help-28

[onyx glen]

cot???

[torn jolt]

Yeah

[onyx glen]

you miscopied the problem statement...

[terse karma]

instead of cos

[torn jolt]

Yeah I miscopied sorry

[onyx glen]

copy it correctly or send a picture

[onyx glen]

okay

rewriting everything on the LHS in terms of sin(θ) and cos(θ) should be a good idea.

[torn jolt]

Well we could write it as
sin0/cos0+ cos0/sin0

[onyx glen]

NEVER use zero as a substitute for theta.

also "it" is the numerator.

[torn jolt]

Oh okay

How would you write it out tho

Cause there’s a divide by secant

[gritty rose]

t for theta

[onyx glen]

write what out

[torn jolt]

Like the left side

Since there’s a divide

[onyx glen]

so you're confused on how to follow my instruction of

rewriting everything on the LHS in terms of sin(θ) and cos(θ)
because of the division symbol that is present.

did i understand you correctly?

[torn jolt]

Yeah

[onyx glen]

do you know how to write csc(θ) in terms of sin(θ) and/or cos(θ)?

[torn jolt]

1/sin t

[onyx glen]

great

the fact that csc(t) = 1/sin(t) is entirely unaffected by the presence of the fraction bar

you can and should write the LHS as $\frac{\frac{\sin(\theta)}{\cos(\theta)} + \frac{\cos(\theta)}{\sin(\theta)}}{1/\sin(\theta)}$ as per my instruction

[glossy valve]

Ann

[torn jolt]

Can’t everything cancel out, but 1/sin

[onyx glen]

no, it is not true that "everything" cancels out.

gotta be careful.

[torn jolt]

Oh ok

[onyx glen]

if you want some lubricant: division by 1/sin(θ) is the same as multiplication by sin(θ),

so your expression transforms into $\sin(\theta) \paren{\frac{\sin(\theta)}{\cos(\theta)} + \frac{\cos(\theta)}{\sin(\theta)}}$

[glossy valve]

Ann

[torn jolt]

Okay makes more sense

I believe the sin can cancel out only so you’re left with cos+cos

[onyx glen]

incorrect.

no cancellation can occur here.

would you like to try to do something on your own or do you want me to give further pointers? @torn jolt

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[onyx glen]

👻

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[calm tinsel]

How am I suppose to do these questions?

[onyx glen]

binomial distribution

[calm tinsel]

do I use this formula specificly?

[wispy hamlet]

yeah

n=10

[calm tinsel]

what would p be in this specific senario

[onyx glen]

curious how the left-hand side is missing

[calm tinsel]

there is no left hand side to this, I didn't crop anything out

[wispy hamlet]

obviously 1/2 ... was a typo .. sorry

[onyx glen]

p is the probability of success; in this case, a success is heads xor tails (pick one and stick with it), and in either case its probability is 1/2

[calm tinsel]

oh so P represents the success of that senario specificly

not the senario when it's been flipped 10 times

I will take the silence as a yes

thanks lads

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[wicked nacelle]

does question 5 mean that to make h as the subject and then write an eqaution for cd?

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@wicked nacelle Has your question been resolved?

[quaint prawn]

It means to write CD=f(h) (using whatever the actual function of h should be)

so not solving for h alone, but solving for CD alone, in terms of h

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[craggy sandal]

Want to prove that if $0 \neq \pi \in \mathbb{Z}[i].$ The following are equivalent:

[glossy valve]

LeftySam

[craggy sandal]

$\pi$ is a Gaussian prime

[glossy valve]

LeftySam

[craggy sandal]

If, for some $\alpha$ and $\beta \in \mathbb{Z}[i]$ we have $\pi = \alpha \cdot \beta,$ then either $\alpha$ or $\beta$ is a unit.

[glossy valve]

LeftySam

[craggy sandal]

I'm stuck on the latter implies the former.

