help-10

[tawny fog]

The Organic chemistry tutor

Bluepenredpen

Maybe you could do that

My god how did I got this role?

[cold linden]

because you are helpful : )

[tawny fog]

🫂

[young solar]

Alright thank you !!

[tawny fog]

Welcome

,w plot x^2 + {y -3/4(x^2)^(1/3)}^2 < 1×1

[warm shale]

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[young solar]

❤️

.close

[obtuse pebble]

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[toxic hollow]

5 cards are dealt from a standard shuffled deck, what is the probability that all are from the same suit

[toxic hollow]

Is my answer correct?

Total ways of choosing 5 cards is 52 choose 5

There’s 4 suits to choose from

After choosing what suit you get, there’s 12 remaining cards in the suit and you would need to choose 4 of them

[calm mesa]

Just confirming: there are 52 cards on the deck, 4 suits(Spade,Heart etc) I think. You have to pick 5 cards each of the same suit and it is random?

If it is like that then it becomes ```
P(Suit1 and Suit1 and Suit1 and Suit1 and Suit1) or P(suit2 or suit2 or suit2 or suit2 or suit2) ...
It becomes p(13/52 * 12/51 * 11/50 * 10/49 * 9/48) * 4 --> This is due to the suitn being equal in probablility
Hence the solution of chosing from the same suit is p(33/66640) * 4 --> 33/16660. That is the probability of chosing 5 cards from the standard shuffled deck of the same suit

Just following the laws of probability is all

[obtuse pebble]

@toxic hollow Has your question been resolved?

[toxic hollow]

I do get your method, and I think I’ve realised where I’ve gone wrong

The numerator should 4 choose 1 multiplied by 13 choose 5

However I still can’t wrap my head fully around why my initial attempt was wrong

[calm mesa]

I dont even think I understand your method. It seems new but thats how i learnt probability if you think you are in the scenario. It also helps.

[obtuse pebble]

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[timid silo]

Kind of lost on this

[timid silo]

I did the following

so for the angles, i presume just convert everything to polar

but for the radius?

[elfin burrow]

how did you get -19x on the LHS?

[timid silo]

i completed the square i guess

[cold linden]

13x to 19x?

[timid silo]

oh sorry

mistyped it, it is 19x

the question just auto-generates

so the numbers rotated

[elfin burrow]

completing the square gives (x - 6.5)^2 + y^2 = 6.5^2 i believe

[timid silo]

here, fixed

[elfin burrow]

okay, so circle with radius 6.5 centered at (6.5, 0)

[timid silo]

right

[elfin burrow]

for a given value of theta, you want the intersection of the line y = tan(theta) x and the circle (x - 6.5)^2 + y^2 = 6.5^2

[timid silo]

[
\tan(\theta) \cdot x
]?

[warm shale]

♡LexQa♡

[elfin burrow]

nvm i think that's wrong

gimme a sec

[timid silo]

alright

[timid silo]

also does my preamble reset or something? @trail portal

where is fleqn

nvm im blind lmao

[vocal temple]

where are you actually struggling? (x-a)^2 + (y-b)^2 = r^2 where r is the radius

[elfin burrow]

actually no i think that's correct

[timid silo]

i am trying to find the minimum and maximum values r and theta could take

as my question asks

No

[vocal temple]

oh ok i see

[timid silo]

the circle is like this

[lost tree]

r goes from 0 to 13/2

r is the length of a chord starting at the origin drawn t some point on the circle

right

[elfin burrow]

they are seeking a parameterization from the origin

i.e. at theta = 0, r must range from 0 to 13

[lost tree]

yeah

theta must vary from -pi/2 to pi/2

[elfin burrow]

yes

[lost tree]

and r given some such theta may be found in terms of theta

thats what they want right

[timid silo]

wait wait, i am very rusty in my polar coordinates

how can you ascertain that?

just check whenever r = 0? i suppose?

[lost tree]

pick any point on the circle

and draw a line from the origin to that point

the length of this line is r

and the angle it makes with +ve x direction is theta

write an equation relating theta and r

?

and also state the range of values theta and r are alloewd to vary through

[timid silo]

[elfin burrow]

pretty sure r goes from 0 to 13cos(theta)

lemme check with desmos

[lost tree]

oh wait u want the region inside it

[elfin burrow]

that's just 0 to the boundary everywhere

[lost tree]

as u vary this red point, that blue angle varies thru -pi/2 and pi/2 right

[timid silo]

how did we figure that out btw?

[lost tree]

what i said?

[timid silo]

I assume you were explaining the process of how to find r above, and before that you said it must vary from -pi/2 to pi/2

I didn't quite get that i guess, sorry if i missed something in what you said

[elfin burrow]

yeah this seems correct

[lost tree]

well i guess u do understand frm the picture that the angle must vary through -pi/2 and pi/2

[lost tree]

yes for some given theta

[elfin burrow]

which is what they want, right?

