help-0

[hollow bear]

I don't think it won't be 6 or 10

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[summer arch]

am i on the right track?

[verbal blaze]

you can just differentiate f(x) and then solve.

$$f'(x) = \frac{d}{dx} f(x) = \frac{d}{dx} x^{\frac{2}{3}} = \mbox{???}$$

[ocean seal]

Shuba

[summer arch]

quite honestly i was just plugging that stuff in to the equation that my notes have

[fresh field]

The limit is the definition of derivative

There exist standard formulas. Results that hold true throughout domain

So, are you required to do it be definition?

[summer arch]

uhhh i dont think so

we literally just started derivatives so im a little lost

[fresh field]

In that case you are required to do it that way

[summer arch]

this is the notes i was basing my work off

[fresh field]

Can you rewrite the denominator

Hold up

[fresh field]

What's f(-8)

[summer arch]

-4 right

[fresh field]

How?

Can you show that

[fresh field]

You know it's (-2)² correct?

[scenic eagle]

Some 1 help me o

[summer arch]

oh my god lol

i might have forgot that

[fresh field]

Anyways

Now rewrite it correctly

[fresh field]

Do you see some a² - b² form somewhere

Some a³+ b³ somewhere?

[summer arch]

in the numerator

[fresh field]

Break it

[summer arch]

uhhhh idk if this is right or not

[glacial gyro]

not quite

[summer arch]

the -2 would become positive from it right

in the numerator atleast

[glacial gyro]

do you know the difference of squares identity

and difference of cubes

[summer arch]

yeah i think so

[summer arch]

ohhh i see

[glacial gyro]

you cant put the cube on the square and cube on the outside

and from here you can use your identities and try and cancel something

[summer arch]

would the x^1/3 cancel out on top

yeah idk where to go from here

i kinda just want to solve the problem and get past it

[glacial gyro]

you would do difference of squares on the top

which would get you (x^1/3 + 2) as a term

and on the bottom do difference of cubes which also gets you (x^1/3 + 2)

and you can cancel them

[lone heart]

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[alpine sable]

need help with this

[lone heart]

@alpine sable Has your question been resolved?

[tacit arch]

Did you try using sin^2 = 1-cos^2

[alpine sable]

yeah i did

i got stuck with 3+ 5cos^4x - 6cos^2x = 6/5 after it

[livid sage]

this is a quadratic. you can solve for (cosx)^2

[alpine sable]

okay i'll try

[alpine sable]

i got it!

option a
cos^2x is 3/5

thank youu

.close

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[livid sage]

you need to make sure the other answers are wrong

but ur welcome

[alpine sable]

well the only value of cos^2x is 3/5. Under root term came out to be 0 and it's a single correct question

[spark shard]

sry for the delay

oh u alr got it nvm

[livid sage]

this question has 2 correct answers

[alpine sable]

oh

[alpine sable]

my bad. I checked now and option d is correct as well

[livid sage]

can u show?

[livid sage]

hm u wrote ((sinx)^2)^3 instead of ((sinx)^2)^4 for example

[alpine sable]

oh wait my bad. i made another silly mistake. In my mind i thought sin^8x so that must mean i have to cube it totally forgetting that the powers get added--

[livid sage]

yee fair enough

[alpine sable]

noo it's okay. Thank you for taking out your time to help me

[spark shard]

ywyw

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[neat folio]

I'm a bit confused bout part 2B here

I thought that given the cumulative distribution function F(x), the area between 1<=x<2 would be (x^4+5)/20

instead of (x^4+4)/20

given that the total cumulative probability will be 1/4 from 0 to 1

I'm just wondering what I am misunderstanding or whether the answer is wrong

[hushed locust]

if you plug in x = 1 to the given function you should get 1/4

so (1+4)/20 = 5/20 = 1/4 as expected

[neat folio]

that was my assumption

but the answers use 4/20 instead

are the answers wrong in that case.

[hushed locust]

therefore (x^4 + 4)/20 is a function which returns 1/4 at x = 1

[neat folio]

ah yep

I see

.close

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[lone heart]

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[alpine sable]

do you want the solution?

[viral loom]

yeah

i just want to check

[void nymph]

send it

because !nosols

lmao

[viral loom]

what?

[void nymph]

if you want your solution checked

you show the solution

[void nymph]

!nosols

[lone heart]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[void nymph]

!status

[lone heart]

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

[void nymph]

you're 4 right

so show your work

[viral loom]

Oo

so 0<=i<=4, plug all of it into the prod:
prod(0j) + prod(1j) + prod(2j) + prod(3j) + prod(4j) = 0+120+120+60+20=320

[void nymph]

you're doing it manually hm

[viral loom]

whats the fast way

[lone heart]

@viral loom Has your question been resolved?

[void nymph]

I'm not sure actually this seems fine enough

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[broken falcon]

guys, am i wrong for the yellow highlited part?

[broken falcon]

my friend said the standard deviation is supposed to be 1.46

[lone heart]

@broken falcon Has your question been resolved?

[pseudo ice]

Where did you get 93.6 from?

[pseudo ice]

(fixing that should get you this)

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[delicate cape]

What is the value of x if sin x=12/13 and cos x=5/13 when x is between 0 and pi/2 ?

