help-39

[brisk nacelle]

Yes

So its right?

[cloud zephyr]

because 15≠30

[brisk nacelle]

I wrote no solution

[cloud zephyr]

Correct

[brisk nacelle]

Okay

.close

[pearl pond]

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[brisk nacelle]

Is this right?

???

[pearl pond]

@brisk nacelle Has your question been resolved?

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[plucky cove]

I need help figuring out each one

[pearl pond]

Please don't occupy multiple help channels.

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[plucky cove]

I need help solving each problem

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[gusty prism]

How do u know when to use these equations really

They said it is all for constant acceleration but didn't rlly say when to use which

I guess if i do practice ill realize but

For the constant acceleration motion

[sharp vigil]

all of them have a subset of all the variables, so you pick one based on which ones you have and which one you need to solve for

[pearl pond]

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[gusty prism]

perfect

thanks

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[toxic tulip]

Hello I need finding the mistake in a proof

[iron stream]

Bro even copied the logo lol !!

[toxic tulip]

Let 𝐴, 𝐵 ⊂ 𝐶. If 𝑝 is a limit point of 𝐴∪𝐵, then 𝑝 is a limit point of 𝐴or 𝐵. Proof Suppose p is a limit point of A union B. If I is an open interval containing p, then by definition of limit point, I intersect (A union B) minus {p} is nonempty. Therefore, it contains a point x. Then x is in I and x is in A union B, and x is not equal to p. Since x is in A union B, either x is in A or x is in B. In case 1, x is in A. Then x is in I intersect A minus {p}, so I intersect A minus {p} is not empty. Therefore, p is a limit point of A. Similarly, if x is in B, p is a limit point of B.

and yeah I made this account in Middle School it was dark times

[pearl pond]

@toxic tulip Has your question been resolved?

[tropic saddle]

for each interval I you get a different x

sometimes that x may be in A and sometimes it may be in B

but it would always need to be in the same one

[toxic tulip]

could you maybe elaborate a little bit

[tropic saddle]

say p was a limit point of A. then x would always have to be in A

but you have only shown that it is sometimes in A or sometimes in B

[toxic tulip]

I see what your saying since x is in A union B it could be in either A or B regardless of where the limit point is

How could this be accounted for in the proof?

[tropic saddle]

tbh not sure right now

[toxic tulip]

I think I have enough to go off now thank you

.clsoe

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[brazen swift]

https://i.imgur.com/rBYO4VM.png

[brazen swift]

The question asks to prove this identity

i mean its just cos2x = cos^2(x)-sin^2(x)

but that seems too easy

[acoustic path]

that's correct

[brazen swift]

i mean if i just use that it gets me the answer immediately but that seems too easy

[acoustic path]

but it is valid proof

[brazen swift]

im not sure if they mean without using that identity

like its not said anywhere

but its too simple

[acoustic path]

cosine double angle formula is a staple identity that should be allowed to be used aside from (maybe) first encounter

[opaque lantern]

I guess u have to use cos2(2x) then cos2(x+x) formula that'll be cos²2x - sin²2x

man i hate trigonometry

[acoustic path]

if sm1 asked me to prove this i would just write "apply cosine double angle formula" and be done

[brazen swift]

i mean the question just says "prove these identities"

https://i.imgur.com/wbt4Jq4.png

butl ike

the question before

number 5

[opaque lantern]

ez

[brazen swift]

says
prove these identities of cosine with a double angle

[brazen swift]

ye i did it

[opaque lantern]

W

[brazen swift]

so i assume i dont need to repeat ?

also the case at question 6

i just apply it and thats it

🤔

[opaque lantern]

yupp u gotta apply the same identity

sin2(3x) this time

[brazen swift]

yeah im just overthinking it

[opaque lantern]

try it

[brazen swift]

thinking its supposed to be harder

[opaque lantern]

nuh

take it ez

[brazen swift]

ig

alr

thanks

[opaque lantern]

yw

[brazen swift]

any help on 14?

[opaque lantern]

lemme see

bro idk abt this one 😭

sorry abt tht i suck at trigono

[brazen swift]

np

@acoustic pathany help?

[opaque lantern]

thanks for understanding

[brazen swift]

nah its ok

the thought is what counts

[opaque lantern]

have a good day!

[brazen swift]

u2

[acoustic path]

did u do 13

[brazen swift]

no

didnt notice it

i was looking at the right side

[acoustic path]

ok

[acoustic path]

in any case, use the cosine double identity to write cos(a) in terms of sin(a/2)

[noble abyss]

can any1 help

[brazen swift]

and how do i proceed from there?

[acoustic path]

[brazen swift]

1-cos^2(0.5a) + sin^2(0.5a) / 2

after using the cos(2a) identity

[acoustic path]

that is not in terms of sin(a/2)

[brazen swift]

Oh

i thought

opening the

-cos(a)

is what you meant

but

makes more sense

give me a min

[acoustic path]

it is what i meant

[brazen swift]

huh

[acoustic path]

but the expression (1-cos^2(0.5a) + sin^2(0.5a)) / 2 is not in terms of sin(a/2)

it is in terms of sin(a/2) and cos(a/2)

[brazen swift]

in other words?

