help-26

[sick hawk]

Here is the mark scheme for you

no im off

sorry 4P2

6!/3!

2! * 4P2 * 6!/3!

.close

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[naive snow]

i have a conceptual doubt related to conic section
in conic section we have a directrix, so is it a random line or the line along which we cut the cone?
what exactly does the directrix define?

[prisma mesa]

the cone is cut with a plane

not a line

directrix lies on that plane (since it must lie in the same plane as the conic section)

[naive snow]

so basically it can be any line in the plane?

is the directrix just there to define eccentricity?

[prisma mesa]

not really, im trying to find the graph to demonstrate it...

there is a pretty interesting connection

https://www.geogebra.org/m/EcFmZR24

press the set alpha = beta for parabola button

oh wait they are missing the cone

ill edit it

https://www.geogebra.org/classic/ttwpnr9j

okay now this should work

so we have a plane that cuts the cone parallel to the cones slope

that results in a parabola

now the neat thing is that when we put a ball into the cone, such that it touches the sides of cone and the plane (i.e. is tangent to all of them), it touches the plane exactly at the focus of the parabola

[naive snow]

ok so that's how the focus if found...

[prisma mesa]

then what happens is, those red lines are equal because they are both tangent to the sphere

its kinda like this in 2D, the red lines here are equal

and it can also be proved that the blue line equals the red line

hence why its the directrix

[prisma mesa]

ill show this in few mins, gtg now, ill be back soon

[naive snow]

so can i say that the directrix is the line perpendicular to the point of intersection of the two tangents of sphere which also happens to intersect the plane along which the cone is cut?

[hollow ginkgo]

Hii

I need some help with bearings

[prisma mesa]

u still here?

tick the additional constructions box in the graph

[naive snow]

yes

[naive snow]

ok

yes i ca see planes

[prisma mesa]

this directrix lies in the horizontal plane at which the ball touches the cone

now we will prove that the parabola property indeed holds, i.e. that these 2 lines are equal in length

youve already seen that this red line is same as the other red line (because they are both tangents to the sphere)

so it suffices to show that the other red line is equal to the blue line

we can first move the other red line like this to the back of the cone. That doesnt change its length

[naive snow]

like vector shifting?

[prisma mesa]

now that blue line is same as this orange line (its actually drawn as blue in the graph)

[prisma mesa]

we're only moving it along the cone. We dont change its length

[naive snow]

ohh right mb

[prisma mesa]

these are the intermediate steps of the move

so by moving it like that, the length stays the same

[prisma mesa]

after that, we can again shift it from the back of the cone back to the front

and finally, these 2 blue lines are the same length

so we've proved that the red line = the blue line

i.e. the directrix property works

[naive snow]

what does the other blue line signify

the one passing through focus?

[prisma mesa]

it doesnt have a name, it's there only as a part of the proof

[prisma mesa]

it's the one that was shifted from the back of the cone

[naive snow]

ohk

and why did we choose a sphere ?

was there like an specific reason

?

[prisma mesa]

because as it turns out, it works. It's called a "dandelin sphere"

it works for ellipses and hyperbolas too

I too wonder how it was discovered

i guess we will never know

https://www.youtube.com/watch?v=pQa_tWZmlGs

pretty cool vid on similar topic btw, just ellipses

it too uses the sphere

[naive snow]

what about circle? it is too part of conic section

[prisma mesa]

circle has a very boring dandelin sphere, it touches it at its center

but for ellipses and hyperbolas, the dandelin spheres touch it at the focus points

[naive snow]

and so we can say the same for pair of intersecting lines too then.. from this one earlier

[prisma mesa]

this fact is unrelated to conic sections, it's just used a lot in the proofs

that fact has more to do with symmetry / congruent triangles

[naive snow]

isnt intersecting lines also part of conic section?

like if we cut the cone straight down.. we have 2 lines?

[prisma mesa]

yeah, thats true

there actually is a connection between that picture and hyperbola

but its too deep to be uncovered in discord chat

[naive snow]

hmmm

[prisma mesa]

A really cool book which includes a very interesting section on conic section is Measurement by Paul Lockhart

probably my favorite math book

if u wanna learn more about this and perhaps geometry in general, that book is really great

[naive snow]

well thanks a lot, ive always preferred calculus over geometry
but this one was actutally really really cool and intresting to learn about.

[naive snow]

thank, sure will try that out now

[prisma mesa]

this has a section on calculus too btw

[naive snow]

it does?

[prisma mesa]

yeah

and it probably offers one of the best takes on calculus ive ever seen

it makes it look incredibly easy

[naive snow]

btw this is actually really cool.. mind blowing

[prisma mesa]

Yeah, its quite cool

that vid was too inspired by the book

[naive snow]

then i need to definitely check that out

[prisma mesa]

You wont regret it if you enjoy math and getting your mind blown

[naive snow]

could you just tell me just a bit more about this?

[prisma mesa]

sure

well, the main difference between that book and standard textbooks is that the book doesnt care about rigor. It only offers intuition. And instead of writting stuff like d / dx x^3 = 3x^2, it writes it like d(x^3) = 3x^2 dx

and it has couple of advantages

in general, d(f(x)) ~ f(x+dx) - f(x)

here is a super quick derivation of the product rule:

d(xy) = (x+dx)(y+dy) - xy = (xy + xdy + ydx + dy dx) - xy = xdy + ydx + dy dx ~ xdy + ydx

its literally just algebra. Unlike the standard way where you have to randomly add +f'(x)g(x) - f'(x)g(x) or sth like that

and chain rule with this setup is literally so easy you dont even notice you are applying it

[naive snow]

yeah, this is quite better than the standard proof and quite amazingly it is following all the properties too

[prisma mesa]

d(u^2) = 2u du
d(u^3) = 3u^2 du
d((x^3)^2) = 2(x^3) d(x^3) = 2(x^3) * 3(x^2) dx = 6x^5 dx

I just applied chain rule (actually, i just did normal algebra with the d-operator)

[prisma mesa]

It's so good it makes me wonder why its not taught like this in schools

if it was taught like this, maybe students would actually understand what theyre doing

[naive snow]

i am thinking the same damn thing, this could save so much time in competitive exams too.
and the fact it is always vaild too, almost feels illegal now

[prisma mesa]

it actually only works on nice enough functions (but so does normal calculus)

pretty much every rule of calculus breaks if applied on ugly enough function

[naive snow]

maybe that's why 😅

anyways thanks a bunch!

