help-26

[mortal mirage]

oh

x is

180-58?

[paper robin]

yes

[mortal mirage]

i have a question

its knda tricky

[paper robin]

because that would give us the sum of 180

sum of obtuse angle

cuz its an obtuse angle

idk if thats why it works but yeah

[paper robin]

i have to go to bed

but there will be more ppl that can help

<@&286206848099549185>

[mortal mirage]

oh alr

[paper robin]

i have time to solve one more question

ill help u out

ok so basically similarity

if the triangles are similar that means the side lengths must be equal

AC/BA = CD/BE

9/5=p/4 (cross multiply)

[mortal mirage]

oh

[mortal mirage]

didnt get this part

[paper robin]

thats basically what similarity means

i jst did a bit of research

cuz where im from australia i think its called congruence? but i dont remember

[paper robin]

so for the first side with cross multiplication it should be 5p=36 p=7.2

so p is 7.2

and same thing for q

with cross multiplication and all

[mortal mirage]

from where to get 36

[paper robin]

9/5 =q+8/8

[paper robin]

9x4

9/5=p/4

[mortal mirage]

where do u get 9 from tho

[paper robin]

5+4

=9cm

9cm is the entire length of side ABC

[mortal mirage]

aha..

ahem]

[paper robin]

lmao

dw

9/5 =q+8/8

9x8=72

5 x (q+8) = 5(q+8)

14.4=5(q+8)

wait

mb

72 = 5(q+8)

[mortal mirage]

oh'

[paper robin]

then that will give u 14.4=q+8

14.4-8 = 6.4

6.4 = q

so therefore p=7.2, q=6.4

[mortal mirage]

ohh

tyy

[paper robin]

in summary, ur jst cross multiplying because similarity means that the side lengths are proportional

[paper robin]

all g

happy to help

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[patent light]

Astronauts who have just landed on a planet in the solar system observe that
a body thrown vertically into the air with an initial speed of 14.6 m/s
it takes 7.72 s to return to the ground. What planet are they on? (N.B.:
gravitational acceleration measured at the equator)

[fiery jetty]

q=6.4
p=7.2

[patent light]

$x(t) = x_0 + v_0 t + \frac{1}{2} a t^2$ \
Since the body leaves and returns to the ground, $x(t) = 0$ when $t = 7.72s$ \
$0 = 0 + 14,6 \cdot 7,72 + \frac{1}{2} a \cdot (7,72)^2$ \
$0 = 14,6 \cdot 7,72 + \frac{1}{2} a \cdot (7,72)^2$ \
$0 = 112,712 + 29,7992 a$ \
$-112,712 = 29,7992 a$ \
$a = \frac{-112,712}{29,7992}$ \
$a \approx -3,78 \text{ m/s}^2$ \

[thorny flame]

alee

[patent light]

so we are on mercury right?

but its normal that ive got the gravitational acceleration negative ?

[ruby mural]

,calc 14.6 * 2/7.72

[thorny flame]

Result:

3.7823834196891

[patent light]

yep

but i used another formula

[ruby mural]

Yeah I got thay

That

[patent light]

its the same?

or mine is wrong?

[ruby mural]

No yours isn't wrong

It's lengthier tho

[patent light]

but why

a is negative

in your calc is not

[ruby mural]

Coz I assumed the direction of a to be positive, as a result I had to take direction of initial velocity to be -14.6
In your case, you took the velocity to be 14.6

The - is only to indicate the direction, so dw

[patent light]

how i velocity negative in your case

im a begginer sorry

[ruby mural]

No it's ok

Do you know the formula
v= v_0 + at?

[patent light]

yes

[ruby mural]

Yes so I took the final velocity to be 0, the initial velocity to be -14.6 and time to be half of 7.72

I assumed "downward" to be positive

[patent light]

why you can do *2

[ruby mural]

Coz time taken to go up= time taken to come down

[patent light]

oh ok

thanks!

[ruby mural]

But anyways
Yours is correct

Just a matter of direction

[ruby mural]

Np

[patent light]

yes

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[minor sand]

I have a function C(t)=C_0e^(kt) and if I find the derivative, I get C'(t)=kC_0e^(kt) or C'(t)=C(t)k, why is this? Why is it not C'(t)=ktC_0e^(kt)

[shut obsidian]

The chain rule tells you that you multiply by the derivative of kt w.r.t. t

That's k

Not kt

[minor sand]

can you explain in a simpler way

[shut obsidian]

Let's simplify your problem: [C(t) = e^{kt}] and you think that [C'(t) = kt \cdot e^{kt}] instead of [C'(t) = k \cdot e^{kt},] right?

[thorny flame]

Kepe

[minor sand]

Yes if its without C_0, then yes

[minor sand]

oh is it because the t goes away when we find the derivative? for example if we say k=2 and t=x, and we find the derivative of 2x, then we get 2? is it the same principle

[shut obsidian]

Yep

[minor sand]

but why is t defined as x

oh its our variable

[shut obsidian]

Our entire function is in terms of t here

Yeah

[minor sand]

yes i get it now, thanks

[shut obsidian]

np

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[mystic siren]

i need help to figure out how to prove this for a in R \ {0}

[mystic siren]

i know this works for a > 0, and for a < 0 it doesnt - but because there's an absolute value for the argument on the left hand side it does work here

i dont know how to do the proof without getting a conflict

a contradiction of some sort

[shut obsidian]

Break it up into cases

Case 1: a > 0.

