help-28

[olive saffron]

Yeah so this is the right idea if there was 5% compound interest per month

but the question is worded strangely

[red frost]

it says it's the same as the 5% compound intrest one

[olive saffron]

Yes

So the monthly interest rate is 5%/12

[red frost]

ok

[olive saffron]

Have you seen the formula for compound interest

[red frost]

yes

money x multipler ^ number of years

[olive saffron]

Yeah that'll work

But it doesn't have to be years

it can be any random unit of time

[red frost]

ik

[olive saffron]

Alright good

So you know the number of months

you know that the monthly interest rate is 5%/12

[red frost]

0.00416666666

i asked google

0.05/12

[olive saffron]

alright yeah a calculator would work also

do not rely too heavily on Google and the like

they will make it easier but you will not have them in the exam

you should also know that sparx works by giving you harder and harder questions depending on how well you do

if you don't know how to do a question and brute force it with online tools etc you're only going to dig yourself into a hole of very difficult questions you don't know how to do

Trust me; i am a teacher and I use sparx for my students and I have to manually reset their account level all the time because they "cheat" at sparx and then it gets too difficult for them

[red frost]

oh ok

i know

this one

i've literally been one for so long that i cant move

that is why i am asking for help here

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[unique dagger]

I think I found the area outside the snowflake but I cant find the area of the hexagon because I cant find its radius

[torpid stag]

The radius of a regular hexagon is two times the length of one of the sides.

(You can split the hexagon into 6 triangles of the same size to see why)

[unique dagger]

oh ok cool thanks

if thats the case I have got the outside areas wrong

the areas of the little triangles should be sqrt(3)/4

but the spikes idk how to find their area

[full forum]

@unique dagger Has your question been resolved?

[tender tapir]

Did you try to split it into smaller shapes?

[unique dagger]

ye tried splittinging into a triangle and 2 trapeziums but I dont know the length of the parallel sides of the trapeziums

[tender tapir]

the bigger trapezium should be the same as the triangle, but with the tip removed. The tip being also an equilateral triangle, but with a side of 1.

[unique dagger]

oh ye

then what about the small trapezium?

[tender tapir]

My guess would be also an equilateral triangle with the tip removed, but this time the full triangle having a side of 2 instead of 3

[unique dagger]

so the small trapezium would be 3sqrt3/4

?

[tender tapir]

Ye

[unique dagger]

then would that mean that the area of the regular hexagon is 768sqrt(3)

I got that from 6(16^2)sin(pi/3)

[tender tapir]

You mean the hexagon that circumscribes the snow flake?

[unique dagger]

yep

[tender tapir]

What's the length of a side of the hexagon?

[unique dagger]

8

[tender tapir]

Ye

[unique dagger]

I think

I got P as 95.7 but that seems rlly big

[tender tapir]

,ask area of regular hexagon

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[tender tapir]

,ask 3/2*8^2

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[tender tapir]

How the hell did you get 768?

[unique dagger]

3r^2sin(60)

and r is 16 because the radius of a regular hexagon is 2 times the length of its side

[tender tapir]

no, it's not

[unique dagger]

oh

[tender tapir]

The radius of a circle that circumscribes it has the same length as a side

[unique dagger]

ohhhhh

thank you

65.6 seems more realistic

[tender tapir]

you can divide it into equilateral triangles, which makes it more obvious what the radius is

[unique dagger]

ohhh

if u use cosine rule u get s^2=r^2+r^2-2r^2cos(60)

which gives s=r

[tender tapir]

Or you know that 60° makes it equilateral 😅

[unique dagger]

some dude before told me the radius of a regular hexagon is twice the length of its side length so I just went with it

[unique dagger]

oh ye ;-;

well thanks for ur help bro

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[elder silo]

Hey guys, whole class got stressed by this and we can't solve it for some reason. Could anyone help?

[elder silo]

I don't really know math in English so a written solution would help a lot

On the second equation we simplified it to
lg(3x-y) = lg10
3x-y=10

3x=10+y

I believe we would need to substitute x with y's

[full forum]

@elder silo Has your question been resolved?

[tender tapir]

Did you find a way to simplify the first equation? Second looks fine, although whether to substitute with x or y depends on what will make the first one easier.

[elder silo]

No, haven't found a a simpler for the first

[tender tapir]

what if you take lg on both sides?

that's the usual approach when dealing with variables in exponent

[elder silo]

That would only make it more complicated imo

[tender tapir]

Did you try to apply logarithmic identities to it though?

[elder silo]

sir, I do not even know these words

To be real I haven't paid attention in math class since 9th grade

I just use common sense

[tender tapir]

$\lg(x^y)=y\cdot\lg(x)$

[glossy valve]

Kex

[tender tapir]

looks familiar?

[elder silo]

ohh

right

[tender tapir]

so what do you get for the first equation?

[elder silo]

(3x²+7x-6)lg y=lg 1 ?