[acoustic path]

look at what u did in q 5

use the identity at the very right

[brazen swift]

👍

Oh

howd you know which one to pick?

Assuming because it's the only one to not include cos?

@acoustic path?

[acoustic path]

yes

[pearl pond]

@brazen swift Has your question been resolved?

[pearl pond]

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[midnight haven]

first time seeing this function can someone nice pls explain wth is going

on

[muted inlet]

its called a piecwise and it's really just saying for all numbers x is less then or equal to -1 use the top eqaution to plug in numbers, and for the x values greater then -1 use the bottom equation

[midnight haven]

yeah but first time seeing this functions graph

[muted inlet]

ex: if x=-2 (plug into top) 5(-2)+1 =-9. x=0 0^2 -5 =-5

[midnight haven]

not really but this specific graph

[muted inlet]

ok It's spilt into two graphs

one graph is just 5x+1 which is a line

[midnight haven]

i understand this part but im not sure how to tell if it's decreasing for this one

[muted inlet]

k. Just look at the graph and see when the Y value is going down and then see what x values coordinate with the range that the y value is decreasing

[midnight haven]

i think it goes down from -4? but idk

[muted inlet]

yes that is the y value when the graph starts to decrease, but you always want to look at the x values because there could be many y=-4 on the graph, but there is only one x=-1

so the graph starts the decrease at x=-1

When does it end the decrease?

[midnight haven]

dont think it ever does but thats my guess

[muted inlet]

when the graph is going down it is decreasing and when it goes up it increases. when the y value goes up it increases and when it goes down it decreases

[midnight haven]

okay

[muted inlet]

make sure you are always looking left to right when seeing if its increasing or decreasing

[midnight haven]

so u said it decreaces at (1-,-4)?

[midnight haven]

ok

[muted inlet]

yes that is when it starts to decrease

but then it starts to increase again

when does it stop decreasing?

[midnight haven]

the same points?

like (-1,-4)

how do i answer it

[muted inlet]

just pretend your drawing the graph from left to right

[midnight haven]

theres no option for (-1,-4)

[muted inlet]

when you pensil starts going down it is decreasing and when it goes up your increasing

[midnight haven]

is it (0,-5)?

[muted inlet]

yes!!

(-1,-4) and (0,-5) is in (x,y) it wants you to answer from a range. (x-x) because graphs often cross the same y value more then once but never the same x value

[midnight haven]

oh

they should've wrote that

[muted inlet]

Yeah, but they dont

[midnight haven]

so is the 2nd option correct

[muted inlet]

no

You said that it starts decreasing at (-1,-4) and stops decreasing at (0,-5) but it only wants a range of X VALUES

so you dont care about the y values and just put the two x values together

[midnight haven]

oh ok

so 3rd option

[muted inlet]

yep

[midnight haven]

?

okay

tysm <33

[muted inlet]

ofc

[midnight haven]

.close

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[gloomy kraken]

is this correct?

[dawn sand]

looks good to me

[gloomy kraken]

sorry i think theres supposed to be an 8 in place of 16

[gloomy kraken]

would 8 be correct

cause k was 12

j-4 = k, 12-4

[dawn sand]

k=12 implies j=12+4=16

[dawn sand]

you've subbed 12 for j instead of k

[gloomy kraken]

oh yea youre right

im slow today

thanks again

[dawn sand]

mondays amirite

[gloomy kraken]

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[orchid cobalt]

process so far

i rlly dont know what im doing

[old marsh]

ok well u showed some good logic

but just start over

keep it in radical form

rewrite sqrt(15) as a product of radicals

what do u get

[orchid cobalt]

alr alr

[orchid cobalt]

square root of 3 and square root of 5-?

[old marsh]

$\frac 1{\sqrt 3} - \frac {2\sqrt 3}{\sqrt 5 \cdot \sqrt 3}$

[old marsh]

ok now what

[orchid cobalt]

guys im sorry im kind slow at this

does the square root of three on the numerator and denominator cancel out-?

[old marsh]

we can do that, but i suggest another path

how about you make common denominators

[orchid cobalt]

oh how would i do that?

[old marsh]

$\frac ab - \frac cd$

[old marsh]

how would u make common denominators here

[orchid cobalt]

ad/bd - cb/db

?

[old marsh]

but those arent common denominators

how about this

[orchid cobalt]

oh damn ure right

[old marsh]

$\frac 12 - \frac 23$

[old marsh]

how would u make common denominators for this

[orchid cobalt]

id just make them both 6

[old marsh]

ok but howd u get that

[orchid cobalt]

by multiplying 1/2 by 3/3

and 2/3 by 2/2

[old marsh]

exactly

and how did u get 3/3

and 2/2

ur on the right track, just tell me ur reasoning

[orchid cobalt]

would i do this-?

