precalculus

[opaque olive]

nice good luck

[hushed sorrel]

i think i messed up s2 but should be getting a high 1 in s3

[still meadow]

(just for reference, step maths would fall under calculus)

[opaque olive]

im thinking of taking a gap year and applying for camb math though

i think i messed up s2 but should be getting a high 1 in s3
@hushed sorrel ah nice

[hushed sorrel]

o

[opaque olive]

(just for reference, step maths would fall under calculus)
@still meadow i just thought since the content was pre calculus so it would go here

[hushed sorrel]

my insurance is warwick so tbh im not too annoyed if i miss it

[opaque olive]

ah yeah

[hushed sorrel]

warwick is gr8 too

[opaque olive]

yep it is

[still meadow]

Precalc don't do calc. ;p

[hushed sorrel]

you thought about unis yet or nah

[opaque olive]

yes ive applied this year and got offers

i applied for compsci

[hushed sorrel]

oh nice

[opaque olive]

turns out i dont really like compsci

[still meadow]

congrats!

[opaque olive]

lol

thanks

so i may reject my offers

and reapply for math

[hushed sorrel]

ye dont waste ur life on a course ur half into

[opaque olive]

yeah

[hushed sorrel]

much better to reapply imo

[opaque olive]

so on results day if i miss my firm

im gonna reapply

gonna try my luck at cambridge

lol

[hushed sorrel]

its messy bc the thing is once you agree to UCAS choices, I think u are bound by a contract

technically

so im not sure exactly how it works

[opaque olive]

hm

i thought i could still reject on results day

[hushed sorrel]

its a bit weird idk if you were told that

[opaque olive]

i wasnt no

[hushed sorrel]

im not too sure but I know that by clicking ur options, ur basically in a contract

[opaque olive]

lemme check ucas quickly

[hushed sorrel]

so u have to think rlly carefully

you probably can refuse with some effort

[opaque olive]

yeah

[hushed sorrel]

its 3.52 am lol u sleeping soon?

[opaque olive]

gonna finish this last STEP question lol

[hushed sorrel]

o nice

what year is it

s1

[opaque olive]

hmmmm

[hushed sorrel]

?

[opaque olive]

its from the STEP support programme

S1

[still meadow]

Looks like s1?

[opaque olive]

yes s1

[hushed sorrel]

oh ok

[opaque olive]

ok

its 2009 Q8

S1

[hushed sorrel]

this years s3 was a calculus fest which was rlly nice

o ok

I think calculus qs are the easiest

[opaque olive]

i like calculus too

[hushed sorrel]

not so good on the number theory/sequences

[still meadow]

Dumb question - S3 is for A-level maths and further maths right?

[hushed sorrel]

yes

[still meadow]

And S2 is A-level + AS FM.

[hushed sorrel]

ye now it is

[opaque olive]

yeah

Results days is soon lol

[hushed sorrel]

.

bit scary but ok

goodluck

[opaque olive]

you too

[still meadow]

good luck to both of you.

[hushed sorrel]

i mean it is down to luck in a way

[opaque olive]

thanks

whats your predicted ?

[hushed sorrel]

4A* and A* EPQ

[still meadow]

I hope it wouldn't be like IB.

[opaque olive]

oh nice

[hushed sorrel]

probably wont acc get that

lol

[opaque olive]

I hope it wouldn't be like IB.
@still meadow yeah i heard the IB storys

[hushed sorrel]

i think ill get 3A* and A in biology

[still meadow]

4A* and EPQ? Welp that's rough.

[opaque olive]

oh

so did you take M FM chem and bio?

[still meadow]

IB has coursework though.

[hushed sorrel]

ye but I mean its mainly downt o my step results

yes

[still meadow]

So that's uh a hot fucking mess.

[opaque olive]

whats your offer?

1 ,1?

[still meadow]

Because they have to moderate coursework and w/e.

[hushed sorrel]

2A*, 2A, and 1, 1 in s2 and 3

yes

its a bit of a tough cookie

[opaque olive]

yep

[hushed sorrel]

only rlly worried about the 1 in s2

but a third make it in missing the offer

[opaque olive]

are you in the STEP server?

[hushed sorrel]

so theres a chance

[opaque olive]

theres other camb math applicants there

[hushed sorrel]

nah I think it would distract me from acc doing qs

so I left

[opaque olive]

oh lol

[hushed sorrel]

yeye I know

I meet a few in other servers anyway

[opaque olive]

ahh

[hushed sorrel]

so its not a big deal

anyway gn i think ill sleep now and I will make yakisoba for breakfast

should be fun

[opaque olive]

oh wow

have fun, and good night!

[hushed sorrel]

good luck on ur maths x

[opaque olive]

ty

[obsidian monolith]

nebula:

[fleet yew]

the coefficients of the polynomial are all positive integers, so that means for any positive integer n there is some P(n) that is also a positive integer

so i guess you can prove that there is an injection from Z+ to Z+

and then just use the fact that there are infinitely composite integers

[ocean cove]

you can look at the constant term for a construction

[viscid thistle]

could somebody explain how to do this

[uncut mulch]

consider: $0.\overline{22} = \frac{2}{10} + \frac{2}{10^2} + \frac{2}{10^3} + \dots$

[obsidian monolith]

ramonov:

[willow bear]

ramonov i feel like that solution may be rejected by a particularly anal and overscrupulous grader

[uncut mulch]

i'm commiting crimes dont hurt me

[viscid thistle]

Why 0.22 and not 0.2

[willow bear]

for the simple reason of not doing $\frac{22}{100} + \frac{22}{100^2} + \frac{22}{100^3} + \dots$

[obsidian monolith]

Ann:

[uncut mulch]

gotta love the word consider though

so much leeway

[viscid thistle]

Oh I understand now

i saw that in my notes but i never understood it

thank you

[wide ocean]

i tried 2-2y =x

for first equation then plugged in for second

but getting -2/3

idk how to get "2"

[harsh smelt]

@wide ocean is log in your book defined to be log base 10 or what

[wide ocean]

yes log base 10 @harsh smelt

[harsh smelt]

ok, so what than first log

[obsidian monolith]

Commander Vimes:

[harsh smelt]

is*

[wide ocean]

2

[harsh smelt]

ok and log0.0001?