[lost tree]

yeah ig

[lost tree]

are u allowed to type theta in those boxes

[elfin burrow]

i did it by substituting y = tan(theta)x

[timid silo]

yeah you are

and yeah it is that

[lost tree]

i did it like that

[elfin burrow]

prettier

[lost tree]

r = 2 * 13/2 * costheta

[elfin burrow]

imagine a ray from the origin making an angle theta with respect to the positive x-axis

for an arbitrary ray, the question asks you to provide a lower bound and upper bound for the length along that ray that you should travel so that you will trace out the given region as you vary over all rays

only rays in the first and fourth quadrants should be considered, as only those intersect the required region

hence you want to restrict theta from -pi/2 to pi/2

[weary delta]

0 thru \pi also works. I usually think about these manually. Not quite sure if there is an analytic way people do these

[elfin burrow]

i recommend finding r using silver's method

[timid silo]

I see. Thank you, I will continue thinking about it

have a nice day!

[elfin burrow]

you can trivially shift to any interval [a,b] everything once you obtain one parameterization

[weary delta]

(cos is periodic afterall, shifting around doesn’t matter indeed)

[elfin burrow]

connect the red dot to the point (13,0)

then the angle at the red dot must be 90 (because of circle theorems)

so you can easily see that r = 13cos(theta)

that's probably the cleanest way to solve this problem

@timid silo see this

[timid silo]

yeah i see it

thank you

[elfin burrow]

yep that

[lost tree]

yes nicer

[elfin burrow]

i wish Mac had a MS paint equivalent

[timid silo]

ohhh okay okay it just clicked thank you

have a nice one

.close

[obtuse pebble]

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[next wagon]

Hey guys could anyone help with this ? I have to prove whether or not this integral converges

[next wagon]

I most certainly need to use the absolute value

And i managed to prove that the integral converges from 0 to 1/2

Now i’m stuck from 1/2 to 1

[rigid lintel]

is it not symmetric around 1/2

and its all positive so you dont need any absolute values

[next wagon]

The other question asked the same question but with bounds from 0 to 1/2

And i proved that it converges so that’s done

But i have a problem in 1 now

And can’t use taylor’s "expansion" i think it’s called in english ?

[next wagon]

No i don’t think it is

Ln(1-x) with x between 1/2 and 1 is a negative number

Because value is <1

[rigid lintel]

and so is ln(x)

[next wagon]

Ah

Yeah you’re right woops

[rigid lintel]

besides

try proving symmetry around 1/2

[next wagon]

You said it isn’t though ?

[rigid lintel]

poor wording

it was meant as a question

[next wagon]

I see

[rigid lintel]

i think it should be symmetric around 1/2

[next wagon]

How though ?

[rigid lintel]

show that f(1/2-x) = f(1/2+x) for 0 <= x < 1/2

[next wagon]

Damn never used this kind of reasoning i love it ! I’ll try

Ans yeah if i prove it’s symmetrical i can just conclude it converges right ?

[rigid lintel]

exactly

[next wagon]

@rigid lintel damn not sure how to prove it

Wrote that wrong

[rigid lintel]

just write out f(1/2+x)

that will look the same

[next wagon]

@rigid lintel

I’m supposed to find x=x’

?

[rigid lintel]

you made a mistake here

[next wagon]

Yep

[rigid lintel]

wait is it to the power 1.8

not 1.5?

[next wagon]

But i end up with this if i want it to be equal

[next wagon]

Yep

[rigid lintel]

oh lord my bad

]its not symmetric then

[next wagon]

😦

Didn’t think so but you might be right

But not for 1/2 so wtf ?

[rigid lintel]

no it shouldnt be symmetric

since the powers are different

if it were both 1.5 it would be symmetric

[next wagon]

Yeah

[rigid lintel]

$\int_0^1 \frac{\ln (x) \ln (1-x)}{x^{1.8}(1-x)^{1.5}} \dd x$

[warm shale]

Gijs

[rigid lintel]

this is the problem right

[next wagon]

How tf can i conclude that it converges then 🙁

Yes it is

[rigid lintel]

and youve proven its convergent for 0 to 1/2 right

[next wagon]

Yes 👍

So the issue is when x —> 1

[rigid lintel]

alright

well that should be alright actually

[next wagon]

We can take out the x^1,8

Cuz it ~ to 1

[rigid lintel]

well its still not that close to 1

,calc 0.5^(1.8)

[warm shale]

Result:

0.28717458874926

[next wagon]

No i mean when x —> 1

But as you pointes out i don’t need to take the absolute value

I know that the integral converges everywhere for 0 <= x < 1

So we just need to study the problem when x approches 1

But that doesn’t help us much

[rigid lintel]

yeah okay fair

i guess try bound it by something

[next wagon]

I’m open to suggestions 😂

[rigid lintel]

how did you manage from 0 to 1/2?

[next wagon]

I’ll send it to you

I managed using taylor expansion

But now that x —> 1 i can’t use it

I tried considering u = ln(x) so that when x —> 1, u —> 0 but i don’t think that works

[obtuse pebble]

@next wagon Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[grizzled wind]

I forgot how to do it

[ornate karma]

what have you done so far?

[grizzled wind]

<D is 85 and <C is 61

[ornate karma]

AD bisects <CAB so it divides it into two congruent angles

[grizzled wind]

so next to 34 is also 34?

[ornate karma]

<DAB

[grizzled wind]

.close

[obtuse pebble]

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[limber eagle]

<@&286206848099549185>

I can't even find a 5 step path that ends with a banana fickle

i don't understand how to do it

[cloud berry]

CHOOSE ONE OF THEM BELOW


1. don't know to add, subtract and multiply
2. don't know english
3. don't know math
4. don't know your name
5. i'm dumb
6. all of the above
7. none of these
8. write custom range between 1 to 7

can you write your question then I can solve

[untold scaffold]

8

[cloud berry]

@limber eagle

[cloud berry]

an error occured

[obtuse pebble]

@limber eagle Has your question been resolved?

