help-10

[rough bough]

but then the solutions have Ims

[strong vale]

If you let $\omega = \frac{-1+\sqrt{3}i}{2}$ then $im(\omega) = \sin(\frac{2\pi}{3})$. $\omega^2 = \frac{-1-\sqrt{3}i}{2}$ then im(\omega^2) = \sin(\frac{4\pi}{3})$. $\sin(\frac{2 \pi}{3}) > 0$ and $\sin(\frac{4 \pi}{3}) = -\sin(\frac{2\pi}{3}) < 0$. $2 \im( \omega) -3 \im(\omega) < 0$ hence this will be the $-\frac{\sqrt{3}i}{2}$ part of the root

[warm shale]

azeem321

If you let $\omega = \frac{-1+\sqrt{3}i}{2}$ then $im(\omega) = \sin(\frac{2\pi}{3})$. $\omega^2 = \frac{-1-\sqrt{3}i}{2}$ then  im(\omega^2) = \sin(\frac{4\pi}{3})$. $\sin(\frac{2 \pi}{3}) > 0$ and $\sin(\frac{4 \pi}{3}) = -\sin(\frac{2\pi}{3}) < 0$. $2 \im( \omega) -3 \im(\omega) < 0$ hence this will be the $-\frac{\sqrt{3}i}{2}$ part of the root
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.55 ... = \frac{-1-\sqrt{3}i}{2}$ then  im(\omega
                                                  ^2) = \sin(\frac{4\pi}{3})...
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.```

[rough bough]

@strong vale thatr doesn't seem to make sense consider sin(4pi/3) = -sin(2pi/3) < 0

[strong vale]

Idk how else to explain it other than

[rough bough]

@strong vale is w and w^2 conjugate pairs?

[strong vale]

yes

[rough bough]

okay thanks i think iget it

and oh yeah, what do you do to figure conjugate pairs

do you do like divide everything by w^3

in which it can w and w^_1

w^-1

[strong vale]

$\omega^3 = 1$. If $\arg(z) = \theta$ then the $\arg(\overline{z}) = -\theta$

[warm shale]

azeem321

[rough bough]

yeah and we can find that out through dividing either ends by w^2

so it would be w = w^-2

[strong vale]

yup

[rough bough]

thanks \

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[timid silo]

Hi, I need to find the tangent line (parametric form) to a curve C at point P.

The curve is C: y=x², P = (3, 9)

I know how to calculate the tangent line equation (6x - 9 - y = 0), but cannot figure out the parametric form of it.

Answer is supposed to be x = 3 + t, y = 9 + 6t

[strong vale]

notice that y = 9+6x

they replace x with t

so if y=9+6t can u figure out x

as a function of t

no im just guessing it was a typo

on his part

@timid silo

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Ohhh that's clever, thank you!

But I suppose this solution would not work with the first two questions, it only works for R²?

Right, thanks for the help!

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[frigid oracle]

relatively simple question, but i couldn't find it on the web, nor do i remember some square-root properties to explain it

[frigid oracle]

but how does $\sqrt{4 + 2\sqrt{3}} = 1 + \sqrt{3}$

[sage geode]

$\sqrt{4 + 2\sqrt{3}} = \sqrt{1 + 2\sqrt{3} + 3} = \sqrt{(1 + \sqrt{3})^2} = 1 + \sqrt{3}$

[warm shale]

dulg

Touch Our Beans

[frigid oracle]

oh right!!

thanks @sage geode

.close

[obtuse pebble]

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[sage geode]

You're welcome

[frigid oracle]

go on

im all ears

[kindred silo]

alternatively
$\sqrt{4 + 2\sqrt{3}}=a+b\sqrt{3}\implies 4 + 2\sqrt{3}=a^2+3b^2+2ab\sqrt{3}$ and solve for a and b lol

[warm shale]

uselessleaf

[sage geode]

Yeah that works out as well

[obtuse pebble]

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[lusty vigil]

how do I write a function for BMI and weight for girls who are 161 cm?

[marsh juniper]

Can't you look up what they use for bmi calculations. I'm pretty sure they tell you the function for it

[obtuse pebble]

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[lusty vigil]

.reopen

[obtuse pebble]

✅

[lusty vigil]

I tried but they didnt give anything @marsh juniper

[dapper geode]

@lusty vigil in python?

[lusty vigil]

oooooh

[dapper geode]

shouldn't it just be

def(weight):
  bmi = weight/(161)**2
  return bmi```

[obtuse pebble]

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.reopen

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✅

[lusty vigil]

like how?

is it to the power of 2?

[rustic ginkgo]

convert the height to m

not cm

[obtuse pebble]

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[next violet]

i need help on my nt homework: for positive integers $n$ define $d(n) = n − m^2$, where $m$ is the greatest integer with $m^2 \leq n$. Given a positive integer $b_0$, define a sequence $b_i$ by taking $b_k+1 = b_k + d(b_k)$. For what $b_0$ do we have $b_i$ constant for sufficiently large $i$?

[warm shale]

SofaCouch

[next violet]

ok i think from testing some stuff out onl perfect squares work

so we need ot prove that non perfect squares dont wor

so

let our b_k=a^2+k

where a^2+k<(a+1)^2

b_(k+n)=a^2+(2^n)k

so we need to disprove

(a+1)^2-a^2 != k(2^n)

for any power of 2

which must be true?

because squares increase in incrememnts of odd powers

is this right?

<@&286206848099549185>

[simple dove]

hello

i need help with rational equtions thingy

i have this

but

it syas this

[next violet]

#❓how-to-get-help

this chnnel occupied

[next violet]

this only proves the a to a+1 case

there are different cases right

a^2+k might jump to (a+2)^2

no wait

thats not possible why am i thinking that

[obtuse pebble]

@next violet Has your question been resolved?