[wide ocean]

-4

[harsh smelt]

yes

so what stopping yu now solve system

2 = x+2y and -4 = x-y

[wide ocean]

are the questions separate

oh isolate y

should both y's be the same?

[harsh smelt]

yes

i mean it is the system

[wide ocean]

what is x?

[harsh smelt]

solution should satisfy both

do u know how to solve system of lineq?

[wide ocean]

not sure

[harsh smelt]

ok, you have
2=x+2y
-4=x-y

you want to find y

what would you do?

[wide ocean]

make both equal

to each other

?

[harsh smelt]

no

use one equation to express one variable in terms of second one

and substitute

[wide ocean]

first onex=2-2y

so second, -4=2-2y-y

-3y=-4

-3y=-6

y=-6/-3

y=2

[harsh smelt]

ye

[wide ocean]

woah that worked

thanks @harsh smelt

helpful!

[harsh smelt]

np

[wide ocean]

I tried 5.6=log(Ao/ (1/4)
10^5.6 =Ao/ (1/4)
Ao = 99,526
but the answer is 5.0
I tried reversing Ao and 1/4 but can't seem to get 5.0

does anyone know which part i'm doing incorrectly?

<@&286206848099549185>

[fossil crown]

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

[uncut mulch]

pretty much all of it, where did your A go

consider starting with the magnitude of the aftershock (denoted with subscript a)

$M_{a} = \log_{10} \br{\frac{A_{a}}{A_0}}$

[obsidian monolith]

ramonov:

[uncut mulch]

and another equation for the initial quake ( i)

$M_{i} = \log_{10} \br{\frac{A_{i}}{A_0}} = 5.6$

[obsidian monolith]

ramonov:

[uncut mulch]

express A_a in terms of A_i and apply the appropriate log laws (and substitutions)

[still quarry]

can anyone give me a hint with this.

easy enough to do when actually given values but...

[uncut mulch]

apply the same methods

your asymptotes will just be in terms of a,b,c,d

[proven marten]

ye do you know the formulas?

[still quarry]

normally when given a quadratic for the denominator wouldn't you just factor it?

[willow bear]

but the denominator isn't a quadratic here

[still quarry]

exactly, so how can I approach cx + d for an asymptote if there are no values?

[willow bear]

when does cx+d equal zero?

[viscid thistle]

cx+d can still be zero

[still quarry]

isn't it equal to zero when not increasing or decreasing

[uncut mulch]

huh?

wdym?

you you mixing first derivatives into this?

[still quarry]

nvm not relevant

[uncut mulch]

eg how would you find the vertical asymptote for: $y = \frac{2}{x+4}$

[obsidian monolith]

ramonov:

[still quarry]

well I guess you could find the domain and then find the limit

-4??

[uncut mulch]

and what exactly are you doing to get -4?

and why not apply the same idea to your question

[still quarry]

hmm ok I will try it

[viscid thistle]

just solve for the denominator... roots are the asymptotes.

[uncut mulch]

technically this is a bit of a dodgy question

[rapid lance]

isnt this just (x+4)/2 = y^-1, so the assympotote is -4

its shifting x = 1/y

[uncut mulch]

if ax+b = k(cx+d) you'd have a hole not an asymptote

[rapid lance]

@uncut mulch isnt a hole just when something cancels, if nothing cancels you get an assymptote

its late, so i might be making a stupid mistake

[uncut mulch]

depends

[still quarry]

so x = -d/c and y = a/c

[uncut mulch]

yes, assuming this isn't the case

technically this is a bit of a dodgy question
if ax+b = k(cx+d) you'd have a hole not a vertical asymptote

[still quarry]

that would be confusing

[rapid lance]

@uncut mulch the definition ax+b = k(cx + d) is just saying if you can factor a k out from a and b, then you have a hole, here (a,b) = 1, so its an assymptote, basically a= k*c and b = k * d is a hole

there is also a horizontal assymptote of y=0

[uncut mulch]

whut?

[rapid lance]

isnt that the definition of the hole, ax+b = k(cx+d)

a = kc, b = kd?

i mean you could also verify by taking the limit at +/- inf

[uncut mulch]

if the variables had that property you'd have a hole, that's all i said.

[rapid lance]

i thought you implied that its a hole

[uncut mulch]

your use of (a,b) = 1 not sure what that means
and "is a hole" is inappropriate

[rapid lance]

mb

(a,b) = 1, means greatest common divisor

[uncut mulch]

if you intended gcd you should write gcd

also a and b don't necessarily need to be relatively prime for there to be a vertical asymptote

[rapid lance]

so (a,b) is correct form

[uncut mulch]

not really

you'd need to be in the specific context of dealing purely with gcd and even then gcd is pretty much always written

otherwise it'd most likely be interpreted as a point or interval

[rapid lance]

since i was talking about factors, it would be taken as gcd, note that i wrote if a = kc, b= kd, but that is untrue since (a,b) = 1

[uncut mulch]

uh still very bad and unclear especially since it wasn't quite true

and its best to remove all ambiguity

[viscid thistle]

I got this as my equation, is it right?