[prisma mesa]

np, have a great day :)

[naive snow]

now i just close it?

idk im new

[prisma mesa]

yeah, u can use .close

[naive snow]

.close

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[weak kestrel]

My answer doesn't match even with options

[drifting swift]

,rccw

[weak kestrel]

Shouldnt answer be cos(beta)

[weak kestrel]

Yet it was told wrong

[drifting swift]

what does this say

it's too mangled to read

[weak kestrel]

I mean it can be an answer but it's not in options so it should be one of the option that I can't derive

[wintry meteor]

,rotate 30

[weak kestrel]

Just doodled writing whole giving exam to solve

[drifting swift]

ok, so it's meaningless. got it.

from my perspective, it looks like there's no correct answer...

[weak kestrel]

Will hit and trial work?

That's meaningless ig

[radiant tapir]

Is this a live exam?

[drifting swift]

ah wait no

[drifting swift]

yeah, ok, you have to go through 2(alpha+beta).

turns out the question was not wrong

but it is quite evil

[weak kestrel]

Oh got it

[drifting swift]

,rccw

[weak kestrel]

I mean it's npi/2

[weak kestrel]

Why not

Oh I see it might become -ve(negative)

[drifting swift]

npi/2 describes too much

cot(x) = 0 when x is an odd multiple of pi/2 only

[weak kestrel]

(2n+1)pi/2?

[drifting swift]

npi/2 allows for n=38 which gives 19pi and cot(19pi) is undefined not 0, so your disappointment will be immeasurable and your day ruined

[weak kestrel]

I have another doubt question no 60 wait it's on path
It's evil

60
No idea where to begin

,rccw

[drifting swift]

this reads less evil and more laborious unless you know dirty trickery

[weak kestrel]

I hate laborious questions

[drifting swift]

unfortunately i am not a JEE nut and so do not know the right trickery.

[weak kestrel]

JEE is freakingly asking for tricks
Meanwhile Olympiad 💕 which I haven't prep for

[weak kestrel]

<@&286206848099549185>

[sharp pagoda]

so you have no idea?

try to use sin^2(x)+cos^2(x)=1 to simplify the first and second equation

[placid swift]

There's an a and b multiplied too

[sharp pagoda]

yea so its (a-b)(sin^2(x)+cos^2(x))=a-b

[weak kestrel]

U r right but b-a or a-b?

[sharp pagoda]

doesnt matter

a can be b but then c is d so its just how you look at it

[torpid trench]

@weak kestrel the answer at the back of the book turns out to be ||b||?

i solved it rn

[weak kestrel]

Book?

Oh solution

Lemme check

Idk

[sharp pagoda]

!nosols

[topaz sinew]

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

[torpid trench]

i knew this will happen

[wintry meteor]

!nosols

[topaz sinew]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[wintry meteor]

Not noans

[torpid trench]

mb sir was conforming

[weak kestrel]

Bro just don't use anything anywhere
Why was it made?
Paste it there, for what was it made

[torpid trench]

if more help is needed on how we deduced the answer by helpee then i am free enough to give it without hesitation

no issue there

[sharp pagoda]

oh yea mb

[torpid trench]

dms

[weak kestrel]

No need now

Tho if u wanna confirm or verify answer then dm me

.close

[topaz sinew]

Channel closed

Closed by @weak kestrel

Use .reopen if this was a mistake.

[torpid trench]

dmed already

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[royal sky]

hello, here i suffice to model the ODE. but i have no idea of what even to start from, since the weight would oscillate right?

[loud oasis]

it would oscillate indeed

the oscillation is a consequence of how the ODE is set up though, all you need to do for the set up is use Newton's second law

[royal sky]

hmm so far i have the equation -ky = mg

but i have no idea how to make this like

[wintry meteor]

By y do you mean x

[royal sky]

include the oscillation factor

[royal sky]

yes sry

x here

[loud oasis]

Newton's second law is $\sum F = ma$

[thorny flame]

κλαουντ ☁ (cloud)

[royal sky]

is

yes

but would be mg here no?

[loud oasis]

no, g is not the acceleration here

mg is the force of gravity

in general though the acceleration is always the second derivative of position

[wintry meteor]

Resultant force = mass * x’’(t)

[royal sky]

wait im a bit confused here because why wouldnt we just put mg, since the only force acting on the cube downward is gravity

[loud oasis]

there are two forces, gravity and spring

[royal sky]

yes

[loud oasis]

so a = g is a consequence of F = ma in the specific case when there is only one force, gravity. then you have ma = mg therefore a = g

but that logic is untrue whenever you add any other forces

[royal sky]

yes, so uh..in this case it would be right to take newtons second force ma and make it mg right?

i thought earlier you were against me using mg instead of ma here

[wintry meteor]

We want to find the total force right now

[loud oasis]

so we have newton's second law is
[ \sum F = ma ]
the gravitational force $F_g = mg$ should appear on the left side as one of the forces being summed, it should not appear on the right side replacing the mass*acceleration term

[thorny flame]

