help-23

[fierce gust]

so do i first factor x^2-9 ?

and set it > to 0?

[pure agate]

You don't need to factor it, just determine when x^2 - 9 > 0

[fierce gust]

wouldnt it be > 0? since it cant be negative?

[pure agate]

Yes.

I was thinking in terms of what it cannot equal.

And as enclave mentioned, the denominator cannot equal 0.

[fierce gust]

what do i do after this?

[foggy salmon]

um

from here actually

its

x2-9>=0

they can take on the values of 3 or minus 3

sqrt0 is valid

[fierce gust]

ok so x can be -3 and 3

[foggy salmon]

yes

just that

[fierce gust]

what do i do with the 4 - ?

[foggy salmon]

x2-9 cant be <0

um

so

the sqrt bit cant be =4

else denom will be 0

so

4-sqrt(x2-9) cant be 0

[foggy salmon]

this one, x cant take values inbetween -3 and 3

if it takes on say, 2

then x2-9=4-9=-5<0

so its just

R excluding when

[foggy salmon]

the sqrt is negative

[foggy salmon]

and the denom is 0

[safe radish]

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[frozen harbor]

hey quick question

[frozen harbor]

i worked this problem out and towards the end i end up with the square root of 225

but the answer to the problem is -15 and not 15 and i dont get why

cause the square root of 225 is 15

[final vortex]

It is because the $x$ value is negative

[flat frigate]

ninjahuman

[nova creek]

√(x²) = |x|

[final vortex]

As you take the limit

[frozen harbor]

ok that makes sense

but then i think it would be more difficult for me to solve since when im solving, i get rid of the x

so im not really left with an x in the end

[dense galleon]

how did all these denominators just appear

[nova creek]

They divided the top and bottom by x

[frozen harbor]

i solve by dividing by the highest power in the denominator, so divide bottom by x, and top by x, except inside the square root its by x^2

[nova creek]

Consider if you had 1/(-1) * √(4)

That's clearly -2

[frozen harbor]

makes sense

[nova creek]

But if you move it in the square root, you get √(4/(-1)²) = √4 = 2

So you need to be careful about the sign

[frozen harbor]

i dont think i get it

[spice grove]

Use |x| in place of x, for your troubles to end.

(for division)

[frozen harbor]

i dont understand

im sorry

[nova creek]

x ≠ √(x²)

[spice grove]

You divided by a negative number in the denominator, while a positive in the numerator.

[nova creek]

If x is negative, then x = -√x²

[frozen harbor]

oh so

when the problem is approaching negative infinity

i need to put a negative outside the radical

[nova creek]

If you intend to move it in the square root

[frozen harbor]

what?

[nova creek]

Well, you don't need the negative if you're not writing x as a square root

[frozen harbor]

what do you mean writing x as a square root

[nova creek]

As √(x²)

So that you can move it in the big square root and cancel out the xs

[frozen harbor]

i dont understand what big square root is

[nova creek]

√(225x² + x)

[frozen harbor]

how can i move it

i dont really understand whats going on

[nova creek]

You already moved it in

That's how you got √(225x²/x² + x/x²)

I'm not saying anything you haven't already done

[frozen harbor]

oh ok

is that how i should do it?

[nova creek]

So long as you remember the negative sign

[frozen harbor]

is there another way?

[nova creek]

Sure

You can recognize that 225x² grows much faster than x

So √(225x² + x)/x approaches √(225x²)/x

Which is 15|x|/x

[frozen harbor]

howd you go from x to absolute value of x

[nova creek]

√(x²) = |x|

[frozen harbor]

oh ok

but then if you plug in negative infinity

wouldnt your answer be negative infinity too

[nova creek]

No

|x|/x is 1 if x > 0 and -1 if x < 0

[frozen harbor]

okay

this is a lot to think through for me

i appreciate you explaining it to me

.close

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[junior wagon]

Is this correct?

[safe radish]

@junior wagon Has your question been resolved?

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@junior wagon Has your question been resolved?

[white umbra]

lemme see

1. is not autonomous

but also it doesn't ask that lol

autonomous just means that the independent variable (so in this case x) is not in the equation, basically

here we have 3x^2, so not autonomous

other ones look fine

well, 3 isn't autonomous either

because there is a t

sin (2pi t/P)

[safe radish]

@junior wagon Has your question been resolved?

[junior wagon]

okok ty

got it

[safe radish]

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[spark merlin]

Um I don’t understand lol

[idle sage]

they factored out (x^2 +3)^-4

[split ether]

View (x^2 + 3)^3 as (x^2 + 3)^7 * (x^2 + 3)^(-4)

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@spark merlin Has your question been resolved?

[spark merlin]

OHHH thx

mb lol

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[honest slate]

how do u factorize (a+b)^2 -c^2?

[thin bridge]

hint:
(a+b)^2 is a square
c^2 is a square

[honest slate]

so (a+b+c) (a+b-c) ?