1=0.53^(52)
?!

1-0.03^52 *

[rapid lance]

i dont think that (1-0.03^52) is right

since that is really close to 1

[viscid thistle]

would it be 1-1.03^52 ?

[rapid lance]

after one year, what percent of the original is it?

then after the next year, what do you do to that?

[viscid thistle]

97%?

[rapid lance]

so you multiply by 0.97 each time

[viscid thistle]

oh

how does that fit into the equation

[rapid lance]

if youre multiplying by 0.97 every year where does than go?

[viscid thistle]

it's the r value?

[wide ocean]

I think it's 0.97/52 on bottom @viscid thistle

and the exponent is 52n

and let the left side a value < or=0

[opaque olive]

$\frac{(n-0)!}{(n-1)!} = n$

[obsidian monolith]

Yes:

[opaque olive]

it does right?

[acoustic harbor]

yes

[unreal reef]

woah lemme try to do the fancy scriptwriting thing

x^5 = 5

oh

lmao

$x^5=5$

[obsidian monolith]

Dirty Dan:

[unreal reef]

yay

[opaque olive]

yay

[rapid lance]

im proud

[still quarry]

how would you describe the end behaviour of a rational?

[wide ocean]

what is end behaviour

y=1/(x-4)

but there's still a horizontal assymptote at y=5/2

y=5/2(x-4)

[uncut mulch]

HA looks like its just 2

$\lim_{x\to \infty} f(x) = 2$

[obsidian monolith]

ramonov:

[uncut mulch]

$\lim_{x\to -\infty} f(x) = 2$

[obsidian monolith]

ramonov:

[unreal reef]

$\lim_{x\to \infty} f(x) = 2$

[obsidian monolith]

Dirty Dan:

[unreal reef]

how come the theme is different

[willow bear]

,tex --color white

[unreal reef]

o cool

got an asymptote at y = 2

[still quarry]

End behaviour.....like as x approaches a value, what value does y approach

[rotund lagoon]

Alright

[merry sphinx]

In response to your question

[merry sphinx]

Try to find f^-1 first

[rotund lagoon]

One sec

I believe this is the inverse

[merry sphinx]

Yep

Now you should know what to do

When you finish after subbing in g(x), shoe me your work and I can check it

[rotund lagoon]

I didn’t get x+1/b

Is there a step I missed?

[merry sphinx]

How did you

Get from step 1 to 2

You shouldn't skip steps

Wait

Nvm I see what you

Evaluate your answer

Distribute the 1/b and simplify @rotund lagoon

[rotund lagoon]

I just applied the sum/difference log laws

Got it @merry sphinx , thank you!

[merry sphinx]

Yeah np

[rotund lagoon]

I had tried to get the inverse of the whole composite function at first, which was really messy

[hearty vector]

@still quarry you can always use a sign line

And use arrow notation

and see what the behaviour is towards the vertical asymptote

i believe

[south flare]

1st page of a wrong solution

2nd page of the wrong solution

can someone point out my mistake?

[harsh smelt]

domain is fine

[void thorn]

x^2+y^2=16

-4 <= y <= 4

(-4)^2 <= y^2 <= 4^2

which is wrong

-2 < -1 doens't imply (-2)^2 < (-1)^2

question has been answered, we were discussing in voice chat

[south flare]

is right to say we cant square both sides in an inequality?

[willow bear]

not always

[viscid thistle]

@south flare yes we can, as long as both sides are positive

[azure junco]

Question 11

For this word problem I don't understand what im meant to be doing, I'd assume finding the y value of the maximum point? I've found the area in terms of X and y and substituted the value in to get A=4x((49-x²)/4)½

[proven marten]

Cant read it

[azure junco]

Sorry

Here

Is that better

[viscid thistle]

venimental

i am going to try solving this in my head

do you have the answer

let me know if it's correct

actually nvm lol

can't solve it in my head

[fading token]

[viscid thistle]

@fading token i thought the numbers were nicer at first 😦

I solved it in my head, show the answer key and I'll tell you whether my number is correct

[fading token]

[viscid thistle]

lmfao

no

i was going to ask him to confirm it

that'd be funny though

you gorilla @viscid thistle

Looks like a boring problem

[opaque olive]

https://gyazo.com/0ce8ff0ab6cbbc0a79dd80deca3c5eeb

Stuck on part 3

I could integrate $(1+x)^n$ but idk how i would get that -1 in $\frac{1}{n+1}(2^{n+1}-1)$

[obsidian monolith]

Yes:

[opaque olive]

only hints please

am i on the right track by doing $\frac{1}{n+1}(1+x)^{n+1} - \frac{1}{n+1}$

[obsidian monolith]

Yes:

[opaque olive]

the left hand side has clearly only been integrated

[viscid thistle]

@opaque olive any particular reason why you wouldn't use induction?

[opaque olive]

i dont need to?

[viscid thistle]

i'm not sure what you mean. induction is very standard in proofs

[opaque olive]

i did the first two without induction

[viscid thistle]

nice chain logic

[opaque olive]

lol

[viscid thistle]

it's like chain rule, except useless

[opaque olive]

wow

=/

[viscid thistle]

anyway

that kind of logic doesn't make any sense

[opaque olive]

so would you have did the first two with induction?