[tender tapir]

what's lg(1)?

but ye

[elder silo]

is it lg 10?

[tender tapir]

lg(1)=0

[elder silo]

yeah

but on the right side am I supposed to do lg 10 or lg 1?

cause if it's lg 1 then I did in fact messed up the second equation

[tender tapir]

You applied 10^() to second equation, not log10()

10^1=10 and log10(1)=0

[elder silo]

3x-y = 0 is what would be correct then?

[tender tapir]

no 3x-y=10

[elder silo]

but then on the right side of the first equation, why isn't it 10 too?

[tender tapir]

because you took log10()

log10(1)=0

and in second you took 10^(1)

[elder silo]

alright I really don't get this part

both right sides were 1

[tender tapir]

just leave out the second equation for now

[elder silo]

and I put lg on both

[tender tapir]

let's just focus on the first

(3x²+7x-6)lg y = lg 1 = 0 is fine?

[elder silo]

yep

but multiplication can only be zero if one of the members is 0

or whatever you call them in English

[tender tapir]

exactly

that's how you solve it

[elder silo]

so either lg y is 0

[tender tapir]

nope

lg(y) = 0 <=> 10^y = 1

[elder silo]

huh

[tender tapir]

Exponential function is the inverse of the logarithm

[elder silo]

yep

but why can't lg y be zero?

if y=1 it would be zero

[tender tapir]

yeah, you're right

I though you meant y is 0 🙈

for y=1 is lg(y)=0

what about the second term?

[elder silo]

that would be a quadratic equation which I am perfectly capable of solving of course

is that what it's called in English?

[tender tapir]

yeah it's a quadratic equation

[elder silo]

so I didn't even need the second equation to solve it?

[tender tapir]

You do to get the other variable, since the solutions are always a pair of x,y values

[elder silo]

I have to put the newfound y values in

[tender tapir]

Is 3x-y=10 fine for the second or still confusing?

[elder silo]

well I think I have enough knowledge to do it now

yep I got it from here

thanks man!

how'd I close this?

[tender tapir]

.close

[elder silo]

.close

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[swift fulcrum]

Hello

[swift fulcrum]

I'm sort of confused of what this is saying

Does both of these matrices have the same rank?

[torn jolt]

no the situation you described applies to point 3 in what they wrote

[swift fulcrum]

But even if n ≠ 1?

[torn jolt]

yes

point is that, rank(A) = 2, rank(A|B) = 3, 2 < 3, and also 0 is not equal to 1 (from the last row), so the system is inconsistent

[swift fulcrum]

So if there's any number other than 0 in the b column, it has a rank in that column?

[torn jolt]

it could still be linearly dependent

[swift fulcrum]

Linear dependence?

Isn't that related to span?

[torn jolt]

take the situation where you have b = (0, 1, 0) here. This then falls under point 2

[torn jolt]

sure, but all of these statements are equivalent once you consider what they mean

[swift fulcrum]

Ohh ic

Tysm!

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[white cradle]

find least solution of the equation

[jovial lagoon]

bring the powers down cancel out the fractions add up all the logs and divide by the coefficent then exponentiate to get x

[white cradle]

thanks

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[feral escarp]

was this done incorrectly? i have a different answer from the solution

[tender tapir]

You aren't done yet though

[feral escarp]

not RREF, just REF

wait leme try

yea it only wants row echelon , not reduced

[tender tapir]

So you're not supposed to calculate the solutions of the elimination?

[feral escarp]

nope, just put it in row echelon form

and determine the rank

which is 3

[tender tapir]

,rotate

[tender tapir]

Well, let's see

,ask gaussian elimination {{1,1,0,1},{2,2,1,4},{1,2,3,4}}

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[tender tapir]

,ask gaussian elimination {{1,1,0,1},{0,1,3,3},{0,0,1,2}}

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[tender tapir]

,ask gaussian elimination {{1,1,0,1},{0,1,1,2},{0,0,2,-1}}

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[tender tapir]

Curious. First one is the original matrix, second is your solution and third is the supposed answer

which doesn't match with the original

[feral escarp]

hmmm

[tender tapir]

Either there's a typo in the original matrix or the answer is wrong

your solution is correct for the given matrix, as you see

[feral escarp]

okay thank you! I’ll consult with my TA to find the right answer :)

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[thin scroll]

What do these symbols mean and how do I complete the behaviour statement

[umbral dome]

• means 'from the right' (along the number line), i.e from higher values

and vice versa for -

[thin scroll]

number line as in x axis?

[umbral dome]

if you want to think of it that way, yes

[thin scroll]

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[clever helm]

One message removed from a suspended account.

[clever helm]

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[shrewd hamlet]

correct

[clever helm]

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[clever helm]

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[shrewd hamlet]

whats there to move on

simplifying?

[clever helm]

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[shrewd hamlet]

ok any common terms in numerator?

