help-0

[ocean seal]

riad

[balmy warren]

ok so now you have 2 equations

I suggest you solve for n in both

[dusky blaze]

ok

$200000 < {n+1}(34 + 11n)$

oops

no fraction*

[ocean seal]

riad

[balmy warren]

make sure it is (n+1)(34 +11n)

all of n+1 is multiplied by all of 34+11n

[dusky blaze]

$200000 < (11n^2+45n+11)$

oops

without the n+1

forgot to remove

[balmy warren]

incorrect

I don't think you have expanded that correctly

[dusky blaze]

sorry

the 11n

not meant to be there

[ocean seal]

riad

[dusky blaze]

im so bad at latex 😭

still learning ig

[balmy warren]

are you sure that 11 is your constant term?

what is 1 x 34?

[dusky blaze]

34*

yes

[balmy warren]

can you now put all terms on one side so the quadratic is > 0

[dusky blaze]

i got

-137<n<133

rounded*

[balmy warren]

nope

think about when

$11n^2+45n-199989 > 0$

[ocean seal]

IntelligentCake

[balmy warren]

draw the graph

for what n values is this function above the x axis

thanks @alpine sable

[dusky blaze]

x<-137 133<x

?

x<-137 x>133

the decimal i got is quite large

so i round?

[balmy warren]

this is correct

your other one is not

ok so you solved the second equation

[dusky blaze]

ok

[balmy warren]

now solve the first inequality and see what value of n satisifes both

[dusky blaze]

i see

@balmy warren 134

would it be that?

[balmy warren]

i don't know

I did not solve it

plug in the value and see if it satisfies noth?

[dusky blaze]

ok

so 133?

i plug x=133

i get 100,299

i plug x=133 into Sn

i get 99 thousand something

oh shit i used the wrong a value

a=12

are these correct

if a is 12

and d=11

@balmy warren

-136<x<134

for the first one

x<-137
133<x

wait how would i know

which works

[lone heart]

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[lone heart]

@dusky blaze Has your question been resolved?

[lone heart]

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[alpine sable]

can a function be defined by itself like f(x) = g(f(x)) where we know what g(x) is?

[mortal trellis]

sure

but in general there might not exist a solution to f(x)=g(f(x))

or it is not unique

[alpine sable]

k

.close

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[lone heart]

@little hornet Has your question been resolved?

[little hornet]

<@&286206848099549185>

[high rapids]

Use binomial theorem

[little hornet]

ye ik but im unsure about the n+1 part

[mortal trellis]

well could you solve for $k$ if in the expansion of $(3+x^2)^k$ the coefficient of $x^4$ is 20412 ?

[ocean seal]

Denascite

[high rapids]

Let's see, n+1Cr * 3^(n+1-r) * (x^2)^r

To get x^4, it must be r=2

So

(n+1)!/2!(n+1-2)! * 3^(n+1-2) = 20412

[lone heart]

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[blissful drift]

kinda confused on the different types of symmetry

[blissful drift]

i think it has symmetry with the y-axis

[lament glen]

yeah seems right

[blissful drift]

i just graphed it on desmos so thats why i think its y

what does symmetry on x axis and origin means?

[balmy warren]

symmetry on x axis means that it has reflective symmetry about the x axis

[lament glen]

some examples, hope I got it right

[balmy warren]

yes

[blissful drift]

oh yea then its neither both then

just the y axis

[lament glen]

x axis is like if you have (x, y) on the curve you also have (x, -y)
origin is if you have (x, y) on the curve you also have (-x, -y)

[blissful drift]

ohhh

thanks for the help

.close

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[woeful gulch]

Could some one point out my error in this question

[woeful gulch]

Despite having a different solution to the markscheme my working out seems reasonable and I cant figure out how the answers are different

Top picture is the mark scheme, bottom is my working out

[balmy warren]

they're asking for angle

you worked out a length

[woeful gulch]

Im an idiot

Thanks

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[rocky atlas]

How do you write energy in terms of length?

[rocky atlas]

I think there might be a formula I've read somewhere

.close

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[visual plinth]

Hi

[visual plinth]

$Travelling \ at \ 60000m\cdot h^{-1}\
Distance \ in \ 0.8\ seconds \
0.8/60 ,ans/60 =\frac{1}{4500} \ to \ convert \ into \ hours \
60000\cdot \frac{1}{h} \\
60000 \cdot \frac{1}{\frac{1}{4500}}$

[ocean seal]

kenZ

[visual plinth]

Units always mess me up can someone help me understand where the math error is here

[high rapids]

Distance in seconds?

[visual plinth]

How much distance he would travel in 0.8 seconds

[high rapids]

Ok let me rephrase the question better

How much distance the object will travel in 0.8 seconds at a constant speed of 6000m/h?

Is this true

[visual plinth]

u have to convert seconds into hours

[high rapids]

Ok

[visual plinth]

which is 1/4500 i think

however like the texit ,I feel like im following the rules correctly but it just breaks

[high rapids]

1 second = 1/60 minute

1 minute = 1/60 hour

[visual plinth]

?

is that not doing this?