Case 2: a < 0.

[mystic siren]

how would i write it for case 2?

[shut obsidian]

In that case the expression in the absolute value is negative, so it'll turn out to be -(a + 1/a)

[mystic siren]

i see, and i get what you're saying, but how do i write it? as in,
case 2: let a < 0:
...

because i cant say "let a = -a"

[shut obsidian]

Yes, just "let a < 0". No need to do anything else

[mystic siren]

but i do need to take the minus sign out of a to do |-y|=|-1|*|y|=|y|

[shut obsidian]

If a < 0, then |a| = -a

You can leave out the absolute value in place of a -

If $a < 0$, then $\big|a + \frac1a\big| = -\big(a + \frac1a\big)$

[thorny flame]

Kepe

[mystic siren]

i understand now

but wouldnt it be confusing for the reader?

i think there has to be a simpler way to show that

[shut obsidian]

If you really want to, you can say that $a = -|a|$ if $a < 0$

[thorny flame]

Kepe

[mystic siren]

this is not what i mean, i meant something like if a < 0, let a = -b

but i guess your way is better

thank you mate

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[silk spire]

can someone check for me if this is correct?

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[hybrid pasture]

.close

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[knotty heron]

just a quick question about u substitution, when taking something like 5x out of would that also take any x in the equation?

[knotty heron]

Like is this right?

[craggy haven]

well, no

u = 5x

u does not equal x, so you can't substitute u in place of x

but if you were to solve that equation for x

you'd get u/5 = x

and you could substitute that in

[knotty heron]

So like this

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[cyan mesa]

Close, you forgot to replace the x.
usec(u)/5

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[lilac furnace]

,rotate

[lilac furnace]

Hey im stuck on question 9

I think thats what you call provint something by induction in english

Im french

Here is what I tried

But I cant go to the point im looking to demonstrate

[lilac furnace]

Just rotate it maybe

[ivory trench]

ahh much nicer(maybe)

[lilac furnace]

So now I bet you can help me ?

😃

[ivory trench]

i cant read french
pls wait

[lilac furnace]

I need to démonstrate the relation

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@lilac furnace Has your question been resolved?

[lilac furnace]

@ivory trench 🙂

<@&286206848099549185>

[eternal olive]

bro the image aint clear

[tough nest]

the question is to find I_n without recursive

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[paper elk]

uh im gonna send an image of the question and give $ to whoever assists me

[paper elk]

gimme a nice juicy answer and i give u money and a cute anime gif

preferably a correct one with a lil reasoning

(can i please say chop chop)

chop chop!

oh and the question just asks

"To the nearest tenth of a degree, what is the measure of angle PQJ?"

[reef fjord]

!noans

[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.

[paper elk]

ohhhh

okay help me use my brain pls

thanks!

help me from the ground up :c

basically hold my hand

PLS PLS PLS PLS

[reef fjord]

this may help

[paper elk]

is not enuf, my mind is an enigma

i will try tho thanks

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@paper elk Has your question been resolved?

[keen raptor]

Can you calculate the length of PJ?

using triangle NPJ

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@paper elk Has your question been resolved?

[paper elk]

sure :c

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[exotic pike]

Can someone help me simplify it

I got 1sqrt(7)

.help

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[half edge]

,calc 144*7

[thorny flame]

Result:

1008

[pearl fog]

you mean 1+sqrt7?

[exotic pike]

Yeah

[pearl fog]

i got the same

[exotic pike]

So fast ?

Can u show me how you did it ?

[pearl fog]

uh

[exotic pike]

I spent 10min

[pearl fog]

do you know $$\sqrt{(a+b)+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}$$

[thorny flame]

Skill_Issue

[exotic pike]

😳 No

But can you explain ?

[pearl fog]

like how i got that?

[exotic pike]

Yeah if possible

[pearl fog]

like how i got this or the answer?

[exotic pike]

Answer if possible

[reef fjord]

this can be derived from binomial theorem

[exotic pike]

Can you guys explain it so that A 14y old can understand

(Me)

[reef fjord]

well

do you know (x+y)^2=x^2+2xy+y^2

[pearl fog]

you just find a and b in each of them and plug into the thing that i showed you

[exotic pike]

Ok Ig

Finally

I have another question if you can help me ?

[pearl fog]

take both sides and square then, so you get (a+b)+2sqrtab=a+b+2ab

[exotic pike]

Ok

Skill_Issue ?

U online ?

[pearl fog]

ye

[exotic pike]

I gimme a sec

How do u solve this one ?

I am studying for a exam in Australia

For a school entrance

And was wondering

About the question

[ember lodge]

First, I suggest to write down the information you have

[exotic pike]

Ok

[ember lodge]

E is the mid-pt. of AD, AE = ED = 1

[exotic pike]

Yeah

[ember lodge]

and think which line you can find the exect value

which is BE and CE

[exotic pike]

Using Pythagoras

[ember lodge]

through those line you can fine CF since you have all the lines you got

[ember lodge]

ys

[exotic pike]

Ok

Then

[ember lodge]

however, I'm still thinking of this, lol

[pearl fog]

i have no clue, but you can graph this question

[exotic pike]

Ok

This is a Year 9 Question

[ember lodge]

I get it

[exotic pike]

I was shocked when

Preparing

[ember lodge]

you may want to find out the angle

[exotic pike]

Ok

[pearl fog]

im only year 8 :p

[exotic pike]

Actually?

