precalculus

[willow bear]

the terms go up by 2 here

(2n-1) + 2 = 2n+1

1 + 3 + 5 + ... + (2n-1) + (2n+1)

[velvet blade]

then I should add n+2 right?

[willow bear]

??

[velvet blade]

okay

[velvet blade]

Okay I just need to prove that adding 2 to the LHS gives me (n+1)^2 in RHS?

You can use 2n+1 too?

oh wait, when n starts from 0

[viscid thistle]

2(n+1) + 1

[velvet blade]

2n+3?

wait this is just confusing me

[viscid thistle]

what is your question

[velvet blade]

#precalculus message

[viscid thistle]

next term is 2(n+1) -1

you assumed next term was 2n

which is not true because you only want odd numbers

next term is adding + 2

so 2n + 1

[velvet blade]

okay

so it's 2n-1 + 2n+1 in the end?

got it

I'm not getting the same result when adding (n+1)^2

the RHS should be $(\frac{1}{6})(n+1)(n+2)(2n+3)$ right?

[obsidian monolith]

Researcher in Pre-algebra

[willow bear]

yes

[velvet blade]

when just adding $(n+1)^2 $to RHS, I should get the equation above equal to $(\frac{1}{6})(n)(n+1)(2n+1) + (n+1)^2$ right?

what had just gone wrong with my syntax?

[willow bear]

the (n+1)^2 wasn't wrapped in dollars.

[velvet blade]

okay

[willow bear]

yeah so your goal now is $$\frac{1}{6}n(n+1)(2n+1) + (n+1)^2 \overset{?}{=} \frac{1}{6}(n+1)(n+2)(2n+3)$$

[obsidian monolith]

Ann

Researcher in Pre-algebra

[velvet blade]

okay

busy work

[viscid thistle]

not how to prove it

[velvet blade]

@viscid thistle bruh sorry, I just heard that book is too long and it's fine to learn from here or amann escher where induction is just 50 pages

[viscid thistle]

who told you that

[velvet blade]

lochverstarker

[viscid thistle]

the books barely 300 pages

[velvet blade]

@willow bear I got an inequality. Does that mean the assertion is false?

velleman?

[viscid thistle]

how to prove it has 40 pages on induction

[velvet blade]

But it's said he starts induction after 250 pages or something

loch said that

[viscid thistle]

yeah

[velvet blade]

let me check

[viscid thistle]

but you lack the knowldeg of that 250 pages

which is like the fundamentals of proofs

(logic and set theory)

@near kiln yo upcat please explain my friend how good H2PI is

[velvet blade]

umm.....

Can I call in loch here to talk about this? We must take it to general chat to keep this channel appropriate

[viscid thistle]

no its ok

here

[willow bear]

what did you get @velvet blade

[velvet blade]

how do you get question mark on equal symbol?

[willow bear]

\overset{?}{=} is the latex code for it, if that's what you're talking about.

did you somehow get $$\frac{1}{6}n(n+1)(2n+1) + (n+1)^2 \neq \frac{1}{6}(n+1)(n+2)(2n+3)$$ or what?

[obsidian monolith]

Ann

[velvet blade]

$3n^2+3n+1 \overset{?}{=} 2n^2 + 7n+6$

[obsidian monolith]

Researcher in Pre-algebra

[willow bear]

how did that happen now

what did you do

[velvet blade]

$\frac{1}{6} (n^2 + n)(2n+1) + n^2 + 2n +1 \neq \frac{1}{6} (n^2+ 3n +2)(2n+3)$

[obsidian monolith]

Researcher in Pre-algebra

[velvet blade]

?

[willow bear]

how did you get that those two aren't equal exactly

am i to assume that your work went in some way that ISN'T starting with one side and simplifying as much as you cna

can*

[velvet blade]

Sorry, give me a min

yeah I'm not getting both are equal

$3n^2 + 3n +1 \neq 2n^2 + 7n+6$

[obsidian monolith]

Researcher in Pre-algebra

[velvet blade]

I don't understand what's wrong with it

Is this just disproving it?

[viscid thistle]

can't you just factor an (n+1) ?

[velvet blade]

I did it

wait, I cancelled out 1/6 and n+1

wait

fuck

HOLY shit I'm dumb. I'm sorry

no I didn't get it

$2n^2 + 2n + 1 \neq 2n^2 + 7n +6$

[obsidian monolith]

Researcher in Pre-algebra

[willow bear]

youre showing me bits and pieces of your work

[velvet blade]

Okay sorry

[willow bear]

aifght so like

you should not start with $\frac16 n(n+1)(2n+1) + (n+1)^2 = \frac12 (n+1)(n+2)(2n+3)$

[obsidian monolith]

Ann

[velvet blade]

?

[willow bear]

i specifically put a ? over that equals sign to mark it as a goal and not an assumption.

[velvet blade]

Okay

So it's disproved?

[willow bear]

what no

its just bullshit and im not reading it any further cause it should be scapped.

scrapped*

you should start with one of the two sides and simplify it step by step until it transforms into the other.

[velvet blade]

Can't you just divide the common factors?

cancel out?

[willow bear]

you. are. not. starting. with. this. equality. as. an. assumption.

[velvet blade]

Okay

Ohh....

[viscid thistle]

just work on RHS

[velvet blade]

Yeah I didn't get that both are equal

RHS = $\frac{1}{6} (2n^3+ 9n^2 + 13n + 6)$

[obsidian monolith]

Researcher in Pre-algebra

[velvet blade]

LHS = $\frac{1}{6}( 2n^3 + 4n^2 + 3n +1)$

[obsidian monolith]

Researcher in Pre-algebra

[willow bear]

are you sure

how did you get LHS as that

[viscid thistle]

how is the 1 there

[willow bear]

are you sure you aren't under the impression that $\frac16x + y$ is the same as $\frac16(x+y)$

[obsidian monolith]

Ann

[velvet blade]

fuck

[viscid thistle]

once again try to work on RHS only

[velvet blade]

I have work I'll be back in a while.

