help-27

[restive river]

Its like with chess

Don't give checks everytime u see one

[sharp dagger]

I think it's a good idea

How can I find out

[restive river]

Probably need some intuition, that's all

And outside of the box mentality

[sharp dagger]

Anyway, thank you

.close

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[burnt thunder]

Question 1: i need help to get the rest of the Euclidean division of P ( a polynomial in K[X] ) by (X-a)(X-b)

The rest should be written in function of a, b, P(a), and P(b).

<@&286206848099549185>

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[stone crater]

Why is the limit of proportionality 92 and not 80?

80 is where the line bends

[dim spindle]

all I can say is, that it must be false.
as the source states, the "Proportional limit" is the point at which no plastic deformation has occurred yet.
sciencedirect.com/topics/engineering/proportional-limit#:~:text=The proportional limit is the,returns to its original dimension.
In that case, you would be right to say, that the Proportional limit is at 80 mm

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[ashen quail]

how did they get to that result that I highlighted at the bottom?

im not sure what identity they used to get that

[wispy solstice]

did you read the first tip that is in red..?

[ashen quail]

i dont understand it

[wispy solstice]

what specifically do you not understand

[ashen quail]

how did they get the 1+

[wispy solstice]

like do you not see that x+xy = x(1+y)

?

[ashen quail]

is that just a method they use to get that 1+y?

cuz im just basing this stuff off a formula sheet but i didnt know you can just do that

[eager quiver]

develop sinx(1+cosx) and you will get the original sinx+sinxcosx

[wispy solstice]

it’s factoring

like when you factor 4x + 2 = 2(2x+1)

[ashen quail]

ohhhhhhhhh

man my teacher did not go over this

gave us a formula sheet and said gl lmfao

thanks guys

[eager quiver]

he doesn't have to thats some thing that can be used in any situation

[ashen quail]

he assumes a lot

compared to other teachers he doesnt really demonstrate examples like this

oh well

.close

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[knotty bolt]

ok

[quaint cove]

5

Tf

[knotty bolt]

will you close this yourself?

[zenith jacinth]

<@&268886789983436800>

[celest dock]

.close

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[lusty sapphire]

https://tenor.com/view/bbkeesha-bb10-troublemaker-keesha-smith-hes-a-troublemaker-gif-13574402

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[undone walrus]

okay.
Topic: tangents

given is the function f with f(x) = x^2 - 3x +2 with a graph K(no graph in the book btw)
Determine the equation of the tangent on K in point P( 4 | f(4) )

[undone walrus]

do I have to show a graph here? even though the task doesn't specify

solution:
tangent equation in PSF: y = 5(x-4)+6
tangent equation in main shape: y = 5x - 14

what do I even have to do?

[lusty sapphire]

The derivative will give you your slope

And you know one of the points the line will have

[undone walrus]

I made a mistake somewhere :/

[undone walrus]

^

[devout snow]

@undone walrus Has your question been resolved?

[warped relic]

Heyoo

Oki what I understand is

You have a function

And you want to find the tangent line at P(4, f(4))

Right?

@undone walrus

[undone walrus]

uh

[undone walrus]

^

Determine the equation of the tangent on K in point P( 4 | f(4))

@warped relic

[warped relic]

Graph K?

[undone walrus]

yea

[warped relic]

Hmm

I guess that's like "Figure 10" or something

Anyways

You have your f(x) = x^2 -3x +2

And you want your tangent line at x=4

Take the derivative of f(x)

[undone walrus]

$2x - 3$

[woven radish]

TheLegend27

[warped relic]

Oki

[undone walrus]

then I calculate the slope at x=4

[warped relic]

Now plug in x =4

Yep

[undone walrus]

$f´(4) = 2 * 4 - 3 = 5$

[woven radish]

TheLegend27
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[warped relic]

Oki

[undone walrus]

$y = m * x + b = 5 * x + b$

[woven radish]

TheLegend27

[warped relic]

Nope

[undone walrus]

?

[warped relic]

You use a formula

$y - y_1 = m(x - x_1)$

[woven radish]

VulcanOne

[undone walrus]

we didn't learn it that way

1 sec

[warped relic]

Or another way

$m = \frac{y-y_1}{x-x_1}$

[woven radish]

VulcanOne

[undone walrus]

this is how we did it

[warped relic]

Oooo

Well same thing

[undone walrus]

so, this is correct?

[warped relic]

You just find your b by plugging in x=4 and y = f(4)

[undone walrus]

?

[warped relic]

Yep

[undone walrus]

okay

so

then I did:

$P( 4 | f(4)) = f(4) = 4^2 - 3 * 4 +2 = 6$

[woven radish]

TheLegend27

TheLegend27

[warped relic]

Hmm

Why did you divide by 5 and 4?

[woven radish]

TheLegend27

[undone walrus]

to isolate b

[undone walrus]

we also did that here

[warped relic]

5(4) + b = 6

[undone walrus]

u get -1

[warped relic]

20 + b = 6

| -20

[undone walrus]

wtf

hold on

thats correct

but wait 😄

[undone walrus]

why do this

when the task is:

[woven radish]

TheLegend27

[warped relic]

Same thing

[undone walrus]

no?