[high rapids]

1 second = 1/60 * 1/60 hour

1 second = 1/3600 hour

0.8 second = 0.8/3600 hour

0.8 = 8/10 * 1/3600 hour

[visual plinth]

im getting confused is the conversion wrong

[high rapids]

0.8 second = 1/4500 hour

[visual plinth]

yes

So I sub that into hours

[high rapids]

Yes

[visual plinth]

but it comes out as like millions

[high rapids]

distance = speed * time

[visual plinth]

so something im doing is incorrect with this working

[high rapids]

6000 * 1/4500

60/45 meters

[visual plinth]

So working out speed you cant do what I did

u have to use distance x time

(with correct units)

[high rapids]

4/3 meters

Speed = distance / time

distance = speed * time

[visual plinth]

ahh i see

thats why it's km/h lol

[high rapids]

Your conversion is correct but your formula is wrong

[visual plinth]

Ye thanks for the help

: )))

.close

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[vocal rose]

I redrew the graph so that it’s correct

And IK c is pi/2

[lone heart]

@vocal rose Has your question been resolved?

[north adder]

@vocal rose hint: b should indicate your period

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[olive oar]

Can someone explain this in more detial for me: chose any point on the sphere where the vector is not zero. remove a small disk about that point. a bit of thinking will convince you that the degree of the map defined by the vector field on the boundary of the complementary disc is 2, hence on the complement of the disc around that point, there are 2 zeroes counting multiplicities.

Reference: https://www.physicsforums.com/threads/another-proof-that-a-vector-field-on-the-sphere-must-have-a-zero.404498/

It's trying to show that a sphere cannot have a non-zero vector field. I'm getting stuck at the boundary of the complementary disc.

[lone heart]

@olive oar Has your question been resolved?

[olive oar]

<@&286206848099549185>

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@olive oar Has your question been resolved?

[shrewd elbow]

yikes feels bad man

[lone heart]

@olive oar Has your question been resolved?

[lone heart]

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[slate shard]

i dont know what to do after

[placid zinc]

Factor sin(x)

[crystal grail]

Then you will have 2sinx =0 or sqrt2cosx=1

Solve both

Using arcsin and arccos

[slate shard]

yeah how do i get to that point

[wary stream]

Factor

[crystal grail]

Factor sinx first as Kaynex said:
2Sinx(sqrt2cosx-1)=0 do you understand this?

[slate shard]

nope

[crystal grail]

I’m pretty sure that if you look on YouTube on a factorization tutorial you will understand much faster than if we write here

But it’s not hard

[wary stream]

Do you know factoring in general? Like 2x^2 + x, do you know how to factor?

[crystal grail]

Basically you take the termos that are repeating on a sum and multiply it by the the sum of the things that were not reapeting

[slate shard]

@crystal grail what should i search for on yt abt this?
because im totally lost with what you guys are trying to tell me

[wary stream]

Factoring by GCF

[crystal grail]

Search factoring

[slate shard]

ok ill give that a shot

[crystal grail]

I’m pretty sure you will get it fast it’s not hard

[slate shard]

.close

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[silent zephyr]

confused where i fucked up, i need to find the standard form equation given the center and a point on the circle.

[tacit arch]

(-3)^2 is what you're supposed to calculate. Not -3^2

[silent zephyr]

oh okay thank you

[silent zephyr]

i get -6+1=-5

but it still dont work

nvm i got it

i forgot to put parantehsis for sumn in the calculator

.close

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[magic terrace]

The original function is indeterminate if you plug in 0

So I took the derivative of numerator and denominator as in L'Hopital's rule

And got

$\frac{sec^2(x)}{-\frac{1}{2\sqrt{x}}}$

[ocean seal]

Eyesonjune

[magic terrace]

Which is indeterminate on its own, but I can simplify to

$-2\sqrt{x}sec^2(x)$

[ocean seal]

Eyesonjune

[gilded nova]

use one sided limits

[magic terrace]

what

it's asking me to use L'Hopital's rule

[gilded nova]

u've used l'hopital already

[magic terrace]

If I put in 0 here, it is determinate, but I'm concerned that it was too simple

[gilded nova]

u can solve this limit

sec^2(0) = 1

[magic terrace]

Do I just plug in the value?

[gilded nova]

English isn't my first language, so I don't know how to explain this with the correct terms

Take a look at this

https://en.wikipedia.org/wiki/One-sided_limit

If u do x --> 0+, the limit will equal 0

If u do x --> 0- the limit won't exist, cause you can't approach sqrt(x) from the left side

So the final answer would be that the limit does not exist

[oak chasm]

The limit does exist.

[ocean seal]

Chai T. Rex

[oak chasm]

But after differentiating the top and bottom, convert sec²(x) to 1/cos²(x).

Then you have another indeterminate form.

So, you can use L'Hospital's again.

[magic terrace]

wouldn't x^(-1/2) be -sqrtx?

[oak chasm]

No, negative power means reciprocal.

[magic terrace]

so it turns into

$\frac{1}{2}\frac{1}{\sqrt{x}}$

[ocean seal]

Eyesonjune

[oak chasm]

Yes.

