help-23

[craggy sedge]

you use this

fact and apply this limit on it

[junior sundial]

OH
wait is this derivatives? we haven't learn that yet

[safe radish]

@junior sundial Has your question been resolved?

[obsidian oracle]

do you know what a^3 - b^3 is

[junior sundial]

uh the differences of cubes? yeah

[safe radish]

@junior sundial Has your question been resolved?

[junior sundial]

denominator? you mean the x?

[safe radish]

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[fallen blaze]

i just started calculus, how do you graph this guys

[fathom jewel]

Do you know how to draw linear functions in the first place?

[fallen blaze]

yeah

[fathom jewel]

Now g(x) is just

a variety of linear functions

just at different intervals

draw 2x-3 from left until x = 0

then between x = 0 and x = 2 you draw -x-3

and from x = 2 to right 3x

[fallen blaze]

ohh ok tyy

[fathom jewel]

do you got it by now?

you will need that drawing

[fallen blaze]

yeah i do

[fathom jewel]

or at least it's something that you can use for orientation

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[lean basin]

hi

[lean otter]

is that the work you did?

[lean basin]

yeah, the solution below is from mine

and I don't know what to do next

I can't think of any ways

help

[rotund turtle]

Have you learned exact differential equations

[lean basin]

I'm still not good at it, I'm stuck with simplifying (I'm also not good at it)

should I divide by y?

please

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[real turtle]

what i've done so far

mistake somewhere, idk where

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[real turtle]

<@&286206848099549185>

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[valid geyser]

@junior smelt Idk my final answer seems a little too simple?

[junior smelt]

For the integral?

[lean otter]

i need help please please please dm me

[valid geyser]

Ye

[main mural]

#❓how-to-get-help

[junior smelt]

I mean (of course assuming n is a nonzero integer) the integral does end up working out like that (see the wolfram evaluation)

[valid geyser]

So was my I final answer for the definite integral correct

Cuz I also have to get the integral from 0 to +pi

But sine is then multiplied by (-t+pi)

[junior smelt]

Yep, it was fine

[valid geyser]

So does that mean my entire value for Bn would be 0?

Because for the function from 0-pi is -pi/n

[sick scarab]

i told you yesterday that even functions have bn=0

and odd functions have an=0

so yes, bn=0

for a repeated triangular gate function

[sick scarab]

+pi/n

[safe radish]

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[lean otter]

In how many points the following function intersects with the x axis

[deft pivot]

what is sen ?

sec ?

secant ?

[lean otter]

no, is sine

seno is sine in spanish

[deft pivot]

ik

ok

[vague slate]

Yeah then pls explain what the question tells to do also pls

[lean otter]

I need to simplify it, graph it and then "In how many points the following function intersects with the x axis"

[vague slate]

Ah ok

[deft pivot]

i guess we would write this equation in terms of cos only

so i guess we use some trig formula to change sinx/2 with cos

do you know such formula?

[lean otter]

I have tried that

[vague slate]

You know the double angle formula right?

[lean otter]

I was thinking of this, but I would need to have just x or just x/2 maybe idk

[lean otter]

yes

I have tried the common ones

but idk how to get a simple expression with one sine or cosine to graph it like I was taught

[vague slate]

I tried this: write cos2x as 1-2sin²x, then sin²x can be written as (2sin(x/2)cos(x/2))^2

When you will put this in place of cos2x, you will get your equation turned into terms of sin²(x/2) and cos²(x/2)
You can change cos²(x/2) to yk (1-sin²(x/2))
Then you have your whole eqn in sin(x/2)

[lean otter]

y = Asin(Bx + C) + D

[vague slate]

Then take sin(x/2) to be let us say t, you get ig a 4 degree eqn

Try this

[red delta]

[lean otter]

mm idk

[lean otter]

do you get 16sen^2 (x/2) - 16sen^4(x/2) + sen(x/2) - 2 ?

[vague slate]

Wait a min , what's it equal to ?

[lean otter]

nothing

is a function

[vague slate]

Ah = 0
Yes

[lean otter]

f(x) = 16sen^2 (x/2) - 16sen^4(x/2) + sen(x/2) - 2

did I do some calculation wrong?

[vague slate]

No see it has asked number of times f(x) cuts x axis , that basically means that number of value of x for which f(x) = 0

[lean otter]

yes oh, so 4

?

[vague slate]

Nah it might be that only 3 or 2 or 1 , r u allowed to use calculators?

[lean otter]

just the basic ones I think

bc I am actually supposed to graph

it

y = Asin(Bx + C) + D

[vague slate]

Oo

[lean otter]

I think

[lean otter]

The exercises of this kind I had to graph it to get the answer

so idk

[vague slate]

Isn't it going to be hard to graph this .....

[lean otter]

If there was a way to simplify it no

necessarily

[vague slate]

Wait let me try

[lean otter]

m okay

[lean otter]

Yeah you can use trigonometry formulas to simplify it

[vague slate]

Which ones ....?