[viscid thistle]

not the first one

probably not the second one either

the first one takes 2 seconds to prove

i would probably do induction on the third one

[opaque olive]

hm

[viscid thistle]

i mean maybe there is some kind of nice combinatorial identity you can use

[opaque olive]

i dont think we are using induction in this question tho

[viscid thistle]

but since you're asking here, then you clearly don't know

what do you mean man.

the binomial expansion is literally proved with induction.

[opaque olive]

oh

ok

ill try induction

:/

ffs

i forgot the +c

thats why it wasnt working ...

[lime bolt]

use the identitiy 1/k+1 (n choose k) = 1/n+1 (n+1 choose k+1) @opaque olive

then it is trivial

[opaque olive]

isnt the obvious way just to integrate

set x = 0

[lime bolt]

no, by using this

[opaque olive]

ive just done it in 30 sec

[lime bolt]

well it depends on where u get this from

is this school stuff

[opaque olive]

no

[viscid thistle]

@lime bolt ah look

the identity i was talking about

lol

[lime bolt]

the identity i mentioned instasolved this

[viscid thistle]

i knew there was an identity

too bad i'm not big enough of a nerd to memorize these

how would you even remember that

[opaque olive]

whats the identity for

[lime bolt]

literally expand it it takes <1 seconds, polynomial

[opaque olive]

ok

[viscid thistle]

kane

tell me

what is the combinatorial argument for that

i'd love to hear

[willow bear]

the combinatorial argument for what

@viscid thistle

[viscid thistle]

@willow bear

use the identitiy 1/k+1 (n choose k) = 1/n+1 (n+1 choose k+1) @opaque olive
@lime bolt

this fucking identity lol

[lime bolt]

bruh try expanding

then u get it

[viscid thistle]

no that's not the same thing

i know the algebra makes it more intuitive

but the whole point of a COMBINATORIAL argument

is for it to be COMBINATORIAL

not "expand with algebra, simplify, and then it's obvious"

[lime bolt]

but like there isnt any simplyfing involved

[jade heron]

Tru

[opaque olive]

you guys sure like arguing

[viscid thistle]

it's not an argument lol

i'm just saying

there are like no combinatorial arguments for some of these "combinatorial identities"

[willow bear]

so (n+1) * nCk = (k+1) * (n+1)C(k+1)

this form admits a combinatorial argument!

you have n+1 people from which you need to pick a team consisting of k players and 1 captain (i.e. k+1 people in total)
both sides count the number of ways to do so

(n+1) * nCk counts it like this:

• pick one of the n+1 people to be the captain
• pick k players from the remaining n people

(n+1)C(k+1) * (k+1) counts it like this:

• pick k+1 of the n+1 people to be on the team
• pick one of the k+1 people on the team to make them captain

@viscid thistle that good enough for you?

[viscid thistle]

yes @willow bear

that's pretty nice

i'm very bad at making those arguments 😦

[willow bear]

shrug

guess that's just sth that comes with experience

doing a bunch of combinatorics problems, seeing patterns etc

[viscid thistle]

anyone know how to do this on desmos

[willow bear]

do what on desmos

a table like this?

[viscid thistle]

yeah

[willow bear]

oh

[viscid thistle]

and this i think

[willow bear]

that's actually fairly easy hold on

right so

equations for graphs can be entered as-is

like straight up write P(x) = x^4 + 11x^3 + ... on a line and it'll plot your thing

[viscid thistle]

Yeah

[willow bear]

click the grey plus icon in the top left, choose table, replace $y_1$ with $P(x_1)$ and then enter your $x_1$ values in the left col

[obsidian monolith]

Ann:

[viscid thistle]

Alright

it works thanks

[still quarry]

I need a second opinion: I think it might be a

[willow bear]

no

if f'(x) were 0 everywhere f would be a constant

@still quarry

[still quarry]

rough. well that cancels out a and d

[willow bear]

nope.

the correct answer is d. you can have a strictly-increasing function with f' not strictly positive everywhere.

for example, x^3

its derivative is 3x^2, which is zero at x=0, but the function x^3 is still strictly increasing.

[still quarry]

hmm ok so it can't just equal zero

[willow bear]

it can't be identically-zero

but it's not required to be positive everywhere

[still quarry]

i see, thx

also, would this be an inflection point?

[open perch]

hello can anyone help me on this problem, my friend and I tried and really don't quite understand, so we took a guess

[willow bear]

why not just add those fractions and simplify

[open perch]

I did try that but my answer was not one of the choices

I checked my answer with photomath too :")

[willow bear]

aight lemme see

what answer did you end up with

[open perch]

I got this

[versed jasper]

I guess you used tan half-angle identities at some point?

try cross-multiplying the fraction directly without using any half-angle substitution

[open perch]

should I start from the top, and just cross multiply the original problem

[versed jasper]

yes from the original problem

[willow bear]

uh

i

think you overcomplicated this

[open perch]

ahhh

I got this from cross multiply

[willow bear]

what

[open perch]

UH

[willow bear]

can you show all of your work bc this looks weird

[open perch]

I thought that too

okay

i don't think I'm supposed to cross multiply lol

[versed jasper]

please show your work

also make sure that (a + b)^2 = a^2 + 2ab + b^2 and not a^2 + b^2

[open perch]

I felt like I was not supposed to solve the problem

:"))

[versed jasper]

can you work out $(1 - \cos u)(1 - \cos u)$ ?