[magic terrace]

ok

So

if we have

$\frac{sec^2(x)}{\frac{1}{2\sqrt{x}}}$

[ocean seal]

Eyesonjune

[magic terrace]

Then why can't I say that this =

$2\sqrt{x}sec^2(x)$

[ocean seal]

Eyesonjune

[magic terrace]

and then just plug in 0

[oak chasm]

Because sec²(0) is ∞.

Move that to the bottom by converting it to 1/cos²(x).

[magic terrace]

Does that make it determinate

$\frac{\frac{1}{cos^2(x)}}{\frac{1}{2\sqrt{x}}}$

[ocean seal]

Eyesonjune

[oak chasm]

It makes it indeterminate.

[magic terrace]

is cos^2(x) infinity

or something

[ocean seal]

Chai T. Rex

[magic terrace]

cos^2(0)*

[oak chasm]

Now you have 0/0.

[magic terrace]

then I have to do it again

and get like

[oak chasm]

Right.

[magic terrace]

umm

$\frac{\frac{1}{\sqrt{x}}}{-sin^2(x)}$

[ocean seal]

Eyesonjune

[magic terrace]

is that correct?

[oak chasm]

No, the bottom needs to use the chain rule, since you have squaring and then cosine inside that: cos²(x) = (cos(x))².

So, you use the power rule first.

With the chain rule.

[magic terrace]

$-2sin(x)$

[ocean seal]

Eyesonjune

[magic terrace]

actually no

wait no yes

I am confuze

[oak chasm]

No, it's a bit different.

First, you pretend that the part inside is a variable.

x² → 2x, right?

[magic terrace]

Well I know chain rule

[oak chasm]

So, (cos(x))² → 2 cos(x).

Then, you multiply by the derivative of the inside.

2 cos(x) (-sin(x))

[magic terrace]

ok

what about the top

did I get that correct

[oak chasm]

Yes, that's right.

[magic terrace]

so, now I have

$\frac{\frac{1}{\sqrt{x}}}{2cos(x)(-sin(x))}$

[ocean seal]

Eyesonjune

[magic terrace]

is that correct?

[oak chasm]

Yes, if we do it this way.

[magic terrace]

ok well what way should I do it

[oak chasm]

Well, let's see what this way gives us.

[magic terrace]

well this is indeterminate still

[oak chasm]

No, it's not.

[magic terrace]

because -sin(0) is still 0

[oak chasm]

You don't have 0 on top.

[magic terrace]

but anything /0 is indeterminate right

[oak chasm]

,calc cos(0)

[ocean seal]

Result:

1

[oak chasm]

Oh, wait. I think I miscalculated.

[oak chasm]

Back to this.

[magic terrace]

there is a 0 in the denom

so it dont matter

[oak chasm]

No, wait, go back to that.

[magic terrace]

ok what do I do now

[ocean seal]

Chai T. Rex

[oak chasm]

The top is 0, the bottom is 1.

I made a mistake and thought the bottom was 0.

Sorry.

So, the limit is 0.

[magic terrace]

So cos^2(0) is 1?

[oak chasm]

,w cos^2(0)

[ocean seal]

Results provided by WolframAlpha

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

Requested by Chai T. Rex#7323

[native cloud]

Do you know that cos^2(x) = (cos(x))^2?

[magic terrace]

yes

[native cloud]

Don't use calculators for this

[native cloud]

You're teaching him ofc you do xD

[ocean seal]

Chai T. Rex

[oak chasm]

So, the limit from the left is also 0.

[magic terrace]

what I'm the one being taughted

[native cloud]

Ohhhh

mb

sorry

[oak chasm]

So, you only needed L'Hospital's once, though there are some problems where you need it more than once.

[magic terrace]

thanks for the help it just seemed too easy

my teacher usually gives us tedious problems

[oak chasm]

No problem.

[magic terrace]

.close

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[night geyser]

test

[night geyser]

.close

[lone heart]

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[viral terrace]

I need help computing this limit $\lim_{x\to{3}} \frac{2-\sqrt{x+1}}{x-3}$

[ocean seal]

gabbles

[gray isle]

what have you tried?

[viral terrace]

taking the conjugate

classic 1st year move, and idk what to do aafter that

[gray isle]

what do you have after using conjugates

[viral terrace]

$\lim_{x\to{3}} \frac{5-x}{(x-3)(2+\sqrt{x+1})}$

[ocean seal]

gabbles

[viral terrace]

oh wait

[gray isle]

you didn't expand numerator correctly

[viral terrace]

yeah i saw it

haha, i only distributed the - to the x not the one aswell

it should be 3-x on the numerator aye

[alpine sable]

u can use lhoptal rule as well but rationalizing also works

[viral terrace]

then its just -the first bracket, so theycancel with a negative

[gray isle]

yes

[viral terrace]

believe me, i wouldve done that if i was able to,

[wary stream]

Not everything needs l'hopital

[viral terrace]

seeing that its x/0

[ocean seal]

ℝamonov

[alpine sable]

ik

this can be done using rationalization ik that thing

[viral terrace]

for my course, they explicitly stated no use of lhr

it was written in bold

thats how explicit it waasa

but ty for pointing out the mistake

[alpine sable]

lhoptal rule is like 2 min solution

[viral terrace]

im awarre

[wary stream]

But if you can't use it, there's no point to use it

[alpine sable]

ik ik?

