help-17

[viscid bridge]

Alsoo thankss kizzyy

Soo do we just ignore the >= sign?

[hybrid flicker]

I mean it's true that 39 >= 5k-1

but that's not enough to find THE value of k

you also need the info 39 <= 5k-1

(so 39 = 5k-1)

It's better to follow this reasoning immediately

[viscid bridge]

Okokk i seee

[viscid bridge]

Thankuu so much

Guys i have another question

Am i doing it wrong? Why do i get decimal

[dull bear]

Isn’t your p supposed to be 15a^2?

If you want everyone, you should include the <@&268886789983436800>

[dull bear]

@viscid bridge

[terse forum]

what happened

[dull bear]

Spammer with a link, they’ve already been packed

[terse forum]

oh I see

[viscid bridge]

Oh ure right

[viscid bridge]

Thank you so muchh

.close

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[viscid bridge]

.reopen

[vocal sleet]

✅

[viscid bridge]

Oops nvm

.close

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[sudden peak]

Can someone pls explain this and work it out

[dense perch]

you start with centres yeah

[sudden peak]

Yeah?

[dense perch]

holdup imma draw it

[sudden peak]

💛

[dense perch]

so now we see

the dist between centres is 10

and sum of radius is 14

so they must overlap

[sudden peak]

Yup

[dense perch]

yeah so now we subtract to find the value of length which is common to both

it will be clearer to draw on desmos

[sudden peak]

Ummm I’m js gonna say Yh for now….

[sudden peak]

Ohh yhhh

[dense perch]

something like this

[sudden peak]

Yep, so now what?

[dense perch]

now we have been asked what is the area of overlapping region right?

[sudden peak]

Yeah

[dense perch]

think for a minute how so and then ill help you :))

[sudden peak]

I did this before but my answer was wrong but I made a triangle

Wait I gotta draw this

[dense perch]

hmm

[sudden peak]

It’s so bad because I’m on my phone but that’s what I had

And then the AQ was 8 and BQ is 6 bec they’re both radio

[dense perch]

yeah thats pretty much correct approach although i gotta check the triangles

[sudden peak]

Radii*

[dense perch]

we have to find area of this triangle approximately

[sudden peak]

This is what I got before

But idk I tried it twice and they were wrong

I got one of my friends to do it and it was also different to what I got

So idk if I’m js dumb in the head or Smt

[dense perch]

no one is dumb its just that things dont click lol

ill try one minute

is the answer 15.6?

[sudden peak]

I have no clue anymore, the book thing I got it off from said 26.56

[dense perch]

oh

[sudden peak]

Yh

[dense perch]

well that is a problem
i used cordinate geometry to come to a solution

[sudden peak]

I used sin rule, and the book used fan

Tan*

sooo idk what to do anymore

[sudden peak]

I also don’t know what coordinate geometry is

[dense perch]

plotting on graph

well this is embarrasing

wait ill try again

[sudden peak]

Oki

If you can’t solve it’s it’s all good, I do have another question I need help with

[dense perch]

i am getting my answer as 8 root7

which is near to 21.2

[dense perch]

yea sure

[sudden peak]

Yeah…idk anymore, the book says 26.6 Smt

But it doesn’t show the working out

How would I do this? Like can you explain?

[dense perch]

ill try

oh i got it

wait ill draw

[sudden peak]

Just a small heads up, this is year 11 methods…so um…I beg you please don’t over complicate it

My brains already fried

[dense perch]

sorry for the bad drawing
see we know that every angle will be 60
angle bisector will be perpendicular so 30
and the using cos 30 we can find the height of triangle and that would be root3r^2/8
and we will subtract this value from pi r^2/6

[dense perch]

i am in year 11 lmao i just finished

[sudden peak]

So um…I got about 70% of that

[dense perch]

ok wait ill draw better

see if you get it

[sudden peak]

Okay, so the cos 30 exact value is the root3/2…okay I get the diagram

[dense perch]

yeah so now we have height of this small triangle you see?

[sudden peak]

Yeah

[dense perch]

we can represent the area of shaded region as total circle region - the region of triangle

[sudden peak]

Wait sorry small question

Where did the 60degrees come from?

[dense perch]

see hexagon has 6 sides and total 360 in middle so 360/6 gives us 60 degree

[sudden peak]

Ohh right factssss

Okay please continue

[dense perch]

yeah so i drew a line to make a 90 degree and it bisected 60 to 30

[sudden peak]

Yuppp

[dense perch]

yeah and then trigonometry to find sides of triangle and then its area

[sudden peak]

So I’d use the area of a segment minus the area of the triangle…?

[dense perch]

yess

total - white part

[sudden peak]

So….um give me a second, let me process rq

Okay so…I might be a little lost again

I get the diagram

[sudden oriole]

do u have to work out the area of the segment 😭

[dense perch]

see
area of circle = pi r^2

so 60/360 degrees area or 1/6 area is pi r^2/6

and then subtract the area of triangle which comes out at roo3r^2/8

[sudden peak]

Why is there an r after the root 3

[sudden oriole]

u dont know the radius

[dense perch]

because the height and base are in terms of radius

[sudden peak]

Huhhh

[sudden peak]

Huhhhh

[dense perch]

🥲

[sudden peak]

I’m Cole’s

Cooked*

[sudden oriole]

is this the formula for that

[dense perch]

i just used trigonometry to find theis length

[dense perch]

idk man its been few month
although i know how to find it

[sudden peak]

So I’m cooked…

[dense perch]

you still dont get it?