[pearl fog]

ye

[exotic pike]

Funny one

[ember lodge]

The strategy is that when you find yourself having trouble to find the sides

Try to use the angle

[pearl fog]

wait

[ember lodge]

Have you learn sin, cos, tan?

[pearl fog]

i have an idea

[exotic pike]

Yeah

[ember lodge]

you first find the angle of AEB

[exotic pike]

SOHCAHTOA ?

[exotic pike]

Ok

[ember lodge]

through alternative angle, you can find the angle of EBC then use trigo identites to find the length of FC and BF

[pearl fog]

BEG is simmilar to BFC

[ember lodge]

Using substitution, you find the length of FE, by using area of triangle to calculate CEF and EDC and then sum up the areas, you get the area of CDEF

[pearl fog]

dont think you should use trig st class 9 :p

[ember lodge]

How do you proof

[pearl fog]

using this, you can find the area BFC, and it becomes trivial to find CDEF

[pearl fog]

they share a common angle B and they are both 90 deg triangles

[ember lodge]

You proof that those triangles are similar but not conguent

[ember lodge]

how

[pearl fog]

yes exactly

[ember lodge]

Those triangle are just similar, their sides may not be the same

[pearl fog]

CDEF=ABCD-ABE-BFC

[pearl fog]

yes

exactly

[ember lodge]

what's the next step then

[pearl fog]

we can find BE with pythag, which is sqrt5, so the ratio of the sides of BEG and BFC is sqrt5:2

[exotic pike]

Bruh the answers need a answers to understand

[pearl fog]

wtf??

why not just uae simmilar trianglea tho

[exotic pike]

Idk

[ember lodge]

how

[exotic pike]

Exactly

[ember lodge]

how to use similar triangles

unless you find the sides' ratio

[pearl fog]

this

[exotic pike]

Do u understand the solution ?

[pearl fog]

i already found it dude

look at what i wrote

[ember lodge]

oh ys

that's true tho

[ember lodge]

I'm lookin rn

[pearl fog]

the ratio of the area is just the ratio of the sides squared, you get 5:4

[exotic pike]

Ok

[ember lodge]

What is 'Area of triangle CDA' ???

[pearl fog]

so thw area of the smaller triangle is 4/5

so 4-1-0.8=2.2

[pearl fog]

huh

[ember lodge]

Note that this question has multi-solution. So the answer is just an example for solving this question

[pearl fog]

i am anti trig

[ember lodge]

ye, I think the answer is more complicated

[pearl fog]

tbh while yes trig is a groundbreaking tool, it kinda sucks the life of any geometric problem thats on the borderline edge of not before trig

where did op go :(

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[vapid linden]

Would saying 2 would produce even values only while 3 will produce only odd values with any given P sufficient ?

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[worn gorge]

any help w part iii)?

[gleaming reef]

Yes

[worn gorge]

thank u 😭😭😭

its good to see u aain

also any help w part iii) of this one asw

[gleaming reef]

Yes of course

[gleaming reef]

Can you write out 3

[worn gorge]

wait just

write it out like nromal?

cos im a bit confused for part iii

like how come there isnt a n-r or to the power of r

[gleaming reef]

It’s a hence

So you want to use the results from part I and II yes?

[worn gorge]

hmym yes

[gleaming reef]

So what do you think

[worn gorge]

umm

[gleaming reef]

Dude expand the sum

[worn gorge]

😭😭

is that all

[gleaming reef]

And then distribute the n choose r

[worn gorge]

wait

[gleaming reef]

Yes it follows immediately

[worn gorge]

ok lemme try

SORRY BRUH

where do i go from here

[gleaming reef]

Can you just make it into two sums

[worn gorge]

huh

like sum of ncr * sum of 2r-n

[gleaming reef]

The sum of ncr * 2r - sum of ncr * n

[worn gorge]

OHHH

wait

okok lemme try

[gleaming reef]

Factor out the 2

From the first sum

[worn gorge]

okok

AHAHHAHAHA

YAY

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[fervent briar]

should i use hopital here or is there some easier method

[fresh ridge]

use the expansions

[fervent briar]

uuuhh how

[fresh ridge]

,w taylor series of e^x

[thorny flame]

Results provided by WolframAlpha

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[fresh ridge]

replace x with the term there ^

[fervent briar]

ah

[primal tulip]

I was thinking somehow constructing a form of e^x - 1 all over x

But well

[fresh ridge]

this way youll get the answer by using the first two terms

itll be short

[fervent briar]

also i do have a doubt
in the expansion series
why do we sometimes not use the other terms
like sometimes we use till the power of 5
sometimes we dont

[fresh ridge]

how many terms are used depends on the question

[fervent briar]

why tho
isnt it going on till infinity

[primal tulip]

But nevermind, don’t think that’s the right direction

[fresh ridge]

in some questions there are terms with higher powers that we have to use the 5th term as well

but usually you wont have to

[fervent briar]

so we have here in the denominator x²
so well use the expansion till x² right?