[wary stream]

So there's no need to suggest it, if the OP can't use it

[viral terrace]

we can use desmos to check out limits aswell

only to check

[alpine sable]

wolfram is better

[viral terrace]

Wolfram: "Solve by using L'hopital's Rule"

[alpine sable]

I mean checking only

not the steps

[viral terrace]

lmao we can use everything

in our exam

my lecturer said "Its closer to real life that way"

[alpine sable]

I wasenr aware he cannot use it

when he posted the question first

[wary stream]

Then pay attention, because it was mentioned

[viral terrace]

but ty

how can i close this

[trim thistle]

.close ig

[viral terrace]

.close

[lone heart]

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[dawn quail]

Is this legal to do?

[outer swallow]

No?

[dawn quail]

Haha OK I think I am confusing it with dividing fractions

[outer swallow]

You have to multiply by 1/x because the x is in the denominator

[dawn quail]

My answer is way different

I have circled the correct pathway here?

[tacit arch]

What question are you even trying to solve

[tacit arch]

Plug in your answer to the original equation and see if it's satisfied

[dawn quail]

I started with this but feels like more work to check answer than to solve lol

[tacit arch]

Get used to verifying answers as part of solving

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[eternal shale]

what is a own vector subspace?

[eternal shale]

$\subsetneq$

[ocean seal]

albertoales

[eternal shale]

with this symbol

it means that the second subspace contains vectors that certainly aren't contained in the first subspace?

[native granite]

That's right. That's just like for sets

$A \subsetneq B$ means that A is included in B, and that there are elements in B that are not in A

[ocean seal]

Twenty

[eternal shale]

so for example kernell is $\subsetneq$ of codomain?

[ocean seal]

albertoales

[eternal shale]

if dim(im F)>0

[native granite]

Not of codomain but rather of domain
But yeah, if $dim(im F) > 0$ then the kernel is $\subsetneq$ of domain

[ocean seal]

Twenty

[eternal shale]

oh yes sorry

thanks @native granite

.close

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[rose sable]

I dont know how to work this out

A label is put around the tin and overlaps by 5mm. The label is also 2mm short from both the upper and the bottom rim.

a) What is the height of the label? (This is the only one I know, 5.8cm)
b) Calculate the total length of the label.
c) Work out the area of the label.

[fierce scarab]

I am off by a single n on this proof by induction, not sure what Im doing wrong

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[wintry skiff]

Okay, available now.

[wintry skiff]

So, I am looking for a symbol that looks like the opposite of ∟

Found it here, but bad font support https://graphemica.com/⯾

The problem is, fonts not showing all available symbols.
So, you see a square instead. Hopefully, the problem can be addressed better later.

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[outer frost]

can someone explain how -2/3 becomes -3/2 (why is it flipped and why its still a negative?)

[high rapids]

because they multiplied both sides by -3/2?

[alpine sable]

multplied both sides by $-\frac{3}{2}$

[ocean seal]

sqrt(-1) is approx -30

[outer frost]

multiplying -2/3 by -3/2 cancels out the -2/3?

[high rapids]

yes

try it yourself

[outer frost]

it gives me 1 not 0

[high rapids]

yes

so?

a * 1 = a

a * 0 = 0

[outer frost]

oh i see

[high rapids]

this is algebra 101

[outer frost]

i always get stuck on the most simple things lol

.close

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[marsh rapids]

@signal jackal

[marsh rapids]

as I said, one faster strategy is to try and locate the correct answer. But here, since they're independant (for each one, you have to determine whether it's true or false), you have no other option than to look at each answer and work it out

you deleted your original message so it closed the channel. Hence I opened this one so we can continue

[ocean seal]

Gaul Soodman

[marsh rapids]

what do you call trial and error ?

looking at the first elements until you find one that's in both sets ?

[ocean seal]

Gaul Soodman

[marsh rapids]

let's try to do a rigorous proof

suppose a != b and they're both nonzero (just for simplicity)

then if their intersection is nonzero, it means there's prime numbers p and p' such that ap = bp'

I was talking about the fundamental theorem of arithmetic

but I got a bit ahead of myself and said something wrong, so I deleted it

a unique product. Order matters to say what is unique

writing n | m to say n divides m for simplicity

if a | b (wlog, we'll ignore b|a), let's write b = ac. Then ap = acp' and p = cp', so c = 1 and a=b

we also wouldn't get an integer for p'/p, but now you lost the simplicity of working with integers

otherwise, let's write d = a^b their gcd. let's write a = a'd and b = b'd. We have ap = bp' so a'p = b'p'. Hence we only need to consider the case where a and b are coprime

then this becomes an equality between irreducible fractions pretty much (baring a few special cases)

what's the intersection of 2 numbers ?