[orchid cobalt]

oh cuz the the lcm

[old marsh]

yes lcm

,rotate

[old marsh]

ok well yes, that looks correct, but theres a better way

mult left fraction by sqrt5 / sqrt5

[orchid cobalt]

is it simplifying-?

would they cancel out?

[old marsh]

no we dont want to cancel anything out yet

$\frac 1{\sqrt 3} - \frac {2\sqrt 3}{\sqrt 5 \cdot \sqrt 3}$

back to this

[old marsh]

see how the denominator of the right fraction is sqrt 3 * sqrt 5

[orchid cobalt]

yes

[old marsh]

mutliply the left fraction by sqrt5 / sqrt 5

so u will get sqrt3 * sqrt 5 in both denominators

[orchid cobalt]

ohhh!!

ok ok i got it

so after this part

[old marsh]

$\frac {\sqrt 5}{\sqrt 5 \cdot \sqrt 3} - \frac {2\sqrt 3}{\sqrt 5 \cdot \sqrt 3}$

[orchid cobalt]

how do we simplify it further?

[old marsh]

now what do we do after we get common denominators

[orchid cobalt]

subtract?

but like im not rlly sure how to subtract ;-;

[old marsh]

$\frac ac - \frac bc = \frac {a-b}c$

[old marsh]

thats all theres to it

[orchid cobalt]

but how do you subtract roots-?

im sorry for wasting so much of ur time

[old marsh]

nah ur good dw

so yes that is correct

[orchid cobalt]

is this the final answer?

[orchid cobalt]

ty kind soul

[old marsh]

not yet, we have to still rationalize

do u know what that is

[orchid cobalt]

im sorry but no ;-;

ive been kinda absent from class for a short while..

[old marsh]

meaning, we cant have any radicals in the denominator

how can we get rid of that sqrt(15) in the denominator

[orchid cobalt]

oh could the square root of 5 cancel out and the square root of 3 cancel out?

btw this is the answer ;-;

[old marsh]

not quite

so we have

$\frac {\sqrt 5 - 2\sqrt 3}{\sqrt {15}}$

[orchid cobalt]

mhmm

[old marsh]

now to rationalize

we have to multiply the numerator and the denominator by sqrt(15)

lmk what u get when doing that

[orchid cobalt]

i got (5 -6)/15 ;-;

how do they still have squares?

[old marsh]

ok so ur denominator is correct, but we have to distribute the sqrt(15) to the numerator

$\frac {\sqrt{15} (\sqrt 5 - 2\sqrt 3)}{15}$

[orchid cobalt]

oh ok

[old marsh]

can u distribute that

[old marsh]

when u distribute u get

$\frac {\sqrt{15} \cdot \sqrt 5 - \sqrt {15} \cdot 2\sqrt 3}{15}$

[orchid cobalt]

is this it-?

[old marsh]

yes

good

and theres ur numerator

and we have 15 as the denominator

[orchid cobalt]

OMG TYSM!!

[old marsh]

so theres ur answer

np

[orchid cobalt]

youre a hero sent from the heavens

ok ok ill quickly take notes from everything!!

[old marsh]

👍

cya

[orchid cobalt]

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[deep thistle]

How does the top group of cells simplify to A'?

[plush bramble]

Not enough context

What are A and Abar

[cosmic charm]

its a k-map

[feral leaf]

Because A'B' + A'B = A'(B' + B) = A'(1)

[deep thistle]

I don't think I you more context for this tbh

[deep thistle]

Wait, I think I get it now. If there's a group of minterms in a row or column, does the simplified Boolean expression correspond to the variable for that row or column?

[feral leaf]

Not always, it depends on the number of terms

[deep thistle]

I meant like for example in this case

2 terms

[feral leaf]

Then yes

[deep thistle]

If it were this, it would be B'

?

[feral leaf]

Yes

[deep thistle]

ahh gotcha!

okay thanks!

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[opaque hearth]

How could you find two vectors u and v given their cross product?

[west sapphire]

there's no unique answer

[sharp vigil]

that wouldn't have a unique answer. for example
i x j = k
j x (-i) = k

[west sapphire]

more generally, take i and j, and rotate them both by the same angle (any angle), cross product is still k

[pearl pond]

@opaque hearth Has your question been resolved?

[pearl pond]

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[glad glen]

I'm having trouble with 31. i dont know how to even start the equation

[versed ledge]

AOC = AOB + BOC , maybe start with that?

[glad glen]

ty

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[outer hare]

Hey, can anyone explain what this k represents:

[outer hare]

here is the 2.5 they are referencing:

im having trouble understanding how to find a max bound

[pearl pond]

@outer hare Has your question been resolved?