[lean otter]

let me see

[red delta]

What is the question

[lean otter]

In how many points the following function intersects with the x axis

https://cdn.discordapp.com/attachments/903481106819612714/1286029252642213960/Screenshot_2024-09-18_at_1.20.26_PM.png?ex=66ec6b93&is=66eb1a13&hm=c392b8a82507be3e7dbbe95f43f69c799477513d2aebeb1ac5021bc97042ebf2&

[red delta]

So in other words how many zeros does the function have for that range

[lean otter]

Yes

[lean otter]

We got to this expression

[red delta]

What happens when sin(x/2)=2cos2x

[lean otter]

we get 0

[red delta]

Yes

Now you can try double angle

[lean otter]

Answer is 1

[safe radish]

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.

[lean otter]

it's not

I have the answer

[lean otter]

Really maybe 2 or 1 i am sure

nop

[lean otter]

Ok dm me

but the thing is that having the answer doesn't tell you much

[lean otter]

Ok let me try

bc I still need to now how to solve it

like the process

it's what's important

[lean otter]

we get sm similar to what we did before

[red delta]

Okay

[lean otter]

0 = 16sen^2 (x/2) - 16sen^4(x/2) + sen(x/2) - 2

Ah, I think I know how, perhaps I need to graph each one

[red delta]

Quadratic

[lean otter]

and then see the intersections

[red delta]

I thought graphs aren't allowed

[lean otter]

Yesss

It's indeed 4

(: thx

[red delta]

,w plot sin(x/2)-2cos2x

[flat frigate]

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[lean otter]

.close

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[real crater]

i'm not sure how it went from sin(bt) to sin(u)1/b du

[real crater]

i subsituted with b and c for the prior equations outside of the integral to make it easier for me

[strong garden]

Unless ive understood your question wrong i think thats just how u subs work

could you elaborate further what you mean?

[real crater]

if you look at the screenshot where it explains it, i'm not sure how exactly it goes from sin(bt)dt) to sin(u) 1/b du

[strong garden]

well like i said you do a u sub
youget a new variable u and let it equal bt

you take the derivative of both sides to get that
du = bdt
so dt = 1/b * du

so sin(bt)dt becomes sin(u) 1/b du

[real crater]

so whatever is inside sin we use?

[strong garden]

yeah whatevers most convenient

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[olive leaf]

hii i know this is a bit silly but my cousin got assigned this homework and i have been helping her but the program doesn't recognise it as correct.

[olive leaf]

genuinely this is year 2 maths but I think i'm going mad because it literally won't recognise my solution as valid.

[zinc stirrup]

Uhh from what I see here the solution should be from 8 to 12 so if you got that the system is wrong

[olive leaf]

i did get that :/

[zinc stirrup]

It's possible that the system has the wrong solution put in

[olive leaf]

surely the teacher would check

[zinc stirrup]

Errors can slip in oftne

Maybe the watch video part has a hint?

[olive leaf]

thats very true I will email her

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[raven dome]

i just need to know if i did it right

[zinc stirrup]

How did you conclude in a that there are no solutions for x-intercepts?

[raven dome]

tbh i dont really remember just says on my note that if i can simplify it even more then there is no solution

[zinc stirrup]

Ok but looking at your graph

Doesn't it seem like solutions should exist?

[raven dome]

ohh yeahh

so i can simplify it by doing the absolute value of x+1?

[zinc stirrup]

You can get two cases

That either x+1 = 3 or x+1 = -3

There would be no solutions if it was -3 because absolute values can take only positive values

[raven dome]

ohh okay okay i didnt really understood my notes

i get it now

i thought its because i cant simplify it even more

so my x-int would be x+1 = - 3 and x+1 = 3

[zinc stirrup]

Yeah

[raven dome]

how would i write it?

[zinc stirrup]

I mean you calculate x to be either -4 or 2

So (-4,0) and (2,0)

[raven dome]

alright thank you!

so i got this for all of it

wait

i need to add x int

[zinc stirrup]

Your domain is good but check the range again

[raven dome]

yeah i fixed the signs

oh nv

nvm

yeah its supposed to be 4. -infinity

[zinc stirrup]

It should be 6 to -inf

[raven dome]

ohh yeah

[zinc stirrup]

And it shouldn't be included

It's a debated topic should it or not but like better not

[raven dome]

i would include it just incase

my professor taught us that it would be better to include it

thank you so much

.close

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[raven dome]

last question, im pretty sure i got it right, but just making sure

[zinc stirrup]

You can write it in wolfram alpha to check the answers

[raven dome]

i think i got the brackets wrong

but thank you

[zinc stirrup]

You can check the brackets with it too

[raven dome]

alrightt i got it right

thank you!

.close

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[pliant zephyr]

So I was wondering if I was doing this right

We set up systems of equations for the flow of traffic and then create an augmented matrix of the system of equation

make it into RREF

and then solve for each vector

but what im confused about is the realistic range of solutions

x1 is easy since that is just 0 <= x1 <= 200

I also said that 0 <= x2 <= 600

but technically x3 can have flow from x2 but im not sure what the range should be

[safe radish]

@pliant zephyr Has your question been resolved?

[pliant zephyr]

@raven vessel

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@pliant zephyr Has your question been resolved?