I asked you because you said "the intersection of a and b" and that's not defined unless you give it a definition

a/b = p'/p means (since a and b are coprime) either a=b or both fractions are irreducible.

if it's irreducible, then a=p' and b=p

i.e. a and b are prime numbers

reciproquely, if a and b are prime numbers, then ab is in Pa and Pb

hence the intersection is non_empty iff a and b are equal or both prime

with this result, the original question is easily answered

[marsh rapids]

wait

because I limited myself to a and b coprime but didn't look at how to arrive back at the general case, the result is a bit more complicated yes

[marsh rapids]

we had this

for 12 and 20, their gcd is 4 and after dividing by it, we get 3 and 5, two prime numbers

hence nonempty intersection

final result: the intersection is non-empty iff a=b or a/(a^b) and b/(a^b) are prime numbers

[marsh rapids]

with the usual notation that a^b is their gcd

since you have the original problem and the answers, does that match with my result ?

and what was their "elegant" solution ?

for 4 cases, it's definitely faster

but much less satisfying

a^b is a common notation in undergrad I believe

unexplained notations go a long way in hindering understanding

arithmetic notations you'll probably see in the future:
a|b : a divides b
a^b : gcd(a, b)
a v b : lcm(a, b)

but they're the same size as the letters and don't go above

also, $v_p(n)$ = the power of p in the prime number decomposition of n

[ocean seal]

themateo713

[marsh rapids]

example: v_5(500) = 3

[marsh rapids]

called p-adic valuation

yes

lets you write (fundamental theorem of arithmetic)
$n = \prod_{p \in \mathbb{P}} p^{v_p(n)}$

[ocean seal]

themateo713

[marsh rapids]

with P the set of prime numbers

[marsh rapids]

the p-adic valuation is thus only well defined thanks to the uniqueness of such a decompositon (FTA part 2)

also big pi means product

like sigma means sum

precisely

that's why in LaTeX it's \prod

and sigma is \sum

.close

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[tidal trout]

welp

[tidal trout]

$$1/cosec+cot = cosec-cot$$

[ocean seal]

Nayan RBLX

[tidal trout]

anyone?

[lime bobcat]

What?

[tidal trout]

sorry ill type it again

[lime bobcat]

$\frac{1}{\text{cosec } \theta+\cot \theta}=\text{cosec }\theta -\cot \theta$

[tidal trout]

$$prove:\frac{1}{cosec\theta +cot\theta }=cosec\theta -cot\theta$$

[ocean seal]

Nayan RBLX

Categorist

[lime bobcat]

$1=(\text{cosec }\theta-\cot\theta)(\text{cosec }\theta+\cot\theta)$

[ocean seal]

Categorist

[lime bobcat]

$$\text{cosec }^2\theta+\cot(\theta) \text{cosec }\theta-\cot(\theta)\text{cosec }\theta -\cot^2\theta$$

[ocean seal]

Categorist

[lime bobcat]

$$\text{cosec }^2\theta-\cot^2\theta$$

[ocean seal]

Categorist

[lime bobcat]

$$\frac{1}{\sin^2\theta}-\frac{\cos^2\theta}{\sin^2\theta}$$

[ocean seal]

Categorist

[lime bobcat]

$$\frac{1-\cos^2\theta}{\sin^2\theta}$$

[ocean seal]

Categorist

[lime bobcat]

and so

$$\frac{\sin^2\theta}{\sin^2\theta}=1$$

[ocean seal]

Categorist

[lime bobcat]

as we wanted to proof

That's the Fundamental Identity of Trigonometry

[tidal trout]

wait what did you do @lime bobcat

I wanted cosec-cot as the ans

[lime bobcat]

It's from Pitagoras Theorem

[tidal trout]

pythagorus..?

[lime bobcat]

The proof of $\sin^2\theta+\cos^2\theta =1 $ is straight-forward from Pythagoras

[lime bobcat]

What do you mean? I've just solved what you asked

[tidal trout]

wait ill try it myself once

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[solid mural]

How to solve the limit question above?

[tender dew]

lim k -> inf (1/k)×(2k!!/((2k-1)!!))

[solid mural]

Then?

[tender dew]

one moment

with just single factorial

ratio of next to previous would just be 2k

so limit would equal 2

[solid mural]

I don't understand

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[lone heart]

@solid mural Has your question been resolved?

[lone heart]

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[whole gulch]

Okay so, here is the problem I’m stuck at. I do not know how to start. I’m going into geometry next year and I want to get ahead. So how should and where should I start with that problem?

[prime badge]

i don;t understand the premise 🤣

you just need to make it fall? why is it called "through 90°"

also it doesn't require energy

[whole gulch]

I have no clue.

[prime badge]

it gives energy if you do that, it would be negative

[balmy warren]

it does require energy

a force is required

therefore energy is transferred into kinetic

[prime badge]

you can recover more than you spend

[balmy warren]

the centre of mass is changing

therefore there is a change in gravitational potential energy

also if it has rolled, there is rotational kinetic energy

[whole gulch]

Oh wow I forgot about those types of energy’s lol

[whole gulch]

So would you have to find the area of it?

[prime badge]

you need to figure out how you do it

you probably push at one point, but which and in what direction

[whole gulch]

I think it’s going from left to right.