[thin bridge]

yes

[honest slate]

okok ty

.close

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[lime oyster]

Solve it

[quasi bison]

Solve it
no

[lime oyster]

Why?

[quasi bison]

because we don't give answers here.

[lime oyster]

So tell me how to solve it

[lime oyster]

Dude do you even know how to do it

[quasi bison]

,rccw

[onyx birch]

dude ofcourse she does

[thin bridge]

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

[lime oyster]

Im getting the value 4alpha.beta.gamma

[onyx birch]

show

your

work

[lime oyster]

But i cant evaluate the value of alpha.beta and gamma

[onyx birch]

now @quasi bison @thin bridge can help LMAO this is outta my range

[quasi bison]

But i cant evaluate the value of alpha.beta and gamma
i mean you're not supposed to do that are you

also there ought to be a cleverer way to do that...

[lime oyster]

Howw?

[quasi bison]

wait are you sure you copied it correctly and the bottom-right entry is beta+ALPHA and definitely isn't beta+GAMMA?

i was copying it down to demonstrate an algebraic manipulation i had in mind

[lime oyster]

The bottom right entry is beta+ alpha

It should be beta +gamma?

It should be gamma i guess

[quasi bison]

it would be gamma if the matrix were to be as symmetric as it almost is

[lime oyster]

Okk

Can you please tell how to solve this polynomial

I cant solve it

[quasi bison]

i can't until you show us a picture of the problem as stated originally so that we can be sure we are solving the right problem

[lime oyster]

This is the right ques

[quasi bison]

,rccw

[quasi bison]

what's that in the bottom right now

i can't read it

anyway, if that's meant to be beta+gamma (and not beta+lambda as it reads rn) then

let S = alpha+beta+gamma

you have that your matrix is $\bmqty{S - \gamma & \gamma & \gamma \ \beta & S - \beta & \beta \ \alpha & \alpha & S - \alpha}$

[flat frigate]

Ann

[quasi bison]

does that help us at all

the determinant is $\alpha\beta\gamma \vmqty{S/\gamma - 1 & 1 & 1 \ 1 & S/\beta - 1 & 1 \ 1 & 1 & S/\alpha - 1}$

[flat frigate]

Ann

[lime oyster]

I cant understand it

But thankyouu i'll try it again maybe i have written the wrong ques

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[hasty sage]

$2(a^{3}+b^{3}+c^{3}) \ge (a+b)ab+(b+c)bc+(a+c)ac$

[flat frigate]

PryingOpenMy3rdEye

[hasty sage]

$\forall a,b,c \in \mathbb{R}^{+}$

[flat frigate]

PryingOpenMy3rdEye

[hasty sage]

I have to prove this

if my calculations are right, I got it to a much simpler form (i think)

$a^{3}+b^{3}+c^{3}+abc \ge \frac{(a+b)(b+c)(c+a)}{2}$

but I still have the third powers

[flat frigate]

PryingOpenMy3rdEye

[safe radish]

@hasty sage Has your question been resolved?

[safe radish]

@hasty sage Has your question been resolved?

[hasty sage]

<@&286206848099549185>

[fresh garnet]

Doesnt AM GM help?

[hasty sage]

how?

[fresh garnet]

Havent tried it yet lemme try it once

[hasty sage]

i thought of AM only when I saw (a+b)(b+c)(c+a)

but idk what to do with the LHS

[fresh garnet]

Ok so

You'll have to use AM GM multiple times

Try to expand the RHS in the question and try to find a AM GM for each term

@hasty sage

[hasty sage]

yeah i saw

i'm doing that now

[fresh garnet]

Kk

[hasty sage]

i don't think i'm doing it right

i'm not getting anywhere

[fresh garnet]

Ok so when you expand the RHS one of the terms you get is a^2b

So for this term you can use the AM GM for (a^2 , a^2, b^2)

So for 6 such terms you will have to make 6 such equations

[hasty sage]

ah

i got it now

[fresh garnet]

K nice

[hasty sage]

where do I go from there?

[fresh garnet]

After you get 6 equations for the 6 terms you can just add those equations to get your answer

[hasty sage]

ah ok

thanks!

[fresh garnet]

Np

[hasty sage]

.close

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[noble sail]

Why the an coeficient in a fourier series is equal to: an=2/t int_0 ^t (f(x)cos(nwx)) dx

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[latent elm]

The relation R is the smallest equivalence relation in terms of the number of elements for which R′⊆R holds.

Define the following equivalence classes of the equivalence relation R when

R′={(j,a),(b,d),(j,h),(l,k),(f,g),(b,i),(g,k),(h,c)}.

[c] = {}
[b] = {}
[k] = {}

[latent elm]

Any help would be appreciated, I've tried to figure this one out for a while now. Not necessarily looking for a strict answer, but more of a nudge forward.

The issue is, I don't understand what the set R contains in this case.

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[fluid hare]

How do I find the magnitude of three vectors?

[foggy salmon]

vector with 3 components?