[obtuse pebble]

@next violet Has your question been resolved?

[autumn adder]

Hi @next violet

[next violet]

hey

[autumn adder]

How're you

Whats your question

[timid silo]

@autumn adder

[obtuse pebble]

@next violet Has your question been resolved?

[next violet]

is my reasoning above correct

?

[obtuse pebble]

@next violet Has your question been resolved?

[obtuse pebble]

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[timid silo]

Anyone

That's c btw

[slim leaf]

[a^(b)]^c = a^(bc)

[timid silo]

What's the solution?

And this

It's c

Lmao

Yep

[slim leaf]

cant just give the solution out

[timid silo]

Uhhh no idea

[timid silo]

I need the solution for this

[warm shale]

Andreww

[obtuse pebble]

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[cursive phoenix]

I solved this question this way, but im not sure if my answer is correct, is my way of solving correct

[clear inlet]

the text to me is very unintelligible but I think with your first line I believe you have made a mistake

when you equate the second derivative to the y co-ordinate

[cursive phoenix]

hmmm

i know i made mistake

but i wana how to actually solve it

[clear inlet]

ok so gimme some time

[cursive phoenix]

okay thanks

[clear inlet]

I am in a talk atm

[cursive phoenix]

okay

[clear inlet]

ok so @cursive phoenix

[cursive phoenix]

okay

im here

[clear inlet]

do you know what point of inflection means?

[cursive phoenix]

yes

in the 2nd derivative when the sign changes

from the negative to postive

[clear inlet]

yeah cool

[cursive phoenix]

or the opposite

[clear inlet]

so we move from here

[cursive phoenix]

okay

[clear inlet]

when you calculate the second derivative for this problem at x=-1

what should it be equal to?

[cursive phoenix]

like replacing x with -1

that would just be -6a+2b

[clear inlet]

yeah but what would this value be?

[cursive phoenix]

that value aint given tho?

slope?

at -1

[clear inlet]

given at x=-1 there lies a point of inflection

continuity is implied yes?

[cursive phoenix]

yes

[clear inlet]

and when you get sign change in continuity

what do you get where the sign changes

what is the value of function at that point

[cursive phoenix]

0?

[clear inlet]

yea

[cursive phoenix]

oh i get it

so when u try to find -1

which is a crital value

u solve that 6a(-1)+2b=0

but we still have 2 unknowns

to move forward

[clear inlet]

it isn't a critical point tho

[cursive phoenix]

[clear inlet]

critical points have first derivatives=0 or undefined

[cursive phoenix]

oh

but for 2nd

diff

[clear inlet]

its a point of inflection

[cursive phoenix]

yes

[clear inlet]

anyway

next bit of information is tangent

[cursive phoenix]

tangent is the 1st derivative?

[clear inlet]

you can find first derivative and match the slope

and you can find the point and match with the point on line

so you have 3 equations

[cursive phoenix]

yws

-6a+2b

first equation

2nd is the tangent which is -3x-1

we find first derivative of the function

which is

3ax²+2bx

so we make 3ax²+2bx=-3x-1 ?

[clear inlet]

you do this while knowing that one of the solutions is x=-1

and not first derivative

[cursive phoenix]

yes for the 2nd

[clear inlet]

you match first derivative with slope

which is -3

so 3ax^2+2bx=-3

[cursive phoenix]

oh

[clear inlet]

since tangents will have same slope as the curve

[cursive phoenix]

but like what unknown are we solving for

a or b

or even x at 1st derivative

[clear inlet]

we know x

since you know its tangent at x=-1

[cursive phoenix]

so thats gone

a or b now

[clear inlet]

we are solving for a,b,c as asked in the problem

[cursive phoenix]

c wont appear in the 1st or 2nd derivative

[clear inlet]

you have 3 equations

[cursive phoenix]

so we need to find a or b fisrt

then find c last

[clear inlet]

each corresponding to values of second derivative, first derivative and the function itself at x=-1

yes so the function value is where c appears

[cursive phoenix]

so what i have done now

i equalized the first

derivative to -3

and found a

a = 2b-1

[clear inlet]

combine that with your last equation to evaluate both a and b

[cursive phoenix]

the -6a + b = 0

b=6/11

replace b in a

so a is 1/11

and now we replace a b and x in the primary function to find c?

so c is -5

[clear inlet]

ok so your values are incorrect

[cursive phoenix]

uh

[clear inlet]

and this is incorrect

you should get

a=2b/3-1

[cursive phoenix]

Oh

[cursive phoenix]

i got - b/3

6a+2b=0

[clear inlet]

I have to leave to attend a lecture now but I think you have enough information to do it on your own now whats left is algebra, hope you get it right

[cursive phoenix]

okay

thanks for the help

.close

[obtuse pebble]

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[hot gust]

what does this mean this general form a polynomial

like the c_i is not specified really in the explanation

[royal basin]

the c_i are coefficients just like a_i

do you happen to have taken any linear algebra? if you have it might simplify matters somewhat

[hot gust]

yeah i do linear algebra @royal basin

[royal basin]

the canonical form of a polynomial is essentially its expression as a linear combination of the basis {1, t, t^2, ..., t^n}

[hot gust]

right yeah im just confused as b_i(t) is a polynomail being scaled by c_i what does this mean is it just related to the basis

[royal basin]

you're overthinking it

(15.5) is basically just f being written in terms of a new basis {b_0(t), b_1(t), ..., b_n(t)}

[hot gust]

it says the t_i are the b_i(t) of between 15.5 an 15.4

what so like b_0(t) multiplied by c_0 is just a function of degree 0 scaled by its coefficient

and so on

[royal basin]

nothing is said about what the b_i are

and if something is said about it then you haven't shown it here

and again i think you might be overthinking it

c_0 b_0(t) means c_0 b_0(t)

[royal basin]

okay, then you aren't told that b_0 is a polynomial of degree 0

b_0 can be just about anything

[hot gust]

but its b_i ?

i was taking example of n = 0

[royal basin]

what so like b_0(t) multiplied by c_0 is just a function of degree 0 scaled by its coefficient

you said this

you asserted deg(b_0) = 0

and i am merely saying the text makes no such statement

[hot gust]

yeah it doesnt so the 15.5 is a some general polynomial scaled by a single scalar?