[velvet blade]

hey I'm back

Ann's right

you're right

did that

LHS becomes RHS

Sorry about that

[worthy seal]

Any ideas?

[sick steppe]

@worthy seal DMT

[worthy seal]

Hm?

[sick steppe]

De Moivre

[worthy seal]

👍

[warm token]

Need some sort of a genius to help me out with these xD

[sour hemlock]

what have you tried so far

[warm token]

I dont know where to start dude thats the issue xD

[sour hemlock]

the formula is for the instantaneous value of the current

so you use it to find the time for 1 A

and then use it to find the time for 1.5 A

subtract the two time values

[warm token]

So how do I go about showing how I did it

if im just typing it into a calculator

[sour hemlock]

what?

solve it on a piece of paper

[warm token]

-_-

Im a dumbass bro

[sour hemlock]

you don't know how to write?

[warm token]

I dont know how to do this equation bro

do I need to rearrange it or na?

[sour hemlock]

what grade are you in?

[warm token]

bro

[sour hemlock]

I want to know whether you know basic algebra or not

[warm token]

Im doing my level 4 HNC, havnt done any maths since I left school

Now im doing it at level 4

[sour hemlock]

so I take it that you don't know basic algebra?

[warm token]

I know basic algebra yes

[sour hemlock]

ok

[warm token]

I just dont know logs

[sour hemlock]

do that here

[warm token]

and exponentials

[sour hemlock]

huh?

[warm token]

this is a logarithm question

[sour hemlock]

what do you mean you don't know logs

they haven't been taught to you?

[warm token]

they have but I dont understand bro

[sour hemlock]

please review your lecture notes on log and you can ask here about anything you are confused with

[warm token]

think I havnt reviewed my lecutre notes?

[sour hemlock]

if you have reviewed your lecture notes and don't understand a specific topic about log, you can ask it here

[warm token]

I just dont understand what im supposed to do in the equation

[sour hemlock]

well you see exponent

[warm token]

yeah e

[sour hemlock]

so that means you need to use log

the natural log

[warm token]

hieroglyphics

your speaking another language my man

appreciate your tryna help but

[sour hemlock]

please don't waste my time like this, if you don't understand something then ask me exactly what you need help with

[warm token]

I said I dont understand waht to do in the question

[sour hemlock]

I told you what to do

[warm token]

your help was pretty much "do the question"

[sour hemlock]

natural log

if you don't know what log is

review your lecture notes

[warm token]

dude

i reviewed them

[sour hemlock]

if you are confused about something related to log

[warm token]

they arnt helpful

[sour hemlock]

then ask it here

if you didn't take proper notes in class

ask your friend for notes

or watch a youtube video about log

[warm token]

done that

[sour hemlock]

it's not a complicated concept

[warm token]

Ive literally tried everything dude

its like im reading chinese

[sour hemlock]

why don't you tell me what you didn't understand about log?

[warm token]

because I dont understand anything

[sour hemlock]

you said you did everything

[warm token]

What from log do I apply to this question and how do I apply it

[sour hemlock]

implies you must have done a lot of research

on log

you apply the natural log (base e)

because you have e in this equation

[warm token]

Im allowed a calculator

[sour hemlock]

and?

[warm token]

Dont have a clue how to put this equation in

[sour hemlock]

you need to rearrange this equation to make t the subject of the formula

for that you need to use log

I suggest you do that

https://mathinsight.org/logarithm_basics

[warm token]

thank you

but I dont know how to do that

but knowing what do to is a start

[sour hemlock]

please have a look at these online notes on log

[warm token]

Appreciate it

[sour hemlock]

if you're still unable to understand log, then ask your teacher

[warm token]

asked my teacher

she said its too late for help

my assignment is due in 2 hours

guess ive failed it

[sour hemlock]

I guess you should've started this earlier

very unfortunate

time management is important

[warm token]

Dude ive done absolutely every other question

had other assignments to do

worked all day yesterday

all day today

[sour hemlock]

when your teacher was talking about logs

and you were confused

why didn't you ask her right then

[warm token]

The stuff we were taught looks nothing like this question

absolutely nothing like it

we were taught the basic principles

[sour hemlock]

and you understood the basic principles?

[warm token]

nope

[sour hemlock]

you are telling me you didn't understand a single thing about logs when your teacher was talking about them?

[warm token]

If I did then I wouldnt be asking for help in here

[sour hemlock]

you should've asked her right then

why didn't you?

[warm token]

But I didnt"!

I dont dwell on the past

I fucked up, ive accepted that I fucked up

[sour hemlock]

https://mathinsight.org/logarithm_basics

some reading on logs

[warm token]

but I still have time to resolve the issue, I just need help

still no help

appreciate you trying

[warm token]

https://cdn.discordapp.com/attachments/782737392774807562/782751111873822730/unknown.png

Am I wrong thinking I divide the 20 by the 8, then do log 10 of 2.5 (20 divided by 8)

[sick steppe]

Just isolate for the P ratio @warm token

$20 = 8\log{\frac{P_2}{P_1}}$

[obsidian monolith]

moshill1

[warm token]

ok thanks 🙂

[hard mulch]

Can someone

Tell me how to solve this

[viscid thistle]

Hello, can someone help explain why my answer was wrong?

I found P, and it was 7. So wouldn’t it be right to multiply it by 4? Because after finding p you plot it into the equation: (x-h)^2 =4p(y-k)

[astral mountain]

How do I tell if a irrational function has sloped asymptotes at infinity?