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[loud dune]

im confused

whats the given angle

[manic radish]

that's exactly what I was going to ask you XD

I think the "22" is the given angle of 22 degrees, and the others are just labels

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[lament cobalt]

Hi

[lament cobalt]

I'm desperate so I'll try asking here, because I believe it's more related to calculus than physics concept

've wasted about 5h trying to solve this.
Besides the method I'm trying (for learning purposes) I've calculated by using the area of the triangle (bh)/2 multiplied by the area of the trapezoid. ((B + b)h)/2 which resulted in 1.5J.

What am I doing wrong that I don't get the same result with integrals?

've wasted about 5h trying to solve this.
Besides the method I'm trying (for learning purposes) I've calculated by using the area of the triangle (bh)/2 multiplied by the area of the trapezoid. ((B + b)h)/2 which resulted in 1.5J.

What am I doing wrong that I don't get the same result with integrals?

[safe radish]

@lament cobalt Has your question been resolved?

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@lament cobalt Has your question been resolved?

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@lament cobalt Has your question been resolved?

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@lament cobalt Has your question been resolved?

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[ocean jewel]

If f(x)= x^4+6, g(x) = x-2, h(x) = sqrt(x), then (f x g(h(x)) =

[ocean jewel]

Hi

If f(x)= x^4+6, g(x) = x-2, h(x) = sqrt(x), then (f x g(h(x)) =

??

.close

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[haughty lodge]

the sum of the first 8 terms of a geometric series is 200. the first term is 5.75 how do i find the ratio

[haughty lodge]

basically im using the formula a(1-r^8)/(1-r)

but i dont know how to solve from there

[lean otter]

Are you allowed to use some computer algebra system to solve an equation? Because thats what it comes down to, as I understand it

[safe radish]

@haughty lodge Has your question been resolved?

[haughty lodge]

nope

[lean otter]

First step: Set up an equation. In the formula described, you know the value of "a"

[haughty lodge]

yes

a is 5.75

[lean otter]

so what is the equation you have?

[haughty lodge]

i have

5.75(1-r^8)

divided by 1-r

[lean otter]

equals

[haughty lodge]

umm

200

[lean otter]

yes. So basically this

$5.75\frac{1-r^8}{1-r} = 200$

[flat frigate]

gautamdb

[haughty lodge]

yes

exactly

can i divide by 5.75

[lean otter]

yes sure

[haughty lodge]

its a long decimal

34.7826087

[lean otter]

yeah

[haughty lodge]

ok

1-r^8 is difference of two squares? does that matter?

[lean otter]

oh yes that will help

[haughty lodge]

so i can just keep

simplifying

until i reach 1-r

as one of the terms

[lean otter]

yes

[haughty lodge]

ok let me do that rq

[haughty lodge]

it cancels

lemme take pic

[haughty lodge]

those two 1-r cancel

34.7826087 = (1+r^4) (1+r^2) (1+r)

[safe radish]

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[haughty lodge]

.reopen

[safe radish]

✅

[lean otter]

okay, and that is also the farthest I could get by hand 🙂

[haughty lodge]

lmaoo alr

tbh i have an idea

what about using the formula for a geometric sequence

hmm nvm wouldn't work

[lean otter]

Are you sure you're not allowed to use wolframalpha or sth?

[haughty lodge]

well the question doesn't specify

im pretty sure my teacher just gets these online

[lean otter]

I guess you cannot simplify this further

[haughty lodge]

lemme try use

wolframalpha

and see

cause the marking scheme has the answer but

no way of solving

[lean otter]

I may be wrong, but I don't think there is an easy way to solve this. This is a 7th degree polynomial

[pine horizon]

,w 5.75 r^8 - 200 r +194.25 = 0

[flat frigate]

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Requested by astar777

[haughty lodge]

yeah

that's what

the marking scheme has

the second real solution is the one given by the marking scheme

[lean otter]

yeah you should use the second equation you got

[haughty lodge]

ok

ty guys

[lean otter]

,w 34.7826087 = (1+r^4) (1+r^2) (1+r)

[flat frigate]

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[haughty lodge]

yea

thanks

.close

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[void cosmos]

just want to clarify if I’m right or wrong.

[void cosmos]

I have been trying to gauge a scenario to contradict this. One where we have a continous function in its absolute value form, but when we revert its not continuous anymore.

[median vigil]

you can consider piecewise functions

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[mental mauve]

Hey, so I did this problem and got only one solution for x, when I put it into my calculator I got more. What am I missing?

[safe radish]

@mental mauve Has your question been resolved?

[mental mauve]

<@&286206848099549185> Excuse the ping but it says I can after 15mins lol

[toxic plinth]

have u learnt the graph of sin(x) yet?

[mental mauve]

Yes

[toxic plinth]

u know how it oscillates for ever

between 1 and -1

[mental mauve]

Yeah but the question stated a domain

[toxic plinth]

yeah

forget the domain for a sec

[mental mauve]

Ok

[toxic plinth]

without it there would be infinite solutions where sin(x) =1

right?

[mental mauve]

Yeah

[toxic plinth]

okay

now what does putting the 3 infront of the x do to the graph

[mental mauve]

changes the period

2pi/3

[toxic plinth]

yeah

yeah

with sin(x)=1 between 0 and 2pi we get one solution right

now that u changed the period we will get more solutions between the period

[mental mauve]

So where in my working did that not give me more than one answer?