[royal basin]

why are you still trying to overthink this

[hot gust]

not sure just a bit odd the notation thats all

i think ill need to see examples tbf to make sense of it

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[obtuse pebble]

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[austere juniper]

how do i factorise this?

$8h^2 - 18$

[warm shale]

leewchaha

[austere juniper]

is $2(4h^2 - 9)$ correct?

[warm shale]

leewchaha

[zinc flax]

2(2h+3)(2h-3)

[zinc flax]

it can be simplified more

[austere juniper]

oh, how'd you manage to do it

i have zero idea

[zinc flax]

a^2-b^2 = (a+b)(a-b)

a=(2h)

b=3

[austere juniper]

how do you know to replace a with 2h and b wth 3 in the first place?

are there some sort of formula to follow?

[zinc flax]

ya a^2-b^2=(a+b)(a-b) this is a formula

[twilit loom]

Notice how both of your terms inside the parentheses are perfect squares?

[austere juniper]

18 is also a perfect square？

[twilit loom]

No no, but 4h^2 and 9 are

[zinc flax]

no u took out 2 right ?

[austere juniper]

ohh, so i start from 4h^2 and 9

[zinc flax]

ya

$(4h^2-9)=(2h+3)(2h-3)$

[warm shale]

Playmaker

[twilit loom]

Right, whenever you notice that you have two perfect squares and you're subtracting one from the other, this is called the 'difference of two squares' and you can use this handy trick

It's just something you learn to recognise

[austere juniper]

thanks alot dino dik and Playmaker, now i get it!

[twilit loom]

np ^^

[austere juniper]

.close

[obtuse pebble]

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[zinc flax]

if u hv doubt on this u can dm too

[austere juniper]

aight gotcha

[obtuse pebble]

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[feral swallow]

Areaaa how to find it? The answer should be 35.6 but i got 32.56

[autumn adder]

Hi

@feral swallow

Start dividing the shape

Into circles and rectangles

[feral swallow]

I did

[autumn adder]

Show me

How you made it

[feral swallow]

Radius is 2 since 11-7 is 4 and there is 2 circles 4/2 is 2 right?

[autumn adder]

Yes

Nice

[feral swallow]

The same goes for the below part

[autumn adder]

Nice

So radius of all the circles

quarter circles is 2

[feral swallow]

Is 2

Yesss

[autumn adder]

Nice

So cant u find

[feral swallow]

90/360

Is 0.25 so 0.25 x pie x 2²

then the answer multiply by 4

Since there is 4 parts of a circle

And we find area for the rectangles

And both answers add them together

[autumn adder]

Yeah

And do it

[feral swallow]

I did and got

32.56

[high lily]

you didn't round properly but where's 35.6 coming from

[feral swallow]

The answer supposed to be 35.6

Book answer scheme thing

[autumn adder]

Wait

Are you looking for the right one?

Wait

You did wrong I guess

[feral swallow]

Yesss

[autumn adder]

Circles thing.

Area of a quarter circle...

[feral swallow]

Area of quarter circle is 90/360 which is 0.25

And so since we need area of the quarter circle thenn 0.25x2² (2 is radius right?) X pie

[autumn adder]

Hmm

Then

The book answer

is wrong

[feral swallow]

Idrk is it me or thr book😭

Okayojay

Loook at this as well

Sorry bro phone died

[feral swallow]

Area supposed to be 19

I got like 17 or something

[autumn adder]

Wait then

That means

Somethings up

[feral swallow]

For sure

[autumn adder]

19?

[feral swallow]

Yesss

[autumn adder]

Which area

Whole figure/

[feral swallow]

Thiss

Yees

[autumn adder]

How so

Its 3.2cm

What is the area of the circles

[feral swallow]

Dw bro i aint blind

Area of circle iss

[autumn adder]

Ohh didnt say that way

[feral swallow]

Noo i meant

[autumn adder]

Oh you mean only the middle part

[feral swallow]

Dont think that im looking at another questions

Like for another answers

Area of circle iss

3.2/2 for radius which is 1.6

1.6² x pie x 0.5 (180/360)

then x 2 since there is 2 semicircles

And for the middle part

[autumn adder]

Hmm

[feral swallow]

100/360) x pie x 3.2²

Then we add them

The answer is 16 17 i dont remember but supposed to be 19

[autumn adder]

so its 17

Not 19

[feral swallow]

Yesss

And like i opened the same question yesterday here

I dont rememebr which help number

[autumn adder]

Do you know the name of this book

So I can check

[feral swallow]

Cambridge IGCSE mathematics

[autumn adder]

Class?

[feral swallow]

Second edition

Core and exteneded

[autumn adder]

Yeah yeah

Wait

dw

Ill help you out

[feral swallow]

Take ur time dude

Thanks man

[autumn adder]

Thanks for that

Chapter name

[feral swallow]

No need to thank me dude

[autumn adder]

Sorry I need it to search

[feral swallow]

Perimeter area and vlume

Chapter 7 page 148

Take ur time

Give me few mins if u dont minf

[autumn adder]

No issues

[feral swallow]

Tyty

U need book link?