κλαουντ ☁ (cloud)

[royal sky]

okay so what youre trying to say is that i shouldnt replace ma with mg

both should be included

[loud oasis]

yes

[royal sky]

ah sorry i misunderstood earlier

uh

okay so

what force would ma even be

if the only force here are the spring and mg

[loud oasis]

ma isn't a force

Newton's second law says that the sum of the forces on an object = that object's mass * its acceleration

it relates cause (forces, on left) with effect (acceleration, on right)

[royal sky]

okay yes i see, so uh yes

okay im just a bit lost because

i see what you mean but i fail to see the connection to the overall problem here

like i cant make the link in my head yet

[loud oasis]

well there are two parts to setting up a newton's second law problem:

1. find which forces are acting on an object
2. find the formula for each force
3. add them up (this is the ΣF part)
4. find the acceleration (from definition) and put that in the ma part
5. set them equal and you're done

[royal sky]

1. there is the force of gravity, opposin the force of the spring's resistance 2. -kx and mg, okay yeah this is where im stuck, because i dont see where ma comes into play

[loud oasis]

ma doesn't come into play until step 4

[royal sky]

okay then 3. -kx = mg 4.

[loud oasis]

for step 3 you should only be adding them, not setting them equal. this forms the left side (the ΣF part) of the ΣF = ma equation

[royal sky]

ah okay so overall force F net would be mg - kx? im not too sure whether it should be mg + kx or mg - kx because they do exert their force in opposite directions

[wintry meteor]

Minus although I only know that knowing the answer

[loud oasis]

well we know that mg should be positive because downward is the positive direction

and the spring force should be upward (in the negative direction) when x > 0

[royal sky]

okay yeah so my logic is that Fnet = mg -kx because when the spring exerts force F net towards the positive direction would be less

okay so here ma = mg-kx

[loud oasis]

indeed

[wintry meteor]

What is a in terms of x and t

[royal sky]

second order

x

over dt

oh yeah thing is

im not too sure how i can incldue the oscillation here

[wintry meteor]

Then solve the differential equation

Oscillation comes out of it

As the solution

[loud oasis]

the oscillation comes completely free from the solution of the ode you set up

it doesn't need to be included explicitly in the setup

[royal sky]

okay uh just a slight problem here so

in the solution they gave here

they kinda just disregarded g?

was it another way they did or

[loud oasis]

oh you can do that by a change of coordinates

[royal sky]

change of coodinates?

[loud oasis]

so given F_spring = kx, at what coordinate x would the sum of the spring force and gravity be 0?

[royal sky]

exactly at x=0

[loud oasis]

-k(0) + mg = 0?

[royal sky]

i think so yes

[loud oasis]

k(0) = 0

so mg = 0??

[royal sky]

oh wait jno

uh

okay you see how for them

x= 0 is when the spring is at natural

like balanced

[loud oasis]

yes so we set up the problem in a coordinate system where x = 0 is the position of the unstretched spring, whereas they set up the problem where x = 0 is the position where spring force and gravity force cancel

so let's revise our forces to match that coordinate system

so we know that spring force = -k * (distance from unstretched position). using their coordinate system, what is that distance in therms of x?

[royal sky]

it would be k * ( -x) until when it surpasses the equilibrium point where it then would be k*x

[loud oasis]

no, because x is not the distance from the unstretched length point

we know that s is the distance between the unstretched length and the equilibrium position, and that x is the displacement from the equilibrium position

so in total, what is the total displacement of the mass from the unstretched point?

[royal sky]

ah so x = k+s

[loud oasis]

k isn't a length

[royal sky]

sry i meant

x = unstretched + s

[loud oasis]

is it?

if x = 0 then we should have (distance from unstretched position) = s

[royal sky]

x-unstretched = s

uh

[loud oasis]

but if you plugged in x = 0 would you get (elongation of spring) = s ?

[royal sky]

yea that wouldnt make any sense

x - unstretched = s?

wait no

[loud oasis]

imagine the mass started out in the unstretched position and then moved to x = 0. then it would travel distance s. it then moved to some position x, which is the additional distance it travels beyond s

[royal sky]

it would make sense if abs (s)

[royal sky]

ohhh thats what you meant

okay its just s + x the elongation

[loud oasis]

yes, so the spring force = -k*elongation

that's the only thing that really depended on where we place x = 0 in our ode

[royal sky]

yeah i was calculating smt else i got confused

okay so

i basically replace my x before by x+s

[loud oasis]

in the spring force specfically, yes

[royal sky]

so now im at the step where i have ma = mg-k(x+s)

and i ahve to make it in terms of displacement x over t

[loud oasis]

so plug in the definition of acceleration

[royal sky]

as in i put everything to second order?

[loud oasis]

wdym?

[royal sky]

oh no sorry like

put the acceleration in terms of x

[loud oasis]

yes

[royal sky]

it would just be d^2(x) / (dt)^2 = (mg-k(x+s))/m

[loud oasis]

it should be just dt^2 on the denominator but yes

because it's meant to be read as (dt)^2 rather than d(t^2)

[royal sky]

ah

okay so

what did they do different here

[loud oasis]

you can simplify the right side

using the fact that the sum of forces is 0 when x = 0

[royal sky]

d^2(x) / (dt)^2 = (mg-k(x+s))/m im not too sure how to simplify from this

[loud oasis]

just looking at the sum of forces mg - k(x+s)

we know that when x = 0, this should be 0

[royal sky]

okay lets put x=0, aka at equilibrium then (mg-k(x+s)) = 0, it just turns a = g

which uhh

[loud oasis]

where did the a = g come from

[loud oasis]

also you didn't actually plug in x = 0 to the equation here

[royal sky]

like assume this is a case where x = 0, so (mg-k(x+s)) = 0 then we are left with a = 0/m okay wait uh

i think i got confused somehwere

[royal sky]

so i only plug x=0

[royal sky]

actually this makes sense cos acceleration would just be 0

if x is 0

[loud oasis]

we aren't plugging in x = 0 to our entire equation $\sum F = ma$, we are making a separate equation $\sum F(x = 0) = 0$

