help-0

[teal orbit]

you know what angle of elevation is or nah

[shell acorn]

yes

[teal orbit]

k

what is it

[shell acorn]

the angle looking up at something

and depression looking down

[teal orbit]

k

so draw the 41 degree angle first

with the base and hyp

[shell acorn]

hol up

@teal orbit

[teal orbit]

good

so what is your base now

[shell acorn]

444

[teal orbit]

bet

[teal orbit]

so what is ur height

and/or hyp

[shell acorn]

need to solve for it

[teal orbit]

use trigo

hint: ||tan(41)||

[teal orbit]

ah wait now boy

that aint how it works

41+27 is not 90

[shell acorn]

tf

so why is it saying the angle of depression

do i ignore that?

[teal orbit]

for now

[shell acorn]

41 tan is 0.8692

[teal orbit]

lemme see if i can draw it accurately

[shell acorn]

okay

[shell acorn]

so I need the opposites

of each triangle

correct

then for how high is the window i do 444 - the opposite of the other triangle

@teal orbit

can you walk me through this

oh wait

428.76582 + 226.2293 = Height Of Tower

226.2293 = Height Of Window

okay i got the height of window correct

i need the first one

i got it wrong

[mild steppe]

you rounding

[shell acorn]

if my answer is 654.99512 and i need to round

right

what would my answer be

i input 654.1

is it 655.10

im a bit tweaked here negl

[mild steppe]

it should be 428.77+226+23

[shell acorn]

where r you getting 23

did i miss somethng

[mild steppe]

it its .229 which is .23

[shell acorn]

ohh

.23

[mild steppe]

yeah

[shell acorn]

655

still wrong

[mild steppe]

does it work?

[shell acorn]

655.00?

[mild steppe]

maybe

[shell acorn]

wtf?

fam it was wrong

the answer was

612.19261116593

how tf they get that

[mild steppe]

did you do you first caluction right?

[shell acorn]

yes

[mild steppe]

the 428

[shell acorn]

double check it

[mild steppe]

ok

[shell acorn]

i even put it through a calculator

[mild steppe]

ok

i check

[shell acorn]

ai bet

[mild steppe]

the window height should be 226.23

[shell acorn]

i did that

check again

at my work

i just didnt round

[mild steppe]

385.96

should be other ehight

[shell acorn]

OHH

I DID 44

INSTEAD OF 41

[mild steppe]

ok

[shell acorn]

well fuck

i got it now ima do another question

[mild steppe]

is should be tan(41)=444 over x

[shell acorn]

can yu help w one more thing

[mild steppe]

ok

[shell acorn]

how do i draw this

[mild steppe]

uhh

ok

[shell acorn]

this my last question

the angle 240 shit confusing

[mild steppe]

yeah

ok

you placed you thing in the third quadurant right?

[shell acorn]

sure

[mild steppe]

ngl

i have done this question yet i am still tryna understand it

[shell acorn]

this that bs

they got me fked up

fk wrong w em n dis math

cant een b a hvac technician in peace

[mild steppe]

lol

[shell acorn]

yu good bro ty for helping me

w that one

<@&286206848099549185>

anybody know how i would start this drawing?

my magnitude is 790.8 so my HYP is 790.8

OH IS IT 240-180

[mild steppe]

yeah

why would you want to draw this thou?

[shell acorn]

doesn't it make it easier to solve for X and Y

[mild steppe]

cant you just do cos240x790.8

[shell acorn]

its just the two other sides

[shell acorn]

idk..

[mild steppe]

for the x

[shell acorn]

i can just go straght to that?

[mild steppe]

it should be −395.4

[mild steppe]

dude i have no idea

[shell acorn]

ima input your answer

ohh

your right

[mild steppe]

ok

[shell acorn]

the other one is

sin

[mild steppe]

you do the other one

[shell acorn]

ahlie

[mild steppe]

yes

[shell acorn]

sin60?

[mild steppe]

no

sin 240

[shell acorn]

its the same

thing

or am i bugging

[mild steppe]

no

one is negitvie

i think

[shell acorn]

oh yeah

240 is negative

60 is positive

[mild steppe]

yeah

[shell acorn]

but its negative since its downward

or sum shit

[mild steppe]

cause sin in negitve in 3 and 4 quad

[shell acorn]

idek

all done

lowk still confused but my exams in 3 weeks

ill restudy

thank you gng

[mild steppe]

ok

[mild steppe]

np

[shell acorn]

hope you get the girl y want soon

[mild steppe]

lol

[shell acorn]

or whatever you want

[mild steppe]

ok

[tawny hornet]

What if I want you….

[shell acorn]

ain none wrong w that

yu got a one piece pfp we already vibing

mans watch solo leveling

ts peak

[tawny hornet]

Real but we should have this discussion elsewhere, also close the channel if you’re done

[lone heart]

@shell acorn Has your question been resolved?

[lone heart]

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[shy cedar]

the only part im lost on is why 177.2=180- the measure of arc ZX

[lone heart]

@shy cedar Has your question been resolved?

[winged pawn]

the quadrilateralX WXYZ is cyclic as angle YZW = angle WXY = 90

[shy cedar]

what is cyclic

[winged pawn]

ABCD is cyclic... i.e. the points lie on a circle

there's theorems about it

but the key is taht the opposing angles add up to 180

[shy cedar]

yeah im a dumb fuck didnt realize that

[winged pawn]

happy mathing then

[shy cedar]

.close

[lone heart]

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[wanton beacon]

For part b here, shouldn’t it be increasing for the interval pi to 4pi/3? Why does it say decreasing on the answer key?