[viscid thistle]

not sure if functions can have asymptotes at x = +-∞, unless i'm crazy

[velvet blade]

How do I prove that $(n+1)(n^2+2n+6) | 6$?

[obsidian monolith]

Researcher in Pre-algebra

[viscid thistle]

opposite

6 | dadadoinaid

[velvet blade]

?

[viscid thistle]

6 divides that thing

not that thing divides 6

[velvet blade]

Okayy that's the notation

got it

So how do I show that?

[viscid thistle]

try to manipulate that thing so you get n(n^2+5) + 6k

[velvet blade]

ok

wait the highest degree you'd get for n is 3 here

How do you make it n^5?

[viscid thistle]

typo lol

[velvet blade]

factorising n from the terms gave me $n(n^2 + 3n + 8) +6$

[obsidian monolith]

Researcher in Pre-algebra

[velvet blade]

got it

This was a little tricky

[atomic ibex]

how do i find the max speed in a distance time graph thats curved?

using derivatives?

[sour hemlock]

yes

[atomic ibex]

how exactly?

what if my function is a quintic

[sour hemlock]

you need to find maxima

[atomic ibex]

of my derivative function?

[sour hemlock]

d(t) is your distance function

take its derivative wrt time

you get v(t)

now you want to know where v(t) is max

so compute maxima of v(t)

[atomic ibex]

so v(t) becomes a quartic

and it starts increasing to like infinity after x =6

but the distance was up to 50...

nvm

im dumb

[mild scarab]

Am I solving number 8 right? I used the compound angle formula chart

[willow bear]

,rccw

[jagged glade]

Hello there @mild scarab

Determine a value of x...
So the final answer should x= something

[velvet blade]

I have trouble with proving 9.(b)

[mild scarab]

Owww I see, so for example on the first question do I have to decide both sides by sin and go from from there. I am not sure how to solve for it

[jagged glade]

E.g. for 8a, you can just write
x=54π/23
Under your last statement.

@velvet blade where are you stuck?

[velvet blade]

I know how to write them down and have a least common multiple. But I don't know how to add factorials up

BTW

x=54π/23
how's it true?

[jagged glade]

I see... Any progress so far?

[velvet blade]

Nope...

[jagged glade]

Why not? The arithmetics seems right.

$\frac{k}k \cdot \frac{n!}{(k-1)!(n-k+1)!} + \frac{(n-k+1)}{(n-k+1)} \cdot \frac{n!}{k!(n-k)!}$

[obsidian monolith]

Biscuit

[jagged glade]

$= \frac{n!}{k!(n-k+1)!} (k + (n-k+1))$

[obsidian monolith]

Biscuit

[velvet blade]

[velvet blade]

I didn't get you

[jagged glade]

Sometimes why you wanna find the LCM for denominator, don't worry and just use the plain old 'multiply both numerator and denominator by the same number' will work.

[jagged glade]

Oh, I get it now, the 8a is not for your question.

[velvet blade]

Okayy, let me try on my own. BTW I didn't get that sin 54pi/23 = sin 8pi/23

[jagged glade]

,w is sin(54pi/23)=sin(8pi/23)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Biscuit#5208

[velvet blade]

Yeah I didn't get how...

did you add 2pi to the LHS?

[jagged glade]

Yea, 8π/23 + 2π = 54π/23

[velvet blade]

okay

yup, proved it. Thanks

@jagged glade Did you get the cookie? 🙂

[jagged glade]

🍪

[velvet blade]

nicee

[velvet blade]

I am not understanding how to prove binomial coefficient using induction

9.c

How to write summation and matrices in latex?

$\sum_{k=0}^{n} \begin{pmatrix} n\ k\end{pmatrix} x^k y^{n-k}(x+y)$

[obsidian monolith]

Researcher in Pre-algebra

[velvet blade]

I don't know how to expand (x+y) here

[steep quartz]

$x^k y^{n-k}(x+y) = x^{k+1} y^{n-k}+x^k y^{n-k+1}$
Distribute

[obsidian monolith]

Apopheniac

[velvet blade]

okay got it. I just got a little confused

[viscid thistle]

a polynomial p(x) has reminader 3 when divided by x-2 and a remainder of -5 when divided by x+2 what is the reminder when divided by x^2 -4

<@&286206848099549185>

x^2-4=(x+2)(x-2)

so is the remainder just -15 then?

@viscid thistle

[fading lodge]

if the activity says that i need to prove by mathematical induction

does that mean even if the statement is false i need to prove it anyway?

[viscid thistle]

I mean if it's false you cannot prove it holds because it does not

[ancient rampart]

is cot^-1(x) the same as 1/tan^-1(x)

[sour hemlock]

no

[ancient rampart]

what is it then

[stuck lark]

arccot is the inverse of cot restricted to (0,pi)

[sour hemlock]

x = cot(a)
x = 1 / tan(a)

take inverse and see for yourself

[ancient rampart]

idk how to take the inverse of cot

[stuck lark]

in restricting the domain of cot to (0,pi), cot becomes invertible and we call its inverse arccot, so arccot(x) is defined as the unique value y in (0,pi) where x=cot(y)

[mighty stone]

Can someone dm me? I really need to understand a concept

Like half angle and double angle identites

[sour hemlock]

please post your question

[mighty stone]

Kinda don't know how to post it (its a picture)

I got it to work i got a new photo

Cna y explain?

How to do it?

@sour hemlock

[sour hemlock]

are you asking me to derive the half angle identities?

[mighty stone]

Yes

To explain

[sour hemlock]

detailed derivations can be easily found online

please look them up

[mighty stone]

Wait what do u by diviations?

[sour hemlock]

if you have specific questions about a derivation, you can ask them

[mighty stone]

What do u mean by derivations?

[sour hemlock]

what's your mother tongue?