[sudden peak]

Nope 😔

I need it to be dummafied x1000 before i understand

[sudden peak]

I get the diagram though

But everything else

[sudden oriole]

are we finding the segment

[dense perch]

yesir

[sudden oriole]

idk

[sudden peak]

Okay…so rq question, let’s just change up the question entirely, how would I find the area of that triangle

[sudden oriole]

sector - segment??? or ishdajsafasj idkdk

[dense perch]

that is correct

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[ruby heart]

could anyone explain why this is true, it was given in a book without any explanation and I dont see why it holds

[ruby heart]

I came across the limit tany/y

but that dfoesnt seem to be helpful

[atomic jasper]

try using lhopital

[ruby heart]

I thought im just not seeing something

i cant

its not um legal

[atomic jasper]

why not?

[ruby heart]

cuz we havent covered that yet

[atomic jasper]

okay

[atomic jasper]

i think this is just a definition the give you then

or is it smth they want you to prove?

[ruby heart]

this is the solution

and im trying to udnerstand why this limit is -1

[atomic jasper]

im saying that its just a given solution

they give you that this limit is equal to -1

,w graph (t-pi/2)tan(t)

[twin meteor]

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[ruby heart]

i am to prove this

[winter hawk]

use tan=sin/cos

[ruby heart]

alr ill try that

[winter hawk]

oh hmm what im thinking uses lhopital if youre allowed

[ruby heart]

sadly no

[winter hawk]

oh it turns out we dont need lhopital

you can rewrite part of the expression as derivative of cos

let $t_0=\pi/2$

[twin meteor]

𝓡𝓞𝓚𝓔𝓣𝓣𝓞─୨ৎ─❥ ♡ <𝟹❤

[ruby heart]

oh from sin(pi/2-x) = cosx?

[winter hawk]

$(t-t_0)\tan t=\sin t\frac{t-t_0}{\cos t}=\sin t\frac{1}{\frac{\cos t-\cos t_0}{t-t_0}}\app[t][t_0]\frac{\sin t_0}{\cos'(t_0)}$

try rewriting the fraction using derivative of cos

[ruby heart]

okay

[winter hawk]

nvm ill just do it

now we're good to take limit

[ruby heart]

oooh

[twin meteor]

𝓡𝓞𝓚𝓔𝓣𝓣𝓞─୨ৎ─❥ ♡ <𝟹❤

[winter hawk]

,w sin(t)/cos'(t) at t=pi/2

[twin meteor]

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[winter hawk]

easy

[ruby heart]

much thanks

[winter hawk]

no prob

[ruby heart]

smart idea to use the derivative

[winter hawk]

yeah i got that idea from seeing t-t0 earlier

but didnt know right away if it would work

[ruby heart]

oh cool

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[hardy crater]

Please could I get some help on this question if possible? Im having a bit of trouble getting started

[hardy crater]

Aside from substituting the formula for z

X

[flat whale]

show what you get after substitution

[hardy crater]

I doubt the cos is right, I was just seeing what I could do

And I missed the oiwer

Power of 1/2

[flat whale]

yea fix your 1/2 power then use double angle formula

,tex .double angle/sin

[twin meteor]

riemann

[hardy crater]

Would the power of a half get the sin2x??

Sorry, trig identities are my weakest subject

[dull bear]

(did you find dx/dtheta as well?)

[sly sierra]

^ aka the best trig identity

[hardy crater]

I’m supposed to be differentiating?

[dull bear]

As part of integration by substitution, yes

[hardy crater]

Ohhh it’s that thing

[hardy crater]

I still don’t really understand where I’m supposed to use this though

[dull bear]

Do this first, it'll become more apparent when you do

[hardy crater]

Would i not be integrating by parts?

[dull bear]

(also what did you get when you fixed this?)

(you won't need integration by parts here)

[hardy crater]

,rotate

[dull bear]

Maybe try simplifying that a bit more?

[hardy crater]

I have no idea what to do with the power, is that what I should be simplifying now?

[dull bear]

Remember that you're effectively square rooting of course, and then surd rules and all that

[hardy crater]

So it’s like this now

.rotate

,rotate

[hardy crater]

Wrong one

Sorry

,rotate

[dull bear]

Not like this sqrt{a + b} is not sqrt{a} + sqrt{b} in general

May help to first factor out $a$ inside that square root, so $\sqrt{a (1 - \sin^2(\theta) ) }$, and hopefully that becomes more apparent what to do with it...

[twin meteor]

@dull bear

[hardy crater]

So that’s better?

Oh you just did that

[dull bear]

Hehe, great minds

[hardy crater]

I don’t think I have one of those

But I’d change that to cos now

[dull bear]

You're not allowed to say that, considering you've figured out what I wanted without me saying

[hardy crater]

So now I’ve got asinxcosx and I should be using that silly double angle rule

[dull bear]

You can use that now, yep

[hardy crater]

Would it be 1/2 for me

Since I don’t have the 2

[dull bear]

Yep if there's no 2, you could "introduce one" from (1/2) * 2

[hardy crater]

So now I’ve got this, I need to do the dx thing you mentioned?

Sorry about how long I’m taking, I must be very annoying

[dull bear]

You do need to do the dx thing, and you aren't annoying at all

[hardy crater]

Sin^2 differentiates to sin2x?