[fluid hare]

do I just do like |v| = sqrt A+B-C

[white swallow]

Find the vector C in x and y component first, add those vectors along the x and y component and find the magnitude of that resultant vector

[fluid hare]

Oh I think I understand what you mean

[white swallow]

Try doing it and send your solution here

[fluid hare]

Okay I am lost lol

[foggy salmon]

um

u can break C into x and y using trig

[fluid hare]

ah

thanks

[foggy salmon]

remember they are kinda negative

so consider that when doing the last bit

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[lean otter]

Guys can anyone answer this

[lean otter]

How do i answer this

Help

How doni so this question

Its related to sets

Heeeeeeeeeeelp

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@lean otter Has your question been resolved?

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[slim spruce]

"Using only the properties, calculate: " really confused on the solution of this one, can someone enlighten me on the steps?

[slim spruce]

the solution is most likely det(A)=0 but i'm not really sure how i can reach that part

(i say most likely because the other matrices under said question all had the same answer it was just a matter of finding the steps to that answer)

[peak estuary]

what's the angle addition formula sin(x+y) ?

[slim spruce]

you mean sin x cos y + sin y cos x?

[peak estuary]

what happens if you use that on the third column

[slim spruce]

i have a sneaking suspicion this this isn't the correct direction to go

that or i'm missing something really obvious

ok wait i figured it out

took a minute

:)

.close

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[lean otter]

How to solve 1kg * m/s to g * cm/s

[halcyon carbon]

do you know how much 1 kilo is

Kilo as a prefix

[lean otter]

The last 3 I need help with

Yes 1000grams

[halcyon carbon]

Ok and what is centi?

[lean otter]

0.01

[halcyon carbon]

Yea

Just use this

[lean otter]

Ik but it's kg times m /seconds that's like 3 things going on

[halcyon carbon]

when changing from m to cm you need to multiply by 10^2

You’re just changing the prefixes

If you add a prefix times by the opposite power of ten to balance it out

If you’re removing the prefix times by the associated power of ten

[lean otter]

So its just Grams to meters

[halcyon carbon]

Grams and meters measure two different things

https://youtu.be/LaLXN2DxQN4

I recommend that you watch this

It goes over the basic of converting between prefixes

[lean otter]

Ik how to convert between prefixes but how do I do double because I'm changing the prefix infront of grams and meters

.close

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[pallid nymph]

Is $C_{p^2}$ a vector space in Q?

[flat frigate]

ohNoiAmHere

[quasi bison]

what is $C_{p^2}$?

[flat frigate]

Ann

[pallid nymph]

cyclic group order p^2

[quasi bison]

...then im pretty sure that no, it isn't a vector space over anything.

aside from maybe the finite field of order p or p^2, if that.

[pallid nymph]

cause like p on p maps to 0 right

[quasi bison]

do you claim to have a scaling operation defined on this thing?

[pallid nymph]

lemme show you rlly quick, i typed something out

So what i was thinking is since $$|Aut(\mathbb{Z}/p^2\mathbb{Z})|=p(p-1).$$

[flat frigate]

ohNoiAmHere

[quasi bison]

Aut in what category?

[pallid nymph]

its like taking $\phi \in Aut(Z_{p^2})$ to determine a generator. like im trying to make a $C_p$ module

idk if thats possible

[flat frigate]

ohNoiAmHere

[quasi bison]

a what module

??

do you mean a (Z/pZ)-module?

modules are over a ring not a group

[pallid nymph]

yea

[quasi bison]

ok so lets start at the beginning

what set do you want to make into a (Z/pZ)-module?

(or rather since Z/pZ is a field i should say (Z/pZ)-vector space)

[pallid nymph]

the elements of $C_{p^2}?$ so ${0,1,2,...,p^2-1}$

[flat frigate]

ohNoiAmHere

[quasi bison]

sus how you are saying "the elements of C_{p^2}" and not just "C_{p^2}"

but ok

C_p^2 it is

i suppose we already know what your addition operation is

now what is your scaling

given k ∈ Z/pZ and n ∈ C_p^2, what is kn

[pallid nymph]

adding n k times

[quasi bison]

i think this fails distributivity

take p = 7 for concreteness

i'm going to denote the elements of C_49 as [0], [1], [2], ...

4*[1] = [4] and 3*[1] = [3] based on what you said,

but [4] + [3] = [7] while (4+3)*[1] = 0*[1] = [0]

[pallid nymph]

wait maybe im being really dumb but why is (4+3) = 0?

[quasi bison]

these are scalars in Z/7Z

[pallid nymph]

oh shoot yea

could we use the euler totient function?

[quasi bison]

idk how youre planning to use it...

i think it is kind of a lost cause

i think C_{p^2} cannot be made into a Z/pZ vector space at all

not least because adding together p copies of an element generally does not bring you to 0 like it would in such a space

[pallid nymph]

i came up with this: j-th way to make Z/p^2Z a Z/pZ module is r•x=φ_j^r(x) where φ_j(1)=jp^2, but it seems sort of pointless now that i think about it

idk its just something i thought of playing with it after reading about groups

anyways thanks

.close

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[pliant moon]

can someone please help me with this

[pliant moon]

Using this channel rn...