[subtle birch]

arcsin(12/13)

[delicate cape]

And how do I get to that answer?

[mystic swallow]

it's not really something you can generally calculate...

it's not a very neat number

arcsin(12/13) just means the value of theta b/w 0 and 0.5pi for which sin theta = 12/13

i.e. sin(arcsin x) = x

[void nymph]

yep

well

more like $\arcsin(12/13) + 2n \pi, n \in \mathbb{Z}$ but yes

[ocean seal]

Percy

[mystic swallow]

shhhh

[delicate cape]

I was thinking x has a specific value

[mystic swallow]

it does have a specific value

[void nymph]

it was an infinite amount of values

[mystic swallow]

it just can't really be calculated

,w arcsin (12/13)

[void nymph]

all are specific though

[ocean seal]

Results provided by WolframAlpha

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

Requested by justarenegade

[mystic swallow]

1.176 radians

[delicate cape]

Thank you

[mystic swallow]

np

[void nymph]

lmaoo

[mystic swallow]

!done

[lone heart]

If you are done with this channel, please mark your problem as solved by typing .close

[void nymph]

also yes shh but for more exactness its this

[calm reef]

are photos not allowed here

my photo got token down immediately

[mystic swallow]

no they are

[void nymph]

yeah they are

[calm reef]

any chance you guys can help me out with this

[void nymph]

!occupied

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

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[calm reef]

whoops

i thought 0 was the main thing to ask my bad

[lone heart]

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[void nymph]

now you can use it ig

[void nymph]

oh no.

!close

@calm reef

.close

[lone heart]

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[calm reef]

here right

[void nymph]

occupy it after it becomes avaliable

[calm reef]

oh okay

[mystic swallow]

#help-8

@calm reef here

use that one

[void nymph]

or do that

lmao

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[floral ocean]

can someone help me with this please

[mystic swallow]

,rccw

[mystic swallow]

jesus chr7st uh

hang on one sec

[void nymph]

not a fan

[ocean seal]

ren

[mystic swallow]

use this

[floral ocean]

isn’t that the first step

[mystic swallow]

can't really see what you've done there

[floral ocean]

i was told that this is the first step towards solving it

[mystic swallow]

well
then use it

[floral ocean]

ok thanks

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[limber depot]

how do i get the third equation... I know its 3 simultaneous

The 2 ones that are circled are the other 2 equations.

I know i have to do the dy/dx of ax^3 + bx^2 + cx -5 = (12)
when i do that i get 3ax^2 + bx + c = 12
but i have to get rid of the x for it to be a simultaneous? help ty

[limber depot]

or am i somehow supposed to deduce c

from the equation given?

like first point in ms is this buttt how do i get it

.close

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[viral falcon]

Here is the translation to English:

1. Let ( A ) be the ( 3 \times 4 ) matrix

[
\left[\begin{array}{cccc}
1 & 2 & 3 & 4 \
5 & 6 & 7 & 8 \
9 & 10 & 11 & 12
\end{array}\right]
]

a) Solve the equation ( A x = 0 ). \
b) Express the solution set as a linear span of vectors in ( \mathbb{R}^4 ), and determine whether the vectors

[
\boldsymbol{y} = \left[\begin{array}{r}
1 \
-3 \
-3 \
1
\end{array}\right]
\quad \text{and/or} \quad
\boldsymbol{z} = \left[\begin{array}{r}
6 \
-9 \
2 \
2
\end{array}\right]
]

lie in the span. \
c) What is the general solution to ( A \boldsymbol{x} = \boldsymbol{b} ) for a given vector ( \boldsymbol{b} \neq \boldsymbol{0} ), if you are given that

[
\boldsymbol{x} = \left[\begin{array}{r}
1 \
2 \
-3 \
0
\end{array}\right]
]

is one solution?
d) Find what ( \boldsymbol{b} ) in the previous part must be, and express ( b ) as a linear combination of the columns of ( A ).

[ocean seal]

Michael

[viral falcon]

What am I even supposed to show in c? And how do I do it?

Musn't $\vec{x}$ be expressed with $\vec{b}$?

[ocean seal]

Michael

[viral falcon]

Oops, forgot to remove "Here is the translation to English" lol (I only translated it from my language with ChatGPT)

[lone heart]

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[lone heart]

@viral falcon Has your question been resolved?

[lone heart]

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[warm osprey]

Hi

[gritty bramble]

e?

[warm osprey]

I need help on this math question

My wifi is laggy so bear with me

Im sending a photo

I need help question 1 A)

[alpine sable]

!da2a

[lone heart]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[dull lintel]

,rotate

[warm osprey]

I already solved and found out the Ts by substituting it in for all 3 times

But idk how to get m

I can do question 1 b and c cause they are easy, but idk how to do average velocity with 3 different times

Is there a specific formula?