[clever helm]

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[shrewd hamlet]

just x or x to the power of something

[clever helm]

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[shrewd hamlet]

yes

[clever helm]

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[shrewd hamlet]

ok now what

[clever helm]

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[peak compass]

The point (-1,3) bisects a rope in the circle x^2 + y^2 = 20. Find the equation and the lenght of the rope

[peak compass]

!status

[full forum]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[peak compass]

1

[full forum]

@peak compass Has your question been resolved?

[peak compass]

<@&286206848099549185> i just need something to start with

pls

.solved

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[potent swan]

How exactly would i solve this?

[potent swan]

I know it has something to do with maybe the gradient vector but my lecturer isnt the biggest on actually explaining.

[celest bay]

why dont u first compute the gradient vector at a general (x,y)

[potent swan]

ok

[potent swan]

Gradient vector of f(x,y) = (2x+y, 2y+x)

[celest bay]

yup

so now maybe plug in (x,y) = (-1,1)

into this

[potent swan]

so (-1,1)

[potent swan]

.

[celest bay]

yup

so the direction the function increases the most in is the gradient

and the direction it decreases the most in is negative of the gradient

[potent swan]

Ok you kind of lost me so

[potent swan]

so are you saying that the function increases towards this vector because its in the gradient?

or?

[potent swan]

or so do you mean that it increases the most along this vector

[celest bay]

yes

[potent swan]

oh ok thanks

Then this is one more that i dont get

[celest bay]

hmm there's a formula for this they must've taught you

we can also derive it from first principles if they didnt

[potent swan]

Im not sure we dont have anything mentioning distance

[celest bay]

ok

I keep missing this so you should just reply to my message if I miss it

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[inland hound]

can someone help me with this

[inland hound]

i dont know how to start

[silver thorn]

You want to plug in the line equation for the x y and z in the plane equation and solve for t

So for example replace x in the plane equation by -2+t

Do that for y and z as well

[inland hound]

ohh okok

[silver thorn]

Then just use algebra and solve for t

[inland hound]

okay ill do that right now

Is this correct?

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[real silo]

Guys, i was asked to solve the simultaneous equations (highlighted yellow).

[real silo]

the answer on the mark scheme is x = −1, y = 3 or x = −5/2, y = 15/2 (but the mark scheme might be wrong)

i don't understand why the x is not x = 1

[devout valley]

You can't have x = 1, as absolute values must always be nonnegative, to make -3x nonnegative, you'd need x <= 0

in particular, x = 1 would imply y = -3, but |2x^2 - 5| can't be negative

Whereas x = -1 implies that y = 3, and that actually is a valid solution, 3x + y is 3(-1) + 3 = 0, and |2x^2 - 5| is |-3| = 3 as it should be

[real silo]

ohhhhh

i get itt

[real silo]

thank youuu so so much!

[devout valley]

always a pleasure

[real silo]

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[graceful peak]

how would you start solving this?

[compact briar]

does x^(-1) count?

[valid sorrel]

Try to construct that polynomial

[sharp flame]

you can only have non-negative powers of x for a polynomial

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[brittle steeple]

That's not a polynomial

So no it doesn't make sense

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[valid sorrel]

Oh yea ur right

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[spring salmon]

I am lost

[brittle steeple]

Shouldn't it be 12/5 instead of 13/5

[spring salmon]

could you plese kill me 😭🤣🤣🤣🤣🤣

god damn

thank you

answer is 4 thanks

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[iron sand]

How to solve this systém of equations in Z_11

3a+b=0
a+4b=0

[iron sand]

I’ll get 3(-4)b+b=0

And that’s -b+b=0

So 0=0

?

[proven dirge]

this step is incorrect

3(-4) <-- this is a multiplication not a subtraction

it's more 3 * (-4b)

[fast peak]

-12=-1 in Z_11. thats correct

[iron sand]

Yes

[fast peak]

your steps are correct

[proven dirge]

oh is this a particular system?

base 11?

[iron sand]

Hm I’m trying to determine whether set of vectors {(3,1),(1,4)} I’d linearly dependent in Z_11

Then what am I doing wrong

[fast peak]

nothing

why would you think that

[fast peak]

modulo 11

[iron sand]

So what does this mean then?

For linear dependence?

[fast peak]

well if you had a system over the real numbers
3a+b=0
12a+4b=0
and you did some stuff and one of the equations got turned into 0=0, what would you conclude

[iron sand]

Hm that this holds for any a and b?

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[green merlin]

Yo

[hoary ember]

ABC is a triangle where AD is perpendicular to BC, BE is the angle bisector of ABC, F is the intersection of AD and BE, if AFB=BEC then find BAC in degrees

[green merlin]

Wow wonderful diagram

[hoary ember]

ikr

if someone can draw a better diagram would be much appreciated i cant draw shit

[green merlin]

Same

Even after doing so much olympiad geo

Anyway ngl D, F, E, C pretty concyclic

[hoary ember]

ugh olympiad geo is the WORST

[green merlin]

ikr

[hoary ember]

proof?

