, you have 3 pivot points in your RREF

Dim Col B = Rank B = 3

[still temple]

yes?

[sacred kestrel]

yes lol

[still temple]

yes, but are you on about the image or kernel

[sacred kestrel]

in your last image, you had a fully reduced form that resulted in the identity matrix. n x n matrix with n pivots means the image generates the entire space. that is, Col B = R^n

[still temple]

yes, so the image is the entirety

[still temple]

yes, but here

x1 = 0

x2 = 0

x3 = 0

so x1=x2=x3?

[still temple]

ok, but m sitl not 100% dollwing

[sacred kestrel]

when you have no free variables then Kernel = {0}

no free variables means the null space is only the 0 vector

it also means there is only the trivial solution to Ax = 0

when x = 0

number of free variables = dimension of kernel

[still temple]

So the solution is [0,0,0]

[sacred kestrel]

for the kernel of B, yes its = {0}

[still temple]

and the image is the entrie R^3

[sacred kestrel]

yes

image(B) = R^3

[still temple]

I think the first 3 are correct, idk about the last one tho

[sacred kestrel]

for a linearm ap to be one to one
don't quite understand

[still temple]

linear map

[sacred kestrel]

ah

the first two are the same problem aren't they

[still temple]

one is polynomials

other is R^4

[sacred kestrel]

oh

i actually haven't covered what happens when the mapping is from R^n to P_n, so i can't tell you.

hopefully someone else can help with this one

[still temple]

that is fair

you know anyone

[sacred kestrel]

someone will eventually see the channel

[still temple]

chatgpt says im right

idk how trustable that is

well a linear transformation from R5 -> p^4

would requite that each 5d vecotr in R5 maps uniquely to a polynomial in P^4

sure that isnt possible

Since the dimension is higher

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[paper raptor]

Ok first genuine question tonight, where do I even begin? :)

[remote grove]

hmmm

[cobalt sentinel]

Wait

[paper raptor]

apparently it was $\sqrt{\sum_{n}\left(\mu-x\right)^2P(x)}$

[elfin berry]

PajamaMamaLlama

[cobalt sentinel]

Yea the inside is the variance

Use that formula instead

[paper raptor]

now I gotta redo the problem with different numbers alrighty, thanks for trying Steve, I'll be in the help channels for the next couple hours suffering immensely so feel free to drop by once in a while if you want

.close

[marsh citrus]

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[paper raptor]

is the first box not just 1/5

[paper raptor]

the teacher rolls a fair five sided die

the prob that it's a 3 on the first roll

there's possible cases, one of them has a 3 being rolled

to 1/5? 😭 it's saying that's wrong

then (1-1/5) case they don't get a 3 on the first roll

then a 4/5 chance they get a 2+ on the 2nd roll

that makes it 16/25, no?

on the second roll the prob of getting a number less than 2 (the only case is 1) is 1/5

so that makes the prob just (1-1/5)(1/5)=4/25, no?

1/5+16/25+4/25=1, where am I going wrong?

it wanted decimals...

sigh AP Stats is nothing but inconveniences...

.close

[marsh citrus]

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[marsh citrus]

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[signal apex]

Are the critical numbers of the derivative the min/max of the original function?

[fervent rampart]

"critical points", being values of x where f'(x) = 0 or f'(x) doesn't exist, are the only places where local minima or maxima can occur on a function. however, any given critical point may or may not be a maximum or minimum

[signal apex]

i see

[signal apex]

but it determines those points for the original function right

and not the derviative?

[fervent rampart]

yes. the local minima/maxima of the derivative would be the critical points found by the second derivative, etc.

[signal apex]

ok

thanks a lot

.close

[marsh citrus]

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[marsh citrus]

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[summer gazelle]

The answer provided in the textbook is AB = 0.620 cm but I dont get how we got this answer
Here are my steps:
(7+r)^2=(6+r)^2+5^2
r = 6
sin^-1(5/13) = 22.6 deg
(22.6/360)(2pi(6)) = 2.36 cm

Can anyone correct me?