[gritty bramble]

find how much it falls in that time

and then divide by that time to get velocity

for example it falls 5m in the first second, so its average velocity is 5/1 = 5m/s

[alpine sable]

i mean

[gritty bramble]

oh wait

for the third second

[warm osprey]

Hnmm

Hmmm

[gritty bramble]

its average velocity would be (s(3)-s(2))/time

similarly for eighth second it would be (s(9)-s(7))/t

[alpine sable]

s_n = u + a/2(2n-1) ((displacement in the eight second

[warm osprey]

Hmmmm

So for second 1, after subbing it in the equation i get like 315

[gritty bramble]

nth* second

[warm osprey]

What do i use

The 315 for

[alpine sable]

ohk

[warm osprey]

Do i subtract it by smth /1- smth

[gritty bramble]

well

for the third it was s(3)-s(2)

what will it be for first?

s(1) -s(?)

[warm osprey]

0?

[gritty bramble]

yes

[warm osprey]

Tysm

[gritty bramble]

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[lone heart]

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[round forge]

Question: In how many ways can 4 novels, 2 mathematics books, and 1 chemistry books be arranged on a bookshelf if the novels must be together, but the other books can be arranged in any order?

[round forge]

My work thus far

Somehow by my logic I am doubling the correct answer but I don’t know why

[twin nimbus]

I am assuming all of the books may be distinguished from each other. So really we have two types of books: novels and other. Novels must be together, so we have 4 "items" (set of novels + 3 other books) we can arrange in total. And the novels themselves have 4 total orderings, so that is 4!(4!)

@round forge ^

Essentially, the your answer double counts the arrangements of the math books

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@round forge Has your question been resolved?

[round forge]

Thanks! Mathematically I get it, I still can’t quite come up with my own quantitative justification but I get how you got that answer. Thank you!

[lone heart]

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[wraith valve]

what i did may be completely pointless

[wraith valve]

!status

[lone heart]

What step are you on?
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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

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

[twin nimbus]

@wraith valve if you have |f(x) - f(y)| ≤ |x-y| what does this imply about the derivative?

[wraith valve]

AHA I DID NOT SEE THAT 🙏

it implies the that the f'(y) <= 1

wait

no

[twin nimbus]

Well |f'(y)| ≤ 1

[wraith valve]

yeah that

[twin nimbus]

In other words it is Lipschitz continuous

[wraith valve]

ight thank you

.close

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[mellow cape]

Hi can someone explain to me how to represent a linear inequality as a graph because I'm having trouble understanding it.

[lone heart]

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[quiet delta]

Am I thinking correctly,$f_X(x)$ is the probability density function. $\int_{y}^{\infty}{f_X(x) dx} = \int_{0}^{\infty}{f_X(x) dx} - \int_{0}^{y}{f_X(x) dx} = 1 - F_X(x)$ ?

[ocean seal]

Delusional J

[tiny sky]

i think you mean $F_X(y)$ at the end

[ocean seal]

LY

[tiny sky]

but yeah

(also here i think ur intending for X to be a non-negative r.v.)

[quiet delta]

Yeah

[quiet delta]

Oops, but yes!

I'm trying to use this in an attempt to prove $E(X) = \int_{0}^{\infty}{(1-F_X(x))dx}$ using the classical definition of $E(X) = \int_{-\infty}^{\infty}{xf_X(x)}$for a positive continous random variabel X

[ocean seal]

Delusional J

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[wise jackal]

why is a * b + a * b = a(b+b)

[gray isle]

distributive property

[wise jackal]

Oh i understand

thanks

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[sinful vortex]

How did he get the length as 1.0785

[sinful vortex]

I have calculated it and got 1.1985

And it’s wrong

The problem is if you could not find the right length (dimensions) all the solution would be wrong

[tacit arch]

,calc 107 * 85

[ocean seal]

Result:

9095

[tacit arch]

,calc (25 - 2 * 3/2 + 35 - 2 * 3/2) * 2 - 0.15

[ocean seal]

Result:

107.85

[tacit arch]

oh that's a decimal

[tacit arch]

107.85cm is the correct arithmetic

[lone heart]

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[sage rose]

hloo

[alpine sable]

Yeah

[sage rose]

solve this?

[alpine sable]

solve

[small lance]

What’s the question?

[alpine sable]

for what

x and y?

[sage rose]

yep

i'm done with everthing but i'm stuck here

[azure needle]

||Vertically opposite angles||

Or whatever it's called

[azure needle]

Have you learned this

[sage rose]

ouh thxs but wouldn't it be x instead of y?

[azure needle]

It means x=y

[sage rose]

ouhhhh thanksss

mb i'm weak in maths

[lone heart]

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[twilit nebula]

can somone help with this please i got the right answer and i didnt guess but it jsut made the most sense btu i cant find a way to work it out to the same answer

[tacit arch]

,tex .exp rules

[ocean seal]

[tacit arch]

power of power

[twilit nebula]

yea

but when i do that

as in ((7/3)^1/2)^2

i dont get 7/3

[modern kernel]

What is 1/2 × 2?

[twilit nebula]

is it somthn do with my calcualtor

[modern kernel]

Might be, How are you inputting it?

[twilit nebula]

2 ways

ive tried doing it one by one

[twilit nebula]

and ive tried doing it as a whole like this one

[modern kernel]

Try doing it

(7/3)^(1/2×2)

[twilit nebula]

oh

wait why do i have to do it like that?

[modern kernel]

I'm not actually certain

[twilit nebula]

yea that works

what?

why doe sit only work some ways?