[obtuse pebble]

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[versed turret]

Is the last part with the 45º the angle between a and b?

[kind hawk]

Multiply out the dot product

Remember the relationships between dot product, angle and lengths of vectors

a^2 doesn't make sense when a is a vector

Where is the cos term from

[fierce lagoon]

You don't multiply the whole thing by cos(45)

Also it's not a^2

You don't square vectors

@light brook understand the significance of a•a and b•b

And look specifically what the 45° is between

I saw that...

[warm shale]

Mehdi_Moulati

[spiral knot]

cos(45) = sqrt(2)/2

[warm shale]

Mehdi_Moulati

[spiral knot]

where did 4*(sqrt(3))^2 came from ?

yeah it's true @light brook

[obtuse pebble]

@light brook Has your question been resolved?

[obtuse pebble]

@light brook Has your question been resolved?

[fierce lagoon]

,w calc 3(4) - 3(3) - 10(2)(\sqrt(3)) cos(150 deg)

Ha I'm blind

,w calc 3(4) -3(3) - 8(2 sqrt(3)) cos(150 deg)

[warm shale]

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[fierce lagoon]

Why is it -4(sqrt(3))^2

-3 • 1 = -3

[fierce lagoon]

I get 27 still

[obtuse pebble]

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[obtuse pebble]

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[oak knot]

Hi there, the unit I am working on is Transformations of Functions.

See below the 2 examples that I was working on.
In part b, I am asked to sketch the original function with a reflection across the x-axis.

I do understand that both the A and the C value ( af[k(x-d)]+C ) would have to be negative in order to make this reflection.

However , I dont quite understand why the positive 2 was grouped together inside the brackets when C values are outside the brackets in the general formula af[k(x-d)] + c, for when there is a negative outside f(x), due to a reflection across the X-axis.

Originally, I thought the equation for r(x) should be r(x) = -√x + 2

Is someone able to explain the reasoning behind this to me, is it wrong to think of this in terms of the general formula? Or am I just totally off...

[lost tree]

omg is this ur handwriting 😲

[timid silo]

that sure is very pretty

[cedar lichen]

The reflection turns f(x) into -f(x)

Since f(x) = √(x) + 2, that means -f(x) = -(√(x) + 2)

[oak knot]

OH Shoot

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh the whoel thing

.close

[obtuse pebble]

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[oak knot]

Thanks

im stupid Lol

[oak knot]

yes it is but its not good

but thanks anyways

[obtuse pebble]

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[stray ridge]

why cant you work this out using turning points

[cedar lichen]

How would you use turning points?

[stray ridge]

because quadratics are symmetrical, wouldnt the turning point be the midpoint?

[cedar lichen]

It would be, only if the arrow is landing at the same height as the archer (which it is)

Nvm

It is not

[stray ridge]

i know its wrong, but i cant quite visualize why

[cedar lichen]

Consider if the archer was on a trillion story building

[stray ridge]

ok

[cedar lichen]

The arrow would reach the turning point WAYYYYY before it hits the ground

Not at the midpoint

If you found the turning point, then doubled it, you'd get the time it'd take for the arrow to become level with the archer

But it still has a trillion stories to fall before hitting the ground

[stray ridge]

wait il re read the question and the chat

[stray ridge]

oh right thanks

[stray ridge]

thanks

.close

[obtuse pebble]

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[dry egret]

This is what im thinking? Idk how to go on from here

[cedar lichen]

πr²h is volume, not area

[dry egret]

im so dumb

[teal turret]

So F/A

And it’s force of gravity correct?

[dry egret]

yes

[dry egret]

area is 2 * pi * r *h right?

[teal turret]

Remember that m/V = density

And that gravitational force is given by m*g

[dry egret]

you mean weight

[teal turret]

See if u can go from there

[teal turret]

Same thing

[dry egret]

so the weight is a force?

[cedar lichen]

What area is acting on the surface?

[dry egret]

ah

its like F = ma lmaoo

[teal turret]

Yes

[teal turret]

Not rlly

[dry egret]

the area of the circle

[cedar lichen]

What's the area of the circle?

[dry egret]

the cylinders cross section

pi * r^2

[cedar lichen]

Thats A

[dry egret]

mass*acceleration (of some sort, like gravitational constant)

[dry egret]

why do I need to include density tho

[teal turret]

How else will u get to mass

[dry egret]

i gotta divide density by weight then?

nvm

Im tripping badly

[teal turret]

U figured it out?

[dry egret]

no

[teal turret]

What are u stuck on

[dry egret]

finding the numerator 😢

[teal turret]

So the force = weight

Weight = mg

[dry egret]

yeah i get that

but where does density come into this

[teal turret]

How can we get mass from density

Is there a formula relating the two

[dry egret]

density * volume = mass

[teal turret]

Yep

So if weight = mg

That means

[dry egret]

ah

[teal turret]

F = weight = mg = density * volume * g

U get it now?

[dry egret]

yeah

but where does the h come in

[teal turret]

So now we have

$p_c = \rhoVg/A$

[warm shale]

Stephen

[teal turret]

Correct?

[dry egret]

yep

[teal turret]

And what is A = ?