[mighty stone]

English

[sour hemlock]

huh?

[mighty stone]

Here i want to do this but in the half angles

[sour hemlock]

wait

stop

[mighty stone]

?

[sour hemlock]

you need to phrase your question properly

what is it you want to know?

[mighty stone]

Oh sorry

I want to know how to solve the half angle

If that makes sense

[sour hemlock]

show me a question that you are having a hard time with

[mighty stone]

Ok

[sour hemlock]

ok, so we are back to this

[mighty stone]

These are the fromulas

Wait its loading

[sour hemlock]

I see

[mighty stone]

I am having a really hard tike with this

[sour hemlock]

why don't you just use a calculator?

[mighty stone]

Well i have to do this for my teacher...

[sour hemlock]

even if you use the formula, you will eventually have to use a calculator to get the proper numerical answer

[mighty stone]

She wants it to be answered in fractions

So not fully simplifyed

Like in the dpuble angle picture i sent

[sour hemlock]

why don't you start by solving for theta

and then use the half angle identity

for sin, cos, and tan

[mighty stone]

Is theta 210?

[sour hemlock]

nobody in this discord channel will confirm your answers until you show your work

[mighty stone]

I'll send my work

Is this right?

?

[sour hemlock]

wait

how did you get cos(210)?

[mighty stone]

Because since its a half angle i need to double it to find theta

[strong ermine]

why did you multiply by 2/2

[mighty stone]

Tbh i am not sure i am confusing myself

Can u help me step by step?

?

Do u know what i should do?

[topaz flint]

theta/2 = 105?

[mighty stone]

Yes so i need to find theta, so its 210

I need to use the chart then by using 210-180 i belive because its in the tan quadrsnt

Thats what i think at leadt

[topaz flint]

u need to ask your self what are you given?

an angle or a function

[mighty stone]

An angle?

[topaz flint]

secondly

is the angle given in radian, degrees or pie radians

[mighty stone]

Degrees?

[topaz flint]

now wuts the angle?

[mighty stone]

210

Right?

Hello?

[topaz flint]

I am confused, isn't the angle that you are trying to solve?

[mighty stone]

Thisbis the double angle, i need to do this for the half angle

Do u know what i mean?

[topaz flint]

did u solve 210 or were u given it?

[mighty stone]

I found it myself

This is all i got from the problem

And i need to use the equations:

To find the blanks on the first picturep

Wrong picture waitp

[topaz flint]

because theta/105=105

we can say sin 105,cos 105 and tan 105?

[mighty stone]

Nevermind i don't think we can communicate this way

Just forget it

[topaz flint]

ye ping helpers

[viscid thistle]

what's your problem?

[viscid thistle]

[upbeat moss]

https://gyazo.com/387450afa992f96398c5a8774b951b2c

first time using sigma, and i have no idea what im looking at

[sick steppe]

so in the sum is the general term

what's the thing that changes in the terms?

[upbeat moss]

numerator

[sick steppe]

yeah, so instead of the numerator being a number, it'll be the variable you're summing over, which is indicated under sigma

$\sum_{i=1}^n \frac{i}{2}$

[obsidian monolith]

moshill1

[sick steppe]

for example

[upbeat moss]

so for problems like this, Im aiming to replace the changing term with a variable?

I understood what you just said tho, just unsure what exactly im trying to find

[sick steppe]

So the term doesnt change, a thing in the term changes

in this case, the numerator of the fraction

What the question is asking for is: When do you stop summing and what general thing are you summing?

[upbeat moss]

so im summing the numerator which increasing by 1 up until 7

[sick steppe]

yep

$\sum_{f=1}^7 [something]$

[obsidian monolith]

moshill1

[upbeat moss]

ah ok this was actually simple, i got it. thanks a lot

or at least this particular problem

[glass breach]

Having trouble verifying trig identity

any clarification would be appreciated

[patent beacon]

@glass breach Multiply left by sin²(x)/sin²(x)

Get right

[glass breach]

wait why is it multiplied by that

[patent beacon]

Try it and see!

[glass breach]

I'm going to learn this a bit better and ask later thanks @patent beacon

[abstract folio]

how do i do this

[viscid thistle]

anyone know how to do this?

[mystic umbra]

@viscid thistle https://www.mathsisfun.com/calculus/implicit-differentiation.html

[viscid thistle]

😦

[mystic umbra]

I think I gave you the wrong link actually

I will do it myself then show you

[viscid thistle]

ty ❤️

[mystic umbra]

$3x^2+10xy+5y^2+8=0$

[obsidian monolith]

galois

[mystic umbra]

you understand that so far right?

$3x^2+10xy+5y^2+8=x^2+3xy+2y^2$

[obsidian monolith]

galois

[mystic umbra]

you understand this part?

$2x^2+7xy+3y^2=0$

[obsidian monolith]

galois

[viscid thistle]

whoops sorry

hows you make them equal to each other so quickly @mystic umbra

[mystic umbra]

$3x^2+10xy+5y^2+8=0=x^2+3xy+2y^2$

[obsidian monolith]

galois

[mystic umbra]

start with $3x^2+10xy+5y^2=-8$

[obsidian monolith]

galois

[mystic umbra]

+8 to both sides

https://www.desmos.com/calculator/m0gpzpj5a6

[viscid thistle]

can anyone help with these?