[cinder sequoia]

the other terms vanish faster so they don't affect the problem, but for the first few terms, the speed at which they vanish is compared relative to other terms in the problem

[fervent briar]

rightt

that makes sense

one sec lemme try

[fervent briar]

also is this correct

@fresh ridge

i got this

wait sorry

therell be a square in the quadratic

so do i now apply hopital here @fresh ridge

[topaz sinew]

@fervent briar Has your question been resolved?

[fervent briar]

i applied hopital

got this as final answer

but now how do i match the options

x tends to beta

if i put beta as the quadratic formula

its way too big of a calculation

@fresh ridge

sorry if the pinging is too much tho

need this answer

TwT

@cinder sequoia any help man?

[cinder sequoia]

honestly i would've just done l'hopital from the beginning

i'm not sure if the answer you've got here is right

[fervent briar]

yeah
i checkes the solution
we both arrived at this step

in the end

[cinder sequoia]

oh

[fervent briar]

yeah so

now beta is a root

how do i simplify this

x tending to beta

the answer is 2(b² - 4ac)

[cinder sequoia]

so you'd get 2(2*beta + b)^2

[fervent briar]

i mean a here is 1

[cinder sequoia]

maybe try writing beta using the quadratic formula?

[fervent briar]

yeah thats what i was asking

do i write that and simplify all that

[cinder sequoia]

yeah and it simplifies quite easily actually

[fervent briar]

huh

i cant do it

[cinder sequoia]

the -b/2 gets multiplied by the 2 and the -b cancels the +b

beta = -b/2 +/- sqrt(b^2 - 4c) / 2

[fervent briar]

can you show pls 🙏

mhm

[cinder sequoia]

from the quadratic formula

then 2beta = -b +/- sqrt(b^2 - 4c)

2beta + b = +/-sqrt(b^2 - 4c)

squaring that gives you b^2 - 4c

and then you double it

[fervent briar]

GOD

IM DUMB

i was expanding the square

THANK YOU

SO MUCH

[cinder sequoia]

happy to help lol

[fervent briar]

.close

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[brittle parcel]

Dihedral group (Dn) is a group of symmetries of the regular n-gon with order 2n (my class's definition). Recently, I remembered our lecturer mentioning something about a subgroup of Dn consisting of rotations in Dn and another consisting of reflections (not sure with this one). Is this true or did I mishear it? Also, if this is true, what are the standard symbols used to represent these subgroups?

[lean pebble]

this is true, but not the whole story, the dihedral group can be defined in terms of generators which are rotation and reflection

so in symbols $D_n=\langle r,s : r^n = s^2 = (sr)^2 = 1\rangle$

[thorny flame]

Bair

[lean pebble]

you can also interpret this as group actions acting on the state of the polygon

[brittle parcel]

Oh that makes sense. So since these subgroups exist, do they have like a standard name to refer to them? (I.e. rotation subgroup). We're currently compiling a bunch of groups of different orders for an assignment. This one struck me since none of my peers also had an idea as to what to call it or what symbol to represent the subgroups.

[lean pebble]

well if i understand correctly, the two subgroups that you are referring to is the one generated by r and the one generated by s, and in that case if im not mistaken, the one generated by r is just the cyclic group of order n, and the one generated by s is the cyclic group of order 2

[brittle parcel]

Oh yeah, now that you mention it, i almost forgot about that notation. Thanks!

Close

.close

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[coarse tusk]

what's the question?

[stuck pawn]

How to get the volume of the green area After rotating around the x-axis

Using shell methods

Y = 3
Y = √(25-x²)

[rigid ivy]

Do you know how to set up the integral?

[stuck pawn]

That's the problem i know it but i get it wrong

[rigid ivy]

!show

[topaz sinew]

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

[stuck pawn]

2π × integration ( 3√(25-x²) ) from 3 to 5

[rigid ivy]

$2\pi\int_3^5 3\sqrt{25-x^2}dx$?

[thorny flame]

SWR

[stuck pawn]

Yes

[coarse tusk]

that doesn't seem right

[rigid ivy]

Why are you multiplying 3 and $\sqrt{25-x^2}$?

[thorny flame]

SWR

[stuck pawn]

Is it wrong ?

[rigid ivy]

Yes

[coarse tusk]

$p(y)\ne 3$

[thorny flame]

kheerii

[rigid ivy]

You should look back to the integral setup for shell integrals

[stuck pawn]

So , what's the setup

[rigid ivy]

I believe it is $2\pi\int_a^b xf(x)dx$. I imagine your $h$ means "height", which would make it $f(x)$ in my integral, but I have no idea what $p$ is supposed to mean in your image

[thorny flame]

SWR

[stuck pawn]

So how to solve?

[stuck pawn]

P is radius or something

[rigid ivy]

Oh, correction:

$V=\pi\int_a^b ([f(x)]^2-[g(x)]^2)dx$

[thorny flame]

SWR

[stuck pawn]

No no

I know this foemula

I can get the answer with it

[stuck pawn]

I mean how to solve with this

Anybody here ?