[thorny flame]

κλαουντ ☁ (cloud)

[royal sky]

and this seperate equation

we assume it to be

at equilibrium

and not moving?

since Fnet =0

[loud oasis]

it's at the equilibrium position. it may or may not be moving since it says nothing about velocity

if the velocity is 0 then it would stay there forever. if velocity is not zero then it just passes through and swings to the other side

[royal sky]

okay uh well

if its at eq

then mg -ks = 0

thing is

im not too sure wehre im supposed to reach

[loud oasis]

ok so we know that mg = ks. this information can be used to simplfy your ODE

[fleet flax]

This is actually a very standard SHM question

Sorry for interfering

[royal sky]

okay if i determine a condition, or like simplify an equation from equilibrium, can i plug this equation back into my original equation where it is not in equilibrium

[royal sky]

yeah i dont usually get sHM questions haha

[fleet flax]

Basically, think about the restoring force when you nudge the block slightly by x from its mean position (it'll oscillate). Then think about the equilibrium condition.

Rest is just Math, honestly.

[royal sky]

this is waht im not sure about

cos if i can just take an equation from equilibrium

and plug it back into my original eqaution

i can just easily simplify it

but idk if i can

[fleet flax]

What's your equation

(or equations)

[loud oasis]

you can do that, it's completely valid

remember that mg and ks are constant, they don't depend on position at all

[royal sky]

d^2(x) / (dt)^2 = (mg-k(x+s))/m

[royal sky]

okay then

yeah its just that

i just didnt know i could substitution an equation from eq into one that isnt at eq

it feels kinda counter intuitive

okay tytyty

.close

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[royal sky]

when they ask instantaneous velocity, is it acceleration?

[topaz sinew]

@royal sky Has your question been resolved?

[loud oasis]

instantaneous velocity means velocity

i.e. dx/dt

[royal sky]

ok ty

.close

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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

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[twilit eagle]

i need help with this because it is about WRITING EQUATIONS WITH VARIABLES ON BOTH SIDES

[placid swift]

Well, what would be the equation for Jaycie?

[twilit eagle]

uh

like the formula?

if its the formala i think its y = mx + b

[placid swift]

Yes

wait what

No, okay tell me

how much would it cost Jaycie to watch a movie after buying a membership, like collectively

[twilit eagle]

it woud cost like hmm

let me solve it and put it into desmos

[placid swift]

Dude no

[twilit eagle]

oh

well

[placid swift]

Like imagine Jaycie's a rich gal

[twilit eagle]

ok

[placid swift]

So she brought the theatre membership

It cost her $27

Now to celebrate this, she decides to watch a movie, which costs her $6

So how much did she spend collectively on that theatre?

[twilit eagle]

like 6.00m + 27

[placid swift]

Well, alright then, you've found your equation I see

Now, what would be the equation for Claire?

[twilit eagle]

like 8.25m because i cant seem to find another number or value to make like another mx + b

[placid swift]

Yes, because there is no starting value

[twilit eagle]

ohh

[placid swift]

Now you're asked when is 8.25m = 6m + 27, so just evaluate them, find the value of m, and that's your answer

(Or just pop LHS and RHS on Desmos and the point where they intersect is your answer)

[twilit eagle]

whats LHS and RHS on desmos?

[placid swift]

like input y = 8.25x

and y = 6x + 27

The x value of the point where they intersect is your answer

[twilit eagle]

wait so what does evaluate mean

[placid swift]

Well, subtract 6m from both sides

then divide the 27 with 2.25, that's your answer

[twilit eagle]

wait show me a math example of what that looks like

[placid swift]

Alright, 6.25m = 5m + 5

(Subtracting 5m from both sides)

1.25m = 5

m = 5/1.25

m = 4

Like that

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[delicate topaz]

65

[delicate topaz]

i got to (k-x)/x^2 = -3/4

dont know what to do after

[maiden plank]

what's the original question? please take a screenshot with the instructions too

[delicate topaz]

sorry instructiions got cropped out

[grand sierra]

which question are you doing

[mighty geyser]

65

[grand sierra]

sorry I misread

[delicate topaz]

655

65

i got to (k-x)/x^2 = -3/4

[grand sierra]

do you know derivatives?

[delicate topaz]

i alr did that part

[grand sierra]

that doesn't see quite right

[delicate topaz]

its quotien trule

i did d/dx = to the slope

[grand sierra]

is a k a function of x?

or is it a scalar?

[delicate topaz]

huh

k is a number

that i need to find

[grand sierra]

you do not need qoutient rule here

[delicate topaz]

k/x

take derivitave

[grand sierra]

= k(1/x)

[delicate topaz]

oh ig u can do that way aswell

[grand sierra]

d/dx(k/x) = k((d/dx)(1/x))

qoutient rule only works when you have two functions

[mighty geyser]

quotient rule applies to when you have something like:

f(x)/g(x)

in this case k is just a scalar(number) so you can do it as greaterthan wrote it

[grand sierra]

and even then it's just product rule in a slightly different form

[delicate topaz]

does quotient rule still like work

[mighty geyser]

yeah but its just more work for no reason

[grand sierra]

you can do (x)(1/(x^2)) if you really really wanted to i guess

[delicate topaz]

ok

[eager pasture]

You can do quotient rule, but you'll find out a lot it cancels out to give k((d/dx)1/x))

[delicate topaz]

ok let me do the d/dx one more time to get the right answer

[eager pasture]

Since f(x)=k has f'(x)=0

[grand sierra]

use Power rule 👍

[delicate topaz]

-k/x^2

ok so

-k/x^2 = -3/4

now what

[grand sierra]

k/x = y = -3/4x + 3

to get x in terms of k

[exotic nest]

You’re c-cheating on me @delicate topaz!? who is this man?