[warped relic]

6 = 5(4) + b

6 = 20 + b

| -20

[undone walrus]

why bracket 4

[warped relic]

Because it's multiplication

It's another way to write 5 is multiplied by 4

[undone walrus]

OOOOOOH

[undone walrus]

this was also done -->
0.5 * 1 = 0.5 | /-0,5 = -1

now I see my mistake

okay okay

thank you :))))

[warped relic]

Glad to help

[undone walrus]

but how do I know

$6 = 20 + b$

[woven radish]

TheLegend27

[undone walrus]

if I have to do -20 or (divide)20

[warped relic]

Because you're isolating b

And 20 isn't multiplied by b

It's added to b

[undone walrus]

when would I need to use division?

[warped relic]

When 20 gets multiplied by b

3x + 5 = 10

Solve for x

[undone walrus]

-5
3x = 5

[warped relic]

Okay

Now?

[undone walrus]

oh

xd

uhh

| /3 because 3 * x
x

x = 5/3

[warped relic]

Yep now let's get back to our 6=5(4) + b

Solve for b

[undone walrus]

6 = 5(4) + b
6 = 20 + b | - 20, because 20 PLUS b
-14 = b

U are a life savior

[warped relic]

Please get your own help channel

I'll check it out but please wait

[undone walrus]

.close

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[warped relic]

I think it is defining the delta epsilon

[long kettle]

Basically, it's saying that, if you give me a number, no matter how small (ε), I can pick an interval around x = c (δ) where every single point in that interval (including at c) is within ε units from f(c)

This is identical to the definition of a limit, except we also look at x = c, while in limits, we don't

This manifests in the continuity definition saying "|x - c| < δ" while the definition of a limit says "0 < |x - c| < δ"

[devout snow]

@reef tendon Has your question been resolved?

[long kettle]

Do you intend on speaking or are you gonna leave the channel open without saying anything

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[warped widget]

how do i find D coordinates

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@warped widget Has your question been resolved?

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[quasi yoke]

hi!

[craggy dagger]

Solve the homogeneous one and guess for a particular solution

[quasi yoke]

yes I have gotten to particular solution

x = t^2/4 -t/8 -1/8

do I just find general solution after this and solve?

[warped relic]

Yep

You add up your complementary and particular solutions

[quasi yoke]

got it! tysm!

[warped relic]

[quasi yoke]

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[vivid cliff]

sup my little cupids

[vivid cliff]

guide me for i and ii

roots

where is my little white pie

[quasi yoke]

why are you everywhere 👻

first one u can expand (z-a)(z-b)(z-y)

pretend the a b and y are the Greek symbols

[vivid cliff]

alr

[vivid cliff]

lmao

ii?

[quasi yoke]

maybe u can use cosine rule and interpret through graphically

[brisk totem]

sub the identity into the expression then expand

the result is i think immediate

[vivid cliff]

uh

alr one min

[quasi yoke]

ohh yes u can do |w-a|^2 = (w-a)* (w*-a*) = |w|^2 - aw* - a*w + |a|^2

and use a+b+c= 0

and similarly for a* + b* + c*

[vivid cliff]

bro how did you see the cosine graph?

ahh wait

nvmnvm i see it

alr cool cool

[quasi yoke]

yaey

[vivid cliff]

missing my little white pie rn

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@vivid cliff Has your question been resolved?

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[marsh agate]

Hey guys, can you help me, i dont understand with half bottom of this proof, i dont understand how can b defined out of nowhere

[craggy dagger]

the motivation is that it disagrees with each number in at least one digit

[marsh agate]

What if b is a repeating number, since a1,a2,... can be anything?

I mean how can we be sure that b != an

[long kettle]

How would b = a_n given the definition?

[marsh agate]

Maybe, Since a_n can be irrational, there could be one that has the same value as b

[long kettle]

Consider if they had the same value

Then that means, when a_n has a 5 in the decimal place, b does too

But, by definition, when a_n has a 5, b has a 4

[craggy dagger]

b doesn't have to be irrational

[long kettle]

And if a_n has something other than 5 in the decimal place, b must have that same number, which is different from 5

But when a_n has something other than 5, b has 5 by definition

No matter what, it's different

[marsh agate]

Oh yeah, it makes sense now, thx guys

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[ember palm]

What's the difference between a collinear vector and a parallel vector?

[ember palm]

[long kettle]

None

[ember palm]

Wut da phuk?

[long kettle]

Different names for the same thing

[ember palm]

Then why "show that these two vectors and collinear" Can't be proved it I prove it through parallel

By having their vector product 0

[long kettle]

You can

[ember palm]

They cancelled the answer tho

[long kettle]

Can I see?