[modern kernel]

Because you're squaring 1/2

That becomes 1²/2²

I believe

Wait maybe not

[modern kernel]

I tried to do (7/3)^(1/4) and got the same result as this

[twilit nebula]

so its not just my calculator?

[modern kernel]

Yeah

[twilit nebula]

thx

ill have to note that down somwhere

do u know any websites with a lsit of all these weird input ways?

[modern kernel]

You're missing a bracket here

1/2 will have additional brackets

It solves the issue

[twilit nebula]

yea

but is there any ways to find these other than trial and error?

[modern kernel]

I guess

I understand

Why it happened

The operation you put

It actually looks something like this

Wait

As

By not giving the brackets at 1/2

The calculator does

(7/3)^1 and divides the whole by 2

[twilit nebula]

what

[modern kernel]

If you give this input : ((7/3)^1/2)^2

The computer won't understand that it's 1/2 if you don't specify it by giving brackets

You only want the 2 to divide the 1 so you gotta specify it

If you don't, the computer will think we will divide the whole with 2

[twilit nebula]

ooooh

so its reading it as

(7/3)^1 divided by 2

[modern kernel]

Precisely

[twilit nebula]

also im not allwoed to post mutiple questions in the same help channel right?

[modern kernel]

Absolutely

[twilit nebula]

kk

.close

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[twilit nebula]

by absolutely u ment im not aloowed to right

[modern kernel]

You're absolutely allowed

My bad

[vague agate]

Hi this is what I did. Since we want to find A, I multiplied both sides with an exponent of 1/2.
The x value will be the same on both sides so I just took it out :D

[modern kernel]

I read it wrong

[twilit nebula]

no i got the answer emidiatly but i could find a way to make my calculator like it

[vague agate]

ohhh ok

[lone heart]

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[storm marten]

yeah seems right

[alpine sable]

yeah

z = y+z
180 = 2z
z = 90

similarly if z>y+z
z>90
and if z<y+z
z<90

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[alpine sable]

welocme

,iamstudying

[ocean seal]

Gave you the studying! selfrole.

[mystic swallow]

#help-5
post this there

[lone heart]

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[real pier]

how can i find the 6

[mystic swallow]

6 is not equal to either 0 or 7

therefore x is not equal to either 0 or 7

therefore, the third expression

[real pier]

what does the third one thing mean, the = with /

[alpine sable]

not equal

[real pier]

oh okay

[alpine sable]

that x can't take that expression for x = 0 , 7

[real pier]

alr thanks guess i understand now

.close

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[dense lagoon]

ans and question,
for part C whys the theta value for max 2pi and not 0?

[gray isle]

they specified positive values in the question

[dense lagoon]

ohh

yea mb

thx

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[fluid iris]

i need helping understanding to solve for the slope. i under stand that it is asking for the rise which is x

[gray isle]

slope isn't just rise
and rise isn't x

did you watch the video guide?

[fluid iris]

https://www.youtube.com/watch?v=mmWf_oLTNSQ&t=123s they provide me this and for whatever reason i am not understanding it 100%

[gray isle]

try to follow along with what they're doing

starting with picking two points
(ideally lattice points with easy to read coordinates)

[fluid iris]

right so i did (10,1) and then i did (-10,-3) and for that solution i got -2

here is another q relating to the Q

[gray isle]

how are you getting -2

[fluid iris]

but all was wrong. i am not fully following it seems

oh, should it be -1?

[gray isle]

how are you getting -1

[fluid iris]

i did -3 -(1) =

[gray isle]

whats that equal to?

[fluid iris]

would that be -4?

[gray isle]

yes, -3-1 is -4

and what else are you doing for slope?

[fluid iris]

determine the vertical intercept?

[fluid iris]

i apologies for not understanding your q

[gray isle]

no

look at the slope formula

[fluid iris]

oh

[gray isle]

they did NOT say that slope is
one y coord minus the other

[gray isle]

-10-10 isn't 0

[fluid iris]

^^ this was b4

it was meant to be positive

[gray isle]

can you reattempt your calculations

[fluid iris]

sure 1 moment

[gray isle]

why are you doing
-10+10

and why are you now putting 4 for
-3-1

the way you set it up before was fine
you just made basic arithmetic errors

[fluid iris]

right

but isnt this the idea thats why -10+10

[gray isle]

no

[fluid iris]

they end up canceling out, no~

[gray isle]

they way you chose your first and second point, no

[fluid iris]

would it better to choose a diff point to make it easier?

[gray isle]

doesn't make a big difference

[fluid iris]

i spot a -5,-2

[gray isle]

yiu issue isn't with the choice of points

but rather with the arithmetic and plugging in your values

in your setup, your
x_2 = -10
x_1 = 10
x_2 - x_1 would be -10 - 10

what you had at the start
(-3-1)/(-10-10)
was completely fine
it's just that
-3-1 isn't -1,-2 or now 4,
-10-10 isn't 0

try not to overthink this subtraction

[fluid iris]

edit its -4

*-20

[gray isle]

yes, but thats equivalent to 4/20
so it's technically fine

[fluid iris]

yes. noted

[gray isle]

don't forget the fraction line
4/20 can be simplified further

[fluid iris]

0.2

[gray isle]

yes

for the following question you mixed up your intercepts

you gave the horizontal/x intercept

[fluid iris]

5▶️ and 0 ⬆️

the vertical

meaning the highest point

?