[rigid ivy]

This is for rotating around the y axis, not the x axis

[stuck pawn]

It's not a problem

[stuck pawn]

So how to solve the question with this

[rigid ivy]

Convert your region into functions of y, instead of x, and change your bounds of integration to be bottom and top bounds, instead of left and right

[stuck pawn]

It doesn't make difference

[rigid ivy]

I promise you it does

I'm staring at the correct answer

[stuck pawn]

Y = √(25-x²)
Y² = 25-x²
X² = 25-y²
X = √(25-y²)

It's the same

[rigid ivy]

and change your bounds of integration to be bottom and top bounds, instead of left and right

[stuck pawn]

1 to 4 ?

[stuck pawn]

Or what do you mean

[stuck pawn]

Broooo

What do you mean

[rigid ivy]

What are the lower and upper bounds?

[stuck pawn]

3 and 5 ?

[stuck pawn]

??

[rigid ivy]

yes

[stuck pawn]

Okay but the answer is wrong

[rigid ivy]

!show

[topaz sinew]

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

[stuck pawn]

Wtf

That's the first thing i showed you

[rigid ivy]

But you changed what you did

Show me what you have now

[stuck pawn]

.

[rigid ivy]

That's not the correct integral

[stuck pawn]

I told you it's exactly the same

[rigid ivy]

.

[stuck pawn]

The bounds is the same

[rigid ivy]

you don't want 3

err

$3$ in $3\sqrt{25-x^2}$ is wrong

[thorny flame]

SWR

[stuck pawn]

You mean it's the same integration without 3 ?

??

[rigid ivy]

No it's supposed to be yg(y)

[stuck pawn]

Just the variables is y

[rigid ivy]

there you go

[stuck pawn]

But the same form

?

[rigid ivy]

same form?

[stuck pawn]

Bro you can just show me instead of asking or giving instructions

I really can't get what you mean

We spent an hour for worthless question

[rigid ivy]

$y$ is just $y$ and $g(y)=\sqrt{25-y^2}$. You literally just have to plug that in

[thorny flame]

SWR

[stuck pawn]

Ok so the integration is
2π integration √(25-y²) dy from 3 to 5 ?

[rigid ivy]

no

[stuck pawn]

What's the problem

[rigid ivy]

You're missing the y

it's yg(y), not just g(y)

[stuck pawn]

Alright

The y in our question

[rigid ivy]

is just y

[stuck pawn]

Is just y

[rigid ivy]

✅

[stuck pawn]

Gonna try

[stuck pawn]

Bruh

It's right

[rigid ivy]

yes

[stuck pawn]

So the question here

Why the y in our question is just y

How to determine it

Like you did

I was thinking its y = 3

[rigid ivy]

shell integration is just $2\pi\times$radius$\times$height. If you'r rotating about the x axis, then your cylinder has radius $y$ and height is cylinder height, which is actually "length" when revolving around x axis, and that would be $g(y)$

[thorny flame]

SWR

[stuck pawn]

Ok i gonna make a lot change to the question to make sure that i got it

The volume gonna be

2π integration (y-2)√(25-y²) dy from 3 to 5

[stuck pawn]

Right?

[topaz sinew]

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[indigo scroll]

hello

[indigo scroll]

i have two circles. One has center A and other has center B. I know the radius of the two circles and the coordonates of the center points A and B. I know that AC = radius of first circle and BF = radius of second circle. I also know the length of AB (from euclidean distance), and that triangles ACF and CFB are right angled. I want to find the coordonates of the points C and F

any idea on how to approach it?

if I could find coordonates of point O, I could solve it

[coarse tusk]

you don't need the coordinates of O

notice that there are 4 similar triangles in the picture

[indigo scroll]

yes, they are, ACO, ADO, OEB, OFB

[coarse tusk]

yeah

you know the length of AB

can you write AO and OB in terms of the angle phi, r1 and r2?

[indigo scroll]

not sure I know how

[coarse tusk]

$AO\cos\phi = AC = r_1$

[indigo scroll]

maybe, BF/CA is BO/OA ?

[thorny flame]

kheerii

[indigo scroll]

so AO = AC / cos phi?

[coarse tusk]

yeah, so $AO=r_1 \sec\phi$

[thorny flame]

kheerii

[coarse tusk]

similarly $BO=r_2\sec\phi$, so $AB=AO+OB=(r_1+r_2)\sec\phi$

[thorny flame]

kheerii

[coarse tusk]

you know the value of AB, r1 and r2, which means you can find out the value of sec phi

[indigo scroll]

but angle of phi is not given

maybe because OFB and OCA are similar triangles, then BO/AO = BF/AC ?maybe this way, since i know AC and BF, and I also know AB, I could find AO and OB?

[coarse tusk]

...

[indigo scroll]

oh

I didn't see that message

sorry 😄 😄

I think this will work

so I find out value of sec phi, then I can find AO and OB. How do I find coordonates of C and F in the end?

[coarse tusk]

from sec phi you can find the value of phi

then you can directly find out the coordinates of C

all you need is the inclination of the line AB with the x axis

[indigo scroll]

which i can find because i know coordonates of A and B

cool

[coarse tusk]

indeed

[indigo scroll]

thanks a lot

[coarse tusk]

just out of curiosity, what is this for?