[lone heart]

@wanton beacon Has your question been resolved?

[lone heart]

@wanton beacon Has your question been resolved?

[lavish cave]

ah, because this is a polar function

https://www.desmos.com/calculator/etsjqmjy6n

3 + 6 cos theta is increasing from a negative value

so it's getting less negative, and hence closer to the origin

[wanton beacon]

So the answer key’s correct?

[lavish cave]

yes

the distance of $(r \cos \theta, r \sin \theta)$ from the origin is $\sqrt{r^2} = |r|$

[ocean seal]

south

[lavish cave]

so think of absolute values, whether the absolute value of the function is increasing or decreasing (further or closer from 0)

and you'll know

[fickle musk]

u could think this also in terms of trig form of complex no

if u want

[hushed shale]

u guys doing math of which std?

[lavish cave]

uni

[hushed shale]

h i am just starting my 11th ..can u give me some tips

[wanton beacon]

Wait, I’m still a little confused

Could u explain it differently?

Yk how f(th)’s less than 0 there?

Doesn’t that make it increasing?

[lavish cave]

yes

when a negative number is increasing

its absolute value is decreasing

[wanton beacon]

But what’s negative?

[viscid sage]

i think f(th)

i think that what they are trying to say is
if you have -3, -2, -1, 0 in ascending order
their absolute value counterparts are decreasing

because |-3|, |-2|, |-1|, |0| is
3, 2, 1, 0, which is decreasing

[lavish cave]

thanks for clarifying!

[viscid sage]

np :)

[lone heart]

@wanton beacon Has your question been resolved?

[neat thorn]

https://tenor.com/view/orange-role-jjk-gojo-gif-9686759356024094267

[lone heart]

@wanton beacon Has your question been resolved?

[lone heart]

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[jagged holly]

Fₙ and Lₙ are Fibonacci sequence and Lucas sequence of numbers

prove by induction that
F₂ₙ = FₙLₙ

An identity given in the excercise is
Lₙ = Fₙ₊₁ + Fₙ₋₁

I don't know about any other identities which might've helped me to get to the proof. I've tried all kinds of algebraic manipulation I could think of but don't really seem to be getting to the proof.

Please help!

[jagged holly]

I know about cassini's identity but don't see how to apply it here in the inductive step

[lone heart]

@jagged holly Has your question been resolved?

[knotty sleet]

Do you know strong induction?

[dire plover]

do you know the explicit formula for fibbonaci and lucas?

[knotty sleet]

If you just use the Fibonnacci identity and the one for L_n you can get the result with strong induction. No need for anything else.

The idea is to write F_{2(k+1)} using F_{2k} and F_{2(k-1)}, then apply the induction hypothesis.

[jagged holly]

not really, this was like the 12th induction related problem I was solving, the book I'm doing doesn't mention any distinct strong or weak induction

[dire plover]

how will you do that?
anytime you use
F_n+2 = F_n+1 + F_n, you will always be left with a odd term

[jagged holly]

Nope

[knotty sleet]

You can look it up, it's not hard or anything. It should make sense if you already know how normal induction works

[jagged holly]

Alright, I will!

[dire plover]

||3F_2k - F_2(k-1)||
but how will you prove that besides induction? youll have to induct twice?

[jagged holly]

These are the only questions besides the examples I've solved

[dire plover]

oh, you solved that?

then you dont need two induct twice

[knotty sleet]

For example, F_{2k+1} = F_{2k+2} - F_{2k}

[jagged holly]

no, I could'nt solve the one I sent, it's circled there

Q5 (ii)

[dire plover]

i mean, 4a

[jagged holly]

Yeah

[jagged holly]

Yeah I know that

I tried doing it that way, I got lost in the algebraic manipulation that comes after

[lone heart]

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[jagged holly]

.reopen

[lone heart]

✅

[knotty sleet]

Was replying to shryanjha. But just be careful and methodical and you'll be fine

If it helps, try writing everything in terms of F_k and F_{k-1}. It's not a particularly enlightening bit of work

[jagged holly]

alright

I'll give it a try again

[knotty sleet]

Nice

You can do it, just stay cool

[lone heart]

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[oak wasp]

I am in the process of writing a small program to generate images/patterns based on certain curves. I want to find a function that will produce the appearance of a 'ribbon candy' pattern (see pic). I'm not sure what function would produce this shape though. Some kind of parametric function should work but I need help designing it

https://prostitcher.com/wp-content/uploads/2021/02/Simple-Ribbon-Candy.gif

[proper tangle]

you can use a problem strategy of divide and conquer to divide it into a few subquestions

1. given that you can make a cycle of "S" curve, can you propagate it to as many cycles as you want?
2. is there any symmetry within that cycle
3. how can make simple linear transformations (reflection, rotation, and translation) on the curve

[oak wasp]

yeah this is pretty much what I'm looking for

[oak wasp]

Do you mean a kind of piecewise addition of S curves?

[dire plover]

you can scale this pretty easily

[proper tangle]

you can say so
https://en.wikipedia.org/wiki/Periodic_summation

[oak wasp]

I see, that makes sense, thanks!