[raw goblet]

is the written formula the correct answer for the problem on the screen?

[stray socket]

#❓how-to-get-help

Ain't your channel

[raw goblet]

yeah i posted before but just took a minute to process sorry

[pliant moon]

I've tried 1.52*1.18^4 for my question

[stray socket]

Well do you know your formula for stuff like this

[pliant moon]

ag

[raw goblet]

i will find new channel

[pliant moon]

Yeah but

[stray socket]

Well let's start there first

[pliant moon]

I thought since it says initial quanitty I'd multiply by 1.18

[stray socket]

$$A_n = A_0(r)^n$$

[flat frigate]

Umbraleviathan

[stray socket]

You know your An, r, and n

[pliant moon]

yeah

[stray socket]

So if you rearrange things

[pliant moon]

wait

so do we divide

by (r)^n?

[stray socket]

$$A_0 = \frac{A_n}{r^n}$$

[flat frigate]

Umbraleviathan

[pliant moon]

oh ok thanks that helps a lot

[stray socket]

Np

[pliant moon]

.close

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[pliant moon]

.reopen

[safe radish]

✅

[pliant moon]

So for part b

I tried 3.36*(1-0.18)^x = 1.5

to get how many more years it'd take for the amount of radioactive material to 0.2 g

but the answer is off by 2 years

[stray socket]

Just solve 3.36(1-0.18)^n = 0.2

Oh als

Well wait

Actually

Funny moment on your side

What is 1.52 - 0.2

@pliant moon

[pliant moon]

1.5

wait

[stray socket]

It's 0.2, not 0.02

[pliant moon]

wait

Funny moment

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[calm surge]

why do i check 4 here

its not in the domain or anything

[blazing sable]

could you give us a bit more context?

[calm surge]

it is a svt problem for calc bc

v(t) = s'(t)

a(t) = v'(t)

[blazing sable]

the domain seems to be 0 to 10

so 4 is in

[calm surge]

but why do we specifically check 4

when doing by hand

[blazing sable]

they're checking 4 because it's the endpoint of the parabole

generally if you think about it the highest point can either be at one of the edges (0, 10)

or at 4 because it's the edgepoint

[calm surge]

how would i find that its at 4 by hand then

[blazing sable]

you know the point of a parabole is the middle of the two roots

and you can see in the image they are calculating the roots when they broke down the equation

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[lean otter]

How does addition and multiplication work for sig figs 1.12 +1.8 +10.562 why does this equal 13.5

[worthy hemlock]

For addition/subtraction, the amount of sig figs is the lowest number of significant figures after the decimal point

[lean otter]

So if it was 1.80 it would be larger

1.8 has 2 sig figs so why not 14

[worthy hemlock]

It's number of sig figs after the decimal point

[lean otter]

Ohh

Thank yall

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[lean otter]

Literally an hour ago just took a whole page of notes on it

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[sour blade]

would the limit of cos(2x) as x--> infinity not exist?

[analog flare]

Yes it wouldnt

Because it doesnt approach anything

[sour blade]

if the cos(2x) was over 2x then would it be 0

[lean chasm]

,w cos(x)/x as x->inf

[flat frigate]

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[sour blade]

oooh

nice

[vast obsidian]

Hello, there

[sour blade]

thank

[analog flare]

Huh?

Oh yeah its just C/infinity

[dire crypt]

yh

[sour blade]

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[dire crypt]

ok

[slender wyvern]

What is the third side or angle in 21

I got side ac beavuse of reflexive property

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[supple monolith]

Ignore the fact that it doesnt look like a cube AND vector a is from A to B

just wanted to ask iff to get PQ can both these asnwers work (i can only use vectors a,b and c)

nr1 1/2(-c+a)+1/2(-c+b)
nr2 1/2(a+b)

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[supple monolith]

<@&286206848099549185>

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[arctic plaza]

Hey, I’m in Calc 1 and trying to figure out how to write step one of this problem. I know how to do it with the other instantaneous rate of change formula we were given (x+h), but I’m not sure what to write for x and a in this instance. Any help would be greatly appreciated

[split ether]

Just rewrite (f(x) - f(a))/(x - a) using the expression for f given and take the limit as x approaches a

[halcyon carbon]

a is -3 and x is just x and you have limit as x—>-3

[split ether]

Then set a = -3

Although yeah you could replace a with -3 from the very beginning

[arctic plaza]

Like this?

[halcyon carbon]

Yes but Change the a to -3

On your limit

[arctic plaza]

On both or just the final answer?

Should the work above my final answer even have that limit notation?