[ember palm]

Na

Coz I recall it

[long kettle]

:(

[ember palm]

Ain't got it written

[long kettle]

Either the teachers fault or you made a mistake somewhere

[ember palm]

I remember they made it prove through

|vector ab|+|vector bc|=|vector ac|

[long kettle]

I guess that works too

[ember palm]

O

Oki

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[finite inlet]

If the sequence in a series does not converge to 0, then the series diverges. Does that mean a series never converges to a non-zero value?

[restive river]

pls help this is timed homework

[finite inlet]

[restive river]

u got here first sorry i leave

[finite inlet]

[low holly]

do you mean the value of the series itself, or its terms?

your first sentence is correct

your second statement is incorrect (e.g. geometric series)

[finite inlet]

oh

hm

[low holly]

i'll give you an example to make it clear maybe

$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = \frac{1}{1-\frac{1}{2}} = 2$

[woven radish]

tushar

[finite inlet]

I wonder because if the series can only converge if the sequence converges to 0 then I thought the series must also converge to 0, but I guess the first few terms can give it a non-zero value anyway?

[finite inlet]

ah yeah I see

[low holly]

exactly, or even all the terms

since converging to zero doesn't mean that any term has to take on the value of zero

[finite inlet]

ahhh okay

[low holly]

every term could still be contributing a varying positive amount to the sum

[finite inlet]

thank you 👍

[low holly]

happy to help

[finite inlet]

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[restive river]

I'm not sure how to simplify this properly

could someone assist me

do you know https://www.symbolab.com/

could help

[stark anvil]

If a number is inside the root for example √45 just do the prime factorization of 45 and multiply the number outside of the root that is repeated twice

For exanple 2√45
45 = 5 × 3 × 3

2√5×3×3

2×3√5

6√5

[restive river]

oh ok

[restive river]

alr ty

[restive river]

what happens to the other x3

[stark anvil]

How do you define a number inside the root for example √9

Square root of √9 is 3 right

[restive river]

yes

[stark anvil]

Similiarly

√45 can be written as √9 × √5

So √9 becomes 3

3√5

Since we already have a 2 outside we multiply it with 3

To get 6√5

[restive river]

ah yes

right

alright ty

what about the denominator

how would u simplify √2025

.close

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[restive river]

<@&286206848099549185>

[knotty bolt]

check #❓how-to-get-help

also !15m

!15m

[devout snow]

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.

[restive river]

okayyyy

[supple knot]

.close

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[fair crescent]

lim
x---> infinity (2x+5/2x-1)^3x+4

[fair crescent]

find the first derivative

[narrow plank]

$\lim_{x \to \infty} \left( \frac{2x+5}{2x-1} \right)^{3x + 4}$?

[woven radish]

Shenmaster

[fair crescent]

i considered the entire limit as y and did the logarithmic

[narrow plank]

That limit will turn out to be a number, so its derivative will be 0

[fair crescent]

oh wait, not the derivative

we're supposed to evaluate

[fair crescent]

lny = lim x---> infinity 3x+4 * ln(2x+5/2x-1)

= lim x---> infinity (2/2x-5 - 2/2x+1) /-3(3x+1)^-2

i don't know what to do after this point

[restive river]

oooooooooo

this is seems like it could be an interesting application of the limit definition of e

[restive river]

hmmm @supple knot what do u think

so
[
\lim_{x \to \infty} (\frac{2x+5}{2x-1})^{3x+4}
]
[
\to \lim_{x \to \infty} (1 + \frac{6}{2x-1})^{3x+4}
]

[woven radish]

♡LexQa♡

[fair crescent]

what did you do here?

[restive river]

okay i am right lol

i see

there is no need for l'hopital

[fair crescent]

how come

[timber patio]

You could also do this

[restive river]

do u know the limit definition of e

[fair crescent]

no

[restive river]

okay

do you know this limit?
[
\lim_{n \to \infty} (1 + \frac{x}{n})^n = e^x
]

[woven radish]

♡LexQa♡

[restive river]

its a very popular limit

and how we came to find what e is

[supple knot]

just take the log then find the limit of that

[restive river]

i mean limit definition of e works too but 😵‍💫

[fair crescent]

i dont know that either

this is my second time approaching limits since high school

[timber patio]

Do you know l'hospital?

[fair crescent]

yes

[timber patio]

Try rewriting it like this

[restive river]

anyways from this,
[
\lim_{x \to \infty} (\frac{2x+5}{2x-1})^{3x+4}
]
[
\to \lim_{x \to \infty} (1 + \frac{6}{2x-1})^{3x+4}
]
[
\to \lim_{x \to \infty} (1 + \frac{3}{x-\frac{1}{2}})^{3x+4}
]

[woven radish]

♡LexQa♡

[fair crescent]

wheres the limit?

[fair crescent]

how did we get 1+6

[timber patio]

I didn't write it because i was lazy lol, its supposed to be there

[fair crescent]

so y = lim e^lny ---> (2x+5/2x-1) = e^(3x+4) ln(2x+5/2x-1)?