[mystic umbra]

the first problem is here

the computer generated a 214 page proof but it crashed when I tried to save it.

this is a hard problem I think

question 4 $cos^{-1}(cos\theta)=\theta$

[obsidian monolith]

galois

[viscid thistle]

im scared

[mystic umbra]

are you in pre-calc?

question 4 this is only true for $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$

[obsidian monolith]

galois

[mystic umbra]

because we have in this case by the picture theta > 2/pi we can take the reflection over the x axis I think

I'm also a student I could be wrong

or a rotation maybe I'm not sure

it is easier when I have a number, I can put it in the calculator. with a picture I need to think about it

[viscid thistle]

yea idk

it says im not supposed to use a calculator

so the answer shouldnt be from a calculator

[mystic umbra]

ok I got it

reflect it left to right

the y coordinate is the same as the fourth quadrant

the sin is reflected left to right

the cos is reflected up and down

-(theta - pi)

for sin

[stuck lark]

by definition arccos(x) is the unique value y in [0,pi] where x=cos(y) so for any x not in [0,pi] i'm guaranteed arccos(cos(x))!=x

[mystic umbra]

https://www.desmos.com/calculator/0cpwlinkey

that shows all solutions for all values of x for both functions

pick your quadrant based on the picture

they also give you the range for Q3 pi < theta < 3pi/2

[velvet blade]

Hey, so since yesterday I don't understand how to prove binomial theorem by induction.

I don't understand the solution given in the book

Let me show

9.c

[willow bear]

the biggest hurdle imo is the notation

[velvet blade]

yeah I don't understand the notation enough to alter it

[willow bear]

but the thing can be done with some clever reindexing

[viscid thistle]

expand out the sums if it's too weird for you

[velvet blade]

he puts j instead of k

[willow bear]

yeah as i said, reindexing

[velvet blade]

Okay

Dude I can't write it down. each term is taking too long

[viscid thistle]

you dont need to write down every term

[velvet blade]

then?

[viscid thistle]

go on with the proof

apply property on binomials

[velvet blade]

I don't understand where j comes from

[viscid thistle]

j = k + 1

for the first sum

for the second one j = k

[velvet blade]

why is j different values for each sum?

[viscid thistle]

why not

[velvet blade]

why is the start of summation of different value?

[viscid thistle]

they are just indices

[velvet blade]

won't that cause them not to add up?

[viscid thistle]

if you write down some terms you find out that you have similar terms with different coefficients

[velvet blade]

I expanded, I'm getting last term as (1/(n+1))x^n

Is it right?

No wait

I got x^n

I don't know what to change in the notation when looking at this expansion and multiplying it with (x+y)

@opaque olive take it in #calculus

[opaque olive]

this is pre calculus

[velvet blade]

then why is integration there in answer?

[opaque olive]

?

[velvet blade]

Integration is calculus

[opaque olive]

its not early university

its pre university

[velvet blade]

But calculus here is in the early university category

[opaque olive]

fine, ill move

[velvet blade]

I got till here

<@&286206848099549185>

[velvet blade]

@viscid thistle you free?

[viscid thistle]

not at the moment

maybe look up some videos

[velvet blade]

Ok

[velvet blade]

@patent beacon Let me post it again

[viscid thistle]

yo

[velvet blade]

@viscid thistle hey

It's proof of binomial theorem

[patent beacon]

Okay so you are struggling with that completed proof

[velvet blade]

there's answer below

yeah

I don't understand how those two summation indices join to be the final one

Like how $\sum_{j=1}^{n+1} + \sum_{j=0}^{n} = \sum_{j=0}^{n+1}$

[obsidian monolith]

Researcher in Pre-algebra

[viscid thistle]

its a change of index

i think

[velvet blade]

the both cases have same number of indices in LHS, why is one added up in RHS?

[viscid thistle]

and using b)

[velvet blade]

how does that change the amount of indices?

[viscid thistle]

look at the left one, instead of starting at j=1 you can start at j=0 and instead j-1 write j in the summation

[velvet blade]

[viscid thistle]

I mean $\sum_{i=1}^{n} i = \sum_{i=0}^{n-1} i+1$

[obsidian monolith]

Godel

[viscid thistle]

similiar thing happened there and using part b) I think

[velvet blade]

I didn't get you

[viscid thistle]

ill let kaynex explain he will dddo it better

kaynex not here

ok so lets think again

which line is unlcear?

2nd to alst?

before the final result? @velvet blade

[patent beacon]

It seems as through they are adding term by term, despite the series being shifted

Which makes sense, because the powers on xy match.

And then you can take advantage of b) to get the final line

It's just a little awkward because the series are not the same indices but it's not a huge deal

[velvet blade]

No I'm having trouble adding up indices. Not the factorials

In the end, there's one more index than the two before adding

j goes from 0 to n+1

[patent beacon]

Yeah yeah, so it's a good idea to consider the two terms that don't pair up

[velvet blade]

I didn't get you. If they don't pair up then why are we getting a single expression?

is $\sum_{j=0}^{n} = \sum_{j=1}^{n+1}$?

[obsidian monolith]

Researcher in Pre-algebra

[velvet blade]

if so, then why is $\sum_{j=0}^{n} + \sum_{j=1}^{n+1} = \sum_{j=0}^{n+1}$?

[obsidian monolith]

Researcher in Pre-algebra

[patent beacon]

It's not!

[velvet blade]

see the answer in the book

[patent beacon]

Instead, they are adding the terms like this:
1
2 + 1
3 + 2
4 + 3
+ 4

And it happens to conform to a nice pattern when this happens

The two terms that don't get a pair are when x has a zero exponent, and when y has a zero exponent

[velvet blade]

How am I supposed to know that?