[green merlin]

Nah it probably isnt

Just realised

oh man i just did geo and my brain is alr fried

[hoary ember]

,rccw

[hoary ember]

maybe better diagram

[green merlin]

Ah

[twilit leaf]

AFB ~ BEC

[hoary ember]

oh wait ABC doesent have to be acute

[green merlin]

Oh yeah

[green merlin]

Dosent change much tho

[hoary ember]

AFB=BEC
ABE=CBE
what am i missing

[green merlin]

Nothing

[hoary ember]

oh im stupid

[green merlin]

AA similarity

[hoary ember]

stinky

[green merlin]

Oh ik

Proof by accurate diagram

[hoary ember]

wtf

[twilit leaf]

BAF=90-2EBC
FAC=90-BAF

that might not be helpful

[green merlin]

What

[twilit leaf]

seems you can angle chase

but im lazy

[green merlin]

Hmm

[twilit leaf]

so have fun :D

[hoary ember]

uhh

[green merlin]

im also lazy

[hoary ember]

topic is angle chase

fuckkk

[green merlin]

Well shit

Aiya let an angle = x

And then brute force?

[hoary ember]

prob

[green merlin]

Becos AFB ~ BEC so we have another angle condition

[hoary ember]

ugh whoever made this should be arrested

[green merlin]

Thats putting it lightly

[hoary ember]

executed

[green merlin]

Ah yes

[twilit leaf]

the angle was pursued in the chase

[green merlin]

Pog

Wow my deep fried brain couldnt see that

I guess its pi

[twilit leaf]

half

[hoary ember]

sorry but i dont see anything from this?

[green merlin]

Angle A is 90

The note

[twilit leaf]

you get 90-2T+2T

[hoary ember]

ABD is simmilar to CBA

[twilit leaf]

yes

[hoary ember]

CAD is also simm

[twilit leaf]

seems like

[green merlin]

Kinda

[hoary ember]

also isnt this wrong like that implies for triangle ABD 180=90+2t+2(90-t)=270

[green merlin]

Nah

90-2t + 2t + 90 = 180

💀

[hoary ember]

whar

oh its phrased like that

[green merlin]

Yea

[hoary ember]

i trated it as 2*-t

[green merlin]

...

[green merlin]

Desmos geometry

[twilit leaf]

geogebra geo

[green merlin]

Oh

[twilit leaf]

i should really go to bed

[green merlin]

No wonder you got angle bisectors

[hoary ember]

where did the numbers come from ugh

[twilit leaf]

arbitrary

except for the 90

[green merlin]

No the 90 degrees dosent change

[twilit leaf]

also sorry about th3 labeling

[hoary ember]

arbitrary as in you put some random values or is that really the number?

[twilit leaf]

i made a traingle with matching conditions then measured the angle

[green merlin]

Random

The diagram can be adjusted

But angle BAC is constant

[hoary ember]

doesent that mean that its just the only one?

if BAC is constant then the angles of ABC as a whole are constant

[twilit leaf]

no

[green merlin]

Nah

[twilit leaf]

the other two angles simply sum to 90deg

[green merlin]

Bro the BAC is 90 but the others can change

[hoary ember]

oh wait im fucking stupid

then the answer is 90 right?

[green merlin]

Yea

[hoary ember]

i thought we were finding abc lmao 😭

[green merlin]

💀

[hoary ember]

.close tyyy

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[twilit leaf]

yup np

[green merlin]

np

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[stiff marlin]

Does k*2^i, where k is odd number and i is non negative integer represent every positive integer?

[sharp flame]

sure I guess

[stiff marlin]

Oh yeah, thank u

[covert wing]

Do you need a more concrete proof?

Or were you just looking for a yes/no

[stiff marlin]

Yes, I would appreciate it

I guess I just prove it

[covert wing]

So any number can be even or odd:
-> If it's odd it can be directly represented by k*2^0
-> If it's even you can divide it by 2 as many times as you can until becomes odd

[stiff marlin]

Thank you you guys

Thanks so much

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[gleaming dome]

Guys how do I find boundedness of sequence

[coarse elbow]

Depends on the sequence in question

[magic oxide]

depends, also just use google if you are going for a general answer lol:

Example 1: Direct Inspection

The sequence {1/n} is bounded because 0 ≤ 1/n ≤ 1 for all n.
Example 2: Comparison Test

The sequence {n/(n+1)} is bounded because 0 ≤ n/(n+1) ≤ 1 for all n.
Example 3: Squeeze Theorem

The sequence {sin(n)/n} is bounded because -1/n ≤ sin(n)/n ≤ 1/n for all n.
Example 4: Limit Test

The sequence {1/2^n} is bounded because it has a limit of 0.
Example 5: Monotonic Sequence Theorem

The sequence {1/n} is both decreasing and bounded, so it converges to a limit. Therefore, it is bounded.