This seems like an awful lot for 4 marks 😭

[dull bear]

It does that's your second factor of sin(2x) for you

And it is a tiny bit involved, not the worstest worst but a few steps, it is a show that so they're looking for your steps I guess

[hardy crater]

Would I do anything to the a whilst differentiating asin^2?

or would it be asin2x

[crimson jetty]

use the chain rule

[hardy crater]

Oh ridge

Right

[dull bear]

the $x = a\sin^2(\theta)$ just turns into $\dv{x}{\theta} = a\sin(2\theta)$

[twin meteor]

@dull bear

[dull bear]

a is a constant so you can just leave it

[hardy crater]

Nearly there 😭

Oh yea it’s all done now

Thank you for your help! I’m hoping part b is much more obvious

[dull bear]

Awwww, a pleasure again part b is slightly less painful, probably just one change to notice you can make I think to get there

You also noticed how to change the integral limits, right?

[hardy crater]

Should I have done that for part a?

[dull bear]

Sadly yep, it's part of the show that

[hardy crater]

More!?

[dull bear]

It isn't too challenging though, I hope

[hardy crater]

I may be wrong but it seems like so much for only 4 marks

[dull bear]

If I'd guess, probably 1 is for the limit change, one for the dx/dtheta, one for substituting correctly and one for fully getting to the final step

[hardy crater]

Would you mind if I asked you one more question briefly?

No problem at all if not

[dull bear]

You can, go ahead

[hardy crater]

It’s something to do with newton raphton

If I’ve spelt that correctly

[dull bear]

I can't remember, but I get the one you mean

[hardy crater]

I think I’vegot parts a and b

But I’m struggling a little on using my f dash x value in the equation

[dull bear]

Be careful of your $f'(x)$, that it's $1 + \frac12 \sec^2\qty(\frac12 x)$ rather

[twin meteor]

@dull bear

[hardy crater]

Oh whoops I forgot to include that

[dull bear]

But otherwise, how so are you struggling?

[hardy crater]

I don’t really know how to get the value for the bottom of my fraction

I should put 3.7 into my differentiated equation right?

But how do I get the proper value?

[dull bear]

One way that I generally like to do it, is to just put 3.7 = into the calculator, to set that to be the "Ans" value

Let me show how I do it on my one

[hardy crater]

Thank you, I don’t really remember how to put sec onto my calculator

Or cos squared

[dull bear]

A bit messy if anything you could also use the memory feature too, though I’m too lazy most of the time to do that

On mine, it comes out as 3.67205...

[hardy crater]

Are you in degrees?

I did that at first aswell

Oh no

You aren’t

[dull bear]

Yep, the R

[hardy crater]

I’ll try it now

Hmm I got 3.96

Oh I thought the cos^2

3.65 😓

Bit closer

[dull bear]

yea maybe the storing way might be a bit less pain

Too easy to mess it up in one go

[dull bear]

The 2 isn't inside the for the denominator

[hardy crater]

Oh whoops

Yes, all done now

Thank you so so much for all of your help today

[dull bear]

Awwww, a pleasure to help

Hope you have a good one

[hardy crater]

Yes, you too!

Well I don’t think I will since my maths exam is tomorrow

Oh wait I’ve been studying so long it’s actually today now

Whoops

.close

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[dull bear]

.(awwww best of luck for it, hope it goes good )

[hardy crater]

Thanks to your help I definitely know a little more so it should be alright

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[opaque shell]

why does it specify "where n is a positive integer"

because can't n also be negative

[mighty nacelle]

yes, but the induction only proved it for integers greater than 1

[opaque shell]

oh ok

also

why aren't these 2 equal?

using the chain rule

[outer warren]

you didn't apply chain rule properly

can you state the chain rule

[opaque shell]

ohhh

i get it now, thanks

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[frozen wagon]

anyone know why the answer not 1/3 ?

[frozen wagon]

is guess or trial and error x = 1.5 at the begining is not allowed or wrong idea ?

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@frozen wagon Has your question been resolved?

[frozen wagon]

<@&286206848099549185>

[lofty vapor]

if x in crease by 1 then let y increase by k (k can be negative)

[frozen wagon]

yeah so yeah x = 1.5

[lament ore]

Solve for a

[frozen wagon]

wait

[lament ore]

Did you try this algebraic approach?

From here, you can break down the fractions individually, and keeping in mind that (x/3) + (y/2) = 1, you can solve for a.

[frozen wagon]

hmmm i don't get it

[lone goblet]

I say we make it into a linear equation

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[lone goblet]

2x + 3y = 6

[frozen wagon]

ok so this is not allowed or what ?

[lone goblet]

Such that y = -2/3 x +2

[lone goblet]

So when x goes up by 1

Y decreases by 2/3

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[modern bolt]

If I could please all the possible combinations within these numbers

2100/200

(my example: 2100*200=420000)
(Example possibilities:
2100/199
2100/198
2100/197)

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@modern bolt Has your question been resolved?

[modern bolt]

<@&286206848099549185>

[dark kiln]

i maybe get the assignment, but i don't get what you expect

like someone uploads a million line text file?

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[barren stirrup]

heres where i got stuck

[cunning plaza]

For the positive reals, x^2+1 is also equal to or greater than 2x. Try going down that path

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[jade vault]

I’ve been working on a thought experiment related to the roots of negative numbers and motion equations. While analyzing two runners, Alice and Bob, I came across an interesting result that I’d love to discuss.