[merry carbon]

The k is the[/an] upper bound of the (absolute value of the) derivative, you ideally should be given that before: for example, if you take g(x) = e^x on the interval, say, [-2, -1], then the derivative is at most e^{-1}, which is less than 1

[outer hare]

So for this case the max will always be 1 no

Like taking a look at the theorem before it

[merry carbon]

Well, for here, the k they mention here, is the same as...

[merry carbon]

...the k in the theorem

[outer hare]

Yup

So is k just the abs value of the derivative at p_0

[merry carbon]

What you want is that there's some constant, which is less than 1, such that the absolute value of the derivative is less than that (which is stronger than saying you want |g'(x)| < 1)

[merry carbon]

Not necessarily, you can e.g. take $k = \sup_{x\in (a, b)} \abs{g'(x)}$, and you want that to be forced less than 1

[jolly parrot]

@merry carbon

[outer hare]

Hmm

[merry carbon]

It needn't be its maximum at the point p0 you've chosen, in fact, for the example of e^x in [-2, -1], any point in that interval can be chosen as p0 (but the max happens at -1)

[outer hare]

Ok wait I’m not looking at it rn but I’ll get a paper and try to understand it when I get back

But from what I get, k is just the max value of the derivative within the interval

[merry carbon]

Yep, that's pretty much what you kind of want, that k is the max (or sup) of the absolute value of the derivative

But as long as that k is strictly less than 1...

[outer hare]

Because of the rule of convergence states before

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[naive hemlock]

Need help with this question
Explanation would be appreciated

[pearl pond]

@naive hemlock Has your question been resolved?

[terse bloom]

is this possible

the identity function satisfies all three of the restraints and has a derivative of 1 at x = 0

[west sapphire]

i suspect it should have said f(0) = 0

[naive hemlock]

Idk

That's why I came

@west sapphire

@terse bloom

[west sapphire]

yea?

[naive hemlock]

Can u help

[west sapphire]

as noted, the problem is incorrect

[naive hemlock]

Like I did not get what you all were speaking above

[west sapphire]

f(x) = x is a counterexample

[naive hemlock]

Oh

[naive hemlock]

So I should change it

f(0)=0

[west sapphire]

well where did you find the problem?

[naive hemlock]

In the assignment

💀☠️ oof

[west sapphire]

is that a screenshot of the assignment, or did you write your own version

[naive hemlock]

Ss

Why would I write my own version

[west sapphire]

yea i would confirm with whoever wrote it, what did they intend

[naive hemlock]

Yes tell

[west sapphire]

you would be surprised, people do it all the time in these channels

[naive hemlock]

Yea ik

Back then when I used to remain active 💀

[west sapphire]

anyway i am pretty sure they meant f(0) = 0, not f'(0) = 0

[naive hemlock]

Ok so now for f(0)=0

[west sapphire]

and if that's the case, you can prove it pretty easily using the mean value theorem

[naive hemlock]

Yes

We should do mean value

[west sapphire]

yep

[naive hemlock]

But can I be clarified

Like I apply lmvt

So I get

f'(c)=1

[west sapphire]

you applied it to which interval?

[-a, a]?

[naive hemlock]

[-a,a]

Should I apply till 0

[west sapphire]

yea that's true but it doesn't help much

instead try [-a,0] and [a,0]

[naive hemlock]

Ok applying

Oh

[west sapphire]

assume for contradiction that f(0) is not 0

[naive hemlock]

Oh

I get it

When I apply for first intervap

[west sapphire]

you should find that the slope has to exceed 1 in abs val in one of the two intervals

[naive hemlock]

Interval*

That's fine

But

Like wait let me send pic

Of my work

[west sapphire]

ok

[naive hemlock]

Here

Don't go pls

[west sapphire]

no start by assuming that f(0) is not 0

so there are two possibilities

either f(0) > 0 or f(0) < 0

[naive hemlock]

Wait

[west sapphire]

if f(0) > 0, then do the MVT on [-a,0]

[naive hemlock]

U are right how did I put f(0)=0

💀 tf

[west sapphire]

otherwise do it on [0,a]

(this is assuming a > 0)

btw, they should have stipulated a != 0 in any case, otherwise it's not true

[naive hemlock]

See

I got

f'(c)=[f(0)-a]/a in [-a,0] and f'(d)=[a-f(0)]/a in [0,a]

Now ?

Fr my mind is blocked today

[west sapphire]

yea so if f(0) > 0 then in the interval [-a,0] you have:
[f(0) - (-a)] / (0 - (-a)) = [f(0) + a] / a = f(0)/a + 1 > 1

and so there has to be a c with f'(c) > 1

contradiction

argue similarly if f(0) < 0

[naive hemlock]

Bro

Do we have any simple solution

[west sapphire]

isn't that pretty simple?

it's like 3 lines

[naive hemlock]

Cz I don't think the teacher has told us about assumption in any of the practice problems

[west sapphire]

which assumption?

[naive hemlock]

I need to frame and explain

Is there not any method

Which uses the mean value theorems

And gives somewhat direct result?