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[strange lotus]

Hi, I'm taking Calculus 2, we were given a couple of exercises to practice but I'm stuck on this one. We are asked to evaluate indefinite integrals with the methods we've covered so far: u-substitution, integration by parts, trigonometric substitutions, and partial fractions. Any suggestions on how to start?

[lean tangle]

could you rewrite that by chance

[magic oxide]

is that a 1 on top?

[lean tangle]

i cannot read the numerator

[magic oxide]

is that a negative e^2x?

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[strange lotus]

Yes

[strange lotus]

Better?

[lean tangle]

oh yep thx thats more readable

what have u tried so far bt

*btw

[strange lotus]

I have tried completing the square in the square root and multiplying the numerator and denominator by e^x, then substituting e^x +1/2 so that I'll have something similar to the derivative of sec-1(x)

But I ended up with this

[final ridge]

You could just type stuff like this to make things readable (and get a solution which you can check).

,w integrate 1/sqrt(e^(2x)+e^x) dx

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[strange lotus]

Thanks, but I'd also like to know the steps and methods used to solve it

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[gritty rose]

have you learned derivatives

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[south bay]

so i multiplied by the conguate to remove the radical but i don’t know how to convert the 1+x to x-1 so they can cancel out

or would i use a lim h -> 0 f(a+h)-f(a)/h instead of the alternate version

[long sky]

Do you know how to take derivatives outside of just first principles?

[south bay]

no this is all i'm learning rn

ik this way is longer

[long sky]

You sure they haven’t taught you guys any rules

Like power rule

Or chain rule

[south bay]

no not yet, this is all we know currently

ive heard of the power rule and chain rule but we're a little behind due to doing review in the start of the year

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[keen heath]

arrange the terms a little different

hmm actually wait the square root is there

[placid oar]

Hint, the expression with x‘s can be factored

[winged void]

tout J'ai montre que: f(x) admet une fonction reciproque

[keen heath]

tu ne sais pas anglais?

[placid oar]

Just lmk

[keen heath]

i speak french too

[winged void]

just a little bit

[placid oar]

Ah well, carry on then

[keen heath]

i actually have no idea how to factor

that

[winged void]

apres je dois resoudre l'equation de f(y)=x

[placid oar]

Doesn‘t x^2 -2 (…) + (…) look familiar?

[keen heath]

-2x

[winged void]

autrement c'est de trouver y =en fonction de x

[keen heath]

i see 2 there

[placid oar]

Exact. Mais pour y arriver je te conseille d‘essayer de trouver comment factoriser l‘expression. C‘est pratiquement impossible sinon je pense

[keen heath]

oh nvm

did i just forget (a-b)^2

[placid oar]

:p

[winged void]

on peut factoriser x^2-2x racine de x + x

ça va donner (x-racine de x)^2

but the problem is on the square root of x

we can't simplify the x-racine de x

[placid oar]

Well you‘re trying to isolate y in $x = (y-\sqrt y)^2 + 2$

[winged void]

exacltly

[placid oar]

Right, then carry on

[winged void]

move 2 to the other side

now we're stuck with $(y-\sqrt y)^2$

[glossy valve]

Neo

[placid oar]

And how do you cancel a square?

[winged void]

now I can't because it's difference

not a mulplication

or division

[placid oar]

Non je voulais dire, si tu a $x^2 = 3^2$, comment est-ce que tu enlèves le carré du x?

[winged void]

par un racine carre ,si l'expression est superieur ou egale à 0

[placid oar]

Exact. Tu peux faire pareil ici.

[winged void]

no

[placid oar]

Explique moi ton raisonnement

Parce que je te garanti qu‘on peut prendre une racine carrée ici et annuler l‘exposant 2

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[fast lotus]

How many different ways of seating 4 married couples in a circular table with men and women alternating and no wife is seated next to her husband (2 seating arrangements are the same if each person has the same left and right hand neighbours)

[fast lotus]

A

[rare dock]

you could draw a picture

[fast lotus]

I did that

[wild sleet]

i don't know this one, but i think it's like around 8

so i would try to count them manually

[fast lotus]

But I don’t have the answer so I’m not entirely sure

[rare dock]

it’s a lot more than 8?

[fast lotus]

Rlly???