[dry egret]

pi * r^2

[teal turret]

Yep

So

$p_c = \rhoVg/[\pi* r^2]$

[dry egret]

you multiply it to get the area of the entire cylinder

[warm shale]

Stephen

[teal turret]

This is what we have now right?^

[dry egret]

yes

[teal turret]

Ok and the problem wants us to show that $p_c = (\rho)(g)(h)$

[warm shale]

Stephen

[teal turret]

And we have $p_c = (\rho)(g)* V/[\pi* r^2]$

[dry egret]

h is multiplied

[warm shale]

Stephen

[dry egret]

to get the full area of the cylinder

[teal turret]

Yep

[warm shale]

Stephen

[dry egret]

its just h

[teal turret]

Correct

So there u go

There’s our answer

[warm shale]

Stephen

[dry egret]

just one more thing

How do I do this question??

[teal turret]

Not sure, been a while since I’ve done fluids, gas laws

I’d recommend opening up a new channel and closing this one

It looks like it’s just PV = nRT

With 0.190 = h so A * h = V

[dry egret]

thats what im thinking

[teal turret]

But I’m not sure how to manipulate to get the addition of those densities so I still recommend opening a new channel sorry

[dry egret]

kk

[obtuse pebble]

@dry egret Has your question been resolved?

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[tawny compass]

-rotate

.rotate

[tawny compass]

.rotate

[alpine raven]

,rotate 180

[tawny compass]

Yeah so how do I do it w/o the a^4 + b^4 identity

If you do that it's easy but since its something from school I'm pretty sure you aren't supposed to use that

I've tried everything i can btw, substituted the cosec^2 and sec^2, but i don't see much approach other than simplify and identity

[unreal musk]

Initial thoughts would be on those 1 + tan^2 and the 1 + cot^2 ...

[tawny compass]

i alr substituted those and did stuff, what happens is u get (a^4 + b^4)/ab (a = sin, b= cos) and from there if you use the identity for a^4 + b^4 it works out but im pretty sure thats not necessary

[obtuse pebble]

@tawny compass Has your question been resolved?

[unreal musk]

Ah yeah I see, there doesn't seem to be a simple way directly - easier though would be to convert both into this form you mentioned, which is easily doable...

Converting the RHS to that form just involves common denominator, multiple uses of the sin^2 + cos^2 = 1 identity, and splitting that -2(...) that you'd get into two separate bits

[tawny compass]

oh so you found a way? please share it, i dont get how splitting the 2sincos term helps

[unreal musk]

Right let me typeset what I did for you

[tawny compass]

oh got

i see it

a^2 - b^2 then substitute in for sin^2 + cos^2 right

wait that doesnt work lmao

[unreal musk]

$$
\begin{aligned}
\sec(\theta) \csc(\theta) - 2\sin(\theta) \cos(\theta) &= \frac{1 - 2\sin^{2}(\theta) \cos^{2}(\theta)}{\sin(\theta)\cos(\theta)} \
&= \frac{\sin^{2}(\theta) + \cos^{2}(\theta) - 2\sin^{2}(\theta) \cos^{2}(\theta)}{\sin(\theta)\cos(\theta)} \
&= \frac{\sin^{2}(\theta) - \sin^{2}(\theta) \cos^{2}(\theta) + \cos^{2}(\theta) - \sin^{2}(\theta) \cos^{2}(\theta)}{\sin(\theta)\cos(\theta)} \
\end{aligned}
$$

[tawny compass]

uhm ur thing failed to render

sorry for pinging i thought u forgot and thought it rendered or something

[unreal musk]

(yeah still typesetting it haha)

[tawny compass]

😬

[warm shale]

chartbit

[unreal musk]

Ah sheeittt it cut off

One second

There we go

[unreal musk]

Final line converts to this, therefore you've manipulated both the left and right hand sides into the same thing

It's often an easier way to prove identities, get the same thing on manipulating both

[tawny compass]

i thought about splitting it to use (a+b)^2 but then dismissed that line of thought too quickly i guess 😭

very nice thank you!!

[tawny compass]

yea i tried to do that, dont know whats wrong with me, this is clearly what we were supposed to do

.close

[obtuse pebble]

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[acoustic pawn]

Im supposed to prove that the complex numbers C are not a total ordered field using the assumption that either 0<i or i<0. i know you can show that in both cases this implies that 0<-1 but isnt this only a contradiction in the standard order we're used to in R?

[acoustic pawn]

For example if we take the order relation to be that the standerd order on the radius of a point to 0 (so the radius of -1 is 1) then -1>0 makes sense

And isnt that a total order since then every z can be compared?

I know im wrong but idk where i went wrong lol

[kind hawk]

at 0<-1 you are not yet done for your proof

if that's what you are wondering

[timid silo]

0<-1 implies 0<-1*-1=1 implies -1<0

[acoustic pawn]

oh lol okey that makes more sense

[acoustic pawn]

what about this? I dont see how that counterexample shows up in this order relation

since then -1 = 1 > 0

[kind hawk]

well its easy to define an order on C

but you also want this to play well with addition and multiplication

[timid silo]

any set has an order is equivalent to choice: https://en.wikipedia.org/wiki/Well-ordering_theorem

[acoustic pawn]

so it doesnt act orderly in terms of addition and multiplication? And thats makes it like not a well defined order on C?

[kind hawk]

*C is not an ordered field with that order

the order is still well-defined

[acoustic pawn]

So it is an order on the set C but not an order on the group C?