[indigo scroll]

I try to solve a programming problem

one second

I show you

so, given two points A and B in a 2d space and a set of circles, each with coordoantes of their center and their radius

find out the shortest path from A to B, without touching circles

[coarse tusk]

ah, I see

that sounds interesting

how are you approaching this

[indigo scroll]

so my idea was to find first the distance between A to tangent of each circle. This way I can find a point T, tangent of the circle, and its coordinates, using Pytagora

for example, given A, i can find T1

then, by using the idea you helped me with now, I find T2 and T3 coordonates

so coordonates of points C and F in our discussion

[coarse tusk]

but how do you find T2

[indigo scroll]

and then the arc from T1 to T2

T2 is C

i know coordonates of A and B, i know their radius

i am not really into geometry (i find it very interesting, but my level is not really that good) is my idea wrong?

T2 is C and T3 is F

so after that I apply Dijkstra. I get from A to T1, then from T1 I can go to other points of the same circle. Assuming there is T2( which is C) and T2' (which is D). Then I find the arc. so I have AT1 + arc(T1, T2) + T2T3. Or maybe AT1 + arc(T1, T2') + T2'T3

I am not sure it actually works

does it make sense to you?

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[woeful drift]

Unfortunately simple formulas tend to not work super well with changing interest rates

You have to rely more on the time value of money

Is that a concept you've been introduced to?

Alright

Can you tell me the value of that 20k after 3 years?

,calc 200001.03^21.02

[thorny flame]

Result:

21642.36

[woeful drift]

Alright

Now let's say the amount you're withdrawing is some number X (or whatever). Can you compute the value of these payments at time 3 years?

You withdraw some amount, X, after one year. If you were to invest it into a different account earning the same interest rate(s), what would it be worth in 3 years from today (so after 2 years in the account)?

(In terms of X)

That's not what I'm asking

The equivalent of that, you're not actually doing that

The time value of money states that the value of money in the future is related to the interest rate you could invest it at today

Right, so today the value of that 20k is just 20k

In one year, the value of that 20k is 20600

Sorry, I have to go, I'll try to find someone to step in

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@dark pond Has your question been resolved?

[zealous otter]

Hi

Is anyone available to answer my question

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[neon iron]

The function above has no fixed point

So something from banachs fixpoint lemma is not fulfilled

So by def of the lemma I thought that f cant be a contraction on the interval

But im not sure if thats correct

Or is it because the only possible fixpoint can be 0 and this is not in the interval?

.close

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[stark crater]

hello

[stark crater]

can anyone help me with this vector field problem

I already confirmed it is conserved

then I tried to get the potential by integerating what I got so P=xy^2
q=xy^2+ye^(3z) w=3y1/3 e^(3z)

Am I correct ? if so how I can combine all what I've got to find the final ans of the potential?

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[quiet crag]

how does it converge? if limit is 0 doesnt mean its inconclusive?

[quiet crag]

(question on the right, answer on the left)

[reef fjord]

limit is 0 is inconclusive by divergence series test yes, but

its an alternating series

if you have sum a_n from n=k to infinity for k some integer and a_n is alternating, and lim |a_n| as n goes to infinity is 0, then sum a_n from n=k to infinity will converge

i.e. if its inconclusive by divergence test, an alternating series will converge

[quiet crag]

i see thanks

[reef fjord]

https://tutorial.math.lamar.edu/classes/calcii/AlternatingSeries.aspx

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[oak remnant]

<@&286206848099549185>

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[oak remnant]

What does this command do?

[fallow igloo]

Notification

[oak remnant]

Oh ok

<@&286206848099549185>

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[oak remnant]

<@&286206848099549185>

[oak remnant]

<@&286206848099549185>

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[slim maple]

Please only ping helpers once

[oak remnant]

I was told by the Bot that you can ping them every 15 minutes.

[slim maple]

Please open your own help channel

[oak remnant]

@slim maple

[slim maple]

Right, once.

[oak remnant]

Ok

My apologies I misunderstood the Bot perhaps.

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[grand solar]

if the projection was reversed ( proj u on to v) would cos(theta) = |u| / scalar projection?

[grand solar]

because cos = adj/hyp and the adj would be u and the scalar projection would be on v, the hypotenuse?

[vernal matrix]

The picture would be the same but with the roles of v and u reversed, no?

[grand solar]

yeah the picture would be the same but we would be projectiong u on to v

I believe the formula would still be the same with the variables switched from a video I watched but I dont understand why

[vernal matrix]

As where you have v, you'd put u instead, and vice versa-

[vernal matrix]

That said, I guess you could draw it as per what you've said

[grand solar]

Sorry I'm not sure im understanding my question might be confusing let me try and explain it better, so for example in this problem I followed the formula in green to get the scalar projection to get the correct answer, but im not seeing why that equation works because when I try to derive it I get a different equation which I written in purple

sorry forgot to post the problem

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[wicked bay]

Hey ! I wanted to know how to operate a change of variable in a sum ! For exemple, i wanted to let i = 2k - n in this sum

[wicked bay]

So I changed the lower value and upper value

And got i = -n at the bottome and n at the top

So, this

However, the two summations are not equal, there are 2n terms in the second, compared to n terms in the first

Am I missing something ? Is there an other step to do ?

[vernal matrix]

(2n + 1 and n + 1 respectively, btw)

[odd pagoda]

the problem is that you arent getting all the i-values

you are only getting every second one

for example to get i=-n+1, you would have needed k=1/2

[wicked bay]

Yeah I see, so how do I only select the "possibe" values ?