[rare dock]

you can permute people

[fast lotus]

That’s what I. Thinking too

[queen crater]

Not a lot more

[rare dock]

more than you’d want to enumerate by hand

[queen crater]

Pretty sure it's not

[fast lotus]

Was thinking of classifying the 4 couples as a 4 people and doing 3! X 2^3 or 2^4

[wild sleet]

i get 2

what's the other one

[rare dock]

hm yes you are right

[queen crater]

Permute your circled 1,2,3,4

[wild sleet]

oh i see

[queen crater]

(but keep one in place since rotations don't count)

[wild sleet]

so it's 12

[fast lotus]

Uh

[rare dock]

yea i agree with 12

[fast lotus]

Working?

prtty pls with cherry on top

[icy falcon]

bear with me because it's way early in the morning but wouldn't taking the cardinality of the set of all alternating genders and subtracting the set of the set of wives next to their husbands be a reasonable approach?

idk the answer but that's what I'd do

[rare dock]

seems like more work than the above

[queen crater]

the set of the set of wives next to their husbands
That's much bigger than the complement, and the complement is very easy to enumerate

[rare dock]

there would be some inclusion exclusion involved

maybe

[icy falcon]

it's a reasonable approach but not the most efficient byfar

anwyay don't mind me just thought I'd throw it out there :)

[fast lotus]

I was doing that tho…

😂

[queen crater]

Anyway... say you seat the women first. You start with one that you call A, then going clockwise you call the next one B, then C, then D.

There are only two seats available for A's husband: between B and C, or between C and D

Once you've seated him, there is only one possible configuration to seat the rest

So you have two choices for that part

Then you just need to account for your naming of the wives (A,B,C,D)

(2 seating arrangements are the same if each person has the same left and right hand neighbours)
This means rotations don't count, so you can just pick A to be any of them, and then you need an order for B,C,D

There are 6 such orders

[rare dock]

[fast lotus]

Yea was wondering why they put that

Okay Ty Mr Mel

Nel

how do I do the thing that everyone does when they’re done

[queen crater]

!done

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[fast lotus]

Oh bettt

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[gilded thorn]

confused on what the ' on the p and q mean here?

[torn jolt]

r u stupid

[distant panther]

Is this proof by contradiction?

[surreal flame]

Surely I hope you know this person because if not why would you ask that

[echo marsh]

they're just variables

p' = (p+q)

[gilded thorn]

yes

[gilded thorn]

also yes

[distant panther]

p and q are integers that equals to -1 + root 2

For rational

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[hoary raft]

Hi everyone 👋. I'm choosing calculus book to get betweem calculus one and several variable by Salas, Hille, Etgen and Calculus by James Stewart. Could you please make a small review on these books if you've used them?

[harsh thistle]

calculus by michael spivak, 4th edition is good 🙂

[gritty rose]

help channels aren't really for book recommendations

try #book-recommendations or #discussion

[hoary raft]

Oh, didn't see this. All the channels weren't shown. Thanks

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[silver forge]

Just confused why the it's x-10 for Myha. Why is Liam the base for "x"? Could I also just do x+10 for Liam?

[tender tapir]

Yeah, you could do that for solving it, but the given table already wants you to use x for Liam's speed.

[silver forge]

Ahh I see, I thought there was a reason why I couldn't. Thank you so much!

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[junior lynx]

stupid question but how do i combine the expansion

[glacial pasture]

you can do them separately
but only go up to x^2 in each of them

then just expand the brackets while ignoring higher degrees that come about

[junior lynx]

first bracket i got a^3+3a^2x+3ax^2

second i got 1+-5x/3+-10x^2/9

but what do i do after that

[glacial pasture]

expand out but only write down degree 2 and below terms

[brittle steeple]

solve for a first though

[glacial pasture]

you have a sign issue i believe too

for the second expansion

[junior lynx]

wdym

[junior lynx]

did i do it wrong?

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@junior lynx Has your question been resolved?

[junior lynx]

<@&286206848099549185>

[polar valve]

multiply them.

[junior lynx]

wouldnt that make the 2nd term x^2 and 3rd x^4

[polar valve]

if you multiply (r+s+t) with (k+l+m) you will get rk+rl+rm+sk+sl+sm+tk+tl+tm, do this with your terms ...

[glacial pasture]

'+-10x^2/9'

,w expand (1-x/3)^5

[junior lynx]

ok

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[glacial pasture]

+, not -

[junior lynx]

ok

,w expand (a+x)^3*(1-x/3)^5

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[hollow loom]

Is this right)

[scarlet hill]

it looks right to me

there may be multiple solutions though

nvm it's good i thiunk

[thorny horizon]

So u took z = a + bi?
Shudnt it be | a + bi - 2 - 2i | = | a + bi + 2 + 2i | ?

[scarlet hill]

he's doing number ii i think

not iv

[thorny horizon]

Ah i see

[hollow loom]

Thx can u also verify this

[scarlet hill]

now he did iv :p

it looks right to me

[hollow loom]

also sorry to bug u but how would you sketch Re(z) + Im(z) < 3

[scarlet hill]

think of Re(z) as x and Im(z) as y

so x + y < 3

y < 3 - x

etc

[hollow loom]

oh of course 🙄

yea alright

thanks for the help

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[hoary ember]

q 15

is it correct? i did (sqrtm+sqrtn)^2-n-m=2sqrtmsqrtn

[bright bronze]

yea, looks correct

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[hazy galleon]

Can someone please help me understand why A=-2 on question 31?

I thought so, because it can’t be that without a proper numeric value?