[delicate topaz]

why

[grand sierra]

what is bro doing🗿

[grand sierra]

then you can solve for k because -k/(x^2) = -3/4

[delicate topaz]

why is k/x =y?

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[merry helm]

shouldn't this be A + (NOT(C) . NOT(A))

nvm

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[reef nest]

How was this simplified

[drifting swift]

$V = \frac{4\pi}{3} \cdot \frac{(\sqrt{S})^3}{(\sqrt{4\pi})^3}$

[thorny flame]

Ann

[drifting swift]

would you like further elaboration or do you think you can continue the simplification yourself from here

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[fallow wharf]

Guy help with this one

[hoary burrow]

Yo like can we see the original question

Cuz like I can't read the handwriting man

[fallow wharf]

Sure

See 11

[hoary burrow]

Aight thanks

[fallow wharf]

I tried substituting, but the end answer was not satisfying at all

[inland mica]

What did you sub?

[fallow wharf]

X +4 = t

[inland mica]

If you're doing it this way, you should split the fraction into 2

[fallow wharf]

I got this in end

[fallow wharf]

Yes, I did that

[inland mica]

Great

What is the issue here

[fallow wharf]

Bruh the ans didn’t match

[hoary burrow]

You could also try subbing t=√(x+4)

[inland mica]

Maybe mistakes in your calculations

[fallow wharf]

Yeah, and then use the chain rule, while differentiating

I don’t think that I made any mistake in calculation. Can you see the actual answer? Wait, let me send it.

See 11

Guys?

[hoary burrow]

Right so

I got the answer and checked it on desmos

And it's correct

[fallow wharf]

So is my answer wrong?

[hoary burrow]

You're not far off

I think you just made a mistake in subbing

Just look through your workings carefully again

After that you should be able to simplify to the answer in the answer key

[fallow wharf]

Wait, let me send my work to you

[hoary burrow]

Alright

[hoary burrow]

Oh yea that's correct now

You were missing the ^1/2 in your first answer

@fallow wharf

[fallow wharf]

Oh yeah my bad

But still, the actual answer is something else

[hoary burrow]

They are equivalent

You just have to simplify

[fallow wharf]

Oo

[hoary burrow]

Although in exam I don't think they'd reject this form

But just to see how they're equivalent

Factorise 2/3√(x+4) out of the expression

[fallow wharf]

That let’s do a cubic expansion

[fallow wharf]

Leads *

Something like this

[hoary burrow]

$(x+4)^{\frac{3}{2}}=(x+4)\cdot(x+4)^{\frac{1}{2}}$

[thorny flame]

denzio321

[hoary burrow]

@fallow wharf

[fallow wharf]

Ooo

Got it

[hoary burrow]

Also should multiply 8 by 3/2

Since you factorise it out

[fallow wharf]

Yeahhh

Thanks, I got the answer

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[dry dragon]

Hello, I'm kind of unsure what to do next. So far I have found length BC (arc length) to be 6pi. Could I get some hints on what to do next and how I should be using area?

[drifting swift]

find the radius of the small arc

call OA=r

express the areas of sectors OAD and OBC in terms of r, then take their difference and equate it to 209pi/5

[dry dragon]

thank you very much 🙏

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[wooden kayak]

For Q1, how do I plug x into the original equation

do I need to make it in exponent form then sub it in?

[chilly walrus]

what x's did you find

[wooden kayak]

but idts, question is asking to put it into original equaion

5,6

same as answer key

[chilly walrus]

yeah replace "x" in the original equation with "5"

and see if it still makes sense

$\log_a(5^2 + 30) = \log_a(11\cdot 5)$

[thorny flame]

artemetra

[wooden kayak]

theres no "a" in my calculator

[chilly walrus]

😄 you don't need a calculator for this

simplify both lhs and rhs

[wooden kayak]

this calc im using

[chilly walrus]

simplify what's inside the log

[wooden kayak]

but the question is saying sub it

not simplify

[chilly walrus]

the whole point of this exercise is for you to check that after you solve the equation, you don't get something like log(-5) which is undefined

[chilly walrus]

well yeah you subbed it in

and now you simplify to verify it

[wooden kayak]

I didnt sub it in

[drifting swift]

would you know how to plug x in if there weren't any logs?

[wooden kayak]

yeah

I would

oh so I plug it in just x^2 +30

and in the rhs too

[wooden kayak]

@chilly walrus I just plug it in without the logs,, corect?

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[zinc obsidian]

.close

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[karmic cloak]

Is this correct?

[ruby cove]

Which one?

[karmic cloak]

Both…. Sorry I keep asking I don’t have an answer key for this

I think it may be the same one I did last night I forget tho

[ruby cove]

Why are you being sorry?

[ruby cove]

Yes, its the same

[karmic cloak]

I would check my paper from last night but it’s at home and I’m at school

[maiden plank]

think I've seen these questions before

[ruby cove]

We did it 8 hours ago

[karmic cloak]

I’m est time zone so I went to bed and now I am at school and have a free

[ruby cove]

Cool, btw we checked and it was correct if im not wrong

[karmic cloak]

Kk

I’m confused why the range of f(X) is [0,♾️) and not [1.7,♾️)

[maiden plank]

,rccw

[maiden plank]

gdi

[karmic cloak]

?