So your pushing to to the right

[prime badge]

maybe

[whole gulch]

It’s implying that the cuboid standing is the starting point and it’s ending lying down

[prime badge]

assuming you don't need to figure out anything complex, find where the centre of mass will be when it's about to fall

then your energy is spent on lifting it there

i don;t know how to do it

i guess it would be at half diagonal height?

so 65

[whole gulch]

I don’t know it’s pretty confusing

[prime badge]

then you lift 4000N by 0.05m

which is 200 J and then you;re done

it falls

[whole gulch]

How did you get 0.05

[balmy warren]

it is 0.5m tho

not 0.05

[prime badge]

and at that point you can start pumping enrgy back and recover more than 200 but i guess that's pedantry

[balmy warren]

I get C

[prime badge]

the centre of mass is initially at 0.6

[balmy warren]

as my answer

[prime badge]

it lifts to 0.65

[whole gulch]

How did you get 0.65

[balmy warren]

lifts?

no

[whole gulch]

Cake explain what you did

[balmy warren]

@whole gulch do you know formula for GPE

[whole gulch]

No I do not

[balmy warren]

E = mgh

where m is mass

g = gravitational acceleration

[prime badge]

0.65 is half the diagonal

[balmy warren]

h is height

@prime badge plz stop

you are confusing

[prime badge]

the diagonal is 1.30

[whole gulch]

Okay I got that

[balmy warren]

mass times gravitational accelerationg gives weight

which is 4000

[prime badge]

i'm behaving myself

i will stop when i start rambling

[balmy warren]

but h has changed from one height to another

calculate change in height

then sub into equation

you get E

so E = 4000 * (0.6 - 0.25)

=1400J

do you agree @whole gulch

[prime badge]

that's practically what i meant, that's how much you get

[whole gulch]

Yeah, that actually makes a bit of sense

[prime badge]

you would spend negative that if that was how it worked

[balmy warren]

@prime badge there was no need for the complications you introduced

It is just a change in GPE

So that is the energy required

[prime badge]

you;re confusing it with work or something

work is change

[whole gulch]

So is that all you have to complete is this equation?

[prime badge]

you extracted energy from the system, and got the right answer for the amount, but that's not energy you spend

[balmy warren]

Yes that is minimum

[balmy warren]

Its not a complicated question

You are just working out the energy required to cause a change in GPE

[prime badge]

don't bicker

[whole gulch]

How do you see how much your spending? Its asking for the minimum it needs to roll over

[prime badge]

that's the concept of work

[balmy warren]

There is no energy loss as it isn't stated

[balmy warren]

You're point being?

Work is done on the system to rotate it

[prime badge]

by the gravity

you're not even doing the wrong thing, it's what i said initially

ok, let's not spam the channel

you can dm if you want

[balmy warren]

@whole gulch if your happy with your answer you can close the channel

[whole gulch]

Alright thanks for helping I appreciate it

[prime badge]

i'm certain it's not that, that's the change in energy

[whole gulch]

.close

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[undone basalt]

Hi

[undone basalt]

The Q is 5y^4=y^2(2z-7ax) and to make x the subject

Got to -7ax = 5y^2-2z

<@&286206848099549185>

[knotty spire]

After 15 minutes, feel free to ping Helpers.

[undone basalt]

Oh yeah sorry

[high rapids]

Yes now divide both sides by -7a to leave x alone

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[nova rune]

How do I make a derivative function from this?

[normal ingot]

product rule + chain rule

[nova rune]

how do i apply this to a square root term

[normal ingot]

write the square root as an exponent

[nova rune]

let me try for a sec

so how would I continue from here

[real solar]

You did it wrong. The last part

[normal ingot]

now multiply the second term by the derivative of 12-3x^2 (which is the expression inside the square root)

[nova rune]

oh oop

let me fix that

there

[real solar]

Still missing one key part of the chain rule. You did the first part right but now you have the multiply by the derivative of the inside. For the last bit again

[nova rune]

hmm

ok

you mean this one right?

[real solar]

Yeah but hold on let me send a pic that might help

So the f(x) is the sqrt and g(x) would be the middle portion

You just do power rule on the sqrt and keep the middle the same then multiply by derivative of the middle

Which is just power rule again

Make sense?

[nova rune]

ah so i dont really need that x^2

and that + turns into a *

[real solar]

thats separate

Thats part is correct

[nova rune]

alright let me try that

[real solar]

That is part of the product rule

[nova rune]

alright so this is what i tried

[real solar]

Yes that is correct

👍

[nova rune]

is that the final result or do i simplify it even more?

[real solar]

Simplifying really just depends on if your teacher cares or not. I would’ve left it as the top line

[nova rune]

yeah youre right

thanks for the help

i really appreciate it

.close

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[drifting heath]

How do I calculate the work done by this conservative field on the following line gamma ( t , t² ) with t in the interval [1,5] ?

[drifting heath]

I thought about calculating f(t) and then calculating f(5) - f(1) but it turns that i will have a cos(5) and sin(25) which is impossible

[placid zinc]

f takes two inputs. I don't know how you might calculate f(5)

But you're starting at (1,1) and ending at (5, 25)

[rugged sun]

I think they mean after parameterizing it

[drifting heath]

So how do I do this

I mean the field is conservative, I think there's no point in calculating the line integral

It's just F(B) - F(A)

[placid zinc]

Since it's conservative, then you only need to do f(1,1) - f(5,25)

[drifting heath]

Yeah but f(5,25) doesnt exist

[rugged sun]

How come?