I’m just gonna have the limit notation on the final answer

Thanks for the help

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[dapper sedge]

Can somebody give me an intuitive or other explanation as to why subset is considered equivalent to implication in logic? I simply do not see the relation between the two

[trim swan]

Well I'm not 100% sure I understand your question

but

$A \subseteq B$ if and only if $\forall x, x \in A \implies x \in B$

[flat frigate]

tatpoj

[dapper sedge]

Sorry, let me reformulate it : it it said that in logic, we have either 0 and 1, while in set theory, we have an empty set and U for universal. 0 = empty set and 1 = U

they'Re sadi to be equivalent

in their properties or whatever

my book claims that subset in sets is equvialent to doing an implication in logic

so basically a subset b is like a implies b

@trim swan

[trim swan]

Sorry, I think this is a bit over my head

[dapper sedge]

its ok yout ried at least

ty

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@dapper sedge Has your question been resolved?

[buoyant shadow]

implication "in arithmetic" is "less than or equal"

0≤1 and 1≤1 and 0≤0 but not 1≤0

subsets are similar too

A           B    
has no X    has X       can be subset
has X       has X       can be subset        
has no X    has no X    can be subset
has X       has no X    can't be subset

it has to be correct for all X like it says, to be sure

the equivalence is in this "3 out of 4" shape basically

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[meager star]

Would someone figure out the next step to this, the question is pretty unclear so as the teacher’s example

[glass carbon]

find minimum of x(x-10)

[idle parrot]

Hmm u r on right track

[meager star]

Wouldn’t there be a second number to this question?

[glass carbon]

firstly you'll find x and then y

because you know relation between them

[meager star]

Im just stuck from this point, I have no idea what to do

[glass carbon]

you know how to find vertex of a quadratic?

[meager star]

Yeah

[glass carbon]

so find vertex of x(x-10)

it's enough to find Vx

[meager star]

x^2 - 10x would be general form right?

[glass carbon]

yes

[meager star]

I got the vertex form but after that do I have to find the coord for that

[glass carbon]

Vx = 5

so the product is minimum if x = 5

find y and that's it

but y as 2nd number not as value of the product

[meager star]

Would I have to calculate the vertex form to get the y

Which I got 0 idk if Im right but y = 0

[glass carbon]

no, y = x - 10 as you've written at the beginning

if x = 5 then what is y

[meager star]

-5?

[glass carbon]

yes

so these numbers are 5 and -5

[meager star]

Alright!

Thanks

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[turbid spade]

would anyone be able to help with this question

[trim swan]

Do you recognize that this is RREF?

[turbid spade]

yes

so ik option 4 is off the table

just unsure about the rest since it's not an augmented one

[trim swan]

why do you say that?

[turbid spade]

because there's not 3 unknowns

there's 4

4 columns

[trim swan]

Oh, right, it's just the coefficient matrix, not the augmented. I almost missed that

[turbid spade]

yea no worries

pretty sure option 2 is right

[trim swan]

Yes

[turbid spade]

because technically the last row

could be like

0 0 0 0 1

which means there's no solutions

[trim swan]

Right

It could also be 0 0 0 0 0

[turbid spade]

that's right

although I'm not sure about

unique solution

not really sure how I should approach that

[trim swan]

0 0 0 0 0 would imply infinitely many solutions

Because it means you have free variables

Actually, since this system has more unknowns than equations, there must be at least one free variable anyway, assuming there is a solution

But in general, a row of zeroes implies that there are either no solutions, or infinitely many solutions

[turbid spade]

will there ever be just one unique solution?

[trim swan]

Not in this case

Only if the reduced coefficient matrix is the identity matrix

Because

if the last row is 0 0 0 0 1, that's no solutions, like you said

if the last row is 0 0 0 0 0, then that row contains no information, it essentially can be ignored

In the latter case, your system is

$$x_1+4x_4 = a$$
$$x_2+2x_3+9x_4 = b$$

[flat frigate]

tatpoj

[trim swan]

You can solve those equations for x_1 and x_2 respectively, making x_3 and x_4 free variables

[turbid spade]

ok wait yea that makes sense

so I guess now is just the debacle between 3 and 5

the wording here is really weird

[trim swan]

Well

Basically the two possibilities are

0 0 0 0 1 (or something non-zero), which means no solution

or 0 0 0 0 0, which leads to free variables

Do you agree with that?

[turbid spade]

yes

[trim swan]

If you have one or more free variables, then how many solutions exist?

[turbid spade]

should be infinite

[trim swan]

must be infinite

So this system definitely has either no solutions, or infinitely many

[turbid spade]

ohh

[trim swan]

And that's all we know, there's not enough information to know which one it is

[turbid spade]

so the last one I guess

consistent means 0 0 0 0 0

and it has to be infinite

5 is true then

[trim swan]

You're right that 5 is true, but this is not right

Consistent just means that at least one solution exists

So 5 is saying, "if any solutions exist, there are infinitely many."

[turbid spade]

ohhh alright gotcha

[trim swan]

But

isn't 3 also true?