[obsidian monolith]

Prame_tan:

[open perch]

yeah

[willow bear]

no.

$(1-x)(1-x) = 1-2x+x^2 \neq 1-2x$

[obsidian monolith]

Ann:

[open perch]

so i got the wrong answer form multiplying the cos :"D

[willow bear]

you didn't expand (1-cos(u))^2 properly.

[versed jasper]

what about the fraction, can you proceed from now on?

[willow bear]

wym

[open perch]

I'm so sorry I'm getting more confused

[willow bear]

...

i think i'm gonna go.

[versed jasper]

I mean the rest of the problem

[open perch]

haha I don't know what to do with that fraction,,it doesn't match anything and I can't simplify it more 😂

[versed jasper]

so let's do this step by step, is that ok?

[open perch]

alright

[versed jasper]

after the example by Ann, can you now work out $(1 - \cos u)(1 -\cos u)$ ?

[obsidian monolith]

Prame_tan:

[versed jasper]

remember FOIL

[versed jasper]

yup that's correct

[open perch]

yayyy

[versed jasper]

$\frac{1 - \cos u}{\sin u} + \frac{\sin u}{1 - \cos u} = \frac{1 - 2\cos u + \cos^2 u + \sin^2 u}{(\sin u)(1 - \cos u)}$

[obsidian monolith]

Prame_tan:

[versed jasper]

So now we are here, right?

[open perch]

so u cross multiply and added it?

mind if my friend joins in, she is working on the same problem

[versed jasper]

Yes, by cross-multiplying, I mean
$\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$

[obsidian monolith]

Prame_tan:

[versed jasper]

which is the way to add fractions

btw let them in, we can learn together

[frail imp]

Hello

[versed jasper]

hi

so can you arrive at this point?
$\frac{1 - \cos u}{\sin u} + \frac{\sin u}{1 - \cos u} = \frac{1 - 2\cos u + \cos^2 u + \sin^2 u}{(\sin u)(1 - \cos u)}$

[obsidian monolith]

Prame_tan:

[frail imp]

Yes

[open perch]

that makes sense

[versed jasper]

ok then

does the term $\cos^2 u + \sin^2 u$ familiar to you? Is there any identities related to this?

[obsidian monolith]

Prame_tan:

[open perch]

yes

[versed jasper]

what is it?

[versed jasper]

yes that's it

[open perch]

YAYYYY

thank you so much

I appreciate it ahhh

[versed jasper]

you're welcome 🙂

[frail imp]

Thank you soo muchhhh

[still quarry]

would you say that as x approaches -infinity, y approaches 1 in this case or...

[stuck lark]

not entirely sure, you should show more of the graph to the left of the y axis to get a better look, but y probably does approach 1 in the limit

[still quarry]

for reference its x^2 - 1/x^2 - 4x +4

vertical asymptote is x=2 and horizontal is y=1

[stuck lark]

parentheses @still quarry

[still quarry]

{{x^2} - 1}/{{x^2} - 4x + 4}

[stuck lark]

would you say that as x approaches -infinity, y approaches 1
then yes this is ok

[still quarry]

so in this case how do we view this part:

is it one of those things where we kind of ignore that it dips like that

[stuck lark]

wdym ignore

[still quarry]

well we could still say that the horizontal asymptote for this line is 1 right

[stuck lark]

what's your definition of horizontal asymptote

[fleet yew]

@still quarry asymptotes are specifically about the limit as x->+-infinity

Local behavior doesnt really matter

[still quarry]

oh ok, I didn't know about the local behaviour before thanks

[latent grotto]

Is (b) correct?

there is no remainder?

[upbeat bone]

Yea @latent grotto

[latent grotto]

Thanmks

How about this? Could somebody check my answer

[acoustic harbor]

does not look right @latent grotto

[latent grotto]

How do I fix it

do i place the remainder inside the breackets?

[acoustic harbor]

confused to why your first term is x^2

should be x^3

[latent grotto]

oh lord

FUCK

[acoustic harbor]

2x*x^2 does not equal to 2x^4

[latent grotto]

but remainder is supposed to be on the outside right?

[acoustic harbor]

the remainder should be written over 2x+7

but i’m pretty sure your remainder is wrong since you fucked up the first step

[latent grotto]

Yeah, ok ill go fix it thanks for pointing it out

[wide ocean]

I tried 10=5(1+ (0.06/2) )^2n

2=(1.03) ^2n

but i'm getting 155.7 periods, so 77.8 years

but the answer is "23.4 periods, so 11.7 years"

does anyone know what i'm doing wrong for 8a)?

[latent grotto]

Not really sure what this is asking me to do

where do I put the 2,3 and 4

is it just $x^4 + x^3 + x^2$

[obsidian monolith]

JOHN CENA:

[patent beacon]

What are the zeroes of x⁴ + x³ + x²?

[latent grotto]

Idk

oh zeroes is the same as the roots of x?

$(x-2)(x-3)(x-4)$ ?

[obsidian monolith]

JOHN CENA:

[patent beacon]

Done haha

[latent grotto]

lol thanks

Are these both correct

[upbeat bone]

Please use parentheses next time 😐

But yeah those are already correct @latent grotto

[latent grotto]

Oh yeah

Thank you

[viscid thistle]

Before I even ask, I want to let you guys know that I am bad at Mathematics.
If I were to have $1.50 funds on my account and it was $2.99/1000 tokens, how many tokens would I have remaining? It means that I have spent $1.49 (X tokens).