[drifting heath]

Wdym?

[rugged sun]

Why does it not exist?

[drifting heath]

How do you calculate cos(5)?

or sin(25)

[rugged sun]

Cos is limited in range to [-1,1], but has domain of all real numbers

[drifting heath]

Yeah that's my point, so you can't calculate cos(5)

[rugged sun]

Assuming we are in radians, put cos(5) into a calculator

[drifting heath]

We're not in radians lol

Can i post the entire exercise here? maybe I did something wrong calculating the potential?

[rugged sun]

Go ahead

[drifting heath]

So it says that i have to calculate the potential (assuming that w is exact)

And then find the line integral (work done by the potential) on gamma (t, t^2)

Even tho I'm 99% sure I calculated it correctly, maybe you guys will find out how to do this shit lol

[rugged sun]

Your scalar field is correct

[drifting heath]

Ok, that's great to know

So there's no way to calculate the work done by my conservative field?

[rugged sun]

I think it is literally sin(25) into a calculator, I don't see what else tbh

It would have been good to have an exact value like for your gamma to have its numbers scales by pi

[drifting heath]

Point is, those are not radians

[rugged sun]

Since we have integrated it, it can only be radians

Only trig functions expressed in radians can be intergrated as we have done

[drifting heath]

oh!

Daaaamn

My baaaad

[rugged sun]

The integral of cos in degrees is not sin, there is a scale factor to make it correct

[drifting heath]

Thank you a lot!

[rugged sun]

No problem at all!

[drifting heath]

One question, how did you find out that my conservative field was in fact, correct?

[rugged sun]

Well I could either do the question and integrate the part multiplying dx with respect to x and the same for dy with respect to y and get your answer, or I can find the differential of your f and check if it matches the w in the question

[drifting heath]

No I meant, you calculated it by hand or did use a third part calculator

[rugged sun]

I calculated by hand, but couldn't be bothered to do xsin(x) by parts so used wolframalpha for speed, could have been done by hand I was just lazy

[drifting heath]

I have literally no words

To describe how grateful i am

[rugged sun]

No problem at all! I mean, I hope I haven't missed something, but it is a fact that calculus with trig functions requires radians so I guess...

[drifting heath]

Still, thanks

Do i have to close this or it closes by itself

[rugged sun]

You can close by typing .close

[drifting heath]

.close

[lone heart]

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[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[amber niche]

[] is the integer part, right ?

[normal ingot]

they just look like square brackets

[amber niche]

why are they there

[normal ingot]

consider the Maclaurin series of e^x

[amber niche]

what do they mean

[normal ingot]

just like parentheses

[amber niche]

they would be redundant

they must be something like the floor, or the integer part

otherwise L is just e and that's no fun

[normal ingot]

some people write sin x and some people write sin(x)

meh

[amber niche]

ok lets say L = e

what about the 2nd one

we can probably find a series

so we are looking for a function whose nth derivative at 0 is n^3

[cyan bloom]

@gleb still need help?

@alpine sable

so first off, L = e

can you see that?

cool

just wanted to see that before we began

I'm gonna give this a try too

[normal ingot]

$\sum_{r=1}^\infty\frac{r^3}{r!}=\sum_{r=1}^\infty\frac{r^2}{(r-1)!}=\sum_{r-1=0}^\infty\frac{(r-1+1)^2}{(r-1)!}$

[ocean seal]

lirmirit

[normal ingot]

try to continue from here

[cyan bloom]

yeah that's the way

the problem is just rearranging the inner term so that you can draw out the expression for L in it

[amber niche]

bruh I'm getting negative factorials

[lone heart]

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[nimble bane]

What are you guys getting for this answer? I am gettting 100/399 But im not surwe if its correct

[near hollow]

Yes

[knotty spire]

Not these.

[near hollow]

If we're saying the sequence is just the ratio of two arithmetic sequences

100th term in counting from 1 is obviously 100

[knotty spire]

And the lower term would be $a_n = a_1 + d(n-1) = 3 + 4*99 = 399$

[ocean seal]

Remavas

[knotty spire]

Unless I am having a 3 am moment

[near hollow]

Is that not what they said?

[knotty spire]

indeed.

[near hollow]

😂

[knotty spire]

But in any case

I am of the view that any unknown term of an ambiguous sequence is $\sqrt{\pi}$

[ocean seal]

Remavas

[nimble bane]

so my answer is correct>?

[amber niche]

yes it is

[lone heart]

@nimble bane Has your question been resolved?

[swift shore]

Why would that be true

[lone heart]

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[lapis stream]

did i get this right?