[marsh citrus]

@summer gazelle Has your question been resolved?

[marsh citrus]

@summer gazelle Has your question been resolved?

[summer gazelle]

anyone?

[marsh citrus]

@summer gazelle Has your question been resolved?

[marsh citrus]

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[marsh lintel]

for a smooth function f and a real number n is restricting f then nth differentiating equivalent to nth differentiating then restricting

[devout mauve]

well depends on how you are restricting it

cause you might restrict it in such a way that the new function you get isnt even differentiable anymore cause the domain doesnt work for that

but if we are assuming you restrict to a nice domain, then yes

[marsh citrus]

@marsh lintel Has your question been resolved?

[marsh citrus]

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[ionic owl]

Convert the polar cordinates to rectangular (-3,4)

[ionic owl]

[r= \sqrt{(-3)^2 + 4^2} = 5]
[tan\theta = -\frac{3}{4}]
[\arctan(\frac{-3}{4}) \approx 36?]

[elfin berry]

dopediscorduser

[ionic owl]

What am I doing wrong?

[dim zodiac]

tan = sin/cos

[ionic owl]

Ah

[tight furnace]

Seems like you're converting rectangular into polar instead

though yes the tangent is also flipped

[ionic owl]

Yeah I really messed this one up

MIsread problem

Thank you

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[azure bone]

If y=2^x , is the inverse log_2(x)=y or log_2(y)=x

[azure bone]

idk

[azure bone]

The first is the inverse and the second one is like the logarithmic form right idk

[marsh citrus]

@azure bone Has your question been resolved?

[azure bone]

uhh i graphed it and it looks like it frfr

[marsh citrus]

@azure bone Has your question been resolved?

[marsh citrus]

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[fair flame]

Simplify √75 + 2√50 + 6√125

[vernal forge]

factorize 75, 50 and 125

[fair flame]

I've done like √(5×5×3)+2√(5×5×2)+6√(5×5×5)

And then idk

[vernal forge]

notice the 5*5

under square root

that's 5^2

[fair flame]

I mean I have to express it in a+b√c form

[vernal forge]

meaning you can take it out

a b c are integers right

[fair flame]

5√3+10√2+30√5

[fair flame]

Yeah

[vernal forge]

hm

[fair flame]

But then idk what to do

Maybe factorize by 5?

5(√3+2√2+6√5)

[vernal forge]

hm

are you sure about the problem?

this doesn't seem possible

show original

[fair flame]

I don't have it

[fair flame]

A friend gave this problem to e

Me

And said solve it

[vernal forge]

well it's not possible

unless you relax the conditions that a b c are integers

[marsh citrus]

@fair flame Has your question been resolved?

[marsh citrus]

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[sullen ice]

How do you make a vector equation for a line with the new 3d Desmos calculator

[sullen ice]

How do you do this

The inequality

[jagged relic]

Have you tried typing the expression?

[sullen ice]

Says t is undefined

[brave marsh]

It appears if you explicitly put a times symbol

At least for me

[jagged relic]

I guess you typed it wrong

[sullen ice]

The equation I posted works btw

I found it on Reddit

Idk how you put the inequality bit under the equation

[jagged relic]

Whenever it can interpret t as a parametric variable, it shows you the range parameters (the inequality)

[sullen ice]

Ohh it’s cuz I put t before it

If you put it after it works

[jagged relic]

Yes, like I said, you typed it wrong

t(...) is interpreted as a function

[sullen ice]

Thank you

.close

[marsh citrus]

Channel closed

Closed by @sullen ice

Use .reopen if this was a mistake.