[oak wasp]

and yeah for this particular pattern, playing with the constants will definitely get to what I'm looking for

[proper tangle]

ofc you need to adjust it a bit

[dire plover]

if you are trying to make it more or less bulgy, try raising the cos to a power of the form odd number/odd number

[proper tangle]

your curve is like F(x, y) = 0, and when it's translated ...

[dire plover]

this is cos^7/5

[oak wasp]

ah yes that looks interesting too

[void nymph]

very

[lone heart]

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[strange fractal]

HI

[strange fractal]

so bsasically

i found torque clockwwise

which is just the woman and the bucket

[quick wind]

a irl picture dont look great to model with, best to draw a proper diagram

[strange fractal]

okok

ok s oifirst of all

[void nymph]

you need practice with your fbds man

[quick wind]

ok think of the force the woman and paint should exert

on a flat surface

62.25g

now on an incline it is still 62.25g

it is just broken down into components

like the gravity and the normal reaction

[strange fractal]

waot..

i know wat i did wrong

[strange fractal]

is it cause i didnt multiply the mass by 9.8

[quick wind]

ye u gotta use g

[strange fractal]

(4.25+58)x(0.5/sin(59.69)x9.8

ohh

ok i think thats waht i did wrong

let me check

NOOOOOO I FORGO TMY CALCUIALTOR IN MY LOCEKRE

[quick wind]

gravity is mgsin theta normal reaction is mgcos theta

where m is mass

[quick wind]

😭

[strange fractal]

hi..

I did this..

[quick wind]

could u model the paint and the woman to be a single object

cuz they are connected after all, shes holding le paint

[strange fractal]

yea

i checked and i get same answer

so its alg

[quick wind]

since sin theta is 2.4/2.78

u could just use that value for mgsin theta

for cos theta just use pythag

so cos theta = other side/ 2.78

u dont have to really find angles rn

[strange fractal]

o u rite

[quick wind]

moreover

for x

can u see u made a rectangle

[strange fractal]

yes

[quick wind]

so x = 90 minus theta

so treat x as another angle of ur old triangle

then find sin x cos x

[strange fractal]

wait lamda is the angle

x was the

length of the where the lady was

on the ladder

was this all i needed to do tho?

I LOVE LADDERs

[lone heart]

@strange fractal Has your question been resolved?

[lone heart]

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[drowsy holly]

How can i make sure the area i calculated is correct for my graph in desmos? https://www.desmos.com/calculator/trah674ke4; i got 27,277 sq units

[lavish cave]

if you find the coordinates of each vertex of your shapes

then you can use the shoelace formula to calculate the area

specifically I was thinking for the kite, which is two congruent triangles

for the rest just use A = 1/2 bh

Wolfram can do the shoelace formula for you

[drowsy holly]

Oh okay, Thank you! I'll look more into it

.close

[lone heart]

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[warped topaz]

Part (c)

I thought I knew how to do it but I got the wrong answer :/

Wait

Wrong question 💀

[lone heart]

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[earnest slate]

…

[vale wigeon]

gotta get a new channel then

[warped topaz]

What now bro

Leave me alone

[vale wigeon]

this one is about to implode

[warped topaz]

It was a mistake it can happen

[earnest slate]

sorry man

[lone heart]

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[strange fractal]

i dont get why tension vertical is there

[wheat isle]

bum cihcken

[strange fractal]

water beam

[wheat isle]

14 mark question is crazy work

[strange fractal]

😭

[strange fractal]

where does tension vertical come from

why does doing 6 x Tv = Tcw

[wheat isle]

thats a fantastic question, where do you think it comes from

[strange fractal]

um...........................................................................................

from the force of the rope

[lethal belfry]

Wait, so here the drawbridge is massless?

[strange fractal]

no its 500kg

[lethal belfry]

That's the counterweight, pretty sure

[strange fractal]

wat does that mean

[lethal belfry]

https://en.wikipedia.org/wiki/Counterweight

[strange fractal]

no it says 6m long drawbridge with a mass of 500g\

wait

whats a moat

this is a moat

[wheat isle]

thats a river

[strange fractal]

nO

[wheat isle]

u know those castles

that r surrounded by a body of water

[strange fractal]

yesss

[wheat isle]

thats a moat

[strange fractal]

oh

[strange fractal]

wait im confused

so the 500kg is the counterweight

or is the bridge 500kg

i think its the bridge

@wheat isle WATERBEAM

[wheat isle]

hello

to answer your question

i think you should ask yourself is the bridge 500kg or is the counterweight 500kg

[strange fractal]

the bridge

[ivory igloo]

yes

[ivory igloo]

balancing torque

the tension is at an angle to the drawbridge

but only the component of it perpendicular to the bridge will produce torque

which is T sin 35 degrees or T cos 55 degrees

times 6 meters

perpendicular force times distance from pivot

[strange fractal]

oh

okok

what do they mean by reaction force?

is it like this force

in red

[ivory igloo]

hmm

try drawing the tension and weight

youll see the net force doesn't cancel out

horizontally and verically

so there's a horizontal component too

[strange fractal]

hold up

so its all these forces

• the one going dowen

and they have to equal 0

[ivory igloo]

yes

something like that

[strange fractal]

and then after i have the horizontal and vertical components

i make a triangle?

aand find the hypotenuse?