I just know induction

[willow bear]

induction only transforms your goal into a different, (hopefully) simpler goal

how you prove the transformed goal is context-dependent and does not have to do with induction directly

[velvet blade]

But I don't know about summation properities much. I don't know how to simplify it. It hasn't been taught in this book

this is everything I've been taught about summation before the exercise

[willow bear]

¯_(ツ)_/¯

[velvet blade]

I don't know what shrug means in this context

[willow bear]

intuition is hard to teach

sometimes there will be points where you'll go "wait how am i supposed to know that"

[viscid thistle]

writing out terms helps out

as Kaynek did

[patent beacon]

So for example, let's look at this proof using
(x + y)³
in terms of knowing that
(x + y)² = x² + 2xy + y²

(x + y)³
= (x + y)(x² + 2xy + y²)
= (x³ + 2x²y + xy²) + (x²y + 2xy² + y³)
This is the step where they move the indices to match the powers:
= x³ + (2 + 1)x²y + (1 + 2)xy² + y³
Note the middle two terms are actually two terms paired

Note that this isn't a general summation formula, it just happens to make a nice pattern here

[velvet blade]

okay

[patent beacon]

They're really doing this in that step
= (x³ + 2x²y + xy²)
+ (x²y + 2xy² + y³)

Then adding down

[velvet blade]

Where will I be able to understand better about summation and indices? Do you suggest any book?

[patent beacon]

That works nicely in general haha

This isn't a good example as we often don't work with sums this way. In this case, writing out the terms helps

Any calc book has sum stuff in it

[velvet blade]

I have to figure out all the factorials before writing the coefficients right?

Ohh

Spivak?

[patent beacon]

Yeah it will have it

I think you're looking for something huge and general in this proof, but sadly that's not really here. This just works well in this case

[velvet blade]

Better to just remember it?

Because I can't understand it with one page of explanation

I'm trying to expand it, I'm not getting it

Nevermind, I feel done for the day.

thanks

[viscid thistle]

How do I factor this
x^4 - 26x^2 + 13x + 3 = 0

i would guess synthetic division

@viscid thistle

i think

oh

do I guess the first root

and then

do synethic to find rest?

i would say so - and then use the intermediate value theorem (i think)

to narrow down what possible zeroes there are

alright

i could be wrong with which theorem

but

basically, since your function is continuous, if you end up getting a remainder that is positive, and then try a different number and get a negative remainder, you know that there is a value in between those two chosen numbers in which that number yields a remainder of 0

[velvet blade]

12: best answer I can think of is "billiard balls aren't integers"

Is it right?

[viscid thistle]

not sure if this statement is necessarily true

i could be wrong though

there is another good one

[blissful ridge]

How do we know that n balls from the end have the same colour??

[viscid thistle]

"all horses are white" proven by induction

[velvet blade]

So umm... is my answer right?

Because the book said induction is for integers

[blissful ridge]

Billiard balls are not integers??

[velvet blade]

Yeah... I don't know how else to justify that induction doesn't apply to billiard balls

[blissful ridge]

This

[velvet blade]

Okay

[hard hornet]

wtf is this proof

[willow bear]

nope that is not the issue

[willow bear]

and neither is this

[velvet blade]

Then

[willow bear]

though the wording does need to be adjusted to "we will prove all groups of n balls are the same color"

the issue is that this breaks when you try to go from n=1 to n=2.

[velvet blade]

Okay

[willow bear]

there will be no overlap between the left group of n (1) and the right group of n (1)

[velvet blade]

I didn't get it

Group?

[willow bear]

set

arrangement

whatever

[velvet blade]

Okay

[willow bear]

i don't know what word you want me to use

[velvet blade]

No nevermind

My bad

I didn't get what overlap meant

[willow bear]

sometimes words mean their everyday meaning

[sleek ruin]

Anyone know how to do?

[mystic steppe]

@sleek ruin I'm trying it out rn, any particular method you're learning rn?

[sleek ruin]

Hmm

Sum and Difference Identities.

[mystic steppe]

ahhh

hh I just found the answer online but they explain it kinda weirdly so I'll try to decipher it in a simpler way

[sleek ruin]

Thank you

I tried as well but I do not have a membership

[viscid thistle]

you can do sum and difference identities

as you mentioned before

$tan(a +b) = \frac{tan(a) + tan(b)}{1 + tan(a)tan(b)}$

[obsidian monolith]

Sir Topham Hatt

[viscid thistle]

Stonks

so

let $a = sin^{-1}(x)$, and let $b = cos^{-1}(y)$

[obsidian monolith]

Sir Topham Hatt

[viscid thistle]

if you do that then you should be able to use right triangles

[mystic steppe]

@sleek ruin

yea it's just like 🎩 Sir Topham Hatt 🎩 said!

and honestly residge you can find a good chunk of the webassign questions (or some variation of them) explained by a math teacher online outside of chegg

[viscid thistle]

YEAAAA

SIR TOPHAM HATT

let's go thomas the tank engine!

[mystic steppe]

but yea this server's great for help too I'm not discouraging that :) either is great, google can be confusing sometimes tho

AYYYY 🎩 🎩 🎩

[viscid thistle]

desmos is arguably one of the better math services out there

well

[mystic steppe]

symbolab has saved my grade multiple times

[viscid thistle]

👍

it serves my needs super well, and thus i said that

[mystic steppe]

preach 🙏

[viscid thistle]

👍

Anyone know what equation to use here

I would preferably need the name of the equation so i can look up similar derivatives of it

I need to find a, c, A,

(capital refering to angles)

law of sines or law of cosines

Life saver, thanks!

🙂

[sleek ruin]

ty

@mystic steppe i did that but i was stuck after plugging it in

simplyfying was confusing

[mystic steppe]

did you still have trig stuff? bc if you don't webassign will usually take it

yea the final answer is really funky @sleek ruin

[sleek ruin]

what do you mean by this

[mystic steppe]

like if there's no tan, cos, sin, webassign will mark it as right, no matter how wack it is

[sleek ruin]

ye i have no idea what im doing

What I got so far

[mystic steppe]

no yea that's right

I think you could probably just put that in and it'll be right

you can expand it if you want but it's kinda wack

[sleek ruin]

ye webassign got some difficult questions

yea that did not work

[mystic steppe]

you gotta expand it then rip

[sleek ruin]

you got chegg?

lmao

[mystic steppe]

Oh wait no you did the equation wrong

[sleek ruin]

o wait its plus

[mystic steppe]

Yea numerator is plus, denominator is minus

[sleek ruin]

yes yes

it works

imagine if i had to expand

[mystic steppe]

eeee that's nightmare fuel

[idle moat]

can someone check me

[gaunt mason]

What is this rule? Can you write it down?