[autumn adder]

I got

The book

Just seeing

Whats wrong with it haha

[feral swallow]

No problem dude

[autumn adder]

The perimeter doesnt match either

Is it just a mistake in the answers

Or are we missing something

[feral swallow]

Im not sure myself

[feral swallow]

Might be

Or maybe the guy who made the book was drunk idk

[autumn adder]

Haha

[feral swallow]

Thanks dude really appreciate it

Take care man have a wonderful day

[autumn adder]

Youre welcome

Thanks for bearing with me

Can you

Check the perimeter

For part c

[feral swallow]

Alright

[feral swallow]

No need to thank me dude thanks to you

I gott

15.63

[autumn adder]

This one

?

[feral swallow]

Ysss

What i did was

180/360

0.5 so 0.5 x 3.2 x pie

Waitwait

And then i did

[feral swallow]

x 2 since there is 2 semi circles

And we got the area for both semi circles right?

[autumn adder]

$(2\times \frac{180}{360} \times 3.2 \times \pi )+(\frac{100}{360}\times 2 \times 3.2 \times \pi)$

[warm shale]

Muhammad Hussaini

[feral swallow]

Exactly

[autumn adder]

$3.2 \times \pi + \frac{100}{360}\times 6.4\pi$

[warm shale]

Muhammad Hussaini

[feral swallow]

And so we got 15 . 63

[autumn adder]

,w solve 3.2*pi + (100/360)6.4pi

Nooo wait

[feral swallow]

Sorry sorry

[autumn adder]

See

It is

[feral swallow]

Yess?

[autumn adder]

Yeah

So

The book answers are either wrong

[feral swallow]

Probably

Man u used a bot literally ofc the book will be wrong🤣

[autumn adder]

Haha

I used Wolfram Alpha

[feral swallow]

Nono but thanks for ur help as well

Really appreciate it dude

U need anything?

[autumn adder]

I dont really

But whenever you want

You can ping me

[feral swallow]

Alright man thank you so much bro really appreciate it

I wish u all the best bro thanks man have a wonderful day and take care!

.close

[obtuse pebble]

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[signal meadow]

I'm confused, is 0 not the breakpoint? Idk how to do this task, the 0 is throwing me off

[obtuse pebble]

@signal meadow Has your question been resolved?

[signal meadow]

<@&286206848099549185> D:

[obtuse pebble]

@signal meadow Has your question been resolved?

[plain surge]

the limit doesn't exist at x = 0
right hand limit is infinity and left hand limit is -4 which aren't equal

[signal meadow]

if the limit is -4 then it means that x=-4 then? did i understand that correctly

?

[plain surge]

A function has a limit at that value only if its right hand limit and left hand limit are same.. here I told that right hand limit at x= 0 is infinity and left hand limit at x = 0 is -4.. which means limit doesn't exist at x = 0

[signal meadow]

english hard

[signal meadow]

if the interval notation for the function is (-inf,0)U(0,inf) how can i find the breakpoints?

x?

[plain surge]

by breakpoints you mean the values of x where limits doesn't exist right

[signal meadow]

im not sure

i wasnt taught about specifications like that

i just need to find breakpoints of y=7^{\left(\frac{1}{4x}\right)}-4 ig

for y = ln(x + 3) the breakpoint would be 3 so x0=3, idk how to find it for my function

[obtuse pebble]

@signal meadow Has your question been resolved?

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[thorny wigeon]

How I find the area of this triangle using their vertices?

[shut sundial]

to gauge your knowledge: do you know about the cosine rule? dot product? cross product?

[thorny wigeon]

yes

vectors, I get it

[unique solstice]

Is it a right triangle?

Well I know that is

But does it have to be

Or are you just interested in this specific one

[thorny wigeon]

no, I'm trying to understand the subject of barycentric coordinates, but before I was trying understand how to take the area of the bigger triangle

[unique solstice]

Barycentric coordinates are the ones where the point is defined by the three areas around the point within a triangle right?

Or am I just understanding wrong

[shut sundial]

here's a simple formula that computes the area of a triangle $ABC$ in any dimension: if $a := C-B$ and $b := C-A$, then the area of ABC is $\frac 12 \sqrt{(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)}$

[warm shale]

MagmaMcFry

[thorny wigeon]

yes

[shut sundial]

in three dimensions the area is more simply computed as $\frac 12 |a \times b|$

[warm shale]

MagmaMcFry

[shut sundial]

in two dimensions it'd be $\frac 12 |a_xb_y - a_yb_x|$

[warm shale]

MagmaMcFry

[shut sundial]

will these formulas help you, or are you rather looking for an intuitive understanding of the linear algebra involved?

[thorny wigeon]

Yes, helped me a lot . But now I'm trying to figure out how get value of P inside of the triangle in a randomly way, like this: https://www.youtube.com/watch?v=I9no7u9srKI

[unique solstice]

It sounds like hes talking in a big metal box

[shut sundial]

reverb math 👀

[shut sundial]

so you want to choose a random point inside the triangle?

[thorny wigeon]

yes

[shut sundial]

uniformly random I assume

here's how to do it:

• pick two uniform random numbers u and v between 0 and 1
• if u + v > 1, replace u by 1-u and v by 1-v
• compute P := u·A + v·B + (1-u-v)·C

here's a tricky alternative:

• pick two uniform random numbers u and v between 0 and 1
• compute P := sqrt(u)·(v·A + (1-v)·B) + (1-sqrt(u))·C

[thorny wigeon]

that is u, v and (1 - u -v) will be the area of the three triangle inside the main triangle, right?

thank you a lot for your help

.close

[obtuse pebble]

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[cosmic wyvern]

Say I have some variable epsilon, where epsilon is initializes at 1. It decays at a rate of 0.9/1000 per thousandth timestep of t.