[turbid spade]

3 seemed like just a generalized statement of 5

so I'm thinking yea it must also be true

[trim swan]

Yeah, you're right

[turbid spade]

so 2 3 and 5 it is

thank you so much kind sir

[trim swan]

yep

sure thing

[turbid spade]

have a great rest of your day

[trim swan]

Same to you 😁

[turbid spade]

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[crisp fractal]

'

how do I open a ticket

[worthy hemlock]

You did

[crisp fractal]

oh

[worthy hemlock]

Now you need to do

[crisp fractal]

alright

Im just stuck overall on where to start

and which would be n or U sub n

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@crisp fractal Has your question been resolved?

[crisp fractal]

<@&286206848099549185>

could some1 please help

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[frozen plover]

For this isosceles triangle how does the underlined in the image I sent part find the remaining distance left between the vertex and 6?

[jaunty owl]

does the eq say Eq: b - x^2 or 6 - x^2

[frozen plover]

6

[jaunty owl]

oh ok

I see

well

you know how they got the underlined part?@frozen plover

or is that what you are asking for

[frozen plover]

ye idk how they got the underlined part

[jaunty owl]

oh OK yeah took me a second

ok well if that point at x = b/2 is also on the parabola defined by the equation y = 6 - x^2

we can substitute in for x

so y = 6 - (b/2)^2

and that's how they get that y value

do u see

[frozen plover]

Oh

yeah i see it now

thank u\

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[torn thunder]

dumb question, if sin(x) = y, does arcsin(y) =x?
and does this mean i can call it kind of like a logarithm for sin or something

[steep dove]

well apart from a few domain restrictions, yep

[torn thunder]

much thanks

that explains a lot

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[fiery ibex]

How is this compressed horizontally?

[mild saffron]

if (a,b) is a point on the graph of y=f(x) then (a/6,b) will be a point on the graph of y=f(6x)

[stray socket]

f(ax) is f(x) compressed horizontal by a factor of a

[mild saffron]

the map that sends points (a,b) to points (a/6,b) is horizontal compression by a factor of 6

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[hazy rivet]

What I did was

x^2 + h^2 = 25
dx/dt = 0.8 m/s
x at t=2 = 1.5 + .8*2 = 3.1

(3.1)^2 + h^2 = 25
h = 1.1612

2x dx/dt + 2h dh/dt = 0

dh/dt = -(x dx/dt)/h
dh/dt = -(3.1*.8)/1.612 = -1.53

According to the answer its -0.632 m/s
But idk how, can someone please explain it to me

[safe radish]

@hazy rivet Has your question been resolved?

[hazy rivet]

<@&286206848099549185>

anyone? .-.

:(

[foggy salmon]

sry the

formatting is weird

[hazy rivet]

ayo

[foggy salmon]

could u use triple `

at the start and end

i cant see some *

[hazy rivet]

oh

What I did was

x^2 + h^2 = 25
dx/dt = 0.8 m/s
x at t=2  = 1.5 + .8*2 = 3.1

(3.1)^2 + h^2 = 25
h = 1.1612

2*x*dx/dt + 2*h*dh/dt = 0

dh/dt = -(x*dx/dt)/h
dh/dt = -(3.1*.8)/1.612 = -1.53

According to the answer its -0.632 m/s
But idk how, can someone please explain it to me

[foggy salmon]

u can edit ur original or resend it

ok um

h = 1.1612```

i think this bits weird

@hazy rivet

[hazy rivet]

i just put 3.1 instead of x in the first formula

x^2 + h^2 = 25

[foggy salmon]

no the val of h u got

[hazy rivet]

oh

h^2 = 25-9.61 = 15.39
h = 3.92

umm

yeah i did smth stupid

oh i did

h^2 = 25/9.61

smh, thanks lmao

[hazy rivet]

yeah ok i changed it to that and i got the right answer

gg

thanks

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[clear coral]

im supposed to find a value at a that makes 0 continuous but i tried the normal substitution method of plugging in 0 into the function but it becomes undefined.

[spiral crescent]

Why do you think it becomes undefined?

[clear coral]

because of 1/x

[spiral crescent]

So..?

[clear coral]

am i supposed to rationalize it?

[spiral crescent]

No

What is the range of sin

[clear coral]

1

wait

[spiral crescent]

[-1,1] yes

[clear coral]

yeah

[spiral crescent]

So think again about what form your limit is in

[clear coral]

well the limit does exist at 0

[spiral crescent]

Then?

[clear coral]

well im not too sure

[spiral crescent]

What is the form your limit is in

[clear coral]

what form?

[spiral crescent]

When you substitute 0 in the limit you currently have, what form is it that you get?

[clear coral]

0?

<@&286206848099549185>

[spiral crescent]

Do you understand why its 0

[clear coral]

yes because the limit of function is 0

[spiral crescent]

But do u get why the limit is 0

[clear coral]

i think so

if x = 0, 0 is undefined so theres a discontinuity

and the limit from 0-, 0+ meet, so the limit is 0

just wondering how to find the a value that x=0

[safe radish]

@clear coral Has your question been resolved?

[clear coral]

just show me how to do it.

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[stoic dust]

How do u do C?

[stoic dust]

C?