The Setup
I have two graphs:

The first graph represents how Alice and Bob move in real life, with their respective velocities and accelerations plotted. Their paths are shown, and I derived an equation to determine when they meet.
The second graph is a hypothetical scenario where their paths extend into negative time. This setup suggests they would meet at
𝑡 = − t=−6 seconds, implying a quadratic equation of the form:
𝑡^2 − 6 = 0

The Question
When solving for their meeting time using the first graph (the real-world scenario), I get two solutions:

𝑡 = 0 (which makes sense)
t=−6 (which seems impossible in real life)

However, this negative time solution appears naturally in the second graph. This made me wonder:
Is there a way to derive an equation directly from the first graph that inherently gives both solutions?
Does the equation depend on their starting positions?
Am I possibly working on a problem with an impossible solution, or is there a deeper mathematical reason for the negative time result?
I’d love to hear thoughts on whether the negative time result has physical meaning or if it's just an artifact of solving the equation mathematically. Let me know what you think!

[heavy yoke]

well the negative time solution says that they would have also met 6 seconds before starting had they been following the same trajectories before the time t = 0 (assuming they only actually start moving at t = 0)

[jade vault]

It is very interesting

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[woeful wagon]

how???

[vocal sleet]

@woeful wagon Has your question been resolved?

[sharp lynx]

how what

[woeful wagon]

how do you draw the graph of f'(x)??

if its too stupid o explain pls refer a yt video cause i feel like im losing my mind haha...

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@woeful wagon Has your question been resolved?

[marsh jungle]

ill give you an example of how to get a point on f'(x).

step 1. find f(1). keep in mind you dont have to start with 1, its just an example

step 2. find the slope at f(1)

the slope at f(1) is f'(1). so you plot that point

doesn't need to be every single point, you can take a few major points from the graph and sketch out what it would look like

important points to definitely include are wherever the graph changes direction, changes from + to - slope or vice versa, or horizontal/vertical slopes, etc.

hope this helps :)

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[glad jasper]

help

[glad jasper]

<@&286206848099549185>

[hollow sundial]

(pls only ping after 15 mins)

[hollow sundial]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[glad jasper]

ik tat it have got somthing to do with the turning point by lokking at the completed square

but idk how to get the correct answer

[hollow sundial]

for a?

[glad jasper]

ab and c

[hollow sundial]

well for finding the max width

you know the end points are at h = 0 correct

[glad jasper]

yes

[hollow sundial]

how would you solve for that

[glad jasper]

i open up 6-(x-3)^2 ?

[hollow sundial]

yes

then you can solve the quadratic

[hollow sundial]

yep

[glad jasper]

do i use this now?

[hollow sundial]

yes

[hollow sundial]

so the height is 0 at x = 3 - sqrt(6) and 3 + sqrt(6)

the maximum width would therefore be the difference between the farther and closer point

$w=(3+\sqrt{6}) - (3 - \sqrt{6})$

[twin meteor]

BuilderDolphin

[vocal sleet]

@glad jasper Has your question been resolved?

[glad jasper]

im still processing the height

i dont understand the height

[vocal sleet]

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[nocturne trout]

does the sign switch from -1/2 to 1/2 indicate that it's being moved from the numerator to denominator?

[cunning ingot]

yes

the 1-x/1+x has flipped because of the negative power

[jagged cargo]

$\left(\frac{1 + x}{1 - x}\right)^{-1/2} = \frac1{\left(\frac{1 + x}{1 - x}\right)^{1/2}} = \left(\frac1{\frac{1 + x}{1 - x}}\right)^{1/2} = \left(\frac{1 - x}{1 + x}\right)^{1/2}$

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[nocturne trout]

ty

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[nocturne trout]

why does square root of (1+x) disappear?

nvm got it

.close

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[finite hatch]

Excuse me whats the difference between square bracket and normal bracket

[finite hatch]

for domain and range

[hybrid flicker]

like [a,b] vs (a,b)

one includes its bounds []

[outer warren]

its not specific to domain and range

[hybrid flicker]

the other excludes them ()

[outer warren]

but interval notation

[hybrid flicker]

you can mix them up

[a,b) contains it lower bound a, but not its upper bound b

[finite hatch]

ohh ok

also

why is it not x>=-2

for the last part

because cant u sqrt0

[hard atlas]

the domain of h^-1 is the range of h

which is x>=-1

[finite hatch]

oh ok

is x>=-1 thed same as

[-1, infinity)

[hard atlas]

[-1, infty), yes

[finite hatch]

ohh ok

thanks

[finite hatch]

wait im confused on the brackets now why they using both..

if u use the square bracekt

does that mean equal to?

[hard atlas]

the square bracket means that the relevant number is included in the set

infinity is never included, so its always with a round bracket

[finite hatch]

ohh ok

ok thanks so much for ur help

i wil,l be back..

[hard atlas]

so [-1,infinity) means that x>=-1, (-1,infinity) would mean x>-1

[finite hatch]

ahhh okk

[hard atlas]

and for example [3,4] would mean 3<=x<=4 while (3,4] would mean 3<x<=4

[finite hatch]

wait one more questions

how do i know when a function is defined vs undefined

[hard atlas]

when you compose two functions you need to make sure that the outputs of one function are allowed as inputs for the other function

[finite hatch]

wat

wait

is inputs the domain

and outputs the range

[hard atlas]

yes

more colloquially

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[red crystal]

The length hanging down as a function of time and the speed when the chain loses contact with the table are derived using Newton's second law and energy conservation.