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[ebon depot]

$\int \frac{x^3+x^2+3x+1}{x^3+x^2+2x+2}$

[elfin berry]

Merineth

[ebon depot]

I'm having trouble solving this

My first problem arrives when trying to perform polynomial division

to simplify it

[sullen ice]

Isn’t bottom $x^2(x+1)+2(x+1)$

[elfin berry]

Joshii

[sullen ice]

$(x^2+1)(x+1)$

[elfin berry]

Joshii

[sullen ice]

Idk if that’s useful I’m doing very early into to integrals rn lol

[ebon depot]

Well to solve it i'll need to do partial polynomial division

[sullen ice]

Ah yeah

[ebon depot]

but i def need some help since i've never done that before

[sullen ice]

I did it once for linear algebra i don’t remember tho

But I see the vision

[ebon depot]

Supposed to work like this i guess

[sullen ice]

Mhm

[fervent rampart]

we have to start by doing polynomial long division to reduce the degree of the numerator

since partial fraction decomposition only works with the degree of the denominator being strictly higher

[ebon depot]

I attempted that but it went to hell ;-;

[fervent rampart]

show your working on that

[ebon depot]

,rotate

[ebon depot]

It just went on and on and on

i stopped since i thought it was wrong

[fervent rampart]

just do the 1 term, then take the rest as a remainder

now you have
(quotient) + (remainder)/(denominator)

[ebon depot]

$\frac{1(x-1)}{x^3+x^2+2x+2}$

Like so?

no wait

that was not denominator

[elfin berry]

Merineth

[fervent rampart]

if i divide 7/3, i can end up with a quotient of 2 and a remainder of 1, so i can rewrite it as:
2 + 1/3. use the same idea here

[ebon depot]

Well

$\int \frac{x^3+x^2+3x+1}{x^3+x^2+2x+2}$

[elfin berry]

Merineth

[ebon depot]

If i have this

and perform the division once

i will have x-1 with a remainder of 0

$\frac{1+x-1}{x^3+x^2+2x+2}$

[elfin berry]

Merineth

[ebon depot]

x - 1 is the sub part

and the +1 is the x^3/x^3

As in like this

[fervent rampart]

in this case, 1 is your quotient (the part that was fully divided) and x-1 is your remainder (the part that was leftover). the quotient should not be part of a fraction at all (it was divided out fully), only the remainder:
(quotient) + (remainder/denominator)

[ebon depot]

ooh i see what you mean

$1+\frac{x-1}{x^3+x^2+2x+2}$

[elfin berry]

Merineth

[ebon depot]

Is this what you meant?

[fervent rampart]

yes

now you can do partial fraction decomposition as normal on the second part

[ebon depot]

I'm not to familiar with it to be honest

I can't really tell what the formula means

[fervent rampart]

we start by factoring the denominator

[ebon depot]

Based on this, my a is 1 and b is -1

that is right at least?

[fervent rampart]

yes

[ebon depot]

And now as you said i need it in the form of (x-c)(x-d)

i do believe i can do that at the very least :p

$x^3(x+2)(x+2)$

No idea what to do with x^3 tho

[fervent rampart]

that doesn't look right

[ebon depot]

oh

You are right

$x^3 +(x+2)(x+1)$

[elfin berry]

Merineth

[ebon depot]

This would be my best guess

[fervent rampart]

we need it to be completely in the form of a product of factors

[ebon depot]

Do you have any advice for working with factors like that?

[fervent rampart]

if you scroll up far enough we already found this

[ebon depot]

Oooh

What joshi did?

[fervent rampart]

yes

[ebon depot]

This part?

[fervent rampart]

yes

[ebon depot]

I would love to get better at that

How would you go about factoring like that

Got any advice?

[fervent rampart]

for a cubic polynomial like that, try grouping every 2 terms together, pull out common factors, and see if you end up with something in common. if it doesn't work one way, try swapping the groupings (there are only 2 possible). once you do that, check if any of the new factors can be split up further

[ebon depot]

isn't this wrong

$(x^2+1)(x+1) not equal x^3+x^2+2x+2$

[elfin berry]

Merineth

[jagged relic]

Should be (x^2+2)(x+1)

[ebon depot]

NEL POGGERS

[fervent rampart]

it has a typo. look at the previous message and you can see the proper method

[jagged relic]

I can't help more than that

[ebon depot]

How can a god not help?

$1 + \frac{x-1}{(x^2+2)(x+1)}$

where a = 1, b = -1

[fervent rampart]

we have what's called an irreducible quadratic factor here, so the formula posted above isn't sufficient

[jagged relic]

(x^2+2)(x+1)?