My notes say: We have $\pi$ divides $\alpha \cdot \beta,$ let $\delta$ be a gcd($\alpha,\pi)$ this implies there exists $\gamma$ such that $\pi = \gamma\delta$ which implies $\gamma$ or $\delta$ is a unit. If $\gamma$ is a unit, then $\pi\gamma^{-1} = \delta$ divides $\alpha$ which implies $\pi$ divides $\alpha.$ Otherwise, if $\delta$ is a unit then 1 is also a gcd($\alpha,\pi)$ which implies $\beta$ is a gcd($\alpha\beta,\pi\beta).$ Since $\pi$ divides $\alpha\beta$ and $\pi$ divides $\pi\beta$ which implies $\pi$ divides $\beta$

[glossy valve]

LeftySam

[craggy sandal]

There's certain points of this proof I don't understand

Largely the last two sentences, I don't know why 1 would be a gcd(alpha,pi)

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[calm tinsel]

I'm trying to do the first one but I didn't get the right answer

I did 8C2 * (1/6)^2 * (5/6)^6

[potent shale]

you mean (5/6)^6

?

[calm tinsel]

yes yes

my mistake

[vital wadi]

To calculate the probability of rolling 2 6-sided dice from 8 dice, we can use the binomial probability method.
The probability that Mary rolls 2 sixes dice from 8 rolls is:
P(2) = (8 c 2) * (1/6)^2 * (5/6)^6

[calm tinsel]

I did it right then?

[potent shale]

But that's exactly what he did

[vital wadi]

yes

[calm tinsel]

because I'm getting a decimal :V

answers say 109375/419904

[vital wadi]

0.2939

[potent shale]

Convert to a rational number

Or just calculate: 8 choose 2 * (1/6)^2 * (5/6)^6 = 4 * 7 * 5^6 / 6^8

= 109375/419904

[calm tinsel]

oh is the problem just the calculator can't proccess a number that big hence its a decimal

:V

[potent shale]

Just break it down

[calm tinsel]

what a inconvinience

yea I understand now

thank you guys

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[strong prism]

what is this uprising graphj called

there's linear

there's oscillating

idk what this one is called, graphical or something like that

[gilded quarry]

Well, could you send the equation of the graph if possible

[strong prism]

oh no

theres no equations

well how doi explain this

uhh

theres liek shapes that graphs come in forms of

sin graphs are oscillating

linear graphs are linear

what is this graph in shape of

theres like another word for curved

[gilded quarry]

Could you send or the whole graph as I am not able to see an oscillating

[strong prism]

no- thats not what i mean

it's in topic of sequences

this is a graph of y = e^x

[gilded quarry]

exponential

[strong prism]

but just focus on the part where x is greater than 0

[strong prism]

im not looking for the name of the graph

im looking for names of graph shapes

[gilded quarry]

monotonically increasing ?

[strong prism]

nah something else

so

let's say

in sequences

if you have a sequence of

t_n+1 = 4t_n

what is that called

there's arithmetic

and somehintg else

[gilded quarry]

geometric

[strong prism]

oh shit

ok thank you

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[gilded quarry]

lol

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[wary marsh]

Wouldn’t this just be times 1.24 for all

[onyx glen]

no

the growth factor is only going to be 1.24 for annual compounding

for monthly it will be (1 + 0.24/12)^12

[wary marsh]

And yearly just be (1 + 0.24/365)^12?

[onyx glen]

and daily is (1 + 0.24/365)^365

[clear lily]

,calc (1 + 0.24/365)^365

[glossy valve]

Result:

1.2711488914413

[clear lily]

are you sure

dont you mean (1 + 0.24)^(1/12)

[wary marsh]

Ah alr that seems more simpler

[wary marsh]

That’s monthly

[clear lily]

and (1 + 0.24)^(1/365)

[wary marsh]

And dat daily

[clear lily]

yes

[wary marsh]

Alr ty

Just saw the formula so it helps a lot more

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[cyan lagoon]

I don’t know what a directix is bruh I missed 2 days

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[wheat turret]

Task is:

show with the help of induction that that thing is diviseable by 9 for all numbers equal or bigger to 0

i did it like this

put in the n+1

how do i advance from here

[ember shadow]

try to express the P(n + 1) expression in terms of P(n)

so you can use the assumption that P(n) is true

[wheat turret]

i think i can just P(n+1)-P(n) here right?

since P(n) is true

[ember shadow]

I don't know what you mean

start with P(n+1), see if you can write it as "P(n) + something" and simplify

[wheat turret]

damn how do i do that

would rather be easier to do

minus

[ember shadow]

I don't think subtracting one from the other is correct/valid

[wheat turret]

we already know the latter is true

[ember shadow]

look at the forms of both P(n+1) and P(n). How can you "obtain" P(n) from the expression for P(n+1)?