Length as a function of time:
The differential equation governing the motion is ( y''(t) = \frac{g}{\ell} y(t) ). Solving this with initial conditions ( y(0) = y_0 ) and ( y'(0) = 0 ) gives:
[
y(t) = y_0 \cosh\left( \sqrt{\frac{g}{\ell}} , t \right)
]

Speed when losing contact:
Using energy conservation, the speed ( v ) when ( y = \ell ) is:
[
v = \sqrt{ \frac{g(\ell^2 - y_0^2)}{\ell} }
]

The length hanging down is (\boxed{y(t) = y_0 \cosh\left( \sqrt{\dfrac{g}{\ell}} , t \right)}) and the speed when contact is lost is (\boxed{v = \sqrt{\dfrac{g(\ell^2 - y_0^2)}{\ell}}}).

I'm not sure how this solution works

[twin meteor]

nino

[red crystal]

for this problem

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[north field]

how to solve this

[regal bane]

Know the determinant?

[north field]

yes

[regal bane]

You can also row reduce

But det is easier to reason with. When det ≠ 0, inverse exists

[north field]

sorry, I still cannot solve this

[proud terrace]

you can

What is the determinant formula for a 2x2 matrix?

[gentle nimbus]

a11 x a22- a12x a21

[north field]

ad - bc

[proud terrace]

yeah, apply the formula for A

[north field]

but this != 0, i don' know how to solve, x**2-9x-162 != 0

[proud terrace]

$\Delta= 9^2 -4*(-162)=729$

[twin meteor]

tm

[proud terrace]

and $\sqrt{729}=27$

[twin meteor]

tm

[proud terrace]

you know the solutions then

[north field]

why this need put square root

[proud terrace]

to use quadratic formula

$x=\frac{-b \pm \sqrt{\Delta}}{2a}$

[twin meteor]

tm

[north field]

ok, i finally clear this

thank you

[proud terrace]

don’t forget, it’s $x^2 -9x -162 \neq 0$ 😉

[proud terrace]

np sir

[twin meteor]

tm

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[spark surge]

.close

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[last agate]

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[stray sonnet]

help

[stray sonnet]

how do I solve this

nvm

.close

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[stray sonnet]

.reopen

[vocal sleet]

✅

[stray sonnet]

yea

how do I solve this

[charred fiber]

Do you know how to complete squares?

[vocal sleet]

@stray sonnet Has your question been resolved?

[stray sonnet]

yes

x^2+a^2 = (x+b/2)^2 -(b/2)^2

but when I factored

I got x(x+4)

so im a little confused

[charred fiber]

Nah you wanna make it (x+2)^2 so what are you missing

[gentle nimbus]

(x+2)^2-2^2

then just

use the standard integral

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[faint pumice]

how do i simplify this

[faint pumice]

i got the answer but idk how they got it

[oak magnet]

Factor by arctan(...)

[faint pumice]

ohh

okay

[oak magnet]

You will have arctan(..)[1/12 rt(48) - 1/3 rt(48)]

[faint pumice]

gotcha, i completely forgot about factoring out

[oak magnet]

Thats the only thing you can do actually

[tidal dock]

you can also combine the fractions after that

[faint pumice]

couldnt i multiply the right side by 4 to make the denominators equal and just do that??

or does that not work

i tried it before but i feel like my math was entirely wrong and not just that

[tidal dock]

wdym by "the right side"

[tidal dock]

this isn't an equality

no sides here

you mean $\frac{1}{3\sqrt{48}} \to \frac{4}{12 \sqrt{48}}$ ?

[twin meteor]

artemetra

[faint pumice]

i meant right arctan

[faint pumice]

yesa

[tidal dock]

yes

should be good

[faint pumice]

ah okay

thank you

.close

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[terse berry]

Hi I’m trying to derivate this but I’m having trouble

[vocal sleet]

Please don't occupy multiple help channels.

@terse berry Has your question been resolved?

[fast shell]

Quick tip, u can use https://www.derivative-calculator.net/ if you're ever stuck on a derivative and want a step by step solution to self check. Hope you still get the help you need tho ;p

[vocal sleet]

@terse berry Has your question been resolved?

[dull bear]

,rccw

[dull bear]

You may just wanna simplify this down to $y = -\frac43 x + \frac{20}{x^4}$, then differentiate that expression instead

[twin meteor]

@dull bear

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[red crystal]

Acceleration with Frictionless Incline

The system consists of two masses ( m ). For the hanging mass, the tension ( T ) and gravitational force act. For the incline mass, the force up the incline is ( 2T ) (due to two tension forces from the pulley), and the gravitational component down the incline is ( mg \sin 30^\circ ). The accelerations are related by ( a' = 2a ).

Equations:

1. Hanging mass:
   [ mg - T = ma ]
2. Incline mass:
   [ 2T - mg \sin 30^\circ = m(2a) ]

Solving:
Substitute ( T = mg - ma ) into the second equation:
[
2(mg - ma) - 0.5mg = 2ma \implies 1.5mg = 4ma \implies a = \frac{3g}{8}
]
Thus, the accelerations are:

• Hanging mass: (\boxed{\frac{3g}{8}}) downward
• Incline mass: (\boxed{\frac{3g}{4}}) up the incline

---

Minimum Coefficient of Static Friction

For equilibrium (( a = 0 )), the tension ( T = mg ). The static friction ( f_s ) balances the forces:
[
2T = mg \sin 30^\circ + f_s \implies 2mg = 0.5mg + \mu_s mg \cos 30^\circ
]
Solving for ( \mu_s ):
[
1.5mg = \mu_s mg \frac{\sqrt{3}}{2} \implies \mu_s = \sqrt{3}
]
Minimum coefficient: (\boxed{\sqrt{3}})