[fervent rampart]

here's a couple of websites about partial fraction decomposition, which include information about irreducible quadratic factors:
https://www.mathsisfun.com/algebra/partial-fractions.html
https://tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx

[ebon depot]

Does this formula state that if c is not equal to d, mean we go the top row. if c is equal to d, we do bottom version?

[elfin berry]

Merineth

[ebon depot]

and since c = 2 and d=1 i go with the top one?

[fervent rampart]

to summarize them, we need to set up
(x-1)/(x²+2)(x+1) = A/(x+1) + (Bx+C)/(x²+2)
and solve for A, B, C

[ebon depot]

what value is dx?

Like

The formula dosn't make sense

it says i should write A infront of the integral

but A is unknown, so we just write A

integral and then dx/x-c

but dx is also unknown

[fervent rampart]

dx is part of the integral notation

[ebon depot]

$A \int \frac{dx}{x-2} + B \int \frac{dx}{x-1}$

[elfin berry]

Merineth

[ebon depot]

This is what i'll have

according to the formula

[fervent rampart]

your formula only works when the denominator is made up of linear factors. since the denominator has an irreducible quadratic factor, it won't work

[ebon depot]

That means we are missing a formula..

[fervent rampart]

partial fraction decomposition is an algebra property that's independent of integrals. it just happens to be useful in integration

[ebon depot]

So how does it work? What do i need to memorise

[fervent rampart]

i would recommend you read through both of these

but the general idea is to rewrite it as
(something)/(factor 1) + (something else)/(factor 2) + (another something)/factor 3, etc.
where the "something" is a polynomial of one degree less than the factor below it

[ebon depot]

Accoring to it i did step 1 that is factor the bottom

and now i want to write it as

$\frac{Ax +B}{x^2+2} + \frac{C}{x+1}$

As such?

is it fine to write A2 as B?

[fervent rampart]

not quite. that works for linear factors, but quadratic factors will have something like (Ax + B) or (Bx + C) on top (see "irreducible quadratics" farther down the page)

[fervent rampart]

that's fine

[elfin berry]

Merineth

[ebon depot]

Okay so whenever i have a nonlinear in the denominator i have to go with Ax+B format?

[fervent rampart]

if you have a quadratic in the denominator it's Ax + B, a cubic in the denominator it's Ax² + Bx + C, etc.

it's always a polynomial of one degree less

[ebon depot]

ok! that shouldn't be to hard to remember!

So now we have to figure out A, B and C?

[fervent rampart]

yes

[ebon depot]

And that is done by substituting x with numbers

?

[fervent rampart]

that's part of it, although it can't necessarily get you all the way

the first step is to write:

$\frac{(x-1)}{(x^2+2)(x+1)}=\frac{Ax+B}{x^2+2}+\frac{C}{x+1}$

[elfin berry]

cloud

[fervent rampart]

then, you can multiply both sides by the denominator (x²+2)(x+1)

this gets you:

$x-1 = (Ax+B)(x+1) + C(x^2+2)$

[elfin berry]

cloud

[fervent rampart]

this is pretty much the same procedure every time for partial fractions

[ebon depot]

Alright

And now when we have this we can find A, B and C by sub x with numbers?

[fervent rampart]

now you have 2 options:

1. substitute in values of x such that a factor = 0
   this is easier, but it can only partially solve the problem
2. expand out the whole thing and reason out what A,B, and C must be
   this is more difficult, but unavoidable in some problems

[ebon depot]

would it be better to just learn and do the 2nd option every time ?

Since it works for every problem even if it's harder?

[fervent rampart]

usually i try to do as much as possible with the first method, then do anything remaining with the second method.

the first method will allow us to solve for C, but we'll need the second method to solve for A and B

[ebon depot]

If x is -1

should solve for C

[fervent rampart]

yes

[ebon depot]

C = -2/3

[fervent rampart]

yes

[fervent rampart]

we'll now do the second method, which will involve multiplying out/expanding the entire right side