Could you help me through Q32? I can’t wrap my head around it

I got m=3 but the answer sheet says m= -6

Wait I see what I did wrong

I forgot to do 2x 15

I just left it as 15

Sorry! Thanks for the help

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[naive monolith]

Evaluating the first line

[naive monolith]

answer should be right, apart from the no constant

what have i done wrong?

oh wait

i missed a 2

nvm

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[twin wolf]

trying to show that ln(1+sinx)-ln(1-sinx) = 2ln(secx+tanx)

why are they allowed to just square the inside doesnt that change the whole log?

[lean tangle]

they did not square

they multipleid by (1+sinx)/(1+sinx)

[twin wolf]

huh

[lean tangle]

1-sin^2x = (1+sinx)(1-sinx)

[twin wolf]

ur allowed to do that inside the log?

[lean tangle]

u are multiply a fraction by 1

why would that change anything

assuming sinx is not -1

[twin wolf]

hm ok

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[white delta]

I am struggling with Part b) I will attach their working because I don't get the first step

I don't understand why you can make the assumption that it will cut the x axis

[valid sorrel]

It's a continuous function that becomes 0 after c

Which means it meets 0 at c

Other wise it will have a discontinuity

[white delta]

I don't really get it

why does it have to meet at C if the point of continuity is at 1

[valid sorrel]

Point of continuity?

[lean tangle]

continuous means the entire function, even as a piecewise, does not have a 'gap'

so that means

[white delta]

nah I get it now, it needs to come back down

so that it can continue for all x

[valid sorrel]

It cannot "jump"

[lean tangle]

f(c) should be continuous to f(whatever) = 0

formatitng looks like HSC, i feel like this is slightly harder for math adv tbh

(might be wrong tho)

[valid sorrel]

What's HSC

[lean tangle]

its australian syllabus

[white delta]

it is a purchased trial for HSC yeah

[lean tangle]

CSSA?

[white delta]

nah its called zeta

[lean tangle]

oh i haven't heard of that before but anyhow

i think this is not really what HSC would test per say

u aren't or rarely ever expected to grasp the concept of continuity with language like this

[white delta]

yeah I agree, its definitely on the harder side

yeah I do extension as well and I just haven't seen a question use that application before

[lean tangle]

HSC at least at the adv level should reward a student having grasp over the content at the levle needed, relying on english here is a bit beyond what HSC would test

but if they do test things liek this, it would be somewhere in the 30s for sure

[white delta]

yeah I get you, have you done your hsc

[lean tangle]

im doing 4u this year although i did do math adv last year

math adv scores surprisingly well im ngl

"scaling" wise at least

[white delta]

yeah it does for how easy most of it is

well good luck on monday, thanks for the help

[lean tangle]

no worries, best of luck for u as well

[white delta]

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[coarse wagon]

3cos2x = x

[coarse wagon]

how do i solve for x

[thin flint]

There's no closed form solution afaik

so numerically

[coarse wagon]

how do we approximate the solutios then

[rustic frigate]

In general equation such as those are called transcendental and you’re only lucky if you can solve it explicitly

[coarse wagon]

that does that mean

[thin flint]

you can try taking the power series of cosine

and then take as many terms

and solve the polynomial equation you get

[coarse wagon]

dawg what?

[rustic frigate]

That or other numerical methods like Newtons method

[lean tangle]

from memory u are studiyng math adv and math ext1 right

this is way to difficult for that

[coarse wagon]

yeah

[coarse wagon]

this is yr 11

[lean tangle]

newtons method hasn't been studied here in like 2 decades in highschool lmao

[rustic frigate]

,w solve 3cos 2x = x

[lean tangle]

is there context for the question?

[coarse wagon]

i thought so yeah

[glossy valve]

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Requested by .4slan

[lean tangle]

that wouldn't even be expeceted at extension 2 without context

[hoary ember]

check and guess 🔥

[coarse wagon]

we have a graph

[thin flint]

although when it's x^5 or higher you still get problems as it still probably wont have a closed form solution either

[coarse wagon]

thats about it

[lean tangle]

can u give full question

[coarse wagon]

i thing w use newtons

[coarse wagon]

no its just that

[lean tangle]

newton's approximation is out of syllabus but your teacher may choose to show you it for fun ig

[coarse wagon]

we approximate the soltuoon

[lean tangle]

especially if they are old

[coarse wagon]

dr du has outside syllabus shit

[lean tangle]

oh du

lmao

[coarse wagon]

like integration factor and what not

[thin flint]

you can use newtons method

[lean tangle]

yeah use newtons

[thin flint]

to get approximate solutions

[coarse wagon]

yeah ok fair

[thin flint]

it's as good as you want

depending on how much computer memory you have

[lean tangle]

1000 iterations later

[thin flint]

and how much time you want to spend

exactly

:D

[lean tangle]

honestly i would just do 2 iterations and tell wolfram to do the rest

[coarse wagon]

nah like the solutions give fractions

[lean tangle]

what

[coarse wagon]

the solutions at the back

[lean tangle]

might have to do some value picking then

🫡

[coarse wagon]

ok

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[spring salmon]

I am stuck, I am certain this is similar to something I did before, I just cannot wrap my head around it

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[timber shuttle]

How can I find x such that (16x)%2 == 1 == (9x)%2

[torn jolt]

what does that mean

[timber shuttle]

I want 16x mod 2 to be 1 and 9x mod 2 to be 1. Basically I want 16x and 9x to both be odd integers.