---

Acceleration with Kinetic Friction

With kinetic friction ( f_k = \mu_k mg \cos 30^\circ ), the equations become:

1. Hanging mass:
   [ mg - T = ma ]
2. Incline mass:
   [ 2T - mg \sin 30^\circ - \mu_k mg \cos 30^\circ = 2ma ]

Solving:
Substitute ( T = mg - ma ) into the second equation:
[
2(mg - ma) - 0.5mg - \mu_k mg \frac{\sqrt{3}}{2} = 2ma \implies a = \frac{g(3 - \sqrt{3}\mu_k)}{8}
]
Thus, the accelerations are:

• Hanging mass: (\boxed{\frac{g(3 - \sqrt{3}\mu_k)}{8}}) downward
• Incline mass: (\boxed{\frac{g(3 - \sqrt{3}\mu_k)}{4}}) up the incline

---

Note: Replace (\mu_k) with the given coefficient in part (c).

[twin meteor]

nino

[red crystal]

i did this problem but the solution seems kinda simple and im pretty sure this problem was supposed to be a lot harder

so i feel like there probably is something wrong

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@red crystal Has your question been resolved?

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@red crystal Has your question been resolved?

[north field]

help me solve this, i don't understand this question

[wary hull]

What do you know about properties of a determinant?

[red crystal]

my original question

i think the kinetic friction is wrong

i think it should be going up instead of own cuz the aciton is slipping not slidding up

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[jovial bear]

Hello

[jovial bear]

Does someone understand technical drawing ?

[fervent wasp]

What’s that?

[jovial bear]

I know its not math but maybe someone studied it, i have examn tomorrow and i seached a lot of server but no one understand it

[edgy kayak]

give us more context

[dark kiln]

no, ask the question

[vocal sleet]

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[tiny trail]

can someone guide me ques64a so that i can do the rest

[tiny trail]

how to change the order

[vocal sleet]

@tiny trail Has your question been resolved?

[bitter pilot]

did you try to draw the region

[tiny trail]

yes

[bitter pilot]

and what did you conclude for dxdy order

[tiny trail]

-sqrt(1-y^2), 1-y^2 and -1, 1

i think i just need change x to y then change 2 intergrations

[bitter pilot]

-sqrt(1-y^2) covers just one part of the circle and 1-y^2 is not correct

you need to solve properly the functions

[bitter pilot]

you will need two integrals because you have a transition of functions at y = 0

vertically it's just easily bounded as given in the exercise but horizontally, you have to cover the 2 regions on their own

the first being bounded from blue to purple and the second from red to green

[tiny trail]

so interval are -sqrt(1-y^2) to sqrt (1-y^2) and -sqrt(1-y) to sqrt(1-y) and -1 to 1

[bitter pilot]

the -1 to 1 part is not quiet well worked out

-sqrt(1-y^2) to sqrt (1-y^2) with what exact bounds of y

just look at the picture

[tiny trail]

so just only -sqrt(1-y^2) to sqrt (1-y^2) and -sqrt(1-y) to sqrt(1-y) right ?

sr im clueless

=))

may be -1 to 1 and -sqrt(1-y^2) to sqrt (1-y)

[bitter pilot]

just ignore the bounds for y

you only told me the bounds for x

[tiny trail]

ignore the boundary of y its -1 1 obviously !! maybe 0 to 1 =))

ah

-1 0 and 0 1 right ??

[bitter pilot]

no

yes

[bitter pilot]

else you would integrate more

[tiny trail]

what yess

[tiny trail]

this one

[bitter pilot]

if you just do -1 to 1 respectively

you get out of your region

[bitter pilot]

red and green start at y = 0 and stop at y = 1

they dont start at y = -1

[tiny trail]

cảm ơn

i have to go class now

bye

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[cloud orbit]

um how do i work out the reciprocal of 20

[sly sierra]

how do you work out the reciprocal of any number in general?

[cloud orbit]

idk

wths reciprocal

😭

[sly sierra]

well look it up haha

we are not wikipedia

[brazen sluice]

1/20

[flat whale]

,w reciprocal

[twin meteor]

Results provided by WolframAlpha

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

Requested by rie.mann

[flat whale]

neat

[cloud orbit]

what on earth

[cloud orbit]

oh tysm

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[faint pumice]

is the divergence test just to see if the sum diverges

[faint pumice]

or the function

i dont understand why if the function has a limit that means the sum diverges

[agile ore]

hi! There is a test tmr and there's a question i cant seem to understand. may i post it here?

[noble spindle]

#❓how-to-get-help

[faint pumice]

you can post it in an available chat

or in the help forum

[pale perch]

well, if the limit of the thing im summing is say, 1, then i will end up with an infinite number of 1's at some point so the sum diverges

if its 0 that wont happen

though that doesnt mean it will converge

need another test for that

[pale perch]

in short, yes

[faint pumice]

okay i see

if the sum diverges, you still have to find whether the function of the sum converges or diverges though?

right

[pale perch]

yeah, it going to 0 is an inconclusive result

it either says: yes it diverges (if not to 0)
or: it could or could not (if tends to 0)

[faint pumice]

if there is a limit, the sum diverges and the function converges

[winter hawk]

the only conclusive thing we can say is this

[faint pumice]

and if there isnt, or the lim is 0 then another test

for the sum at least

[winter hawk]

if $a_n$ diverges or has a nonzero limit then $\sum a_n$ diverges

[twin meteor]

𝓡𝓞𝓚𝓔𝓣𝓣𝓞─୨ৎ─❥ ♡ <𝟹❤

[faint pumice]

okie

gotcha

thanks yall

.close

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[winter hawk]

np

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[vast shale]

why is 15x(-2) = -30?