[toxic stratus]

wait a second ive seen this question before

👀

methods nht exam 1 2019

why has the question been cropped out of the exam paper tho

[toxic stratus]

anyway in part b you find the ratio of consecutive terms

so when that ratio > 1

it means the terms are still getting larger

whereas when the ratio < 1

the terms are getting smaller

so you want to find when it switches over

that'll be when the maximum is reached

[stoic dust]

@toxic stratus

Could u send the notation for the first step?

@toxic stratus and why are we using a ratio?

[toxic stratus]

hmmm

im on a phone so i can't type it out

we're using the ratio because that's what you've got from part b

maximising it otherwise is kinda horrible

the pmf for binomial distribution has factorials in it so you can't readily use calculus

check the exam report

on the vcaa website

[toxic stratus]

this is the exam it's from

[toxic stratus]

actually i can

[covert star]

common snow W

[toxic stratus]

oh no

it's Pr(W = k + 1) < Pr(W = k) solve for the least k that satisfies

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[viscid shuttle]

someone help please

[unique bison]

Help with what

[lean otter]

common factor?

take out 2048r^4

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[lean otter]

% is just 1/100

so 0.4/100 * 0.4

so 0.0016 or 1.6 * 10^-3

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[patent copper]

my brain be frying

trynna figure this out

[wild copper]

Express it in a logical format.

[patent copper]

oh

idk what that is maybe i should read till the end of the chapter

[wild copper]

Assume $t_n$ means that troll n is a knight, then we get a boolean equation
$$t_1 = ~t_1 => (t_1 \and t_2 \or t_2 \and t_3 \or t_1 \and t_3)$$
$$t_1 = ~t_1 => (t_1 \and t_2 \or t_2 \and t_3 \or t_1 \and t_3)$$
$$t_2 = ~t_1$$
$$t_3 = ~t_1 \and ~t_2 \and ~t_3 \or t_1 \or t_2 \or t_3$$

[patent copper]

whoa

neva seen that before but ill go do my own looking i wont get u to explain for me

thanks anyway

[flat frigate]

BlokirKominfo
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[patent copper]

--close

.close

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[lethal scaffold]

can someone help me with this

[safe radish]

@lethal scaffold Has your question been resolved?

[safe radish]

@lethal scaffold Has your question been resolved?

[zinc bronze]

Do you know what increasing means

[lethal scaffold]

no

can you just confirm my naswers

part a is a and parrt b iis d ?

[zinc bronze]

B is right, but A is wrong, you gave me the intervals where it's decreasing

[lethal scaffold]

part a is wrong?

[zinc bronze]

Yes

Part A is not the first answer choice

[lethal scaffold]

damn it, can you help me with part a

[zinc bronze]

So increasing, what does that mean in this context?

[lethal scaffold]

increasing

like adding or going up

[zinc bronze]

Cool so why isn't it going up from -infty to -1

Because it's going down

We look at it from the left to right

Pick a point and pick a point slightly to the right of it to see if the value went up or down

[lethal scaffold]

wdym pick a point, like does it have to be on the yellow lines?

[zinc bronze]

Yeah pick a point f(x) on the function where x is whatever

And then look at f(x + a little bit)

[lethal scaffold]

so i could pick -1 and it would be fx) +-1?

[zinc bronze]

Uh no let's see an example

So if the green is my function

I pick some random blue point to see if it's increasing or decreasing

And then I pick the second blue point a little bit to the right

I see that the second blue point is LOWER than the first

So my function is DECREASING on that interval

Roughly speaking

You're looking at whether the function goes up or down

[lethal scaffold]

so if i picked point -2,3 it would be decreasing if i picked -1, 0 ?

[zinc bronze]

Well really you want the two blue points to be super super close to each other

[lethal scaffold]

but in the problem the function goes both up and down

[zinc bronze]

But yes, at x = -2, your function is decreasing

Right

[lethal scaffold]

i think so

[zinc bronze]

But the problem asks WHEN is the function going up specifically

[lethal scaffold]

ohhhhh

[zinc bronze]

So what x values is the function going up on

[lethal scaffold]

and on which part would be infinity

cause the infinity is in like all the options

[zinc bronze]

Infinity is on the positive side and -infinity is on the negative side

Infinity is infinitely far away, you can't see it it's so far off the edge of the picture

[lethal scaffold]

so the part where the arrow goes off the graph is infinite?

[zinc bronze]

Yes, that is what the arrow signifies

[lethal scaffold]

okay so the if the arrow is on the negative side would it still be increasing??

[zinc bronze]

Aha while the arrow may point up, it's also pointing from the right to left

And remember we want to look at the function from left to right

[lethal scaffold]

does that mean i look at the left side of the graph first and then the right?

[zinc bronze]

Well it means you pick a blue point on the left and then another slightly to the right of the first

Regardless of what direction the arrow points

[lethal scaffold]

ohh so we set a point and go to the right to get the next point?