[toxic plinth]

when u did arcsin(1)

u should write the general solution

[mental mauve]

Idk what arcsin means

[toxic plinth]

like inverse sin

sin^-1 (x)

[mental mauve]

oh gotcha

[toxic plinth]

it should read 3x = pi/2 plus or minus 2kpi

have u ever seen that?

[mental mauve]

Wait, why plus or minus?

[toxic plinth]

actually just plus is fine

as long as k can be negative and positive

[mental mauve]

K? 😭

From knowing my exact values, I know sin(x) = 1, x = 90 degrees or pi/2

[toxic plinth]

if u look at a normal graph sin(x) the sin(x)=1 at pi/2, pi/2 plus 2pi, pi/2 plus 4pi and so on

[mental mauve]

I understand that if I were to graph it I'd see the other solutions, however, I'm not sure why it isn't showing in my working

[toxic plinth]

i’m about to get home i’ll write it out

[mental mauve]

Thank you and no rush lol, I do appreciate your help tho

[mental mauve]

Where did the pi/2+2kpi come from?

[toxic plinth]

its called the general solution

it tells u all the possible x values where the graph will equal 1

k is an integer so counting number, 0,1,2,…

sub in different value of k and u get the x solutions

[mental mauve]

So how does the domain limit it?

Bcs if it isn't I'll get infinite solutions again?

[toxic plinth]

yes

the domain basically limits the value of your graph to the lower limit the left value and upper limit the right value

[mental mauve]

Oh I see what you did

[toxic plinth]

yeah

[mental mauve]

You did like the first couple of numbers and selcted the ones that fit into the domain

[toxic plinth]

yeah

at the start u might need to write a few of them out until u can just tell straight away

it better to have more solutions then cross them out because they aren’t the domain then to have too little

[mental mauve]

Alright, thank you for your help and patience.

[toxic plinth]

no all good

it’s hard to wrap ur head around at the start

[mental mauve]

I can definitely do all that

Yeah

Anyway, thanks I've got other problems to do 🙂

.close

[toxic plinth]

all good cya

[safe radish]

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[glacial meadow]

Posted this yesterday but couldn't get any help

[glacial meadow]

(if it seems familiar it's cuz some guy posted it before)
Well, the thing is: I'm fairly sure there exist no such two lines, but it'd be pretty weird for the exercise to have no solution
The second and the third condition state that L, L1 and L2 are all parallel between themselves, thus it then follows that their vectors have the same direction
But the first condition states that L1 is contained within the first plane, which only happens if v(L1) is perpendicular to n(π1) and that L2 is contained in the second plane
If the vectors of L,L1 and L2 are all the same, then The same vector must be contained in both planes, thus v(L) should be perpendicular to n(π1) and n(π2), in other words, it should be their cross product
But when I did the cross product of those two vectors the result didn't have the same direction as v(L)
Oh, in case it is not clear, v(L) is the vector whose direction is that of the line L, while n(π) is the vector normal to a plane π

[short topaz]

i mean you can find the direction vector for L1 and L2

which would just be (1,3,2)x(-3,-1,3)

,w (1,3,2)x(-3,-1,3)

[flat frigate]

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[short topaz]

that'll give you the direction vector for L1 and L2

[glacial meadow]

Thing is, if that was the direction vector, the lines wouldn't be parallel to L

[short topaz]

they're definitely not meant to be parallel to L lol

[glacial meadow]

That direction vector is.not proportional to (2,0,1)

[glacial meadow]

Third condition

[short topaz]

if L1 were parallel to L, then they wouldn't intersect

so L1 intersect L = empty set

unless we're including the point at infinity

[glacial meadow]

Doubt we're working with the extended real numbers

[junior smelt]

(The third condition only says that the line L meets the other two lines, not that they need to be parallel)

[glacial meadow]

The third condition states that the set of intersections of L and L1/2 is an empty set

Don't get how that could mean the lines meet

[junior smelt]

is not the empty set

[glacial meadow]

Ain't no way, it was not equal

Oh my god, guess it's what I get for attempting it half asleep

My bad, god am I stupid

Thanks

.close

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[wild island]

the number of bacteria in a culture increases by the same percentage every hour. The number doubles in 10 hours. By what percentage does the number increase in 2 hours?

[buoyant shadow]

,calc (1.11)(1.11)(1.11)(1.11)(1.11)

[flat frigate]

Result:

1.6850581551

[buoyant shadow]

if it was eleven percent, you would get +68% after 10 hours

so it's more than 11

[wild island]

yeah more then 11 but whats the number

[safe radish]

@wild island Has your question been resolved?

[obsidian oracle]

,w fifth root of 2

[flat frigate]

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[obsidian oracle]

it doubles in 5*2 hours

so

you're looking for an increase a

such that a*a*a*a*a = 2

that's exactly the fifth root of 2

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[void spruce]

would u square both sides and then rearrage for sin double angle and then find cos double angle and then cos on sin

[brave wolf]

that would work

[void spruce]

any other better methods

[brave wolf]

I think that your method is quick enough

[void spruce]

alright

would u not get a + - solution for cos then

what would you do in that case

oh nvrmind the restriction

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[neat tartan]

Can anyone help me with this question. I’ve been stuck on it forever. F(2)=2 how do I find f(3)?