I want it such that it decays until t = 1 mil, where it stays flat at 0.1.

Is there any way to make a function f(x), which goes to 0.1 from x in [0,1 Mil] and then stops at 0.1?

I tried making a logarithmic function:

log_10( x / 100000)* 1/10, but after 1 mil it continues past 0.1

[unique solstice]

So it just stops at exactly 1 million and sits flat? @cosmic wyvern

[cosmic wyvern]

yes, it should do that

[unique solstice]

I mean the easiest way would be a piecewise function

[cosmic wyvern]

Cannot do that.

And by flat I mean that it goes towards 0.1 as it approaches infinity

[unique solstice]

Is there any particular reason

[cosmic wyvern]

Yes, I am trying to run a program where it will run repeatedly.

Can't I make an ODE which will go towards 0.1 at 1 MIL and then just stabilizie at that point?

[unique solstice]

Youd prob not even need differential equations

If you just write some derivative formula that works somehow

I mean differential equations would probably make it easier but

Im not too familiar with them

[cosmic wyvern]

Could you help with making the function still? 😦

[unique solstice]

I could try

[cosmic wyvern]

alright

[unique solstice]

Can you explain the rate of decay again

[cosmic wyvern]

I have epsilon_start = 1.

Epsilon_decay = 0.9/1 mil

Epsi_func(t) = epsilon_start - t * epsilon_decay

When t = 1 Mil

Epsilon should stabilize at (approx.) 0.1

But it has to be linear

Or approx. Linear

[unique solstice]

I mean its pretty hard to write a non differentiable function with continuous functions

There might be a way though

You could use a step function but it wouldnt be continuous at a single point

Are you concerned that a piecewise function would cause performance issues?

[cosmic wyvern]

yes

[unique solstice]

Hmm

I mean most piecewise functions would just be as simple as a single if statement

[cosmic wyvern]

yes

but that is a lot of time if you do it 1 million times

[unique solstice]

It prob wouldnt be as fast as multiplication but it would def be faster than all kinds of complicated square root square craziness

Or other complicated functions like log

Like log is 35 cycles google says

Id like to see a 35 cycle simple piecewise function

[cosmic wyvern]

what?

Once again, I am looking for a function. I would like to use a function. If you cannot help with that, that is okay. But I am going to use a function.

[novel knoll]

And you were told (correctly) using a single if statement is gonna be the most efficient way

[cosmic wyvern]

Alright

but I am still curious as to how to make a function, that can do that

[novel knoll]

If epsilon>0.1
…
Else
epsilon=0.1

[cosmic wyvern]

... A non piecewise function

[obtuse pebble]

@cosmic wyvern Has your question been resolved?

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[carmine bear]

Hi i need help, how would I approach part b of this question? I've done part a and replaced the LHS of part b but every "path" i choose to end up with only 1 trig function in the expression i end up going in circles.

[carmine bear]

i haven't tried dividing by tan x because there's no evidence that tan x ≠ 0

[forest sinew]

there is

look at the excluded x values @carmine bear

[carmine bear]

yeah but tan 90 and 270 asymptotes

[forest sinew]

oh

yea

nvm

well

what if you split it

assume tanx != 0

then solve explicitly the case where it is 0

you should be using the equality above, right?

[carmine bear]

I was taught you can't otherwise you'd get the wrong answer

[forest sinew]

lemme try

i mean theres only two possibilities

yea so 0 and pi are solutions

lemme see if theres others

[hot sonnet]

Hello, I think you can factor out tan(x) and get a quadratic expression in terms of sin(x) as a second factor

[forest sinew]

oo

yea where tanx=0 is a solution

then assume its not

divide it out, identity

you get a quadratic

[carmine bear]

ok ill try factoring out tan(x) first

ok so i tried dividing by tan(x) and safe to say the answer I got was wrong because i subbed it back in and got 2 different answers for the LHS and RHS, but i drew both graphs and I know that 0 and 180 are solutions as you said, but how would i prove that mathematically (the question is 5 marks and says i can't use graphical solutions) ? Like as in how would i rearrange to end up with tan(x) = 0? I definitely can't submit my answer as "assuming tan(x) = 0 ....."

[untold badge]

use part (a) of the problem

[carmine bear]

yeah done that

[untold badge]

and use power reduction formula for sin^2(x)

tan(x) =0 would be a solution indeed (factoring shows that)

[carmine bear]

again I'm not sure I can submit an answer using the power reduction formula cuz we haven't been taught that yet, and id assume the mark scheme is based on another method. The last time i tried using something we weren't taught it was just marked wrong by my teacher even though i got the right answer

The double angle formulae and the trig identities are all i can really use

[untold badge]

Power reduction formula is exactly double angle +trig identities

[carmine bear]

oh

[untold badge]

$[\sin(x)]^{2} =\frac{1-\cos(2x)}{2}$

[warm shale]

Anna-Isabella

[untold badge]

and you have cos(2x), and also sin(x), I hope you're following

by the way, show me what happenes when you factor out tan(x)

[carmine bear]

yeah it's just i don't know how I would get there without just using this formula (as in how can i break down that formula into the double angle rules and trig identities)

[carmine bear]

so i divided by tan(x) and then got cos 2x = 3 sin x

and split that into 1-2sin^2 (x) = 3 sin(x)

and then i just used my calculator for the quadratic

[untold badge]

1. write cos(2x)=...
2. add sin^2(x) to the both sides
3. use sin^2(x)+cos^2(x)=1
4. express sin^2(x)

[untold badge]

1. sin(x)=u
2. find u manually from the quadratic equation
3. x=?