[idle moat]

im trying to find the rule

but im sure, the x-value cannot equal 0

not y

[gaunt mason]

I guess the rule is just the formula for tan(theta) = ...something... what is this something?

[idle moat]

y/x

[gaunt mason]

are you sure?

[idle moat]

yeah

[gaunt mason]

OK, then yes, y can be 0, and x cannot be 0.

So you can use the rule to find the theta for y-value of 0.

Which means False is correct.

[idle moat]

thank you

and my other question is

when converting from rectangular to polar coordinates, should you always plot the point first?

[gaunt mason]

You mean because of the quadrants?

[idle moat]

yeah

[gaunt mason]

You don't have to plot it necessarily, but you need to be careful. So plotting can be recommended as a verification.

You can plot it in your head.

I think you do need to examine the signs, that would be more appropriate procedure than "plot"

tan will give you the angle in range -90, 90 degrees, which does not suffice if the points are in 2nd or 3rd quadrant.

[thorn moat]

I'm having trouble interpreting the problem above, why is the y cord of the vertex considered to be the highest point of the trajectory. Do we consider the x axis to be ground level?

[viscid thistle]

so im doing this question right and i just wanted to make sure i was doing right
so the question is "A box with a lid is to be created from a 52 cm by 33 cm piece of cardboard by cutting x by x squares from the four corners of the cardboard, and at the centre of the two sides, as shown in the diagram. Determine the function that represents the volume of the box in terms of x, and state the restrictions on x. If the box is to have a volume of 1904 cm3, determine the side length of the squares that need to be cut. "
I got v(x) = x(33-2x)((52-3x)/2) such that x cant equal 0 and greater than 33-2x, 52-3x
I got my final anwser to be x = 8 cm
is that right?
please @ me if someone anwsers

[gaunt mason]

@thorn moat Yes, it is pretty hard to explain.

The x-axis is the time.

[thorn moat]

i think i got it

[gaunt mason]

The y-axis is the height.

[thorn moat]

ok

[gaunt mason]

Because y-axis is the height, when the function takes the biggest y the height is the biggest.

So they don't describe the 2D trajectory of the rocket.

You don't care about its horizontal position, you just consider the vertical movement of the rocket.

So you can basically consider as if the rocket has been thrown straight up in the air, reaches max point, and falls back down.

Now, at t=0 (launch time) the height is 318, so rocket has been launched from the platform at that height.

[thorn moat]

i see

[gaunt mason]

It is then falling down to the platform, and after it passes the platform on the way down it goes further to the sea level.

---

Surprisingly, the function of height with respect to time is a parabola.

Same as the spacial trajectory of a rocket launched at an angle.

But this is a coincidence. And this is confusing, and probably confused you.

@viscid thistle Can you sketch it

[viscid thistle]

@gaunt mason

[gaunt mason]

The side is just cuberoot(1904)

@viscid thistle

[viscid thistle]

um how?

@gaunt mason

[gaunt mason]

12.3943*12.3943*12.3943 = 1904

Wait a second

I seeee.

I got it wrong

OK, so the side sides of the top cover just go over the bottom sides, so they are ignored.

@viscid thistle

[viscid thistle]

yea

so is my anwser right of x=8?

[gaunt mason]

A second, the sides are: 33- 2x, (52-3x)/2, x ?

@viscid thistle

Yes, and the volume is v(x) = 1/2*(33-2x)*(52-3x)*x

[viscid thistle]

yea

so i was right? @gaunt mason

[gaunt mason]

x>0, 33-2x>0, 52-3x>0 are constrains.

[viscid thistle]

yeye

[gaunt mason]

1904 = 1/2*(33-2x)*(52-3x)*x

How you solved this equation? It is the equation of the third order

You guessed the solution?

I mean you just put 8 inside = 1/2*(33-16)*(52-24)*8

14*17*8

[viscid thistle]

no its 6x^3 -203x^2 +1716x - 3808=0 then i factored it

and i got (x-8) as one facotr

factor

[gaunt mason]

How you factored it?

[viscid thistle]

(x-8)(6x^2-155x+476)

[gaunt mason]

But how you knew to factor it like that

[viscid thistle]

it looked factorable so i just put it in wolfram alpha

i wasnt gonna take the brain power to do it

[gaunt mason]

Well, if that is allowed 🙂

Any other real solutions for this equation, or only imaginary?

[viscid thistle]

ye theres x= (155 +- root 12601) / 12

[arctic kestrel]

i've been trying to do this for the past hour

[gaunt mason]

@viscid thistle You can just do this: https://www.wolframalpha.com/input/?i=1904+%3D+1%2F2*(33-2x)*(52-3x)*x and get the solutions. Why factor it first, if you anyways going to use wolfram alpha. But you need to check the two other solutions as a solution to your problem, specifically if they satisfy the constrains.

[arctic kestrel]

can someone help out

idk how to find next point

[viscid thistle]

i only use wolfram alpha on stuff i can do but just will take some time

[gaunt mason]

So wolfram alpha gives 3 real solutions.

[viscid thistle]

2 dont work since they dont work in the constrains

[gaunt mason]

I think 2 work.

[lucid hinge]

2 real

[viscid thistle]

but has x has to be 0 < x < 33/2

[gaunt mason]

x= (155 - root 12601) / 12 is

[gaunt mason]

3 real, 2 of which seem to satisfy the constrains

[lucid hinge]

Ah okok

[gaunt mason]

but let Viper to confirm.