[paper depot]

well let's see

is it the negatives that trouble you?

[vast shale]

i guess yeah

[paper depot]

ok so let's just strip those away for a second

15 * 2 = 30

does this look clear to you?

[vast shale]

ye

[paper depot]

ok, and do you know how to work with signed numbers when it comes to multiplication?

"signed" meaning they could be positive or negative

[vast shale]

i haven't been on school since 5y ago, so i don't really remember what was it tbh

[vast shale]

o

like - + - is = +

?

[paper depot]

not like that

negative times negative gives a positive answer.

but negative times positive, in either order, will give you a negative result.

you can recall that any negative number can be thought of as a positive number times -1

so 15 * (-2) = 15 * 2 * (-1)

[hard atlas]

maybe a step slower, what is (-2)+(-2)

[solid sage]

Is it 4

[vast shale]

sorry if i don't understand so well, english is not my native language

[solid sage]

😦

[paper depot]

it isn't, but also the question was directed at OP not you

[paper depot]

ok, what is your native language?

[vast shale]

spanish

[paper depot]

ok, nevermind...

who?

[vast shale]

💀

[vast shale]

what does time stands for in that

[paper depot]

times = multiplication

[vast shale]

oh

i never heard that

[paper depot]

... i don't remember/know the spanish word for it

hold on

[vast shale]

dw i get it i guess

[paper depot]

por, apparently?

[vast shale]

ye

it is

[paper depot]

also yes khanacademy comes in a few different langauges, incl. spanish

[vast shale]

i was using it but i started watching a very basic guide to step a bit slower

since i was already getting into multiplications and i can't remember the numbers so well

[paper depot]

yes then that's definitely a sign to drop back to addition/subtraction

[vast shale]

or idk where i started

[paper depot]

with khan academy if i recall correctly the way it works is that it just auto drops you back to whatever level you actually need

gradually

[vast shale]

i didn't understood

it may be the lenguage

yeah

[paper depot]

deuda

[vast shale]

i feel so stwupid 🙏 😭

hm

[vast shale]

so back to the question

i understand that negative por negative is positive

and the other one

but how do i apply that here

[paper depot]

negative * positive = negative as i already told you

[vast shale]

hm

[paper depot]

when you multiply 15 (positive) and -2 (negative) you get a negative number as the result

thus, -30

[vast shale]

oh i see

understood

thank you guys for helping me out, i definitely need to practice alot again, too many problems that i need to solve

oh yeah i forget to say, i learned english by my own

[paper depot]

your english is pretty good for self-study

[vast shale]

idk how i made it 💀

i just...

started talking

and that's all

i didn't watch a single guide

i feel kinda ashamed for my math level, but i think i can get alot better than i expect if i put a lot of effort on it

so i'll come back whenever i need any help or sum

yeah?

mhm

ye

-30

bc i go down 2 times 15 meters

and it's negative

since i'm below the swimming pool

not on top

https://cdn.discordapp.com/attachments/990004172755726488/1346185306726203442/SnapTwitter.io_viktortaliss_522-ezgif.com-video-to-gif-converter.gif

ye

ye

the order of the product doesn't affect the product

understood

tysm

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[calm jackal]

yo

[calm jackal]

i don't even know how to set up this question

plugging in numbers for C that are less than 9 after being added to 3 is a way to do it, but I just want to know if there's a more efficent approach

[proud terrace]

first, put outside the sum 4c and then express $(\frac{c+3}{9})^{k-1}$ with $(\frac{c+3}{9})^{k}$

[twin meteor]

tm

[calm jackal]

yeh

so you make the starting term 0 yeah i already did that

[proud terrace]

I didn’t think about that 😅

just to make a clear geometric series

[calm jackal]

q

just so u know

they gave us 4 answer choices

i can plug in, but i want to know is there another way to do it

[proud terrace]

yeah, we can find c instead of seeing which values works

$\sum_{k=1}^{\infty} 4c (\frac{c+3}{9})^{k-1} = 4c * \frac{9}{c+3} \sum_{k=1}^{\infty} (\frac{c+3}{9})^{k}$

[twin meteor]

tm

[proud terrace]

you agree ?

[calm jackal]

you can take out the 4c?

[proud terrace]

yeah, 4c don’t depends of k

[calm jackal]

oh yeah

how is it 9 / (c + 3)

u inversed it?

[proud terrace]

yeah

[calm jackal]

tf

you can do that ?

ah i get it

[proud terrace]

$(\frac{c+3}{9})^{k-1} =\frac{(\frac{c+3}{9})^{k}}{\frac{c+3}{9}}$

[twin meteor]

tm

[calm jackal]

yup

then keep change flip denominator

[proud terrace]

do you know the formula for a geometric series ?

[calm jackal]

to find the value it converges to?

[proud terrace]

yes

$\sum q^n$ for example

[twin meteor]

tm

[calm jackal]

if the absolute value of r (in ur example q) is less than 1

it converges

and the value it converges to is

(ar^k)/(1-r)

with a being the inital term

does that make sense or naw

[proud terrace]

hmm i don’t understand this formula 😭

[calm jackal]

damn

how do i make fractions

with texit

[proud terrace]

It’s rather $\sum_{n \geq 1} q^n =\frac{1}{1-q}$ no?