[carmine bear]

wait i realised where i went wrong now

got sin (x) = (-3+sqrt(17))/4 but then to check i put that value in the equation instead of its inverse

so i got x = 16.306.... and when i put that back in it works

[untold badge]

you also had tan(x) =0 remember?

[carmine bear]

I didn't get that mathematically though i got it from the graph

[untold badge]

Can you solve quadratic equations?

like, let's say, u^2+3/2u-1/2=0

With no calculator

[carmine bear]

put everything in the quadratic formula i suppose

[untold badge]

I don't see why not

this one has two real roots

find it

[carmine bear]

ok

yeah it's the same as what i got

i ignored the negative solution though because it's < -1

but the positive solution is what i put as = sin(x)

[untold badge]

sure

[carmine bear]

I know what the solutions are now it's just i don't know how i would get them without just assuming tan(x) = 0 and solving and assuming tan(x) != 0 and solving

[untold badge]

If you would show me how you factored out tan(x) that would be lovely

because we didn't fantasie anything

it's all algebra

[carmine bear]

i don't think i did factor out tan(x) i just divided everything by tan(x) after getting tan(x)cos(2x) = 3tan(x)sin(x)

[untold badge]

bring everything to the one side now

and factor out

to the left

[carmine bear]

and then got to the 2sin^2(x)+3sin(x)-1=0

[untold badge]

tan(x) cos(2x)-3tan(x)sin(x)=0

if A times B =0 it splits into two equations, A=0 and B=0

[carmine bear]

ah ok i never did that

[untold badge]

because if the product is zero, at least one of the multipliers has to be zero

[carmine bear]

yeah

I don't know how i didn't notice that

[untold badge]

tan(x) (cos(2x) -3sin(x))=0
so either tan(x) or that thing you're getting by dividing with tan(x) has to be zero, or both

[carmine bear]

my brain has probably shrunk by a lot because of 2 weeks of not doing any maths and just playing games

[untold badge]

It's a matter of practice, same as games, more you play...

[carmine bear]

ok thanks, I didn't look at it that way so i only got half the solutions

thanks a lot ill be sure to remember that

[untold badge]

when you have something like f(x) h(x) =g(x) h(x) and you're dividing by h(x) both sides you're loosing solutions that way

because of that factoring trick

[carmine bear]

yeah that's what my teacher taught me but i kept forgetting

[untold badge]

but now you know why it's that

[carmine bear]

Should've done more practice on my part, after learning something i just tend to never come back to it until i need it again

[untold badge]

did you solve part (a) tho?

[carmine bear]

yeah

didn't take long as well

[untold badge]

alright

[carmine bear]

went back to my working turns out i did it this way when i was doing part a lol

i don't how i managed to forget something over a span of a couple of hours (did part a this morning)

anyway thanks a lot

.close

[obtuse pebble]

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[stoic merlin]

Hello I’m having a bit of a difficulty in doing some basic mental math... problems such as: if I read 30 pages per hour how much am I reading in 1 minuite? How do I calculate such

[stoic merlin]

What mathematical equation would be written to solve

And how do I know if it’s 30/60 or 60/30

What determines that?

[spiral maple]

$30\frac{\text{pgs}}{h}\cdot\frac{1h}{60\text{min}}$

[warm shale]

Mosh

[stoic merlin]

Why is the 30 there

What does the 30 stand for

[spiral maple]

cause.... you wrote 30??

[stoic merlin]

Yes but then what is pages?

[spiral maple]

the units..

[stoic merlin]

Then what’s 30...

[spiral maple]

the number of pages..........

[stoic merlin]

Yes

So why is it there twice

[spiral maple]

????

It isnt

there's only one 30

[stoic merlin]

Yes

But

When you put 30 for Pgs

Then there’s 2, 30’s

[spiral maple]

there isnt

units and numbers are completely different things

[stoic merlin]

So what’s the difference of the units and pages in this case

[spiral maple]

nothing

pages are units

[stoic merlin]

Oh

Then shouldn’t it just be 30/h

[spiral maple]

No.

cause 30 what?

[stoic merlin]

Pages

[spiral maple]

yes

[stoic merlin]

Yes

[spiral maple]

30 pages per hour

[stoic merlin]

Yes

30/1

[spiral maple]

yes, 30=30/1

[stoic merlin]

Oh so then 30/60

Then we get 0.5

[spiral maple]

yes.

[stoic merlin]

But what’s 0.5

[spiral maple]

0.5 pages per minute

[stoic merlin]

Is it hours or pages

Oh

[spiral maple]

the hour units cancel

[stoic merlin]

Ohhh

Okay what if I were to find out how much words I can read per page

Let’s say I can read at the speed of 30pg an hour

And let’s say each pg is 500 words

Hypothetically

[spiral maple]

you mean words per hour?

[stoic merlin]

Yes

30 x 500

[spiral maple]

so you'll have $\frac{30\text{pgs}}{h}\cdot\frac{500\text{words}}{1\text{pgs}}$

[warm shale]

Mosh

[stoic merlin]

Ah

[spiral maple]

so pages cancel leaving words/h

[stoic merlin]

Gotcha

Thanks bro

.close

[obtuse pebble]

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[timid silo]

Hello

[timid silo]

Will the answer of 7th question be infinite?

[daring parcel]

Nop

What happens if you bring out the x^2 in the top and bottom equations?