[main gull]

Fyi if you do \left and \right before each parentheses, it makes it bigger

[timber patio]

Theres probably an easier way to solve it but this is how I did

[restive river]

just this really

[timber patio]

Oops this was wrong

[restive river]

oh like

[fair crescent]

i dont know how to get to this point

[restive river]

[
\left(x \right)
]

[woven radish]

♡LexQa♡

[timber patio]

I've just never seen that before that why i showed this solution

We werent taught that limit in school

[fair crescent]

whats your solution in the end?

like what is the limit nearing to

[timber patio]

1

[restive river]

no lol

[fair crescent]

no

[restive river]

it is e^9

[fair crescent]

^

but

[timber patio]

Really?

[fair crescent]

i dont know how

[main gull]

anyways from this,
[
\lim_{x \to \infty} \left(\frac{2x+5}{2x-1}\right)^{3x+4}
]
[
\to \lim_{x \to \infty} \left(1 + \frac{6}{2x-1}\right)^{3x+4}
]
[
\to \lim_{x \to \infty} \left(1 + \frac{3}{x-\frac{1}{2}}\right)^{3x+4}
]

[woven radish]

dldh06

[restive river]

oh thanks lots, never knew of it!

[timber patio]

I forgot a +1 in my solution and it all went to hell

[restive river]

this simplifies to (e^3)^3 more or less, but explaining that is going to take so long, u might as well just live with someone else explaining it with l'hopital

[fair crescent]

nobody explained it to me with lhopital

and everytime i asked someone i got a different method

sigh

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[timber patio]

@fair crescent
Do you know how to solve this limit with l'hopital?

[fair crescent]

thats where i got to

lim x---> infinity (2/2x-5 - 2/2x+1) /-3(3x+1)^-2

[timber patio]

Well if you write it like this it becomes 0/0 so you can use l'hopital

[fair crescent]

yes

[timber patio]

You should be able to solve it from there

[fair crescent]

after that i use lhopital and get this

then idk what to do

[timber patio]

You must have calculated something wrong

Because if you take the limit of that its equal to 9 so you get as a solution e^9 which i think you said was the right answer

[fair crescent]

that is equal to 9 but remember the limit is lny

[timber patio]

No the limit is e^y

[fair crescent]

if you did it this way then limit is gonna be lny

cus u did ln of both sides

[fair crescent]

lny= all this (which you said was 9)

y = e^9

[timber patio]

What i did was rewrite whats inside the limit like this

And then set everything thats on e exponent as y

[fair crescent]

ok i got it thank you

[timber patio]

Np

. close

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[wooden island]

i’m not sure where i went wrong

[hybrid elk]

u wrote $$x+4\cdot x^2$$ to equal $$x+4+x^2$$

[woven radish]

light

[hybrid elk]

you can further reduce the top and bottom by x

[wooden island]

i thought when they multiply it’s adding?

[hybrid elk]

$$\frac{x+4}{2x}\cdot\frac{x^2}{3(x-8)}$$

[woven radish]

light

[hybrid elk]

thats essentially what u have

[hybrid elk]

no, its multiply

$$\frac{a}{b}\cdot\frac{c}{d}=\frac{a\cdot c}{b\cdot d}$$

[woven radish]

light

[wooden island]

so how do i get to the correct answer from this?

[hybrid elk]

x^2 is just (x)(x)

$$\frac{x+4}{2\cancel{x}}\cdot\frac{(x)\cancel{(x)}}{3(x-8)}=\frac{x+4}{2}\cdot\frac{x}{3(x-8)}$$

[woven radish]

light

[hybrid elk]

then multiply the numerators together and then multiply the denominators together

[wooden island]

so combine the like terms (x) to get it squared?

[hybrid elk]

yeah you would get x^2 + 4x for the numerator

[wooden island]

wait how do i get that if there are only two x on the numerators?

[hybrid elk]

wdym?

[wooden island]

how do i get x^2 + 4x with x + 4 * x

[hybrid elk]

$$x\cdot x + 4\cdot x$$

[woven radish]

light

[wooden island]

ah i see

thank you

.end

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[verbal robin]

Help me solve this! Please explain as well as you can, and be patient, as I am stupid sometimes

[brazen parcel]

which part do you need help in?

@verbal robin

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@verbal robin Has your question been resolved?

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[urban marsh]

hi

[urban marsh]

can someone help me with a discord server?

[winter hound]

This is a math helping server.

If you are asking for help in non-math related, please close this channel.

[urban marsh]

do you have server that helps with discord server

[winter hound]

No

[urban marsh]

im wait for someone

[winter hound]

Close this channel now

Before I’ll cal a mod to handle you

@urban marsh

[urban marsh]

oof

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[quasi yoke]

hi!

oh wait oops

[devout snow]

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[wet fox]

Can someone help me with this question

[devout snow]

@wet fox Has your question been resolved?

[cobalt mauve]

Okey, so we need to construct a matrix of linear transformation such that goes from $e_3$ to $e_3$, so $[P]_{e_3}^{e_3}$, right?