[gray isle]

no

[fluid iris]

oh my its -1

hmm if m = -4/20 then would those be the nums to plug in?

[gray isle]

why are you now saying that m is -4/20

[fluid iris]

4/20

error

my b

[gray isle]

and you simplified that earlier

[fluid iris]

correct.

[gray isle]

those would be the values you'd use

[fluid iris]

i have not sent the answer yet

gonna give it another shot

first pick any points '

[gray isle]

why does it look like you're multiplying your coordinates

[cosmic mural]

average rate of change is (y2-y1)/(x2-x1) so the top would be only y coordinates and the bottom would be their corresponding x coordinates

[fluid iris]

oh

[cosmic mural]

yes and remember you need to subtract them not multiply

[fluid iris]

so it would be -4/1? which is -4

i am need of help 🙏

so much help 😭

[fluid iris]

would it be -4?

i might need to come back to this prob.

cause -5+9

or it would be -5-9

oh

-5-+9

plus 4

and bottom +9

4/9

[cosmic mural]

yes

[fluid iris]

@gray isle should i open another ticket for the other qs?

for me this has been more exhausting that i thought

[lone heart]

@fluid iris Has your question been resolved?

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[shadow frost]

I'm more focused on the bottom portion.

I have issues communicating with my calc 1 prof.

I just want to know how I know if the bottom goes to negative 0 or positive 0.

[shadow frost]

I know that I can input something like 05 becuase -5<x

That will give me -1/0(positive)

but we were using direct substitution.

[dull lintel]

what do you get when you put -4

[shadow frost]

for the bottom I'd get

16-4-12=0

[dull lintel]

what about for the upper one?

[shadow frost]

the upper one would give me a negative (-7)

[dull lintel]

so you get -7/0, you know the answer will be positive or negative infinity right?

[shadow frost]

Yeah, but I've also seen him put a negative in the 0 portion also.

like he did something like

[dull lintel]

okay now put a value less than -4 (this is what it wants) such as -5 and see the sign of the bottom part

[shadow frost]

bottom would be 8

[dull lintel]

so it is positive right (bottom is)

[shadow frost]

yeah postive 8 on the bottom and a negative on the top

so that would be negative infinity.

[dull lintel]

yes

[shadow frost]

So within a problem like peicewise when am I able to substitute out?

like

[dull lintel]

not sure what you mean by substitute out

[shadow frost]

factor

because if I can factor or when I get to a step I can factor I could just make it

1/|(x+4)|

[dull lintel]

I don't see any reason why can't you do that, not sure tho

[shadow frost]

I got points taken off for doing that.

He said review algebra on that one

[wet linden]

Are you allowed to graph it

Because that would be the easiest way

[shadow frost]

yeah we can do that, but we need to show the process we used to get to the point.

[wet linden]

You just draw the graph of 1/x+4

The asymptote is gonna be at x=-5

[dull lintel]

,w plot 1/(x+4)

[wet linden]

-4*

[ocean seal]

Results provided by WolframAlpha

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

Requested by metumortis

[wet linden]

Ye that’s how I would go about it

[shadow frost]

because

x>0 is positive x.
x<0 is -x within an absolute value expression

[wet linden]

What

[shadow frost]

yeah let me get it that's what's confusing me.

[wet linden]

This id right

Wait why do you have

An absolute value sign

[wet linden]

Here now

I’m confused

[wet linden]

What’s this for

[shadow frost]

Second let me get the full thing he posted on it

okay pause I might have just found out why I'm beyond confused.

In absolute value, if I have something like |x-5| and my x is -6

[wet linden]

Then you select the negative sign

like you + or - (x-5)

Then you would circle the negative sign

[shadow frost]

so on the outside of the absolute value there is a negative?

-|x-5|

[wet linden]

absolute value = + or - negative of a number

[wet linden]

Ye

Or positive

Depending on x value

[shadow frost]

okay so since I'm choosing in my problem above a value less than x, which is anything less than -4

I basically do

[wet linden]

Why do you still have an absolute value on the denom

I don’t quite understand

[shadow frost]

becuase I don't know how to get rid of it

[wet linden]

Oh is it in the problem

I didn’t see it

[shadow frost]

yeah so that's where I'm having the most issue. so in abs value

If my value is less than 0, I'm going to make the absolute value on the outside negative like the above, right

[wet linden]

No

It matters where the center is

You know how the absolute value graph is like a v

And there is a point where it meets

It matters where the x value is at that point

Like for lim as x approaches -2 |x+1|

[shadow frost]

If I chose something like -3

that would be -1

[wet linden]

Uh highkey bad example

Because it’s still 2

Because -(-3+1)

-(-2)

=2

[shadow frost]

Ohh I thought you were doing it for

Like for lim as x approaches -2 |x+1|

-2+1=1

[wet linden]

He

Ye

[shadow frost]