I want to be able to convert my funds into tokens if that is possible.

[still quarry]

can someone please help me break this down

[viscid thistle]

it's giving you 3 values of the function

the points at which the slope of the function is 0 occurs only at x = 2, x = 4

x = -6 and x = 0 are holes

(in the derivative)

then the graph of the derivative is below and above the x-axis for those intervals listed

[still quarry]

so will it be a piecewise?

or no jump

[viscid thistle]

idk lol

[still quarry]

hmm isee

[viscid thistle]

i guess piecewise is one way to look at it

(-2,-1) is a point of inflection

so in the 2nd derivative, it changes from concave up to concave down or vice versa

[still quarry]

so if -6 and 0 are holes in the derivative, how does that contribute to the original function?

[wide ocean]

Since 9 and 8 can't be put to similar base

[viscid thistle]

take log of both sides

use log rules

[wide ocean]

is it 3xlog9 -10log9 = xlog8+6log8

[viscid thistle]

no

you move the whole exponent infront.

but before that

i would suggest combining everything into one log

and just using the definition of a log

[wide ocean]

what do you mean

[viscid thistle]

i mean exactly what i said

[wide ocean]

(3x-10) log 9 = 3xlog9 - 10log9.

[viscid thistle]

you will mess yourself up doing that

dude

are you even listening to me

@wide ocean i'm getting really annoyed

you have $9^{3x - 10} = 8^{x + 6}$

[obsidian monolith]

polynomial:

[viscid thistle]

you take log of both sides $\log 9^{3x - 10} = \log 8^{x + 6}$

[obsidian monolith]

polynomial:

[wide ocean]

Taking the log of both sides

What do you do after that.

[viscid thistle]

sigh

$6x-20 = (3x+18) \log_3 2$

[obsidian monolith]

polynomial:

[wide ocean]

Where is 6x-20 from log 9 ^3x-10

[acoustic harbor]

(6x-20)ln(3)=(3x+18)ln(2)

change 9 to 3^2

3^6x-20

[viscid thistle]

not sure about that one chief

[wide ocean]

@viscid thistle your logic is unclear.

[viscid thistle]

my logic is very clear

you just don't know log properties

actually

i guess your way will work too

but i wouldn't do it that way since i find it longer

but you do you

[odd helm]

how would i do this problem

[cyan prism]

depends on the situation. If it was timed, I'd probably just try answers until one worked. In this case, S_1 is clearly just the first term, so 1/15. There is only one answer choice where S_1 = 1/5.

The more correct way of doing the problem is making sure the formula works in general. Let's just take a as an example

if $S_n = \frac{n}{6n+9}$, then $S_{n+1} = \frac{n+1}{6(n+1) + 9}$

[obsidian monolith]

abnew123:

[cyan prism]

But we also know $S_{n+1} = S_n + \frac{1}{(2n+1)(2n+3)}$ since the general form of a term is given

[obsidian monolith]

abnew123:

[cyan prism]

So then you can plug in the expressions for $S_n$ and $S_(n+1)$ and see if they work

[obsidian monolith]

abnew123:

[viscid thistle]

budget induction

[heady jewel]

It also telescopes and is not really that long to actually find out the sum

Tricks aside

Hint:|| Try to write the 1 as a difference of the two factors in the denominator and break the fraction||

Also , || 2(i+1)-1=2i+1 incase you dont know what im talking about,this would make a pair which would add up to 0||

[chrome geode]

if f(x)=1/3x-1 and g(x) = 1/x what is (fg) x

[uncut mulch]

do you have a pic of how the fg part is written?

[chrome geode]

it is (f times g) (x)

[uncut mulch]

(fg) x is not appropriate function notation

[chrome geode]

composite function

[uncut mulch]

$(fg)(x)$ or $(f \circ g)$ or $(f \cdot g)(x)$

[chrome geode]

third

third

[obsidian monolith]

ramonov:

[uncut mulch]

like a clear black dot between the f and the g?

[chrome geode]

yes

[uncut mulch]

then its not a composition

[chrome geode]

its not a clear black dot

its the last option

the full dot

thats a multiplication sign

[uncut mulch]

$(f \times g)(x)$?

[chrome geode]

$(f \cdot g)(x)$

[obsidian monolith]

ramonov:

smcgea1223:

[chrome geode]

that

[uncut mulch]

that's equivalent to

$f(x)g(x)$

[obsidian monolith]

ramonov:

[uncut mulch]

the product of the two functions

[chrome geode]

yeah

[uncut mulch]

so just multiply your 2 functions together

[chrome geode]

oh ok

1/3*1/x-1?

[uncut mulch]

ugh, can you actually make your expressions clearer with appropriate parentheses

[chrome geode]

sorry about that

[uncut mulch]

$f(x) = \frac{1}{3x-1} , g(x) = \frac1x$

[obsidian monolith]

ramonov:

[uncut mulch]

that?

[chrome geode]

yes

[uncut mulch]

just multiply them

together

[chrome geode]

okay thanks for ur help!

[uncut mulch]

if its a multiplication dot, its not asking for the composition

[chrome geode]

oh okay

got it

thanks!

[viscid thistle]

Hello! I got the A and D but I dont know how or what to do to solve B and C

Not sure what is break-even point is

[blazing parrot]

Break even point is explained in the question as the number of patrons for which there is no profit

b is similar to d in that you set the profit to a number and solve for patrons, but in B they expect you to understand from the question that you need to find when profit = 0

for C, they want you to change the equation for profit in terms of patrons to one of patrons in terms of profit

do you know how to find inverse functions?