[twin meteor]

tm

[proud terrace]

with \frac{}{}

[calm jackal]

well

that could be another way, but we never learned that way

[wary hull]

$\frac{q}{1-q}$

[calm jackal]

and if what you said is true, arguably my version is better (correct me if im wrong) because n doesn't have to be greater than 1 or equal to 1 - it can start at 0

[twin meteor]

mathisfun

[wary hull]

First term is q

[wary hull]

Should just be $\frac{a}{1-r}$ with $a$ first term, $r$ common ratio

[twin meteor]

mathisfun

[proud terrace]

yeah mb

[calm jackal]

that means the starting term HAS to be 0 right

[wary hull]

Which question?

[calm jackal]

for this

[calm jackal]

this is my question

[wary hull]

Nope

I know

[wary hull]

Plug in k=1 into the expression (summand) and see what you get

This is your first term

[calm jackal]

first term is 4c then

[wary hull]

Yeep

[wary hull]

So use this formula now

[calm jackal]

oh shi

what will r be though

oh nvm

i get it thanks bro

[wary hull]

You welcome

[calm jackal]

thank you @proud terrace

[proud terrace]

np sir

[calm jackal]

close

/close

[proud terrace]

.close

[calm jackal]

.close

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[winged urchin]

How do I do this? Currently, I’m just mashing numbers into the calculator incrementally until I find one that matches the radical. It works but is not time efficient at all.

[spiral turtle]

You mean simplifying radicals?

[winged urchin]

Yeah

[heavy yoke]

you have to start by factoring the number

then you can pull out any factors which are perfect squares

[winged urchin]

I don’t have them memorized

[heavy yoke]

you don't have to have the factorizations memorized

[winged urchin]

How do I do them using math?

Rn I just put random numbers time a random number into the calculator until it equals 504 for exam[le

[heavy yoke]

if you can spot any obvious factors you can divide them out

for example even numbers have 2 as a factor, nunbers ending in 5 or 0 have 5 as a factor, numbers whose digits add up to a multiple of 3 have 3 as a factor (common divisibility rules)

if there aren't obvious factors you can try dividing by prime numbers

[winged urchin]

But how do we know that the square root of 405 is the same as 9 square root of 5

Like is there a formula I can use to find it

[wary hull]

Multiply them out. Or find the prime factorization.

[heavy yoke]

by factoring it as 405 = 9^2 * 5

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@winged urchin Has your question been resolved?

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[desert anvil]

how do i solve for the zeros in this situation?

[desert anvil]

or wait

axis of symmetry

nvm i got it

.close

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[jagged kraken]

im a bit confused about how the answers here are wrong.
if im not mistaken, the local max occurs when the func changes from increasing to decreasing and the peak here should be at 2
then the opposite with local min
the global max being the highest point and opposite for the min

[jagged kraken]

for the global part am i giving the x or y value?

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@jagged kraken Has your question been resolved?

[past thicket]

i suppose y

[past thicket]

double check the scale of the x axis

[jagged kraken]

for the first one, using the where it changes from increasing to decreasing logic, would x=3 be right?

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[rose snow]

Is this it for this question or am I missing something

[outer warren]

,rotate

[outer warren]

from your work,
you're doing dy/du at x=1

[rose snow]

so the difference would be I would sub in u prior to deriving y

[outer warren]

here you want to apply chain rule
$$\dv{y}{x} = \dv{y}{u} \dv{u}{x}$$

[twin meteor]

ℝαμOmeganato5

[rose snow]

So like this instead?

[outer warren]

i wouldn't recommend ' notation in the presence of multiple variables

[rose snow]

im not completely familiar with lebiniz though

[outer warren]

what you're differentiating is on the numerator,
what you're differentiating with respect to / considering is on the denominator

derivative laws are the same

[rose snow]

This is my current solution

[rose snow]

i see

[outer warren]

technically you should've had an x here

also try not to use x for multiplication especially if you're writing your x rigid like that

[rose snow]

is the solution itself correct?

[outer warren]

the final value yes,

[rose snow]

ik that the 2 should be simplified out

for the u'

[outer warren]

the issue isn't the 2

the issue is the missing x

[rose snow]

yea i copied it over wrong

other than that its all gud right?

[outer warren]

yes

[rose snow]

ok tank

.close

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[quiet epoch]

Rather than use the standard definitions of addition and scalar multiplication in R^3, suppose these two operations are defined as follows. With these new definitions, is R^3 a vector space? For this I noticed that (-2,-2,-2) satisfies the additive identity, because the additive identity is not (0,0,0) does this mean it isn't a vector space?

(x_1,y_1,z_1) + (x_2,y_2,z_2) = (x_1+x_2+2,y_1+y_2+2,z_1+z_2+2)
c(x,y,z) = (cx,cy,cz)

[quiet epoch]

In other words I guess, does the additive identity for vector spaces always have to be (0,0,0)?

<@&286206848099549185>

.close

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[sudden peak]

how do i make this move to the right more so i connects to that red line underneath

[sudden peak]

i used an exponsential

exponential

[wise drift]

oh very cute btw

[sudden peak]

thankssss

i tried to make miffy

shes a lot harder to make than i though

thought

Her eyes are a little scary too bec i havent figured out how to fill them in yet

[dense perch]

you used circle equation right?

[sudden peak]

yeah

[dense perch]

just replace = with less than equal to

[sudden peak]

really??

[dense perch]

that would colour the region

yess