[odd pagoda]

you could write it as an extra condition. something like i+n even. but ehh

you probably just shouldnt do this variable change

there are other things you could do here

[wicked bay]

Hmm in fact

I simplified it a little bit

[left juniper]

The function $k \rightarrow 2k-n$ is not bijective in the domain ${0, 1, ... n} \rightarrow {0, 1, ... n}$, so it's invalid.

[wicked bay]

It was the sum as k=0 goes to n of e^(ix(2k-n))

[thorny flame]

Xwtek

[odd pagoda]

\to

[wicked bay]

I am trying to arrange it in a way so that a trig expression appears

[odd pagoda]

given that you already have i in your formula, you definitely also shouldnt change the variable sum to i btw

[left juniper]

What is the difference betwen \to and \rightarrow?

[odd pagoda]

one is shorter

[left juniper]

You can split a summation into two.

[wicked bay]

Well yeah, I had it change to t in ly paper but i here because i just deleted the complexe values

[odd pagoda]

ok. just something to keep in mind

[wicked bay]

What are you thinking about ?

Oh like

A sumation that goes from 0 to n/2 and another that goes to n/2 to n, assuming that n is even?

[left juniper]

No, I was thinking about a summation that goes from 0, 2, ... n and another that goes from 1, 3, ... n-1. But wait, what is the original? Because the question as it is, is actually a geometric sum

[odd pagoda]

I would have done $e^{-n} \sum_{k=0}^n e^{2k}$ and then you have a geometric sum. no trig stuff tho

[thorny flame]

Denascite

[wicked bay]

Yeah but that's pretty much the point

[odd pagoda]

!original

[topaz sinew]

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.

[wicked bay]

Okay so

I'm trying to cos^n(x) with a sum of αcos(βx)

I don't know if it's understandable written like that lmao

So I used euler identity, then Newton's binomial

And I'm trying to go back to a cos expression again

[odd pagoda]

oh ok

not sure what you mean with newtons binomial

but roughly you just mean cos(nx)+i sin(nx)=e^(inx)=(e^(ix))^n = (cos(x)+i sin(x))^n ?

[wicked bay]

It's the formula to expand (a + b)^n

[odd pagoda]

although well that solves cos(nx) in terms of cos^k(x). you want the other one

oh so binomial theorem. gotcha

[wicked bay]

Well no, I use the euler formula not Moivre's, so cos(x) = (e^ix + e^-ix)/2

[wicked bay]

Ohh my bad I'm french, we call it Newton's binomial here x)

[odd pagoda]

thats not usually whats meant with eulers formula

in english anyway

e^ix = cosx+ i sin x is eulers formula in english

ok I see what you want

that seems a bit painful

[wicked bay]

Ohh okay we just learned this as the exponential form theorem, and we used "euler's identity" as the cos(x) and sin(x) definitions

[odd pagoda]

meanwhile eulers identity is e^(i pi) + 1=0 in english

[wicked bay]

So I'm stuck with this

[wicked bay]

Okay thank's for the tip

Maybe I can like

Split the sum in two

One that goes from 0 to ceil(n/2)

And the other that goes from ceil(n/2) to n

[odd pagoda]

btw what happened to the binomial coefficients

[wicked bay]

They merged

I had a e^(ixk) and a e^(-ix(n-k)

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[formal jetty]

i am not understanding a and b

<@&286206848099549185>

i dont understand b

how do i know how much is going left and how much is going right

[opal kestrel]

you don't want x to be equal to 5, but as close as possible without equaling 5. Think of x=5 as an open dot on a graph

[formal jetty]

for letter b?

[opal kestrel]

yeah

[formal jetty]

what do i even plug in and where

[opal kestrel]

You're not exactly plugging in any numbers, just analyzing the table. Look at the f(x) values next to x=5. What do you notice about them?

[formal jetty]

they are approaching 5

or is it .25?

[opal kestrel]

thats for the f(x) values, yes.

[formal jetty]

do i hve to look at f(x) value?

or x values?

[opal kestrel]

you can look at both. just don't look at x=5. while x=5 is associated with the limit, it won't be equal to 5. try to see what f(x) is approaching from both sides of the table.

[formal jetty]

so for this

limit woudlnt exist right?

[opal kestrel]

check a value of x and see if the f(x) values on both sides approaches a single value. Is there one single value x approaches?

[formal jetty]

no

xhow do i check what number is being used to approach the x

is left side 3?

[opal kestrel]

you can really check any value of x, but try to check the value that seems most likely to approach a number

[formal jetty]

i got left side 3 and right side -1/3

[opal kestrel]

For example, as x approaches 4, its not clear as to which number f(x) is approaching.

[formal jetty]

yes

so at this we would use the top

and the top isnt reallt approach a number

[opal kestrel]

you're mainly analyzing f(x) to see what number it approaches, yes.

[formal jetty]

oh i see

i think im getting the gist of it but the tables are always confusing

what about for a. i am pretty sure i know how do it

i just plug in the approaching number right?

[opal kestrel]

you have to simplify the fraction first

[frank briar]

Hi

[formal jetty]

oh yes i know about the simplificaiton

it is pretty easy

[formal jetty]

hello

[opal kestrel]

factor out the bottom and cancel out terms

[formal jetty]

and if the thing is different from the limiting function it doesnt exist right?

[opal kestrel]

only if there isn't a set value x is approaching. For example, as x approaches 1, its undefined because the denominator is 0.