@viscid thistle

[viscid thistle]

Ohh

So I have to find inverse function of y=9.5x-1450?

I got B! Thank you

I got it thank you😭😭

@blazing parrot

[blazing parrot]

nice!

[solemn elbow]

Are all rectangles parrallelgrams?

parallelograms.*

[willow bear]

are they?

[solemn elbow]

they must be since all rectangles have 2 pairs of parrallel sides and opposites are equal

[narrow peak]

i didnt know that LOL

just realized tbh

[proven marten]

VSauce moment

[latent grotto]

How do I factor this

[fluid shore]

@latent grotto I mean, you can immediately see that x = 1 is a solution

so what can you do with that?

[latent grotto]

not sure

I need to find all the roots

of x

[fluid shore]

Well, you know that it will have one linear factor and one quadratic factor. You can possibly reduce the quadratic, depending on whether you consider complex roots to be roots. So, let $ax^2+bx+c$ be the other quadratic. Then:

$$x^3+8x^2+5x-14 = (x-1)(ax^2+bx+c)$$

So, what can you do with that?

[obsidian monolith]

Abhijeet Vats:

[latent grotto]

Idk ?

[fluid shore]

..........

Like, what do you know?

[proven marten]

What if you didnt know the 1 though 👀

[fluid shore]

Then you can still just factorize this with grouping:

$$x^3+8x^2+5x-14 = (x^3 + 7x^2)+(x^2+5x-14) = x^2(x+7)+(x+7)(x-2)$$

$$x^3+8x^2+5x-14 = (x+7)(x^2+x-2) = (x+7)(x+2)(x-1)$$

[obsidian monolith]

Abhijeet Vats:

[proven marten]

I-

I thought you would explain tahat to John Cena

but that works too

[latent grotto]

how did u group it

oh

i see

[fluid shore]

Oh it was a little bit of this and a little bit of that.

For me, I tend to try out small numbers like 1 or 0 or -1 in order to see if they yield solutions

If I can't see a way to group it immediately, that is. Then, I'll use that one soluton in order to obtain the others by way of comparison.

But if you get enough practise with this stuff, you sort of start to see how to do it

[latent grotto]

Is this right? i got a=2 so i assumed that was correct

[proven marten]

Is this for a test man

[latent grotto]

but i also think it should be (x+5)(x+1/2)(x+3)

quiz

[fluid shore]

are you doing a test now?

[latent grotto]

Ya a quiz

[fluid shore]

That is not allowed

You cannot ask for help on a quiz or test

[latent grotto]

Oh

[fluid shore]

<@&268886789983436800>

[proven marten]

Thank you for being honest though

[wind igloo]

👀

[latent grotto]

np i guess im banned now

[proven marten]

No, I think its a warning first

[fluid shore]

i think a stern warning will do

[wind igloo]

Depends on the attitude of the person. In this case, I agree. Don't do it again.

[viscid thistle]

can someone explain why i have to take the natural log of both sides to solve for x

[fast slate]

Logs are what let you get rid of exponents

So $\ln(5^{x+3}) = (x+3)\ln 5$

[obsidian monolith]

owomorphism:

[viscid thistle]

thanks

[viscid thistle]

Beautiful

Welcome to the club

[viscid thistle]

Why do we need to rationalize the denuminator just like the solution on the picture? What's the difference if I have square root of 2/3?

[willow bear]

none

except maybe stylistic

[proven marten]

dont 'need' to

[viscid thistle]

For the sake of not having a sqrt in the denominator

[uncut mulch]

easier to work with if the denominators are rational

it just became convention at some point in time

[viscid thistle]

Thanks

[uncut mulch]

(which gets thrown out the window when variables and other things are involved)

[viscid thistle]

Could someone explain why this is clockwise and not counter clockwise?

[charred hull]

within -360 and 360

if you want to find a coterminal domain while going clockwise, the measure will be greater than 360 degrees
if you want to find a coterminal domain while going counter-clockwise, the measure will be greater than 360 degrees

@viscid thistle

[viscid thistle]

What ? Sorry im still confused

if it's counter clockwise, the angle should be a negative?

@charred hull

[charred hull]

yes

wait

sorry, mistyped

[viscid thistle]

is it the other way around

[charred hull]

if you want to find a coterminal domain while going clockwise, the measure will be greater than 360 degrees
if you want to find a coterminal domain while going counter-clockwise, the measure will be greater than 360 degrees

the second is what i meant

[viscid thistle]

Ok I understand thank you

[proven marten]

,rotate

[proven marten]

Not to be rude

but it says

Calculus

So is this really the appropriate channel?

[frail imp]

Well... I was thinking maybe it could go either way, because we learned a bit of it in precalc

But I will put it in calculus, sorry

[viscid thistle]

for this

do I have to change 60 degrees to radians?

[acoustic harbor]

yes

[viscid thistle]

how do you do that?

[rich flint]

You use the conversion that 180 degrees equals pi

[acoustic harbor]

multiply by pi/180

[viscid thistle]

alright

[rich flint]

Sorry didn't see you were already getting help i apologize

[viscid thistle]

Np

Should I keep my answer as 4pi or 12.5

[acoustic harbor]

4pi

never 12.5

[viscid thistle]

Ok

How do you do these? All I know is that sec = 1/cos(x) and csc = 1/sin(x)

no i haven't used special triangles

How?