[kind hawk]

well any set having a well-order is equivalent. just having some order surely is weaker

not sure why you want to call it group now

[acoustic pawn]

idk what that means haha 😅

i ment field sorry

mixing up my words

[kind hawk]

yes

[acoustic pawn]

So its not a well defined order on the field C but it is on the set C?

[kind hawk]

get away from this well-defined stuff

the order is well-defined, there is absolutely nothing wrong with that

but it just doesn't have the properties that would make C with it a totally ordered field

[acoustic pawn]

But you cant define that order on the field C or am i missunderstanding

because its not compatible with the field axioms or whatever (addition/multiplication)

[kind hawk]

you can define whatever order you want on the set C or on the field C

but with whatever order you define, you do not get an ordered field

"ordered field" is a new thing, it does not just mean order defined on a field

[dark stirrup]

Stealing from wikipedia:

[acoustic pawn]

oh thats exactly how my book introduces it tho. "In general we introduce a totally ordered field (F,+,.,<=) as a field (F,+,.) provided with a total order <= so that it satisfies the following properties : ....

[acoustic pawn]

isnt that what im saying tho?

the order isnt compatible with the field operations

so it cant be defined on that field

[kind hawk]

that quote says that "ordered field" is more than just field+order

it's field+order+properties

it can still be defined

it just doesn't satisfy the properties

[acoustic pawn]

OHHHH

i thought defining the order on the field "breaks" the addition on the field and therefor its not a field anymore

i think i got what you mean now

still a bit confusing but i think i got you?

[kind hawk]

yeah I think you do. I am being overly precise but there is a difference between not being able to define certain stuff and being able to define stuff but it not having properties you want

[acoustic pawn]

So if i define this order on the field, is it an ordered field, just not a totaly ordered field?

[kind hawk]

no

[acoustic pawn]

Okey so if i define that order on the field then its just a field with some order that isnt compatible with the field

sorry having trouble making it click

brb tho libraries closing haha

[kind hawk]

yes

[acoustic pawn]

cool, i'll give it some time and come back if it wont click

Thanks alot!

.close

[obtuse pebble]

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[prime crane]

How would a toroidal planet core's magnetic field? Basically, if the Earth was torus shaped, it would have a toroidal core inside, and I'd like to figure out how would the magnetic field look like, mainly to know in what direction would a compass point if it was on it's surface

[tardy epoch]

#old-network for physics

[prime crane]

Oh ok

.close

[obtuse pebble]

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[timber fox]

I believe there is multiple interpretations that all work

[prime crane]

If you know about this dm me

Also, why is the channel still occupied? I already closed it

[timber fox]

It will close soon no worries

Although considering howww the magnetic field is ‘generated’ and assuming planets are held together with voodoo, you would have weird horizontal field lines as well

[obtuse pebble]

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[dire nexus]

i know part a

its just part b im struggling on

[unreal musk]

Mention of decimals suggests quadratic formula (or completing the square)

From those y values, match them with the corresponding x values

[dire nexus]

im not sure how to go about doing that though

[high lily]

do you know the quadratic formula?

[dire nexus]

yes

but thats for part a no?

[high lily]

no

substituting/eliminating x gets you the equation in part a)

apply the quadratic formula to that equation to get the values of y

[neon eagle]

possible*

[dire nexus]

uh

mb but idk how

[high lily]

wdym

you said you know the quadratic formula

[dire nexus]

well ye

but how do i substitute in simultaneous equation

[high lily]

wdym

[neon eagle]

You only have one equation from part a

Part a is a quadratic in y

[high lily]

you said you knew part a)
so supposedly you already eliminated x to get
3y^2 - 24y - 47 = 0
this is the equation you want to solve atm

and this is a quadratic equation

which can be solved using the quadratic formula

[dire nexus]

so 9.6 and -1.6

[high lily]

sounds about right

[dire nexus]

what does that mean tho

theres 2

[high lily]

it means there are (potentially) two possible values for y

and each will give you a respective value for x

[dire nexus]

so subsitute each in one?

for y

x = 67.6 y = 9.6 x = 0.4 y = -1.6 ?

cos i tried that and it didnt work

[high lily]

when determining x,
you should use the exact/unrounded values of y

[dire nexus]

so 9.627314339 and -1.627314339

x = 67.8 y = 9.6 x = 19.8 y = -1.6 ?

[high lily]

how are you getting 19.8 for x

[dire nexus]

oh nvm

0.2

is that right?

[high lily]

yeh

[dire nexus]

tyvm

[obtuse pebble]

@dire nexus Has your question been resolved?