[ivory igloo]

yeah

you know three of the five arrows already

[lone heart]

@strange fractal Has your question been resolved?

[lone heart]

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[cerulean root]

For this frq I got:

a) 0.127
b) 0.561 (with the only theta I found being 0.86 via the calculator)
c) (2-pi)/4

How would I justify part b though?

Also if anyone can, are my solutions correct?

[lone heart]

@cerulean root Has your question been resolved?

[cerulean root]

<@&286206848099549185>

[mental canopy]

hello!

how can I help

[reef grove]

b) 0.561 (with the only theta I found being 0.86 via the calculator)

what do you mean with 0.561

[cerulean root]

that is the greatest distance I founded from any point on the graph of r to the origin

but im confused on how to justify my answer

[reef grove]

do you know what r(theta) represents

[cerulean root]

the polar function?

[reef grove]

r is the distance from the origin defined via the Pythagorean theorem

r = sqrt(x^2 + y^2)

but now you are given that r is only depended on theta so for every given theta r(theta) gives you the distance from the origin to that point on the curve

[cerulean root]

oh I didn't think of it like that

but how would I use that to justify my answer for part b)?

cause I solved for the critical point when dr/dtheta = 0 and then plugged that into r(theta) to find the maximum distance but I don't know how I would justify that it is the maxmium distance

like do I need to prove its a maximum at that point?

[reef grove]

so what final equation did you get by setting dr/dtheta = 0

[cerulean root]

theta is equal to cotangent of theta

and I got theta to be approximately 0.86

[reef grove]

well we have to think about the domain of theta

[cerulean root]

the domain is 0 to 3pi/2

[reef grove]

do you know how cot(x) looks like as a function

[cerulean root]

yeah

[reef grove]

for every interval (k,k+pi) there exists an x in (k,k+pi) such that cot(x) = x

with k in Z

[cerulean root]

I found theta by just using the non simplified equation 0 = cos (theta) - theta * sin (theta) and on my calculator locating the zero on the graph

[reef grove]

well this solution is correct, but its only one out of the infinite possible solutions

[cerulean root]

oh ok

[reef grove]

but for your given Interval you have 2 possibilities

[cerulean root]

so then how would you know which theta?

[reef grove]

the intersection of the green and blue graph are your possible solutions, and if its then also in between 0 and 3pi/2 those are your possible solutions

[reef grove]

you try both solutions and see which one is bigger

[cerulean root]

oh thats it

[reef grove]

not quite

you always have to check for the boundaries

so you have to try r(0) and r(3pi/2) as well

because the derivative can only give you the local minima and maxima

[cerulean root]

so we are looking for the absolute maximum?

[reef grove]

yes and for this you have to check all local maxima and the boundary values

,w cot(3.4256)

[ocean seal]

Results provided by WolframAlpha

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

Requested by tobi_012345

[reef grove]

this is the second numerical solution

[cerulean root]

so overall to justify my answer would I have to plug in the boundary values and both values of theta and then I can conclude the greatest one is my maximum distance?

[reef grove]

theta = 0
theta = 0.86
theta = 3.4256
theta = 3pi/2

[reef grove]

yes you have to try all those 4 values and this is enough

this process will always give you the maximum value for a differentiable function in any interval [a,b]

[cerulean root]

alright that makes sense

I plugged everything in and I indeed got 0.86 as the greatest distance

thanks

[reef grove]

oh btw

3.4256 is a minimum not even a maximum

could have checked that with the 2nd derivative as well

but

actually 3.4256 should be the maximum if we think about this

[cerulean root]

If you don't mind cause I spoke to some other people for part c) would evaluating for dr/dtheta by doing dy/dx = dy/dtheta = dx/dtheta work cause that is how I got (2-pi)/4

[cerulean root]

so then the strategy of plugging in the boundaries and the critical point still works right?

[reef grove]

yes that always works, but I am a bit confused because r(theta) can be negative

for theta = pi for example

[cerulean root]

yeah

[reef grove]

ah it just means how to plot it

but the distance is always abs(r)

[cerulean root]

wouldn't that make 3pi/2 the biggest then?

[reef grove]

cos(3pi/2) = 0

[cerulean root]

sorry I ment 3.4256

[reef grove]

yeah

,w cos(3.4256)*3.4256

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[reef grove]

,w cos(0.86)*0.86

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[reef grove]

If \( r(\theta) > 0 \), the point lies in the direction of \( \theta \).

If \( r(\theta) < 0 \), the point is rotated by \( 180^\circ \) and lies in the direction of \( \theta + \pi \).

[ocean seal]

tobi

[reef grove]

this is what it means atleast thats what i found

sounds reasonable

because r cant be negative

distances are always greater or equal to 0

[cerulean root]

so would that make the greatest distance 3.288?