So if I factored

my orignal problem into |(x+4)(x-3)|

I would put a negative outside as it's less than by limit value (aka -4 and I'm going to the left) so -4.1

well no, it wouldn't be able to be that

!solved

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[fossil pumice]

I need help for this question

[fossil pumice]

not even sure where to begin

and it has to be an exact value

[tacit arch]

https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/06%3A_Applications_of_Integration/6.08%3A_Exponential_Growth_and_Decay

[fossil pumice]

I've read through that but it still doesnt really help, since this question is different thand normal halflife questions

I've checked a few halflife questions but for this one I don't know the initial value

but even if I find it

when I plug it into the formula the application still marks it as wrong

[tacit arch]

show your work so someone can find your error

this example is extremely close to your problem

[fossil pumice]

ok I'll do that

hey thanks alot!

got the right answer thanks to that

I was messing up the k

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[heady charm]

<@&286206848099549185>

<@&286206848099549185>

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@heady charm Has your question been resolved?

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[grand trout]

<@&286206848099549185>

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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.

[verbal blaze]

whats the question?

!da2a

[lone heart]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[grand trout]

Pls help me correct my proof in set theory anyone?

[serene junco]

What is the prompt, exactly as it's written?

Is it exactly that? Prove $A \subseteq A \cup B$?

[ocean seal]

tatpoj

[grand trout]

<@&286206848099549185>

[serene junco]

@grand trout is this the goal?

to prove A is a subset of AUB?

If so, we shouldn't assume anything extra, like A being a subset of B

[grand trout]

Yes sir

[grand trout]

Sir I only remove the what I assumed?

[serene junco]

No, let's actually start over

A lot of what you wrote here isn't quite right

To prove that $A \subseteq A \cup B$, you want to let $x \in A$, and then show $x \in A \cup B$.

[ocean seal]

tatpoj

[serene junco]

Your basic idea is right, but the steps don't really make sense

Just from $x \in A$, how could we get to $x \in A \cup B$?

[ocean seal]

tatpoj

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[grand trout]

Sir I have renew my proof

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[quiet pasture]

dd

[buoyant saddle]

dd

just chilling hbu?

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[pliant estuary]

,rotate 270

[pliant estuary]

hi, can somebody please help me with solving this equation? i have never solved PDE's before, so i tried Fourier transform but that doesnt assume initial conditions, so i think thats an dead end

also this problem comes from plasma physics (electrostatics), so i think that at r=0 the initial condition should be something like 1/4piepsilon_0 but i am not able to find anything online

like a point charge

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[pliant estuary]

<@&286206848099549185>

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@pliant estuary Has your question been resolved?

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@pliant estuary Has your question been resolved?

[radiant scarab]

I need help with algebra
P1/(V1+V2)=P3/V3
i need to find V3

[lone heart]

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[sick lion]

Point O is the centre of the circumscribed circle of the acute triangle ABC, in which AB > AC. Let X base the perpendicular lowered from B to AO, M be the middle of the BC, AA1 is the height of the ABC triangle. Prove that XM = MA1.

[lone heart]

@sick lion Has your question been resolved?

[sick lion]

<@&286206848099549185>

[silent oar]

ill try it

gimem a bit

[silent oar]

did you mean prove XM=MA1

[sick lion]

Oh, yes

[silent oar]

okee

[sick lion]

I know A1X || DC

I think we need to prove that BD || AC

[silent oar]

whats D

[silent oar]

yea but what is it

[sick lion]

AO diameter crosses the circle in A2 for the second time, D is the intersection of XM and A2C

[sick lion]

Because => BDC = 90 => BM=CM=MD

[silent oar]

im not sure how thats enough

unless you proved A1 X C D are on a circle

[silent oar]

nvm its obvious

ok i solved it

@sick lion do you want the solution or a hint

(i didnt use D)

[sick lion]

I think solution, I spend almost 2 hours on it

[sick lion]

I’m not

[silent oar]

okee

[silent oar]

then simple BM=CM=MD isnt enough

anyways

let B1 on AC such that BB1 is perpandicular to AC

A B A1 X B1 are all on the same circle

so angle BB1X=BAX=90-C

also BB1M = MBB1=90-C

so BB1X=BB1M so M X B1 are on one line

power of point from M with respect to (AB1XA1B) gives you MXMB1=MA1MB

and since MX=MB

then MA1=MX

[sick lion]

Lol that’s so easy

Thanks

.close

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[silent oar]

\o/

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[void nymph]

okay so what's this

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[dusk tendon]

I understand that this is the correct answer according to the SFTC but I don't understand why, how does taking the derivative of a definite integral to x change the variable in the derivative?

[mystic swallow]

it doesnt?

you integrate, evaluate, and differentiate

the actual variable is always x

t is just a dummy variable for integration

[glacial gyro]

when you integrate it

[dusk tendon]

but the original integral was wrt t

[glacial gyro]

you plug in the x

so for example

it will be (4/3)x^3+2x+c

technically it will be t not x

[mortal trellis]

no +c

[glacial gyro]

definite

i mean the +cs just cancel with a definite

[dusk tendon]

+c still there but will cancel out, im following klint

[glacial gyro]

but yeah no +c

[dusk tendon]

klunt*

[glacial gyro]

it will have the t

and then you plug in the x for the bounds of integration

and then when you derive it will end up with the x

[dusk tendon]

can you do the same with a lower bound but in reverse? like for instance if the 0 was on top and x was on bottom?