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[obtuse pebble]

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[dim bobcat]

Do the Same thing as earlier

Find displacement along each component

s=ut+1/2at^2

[upper hinge]

hi again

i just cant get my hjead around this topix

[dim bobcat]

Will become normal

Just solve some questions

[upper hinge]

.close

[obtuse pebble]

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[warm wadi]

idrk how to start, I know its moved to the right by 5, stretched by sf 3 and moved down by 2

[dark stirrup]

yup you nailed it

moved to the right by 5

So a moves right by 5, then becomes a/2

[warm wadi]

im kinda lost by what the point is

[muted delta]

What does stretching do

@dark stirrup what does stretching do

I’ve never understood that properly

[dark stirrup]

Stretching (vertically) depends on the scale factor

Say the original y value c

[muted delta]

How does it affect a point

[warm wadi]

it says the answer is a = -10, b=1 or b=2

[dark stirrup]

And say the scale factor is a

Then, @muted delta, the new value is a*c

[muted delta]

Gotcha

Does stretching do anything to the x value

[warm wadi]

idk how to get to the answer from this tho

[muted delta]

You do those operations on the point (a,b)

For example moving it to the right 5 units would get you to (a+5,b)

[dark stirrup]

@muted delta, you can stretch both vertically or horizontally. y=a*f(x) will scale vertically by a factor of a. y=f(b*x) will scale horizontally by a factor of b

[muted delta]

That makes a lot of sense

[dark stirrup]

@warm wadi, you deduced that the graph shifts to the right by 5

That is, whatever the x coordinate was originally, it'll be 5 units to the right now

[warm wadi]

im lost

how do I find the possible values of it so if its a+5,b then the point it does (a+5)/2 how does it equal -10

[dark stirrup]

You were told that the original x-coordinate was a

[warm wadi]

im so dumb

[dark stirrup]

When you shift a 5 units to the right, it says the new x-coordinate is a/2

[warm wadi]

i get it now

ty

.close

[obtuse pebble]

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[nocturne minnow]

Post your work

[obtuse pebble]

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[mossy spear]

Translate the parabola y=3x^2 + 1 along the vector v = (2, -1) what is the new function?

[cedar lichen]

What have you tried

[mossy spear]

x'=x+2
y'=y-1

so then x=x'-2 and y=y'+1 ?

[cedar lichen]

Yes

Then?

[mossy spear]

should I just sub them into the original parabola equation?

[cedar lichen]

Yes

[mossy spear]

ok

I have another problem

which I can't solve

I have to show that for all foci and for all horizontal lines y=k (with k in R) there exists a parabola with focus F and directrix d:y=k

no idea how to do this

<@&286206848099549185>

[trail musk]

The choice of F is arbitrary?

[daring rock]

@mossy spear Do you know what a parabola is?

Like, geometrically I mean

this actually isn't completely true, there's a bit of an edge case if F is on y=k

[obtuse pebble]

@mossy spear Has your question been resolved?

[obtuse pebble]

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[mossy spear]

.reopen

[obtuse pebble]

✅

[mossy spear]

yes

all the points in a parabola are equidistant to the focus and the directrix

[daring rock]

Yeah, exactly

so put that in an equation

(distance from (x,y) to focus) = (distance from (x,y) to directrix)

[mossy spear]

ok one sec

|y-k|=sqrt((x_f - x)^2 + (y_f - y)^2)

where x_f is the focus x value and y_f is the y value

|y-k| is the distance from the point (x,y) to the directrix

[daring rock]

Yeah

Notice if you square both sides, the y^2 terms cancel out

leaving you with something you can rearrange into

a form like y = ax^2 + bx + c

the weird edge case is when y_f = k

kinda falls apart there

[mossy spear]

uhh

[daring rock]

don't worry too much about that part

[mossy spear]

so after I square both sides, I cancel out the y^2

[mossy spear]

I am left with something of this sort

[mossy spear]

and I showed this?

[daring rock]

A quadratic equation (with a != 0) is a parabola

so, yeah

[mossy spear]

okk perfect thank you

.close

[obtuse pebble]

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[bleak skiff]

is there a way to prove that $f_n (x) = x^n - x - 1$ is always increasing, if $n \geqslant 2$ integer and $f: [1, +\infty [ \mapsto \mathbb{R}$; without using derivatives?

[tardy epoch]

[bleak skiff]

how so

[warm shale]

lilisworld

[trail cloak]

I guess you can prove it using reasoning

[bleak skiff]

so it basically means comparing f(x) and f(x+1)

[trail cloak]

Since x varies from 1 to infinity

[bleak skiff]

ah yes

[trail cloak]

Well you can just mention the fact that x^n will outweigh -x-1

And thus will give you a net increase over all the values of x between 1 and infinity

[bleak skiff]

yes im dumb

thanks

[trail cloak]

No you ain't

No worries :)

[wild swallow]

that

[trail cloak]

[wild swallow]

seems kinda

[zenith raft]

[trail cloak]

Really?

[wild swallow]

[trail cloak]

Maybe I missed something

[wild swallow]

you need to prove that x^n - x is increasing for x>1

[trail cloak]

[zenith raft]

couldn't you use the same reasoning to extend it to f: (0,infty) to R?

[zenith raft]

which doesn't work

[wild swallow]

and both factors are increasing

for x>1

[fierce lagoon]

Is the \parens a packet thing

[trail cloak]

Maybe

[wild swallow]

no its a me thing

[fierce lagoon]

$\parens{x+2^x}$

[warm shale]

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

[trail cloak]

Package moment

[fierce lagoon]

Crying

I want :((

[trail cloak]

Preamble

[trail cloak]

Yep

All preamble coding

All I have is green

$lol$

[warm shale]

VulcanOne

[trail cloak]

Anyways let's not stray away

From the question

Am genuinely curious

[zenith raft]

you could also just take any bounded from below function that's not increasing and use the same reasoning and say it will outweigh it's lower bound

[trail cloak]

Hmm

[zenith raft]

sorry i'm beating this reasoning to a pulp

[wild swallow]

[trail cloak]

Well I haven't done reasoning in a while

Civil Engineering moment

Beat it all you want

[tough tulip]

your a civil engineer?