[vast fossil]

With x from what set?

[brittle steeple]

16x mod 2 to be 1?

You want 16 times a number...

[timber shuttle]

x is any real number.

[vast fossil]

Maybe x in Q

[brittle steeple]

to be odd?

[vast fossil]

Right, then you want x to be rational

Namely, its denominator will have to divide 16 and 9

[brittle steeple]

is 1/3 1 mod 2

[brittle steeple]

I don't think it has to divide 9

1/16 for instance

[vast fossil]

How would 9x be an integer?

[brittle steeple]

oh

if 9x and 16x are both integers, then x is

x = 16x*4 - 9x*7

[vast fossil]

The only common divisors of 16 and 9 are 1, -1, so x needs to be an integer tho

So no solutions

[brittle steeple]

!xy maybe

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

[rocky delta]

X=0

[timber shuttle]

I have a grid that I want to be 16*9 but I want it to have a center that lies in the center of a space

[brittle steeple]

what about it?

[vast fossil]

You mean with ratio 16 by 9?

[timber shuttle]

yeah

[brittle steeple]

That's impossible since one dimension will need to be a multiple of 16

[vast fossil]

Not happening I guess

[brittle steeple]

hence even

[timber shuttle]

Alright, thank you qvq

!close

how do I close?

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[timber hull]

I need help with the first and 3rd question

[timber hull]

The moment I get to the 2nd driv I'm just stuck

[umbral dome]

,rccw

[timber hull]

So?

[torn jolt]

show us your work

[timber hull]

I'll show for the first question

[torn jolt]

sure

[timber hull]

After that it's a complete mess

[torn jolt]

i think there's little use in turning this into quotient rule, it's generally easier to deal with products

keep it as 3x = dy/dx * (y + x)

then you have $$D_x\left[3x = \dv{y}{x} \cdot (y + x)\right] \implies 3 = y''(y + x) + y'(y' + 1)$$

[glossy valve]

maxim

[torn jolt]

from which you can obtain y'' through some rearrangement

make sense?

[timber hull]

yes

I'm trying to think of a way to implement it into what he wants

[torn jolt]

well, you have the first and second derivative now

try putting it in that format of $x\dv{^2y}{x^2} + 2 \dv{y}{x}$

[glossy valve]

maxim

[torn jolt]

i have a tingling feeling that this'll reduce into a constant expression

(the answers are all constants lol)

[timber hull]

How would I get the 2nd driv from this as there is 2 dy/dx or do I just get it as simplified as it gets and just pray it works?

[torn jolt]

you would rearrange the equation for y''

$3 - y'(y' + 1)= y''(y+x)$, etc.

[glossy valve]

maxim

[torn jolt]

oh actually

since it's asking for xy''

we can just expand the right-hand side and rearrange again

[timber hull]

How

[torn jolt]

scratch that, I don't think we want the second derivative in the expression

so yeah

divide by (y + x)

[timber hull]

Hmm ight

This is what i got after dividing by y + x now what? I can't seem to think of a way to simplify it

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[timber hull]

not yet

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[vernal wigeon]

Show that if X is a non-zero real number, then X⁸ - X⁵ - 1/X + 1/X⁴ ≥ 0

[vernal wigeon]

How do i do this ?

<@&286206848099549185>

[gritty rose]

Try finding the minima of the function and show that it's nonnegative

[nova acorn]

if you can't use calculus try working backwards and start by multiplying both sides by x^4 (a positive number, so the inequality still holds)

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[crystal ibex]

<@&286206848099549185>

template

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@crystal ibex Has your question been resolved?

[agile holly]

so there are a few mistakes here

(-7x^5)^n

we are subbing this as x^n

(-1)^{n-1} * (-1)^n = (-1)^1

btw

[crystal ibex]

yea i changed the thing on this one cuz i tried multiple

ways

do we need the -1 or +1 sinc ethe first value will be 0 which is even ig

[agile holly]

-1

how did you get x^{5n+5}

[crystal ibex]

i chat gpted it after i tried it 3 times 3 different ways

the actual one i got is -summation (-1)^n-1 7^n x^5n all div by n

its still wrong all of them

[agile holly]

this is somewhat right

the (-1)^{n-1} is incorrect no right?

[crystal ibex]

i dont know if its cuz if we dont need the -1 on the n-

[agile holly]

think about it