[maiden plank]

look at your first image

[karmic cloak]

Yea

[clear python]

Yeah I hate this bot for this reason

,rccw

[maiden plank]

did you list two domains for f^-1

[karmic cloak]

No we only need to find the range

But I’m confused why google says the range is zero to infinity but I think it is 1.7 to infinity

[maiden plank]

so you want to find the range of f^-1

[karmic cloak]

Oh wait

[vocal escarp]

Domain of f^-1(x) is the range of f(x)

[maiden plank]

if this is the case neither range makes sense, because f^-1(0) = (0^2 - 8)/5 = (0-8) / 5 = -8/5

[karmic cloak]

He’s

Yea

[vocal escarp]

X^2 makes any number positive, so we're js going from 0 to inf

Since there's no real number that makes the function undefined

[karmic cloak]

Yea but my teacher told us to use the tables to find range and my table says 1.7 so I’m confused on a test how I would know which to do

[maiden plank]

that's why I'm asking if you're finding the range of f^-1 or f

this is the graph of f, not f^-1

[karmic cloak]

F

[vocal escarp]

Lool

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[rough furnace]

In space, given 2 lines $l_1$ and $l_2$ with equations $l_1:\frac{x-3}{1}=\frac{y}{2}=\frac{z+1}{3}$ and $l_2:\frac{x+3}{3}=\frac{y-1}{1}=\frac{z+2}{2}$. Determine a plane equation (P) contain $l_1$ and maximize angle between that plane and $l_2$

[thorny flame]

Alexis_Fx

[rough furnace]

This's the original question, after doing a bunch of thing now I have to maximize $\frac{3a+b+2c}{\sqrt{14(a^2+b^2+c^2)}}=\cos(\theta)$

[thorny flame]

Alexis_Fx

[rough furnace]

$a+2b+3c=0$

[thorny flame]

Alexis_Fx

[rough furnace]

this's the condition and also a,b,c can't all be 0 at the same time

Well I tried Cauchy-Schwarz but equality hold when a=b=c=0 which can't be true so

Should I just sub a=-2b-3c? that seems lengthy

and also 2 variables

maybe I could use AM-GM after I sub?

idk

[thin girder]

Im guessing theta is the angle between the plane and l2 right?

[rough furnace]

between normal of the plane and l2 yes

oh wait I can't use AM-GM cuz a,b,c can be negative bruhhh

[marsh charm]

hmm sounds very TSA-ey

[rough furnace]

Đề Chuyên ĐHKHTN Huế câu nào thì em ko bt thấy họ ghi nguồn mỗi v( ko cần rep đâu để kênh này tiếng anh đi )

[ruby cove]

I think cauchy schwarz can be applied

Can you send your work? @rough furnace

[rough furnace]

bruhh, $(a^2+b^2+c^2)(3^2+1^2+2^2) \geq (3a+b+2c)^2$

[thorny flame]

Alexis_Fx

[rough furnace]

This's what you meant right?

Then it cancel out the numerator

But in this case equality hold when a=b=c=0 which can't be true

[ruby cove]

Have you tried converting it in vector form?

[rough furnace]

I actually converted it from vector form to that thing

Vector is difficult to work with

[ruby cove]

With vector, applying cauchy schwarz might be easy

[rough furnace]

we have to follow this condition, not sure how we could do that with cauchy schwarz

[ruby cove]

I’ll try to solve myself and send the solution if you wont get any help

Im currently out rn

[rough furnace]

btw vector equation of that plane look disgusting

[ruby cove]

Gotta dig up the cauchy schwarz conditions

[marsh charm]

actually imagine this

[rough furnace]

I'm listening

[marsh charm]

your 2 lines I1 and I2 are seperated now right?

[rough furnace]

yeah

[marsh charm]

Imagine there's a line parallel to I2 and intersects I1

[rough furnace]

hmmm

okay

[marsh charm]

now when does the angle reaches the maximum?

you can try to visualize this with 2 pencils on your hand

[rough furnace]

okay...

[marsh charm]

have you imagined how it'd look like

[rough furnace]

okay đoạn đó thì hiể rr

Nhma bấm mt ntn

[marsh charm]

đc r

thế thì e có đồng ý là lúc đấy vector pháp tuyến n, vector chỉ phương u1, u2 nằm trên cùng mặt phẳng ko

[rough furnace]

cs

[marsh charm]

ok

nghĩa là n sẽ vuông góc với [u1, u2]

mà n còn vuông góc với cả u1 nx

[rough furnace]

à hiểu hiể

À tutu

[marsh charm]

nên ra công thức n = [u1, [u1, u2]]

[rough furnace]

hmmm

ohhhhh

[marsh charm]

bấm máy tính, xong

[rough furnace]

trường chuyên dạy xịn v

Hiểu

;-; Em nghĩ đến việc nếu 2 đường thẳng đó cắt nhau thì tìm hình góc như a

Nhma tính ra ko cắt nhau nên bỏ luôn ý tường, vl

[marsh charm]

má sách ngu

[rough furnace]

Xong còn nhìn gợi ý nx tr ơi

[marsh charm]

đốt sách đi

[rough furnace]

có quyển nào hay ko

[marsh charm]

anh

[rough furnace]

Cho e bt phát

[marsh charm]

I'm the better book

[rough furnace]

lol, a có rảnh để em hỏi 24/7 ko

[marsh charm]

sáng thì ko đc hehe

[rough furnace]

e thật sự rất cần quyển nào hay về Oxyz bruh

[marsh charm]

a ko phải cái phao để e dùng đâu

[rough furnace]

Trường thường có dạy mấy cái này đâu tr🥀

[thin girder]

Rip

[marsh charm]