[woven radish]

marejak023

[zenith jacinth]

thats not what it is asking tho

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[cobalt mauve]

wait what is it the? i thought that they need the matrix that will do transform

[devout snow]

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[zenith jacinth]

its asking for the expression of the matrix in the basis {e_i} (1<= i <= 3)

[zenith jacinth]

@cobalt mauve

[real grail]

$P*e_1 = e_3 \eq \matr[b]{a&b&c\d&e&f\g&h&i}* \matr[b]{1\0\0} = \matr[b]{0\0\1}$

[woven radish]

Mehdi_Moulati

[zenith jacinth]

you dont even need to go that far

[real grail]

i know

but just to give him an idea

[zenith jacinth]

[restive river]

.close

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[restive river]

saved u the time

[devout snow]

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[waxen stump]

doing some desmos art, is there any funtion that does that s like shape in between me two lines htat i already made

[low holly]

maybe a logistic?

[waxen stump]

ahhhh thnx ill take a look

[low holly]

https://www.desmos.com/calculator/3c1pclfbwq

try playing around with that and restricting the domain

[waxen stump]

mhm yea just what i was looking for thnx

how do i close chat?

[low holly]

.close

[waxen stump]

ight thnx

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[boreal sage]

I need help with this question

[boreal sage]

i need to simplify it but i got no clue how to start

tried to use de morgans law first with the double complement law but i just got more confused

the only i thought of doing it is like this

but am not sure if thats correct

[rigid terrace]

wait

im gonna try to solve it now

the answer is 1

• is union

• is intersection

you use the de morgans law first

and then you see that we have A union not A

which is 1

and we also have C union not C

which is also 1

1 union with any other element is 1

[boreal sage]

the answer in the book says U

am guessing its the same as 1

_

?

[rigid terrace]

hm

yeah i think yes

I did the exercise like its a Boolean algebra exercise

you did it the Math way

but its the same thing

only different symbols

[boreal sage]

maybe am not sure

[rigid terrace]

how is that called in your book?

which school lesson is that?

and which subject is that?

[boreal sage]

discreet mathematics simplifying sets

[rigid terrace]

and the answer is "U" ?

only "U" ?

[boreal sage]

tep

yep

legit nothing else

[rigid terrace]

what will happen if we have A U not A?

what does the book say?

[boreal sage]

that one of the inverse laws

[rigid terrace]

yeah

[boreal sage]

A U not A = U

[rigid terrace]

yeah

thats correct

the answer is U

U in math is the same as 1 in Boolean algebra

[boreal sage]

yea no ik your answer is right

but the examples they showed before in the book

[rigid terrace]

are bad?

[boreal sage]

they use a bunch of other laws

to simplify it more and more

[rigid terrace]

do you need help with any other exercises?

[boreal sage]

no i think am good now, ill look into it alone

thx for the help

[rigid terrace]

okay np

[boreal sage]

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[small raptor]

https://gyazo.com/a4c31ee76e9b24fe54275b783e50e8ea

[small raptor]

i swear its impossible to put it in the form alog(7)

u cannot simplify 7log7 - 7log8 to alog7

[topaz beacon]

7log7 is of the form alog7

-7log(8) needs to be put in the form blog2

[small raptor]

hm

made a mistake should be +7log8

so it would be 7log(2^3)

which is 21log(2)

[topaz beacon]

not necessarily

i havent actually done the calculation but it say b is an integer

could be negative

[small raptor]

no no

ive integrated it

i just typed it wrong

But yea

thanks didnt see that

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[coarse current]

For all complex numbers z, z=a+bi is such that z is an element of complex numbers, a is Re(x), b is Im(x)

[coarse current]

How do I write that in set notation?

[hollow pollen]

would you like to write the set of complex numbers?

[low holly]

what set do you want to write?

[tropic kindle]

ℂ

[low holly]

what is x?

[torn wave]

i mean its an exponent of i and itself is real, but bi is imaginary as a whole

[tropic kindle]

the problem is weird, i always use Re z and Im z
cause, how can Im be Im(x)

[torn wave]

yeah, maybe x is a new variable?

[devout snow]

@coarse current Has your question been resolved?

[tropic kindle]

Z∈C
idk

z∈ℂ <-> [∀z=a+ib ∃a=Re & ∃b=Im]

i dont rly understand what do you want

it should be ∈ before Re and Im, instead of =

[worn sphinx]

If $x\in\bC$ is fixed, then

$${z=a+bi\in\bC:a=\operatorname{Re}(x)\land b=\operatorname{Im}(x)}={x}$$

lol

Whoops typo

[woven radish]

Lorago

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[mystic lava]

How to find the limit of (1-e^x)/(ln(1-x)) [x→0]? Without using L'hôpital's rule

[warped relic]

$\lim_{x \rightarrow 0} \frac{1 - e^x}{\ln(1-x)}$

[woven radish]

VulcanOne

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@mystic lava Has your question been resolved?