What happens once you take the limit as x -> infinity of that?

[timid silo]

I divided the entire equation by x (Highest power of denominator)

[daring parcel]

Actually, x^2

So you would get $\frac{4x^2 - 5x}{12x^2 + 6x + 7} = \frac{x^2(4 - \frac{5}{x})}{x^2(12 + \frac{6}{x} + \frac{7}{x^2})}$

Now, the x^2 cancel out right?

[warm shale]

Corn

[timid silo]

I am so sorry, I typed the question number wrong.

Please pardon me. It is question number 6.

[daring parcel]

Oh! No worries 🙂

[timid silo]

Thank you for your understanding.

Will the answer be infinity? @daring parcel

.close

[obtuse pebble]

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[sterile sequoia]

is my working out correct for part a?

[obtuse pebble]

@sterile sequoia Has your question been resolved?

[obtuse pebble]

@sterile sequoia Has your question been resolved?

[junior inlet]

your method seems correct, but you should try

1. replacing the 180deg with pi radians
   or 2. write down the degree symbol

bc you know, in trigonometry, angles are usually measured in radians

[junior inlet]

@sterile sequoia hi?

[obtuse pebble]

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[obtuse pebble]

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[obtuse pebble]

@rough bough Has your question been resolved?

[rough bough]

<@&286206848099549185>

@ruby fulcrum

[obtuse pebble]

@rough bough Has your question been resolved?

[rough bough]

I solved the above

but then I can't seem to do the following

(c)

[sage geode]

Which part of the proof are you stuck on?

[rough bough]

I am stuck on trying to make it divisible by 3 @sage geode

I have it as 12m + 4

but can't seem to havec it divisble by 3

[sage geode]

I assume you've already checked the base case, so I'll start with explaining the P(x) => P(x + 1) part

Assume 2^k - 1 is divisible by 3 for some even k, we need to show that the next even number is divisible by 3 as well, which is k + 2

So all we need to prove is that $2^{k + 2} - 1$ is divisible by 3 using the assumption

[warm shale]

Touch Our Beans

[sage geode]

Using the laws of exponents, we get $2^{k + 2} - 1 = 4\cdot2^k - 1$

[warm shale]

Touch Our Beans

[sage geode]

We can write 4 as 3 + 1

$(3 + 1)2^k - 1 = 3\cdot2^k + (2^k - 1)$

[warm shale]

Touch Our Beans

[sage geode]

3*2^k has a factor of 3 for any k

and 2^k - 1 is also divisible by 3 according to the assumption

[rough bough]

why can't we just sub in 2^k

for which 2^k = 3m+1

by the asssumption

and then we have 4(3m +1) -1

[sage geode]

Surely you could do that here

[rough bough]

Okay thanks

[sage geode]

But that's really not necessary

[sage geode]

This is pretty much the last step of the proof

[obtuse pebble]

Channel closed

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[rough bough]

.reopen

[obtuse pebble]

✅

[rough bough]

@sage geode can you please look at (d) really quick

[sage geode]

Surely, just a moment

Yeah we need the same method here

Assume 3^k + 5 is divisible by 8 for some odd k

We need to show that the same is true for k+2 (the following odd number)

[rough bough]

can't we do 2k+1?

[sage geode]

Surely you could rephrase the question into "Prove that 3^(2n + 1) + 5 is divisible for all natural numbers n", which would be exactly the same statement

But that extra work could be avoided

[sage geode]

So we get $3^{k + 2} + 5 = 9\cdot3^k + 5 = 8\cdot3^k + (3^k + 5)$

[warm shale]

Touch Our Beans

[sage geode]

$8\cdot3^k$ is clearly divisible by 8 and so is $3^k + 5$ by the assumption, and the sum of them should also be divisible by 8; therefore completing the proof

[warm shale]

Touch Our Beans

[rough bough]

what about subbing 8M-5?

[sage geode]

Look

$8\cdot3^k + (3^k + 5) = 8\cdot(8M - 5) + (8M - 5 + 5) = 8\cdot(8M - 5) + 8M = 8\cdot(8M - 5 + 1) = 8\cdot(8M - 4)$

[warm shale]

Touch Our Beans

[rough bough]

okay thanks

.close

[obtuse pebble]

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[timid silo]

cant tell whats going on here

[timid silo]

i mean

how do i find where point p or q is at all?

all i know is q is a point somewhere on the line x = -4

[sage geode]

P is any point on the parabola

[timid silo]

and p has the same y co ordinate as q

oh

[sage geode]

Wait I might have to look at the statement carefully

To check if I'm right

Yes P is any point on the curve

Just get an expression for distance between P and R

[junior inlet]

hi beans

[sage geode]

Then the same with P and Q

And they'll be equal

[timid silo]

oh yes, thats pretty easy then

[sage geode]

Hello

[timid silo]

oh alright, im sorted then

[sage geode]

You reminded me of smth

[timid silo]

i thought it was a fixed point for some reason

thanks 🙂

[sage geode]

Np

[timid silo]

.close

[obtuse pebble]

Channel closed

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Use .reopen if this was a mistake.

[timid silo]

.reopen

[obtuse pebble]

✅

[timid silo]

uh i got problem

[junior inlet]

yeah?

[timid silo]

im getting a -24t^2

its only a little issue but im having a moment i think

(4t^2 -4)^2

[junior inlet]

where?

[timid silo]

16t^4 - 16t^2 - 16t^2 + 16

16t^4 -32t^2 + 16 (then the + OHHHH)

solved it

its (8t^2)

not 8t^2

thanks again :))

[junior inlet]

(8t)^2 = 64t

[timid silo]

.close

[obtuse pebble]

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[leaden vortex]

MLE for a contingency matrix.