à mà

chụp cho a xem cái cách trong sách

[rough furnace]

okay em chụp lời giải luôn, có lời giải nhma e ko đọc xd

[marsh charm]

chụp hết cho a

a check body quyển sách

[rough furnace]

bruhh, em ko có ,này em lấy file thg bn em ra in

In mấy câu nhìn căng căng vs đáp án thôi

[rough furnace]

Nhìn cái đáp án kìa

sợ thật ;-;

[marsh charm]

ok chúng ta sẽ cùng check body cho quyển sách

[rough furnace]

còn 1 khúc nx mà chắc a ko cần đọc

[marsh charm]

úi zồi ôi

bộ trưởng bộ hoàng kim cốt đây r

[rough furnace]

Em ra giúng họ đó

Có mỗi k bt tính max

[marsh charm]

thế này quyển sách phải lọ 36 phát mới ra cái lời giải điên khùng này

[rough furnace]

đc rồi đi kiếm sách mới thôi

Gợi ý gì k

[marsh charm]

tính max à e?

[rough furnace]

um max góc

Đề nói v

oh wait

[marsh charm]

thì max là góc giữa u1 và u2?

[rough furnace]

yeah

hold on

Em cần 1-2p để load

Thông cảm, đang ốm

[marsh charm]

a cx hơi ốm

[rough furnace]

em đang chảy nước mũi đây

[marsh charm]

same

[rough furnace]

Um okay mọi thứ vẫn v

[marsh charm]

bão chó

[rough furnace]

Sắp cs cơn nx

Em cần sách mới

[marsh charm]

chắc ko cần

[rough furnace]

hồi đó anh ôn sách j

[marsh charm]

ko có sách j

[rough furnace]

aw

[marsh charm]

chỉ có tư duy mà làm

thầy giao đề, về nhà làm, xong

[rough furnace]

Làm quen ms tư duy đc

Đề trường chuyên nó khác....

Đề em đi tính max min MA^2+MB^2+MC^2

[marsh charm]

vcl

[rough furnace]

lol

[marsh charm]

thế này thua

[rough furnace]

Làm riết chán

[marsh charm]

làm 1 bài 10000 lần chả có tác dụng gì

luyện 1 võ 10000 lần ms có tác dụng

[rough furnace]

chịu rồi, trường em ntn chứ bt sao

[marsh charm]

trường biến HS thành gà công nghiệp

[rough furnace]

Yeah, mé em cần quyển nào có đề hay để luyện quá tr

[marsh charm]

có một đống đề hay đấy

chứ cách giải nhanh nhất thế nào thì...

[rough furnace]

yeah cái đó ms là cái e cần....

[marsh charm]

tsu có thì cx hiếm lắm

[rough furnace]

Làm bài mà nó ntn thì....

[marsh charm]

thấy cái nào dài quá thì đem lên đây hỏi

hoặc DM anh

[rough furnace]

hmmm, oke luôn

Hay ngồi luyện đề Trường Sở nhẻ

Căng quá k

[marsh charm]

ehh cx đc

mấy bài nào chán quá thì bỏ

[rough furnace]

Hợp lý phết kk

Cảm ơn a nhá, em ốm nên đi ngủ trc đây

[marsh charm]

ok e

[rough furnace]

.close

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[neon iron]

How do I solve absolute value functions?

[sweet shard]

,tex .abs def

[thorny flame]

riemann

[sweet shard]

use ^

[neon iron]

Okay how to solve it on a graph with equation?

[flat pike]

do you have a specific example in mind?

[neon iron]

I'm confused by the signs

I've been struggling with this for a few days now

[sweet shard]

do you see the right side and its signs?

[neon iron]

Yes

[sweet shard]

can you be more specific then?

[neon iron]

How do you know what direction to go?

[sweet shard]

,tex .transformation rules

[thorny flame]

riemann

[neon iron]

Ohhh

That kinda makes sense

Thank you, I think I get it now

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[merry helm]

don't understand how to do this

the "less than" is confusing me

[finite storm]

why so

a quarter of 240 is 1/4 * 240 = 60

so less a quarter is less than 60, no?

[merry helm]

yes

but

it would be a range of answers, how can it be an exact number

[oak skiff]

Hello guys does anyone know comp sci servers that help with java

[sweet shard]

#serious-discussion

[merry helm]

so
let x < 60
day 1: x for $12 each, making $k total
day 2: 2x for $9 each, making $k + 342 total

[finite storm]

why k

cant u write k in terms of x

and perhaps use that to set up a rel

[merry helm]

so day 1 he made $12x and day 2 $(18x + 342)?

leaving him with < 80 spruggles (stock)

[finite storm]

ok so

we know that day 1 + 342 = day 2

[finite storm]

can u find x using this

[thin girder]

Day 2 is 12x +342 and its also 2x×9

[finite storm]

wonderful gng

[merry helm]

12x + 342 = 12x + 342

[finite storm]

bruh

[finite storm]

[merry helm]

ah so it's not day 1 + 342 = day 2

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[finite storm]

what😭

[finite storm]

what u sent me is correct but it js didnt help much w/ the ques

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[merry helm]

only day 2 was needed to form the simultaneous equation

[merry helm]

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[merry helm]

this is the most trippy question

why is it so hard to visualize

i would have to draw it out but the problem is i'm bad at drawing

[thin girder]

I mean i can visualize it quite well
The answer should be 5

Oh wiat

Ye im dumb my bad

[merry helm]

it's A - 1

it's like you have to look at it straight on, rotate it, visualize what numbers are behind what and which side is going to be touching the board

[thin girder]

Ye damn this is hard

[torpid shard]

Its pretty easy to visualize if you consider whats the number on top instead of the bottom

From the start point: you have 5 at the bottom, 4 on top
You move to P, 3 ends up tops, so 6 is on the bottom.

and mind you, the triangles on top and bottom have opposing directions

[merry helm]

so Q would have 5 on bottom?

the only way to get 6 touching P would be to flip the dice upside down (to fit into the triangle outline on the board)

then if you would push that forward, it would fall onto 5?