[trail cloak]

Yeah

[tough tulip]

wow is it hard

[trail cloak]

Student still

Nah

Needs a

A bit of imagination and some uhh

Future thinking

You know

Like thinking ahead

[tough tulip]

do you mostly study calc

[trail cloak]

No

Which is why I hate undergrad studying for Civil

Masters and PhD uses calc a lot

[bleak skiff]

.close

[obtuse pebble]

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[trail cloak]

Let's go to #discussion for more chats

[tough tulip]

okay

[obtuse pebble]

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[wild orbit]

how do you do the last question

[wild orbit]

this is what i have so far

but i cant turn this into a quadratic so idk

[obtuse pebble]

@wild orbit Has your question been resolved?

[rugged wind]

where do you find those questions

[wild orbit]

past papers

for igcse

[mossy spear]

a and b are correct

for c you know that rula buys 4 more sacks than paul

so let r be the number of sacks rula buys and p the number of sacks paul buys

r=p+4

because rula buys the same amount of paul's sacks but gets 4 more

[mossy spear]

so now from this substitute what r and p are using the previous equations

[obtuse pebble]

@wild orbit Has your question been resolved?

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[wild orbit]

thx

i got it

[obtuse pebble]

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[fading birch]

So I know the 1/x^p is equal to x^-p

[fading birch]

oh wait

.close

[obtuse pebble]

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[obtuse pebble]

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[marsh sand]

has anyone here completed Schaums Outlines of Advanced Calculus

[obtuse pebble]

@marsh sand Has your question been resolved?

[marsh sand]

I need someone who has completed the book to help me.

[tardy epoch]

[obtuse pebble]

@marsh sand Has your question been resolved?

[obtuse pebble]

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[sullen tusk]

is there any trick to reform this easier than multiplying all out ?

[kind hawk]

well you can factor out (n+1)^2 at least

in what form do you want to end up

[sullen tusk]

n+2 choose 2 squared

[obtuse pebble]

@sullen tusk Has your question been resolved?

[sullen tusk]

i thought about getting (n+1)^3 to (n+1)^2 , so i could get it into the first brackets and already have the ^2 for the result but not working yet

[obtuse pebble]

@sullen tusk Has your question been resolved?

[austere shard]

((n+2)!/2n!)² is ¼n²(n+1)². the original expression is (n+1)²(¼n²+n+1)... If I didn't do anything wrong, they are completely different

[kind hawk]

$\binom{n+2}{2}^2 = \left(\frac{(n+2)!}{2!n!}\right)^2 = \left(\frac{(n+2)(n+1)}{2}\right)^2 = \frac{(n+2)^2(n+1)^2}{4}$

[warm shale]

Denascite

[sullen tusk]

Thats currently what i am aiming for but i couldnt be able to get to that

[austere shard]

thank you, I'm slowly getting stupid writing math on my phone

[sullen tusk]

Totally relatable. its freaking hard to read plain text math

[kind hawk]

show your work

pulling out the (n+1)^2 like you said is correct

[sullen tusk]

My idea was to get the (n+1)^2 in the first brackets and then continue but i havent had an idea how to do this

[kind hawk]

pull the (n+1)^2 out of both terms

[sullen tusk]

Oh

[kind hawk]

$(n+1)^2 \left[ \frac{n^2}{4} + (n+1)\right]$

[warm shale]

Denascite

[sullen tusk]

why is it n^2/4 ?

if i multiply it again, wouldnt it be n^3 + n^2 / 4 then ?

[kind hawk]

the first term is n^2(n+1)^2/4

I pull the (n+1)^2 out

[sullen tusk]

Ahhh

ok now it makes sense

[kind hawk]

can you finish it from there?

[sullen tusk]

ill give it a try. but should be able to. i think the hardest part is done

[obtuse pebble]

@sullen tusk Has your question been resolved?

[obtuse pebble]

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[karmic hedge]

https://gyazo.com/f377a8834b242feb0ab55dd708e1893e.png

[karmic hedge]

I know the vector 3 over 1 is orthogonal to the line and passes through zero

but idk how to get the intersection point

[obtuse pebble]

@karmic hedge Has your question been resolved?

[viral blade]

Vectors aren't lines, they don't pass through zero

if vector <a,b> is orthogonal to the line and <3,1> is orthogonal to the line then how can you relate <a,b> with <3,1>

[karmic hedge]

its the radius of the circle and the norm of the tangent

.close

[obtuse pebble]

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[mild cipher]

help

[lost tree]

which one do u need help with

[supple apex]

mind zooming the image pls

[mild cipher]

8、9

plss

pls help me

@supple apex

[nocturne minnow]

Don't ping specific people

[obtuse pebble]

@mild cipher Has your question been resolved?

[obtuse pebble]

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[mighty geyser]

what have you tried?

[obtuse pebble]

@mild cipher Has your question been resolved?

[mild cipher]

tq for your help

i have solved it

:)

[obtuse pebble]

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[sleek breach]

you know how when we mean 65432*1 we call it 6 factorial (6!)
so whats it called when we mean:
6+5+4+3+2+1

[warm canopy]

6th triangular number

[sleek breach]

pretty cool, thx

[elfin burrow]

you can write a closed form for such an expression

[sleek breach]

like?