[safe radish]

@neat tartan Has your question been resolved?

[neat tartan]

No

<@&286206848099549185>

[wet mirage]

the pic isn't the best, but i think what you need to do is get f(3) and f(2) (since h = 1), and then just input f(3) - f(2)

(i can't read the bottom lines btw, can you send a screenshot instead?)

[neat tartan]

I got the bottom question done. I just need help with finding f(3). I don’t understand how to find it with just looking at the graph

.close

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[rare olive]

I'm trying to formalise this proof instead of just using pictures. How is this derivation made formally?

[covert yoke]

If you're trying to formalize this, like actually formalize it, you'd have to generate some sort of measure of complexity for any given hypothesis, which is a terribly difficult thing to do.

[safe radish]

@rare olive Has your question been resolved?

[rare olive]

the proof just says assuming:

1. group 1, G1, is equally probable to group 2, G2, and
2. each theory in G1 is equally probable, and each theory in G2 is equally probable, and
3. there are more theories in G2 than G1

then you can derive that:

each theory in G1 is more probable than each theory in G2

the question is how the derivation is made

the definition of complexity is not essential. you could dispense with talk of complexity & the proof would run the same

[zinc stirrup]

Say that the probability of each group is $p = p(G1) = p(G2)$ then assuming that $g1 = card(G1), g2 = card(G2)$ and $g2>g1$ (the number of theories in each respective group). Since all elements are equally probable we can conclude that for any $t \in G1$ must be so that $p(t) = p/g1$ (and similarly for every $t \in G2$). Then let's assume that if $t \in G1, h \in G2$ that $p(t) > p(h)$. Inserting the formula we get $\frac{p}{g1} > \frac{p}{g2}$ and since all numbers are strictly positive we can simplify to $g2>g1$ which is true by our assumptions. Have we taken that $p(t) < p(h)$ or that $p(t) = p(h)$ we would contradict our assumptions so it must be so that $p(t) > p(h)$ or that every theory in G1 is more probable than every theory in G2.

[flat frigate]

Goran

[rare olive]

Since all elements are equally probable we can conclude that for any $t \in G1$ must be so that $p(t) = p/g1$ (and similarly for every $t \in G2$).

How does this go?

[flat frigate]

holy

[zinc stirrup]

Every element is equally probable so you divide the probability of the group by the number of elements

[rare olive]

is that a well known theorem?

(im not disagreeing, just unfamiliar)

[zinc stirrup]

I mean yeah

It's not a hard proof

Although this whole thing is a simplification

This proof which I'm assuming they're using relies on the assumption that the probability of each theory is independent which is reasonable but wasn't stated

[zinc stirrup]

To prove it assume that there are n elements in a group each with a probability q and the group as a whole has the probability p. Then the sum of all the probabilities of the group which will be nq has to be equal to p or in other words nq=p or q=p/n

[rare olive]

makes sense

[rare olive]

how does it assume probabilistic independence?

i thought it assumed mutual exclusivity

that each theory is mutually exclusive

in the group

[zinc stirrup]

Mutually exclusive events are also independent but that also is an unstated assumption which is not necessarily true

[rare olive]

not necessarily true?

[zinc stirrup]

Each theory has it's assumptions so if there are shared assumptions they aren't mutually exclusive but thinking about that leads you to that no answer is possible because of the lack of information

[rare olive]

okay, also wanted to ask about what probabilities attach to. probabilities attach to sets, so doesnt that mean the elements of the groups also have to be sets?

[zinc stirrup]

Probabilities can attach to sets or individual elements

[rare olive]

even if those elements arent sets?

im just asking due to lack of familiarity

my understanding was that probabilities attach to events, and events are sets of outcomes

so they dont attach to outcomes

the outcomes would be the individual elements

[zinc stirrup]

Well yes they do but each of those outcomes has a probability of it's own

If you take the classic example of dice outcomes are 1 2 3 4 5 6 but an event can be an even number is rolled

So for the event the probability is 1/2 but each outcome has the probability of 1/6

[rare olive]

sure, i see how that works

just to make sure im understanding, the principle says:

if everything in a group, G, is equally probable, then

Pr(g) = Pr(G) / cardinality of G, for any g∈G?

[zinc stirrup]

Yeah

If they're mutually exclusive

[rare olive]

added the correction. is this correct?

if everything in a group, G, is equally probable and mutually exclusive, then

Pr(g) = Pr(G) / cardinality of G, for any g∈G?

[zinc stirrup]

Yes

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@rare olive Has your question been resolved?

[safe radish]

@rare olive Has your question been resolved?

[safe radish]

@rare olive Has your question been resolved?

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[balmy cloak]

How to solve this?