[devout valley]

You could do that - that would be equivalent to what you said if you can show what you get is divisible by 9
(In particular, that's P(n+1) - P(n) = "something")

[wheat turret]

yeah you get the answer to that questions by subraction right? you just need to write it again in the form you suggested

[ember shadow]

you can try it, and then I can explain how I'm thinking about it

[wheat turret]

sec i need to find out how to calculate this term

never done ^3 ^4 ^5 in school before

[devout valley]

So you haven't done binomial expansion?

Alternatively you can expand (n + 3)(n + 3)^2

[wheat turret]

we only used ^2 all the time

[devout valley]

That's fair enough, as per before you can do this here

Expand it bit by bit

[wheat turret]

is it n^3+9n^2+27n+27 ?

[devout valley]

Expand (n + 3)^2 then multuply the result of that by (n+3)

,w expand (n+3)^3

[wheat turret]

kk then we have the answer here

[glossy valve]

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Requested by chartbit#3171

[wheat turret]

sec

i will show you

[devout valley]

[wheat turret]

so this is the term from the left

and its the old term +

[devout valley]

Accidentally missed cubes on the RHS but otherwise it's all good

[wheat turret]

so now we know what's missing

what do i do now?

[devout valley]

[Worth noting that this is the same idea - expand the (n + 3)^3 and then you can show that P(n + 1) is P(n) + "something"]

[wheat turret]

i forgot the ^3 after the brackets on the right side ^^

but still you get the point

[devout valley]

You have two things that are divisible by 9 on the right hand side, the n^3 + (n + 1)^3 + (n + 2)^3 [by induction hypothesis] and the 9n^2 + 27n + 27 [= 9(n^2 + 3n + 3) ]

[ember shadow]

Yeah I see that now. Just seemed "backwards" to me at first

[wheat turret]

i am still totally confused why you can't just - the old thing that we know that is already diviseable

[devout valley]

Added them together so therefore you can conclude that (n+1)^3 + (n + 2)^3 + (n + 3)^3 is divisble by 9

[wheat turret]

then we would only need to check it for the rest that is left

[wheat turret]

yeah exactly

so what is the correct notation/process here?

Is my technique with just subtraction allowed?

[devout valley]

Provided you then justify why P(n+1) is divisible by 9 you could

[ember shadow]

I think so, but I'm not sure it would be for all induction problems

[ember shadow]

you just show that since n^2 + 3n + 3 is an integer then the whole thing must be divisible by 9

[wheat turret]

I mean how else would you have found out what you needed to add to get the new term?

[devout valley]

Algebra - see the comment that i had made before

Expanding the (n + 3)^3 in P(n + 1) should reveal to you the P(n) and the "something" [the 9n^2 + 27n + 27]

[ember shadow]

I would have just expanded (n + 3)^3; the other 2 terms of P(n+1) are already "part of" P(n).
then you have (n + 3)^3 = n^3 + 9n^2 + 27n + 27.
substitute "9k" for the "P(n) terms", then you have P(n+1) = 9k + 9n^2 + 27n + 27.
now you can factor 9 on the RHS

[wheat turret]

ah ok that's pretty similar to my idea anyway ^^

[wheat turret]

how do we know this is diviseable by 9 now?

[ember shadow]

what is the definition of divisibility?

[wheat turret]

9(n^2 + 3n + 3) = 9 x A

ahh ok thx

[wheat turret]

^^

getting smarter every day

it's like looking up definitions for vocabulary

i wrote it like this now

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[pure pond]

Hi i did not get how he got this and im really confused how

[sharp flame]

Damn look at that formatting

Do you understand what the slope and intercept of a line is

[pure pond]

yes i think

[dim compass]

And so you know how to find x and y intercepts from a linear equation?

If you really understand what is x intercept and y intercept then you should be able to do that

[pure pond]

nope im confused how to find those

[dim compass]

x intercept is just like its name, the point on x axis where the graph of the equation intersects. And x axis lie at y=0 so tell me, based on this information, can you find x intercept of any linear equation?

like if I tell you to find x intercept of y=3x+2, would you be able to do that?