[west sapphire]

that's the simplest argument i can think of

[naive hemlock]

Bro you are right?

But is there?

Pls try

This question's solution has to go in the assignment In a well explained wau

Way*

[west sapphire]

i already gave you a nearly complete argument haha

i'm not gonna write it out in camera-ready form

[naive hemlock]

Oh

Oof

Bro I don't need the solution

I need some other simple way

Which directly proves f(0)=0

Without taking f(0)>0 and f(0)<0 in the beginning

@west sapphire

[pearl pond]

@naive hemlock Has your question been resolved?

[pearl pond]

@naive hemlock Has your question been resolved?

[naive hemlock]

<@&286206848099549185>

[pearl pond]

@naive hemlock Has your question been resolved?

[naive hemlock]

@west sapphire thanks got it

@west sapphire

Can I be helped with these two questions also

Aah

Fr

<@&286206848099549185>

Anyone?

[rough forge]

Maybe you are better of in #real-complex-analysis or #proofs-and-logic

[naive hemlock]

No

What

This is

Theorems bro

Bacc do you have any idea

How to do

[rough forge]

No

It's like the same like last time lol

You may also ask in calculus

most of the time the pros are there

[plucky python]

are you sure 15 is true?

if f(x) = 1 - x^2, a=-1, b=1

[naive hemlock]

Idk

Honestly

Idk

[plucky python]

then it's twice differentiable with f(a) = f(b) = 0 and f(c) > 0 for all c in (a,b)

[naive hemlock]

I am tired

This question has troubled me too much now

[plucky python]

but f''(x) = -2 so the conclusion is not true

[naive hemlock]

My assignment is due tmrw

That's why I came here now at last to understand

This

One

Question

[plucky python]

this seems to be a counterexample so i think there needs to be extra conditions

[naive hemlock]

Can u try for once

[plucky python]

wdym?

[naive hemlock]

No like

Can it be true for any case

[plucky python]

i mean like you can create a scenario in which it is true

[plucky python]

but this is a counterexample so the statement must be missing something

[naive hemlock]

Ik

Honestly

Speaking

Can u see the 14th one

Once

I will send the pic again wait

[plucky python]

i've had a look at it

still thinking but like that's hard

[naive hemlock]

In 14th one

If u remove the right term from consideration

It becomes cmvt

Cauchy's mean value theorem

[plucky python]

yes

[naive hemlock]

But then we don't have g'(x) not equal to 0

Its given g(x) not equal to 0

The assignment has questions that are too ambiguous

Can u suggest what should I do now

For the 15th one

I need to write an answer for it

Should I modify the question

[plucky python]

probably write up like 1-x^2 as a counterexample

[naive hemlock]

I can't

[plucky python]

and say that the problem seems to be wrong

[plucky python]

do you have to submit all ur problems?

[naive hemlock]

Yes

All 15

[plucky python]

bruh

[naive hemlock]

But I am troubled with this one

Two above troubled me too

But I cross checked and the questions were wrong

So I modified them

But 14. And 15. Trouble me

Too much

[plucky python]

well idk abt 15 cus i'm not exactly sure what the problem is meant to be?

[naive hemlock]

Me too

[plucky python]

also have u tried posting it to real and complex analysis?

[naive hemlock]

Should I?

[plucky python]

oh wait

[naive hemlock]

Isn't it supposed to be calculus?

[plucky python]

for 14

[naive hemlock]

Yes

Tell me

Asap

[plucky python]

try cauchy MVT on f and 1/g

[naive hemlock]

Aah

[plucky python]

i think that works

[naive hemlock]

I m done

Fr I can't unsee it now

[plucky python]

cus G(a)G(b)/(G(b)-G(a) is 1/(1/G(a) - G(b)) or whatever

[naive hemlock]

But

Wait

Its surreal

Yes it works

That's why they gave g(x) not zero for any x

Thanks bro u saved me

[plucky python]

nw lol

[naive hemlock]

Wait now I need to get other s checked.too

[plucky python]

also it seems a little strict that you have to submit for every problem

[naive hemlock]

pls wait

I need to submit soln for each question in assignment file

[plucky python]

like at my uni we don't need to submit all the problems, as long as we've given like a proper attempt at the sheet

[naive hemlock]

And they will be evaluated

Under iA

Internal assessment

Along with if the professor wants

A viva

Sort of

💀

Wait see this

Wait let me send specific one

Check for 11

Is it correct

cz I have assumed that it should be f(0)=0

And done it

@plucky python

[plucky python]

you can prove that

if f(0) > 0 then you get a contradiction by applying MVT to -a and 0

if f(0) < 0 then you get a contradiction by applying MVT to 0 and a

UK

i study at cambridge uni

[naive hemlock]

I have done

F(0)=0

Can u check will it be f'(0)=0 only

[naive hemlock]

Yea did this only

[plucky python]

well i think it'll be f'(x) = 0 for all x on that interval

but technically if you prove that you also prove f'(0) = 0 lol?