[balmy cloak]

I know with three equation

but I don't know where to go afterwards

I have (5-h)^2+(4-k)^2=r^2

(2-h)^2+(5-k)^2=r^2

(2+h)^2+(3-k)^2=r^2

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[balmy cloak]

.close

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[mellow mountain]

am i tweaking

[mellow mountain]

.close

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[desert pasture]

this would just be span(x) right

[desert pasture]

so $a \mathbf{x}$

[flat frigate]

Veni, vidi, perii

[median vigil]

all vector spaces have at least 2 subspaces, themselves and ____

[desert pasture]

{0}

[median vigil]

yes

[desert pasture]

Cool

How would I prove these are the opnly subspaces though

[median vigil]

you can use the dimension

[desert pasture]

not done that yet

[median vigil]

you can probably swing something about any other subset not being closed

[desert pasture]

cool

thanks

.clos

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[versed bridge]

please save me

[versed bridge]

the equations that I came up with were, r + l + i = 12, 160=2.5r+4l+2i, 2r=l+i

<@&286206848099549185>

but I cant understan the 32 or the r part

nvm

.close

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[bright pier]

i understand all of it except the last line. i don’t know what happened to x^1/3

[plucky elk]

,tex .exp rules

[flat frigate]

riemann

[bright pier]

i understand the x^-2/3 goes to the denominator but where does the x^1/3 go i don’t see it

[plucky elk]

Last row here

[bright pier]

yea but then wouldn’t it be cube root x next to the 4

[plucky elk]

No

$(a+b)/c = a/c + b/c$

[flat frigate]

riemann

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[sleek hornet]

Is there an easier way to do this

[sleek hornet]

Factorise it

[turbid obsidian]

isnt that y^3 + y^3/y^6

take out y^3

y^3 ( 1 + y^-6)

nah thats just complicated

[sleek hornet]

Urs is simplifying I think

[compact rampart]

no wait

u can write

x^3/2 - x^(-3/2)

as x^3/2 - 1/(x^3/2)

let y = x^1/2

then this would become

y^3-1/y^3

for which u can use the identity a^3-b^3

@sleek hornet

[sleek hornet]

Is that what I wrote

[compact rampart]

eh its just more presentable

plus it factorizes it furthure ig

[sleek hornet]

Ah alrite

There's no any other way tho?

[compact rampart]

idont think there is anyohter way

[sleek hornet]

Alrite thx alot

.close

[compact rampart]

no worries

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[tardy mural]

y = -2x - 3

[tardy mural]

i need help putting this in slope intercept form

to find m(slope)

[median vigil]

what does slope intercept form look like?

[tardy mural]

y = mx + b

[vagrant ice]

great, so it's already in slope-intercept form

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[wet sentinel]

Hello, I'm kinda struggling with this one, in the way that I don't even know where to start, help would be appreciated

[thin bridge]

identify the set of points that are equidistant from both

[wet sentinel]

well they would all be on a chord equidistant from A and O, is that right?

[thin bridge]

yes

that chord splits the circle into 2 sectors

[wet sentinel]

Yes

the points in the smaller sector are closer to A

[thin bridge]

yes,

sry, mispoke

it's actually segment, not sector

[wet sentinel]

Oh yeah

[thin bridge]

and calculating it's area will lead you to the proportion

[wet sentinel]

Oh yeah

THANK YOUU

[thin bridge]

np

[wet sentinel]

do i just do

.close

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[split dock]

help pls

[split dock]

someone

I need help

help in this math problem

[strange compass]

!da2a

[safe radish]

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/

[split dock]

can someone help me solve 2x+4=8

[split dock]

ok

can someone help me

[split dock]

I took out 4

so 2x=8

[strange compass]

what

[split dock]

I took out 4

[strange compass]

u cant just remove numbers like that

[split dock]

why not

[strange compass]

cuz then id have a billion dollars in my bank account no

just add a billion

[split dock]

oh ok

what do I do with that 4

do i subtrac

[strange compass]

whatever operation u do u need to do for both sides

so subtract 4 on both sides

[split dock]

what is operation

?

[strange compass]

+-x/

[split dock]

ok

[strange compass]

amongst others

[split dock]

so 2x=0

[strange compass]

no

[split dock]

I mean 12

2x=12

[strange compass]

ur subtracting

from both sides

[split dock]

oh

2x=8

x=4

[strange compass]

wait

[split dock]

thanks

[strange compass]

no wrong

2x+4=8

ur only subtracting from one side again

do u know what 8-4 is

[split dock]

4

cause I am holding 8 fingers and I take 4 fingers away and now I have 4 fingers

[strange compass]

that is rather inefficient but at least its correct

ok now 2x=4

[split dock]

ok now u do the complex

u split in half

dividie

so u take out x

x=2

[strange compass]

ok yes

[split dock]

can u help me in other problem

it small one

[strange compass]

say it

[split dock]

this is the anwser key'

I dont get the formula

methord

[strange compass]

what do u not understand about it

[split dock]

so u put it like (2,1)

=

[strange compass]

huh

midpoint formula is just taking the average of two points' x and y vals

[split dock]

OH OK I GET IT NOW

TY

[strange compass]

[split dock]

yo last 1 problem pls help me

PLS

[strange compass]

ok but hurry

[split dock]

the difference of logarithms is the logarithm of the quotient?

@strange compass so how do we write it?

?