[pure pond]

uhmm nope

[dim compass]

If y=0 in y=3x+2, won't that means that the x axis for y=0 will be the x intercept?

[dim compass]

So whenever you're given y=linear function of x, you can find x intercept by putting y=0

And same concept is used to find y intercept, you put x=0 in the equation and the y value which you get will be the y intercept of the linear equation

[pure pond]

ok srry i need to renew evrything since i forgot so How do i find the slope first?

[dim compass]

Okay, have you tapped into calculus before. Like differentiation?

In any case, slope is the tangent of the angle that a line makes with the positive direction of x axis.

By tangent of the angle, I mean tan theta

[pure pond]

i have no idea what tangent mean

[torn jolt]

Slope is the rate of change of $y$ with respect to $x$.

$m=\frac{y_2-y_1}{x_2-x_1}$

[glossy valve]

Masala Oats

[torn jolt]

This equation represents that if you take 2 points on the line, i.e., (x1, y1) and (x2, y2), then the slope is the difference in the y values divided by difference in the x values.

Slope := (Change in y) ÷ (change in x)

[pure pond]

is there an example?

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[shell glen]

I have got an interesting problem here.

[shell glen]

We are given a^x = log (base a) x, a < 1. Find the range of a for which only 1 solution exists and range for which 3 distinct solutions exist.

Ping me if anyone gets any idea.

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@shell glen Has your question been resolved?

[fleet cedar]

i dont think you can get a nice closed form for this (unless you use lambert W)

but ig the general idea is

[fleet cedar]

if not, then it will have only 1 soln

[fleet cedar]

*positive x

[shell glen]

I have been given the solution, it being that 3 solutions exist if a belongs (0, e^-e] and 1 solution if a belongs ( e^-e, 1). Need to figure out the process.

[fleet cedar]

https://math.stackexchange.com/questions/3020913/how-many-solutions-are-there-for-the-equation-ax-log-a-x-where-0-a-1

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[sonic thicket]

Hey guys I need some help

[sonic thicket]

How can I solve this

I don't know what can I do now

I just prepared my matrix with my equations

But I don't know what's the next step

@fleet cedar @shell glen @ember shadow @wheat turret

Please

[fleet cedar]

dont ping random ppl

[sonic thicket]

Okay sorry

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[silk nexus]

guys when writting the domain for a function

[silk nexus]

how do I exclude

and how do I include

using brakets

[ , ]

] , [

idk how to use them

[fast peak]

[] is including

][ is excluding

the second one is also often written as () instead

[silk nexus]

and on infinity

always excluded right?

[fast peak]

yes

[silk nexus]

ty

[1,+infinity[

this includes 1 right?

and that's how i write infinity right?

[fast peak]

yes

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[tidal flicker]

ok so my question in algebra is x-5= something. I know that x=-2 due the the code thing at the top. i was wondering if it would be -2-5=-7 or would you change the minus in the question to a plus so -2+5=3

[light sonnet]

would you change the minus in the question to a plus so -2+5=3
No because it's not minus negative 5, aka - -5

[tidal flicker]

so it would stay the same, -2-5=7 ?

[light sonnet]

It'll be this

-2-5=-7

[tidal flicker]

oh yes sorry i frogot to put in the minus before the 7.

thank you so much

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[astral garnet]

how do I solve this

[astral sinew]

diagonalize or some shit ig

[astral garnet]

do you mind showing me

im lost

[astral sinew]

let's find the eigenvalues of this matrix

[astral garnet]

so 2 and 1

[astral sinew]

check your signs

[gritty rose]

!show

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

[astral garnet]

💀

do i need to show for 2x2 matrix

[astral sinew]

also it'll be easier if we just multiply through by the scalar

so we have

$\begin{bmatrix} \frac{3}{2} & 1 \ -1 & -1 \end{bmatrix}$

[glossy valve]

Mr. Gamer

[astral garnet]

right

wait let me find the eigenvalue again for this

@astral sinew

1 and -1/2

[astral sinew]

yes

let's let $k_1 = -1/2$ and $k_2 = 1$

[glossy valve]

Mr. Gamer

[astral sinew]

now find v1 and v2

[astral garnet]

is that the eigenvector

[astral sinew]

yea

keep track of which eigenvector goes with which eigenvalue (IMPORTANT!)