[gray isle]

which part, there are multiple different things there

[lapis stream]

both of them lol

it’s embarrassing that i never got to learn how three floor divison works

[gray isle]

well you didn't really do anything for the first one, there seems to be a mysterious handing division symbol there

and there's a mysterious hanging cross in the second one

don't use ambiguous notation

write things out clearly

[lapis stream]

that’s the most basic multiplication symbol

let me make things more simple for you

[gray isle]

but what exactly are you multiplying

[lapis stream]

is the answer to the second one correct?

should i multiply the m^2 by the m?

[gray isle]

i mean the end result is correct, the notation is just bad

[lapis stream]

.close

[lone heart]

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[latent dock]

amogus

[latent dock]

alright guys i think the angle of D is 30 since

180 - 90 - 60

is 30

i created this to help

[mossy meadow]

you can find C's angle and use trigonometry to calculate AD

[amber niche]

tip : use the cosine law

sus

[lone heart]

@latent dock Has your question been resolved?

[lone heart]

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[radiant topaz]

can someone help me find the perimeter? i know the answer is D but i cant get to it

[crude rose]

@radiant topaz does the perimeter include the line in the middle?

or just the outside

[radiant topaz]

no

[crude rose]

okok

[radiant topaz]

it just asks for the perimeter

[crude rose]

do you know the pythagorean theorem?

[radiant topaz]

i know it but i dont know how to apply it

[crude rose]

u see this triangle right?

[radiant topaz]

yeah

[crude rose]

u want to get the bottom

agreed?

[radiant topaz]

yes

[crude rose]

do you know what the left is?

[radiant topaz]

x

[crude rose]

correct

and the pythagoream theorem is a^2 + b^2 = c^2

where C is the hypotnuse

[radiant topaz]

c is 2x then

[crude rose]

yes

so you would have x^2 + (the bottom what we need to find lol)^2 = (2x)^2

right?

[radiant topaz]

yes

so it would be

[crude rose]

,,\sqrt{4x^{2}-x^{2}}

[ocean seal]

epiphonically

[crude rose]

this

,,\sqrt{3x^{2}}

[ocean seal]

epiphonically

[crude rose]

which is alsothis

and now

add all the sides

[radiant topaz]

wait

i didnt understand

why is it the square root of 4x^2-x^2

[crude rose]

,,x^{2}+\left(what\ we\ need\right)^{2}\ =\ \left(2x\right)^{2}

[ocean seal]

epiphonically

[crude rose]

,,\left(what\ we\ need\right)^{2}\ =\ \left(2x\right)^{2}-x^{2}

[ocean seal]

epiphonically

[crude rose]

,,\left(what\ we\ need\right)=\sqrt{\left(2x\right)^{2}-x^{2}}\

[ocean seal]

epiphonically

[radiant topaz]

ohh

[amber niche]

good friend Pythagoras

nice pfp

[radiant topaz]

ok so

[crude rose]

,,\left(\sqrt{3}+8\right)x+3

[ocean seal]

epiphonically

[radiant topaz]

this is the square root of 3x^2

[crude rose]

this is final answer

[radiant topaz]

wait what

[crude rose]

oh

i did something wrong

wait

i forgot an extra 5x+3

my bad

ok

,,10x+6+x+2x+\sqrt{3}x

[ocean seal]

epiphonically

[crude rose]

,,13x+6+\sqrt{3}x

[ocean seal]

epiphonically

[crude rose]

we can factor x

,,\left(13x+\sqrt{3}\right)x+6

[ocean seal]

epiphonically

[radiant topaz]

why isnt it the square root of 3x^2 anymore?

[crude rose]

,,\sqrt{3x^{2}}

[ocean seal]

epiphonically

[crude rose]

remember

,,\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}

[ocean seal]

epiphonically

[radiant topaz]

so like

[crude rose]

,,\sqrt{3x^{2}}=\sqrt{3}\cdot\sqrt{x^{2}}

[ocean seal]

epiphonically

[radiant topaz]

yes

[crude rose]

ya

[radiant topaz]

so square root of three multiplied by x

wouldnt that be x√3?

[crude rose]

yea

[radiant topaz]

why is it √3x then?

[crude rose]

they are the same

[radiant topaz]

oh

[crude rose]

,,x\sqrt{3}=\sqrt{3}x

[ocean seal]

epiphonically

[radiant topaz]

i didnt understand well this part

but i think i get it

thank you

[crude rose]

ok

[radiant topaz]

.close

[lone heart]

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[soft needle]

Can someone solve this for me please?

[last ether]

No but I can help you

[soft needle]

That works too

[last ether]

Start with 1a

Assume it's quadratic to spare me the pain of having to do sines and shit

[soft needle]

👍

[last ether]

$$y = a(x-h)^2 + k$$

(h, k) is the vertex.

What's the vertex of that function (in intergers)

[ocean seal]

Umbraleviathan

[last ether]

Actually that doesn't even look like it's vertex can be represented in intergers

[soft needle]

I'm confused

[last ether]

Well

try this

$$a(x-1.5)(x-8)$$
$$[a(x-1.5)(x-8)]|_{x=0} = -1.2$$

Solve for a

[ocean seal]

Umbraleviathan

[soft needle]

How do I solve for A?

[last ether]

Plug in 0 for x

That's what that x = 0 part means

@soft needle what did you get for a?

[soft needle]

12