[reef grove]

yeah

,w solve cot(x) = x for x in [0, 3*pi/2]

[ocean seal]

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[cerulean root]

wait so im a bit confused why is the greatest distance come from 3.4256 not 0.86

[reef grove]

okay of we plug in r(3.4256) we get -3.28837 right

which does not make any sense at all if you think about it because the radius from the origin can not be negative

[cerulean root]

yeah

[reef grove]

but what the negative sign is telling us is where to plot the point in the coordinate system

mainly that if r(theta) < 0 then the point is rotated by 180 degrees in the plane

but the actual distance from the origin to the point is the absolute value of r(theta)

[cerulean root]

so how would you like write a justification then for all of this as to why 3.288 is the largest distance from the origin

[reef grove]

we look at our function r(theta), check for all local minima and maxima via setting the derivative equal to zero and we also check for our boundary conditions of theta and plug all those possible solutions in our function r(theta) for all values of theta that we found.
those are all possible values where r(theta) could have minima or maxima on [0,3pi/2].
(It took mathematicians a long time to understand why its enough to only look at these values)
Then you plug all values in r(theta) and look for the value of theta, where abs(r(theta)) has its maximum value

[cerulean root]

so like would saying at the end as distance can't be negative r(3.4256) must equal 3.288 and thus be the maximum distance from the origin

[reef grove]

so first of all radius and angle always give you a point in the plane.
and in our case knowing theta means knowing r(theta), so lets say we choose polar coordinates and write all our coordinates
using a radius and an angle meaning any point can be written as (radius, angle)

Then
If r(theta) is positive than the corresponding point on the coordinate system is at ( abs(r(theta) , theta )
If r(theta) is negative than the corresponding point on the coordinate system is at ( abs(r(theta) , theta + pi)

[reef grove]

the distance from the origin to your point is always the absolute value of r yes

[cerulean root]

alright that makes sense

are my answers for parts a and c correct though?

I got 0.127 for a) and (2-pi)/4 for c)

[reef grove]

I got the same answer for c)

how did you do a

[cerulean root]

I used the formula 1/2 integral from a to b of r^2 dtheta

I got the limits of integration as 0 and pi/2

and then plugging it into the calculator I got 0.127

[reef grove]

,w Int_{0}^{\pi/2} 1/2 (x cosx)^2 dx

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[reef grove]

yeah there is the closed solution as well if you want it

if you want to integrate this you need a lot of double angle and trig identities I guess

[cerulean root]

does my method of solving for part c) work?

dy/dx = (dy/dtheta)/(dx/dtheta)

[reef grove]

yes thats allowed because of the chain rule

[cerulean root]

alright

kinda a seperate question

do you know when I should use 1/2 integral from a to b r^2 dtheta vs 1/2 integral from a to b of [R^2 - r^2] dtheta

[reef grove]

yeah you use the R^2 - r^2 one, when you want to find the area between two polar curves, where R is the outer curve (large radius) and r is the smaller radius.
notice that this formula also applies here just that the smaller polar curve is just always equal to zero so in that case above its R^2 - r^2 = R^2 - 0 = R^2

or in other words you use 1/2 integral from a to b r^2 dtheta when you only have one curve

and you want to find the area enclosed by that one curve

[cerulean root]

thanks that clears it up

how do you find the correct limits of integration though

cause sometimes I get that wrong

[reef grove]

first of all look at what values are allowed for theta in your given problem.

If you want to find the area that's enclosed between the first loop then set r(theta) = 0
and then you will find values for theta lets say
theta_1, theta_2, theta_3, ...
and lets order them meaning that theta_1 < theta_2 < theta_3, ...
the boundaries for your first loop would be theta_1 and theta_2 and for the 2nd loop it would be theta_2 and theta_3
(I guess this should work for all the problems that you are working with)

But there is a Problem with this!
It could not even be a closed loop in the first place, it is possible to imagine curves where this would not work

[cerulean root]

do you think that would be in the scope of ap calc bc?

I have seen cardioids, rose graphs, and stuff similar to the curves in chapter 8 of calc bc

[reef grove]

actually finding what counts as a closed loop is not inside of the scope of what one does in ap calc bc.

[cerulean root]

so then this method works for calc bc?

[reef grove]

did you have something like

r(theta) = 1 - 2sin(theta) ?

[cerulean root]

yeah I think I have seen that

I got one that was r = 2 + 4sin(theta)

[reef grove]

have you had more than 2 loops

[cerulean root]

no I don't think so

[reef grove]

then it should always work to find the zeros and order them

[cerulean root]

alright thanks

[reef grove]

but most of the time its just very dependend on your specific curve and what area you want to find

[cerulean root]

thanks for helping me I appreciate it

[reef grove]

glad I could help :)

[cerulean root]

.close

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[autumn vigil]

guys, how to check for transitive/reflexive etc on relations of the form {x1,x2}->{y1,y2}?

[median oar]

do you have an example?

[autumn vigil]

like this

what i am asking is like for {x}->{y}
we can check reflexive by {x,x}
transitive is checked by {x,y} {y,z} so {x,z}

but i dont understand how to check in {x1x2} {y1y2} type of questions

[carmine wharf]

reflexive would be {x1x2,x1x2}

transitive would be {x1x2,y1y2} {y1y2,z1z2} then {x1x2,z1z2}

[autumn vigil]

ahhh

i see

thnx a lot

.close

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[vale wigeon]

you cut off the actual instructions

sqrt( n/(n+2) ) approaches 1

no, that would change its value (though by dumb luck the limit would be unaffected in this particular case)

a/b is not equal to a^2/b^2

yes, i just combined the roots into one. if you didn't notice, i put parentheses accordingly.

n/(n+2) approaches 1 and this is easy to see

sqrt(n/(n+2)) approaches sqrt(1)

sqrt(a)*sqrt(b) = sqrt(ab)

and sqrt(a)/sqrt(b) = sqrt(a/b)

what

what 1/n

i did not conjure any 1/n ex recto!

ok you multiplied the top and bottom by 1/n

yeah whatever that works

if you want/need to spell that step out, go for it

if you really really really want to use l'hop then i guess i can't really stop you.