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[wanton harness]

how do I find the two points that can pass a line with the slope of 5/2

[woven radish]

FinnaganFox

$y=\frac{5}{2}x$

[hearty gyro]

you can plug in any number for the x, and the y that comes out with it will be a point on that line.

for example, if you input 0, and get out 0, a point on the line is (0, 0)

[wanton harness]

oh ok tysm

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[onyx basin]

Hello

[onyx basin]

I need help learning how to divide a function into parts

from modulus

Example ^

Can someone explain?

[gentle estuary]

Consider different intervals for values of x

[onyx basin]

wdym

[gentle estuary]

What happens when x = 0? When x > 0 but < 1 (0 < x < 1)? And so on...

You're trying to split the function into cases like the second image, yes?

[onyx basin]

yes

[gentle estuary]

The interval I mention here should help

Think about what happens when x > 1 and when x < 0, the function behaves simply then

[onyx basin]

What about the sign

[gentle estuary]

Sign? I just mean play around with x values in certain intervals until you get a sound idea of how it behaves

[onyx basin]

Like drawing the function you mean

[gentle estuary]

You can get a better idea of the graph too, sure

But, I purely mean putting in x into f(x) and looking at value of the different expressions in f(x)

Like, x = 1/2 => f(x) = 0.5 * e^-|-1/2| = 0.5 * e^-0.5

[onyx basin]

I still have doubts on how this works

is it

[gentle estuary]

So the inside of the exponent is negative here. For what values might it be positive?

[onyx basin]

• | -1 - 1 | or | -(-1-1) |

[gentle estuary]

If x = -1, then it's -|-1 - 1|

[onyx basin]

ok ty

will try

[onyx basin]

1

only for 1

right?

[devout snow]

@onyx basin Has your question been resolved?

[gentle estuary]

I mean the just expression x - 1. But a more useful approach to this would be to consider the values -|x-1| can take when x is between 0 and 1; what values it can take when x < 0 and when x > 1. So, look at the intervals [0, 1], (-infinity, 0), and (1, infinity).

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[modern wind]

how do i find the domain of F(x)?

[wooden zodiac]

Haven't done questions like that, maybe solve the definite integral and find the domain afterwards?

[modern wind]

i just cant seem to grasp their restriction on [0,1[

[versed heron]

you need to respect to the intervals(a,b)

[modern wind]

the part where x need to be bigger than 4 i can get

[modern wind]

the what

[wooden zodiac]

[X,4x]

[versed heron]

oh you not from the uk

yeah

so we call them a,b but they given the vaules x and 4x

[modern wind]

yeah but how does that give back that domain

because

unless i choose and interval which contains 4 everythings good

i have an asymptote on 4

[versed heron]

so definite integrals, the domain of the integral can be found by visually inspecting the lower and upper bounds applied on the integral

[modern wind]

i can definitely tell the domain by just solving the integral and then calculating the domain afterwards

i thought there would be a way of calculating it without going through that

[versed heron]

think its safest to do with integrals

[modern wind]

it works for all?

[versed heron]

im not sure ill be honest with you

[devout snow]

@modern wind Has your question been resolved?

[devout snow]

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[restive river]

y[x]=y'[x-1]

[restive river]

A bullshit solution I found is just the lim of n to infinity of f(nx) where f is just any normal trig function

Other than that I have no idea

Cant think of any function (that doesnt involve infinity anywhere) where its equal to its derivative from one second ago

Generalizing it definitely doesnt help

I probably need a starting value or something

[stone stump]

well most differential equations need a starting value

[restive river]

y(0)=y'(0)=0?

Interval from 0 to 2?

[stone stump]

is there more context or is this just something you came up with?

there are delay differential equations but they are instead of the form y'(x) = y(x-1)

I guess maybe you could try running the time backwards

[restive river]

Tinkering around

[restive river]

Sounds interesting

Do you have any online resource for delay differential equations?

[stone stump]

no sorry

[outer marten]

this doesnt make sense do you mean y' evaluated at x-1? what are you differentiating wrt?

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@restive river Has your question been resolved?

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[placid fjord]

How do you prove this

[dark dawn]

hmm there are a couple ways to get there

[restive river]

equivalently, $x^2 - 2x + 1 \geq 0$

[woven radish]

Ø

[dark dawn]

but I would suggest trying to get a term that is always geq or the likes

[midnight dirge]

which can be rewritten as something ^2

[placid fjord]

ok let me work it out

[restive river]

well the question boils down to, does multiplying two numbers yield higher or lower value than addition ?

[placid fjord]

intuitively it makes sense I just can't seem to formally explain it

[restive river]

Ah, ok

[placid fjord]

and I get (x-1)^2 but I don't know where to go from there

[dark dawn]

👍

try plugging in 0,-1,1,2 see if you notice anything about the result you get

[restive river]

Wouldn't it be equal to 2x for those values then ?