[formal jetty]

i see alright

thank you pizzaboy

for helpinig me, i was trying to understand it for a while

[opal kestrel]

yep

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[quaint badge]

really quick question, are the processes of implicit differentiation and partial differentiation interchangeable (using the formula dy/dx = - partial derivative wrt x/partial derivative wrt y), i don't know how to properly type it

[glacial adder]

not sure if that’s always true

we typically use partial derivatives for functions that take in multiple variables as an input

implicit differentiation is taking the fact that y is a function of x

not really giving us a function that depends on x and y

[quaint badge]

yeah it was mainly something i was thinking about because although it's generally more complex to do, i couldn't think of a counterexample

ie. you can define implicit differentiation problems as a function of x and y and then use partial differentiation

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[hot walrus]

I wrote a National Benchmark Test yesterday for a university, and came across this question. I've been stuck on it since, and to make it even harder (in case the text is not clear) you cannot solve for "x" or "y" is values (if you have to solve them, then I guess the question is flawed). I would really appreciate help with this apparent simultaneous equation!

[rigid ivy]

$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$

[thorny flame]

SWR

[hot walrus]

$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{5}$

[thorny flame]

ZeroNewt

[hot walrus]

I got stuck here, no further

[rigid ivy]

What is $(x+y)^2$?

[thorny flame]

SWR

[hot walrus]

$(x^2 + 2xy + y^2)$

[thorny flame]

ZeroNewt

[rigid ivy]

Yup. There you go

[hot walrus]

Oh my I am stupid haha

Thank you, I can't believe I didn't see it

[rigid ivy]

so what do you end up with?

[hot walrus]

5?

Give me one min

[rigid ivy]

Not quite. So close, but not quite

[hot walrus]

4?

[rigid ivy]

No. You were closer with 5

How did you get 5?

[hot walrus]

I am struggling here, this is what 48 hours of maths does to the brain

[hot walrus]

Well, the sum of this is 25?

[rigid ivy]

Okay, so (x+y)²=25, so what can x+y be?

[hot walrus]

5

whoops

[hot walrus]

whould that mean it is 1?

subbing it into here?

[rigid ivy]

close

[rigid ivy]

What else can x+y be?

[hot walrus]

-5?

This subject is the bane of my existence🥲

Thank you so much 🙏 Have a blessed day/night

.close

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[shut obsidian]

Determine a system of equations that yields the solution $(2s, s + 1, s -2)$ with free variable $s$.

[shut obsidian]

I'd do this by converting 2s to t, then I can deal with t like it's x_1

Is there another way too?

[odd pagoda]

not sure what you mean by that

first I would remove the +1 and -2

so you just have s*(2,1,1)

so you want a homogenous system which has that as solution

now what can you say about the first row of that system

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[shut obsidian]

.reopen

[topaz sinew]

✅

[shut obsidian]

How can you just remove them

You mean write it as (2s, s, s) + (0, 1, -2)?

[odd pagoda]

yes

and then ignore the second part

[shut obsidian]

ok

[odd pagoda]

you get the first part from finding a system Ax=0

and then by choosing b in Ax=b you can get the second part

[shut obsidian]

We want A * ((2s, s, s) + (0, 1, -2)) = 0, so A * (2s, s, s) = -A * (0, 1, -2)

Yeah

If that's what you mean

[odd pagoda]

no

[shut obsidian]

I mean (2s, s + 1, s - 2) is the same as (t, t/2 + 1, t/2 - 2). Now I can directly give you a system to that:

$x_2 = \frac{x_1}{2} + 1$ \ $x_3 = \frac{x_1}{2} - 2$

[thorny flame]

Kepe

[shut obsidian]

We will choose x_1 as the free variable, t

And get exactly that

[odd pagoda]

oh that. yes that works

[shut obsidian]

But for this method, we need one entry with a factor of 1 in the tuple

So I was asking if there is a more direct approach maybe

[pseudo jetty]

i'm not sure why replacing with t was necessary

you might as well say x_1 = 2s, x_2 = x_1/2 + 1, x_3 = x_1/2 - 2 directly

[shut obsidian]

It feels weird to choose 2s as a free variable

[shut obsidian]

Oh, true..

[pseudo jetty]

well it doesn't really matter what is and isn't a free variable

as long as you get the same constraints

[shut obsidian]

Thanks a lot, then I'll just go with this method

.close

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[shut obsidian]

Wait

[pseudo jetty]

id say you should consider what denascite was saying though

[shut obsidian]

.reopen

[topaz sinew]

✅

[pseudo jetty]

because it's important to understand conceptually

[shut obsidian]

How is this a valid system of equations?

Oh, only the last two

[pseudo jetty]

well x_1 is free there essentially

[shut obsidian]

Yeah, you omit the x_1 = 2s

You only give x_2 = x_1/2, x_3 = x_1/2 - 2 as your solution

Thanks

[pseudo jetty]

it should be A(s(2,1,1) + (0,1,-2)) = b

and if (2,1,1) is in the kernel of A, you get A(0,1,-2) = b

in fact it is necessary for (2,1,1) to be in the kernel of A

[odd pagoda]

different way to see it, if you choose s=0 then you need A(0,1,-2)=b

[odd pagoda]

which you can then subtract from both sides of this