[naive hemlock]

What

What

What

@plucky python

I will make myself clear

I assumed the question will be f(0)=0 and proved it

Will in the question it will remain f'(0)=0

Only

[plucky python]

oh right i understand what ur asking now

i think the question is meant to be prove f'(0) = 0

[plucky python]

cus this same argument shows that f(x) = x for all x

[naive hemlock]

Ok

[plucky python]

so f'(0) = 0

[naive hemlock]

Once more pls

Bro but don't you think this prove f(0)=0

?

[plucky python]

so basically apply this same argument for every other x in (-a, a)

this gives you f(x) = x for all x in that interval

i think the question basically wants you to just show that f has to be linear

and it's asking you to do that by showing f'(0) = 0

oh wait ignore me i'm dumb

the derivative of a linear function isn't 0 mb lol

[plucky python]

yeah ok it'll be that then

[naive hemlock]

@plucky python

Bro one doubt

In the 14.

[rough forge]

Your assignment is crazy

[plucky python]

yeah?

[plucky python]

is this about that G'(c) might be 0?

if so i feel like that's an issue that they sort-of brushed to the side

[naive hemlock]

No

I mean

Will there be no problem

Using 1/G(x)

Imma come

I need to make a cover

Wait

[plucky python]

there might be an issue if say g'(c) = 0 but i think they just forgot about this issue

[naive hemlock]

Okok

No it does not say g'(c)=0

Ya wait

You are correct

They have not given that g'(c) is not zero

[pearl pond]

@naive hemlock Has your question been resolved?

[pearl pond]

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[abstract halo]

so in this situation, finding the characteristic equation would be 10r^2+20r+6260=100 -> r^2+2r+626=10

solving that gets me r = -1 +- 25i

for my homogeneous equation, would it be q(t) or I(t)?

I'm thinking its q(t) but the hint makes me go towards I(t), but I'm not sure how I(t) makes sense. If I go the q(t) route, I get C1= 0, C2=0

[merry carbon]

Assuming you put $I = \dv{q}{t}$ into $L\dv{I}{t} + RI + \frac{q}C = E(t)$ to get a second order diffeq in terms of $q$?

[jolly parrot]

@merry carbon

[abstract halo]

yes

[merry carbon]

In which case, the solutions you find would be for q

[abstract halo]

I'm getting my constants as C1 = 0 and C2 = 0, in which case q(t) just = 0

do I have to multiply them by some power of t?

[terse tide]

the differential equation gotta be in terms of i, not q

after you differentiate, put dq/dt = i

[merry carbon]

What constants, the homogenous one?

[terse tide]

dont change everything to q, why

[abstract halo]

hmm I'm not sure if I'm explaining well

I'll send a picture of my work

[merry carbon]

It would be easier that way no? Am I missing something

[terse tide]

no no :p

when he differentiates, hell get the dq/dt (3rd term)

he replaces that with i(t) and there you go

i'' + 2 i' + 626 i = 0

[abstract halo]

hmmm I seee

[abstract halo]

sorry, so just to confirm, I should differentiate the original equation then to get i''?

and i' and i terms

[terse tide]

yes, i' will become i'', and i will become i'

[abstract halo]

thank you guys!!!

[terse tide]

and q will become dq/dt, which is i

[abstract halo]

.close

[pearl pond]

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[pearl pond]

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[dusk otter]

Need some assistance with this natural deduction proof. I have been playing around with it, but unsure of the next step I can take. I have attached the rules that I can use.

[pearl pond]

@dusk otter Has your question been resolved?

[pearl pond]

@dusk otter Has your question been resolved?

[pearl pond]

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[slim rock]

i dont get how im supposed to solve this

[slim rock]

this is all of the instructions

do i not just have to get the x value of that midpoint?

[vestal tapir]

yeah, root is x

but you need to improve the guess 10 times

[slim rock]

right i give the value thats literally on top of that mid point

i just dont understand what its asking in that first part

[vestal tapir]

the midpoint is closer to 1.512 in that picture

it's the average of the 2 side xs

[slim rock]

right the 2 sides are movable

the mid one is stationary

[vestal tapir]

ok that doesn't make sense, y=0 is not even visible

you should zoom out and move the sides so they are around 0

[slim rock]

thats what im confused about

it doesnt zoom out

[vestal tapir]

yeah no idea

[slim rock]

ive been trying to zoom out the whole time

just confusing

[vestal tapir]

maybe you can retry it and it will be zoomed out

[slim rock]

no way to reset it either

been trying to find a way too

[pearl pond]

@slim rock Has your question been resolved?

[pearl pond]

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[quick jewel]

How do I complete number 7

[quick jewel]

I've just been figuring out what kind of equations I could possibly make to get a (first term) or d (difference)

I'm not sure what to start with

[tall flint]

"the 11th term is 53" should give you one such equation

[quick jewel]

What would the equation be though

I have no examples from the teacher to base it off of

[tall flint]

if you have an arithmetic sequence, what is the 11th term in general?

please don't say 53

[quick jewel]

53 = a + 10d?