[dark dawn]

the current setup is (x-1)^2 >= 0

[restive river]

How about contradiction proof, I feel it can be helpful

[dark dawn]

sure but this is quite literary the proof '(x-1)^2 >= 0'

which is true

[restive river]

$\forall x \in \mbb{R} : x²+1 ≥ 2x \
\text{negating the proposition} \
\exists x²+1 ≤ 2x$

[dark dawn]

you can also say x^2 + 1 -2x + 2x >= 2x

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

[placid fjord]

So maybe I should ask how you would formally write the proof

[restive river]

I would definitely go with contradiction, which is what I am trying to think of it since it is a lot for formal ig

Ah, I can prove a smaller version of it to build upon it

$\forall x \in \mbb{R} , x ≥ 2 : x² ≥ 2x$

[restive river]

I can definitely further build on this if I am able to remove x≥ 2 and add +1

[placid fjord]

right

[restive river]

@placid fjord is this proposition formally provable for u ?

[placid fjord]

Yeah

well im not too sure could I have feedback if I send a proof?

[restive river]

Sure

[restive river]

Yep

[placid fjord]

okay now im getting a good idea on how to do the original problem

[stone stump]

where are you guys going with this. why are you abandoning the (x-1)^2 thing

[restive river]

Idk, it just didn't let us much further ig ? Maybe we can proof using that too, but we couldn't think about so we tried many different ways as well as simplifying proof

[stone stump]

well you don't need to go much further

it's only one step more

[restive river]

Oh ok

[stone stump]

what do we know about square numbers with respect to their sign

[low holly]

you can (probably) take for granted that the square of any real number is nonnegative

[restive river]

They are always positive ?

[stone stump]

yes

well non-negative

but you mean the correct thing

so what does this mean for the inequality (x-1)^2 >= 0

[restive river]

AH OK!!

Yeah!

[placid fjord]

ah so it will always be non-negative

[restive river]

Makes sense now ^=^

[placid fjord]

I mean I also wrote this

would this be an acceptable proof?

[low holly]

uhh

[stone stump]

it's fine

[low holly]

why not start from: if $x$ is a real number, $x-1$ is a real number. the square of any real number is nonnegative, thus $(x-1)^2 \geq 0$.

[stone stump]

but implicitly uses more knowledge about quadratics

[scarlet sequoia]

increases only for x > 1

[stone stump]

namely that if a quadratic has a single root and positive sign then it also has the minimum there

[low holly]

positive leading coefficient

[woven radish]

tushar

[placid fjord]

ok so maybe this is a bit cleaner then:

[stone stump]

yes

[placid fjord]

nice thank you guys

[restive river]

I didn't finish it lol

It got confusing with too much text

And also it has ALOT of errors and typos

[stone stump]

not sure what you are trying to do. definitely overcomplicating stuff

[restive river]

Yeah I will write it better

[placid fjord]

how do I mark as solved?

[wooden zodiac]

QED, or write a square.

[placid fjord]

I meant the channel lol

[wooden zodiac]

.close

[devout snow]

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[wicked rover]

(x-1)^2>=0 and "(x-1)^2 is nonnegative" say the same thing

[placid fjord]

ok so I don't even have to say that nonnegative part then

[wicked rover]

a>=0 and "a is nonnegative" are different ways to write the same thing

[devout snow]

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[peak geode]

do i use integration by part to differentiate this

[supple knot]

just hand wave and differentiate under the integral

[magic thicket]

Grandpa was fun while it lasted I guess

Also yellow on the pfp looks weird. I got so used to the green, I preferred it

[hollow pollen]

yellow is hgigher social status

[supple knot]

maybe i'll go back to green in the future. but credit to @copper flower for the change. i'm too old to learn how to do such fancy photo editing

[hollow pollen]

or discord status? idk

[restive river]

i agree with mateo713

[wooden zodiac]

Guys, help Melina instead of having a coffee conservation 🤓

[wooden zodiac]

U have to take the derivative of bestelling?

Bessel*

[peak geode]

so i did and

the result was completely different

[supple knot]

can you show both your attempt and answer?

[peak geode]

yeah integral representation of generating bessel

wait a min please

did they differentiate with respect to t ?

[supple knot]

can you show the whole page

[supple knot]

i'm using this but i'm not getting their answer

$f(x, t) = \cos(nt - x\sin(t))$

[woven radish]

riemann

[supple knot]

so $\frac{\pa}{\pa x} f(x,t) = -\sin(nt - x\sin(t)) \sin(t)$

[woven radish]

riemann

[peak geode]

still different from the answer

[supple knot]

yea i can see that

[peak geode]

i asked the professor about this one before

and he said its done using integration by parts

so maybe the same for this one

[supple knot]

try it with the product of these two functions

[peak geode]

but they're not related

[supple knot]

what do you mean?

you can try to do integration by parts on anything that has a product

[peak geode]

oh im sorry i didn't know what you were referring to

[peak geode]

which one to take u and v'

thats what confuses me

[supple knot]

v' should be the simpler one

[peak geode]

but i will have to work only with -xsin(t)

[supple knot]

v' = dv/dt in this case

[peak geode]

not d/dx ?

eventhough j is a function of x?