[zinc bronze]

Yes, but only a little little bit to the right

[lethal scaffold]

okay so how do we work part a

[zinc bronze]

Because if I go to far the the right in one go, I get something like

Where something is clearly happening between the blue points, but the points are too far apart, so I can't detect it really

[lethal scaffold]

ohhh

[zinc bronze]

So that's why the blue points have to be really close together

[lethal scaffold]

theres an increase and decrease so that makes sense they have to be closer so you get one result

right?

[zinc bronze]

Yeah

When I put the blue points close together, I can get the info I want

I can see that it's increasing first (function goes up) and then decreases (function goes down)

[lethal scaffold]

i get it

so here we are just trying to find when the function increases

[zinc bronze]

Yes

[lethal scaffold]

so would it start at -1?

[zinc bronze]

Yessir

[lethal scaffold]

-1, 0 for the first part?

[zinc bronze]

Mhm

Another way to solve would be to trace the graph from left to right with your finger and ask yourself "when is my finger going up (increasing?)"

[lethal scaffold]

okay so i think i get it

first part would be -1, 0 and the second part would be 1, inf?

[zinc bronze]

Yes exactly

[lethal scaffold]

and thats the answer?

[zinc bronze]

Yep

[lethal scaffold]

omg thanks

you are awesome

[zinc bronze]

Thanks, you did all the thinking so give yourself some credit

[lethal scaffold]

really helped me understand this and tbh you probably one of the best helpers here, most dont help me understand the problems and how to do them

[lethal scaffold]

thank you

[zinc bronze]

Yeah it gets harder for us to explain basic concepts in understandable terms as we progress in math

Like after you take calculus it's hard to tell people what "increasing" is without using calculus haha

[zinc bronze]

I appreciate it man

[lethal scaffold]

fr

[lethal scaffold]

your welcome keep doing what u do man

[zinc bronze]

👍

[lethal scaffold]

ima close this now thanks

.close

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[boreal laurel]

why is the square root of an exponent just half of it?

[boreal laurel]

x^36 to x^18

[peak estuary]

$\sqrt{x}=x^{1/2}$

[flat frigate]

Denascite

[peak estuary]

and then exponent laws

[boreal laurel]

ohhh thank you

.close

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[lean otter]

Is this correct

"Find the standard form of the ellipse given below in general form."

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[cedar quartz]

number 2

i’m a bit stuck

this is physics but dealing with vectors (sin cosine law and Pythagorean theorem)

[safe radish]

@cedar quartz Has your question been resolved?

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[mortal sandal]

!15m

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[hallow blade]

(250•1.29)-(250•1.34)

[urban sparrow]

sorry

[urban sparrow]

wdym?

[hallow blade]

That's how you solve it.

Convert into dollars then subtract for the difference

[urban sparrow]

idk the awnser,thats y i came here

like

[urban sparrow]

idk how to solve that

[hallow blade]

Well he has £250, right? And exchanging them into dollars would give him different amounts depending on where he exchanges them.

[urban sparrow]

yes?

[hallow blade]

Exchanging his money in the post office would give him $0.05 more per £.

[urban sparrow]

ooh

so the awnser is 0.05?

[hallow blade]

Nonononono

You have to convert the £ first on both rates.

Then you'll know how much more you'd have on the post office.

[urban sparrow]

so the awnser is..??

[hallow blade]

Multiply 250 with 1.29.

What's the answer?

[urban sparrow]

322.5

[hallow blade]

Good. Now multiply 250 with 1.34

What's the answer?

[urban sparrow]

335

[hallow blade]

Now subtract the higher number to the lower number.

What do you get?

[urban sparrow]

so 12.50

[hallow blade]

That's how much more dollars he would get if he exchanged his £250 in the post office rather than in the travel agency.

[urban sparrow]

Ok thanks!

[hallow blade]

You're welcome.

[urban sparrow]

.close

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[polar knot]

how to do this

[polar knot]

pls

<@&286206848099549185>

[glass carbon]

plug (x,y) = (a,b)

then we'll what happen

[polar knot]

how does thta help tho

theres still 2 variables

[glass carbon]

how

look

[mortal sandal]

Could find 2 different lines for different k values and find their intersection

[glass carbon]

(k-1)a + b = 9k + 10
ak - a + b = 9k + 10
Comparision gives:
a = 9, -a + b = 10 --> b = 10 + a = 19
a + b = 19 + 9 = 28

[polar knot]

man why did it take u like 500 mins to write how i stared at the screen for lole 5 mins waiting lmao

[polar knot]

yes

isnt thag the same thing

[glass carbon]

wait I'll rewrite it in a right way

(k-1)a + b = 9k + 10
ak - a + b = 9k + 10
Comparision gives:
a = 9, -a + b = 10 --> b = 10 + a = 19
a + b = 19 + 9 = 28

simple

[polar knot]

how did u know a= 9

[glass carbon]

because ak = 9k

[polar knot]

wait

so

ak -a+b = 9k +10

so 9 is a

and rhen -a+b = qp

so b = 10

oh