[split dock]

yo

yo

[strange compass]

stop pinging me

this is a log rule

[split dock]

I see

[strange compass]

its the second one

[split dock]

this as
ln2 
x  1
x  1 

[strange compass]

cant read any of that

[split dock]

In2(x-1)/(x-2)=In28

?

In 2 under and 8

[strange compass]

if u mean

[split dock]

yea I mean that

[strange compass]

$$\ln_ 2 \frac{(x-1)}{(x-2)} =\In_{2} 8$$

[split dock]

yes

[flat frigate]

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

[strange compass]

k i give up

u know how to do it hpoefully

[split dock]

yea I got it

[split dock]

can u help me 1 more question

@strange compass

[strange compass]

im going to sleep but why ru doing this when u cant solve first order algebra equations

[split dock]

I can bro

(after substitution u = y
4 + 1), thus 1
4
ln(y
4 + 1) = sin x + C1 or ln(y
4 + 1)

[split dock]

Bro I am in pre-cal rn

[strange compass]

yea no its not checking out lol

[split dock]

r u good

[strange compass]

separable equations are after calc 3

so i really doubt that its u whos doing this

[split dock]

my brother is here with me

hes in calc 3

hes in colledge

hes smart kid

hes hsarp

sharp

/leave

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[vague flame]

Idk how to start this

[vague flame]

question 63

[frozen marlin]

the equation of a tangent line at "a" for some f(x) is

f(a) + (x-a)f'(a) = 0

and i'm sure you can figure out the normal line using that

[vague flame]

when they mean normal line, it means its perpendicular to the tangent right?

[frozen marlin]

mhm

[vague flame]

so would i just solve for the derivative

[frozen marlin]

for part b i'm sure you can just see that that means f'(x) = 0 i.e. find extrema

[frozen marlin]

and substitute into that equation, yes

[vague flame]

ngl i forgot how to get from the derivative to find the slow of the line

oh

do i just put the x coordinate for x in the derivative?

[frozen marlin]

yes

[vague flame]

oh ok

[frozen marlin]

e.g.

for x^2

if you want to find the slope at x = 2

differentiate to get 2x

then 2(2) = 4

[vague flame]

ok so i got 4(4x^2-8x+3)^3(8x-8) as the derivative

would i just plug in 2 for x?

to get the slope

?

[frozen marlin]

uh

yeah

slope and equations of a line are different though

[frozen marlin]

this is the eq

[vague flame]

wait

so id use the equation?

[frozen marlin]

yes

it says equation of the tangent line

[vague flame]

so u dont plug in the x coordinate

[frozen marlin]

not slope

[frozen marlin]

just use this

[vague flame]

ok cuz i dont remember learning that 😅

[vague flame]

ok but for the slope, id just plug it in right?

[frozen marlin]

yes...

[safe radish]

@vague flame Has your question been resolved?

[vague flame]

i dont get this help

so am i solving for x?

ok wait so how do i do part b

[vague flame]

how do i do that

[undone burrow]

that's the tangent line, so it's just some function of x

[vague flame]

i was able find part a

but how do i do part b

[undone burrow]

you set find the roots of that

[vague flame]

wdym?

[undone burrow]

set it equal to 0
since it's already factored, you can just find where those factors are equal to 0

so like
(8x-8) = 0
and (4x^2-8x+3)^3=0

[vague flame]

ohh

then i use algebra and solve?

[undone burrow]

yea

[vague flame]

ohh ok

oh and one last semi-related question.
how do i find the second derivative

is that just the derivative of the derivative

[undone burrow]

yea

[vague flame]

oh ok tyy

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[quick ibex]

Is it possible to have 2 shapes with a difference in number of sides of 1 be inside each other such that the inner shape perfectly touches all sides of the outside shape, assuming the inner shape is the one with an extra side. If so, which ones can this occur if your range of shapes is square to dodecagon.

Example: Can a pentagon fit inside a square and have 4 of the pentagon's sides be parallel to the sides of the square + touching them?

[eternal carbon]

yes

[safe radish]

@quick ibex Has your question been resolved?

[quick ibex]

How, is there a way to show that?

[safe radish]

@quick ibex Has your question been resolved?

[safe radish]

@quick ibex Has your question been resolved?

[dry badge]

My bad.
Thanks for the tip

[quick ibex]

np

[quick ibex]

Anyway, can anyone explain how I'm meant to show this?

[cold relic]

If you have a shape and draw a line running through two adjacent sides of that shape, you get a shape with one extra side. Would that shape fit your description?

[quick ibex]

Sorry, do you have a visual representation? Also would that still be the standard shape, or would it be irregular?

[cold relic]

Irregular

Like this

For a square/pentagon

[quick ibex]

nonono, you need the whole pentagon inside the whole square

like the image on the right, but all sides of the pentagon must be parallel to the square and touching the square except for 1

[cold relic]

You could place this inside a square

And 4 of the sides would be part of the sides of the square

[quick ibex]

Oh, gotcha, what about if they all have to be regular shapes?

[cold relic]

I’d say then the angles are predetermined such that they can’t all be parallel

[quick ibex]

Great, thanks

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[normal cedar]

Hello

[normal cedar]

I need help simplifying sqrt(x)/x^2

[lean otter]

root(a^2) = a

root(x^2) = x

root(a/b) = roota/rootb

x^2 = root(y)

can you find y?