[sly mantle]

u should be able to see the limit wo lhopital

sure but itd be very dumb

its like using lh on lim n/n

[vale wigeon]

hammer-nail syndrome

[sly mantle]

didnt say it dont work, im saying its dumb af

some lim are easy w/o lh

[vale wigeon]

if you really really really want to use l'hop then i guess [we] can't really stop you.

[sly mantle]

in future u should try more elementary methods with lh as a last resort

[vale wigeon]

hammer-nail syndrome: when you wield a hammer, everything looks like a nail

in this context specifically: when you wield l'hop, all limits look like they would be reasonable to do with it

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[orchid reef]

How can I solve this ( no sub method I haven’t learned it yet )

[shell agate]

This is easy with substitution

[lavish cave]

do you accept something like $\int_{-1}^1 \frac{d(x^2 + x + 1)}{(x^2 + x + 1)^2}$ then

[ocean seal]

south

[orchid reef]

I don’t think so, the only thing we have seen is the integration by parts and classical primitive

[wheat isle]

how have u learnt integration by parts before u-sub

[orchid reef]

Bro I couldn’t tell you ☠️

[weak valley]

it's not uncommon

[weak valley]

remember the e^t problem

it's the same thing

this time sub the whole denominator

[wheat isle]

its also not common ibp basically involves a substitution anyway Lol

[weak valley]

not really

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@orchid reef Has your question been resolved?

[orchid reef]

i didnt get how does it works, moreover it was e^t so derivative is the same i didnt see the difference

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[ocean seal]

dyxn

[weak valley]

@orchid reef

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[orchid reef]

that makes sense

[weak valley]

you should watch the organic chemistry tutor on substitution if your teacher hasn't covered it

it seems all problems that you're doing require it

[orchid reef]

i'll do it thanks

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[dapper root]

hii

[dapper root]

please help me witth probalilty

what is a complementry probalilty

[stiff trench]

the probablity that something is not going to happen

do u mean complementary events?

[dapper root]

yes yes

and like i dont understand the way exprintemal probality is worded

like what

[stiff trench]

complementary events are two mutually exclusive events

[dapper root]

okay okay so things that dont occur?

[stiff trench]

like "my favorite color is red" and "my favorite color is blue" arent bc it could be green

[dapper root]

ohhh okay thank you

[stiff trench]

but "this coin lands tails" and "this coin lands heads" are bc it has to be one of either

(excluding like it landing on its side or something idk)

[dapper root]

yes right now i get it

thank you

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[dapper root]

yes

[lavish cave]

.close

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[mental python]

youre supposed to respond via the reactions ✅ and ❌

[dapper root]

ohh okay tyy

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[alpine sable]

i need help

[keen idol]

i see you are studying functions

what do you need help with

[alpine sable]

the ex one

y=√-4

[keen idol]

what do you wanna do with that?

[alpine sable]

i don't get it cause why is the answer positive and negative

instead

i mean

[keen idol]

actually y=√(-4) is not defined in the real numbers

[alpine sable]

so it should be positive

the answer

[keen idol]

it should not be positive or negative

√(-4) is undefined in real numbers

[alpine sable]

should it be like element of real numbers

[keen idol]

well it is not an element of the real numbers

[alpine sable]

what should it be then

[keen idol]

if you have studied imaginary numbers, √(-4) is an element of complex numbers

[alpine sable]

like ≠0 or ≠4

[alpine sable]

our teacher didn't mention that yet

[keen idol]

well just know that sqrt for negative numbers is not defined in R

[alpine sable]

oh ok

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[cinder sundial]

What is

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[deft thorn]

i need help with this

[jolly notch]

!showwork

[lone heart]

Show your work, and if possible, explain where you are stuck.

[deft thorn]

i know the formula to do it but how can i parameter it?

it's like p(y(t)) y'(t) DT

and i have to intergrate it

[pale kestrel]

how do you parameterize a circle..

[deft thorn]

is it a + R * e^i theta

[fickle musk]

(x,y)=(rcos theta, r sin theta)

[deft thorn]

ah yes

how do i combine those two equations into one

the cos^2 + sin^2 = 1 identity?

[fickle musk]

unsure what u mean

i am also unsure of complex integration from your question i only understand that C represents a circle centered at (a,0) with radius R

[deft thorn]

ah i see

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[deft thorn]

after i do the parameter

how do i get the conguate

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@deft thorn Has your question been resolved?

[deft thorn]

<@&286206848099549185>

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@deft thorn Has your question been resolved?