[tall flint]

that doesn't answer the question i just asked but yes that's one of your equations

[quick jewel]

Sorry lol

[tall flint]

where do you think your second equation comes from?

[quick jewel]

One of the Sn formulas?

How would I get the other missing numbers tho

[pearl pond]

@quick jewel Has your question been resolved?

[cloud zephyr]

what formula(e) do you have right now?

[cloud zephyr]

these 2 equations, right?

[quick jewel]

Yeah

I have these formulas to work with

[pearl pond]

@quick jewel Has your question been resolved?

[pearl pond]

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[covert nymph]

Could someone explain why the answer is not DNE or 0? I've tried multiple times?

[plush bramble]

The slope at 4 is ?

Do you know how to find the slope of a line?

[covert nymph]

yes it's (y2-y1)/(x2-x1)

[plush bramble]

Sure. Do that for the line passing through f(4)

Line segment

[covert nymph]

ok so i got

the slope is 1/2 and if i plug in (using 5,2) y-2 = 1/2 (x-5) i get y=1/2x - 0.5

[plush bramble]

Yup

The slope is 1/2

[covert nymph]

oh ok thank you!

i was thinking too hard about it

.close

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[meager fog]

Back again…. I think I have the math part down. I’m just confused on how I answer.

[cosmic garnet]

what?

where did the middle 100 come from

also

why are you calculating values with both

[meager fog]

d. f(-100) + (100)

[cosmic garnet]

f(-8), you have to calculate f(x) at x=-8, for x < 0 the function is defined as

so you just have to calculate with this

[cosmic garnet]

it is f(-100) + f(100)

[midnight haven]

so basically

f(-100) -100 is your x

which is ofc <0

so you enter case 6x+1

[versed ledge]

Maybe give a comma

[midnight haven]

then f(100) is ofc >0

[meager fog]

Oooo ok my math teacher literally went over this in the last five minutes of class and was like OK guys go do homework

[midnight haven]

so you enter case >=0

[midnight haven]

i think she understand with or without

[midnight haven]

sad

its okay we can help

that's why we are here for

all of us, helpers.

[meager fog]

Ok let’s see if I got it

[midnight haven]

show us

[midnight haven]

can you show me your work

im lazy to do it on my own 😛

[meager fog]

lol ya

[versed ledge]

d one

Check again

Rest are fine

[midnight haven]

d is not 3 yea

[meager fog]

Would it be 8 (-100)+3=-797?

[midnight haven]

no

you dont have the case just f(-100)

if it asked you for just f(-100) that would be the answerr

but it asks you for

f(-100) +f(100)

the result from f(-100) you add with the result from f(100)

brb

[meager fog]

Oh ok

So -797 + 803= 6

Brb let me go change that math

[meager fog]

-599+803=204

[midnight haven]

yea

good job

[meager fog]

Phew I wouldn’t have thought of that. Thank you guys!! Google and YouTube were not teaching me how to do it!

[midnight haven]

no problem

have fun

and have a good day

i gues

[meager fog]

.close

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[hasty crater]

sorry if im in the wrong channel I was looking to get some help with this https://cdn.discordapp.com/attachments/1285669898009317417/1285669898155982848/7127df816e6612f571bfef8732736102.png?ex=66eb1ce6&is=66e9cb66&hm=dde76fd8f6f51b8358ab979dcc8a8d994ef98b333c29491eccc5c527f38f9f3d& I know it's a non right triangle vector I've watched videos but don't understand what i need to do

[pearl pond]

@hasty crater Has your question been resolved?

[pearl pond]

@hasty crater Has your question been resolved?

[pearl pond]

@hasty crater Has your question been resolved?

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[stone imp]

I don’t understand how im supposed to know how much x is worth

Nvm im stupid

.close

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[stone imp]

.reopen

[pearl pond]

✅

[stone imp]

I’m confused again

[wet swallow]

Is this translated

[stone imp]

Unfortunately

[woeful stump]

yeah but it's pretty clear imo?

the equation is coffee + roll = juice + cookie

[stone imp]

Won’t it be the same as

[stone imp]

This

How is it different

[woeful stump]

idk maybe

[stone imp]

They have the same worth and numbers

[woeful stump]

solve and find out

[stone imp]

Okay I try

[midnight haven]

Hello

[stone imp]

It was right

I’m not gonna question it

I’m just gonna go with it.