No

I haven't seen them in my notes

x/y

?

x/r*

[obsidian monolith]

slimvesus:

[viscid thistle]

oh so x would be pi and r would be 4 ?

$\frac{1}{\cos(\x/r)}$

idk how to do it

cos(x/r)

1 / cos(x/r)

oof

pi/4 ?

r/x

4/pi ?

x/r

wut

oh

lol

yeah

Yes

r/x

nope

Yeah

i've never worked with radians

Yeah

So the angle is 45 degrees?

Alright

Yeah

hypotenuse would be 4?

and adjacent is pi

not sure

No

I think it was part of my class from last year but we weren't able to go over it because of the virus

so for the 45 45 90 triangle the sides are always 1 1 and sqrroot 2

so square root 2/1

[obsidian monolith]

alfred:

[viscid thistle]

Alright thanks

What type of triangle is 210 degrees?

the 30 - 60 - 90 one?

Oh

So the angle that I write for the triangle would be 210?

cos(210) =root 3/2

I thought the negative 2 would cancel out?

is the 2 negative?

Yeah I got it now

Thanks for your help i appreciate it

[wide sequoia]

How do this?

[viscid thistle]

shouldn't k be n ? or k=1 instead of n=1

@wide sequoia

[odd helm]

what would i be solving for in this problem

i don’t see a variable i just see combinations

[uncut mulch]

show that the equality holds

well each side i guess

show that they're the same value

and choose that value

its a poorly worded question

[patent beacon]

Give me 8C4
Give me 7C4 + 7C(4-1)

They both should be the same number. What is it?

[odd helm]

i didn’t get the same on both sides on the left i got 20160 and on the right i got 21420

[patent beacon]

,w 8C4

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Kaynex#7520

[patent beacon]

Good

[viscid thistle]

21420

[odd helm]

lol

[patent beacon]

I'm getting 8C4 = 70

[odd helm]

i did 7!/(4)!(3)!

is that right?

[patent beacon]

I used a calculator that has C on it rofl

[odd helm]

oh i’ll try that lol

i forgot that u could do that

[patent beacon]

You may want to check if yours does

[odd helm]

ye got 70

ty

[patent beacon]

Also 8C4 = 8! / 4!4!

Parenthesis is likely the problem you had

[uncut mulch]

()
you could use factorials if you want. the numbers are small enough to do it by hand if you don't have access to a calculator

[narrow peak]

ok boyos ima need someone to check my work

let me test my latex

$log_2(5x+3)$

[obsidian monolith]

Hmm:

[past meadow]

\log

[obsidian monolith]

Commander Vimes:

[past meadow]

your x looks like a chi

[harsh smelt]

no u

[narrow peak]

$log_2(5x+3)-log_(5x+3)²=1\2log_2(5x+3)-\frac{1}{log_2(5x+3)}=1\\frac{4[log_2(5x+3)]²-1}{log_2(5x+3)=1}\4[log_2(5x+3)]²-1=log_2(5x+3)\u=log_2(5x+3);\4u²-1=u\4u²-u-1=0$

[willow bear]

\log

also you messed up in other places

[narrow peak]

mess what up

[past meadow]

you wanna fix this one up

[narrow peak]

oh fuck

[obsidian monolith]

Hmm:

[narrow peak]

i swear i messed up somewhere

oh my god i fucked up latex

[willow bear]

you messed up in a lot of places

[narrow peak]

again

[willow bear]

maybe it's easier if you just write it out on paper

[narrow peak]

u know what

ill take a pic

dog tier handwriting my bad

[willow bear]

so far so good

[narrow peak]

oo alright thx

i just feel like i fucked up somewhere tho

,w factor 4u²-u-1

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Hmm#2554

[narrow peak]

bruh

[past meadow]

\

,w 4x^2-x-1

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Sneaky#0928

[narrow peak]

epik

am doing a practice paper and usually the results are nice

not this time i guess

[uncut mulch]

seems there actually is a mistake

when you were combining the fractions

[narrow peak]

oh fhck

fuck

[uncut mulch]

the coefficent of 2 changed to 4

[narrow peak]

i squared the entire thing

so 2²

oh

uh

UH

gimme a sec

[uncut mulch]

also $\log_2(5x+3)^2$ is a bit ambiguous

[obsidian monolith]

ramonov:

[narrow peak]

i know

like the entire thing squared

so no using power law or whatever u call it

[uncut mulch]

$(\log_2(5x+3))^2?$

[obsidian monolith]

ramonov:

[narrow peak]

yes

i am latexing on phone

hard

[past meadow]

3.75=15/4

[uncut mulch]

in that case you'd have to restart from the beginning

[past meadow]

thats where taking an extra 3.75 off comes from

[narrow peak]

oh the first line u meant?@uncut mulch

that one is correct

ill just put the qn here

[uncut mulch]

$\log_2((5x+3)^2) = 2\log_2|5x+3|$ \
but $(\log_2(5x+3))^2$ doesn't represent the same thing

[obsidian monolith]

ramonov:

[narrow peak]

i know

[willow bear]

5x+3 is required to be positive anyway ramonov

the absolute value bars can be dropped here

[narrow peak]

Solve the equation $log_2(5x+3)²-log_5x+3(2)=1$

fuck

dog keyboard

[obsidian monolith]

Hmm:

[uncut mulch]

force of habbit

take a pic

[narrow peak]

how the fuck do i put a base with linear function