[toxic spruce]

what is the answer if the polynomial is (z^n)

[ocean seal]

peanutbutter

[toxic spruce]

maybe a simplification you could make is that if (P) is a polynomial then (Q(z) = P(z + a)) is also a polynomial and if (C') is the circle with center (0) of radius (R), then (z(t) = Re^{it}) gives us a parameterization of (C') and (\int_{C'}Q(z),d\bar{z} = \int_0^{2\pi} Q(Re^{it}) (-iRe^{-it}),dt = \int_0^{2\pi} P(a + Re^{it}) (-iR^{it}),dt = \int_C P(z),d\bar{z} )

[ocean seal]

peanutbutter

[toxic spruce]

so you might as well assume that (a = 0)

[ocean seal]

peanutbutter

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[fiery pawn]

The rectangle $R$, , represented in the figure, has its sides parallel to the coordinate axes and was obtained from a point $(x,y)$ on the ellipse given by the equation $4x^2+5y^2=20$, in the first quadrant.
The point $(x,y)$ that satisfies these conditions and allows for the maximum area of rectangle $R$ has $2x^2+y^2$ equal to:
\(A) 7
\(B) 9
\(C) $\frac {19}2$\
\(D) $\frac{21}2$\
\(E) $\frac{43}4$

[ocean seal]

pirateking0723

[fiery pawn]

so first of all R doesnt have to be in the first quadrant right ?

only (x,y) must be in the first quadrant

[glacial rover]

@ban @lone heart

@glacial rover

[fiery pawn]

huh

[glacial rover]

@fiery pawn

want to play video games

[fiery pawn]

no ty

[weak valley]

@fiery pawn did you parameteize the ellipse?

[fiery pawn]

no

[weak valley]

do you know how to?

[fiery pawn]

x=5cosθ and y=4sinθ?

@weak valley is that it ?

[weak valley]

yeah

good

so now find the area of the rectangle in terms of these coordinates

[fiery pawn]

the area of the rectangle in the picture is 20sinθcosθ=10sin(2θ)

and now we should maximize this

along with keeping in mind that 0=<θ=<π/2 right ?

then θ=π/4?

ah with this i am not getting any of the choices

something is probably wrong with what i did

@weak valley sorry for the ping but any ideas on what i am doing wrong

or is it just that none of the choices is correct

[weak valley]

You get A right?

all your work seems right

[fiery pawn]

i am getting 33

[weak valley]

$x$ is $\sqrt{\frac 52}$ and $y$ is $\sqrt 2$

[ocean seal]

dyxn

[weak valley]

so 2x^2 + y^2 = 7

[fiery pawn]

isnt it x=5/sqrt(2)

because x=5cosθ

and y=4/sqrt(2)

ah wait

here is my mistake xD

[fiery pawn]

this is wrong

it should be x=sqrt(5)cosθ and y=2sinθ

[fiery pawn]

but something is still bothering me

the process rn maximizes the area of the rectangle that has 2 of its sides being the x and y axes right?

but that isnt mentioned in the given

so the rectangle may have greater area no ?

[fiery pawn]

in fact the maximum possible area of the rectangle is 4xy=40sin(2θ) right

well the angle here wont change

[fiery pawn]

but isnt this the real maximum area ?

[rustic coral]

What?

$4xy=4 \cdot \sqrt{5} \cos \theta \cdot 2 \sin\theta=8\sqrt{5} \cos \theta \sin \theta=4 \sqrt 5 \sin 2\theta$

[ocean seal]

Civil Service Pigeon

[rustic coral]

Note that it rlly matters

You just need to know when $\sin(2\theta)$ is maximised

[ocean seal]

Civil Service Pigeon

[fiery pawn]

well i mentioned that it doesnt matter here because whats important is the angle

[fiery pawn]

but that doesnt mean that i can assume that the maximum area of the rectangle is xy

[fiery pawn]

yea at first i had the wrong parametrization

i wrote it as x=5cosθ and y=4sinθ instead of x=sqrt(5)cosθ and y=2sinθ

and i didnt rework the area with the correct parametrization idk why lol

so to say what i was saying correctly

originally i took the area of the rectangle as xy=2sqrt(5)sinθcosθ=sqrt(5)sin(2θ)

and then i said this is incorrect and i should instead look at the area as 4xy=4sqrt(5)sin2θ

(these areas are after using the correct parametrization)

[fiery pawn]

along with this note too

so ik that it doesnt affect the final answer but i just pointed out that it is a wrong assumption

[rustic coral]

Why not? The length is the distance from (0,0) to (x,0), aka x

And the width is the distance from (0,0) to (0,y), aka y

[fiery pawn]

or rather i wanted to check if this is true or no

[fiery pawn]

the given didnt specify anything about this

[rustic coral]

It’s implied from the diagram

[fiery pawn]

the figure must be taken as a reference just to visualize things to a certain extent no ?

but not to take any given from it

thats what i thought about at least

no i dont intend to say anything like that

just saying that the written given didnt mention that the 2 axes are sides of the rectangle

they are parallel

but need not be confounded with the axes

thats what i am saying

[rustic coral]

my point still stands, but I’ll delete my messages if someone else wants to engage in this conversation

[fiery pawn]

well i am not trying to argue , more like trying to know if i could take this as a fact from the figure

to do that from now on

[fiery pawn]

so i take it that i could take this as a given from the figure from now on ?

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@fiery pawn Has your question been resolved?

[fiery pawn]

well i guess i will just close now that the question is solved

tysm both of you for your help

.close

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[solar hollow]

obviously both the graphs are the same

[north rover]

Have you tried parametrization?

Or perhaps just checking each option?

[solar hollow]

nope

[north rover]

So try it!

[solar hollow]

what's that

[gloomy vigil]

is this correct?

[north rover]

!occupied

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[north rover]

Is this for the SAT?

[gloomy vigil]

no

[north rover]

Not asking you.

[solar hollow]

yea

[north rover]

Then don’t worry about it.

Just choose each option and see if it works.