[woeful stump]

lol okay

[woeful stump]

substitute back into this equation if you wanna check

confirm they're equal

[stone imp]

Wait 12+7 is 21 Right

[wet swallow]

Lol

[stone imp]

Yeah it juys hit me

The way I was so confused when it said it was wrong

Okay we’re done with my stupidity here u will probably see me soon again but bye bye for now 😚

.close

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[stone imp]

.reopen

[pearl pond]

✅

[stone imp]

Hi hi

0.01+0.01x=0.1x

That’s

That’s 0.1111111111 right

Right …

It says answer algebraically

And to answer exactly

So I did that

[versed bluff]

May if you Transform it to a fraction Will be easer to solve dontk

[stone imp]

How do I do that

[versed bluff]

Soo You write in the numerator the Hole number whitout comma and below You put a 1 and add ceros as many decimals the number has

[stone imp]

Okay where do I do that

[versed bluff]

<@&286206848099549185>

[summer prawn]

Oh, couldn't you represent all of these by the same denominator?

[wet swallow]

Wut

[summer prawn]

so have like 1/10 = 100/1000

also this makes no sense lmfao

[wet swallow]

Is this the question

[summer prawn]

oh, it uses x

😭 bro write it down

[wet swallow]

Convert to fractions

[summer prawn]

$\frac{1}{100}+\frac{1}{100}x=\frac{1}{10}x$

[jolly parrot]

Makenna

[summer prawn]

$\frac{1}{100}+\frac{1}{100}x=\frac{10}{100}x$

[jolly parrot]

Makenna

[summer prawn]

$\frac{1}{100}=\frac{10}{100}x-\frac{1}{100}x$

[jolly parrot]

Makenna

[summer prawn]

can you go from here?

[stone imp]

The answer was 1/9 bcs my brother touched the answer button he wasn’t supposed to but I still don’t get it

[pearl pond]

@stone imp Has your question been resolved?

[pearl pond]

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[dusk otter]

I am trying to prove that associativity is present in this binary operation (i.e. (g+h)^3).
I am not sure how to do the g*(h * i )

[torn shell]

you just do the same with h * i first

g * (h+i)^3=(g+(h+i)^3)^3

im not sure how its associative though

[dusk otter]

it may not be, my job is to checkif it is or not

but I don't believe it is

[torn shell]

so you would need those to be equal

[dusk otter]

yeah

so it's not associative

[torn shell]

i think you cna just argue that they are not the same directly from this idk how necessary it is to expand

[merry carbon]

(you could try and also find explicit a,b,c such that ((a + b)^3 + c)^3 and (a + (b + c)^3)^3 are diferent)

[dusk otter]

Like this?

[torn shell]

would comparing terms work too

[dusk otter]

Hmm I am not sure which way would work for a better argument

Is there a point to the k in the first aprt?

Idk If I am seeing the purpose there

[torn shell]

like you would hasve some g^9 term but not on the other one right

[torn shell]

no not needed. but you coped the g * h part in the 2nd part should be h * i for clarity

chartbit counterexample works well and is easyh

[dusk otter]

hmm okay maybe a counter example would be the better option then

[torn shell]

also for the conclusion the operator * is what is not associative

[dusk otter]

Okay thank you!

[dusk otter]

which piece are you referring to ?

[torn shell]

after g * (h * i):

you did g * h again

not h * i

[dusk otter]

oh I see I see

[torn shell]

well h * i would be (h+i)^3

[pearl pond]

@dusk otter Has your question been resolved?

[pearl pond]

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[sand coyote]

help

[sand coyote]

I need to show that the angle is 90 degres

I know its the dot product

But how do i apply it

in this context

[merry carbon]

Which angle do you want to show is 90?

[sand coyote]

B

[merry carbon]

Well, you can effectively rephrase that as saying that the line (segments) AB and BC are perpendicular

...if that's enough to cook with?

[sand coyote]

No i think they want me to proove it with

Calculations

I can do Dot product of a and b =0

But im not sure how to apply that here and which coordinates to use

[sand coyote]

I know this will work as a proof

To find vector BA, (-2-0; 0-4) and vector BC (8,-4)

something like this

[pearl pond]

@sand coyote Has your question been resolved?

[merry carbon]

Yep, and...

[merry carbon]

...you figured that if you showed their dot product to be 0, then they're perpendicular

[pearl pond]

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[deep girder]

can someone explain this funky rule to me?

[frank goblet]

it is called substitution

consider the following:

let u be a function

u(x)=x^2+1

note that this is exactly the denominator

so we get

[deep girder]

ok yup

[frank goblet]

$5\cdot\int\frac{x}{u}dx$

[jolly parrot]

Martin

[frank goblet]

now we dont want u and x

because remember: u depends on x, so we still have to take it into account when integrating

since the x-world was not so nice, we wanna completely switch to the u-world

[deep girder]

lol

[frank goblet]

in order to do so, we have to exchange our dx for a du

the question is how we do that

we know that u=x^2+1

therefore du/dx=2x

which we can write as: du=2x*dx

or dx=1/2x * du

so we get

[deep girder]

ahhhhh

[frank goblet]

$5\cdot\int\frac{x}{u}dx=5\cdot\int\frac{x}{u}\frac{1}{2x}du=5\cdot\int\frac{1}{2u}du=\frac{5}{2}\cdot\int\frac{1}{u}du$