[glacial gyro]

wym in reverse

[dusk tendon]

instead of 4x^2+2 it'd be -4x^2+2?

[glacial gyro]

-2

but yes

you could also just flip the bounds and make it negative

one of the principles

[dusk tendon]

why does this work even if the constant bound isn't 0?

[glacial gyro]

yes because when you plug in a different constant

it will give you a number

and then when you derive it will just be 0

[dusk tendon]

oooooooh

so in a way, this works because the x is unresolved or w.e, im unsure the term

[mortal trellis]

imagine you have an antiderivative of the integrand. lets call it F(x). then you know that F'(x)=4x^2+2 because thats just what antiderivative means.

[dusk tendon]

and we still have to express the integral in terms of x

[mortal trellis]

and you know that the integral is equal to F(x)-F(0). or if the bounds were reversed then it would be F(0)-F(x)

now what happens if you differentiate that

[dusk tendon]

yea that makes sense, you're have to express it in terms of x cause you don't know what x is so in the end you'll have the (4/3)x^3+2x+c and that's what you're differentiating

ok i appreciate it, thank you that makes sense

♥️

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[elfin bison]

First one is the association law but if u make it that the sign in middle is not same as sign in bracket then it's distributive law right?

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@elfin bison Has your question been resolved?

[elfin bison]

No

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@elfin bison Has your question been resolved?

[elfin bison]

Close

Close

Can I use distribution law

To get this

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@elfin bison Has your question been resolved?

[obsidian night]

!occupied

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[spiral pumice]

HELP! i dont understand how inputting a dependent vector into a linear transformation maintains its linear dependence! ):

[spiral pumice]

same thing with independent vectors

in my mind the linear transformation is acting like a scalar?

just a really complicated one

i dont know if that holds up

[hard oasis]

Lets say dim(V) = n. If its dependent then we can deconstruct it into a linear combination of < n unit vectors, we can then map those unit vectors using the transformation and add them together in the image and get what we would have gotten if we has just mapped the original vector

[spiral pumice]

woah man ive been to 3 lin alg classes slow down 💀

what is dim(V) ?

[hard oasis]

The dimension of the vector space

[spiral pumice]

and you are saying there are n dimensions

[hard oasis]

What does it mean for a vector to be linearly independent vs dependent

[spiral pumice]

yeah honestly that explanation confused me more haha

okay so

if its dependent

[hard oasis]

I can simplify sorry

[spiral pumice]

you can make one of the rows up from the other rows

each column has a pivot

doesnt*

have a pivot

NOT EVERY COLUMN HAS A PIVOT

sorry

[hard oasis]

I think there's a slight misunderstanding here

[spiral pumice]

in my mind you are missing a vector that SHOULD be giving you a little more information about the matrix, so that you could map to every point in the vector space. this is because that last vector you needed to complete your collection gives you no new information

totally man ive been paying a tutor cause my professor is shit

i dont understand ANY of this 😭

[hard oasis]

Alright I'll give you a more concrete example

[spiral pumice]

THANK YOU

i like the concrete ones

[hard oasis]

Lets take the matrix $$\begin{pmatrix} 1 & 2 \ 2 & 4 \end{pmatrix}$$

[ocean seal]

Bean Man

[spiral pumice]

yes

thats dependent

because the first row * 2 is the second row

[hard oasis]

Alright so the idea behind a linear transformation T is that if you input a vector, it has these properties alright

T(ax) = aT(x)
T(x + y) = T(x) + T(y)

[spiral pumice]

the vector has those properties?

or the transofrmation

[hard oasis]

the transformation

[spiral pumice]

uhhh

could you show me a transformation?

like an actual one

arent they kind alike functiuons?

[hard oasis]

Linear transformations can be represented as matrices

[spiral pumice]

okay

[hard oasis]

Yea lets just consider a transformation from $\mathbb{R}^2 \rightarrow \mathbb{R}^2$

$$
\begin{pmatrix}
1 & 0 \
0 & 2
\end{pmatrix}
$$

[ocean seal]

Bean Man

[spiral pumice]

syes

that is the initial matrix?

or the transformation

[hard oasis]

that is the transformation

For all vectors $x \in \mathbb{R}^2$

It is the case that $T(ax) = aT(x)$ for all scalars $a$ , and $T(x + y) = T(x) + T(y)$ for all vectors $y$

[ocean seal]

Bean Man

[spiral pumice]

could you give me an actual example? i dont do well with the proof language

like if you had a vector
3
5

[hard oasis]

Gotcha

[spiral pumice]

the transformation of
3
5

would be the same as the two times the transformation of

1/2
5/2

right?

[hard oasis]

What this transformation does is simply multiply the second component by two

[spiral pumice]

what would it do to the vector

4
5

make it
4
10

?

[hard oasis]

We can check this easily by making
$$ x = \begin{pmatrix}

3 \ 5

\end{pmatrix}

Then we have that $$

\begin{pmatrix}
1 & 0 \ 0 & 2
\end{pmatrix} \cdot \begin{pmatrix}
a3 \ a5
\end{pmatrix} = \begin{pmatrix}
a3 \ a10
\end{pmatrix}
$$

Which is what we would get if we multiplied by $a$ after we did the transformation.