[gloomy vigil]

oh

[solar hollow]

huh?

[north rover]

Like, plug in each answer choice.

[lone heart]

@solar hollow Has your question been resolved?

[north rover]

You have the coordinates. Recall that each is in the form (x, y).

[solar hollow]

🫤

😐

.close

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[clear jungle]

Nevermind, made a stupid mistake

.close

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[ivory siren]

help

[ivory siren]

my first thought was that the derivative of (1+sqrtx) is present 1/2sqrt(X)

i mean

kinda

i could multiply by 1/2 and then put (2) to the left of th integral sign

but idk if this works

[hard patio]

you're forgetting something there

[ivory siren]

erm

[hard patio]

with referance to this^

[ivory siren]

hold on im thinking

...

no thoughts

[hard patio]

a good hint would be to put a 'dx' in the integral

[ivory siren]

uhuh

wait im gonna write this down

[hard patio]

you subbed 1 +sqrt(x) = u

how do you convert the variable in the integral from x to u (a transition from dx to du)

[ivory siren]

derivate then get du

dx= 2sqrt(x)du ?

[hard patio]

see if there is a better way to write that considering the problem at hand

[ivory siren]

?

wait no i need to sub u into this

Im not used to the substitution thingy

nvm ignore that

[hard patio]

try writing 'du' in form of the variable x

[ivory siren]

yeah i forgot to write that

the square roots cancel

then integrate normally?

2/u^2

will become -2/u +c after integrating

.

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[silk folio]

does anoyone know how to do angles?

[desert star]

!xy

[lone heart]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[desert star]

wait no thats not what i wanted.

don't ask to ask questions

just ask your question

!da2a

[lone heart]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[fickle latch]

Yes. Be specific with what you mean by "angles."

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[robust geode]

Multiple things : The image where the measurements are are a bit too tiny, A corresponds to what, BH corresponds to what, why is there =A after A=, what does (7) and (2) correspond to ... A lot of things you have to clear out right there

A lot of people didn't get an answer out of people not being clear enough with what they are asking

Thank you very much, and it's okay, just letting you know lmao

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@last pivot Has your question been resolved?

[earnest token]

Can you just explain the problem in word terms for me quick then I'll try the problem

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[lone heart]

Please don't occupy multiple help channels.

@zealous falcon Has your question been resolved?

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@zealous falcon Has your question been resolved?

[quick wind]

think of what the two words mean literally

rise

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[coral zephyr]

if I substitute x with -2

I get this

Do I do the exponent first?

[tacit barn]

Yes. But try rewriting the multiplication symbol in a neat, better way with parenthesis

[coral zephyr]

I love that

what did I d owrong there

[fast tide]

(-2)^2 is 4 cause youre doing (-2) times (-2) so the negatives cancel out

[coral zephyr]

GOT IT

thanks

.close

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[chilly venture]

need help checking if my result is correct

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[chilly venture]

.reopen

[lone heart]

✅

[chilly venture]

<@&286206848099549185>

.close

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[chilly venture]

.reopen

[lone heart]

✅

[chilly venture]

😭

is surface of revolution everyones kryptonite here or what

<@&286206848099549185>

[chilly venture]

,

do you know how i can find solutions to stuff like that

chatgpt does not work and i can only rely on professors irl

studying alone is impossible

oh bro 💀

[tacit arch]

Oh what are you like 11 years old?

[chilly venture]

u deadass spawn in for that and not to contribute helping

[tacit arch]

<@&268886789983436800>

[chilly venture]

he could be a troll, poor guy

[tacit arch]

Right. You're not entitled to help

[chilly venture]

thats funny big bro

[tacit arch]

!vol

[lone heart]

Helpers are just people volunteering their time to help you. Be polite and patient.

[chilly venture]

need help checking if my result is correct

<@&286206848099549185>

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@chilly venture Has your question been resolved?

[chilly venture]

.close

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[full marsh]

"There are 5 people. How many possible pairs (all of them) can be formed? Remember that pairs consist of 2 people."

A) 20
B)25
C)32
D)10

[full marsh]

sorry, its in spanish

[noble notch]

have you learnt combinatorics?

[full marsh]

text 1: FOUR RESULTS ARE ADDED IF THE ROOT (IN THIS DIAGRAM, 1) IS REPLACED BY 2, 3, 4, OR 5.

text 2: THE ONE ON THE LEFT IS DONE FIRST BECAUSE IT IS THE ROOT.

text 3: IT IS MULTIPLIED BY 2 BECAUSE THERE WILL ALWAYS BE ONE LEFT OUT; THEREFORE, ONE RESULT WOULD BE MISSING FROM EACH OF THE VISIBLE POSSIBILITIES, AND THAT IS WHY IT IS MULTIPLIED BY 2

[full marsh]

no

[quaint echo]

here order doesn't matter so use combination formula

5!/2!(5-2)!

[full marsh]

thanks

.close

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[hoary nimbus]

How do I proceed?

[void bear]

What problem art thou facing?

You proved that if it is known that a__n is divisible by 24, it can be concluded that a__(n+1) is divisible by 24. Since a_0 = 1 + 3 + 20 = 24 is divisible by 24, it can be concluded that a_1 is divisible by a_1. We can repeat this process over and over again. It can be shown that this process ensures that every a_n is divisible by 24.

[west rampart]

You have proved it... Good job