[elfin burrow]

1 + ... + n = n(n+1)/2

[obtuse pebble]

@sleek breach Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[waxen hemlock]

Did I set this up correctly?

[waxen hemlock]

The word problem is really confusing me lol

But if I did it correctly then I assume I would use sine

[ornate karma]

yup

[waxen hemlock]

oh really?

yay lol

[ornate karma]

solve for x yes

[waxen hemlock]

I got a weird number tho

[ornate karma]

i got 18.663

[waxen hemlock]

How did you put it in your calculator

[ornate karma]

sinx = 80/250

sin^-1(.32)=x

[waxen hemlock]

oh ok maybe I mistyped

I don’t get how you calculated it

[ornate karma]

Just use a normal calculator

I dont know why its giving you that

[waxen hemlock]

alr

[ornate karma]

you can check it

sin(18.663)=x/250

[waxen hemlock]

Did I set this one up correctly

[ornate karma]

250sin(18.663)=80

[ornate karma]

yes

[waxen hemlock]

Woohoo

What did you get for that one

Because my calculator is janky

[ornate karma]

sinx = 7/15

sin^-1(7/15) =27.818

[waxen hemlock]

okay yeah I got a a totally different answer idk why my calculator is being weird but Ty for ur help

.close

[obtuse pebble]

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[silver plover]

doing a past paper rn

[silver plover]

plz dont tell answer

unsure with that x

is it with the sqrt ?

[robust sleet]

no

its outside

[silver plover]

so is it (e^sqrt2)x

?

[elfin burrow]

no, e^(sqrt(2)x)

[robust sleet]

e^((sqrt(2)*x))

[fickle turret]

What's a past paper?

[robust sleet]

past exam

[silver plover]

i DONT WANT THE ANSWER LOl

hahha dw mehdi

ight cool

thanks

.close

[obtuse pebble]

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[spiral knot]

Sorry i didn't see it

[silver plover]

dw dw man

[versed cave]

(You can always use l'hopital if you are already familiar with it)

[unreal musk]

No need tho - it should be easy to spot what the limit would be with an appropriate sub

[spiral knot]

indeed

or you can use the definition of the derivation

[warm shale]

Mehdi_Moulati

[elfin burrow]

forgot the limit, but yes

[silver plover]

l hopital rules gives sqrt2

for that

question

[versed cave]

What's wrong with it

[silver plover]

i thought mehdi said it was 1

the answr

[spiral knot]

i said :
$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$

[warm shale]

Mehdi_Moulati

[silver plover]

oh ok

kl

[warm shale]

Mehdi_Moulati

Mehdi_Moulati

[obtuse pebble]

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[zinc spruce]

What mistake have i made in sketching this graph

[spiral knot]

where did 8x^3 came from ?

[zinc spruce]

the inside derivative

[spiral knot]

$$g(x) = x²e^{-2x}$$
$$g'(x) = 2xe^{-2x} -2 x² e^{-2x}$$

[warm shale]

Mehdi_Moulati

[zinc spruce]

x is powered to 4

[spiral knot]

ohh ,t hat make sense

[zinc spruce]

i came as far as this

[warm shale]

Mehdi_Moulati

[zinc spruce]

I cant upload my actual pic of my graph because discord is ruining it

but that looks similar to what i have posted which isnt entirely correct

x=1/sqrt(2)

[spiral knot]

you just missed x= -1/sqrt(2)

[zinc spruce]

oh right

so i just have to continue skething on the left side as well

tnx

[spiral knot]

you can say that g(x) is even function :
because g(x) = g(-x)

you need just to check one side

[zinc spruce]

yes exactly

tnx again

[warm shale]

Mehdi_Moulati

[spiral knot]

since e^{-2x} is always positive

[spiral knot]

anytime

[obtuse pebble]

@zinc spruce Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

@violet arrow Has your question been resolved?

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[strange fiber]

If the determinant of a matrix is equal to 3, is the determinant of its second power equal to 9?

[haughty coyote]

det(AB) = det(A) det(B)

It follows for any integer n (including negative if A is invertible) det(A^n) = det(A)^n
Taking n = 2 and det(A) = 3 answers your question

[strange fiber]

Seems to be 9

If we change the order of columns in a regular matrix, the determinant of the matrix will always..."

change sign
none of the other answers are correct
remain the same

Its the “none of the other…” right?

Cuz it can get multplied by either 1 and -1

Since there is the always

[haughty coyote]

Yes. Changing sign can hardly be interpreted as not changing sign

[obtuse pebble]

@strange fiber Has your question been resolved?

[strange fiber]

The number of odd permutations on a five-element set is equal to..."

25
5
60
120

It is 60, correct?

Half are odd

Half are even

[haughty coyote]

Yes

It's always n!/2

[obtuse pebble]

@strange fiber Has your question been resolved?

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[timid silo]

is this right

[nocturne minnow]

Use desmos to confirm

You can plot x = 6 in desmos

[timid silo]

(But yes)

.close

[obtuse pebble]

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[lone harbor]

what i had done recreated in paint

as you can see i have found the inequalities

but apparently there is a third inequality x<-

x<-5

where did that come from

[rigid lintel]

thats the intersection between l and the parabola

x = -5

[lone harbor]

yes but how does that count as an inequality in defining the region

[obtuse pebble]

@lone harbor Has your question been resolved?

[lone harbor]

<@&286206848099549185>