[normal cedar]

What

[lean otter]

well

a = root(a^2)

[normal cedar]

Yes

[lean otter]

so x^2 = root(x^4)

now there is root in both denominator and numerator

you can take root on the whole fraction now

[normal cedar]

So I should square my expression?

[lean otter]

squaring and then taking the root won't change

the expression

so yeah

[normal cedar]

Okayy

so I'd get x/x^4?

[lean otter]

yeah

now it will become root(x/x^4) = root(1/x^3)

or 1/(root x^3)

root(x^3) = x * root(x)

so it comes 1/(x(root(x)))

,, \frac {1}{x*\sqrt{x}}

[flat frigate]

anjali

[normal cedar]

So I gtg if you can send this in dms it'd be appreciated

.close

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[cedar void]

could I get some clarification, is this legitimate or a joke?

[main mural]

it's legit

similar to king's rule for integrals

[cedar void]

okok
got it

tysm

[main mural]

this is true because addition is commutative

[cedar void]

ty comrade

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[quiet jewel]

How to do 1

So basically

I'm like yay 30% sure it's just 21!/26!

<@&286206848099549185>

[quick ibex]

Could you briefly explain where you got the 21! from, I think I know but just want to double check

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[raven veldt]

Need some help improving/checking proofs. Here is the first one. I have already edited it a bit (as seen in red)

[lean otter]

Very blur

[raven veldt]

would you like me to rewrite it?

[lean otter]

Yes

[brave wolf]

Looks fine to me

[lean otter]

Downside is blur

[brave wolf]

What kind of blur?

[raven veldt]

I have to write it in laTex anyways so this is just rough work

[brave wolf]

Are you talking just about the pic?

[lean otter]

Yep

[brave wolf]

it doesnt appear to be blurry to me 🤷‍♂️

[raven veldt]

There is also this one. I am unsure if I should replace what is circled with what I have there in red

Also not sure if the proof is structured in the best way it could be

[brave wolf]

You could probably start directly by presenting the counterexample

but this works as well

the first part shows that if it's associative, then for all g, h, i, [g + (h+i)^3]^3 = [(g+h)^3+i]^3, and later you presented a counterexample to this. So your proof works as well

[lean otter]

Well c) e=0
d)No
e)No

[brave wolf]

!noans

[safe radish]

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.

[brave wolf]

Also, I disagree on c)

[lean otter]

No e is the identity element

[brave wolf]

I actually disagree on more things than just c

[lean otter]

What

[brave wolf]

but let sara do it herself, i think she can do it

[raven veldt]

Here is what I have for them

[brave wolf]

Your answers look correct, let me check the proofs

[short topaz]

yeah i think the proofs look fine

[lean otter]

Yeah e is identity element

[brave wolf]

this line looks incomplete to me

there should be = g as well

since you use it on the next line

but that's a detail

[short topaz]

i mean tbf i think that line's probably redundant

cus u don't need (g+e)^3 = (e+g)^3 to conclude there's no identity element

[brave wolf]

I agree, it just simplifies it to (g + e)^3 which can be done as part of the next step

[brave wolf]

this is true as well

[lean otter]

To find an identity elemente such that
ge=g and e*g=g for all g€ R.

Set ge=g:

This simplifies to:

39e2e+g+392e=939e2e3+ 3g²e = 0

For this to hold for all 9, each coefficient
must be zero:

3e2 0 e 0

e3=0

e=0

3g²e=0⇒e=0

Thus, e = 0 is the identity element.

Result: The identity element is e = 0.

[brave wolf]

is that gpt?

it looks copied smh

3e2

[lean otter]

I guess i am right

[short topaz]

chatgpt is a large language model trained to give responses that look like human language

[lean otter]

E has 0 identity element

[short topaz]

that does not mean it can do maths

[brave wolf]

anyway, even if your proof is correct, you have only shown ge = g -> e = 0

[brave wolf]

Which isnt sufficient to conclude that 0 is an identity element

[brave wolf]

you would need to verify the other direction for that, and that's where you would fail miserably

[lean otter]

Can you defend your statement

[brave wolf]

Ofc I can

[short topaz]

i think that's fine now

[short topaz]

????

[lean otter]

Prove

[short topaz]

i mean i can really easily disprove that 0 is not an identity

[brave wolf]

this is =>, not <=>

[short topaz]

0 * 2 = 8

not 2

qed

[raven veldt]

Do you think this is enough or do I have to show more?

Becuase idk if as indicated in part (c) really can be done : )

[brave wolf]

Btw, i think that it could include a line where you explicitly assume that e is an identity, otherwise it's fine.

For a simpler proof, you could also just note that by setting g = 1, we get that e must be equal to 0. But by setting g = 2, we see that e = 0 doesnt work. Hence there is no identity

[brave wolf]

can you explain why does the existence of inverse for all integers require the existence of an identity element?

[raven veldt]

hmm okay so just saying where e is the identity element would not suffice (like I did at the top there)?

[brave wolf]

oh, i didnt notice that, yeah it's probably good

and commutativity is good as well

[raven veldt]

Okie so just this one then needs to be tweaked a bit?