Oops

[spiral pumice]

whoops

haha

[hard oasis]

Thats better

[spiral pumice]

should the 2nd component in the vector say 2a5

or no

wait

[hard oasis]

it can be 2(a * 5) yea

wait

I'm slow

[ocean seal]

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

[hard oasis]

There we go

[spiral pumice]

i think that makes more sense

[hard oasis]

Can you see that T(ax) = aT(x) or no?

[spiral pumice]

so what was the purpose of having the a's there

why not just have 3 and 5

yeah i can see it i think

[hard oasis]

T(x) = (3, 10)
T(ax) = (a3, a10)
aT(x) = a(3, 10) = (a3, a10)

a is just a random scalar that we're multiplying x by

[spiral pumice]

yeah okay

makes sense

that was very well put that last message

okay so how does this show that dependence is maintained through linear transformation

[hard oasis]

Alright, we can also show that T(x + y) = T(x) + T(y), and you need both of these properties to see why it maintains dependence/independence

Do you need me to show that as well?

[spiral pumice]

yes please

i appreciate it man

that one isnt as intuitive for me

i think we covered that in calc tho

like deriving 2x + y is the same as deriving 2x then adding it to the derivative of y

[hard oasis]

Alright, we'll work with the same matrix for a concrete example.

let $x = (x_1, x_2)$
and let $y = (y_1, y_2)$
$T(x) = (x_1, 2x_2)$

$T(y) = (y_1, 2y_2)$

Then $T(x + y) = T(x_1 + y_1, x_2 + y_2) = (x_1 + y_1, 2(x_2 + y_2))$

$T(x) + T(y) = (x_1 + y_1, 2x_2 + 2y_2) = (x_1 + y_1, 2(x_2 + y_2))$

[ocean seal]

Bean Man

[hard oasis]

Does that make sense or do you need concrete numbers for x_1, x_2, y_1, y_2?

[spiral pumice]

can we do concrete numbers?

thanks man, its just a little abstract for me

[hard oasis]

Do you want to choose x and y

I'll just do $x = (1, 2), y = (3, 4)$

[ocean seal]

Bean Man

[spiral pumice]

that works for me

that linear transformation matrix looked like the one you used before, yes?

the
1 0
0 2

[hard oasis]

?

[spiral pumice]

where it doubles the second component

[hard oasis]

This is still using that same linear transformation if thats what you were asking

[spiral pumice]

here

[hard oasis]

Yea I'm using the same transformation

[spiral pumice]

yes okay just making sure

bet thanks

[hard oasis]

Then $T(x) = (1, 4)$

$T(y) = (3, 8)$

$T(x) + T(y) = (4, 12)$

$T(x + y) = T(1 + 3, 4 + 8) = T(4, 6) = (4, 12)$

So we can see that they are equal

[ocean seal]

Bean Man

[spiral pumice]

ohhh i think i can see it

you are adding up the components then transforming it?

[hard oasis]

yes

[spiral pumice]

and its the same as if you transformed the components then added it

[hard oasis]

exactly

[spiral pumice]

okay very well explained

sweet now onto the actual problem haha

[hard oasis]

Alright now we can show transformations maintain (in)dependence

It will have to get a bit abstract from here for you to fully understand the idea

[spiral pumice]

shit

okay

im ready! D:

[hard oasis]

We say that a set of vectors is linearly dependent if we can write one vector in the set as a linear combination of the others.

In our first example, we had the matrix $$ \begin{pmatrix} 1 & 2 \ 2 & 4 \end{pmatrix}$$

Which is the set of vectors (1, 2) and (2, 4). lets call the first one $x$ and the second $y$, then we can see that $2x -y = 0$ or $2x = y$

[spiral pumice]

yes

[ocean seal]

Bean Man

[spiral pumice]

understood

[hard oasis]

Alright do you have an idea of a basis or no?

[spiral pumice]

basis?

no

unfortunately

wiat

wait

arent basic vectors like

1 0 0
0 1 0
0 0 1

[hard oasis]

Yea

[spiral pumice]

just one unit in each direction, visually

im pretty sure

3 blue 1 brown!

[hard oasis]

Yea, although not all basis vectors are like that but yea thats the canonical basis for $\mathbb{R}^3$

[ocean seal]

Bean Man

[hard oasis]

So the basis tells us that we can represent all vectors in the vector space as a linear combination of the basis vectors

So in our case $x = \begin{pmatrix} 1 \ 0 $ \end{pmatrix} + 2\begin{pmatrix} 0 \ 1 \end{pmatrix}$ right

[ocean seal]

Bean Man

[spiral pumice]

yes yes that makes sense

like getting it into reduced row echelon form

then using scalars

[hard oasis]

Alright, now you should be able to see that via our previous properties of linear transformations, instead of transforming x, we can transform its decomposition into unit vectors and get the same thing right

and likewise for y

[spiral pumice]

wait wait

how would you represent the vector
2
1
or something like that with the basic vectors

[hard oasis]

this