[obtuse pebble]

@leaden vortex Has your question been resolved?

[obtuse pebble]

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[grave gale]

How is it so that the two groups circled in red aren’t the same? And how does the n and m come in in the second example?

[obtuse pebble]

@grave gale Has your question been resolved?

[alpine raven]

Zn, Zm and Z arent the same thing

[grave gale]

That was not exactly my question 🙂

[alpine raven]

what does grp means ? group ?

[grave gale]

It is right that $grp{(1,0),(0,1)} = \mathbb{Z} \oplus \mathbb{Z}$, right?

[warm shale]

Luka835

[grave gale]

Yeah grp means ‘group produced by’ so it is the smallest group that contains all the elements mentioned inside the brackets

Okay maybe they used the notation $(1,0) = ([1]_n,0)$ and $(0,1) = (0,[1]_m)$

[warm shale]

Luka835

[grave gale]

Then I get it

[grave gale]

.close

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[signal cobalt]

this is what I got

is this right

[sage geode]

What did you get for the value of s

[signal cobalt]

24/11

[sage geode]

Yeah that's right

[signal cobalt]

ty

.close

[obtuse pebble]

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[sage geode]

But wait

x should be 94/11 instead

y should be -13/11

And z should be -20/11

@signal cobalt

[signal cobalt]

.reopen

[obtuse pebble]

✅

[signal cobalt]

How

@sage geode

[sage geode]

$x = 2 + 3s = 2 + 3\cdot\frac{24}{11} = 2 + \frac{72}{11} = \frac{22 + 72}{11} = \frac{94}{11}$

[warm shale]

Touch Our Beans

[signal cobalt]

Oh

[sage geode]

$y = 1 - s = 1 - \frac{24}{11} = \frac{11 - 24}{11} = -\frac{13}{11}$

[warm shale]

Touch Our Beans

[signal cobalt]

i for got to add

[sage geode]

Yeah for some reason you seem to have forgotten the constants

Yup

[signal cobalt]

ty

.close

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[frozen willow]

The goal is to find the points where the tangent line to the curve is horizontal (0)

[frozen willow]

Here is the derivative

So, I set this equal to 0 (to find where the slope is 0), and got x = 0, -2/3

[mild ocean]

what’s confusing you

[frozen willow]

I plugged x=0, and x = -2/3 into the original equation

Then got 4 points

Only 2 are correct

The points are:

(x, y) = (0, 0), (-2/3, 2/{3sqrt(3)}), (-2/3, -2/{3sqrt(3)})

[mild ocean]

,w y^2 = x^3+x^2 when x = -2/3

[warm shale]

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Requested by quantum#6320

[mild ocean]

those are the first two points

the last point is just (0,0)

[frozen willow]

Oh yes, 3 points

[frozen willow]

I'll edit this

Sorry, my bad

[frozen willow]

The point (0, 0) is wrong

[mild ocean]

oh wait yeah

[frozen willow]

Do you know why only the other 2 correct?

[mild ocean]

because of the derivative

[frozen willow]

hmm

[mild ocean]

because the values of y you choose for the derivative also have to satisfy the original equation

[frozen willow]

0 doesn't satisfy?

[mild ocean]

no

because y can’t be 0 for the derivative

[frozen willow]

y not?

[twilit loom]

Dividing by zero

[frozen willow]

Do you know what is dy/dx at 0 then?

Just, is not defined*?

[twilit loom]

It's not defined

,w plot y^2 = x^2(x+1)

[warm shale]

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Requested by dino dik#9674

[twilit loom]

I suppose you can plug in y to the derivative to find the two values of the gradient

That works because the x cancels in the denominator, so it's no longer 0/0

[frozen willow]

Looking at the graph, it shows there are 2 slopes when (x, y) = (0, 0)

So we can figure out that the slope there isn't defined

Is there a way we'd know this without the graph (like on a test where no calculators are allowed)?

[twilit loom]

My bad my bad, it is defined

[frozen willow]

Ok

[twilit loom]

It just has two values

[frozen willow]

yeah

[twilit loom]

$\dv{y}{x}=\pm\frac{\cancel{x}(3x+2)}{2\cancel{x}\sqrt{x+1}}$

[warm shale]

dino dik

[twilit loom]

$\implies\dv{y}{x}=\pm 1 \text{ at } x=0$

[warm shale]

dino dik

[twilit loom]

So to answer the original question, the gradient is not horizontal at x=0

[frozen willow]

Oh I get it

[frozen willow]

So, what's wrong with this?

It looks like my logic at the start was whack

Since I got x=0 for when the slope was horizontal too along with x=-2/3

[twilit loom]

So what you did was $\dv{y}{x}=\frac{x(3x+2)}{2y}=0$ right?

[warm shale]

dino dik

[frozen willow]

Yeah

[twilit loom]

Then tried to solve for x

[frozen willow]

Exactly

[twilit loom]

You're right that x=0 and x=-2/3 both satisfy that, but you have to be careful and check the y as well

So the corresponding y values for your x points are y=0 and y=+-2/{3sqrt(3)

But if you plug in x=0 y=0 into your derivative, you don't get 0

[frozen willow]

Oh, you get DNE

[twilit loom]

Yup

Because you're dividing by 0

[frozen willow]

So basically, x = 0 isn't in the domain of dy/dx

[twilit loom]

y=0 isn't in the domain of dy/dx

[frozen willow]

Okay

[twilit loom]

With x=0 we got this

[frozen willow]

Because you can do this

[twilit loom]

It just gets a bit messy when you have both coordinates in the expression

More checking to do

[frozen willow]

Okay

I think I got it