[torpid shard]

You get 3 tops in P, so moving to Q, you have 2 opposed, so ends up with 2 on top, 7 bottoms.

if im not going dumb

[merry helm]

ur right

why can i not see this brah

[torpid shard]

imma hardly try to draw this in paint

wish me luck for a sec

no yeah, theres no chance that i draw this and it makes sense

[torpid shard]

idk, you can kinda see how by rotating it from Start to P, the one side facing to you its the 2, per se, the opposing face.
Once you roll to Q, since you are rotating perpendicular to the 2, it has to end up tops.

And by determining tops, you can know the result in bottom

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[blazing umbra]

Hello.

[blazing umbra]

How does this make ANY SENSE

[blazing umbra]

b* means anti-parallel

no?

matter of fact on the right side he is writing b // b* which means b is parallel to b*

[fallow heart]

I've never seen this notation honestly

[blazing umbra]

well he said it means anti parallel

[fallow heart]

Oh ok

[blazing umbra]

by negating the arguments (with multiplying -1)

[fallow heart]

Exactly, that's the meaning of antiparallel

[blazing umbra]

Dude this is my first day ever at uni and its already confusing

[fallow heart]

But I don't understand your doubt

[blazing umbra]

why did he say b = 2* b*

[loud oasis]

maybe he meant to write b = -2b* and write b* = (-1, 1)

[blazing umbra]

well thats what I thought but doesnt b* also miss a 2

b* = 2 (-1,1)

[loud oasis]

well idk your notation so idk if b* is meant to be any antiparallel vector or specifically -b

[fallow heart]

Indeed, that's what confuses me as well

[blazing umbra]

well even tho its german its readable, anti parallel vector it says

and he drew a and a* parallel even tho they are anti parallel

[fallow heart]

The only thing you have to know regarding this photo is that two vectors u and v are parallel if and only if u = k v (for some number k)
@blazing umbra

[fallow heart]

Yeah the issue isn't the language, just the notation

[fallow heart]

Wait, anti-parallel doesn't mean NOT parallel

If two vectors are anti-parallel, they are parallel as well

[blazing umbra]

it means opposite directions

[fallow heart]

Anti-parallel means parallel with negative value of k

[fallow heart]

So they're still parallel 😅

[blazing umbra]

how are they parallel if they are going away from eachother

[fallow heart]

That's what you need to know @blazing umbra

[blazing umbra]

just because of this definition or what

[fallow heart]

Huh?

[fallow heart]

Sure

[blazing umbra]

one vector going down right and the anti parallel going up left

but I get it doesnt matter

if thats the definition

the b = 2 b* is still stupid

hoping he just missed the -1

[fallow heart]

I think that's probably a typo or something

[fallow heart]

Yeah I think so too

[blazing umbra]

so anti parallel just means if k < 0

when u = k v

[fallow heart]

Mmh I might remember a different definition

Let me check

[fallow heart]

Yes correct

[blazing umbra]

even tho its AI sounds true, no?

my teacher said anti parallel means negating by -1 which would mean magnitude is equal

so I was lowk confused

but alright gotta blame myself

[fallow heart]

Yes, this is correct

[blazing umbra]

dude I should really start asking questions

I was already suspicious mid class ngl

Im just showing the stuff he wrote down meanwhile

whatever thanks alberto

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[dusky ridge]

If P(x) is a monic biquadratic polynomial with P(1) = 10, P(2) = 20 and P(3) = 30, find the value of (f(12) + f(-8))/10

[dusky ridge]

😭

pls help me do this question

[quasi depot]

there are 4 unknown coefficients

you have 3 linear equations

[drifting swift]

ok one moment

what's f

[quasi depot]

P

[drifting swift]

and also when you say biquadratic, do you just mean degree 4? or do you also mean no odd-power terms

[dusky ridge]

Note: f(x) = P(x) is true.

[dusky ridge]

Not x^4 + bx^2 + c but ax^4 + bx^3 + cx^2 + dx + e

[drifting swift]

linear term known absent?

[dusky ridge]

what

oh

my bad

[drifting swift]

aight so we call this a quartic.

or just "4th degree".

[dusky ridge]

We also call it a quartic, yes.

[thin girder]

4 unknowns and only 3 equations damn thats crazy

[dusky ridge]

maybe the variable parts cancel while calculating the sum f(12) + f(-8)

[quasi depot]

that what will happen

[ruby tree]

Yes, this apparently has a unique solution

[ruby tree]

First of all, you said monic, so remove that leading coefficient

Then take the three points, make a system of three equations and four variables, and solve it

You'll have a free variable, it's fine just keep it and calculate the answer

[drifting swift]

f(x)-10x has roots 1, 2, 3 and another unknown one. does that help us at all or no?

[worthy flax]

factor theorem should help

f(x) - 10x is a quartic, right?

a monic one, even

[drifting swift]

i mean yes but is the lack of knowledge of the 4th root an impediment

[worthy flax]

this is where i assume we'll get stuff cancelling

[drifting swift]

f(x) = 10x + (x-1)(x-2)(x-3)(x-r)

[worthy flax]

you can write f(x) - 10x = (x - 1)(x - 2)(x - 3)(x - a) for some unknown a

yeah

[drifting swift]

f(12) = 120+11×10×9×(12-r)
f(-8) = -80+(-9)×(-10)×(-11)×(-8-r)

[worthy flax]

yeah the r parts will cancel

[dusky ridge]

yeah

[drifting swift]

@dusky ridge you can continue from here yourself, can't you?

[dusky ridge]

I can. Thank you very much.

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