[supple knot]

nope. you're doing integration with respect to t

[peak geode]

oh wait

but were asked to differentiate

[supple knot]

the problem involves multiple steps

in order to get their answer, you must first differentiate with respect to x, then you must integrate by parts with respect to t

[supple knot]

if you just plugged this in, you'd have an equivalent expression. but they explicitly want a different one shown, so you do IBP to get to it

[peak geode]

alright i will try

[devout snow]

@peak geode Has your question been resolved?

[supple knot]

Evaluate at the endpoints

[peak geode]

the first term goes to 0

so we're left with the integral

[supple knot]

You lost a minus sign from du

Oh did it cancel the minus sign from IBP

[peak geode]

yeah

we were asked to differentiate but

were going to do it twice

[supple knot]

Do what twice

[peak geode]

this first

then now i differentiate the integral

[supple knot]

You're just plugging this into j'(x)

Then IBP

[peak geode]

yeah and then ?

[supple knot]

You lost an x here

[supple knot]

Compare yours to the image. You're almost there

[peak geode]

yeah but that was a differentiation

a lost x ?

[supple knot]

Fixed

[supple knot]

Chain rule. Your answer is different by an x

[peak geode]

yes i see it now

but i have a minus here

oh no its right

i finally got the result

Thank you so much !

[devout snow]

@peak geode Has your question been resolved?

[devout snow]

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[supple knot]

Where does it say characteristic equation?

[radiant drift]

Above it

I didn’t ss just

[supple knot]

Show the full page

[radiant drift]

I’m taking a bath rn, can’t. Just assume it said it ig

Oh this is just sum and angle identity

weird

[devout snow]

@restive river Has your question been resolved?

[supple knot]

It's the sum identity for tangent

[devout snow]

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[devout snow]

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[split radish]

hello, what happened to the cosine square on the right? I don't see it 🥺

[nova ibex]

they've just factored out -cos(x)

^

you can check this by expanding the bottom

it'll be the same as the top

[devout snow]

@split radish Has your question been resolved?

[split radish]

thank you so much! I'll make sure to do that next time c:

[devout snow]

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[winter patrol]

what have you tried

looks good,
that's part a) done

[spark smelt]

Each chair 5/4 dollar each table 1.25

The numbers were terrible

[devout snow]

@tidal notch Has your question been resolved?

[devout snow]

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[heavy terrace]

Context: BO and CO are bisectors of their corresponding angles. r - radius of the inscribed circle. The trapezoid is isosceles.
My question is how does segment QC become b/2 and QB a/2

[heavy terrace]

oh also angle BOC is 90 degrees

[novel dome]

this is an ellips right?

[heavy terrace]

wdym?

[novel dome]

if you're looking at this image as a 2dimensional image then the circle is just an ellips

it's kinda confusing my head

[heavy terrace]

nothing in the picture is correct

this is paint

[novel dome]

oh ok

[heavy terrace]

the image is just an example

please ping when you send a message, I will be watching YT

[novel dome]

ok

[stuck cipher]

Search up power of a point

Anyways, it's just that when you have a dot, and two tangent line segments to the circle which also pass through the dot, they're equal

It's easy to prove by just noting that a tangent is always 90 degrees to center, and so you construct a right triangle

[heavy terrace]

[stuck cipher]

Hope that helps

[heavy terrace]

yuck

I remember there was something of that nature in pervious topics about circle

I think two grades ago

anyway gotta go

.close

[devout snow]

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[polar bolt]

So solve the system of equations?

[void hemlock]

I'm guessing you want to know how much each box weighs?

let's say a large box weighs a kg and a small box weighs b kg.
Then 2a +3b = 77 and 4a+8b = 181

you have alot of options to find a and b

Substitution for example

[devout snow]

@tidal notch Has your question been resolved?

[devout snow]

@tidal notch Has your question been resolved?

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[coarse fractal]

Need help with q 3c

[coarse fractal]

Q3c^

If one could explain what they done near the end by which they split the integral into separate fractions

Surely there’s other ways to tackle this

[mellow panther]

That's splitting into partial fractions

First thing is writing u^2 = u^2 - 4 + 4

$\frac{u^2}{u^2 - 4} = \frac{u^2 - 4 + 4}{u^2 - 4} = 1 + \frac{4}{u^2 - 4}$

[woven radish]

numbpy

[coarse fractal]

Dayum, I wouldn’t think of doing it that way cl

[coarse fractal]

How’d you type that so fast btw 😂

Tysm, lemme work through it once more so I can get a better gidt

Also another question

[mellow panther]

then you can write 4 = (u + 2) - (u - 2) and do similar as above

[coarse fractal]

In this question type, they didn’t specify what to use for the substitution, in this case would you substitute for u what looks the most tricky

[mellow panther]

Substitutions are almost always tricky so you can only get better with practise

[coarse fractal]

4=?

[mellow panther]

$\frac{4}{u^2 - 4} = \frac{(u + 2) - (u - 2)}{u^2 - 4} = \frac{u + 2}{u^2 - 4} - \frac{u - 2}{u^2 - 4}$

[wooden veldt]

Last equality is incorrect

[mellow panther]

My bad, typing mistake