[viscid thistle]

yea youre right x=8 and (155 - root 12601) / 12 both work

so which one do i pick then?

[gaunt mason]

Both

[viscid thistle]

so x= 8, (155 - root 12601) / 12 ?

[gaunt mason]

yes, if we didnt miss something

[viscid thistle]

okok thank you

[arctic kestrel]

'

how do i find the answer?

i tried subtracting -22pi/3 -2pi

then -22pi/3 +2pi

[gaunt mason]

You need to convert this fraction into whole number and the reminder

[arctic kestrel]

so

[gaunt mason]

22/3=7 1/3

[arctic kestrel]

7 1/3 -2pi

?

[gaunt mason]

So -22/3 pi = -7 1/3 pi

You can add or subtract 2pi any number of times, and still get the same angle (that is coterminal, although I never heard that term).

To get the smallest angle that is positive you should add 8pi, which is obvious.

[arctic kestrel]

15+16pi/2

31pi/2?

[gaunt mason]

No.

[arctic kestrel]

but u can't add 15 and 16pi

[gaunt mason]

-7 1/3pi + 8pi.

You can add 16pi, because it is a multiple of 2pi

You cannot add 15pi, because it is not a multiple of 2pi

You can only add even times pi.

[arctic kestrel]

i'm so confused

so it's not 15+16pi/2?

[gaunt mason]

Obvious amount to add is 8pi, because it adds just enough to be positive.

[arctic kestrel]

wait

22+24pi/3?

[gaunt mason]

Wait a second.

[gaunt mason]

You have the angle -7 1/3 pi

[arctic kestrel]

so

i make it a whole

number

and then after I

[gaunt mason]

You make it a whole and a reminder

as a first step

so -5 1/6 pi is the first step

[arctic kestrel]

-5 1/6pi

[gaunt mason]

For this to get barely positive you have to add 6pi

Because if you add 4pi it is not enough.

[arctic kestrel]

so

-5+6pi

[gaunt mason]

You lost the 1/6pi which was part of the -5

-5 1/6 pi + 6pi

[arctic kestrel]

so

-5pi

since the two cancell out

[gaunt mason]

which is basically, to make it more obvious:
-5 1/6 pi + 5pi + pi
-1/6 pi + pi
which finally is 5/6 pi

[arctic kestrel]

or -10pi?

really?

omg

it gabe me a new question

it's annoying

so first

i convert into whole mixed number

8 1/5pi

omg

[gaunt mason]

Minus. And you can add either 8pi or 10pi, 8pi is not enough, so you add 10pi

[arctic kestrel]

its killing my soul

[gaunt mason]

Can you stop posting random problems, we solved 4 examples already.

[arctic kestrel]

yeah sorry

i'm just getting pissed off with it

so i have 25/3

and i convert it into a whole number?

8 1/3

[velvet perch]

hey there im doing a few prac qs, can someone help me with this one

i don't really know how to do this

i am trying to find the inverse function

[elder horizon]

@velvet perch

[velvet perch]

hi

[elder horizon]

y = √1-x^2 --> switch x's and y's and solve for y

[velvet perch]

that is the right answer yup but i still don't get why that happens

i inverted after solving for x

(Y^2 - 1)=-X^2

SQRT((Y^2 - 1))=-X

Ohhhh did i forget the minus sign

[elder horizon]

x = √1-y^2 --> x^2 = 1 - y^2 --> x^2 - 1 = y^2 --> y = √x^2-1

[velvet perch]

ohhhh

[elder horizon]

switch x and y's first always when finding inverse and the rest is just algebra

[velvet perch]

righto thank you so much, i was stuck on this ik why now

oh

[elder horizon]

(solve for y)

[velvet perch]

oh

[elder horizon]

no problem

[willow bear]

x = √1-y^2 --> x^2 = 1 - y^2 --> x^2 - 1 = y^2 --> y = √x^2-1
second step is wrong

[gloomy bane]

heya everyone, how would I go about solving this problem?

total = $345.64

[jagged glade]

Hello @gloomy bane !
Do you still need help with the question?

[gloomy bane]

yes please

[jagged glade]

Okay, so it's compounded monthly.

We shall let the amount to be invested monthly be $x.

How do we set up the equation then?

[gloomy bane]

need to apply a geometric series I assume, although I'm sure I used the wrong equation

I tried Sn = a(1 - r^n)/(1 - r)

[jagged glade]

I see what's the problem here.

What does the 'total' mean in this context?

[gloomy bane]

The total cost of refurbishing a room

• 20% extra to account for price rises

[jagged glade]

Good!

Now let's start with the first month

We will eventually know why that GS is not working

[gloomy bane]

n = 0 is the first month right?

[jagged glade]

n=1 is the first month

Which is a.

[gloomy bane]

got it

[jagged glade]

What your r then?

[gloomy bane]

0.006

7.2/12

[jagged glade]

Good

So, you wanna try and plug it in again? To see if it's working?

[gloomy bane]

S1 = a(1 - 0.006^1)/(1 - 0.006)

there's two unknowns

[jagged glade]

Sorry, you mean s1

Yea.

S1 is just a.

Now for S2

[gloomy bane]

S2 = a(1 - 0.006^2)/(1 - 0.006)

S2 = 0.999964a/0.994

[jagged glade]

Nice

And for S12

[gloomy bane]

S12 = a(1 - 0.006^12)/(1 - 0.006)

[jagged glade]

So, does that work?

[gloomy bane]

S12 = a/0.994

[jagged glade]

Nope

,calc 1-0.006^12

[obsidian monolith]

Result:

1

[jagged glade]

Oh... It's too small

Need a better calculator