[unreal roost]

I got a question regarding center of dialation

Where is it at exactly?

How do I find it?

[astral sinew]

#❓how-to-get-help

[astral garnet]

@astral sinew letting M=PDP^-1

is $P=\begin{bmatrix} 1 & 1 \ 2 & -\frac{1}{2} \end{bmatrix}$

[glossy valve]

ミョーみ

[astral garnet]

wait fk

[astral sinew]

lets slow down

what's v1 (your eigenvector for k1)

[astral garnet]

is it (1,2)

thats the eigenspace for k1=1

[astral sinew]

not quite

[astral garnet]

hmm

but the rref of 1/2 1; -1 -2 is 1 2; 0 0

[astral sinew]

i think you're getting row and column vectors mixed up

[astral garnet]

oops mb

should be span{(-2, 1)} for k1 = 1

[astral sinew]

yea

[astral garnet]

span{(-1/2, 1)} for k2 = -1/2

@astral sinew

right

[astral sinew]

yeah

then P = [v1, v2], where v1 and v2 are COLUMN vectors

also i think you mean k2 = -1/2 (please keep your subscripts straight it's a pain in the ass if you don't)

[astral garnet]

wait so $P=\begin{bmatrix} -2 & -\frac{1}{2} \ 1 & 1 \end{bmatrix}$

[astral garnet]

ah yea

[astral sinew]

v1 and v2 are COLUMN vectors

maybe it's just the latex that's tripping you up

you should have:

[glossy valve]

ミョーみ

[astral sinew]

yeah there you go

now what's your diagonal matrix?

[astral garnet]

$D=\begin{bmatrix} 1 & 0 \ 0 & -\frac{1}{2} \end{bmatrix}$

[glossy valve]

ミョーみ

[astral sinew]

yep, looks good

now just find the inverse of P

$P^{-1} = \frac{1}{-3/2} \begin{bmatrix} 1 & -1 \ \frac{1}{2} & -2 \end{bmatrix}$

[glossy valve]

Mr. Gamer

[astral garnet]

yep got it

[astral sinew]

$P^{-1} = \begin{bmatrix} -2/3 & -1/3 \ -2/3 & 4/3 \end{bmatrix}$

let me double check that

,w inverse ((-2,-1/2),(1,1))

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by Mr. Gamer#6658

[astral sinew]

mb

forgot to take the transpose of the cofactor. lol

so now we have:
$$M = \begin{bmatrix} -2 & -\frac{1}{2} \ 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \ 0 & -\frac{1}{2} \end{bmatrix} \cdot \begin{bmatrix} -2/3 & -1/3 \ 2/3 & 4/3 \end{bmatrix}$$

[glossy valve]

Mr. Gamer

[astral garnet]

yep let me try doing M^n

$\begin{bmatrix} \frac{4}{3} & \frac{2}{3} \ \ \frac{2}{3} & -\frac{1}{3} \end{bmatrix}$

[glossy valve]

ミョーみ

[astral garnet]

by doing first evaluating $M^n=PD^nP^{-1}$

[glossy valve]

ミョーみ

[astral garnet]

and we know that $\left( -\frac{1}{2} \right)^{\infinity}=0$

[glossy valve]

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

[astral sinew]

\infty

also you should have:
$M^{\infty}= \begin{bmatrix} \frac{4}{3} & \frac{2}{3} \ - \frac{2}{3} & -\frac{1}{3} \end{bmatrix}$

holdon

[glossy valve]

Mr. Gamer

[astral garnet]

oh yea sorry i forgot the negative sign

[astral sinew]

just missing a negative sign

[astral garnet]

perfect thank you!

[astral sinew]

there is something interesting to be gained from this exercise

what do you notice about the magnitude of the eigenvalues?

[astral garnet]

less or equal to 1?

[astral sinew]

yeah. why do you think we can even write M^infinity at all?

do you see what i'm getting at?

[astral garnet]

ah right

cuz less than 1 will equal to 0

if 1 then its just 1

anything bigger will just be inf

[astral sinew]

yep. so M^infinity only makes sense if all eigenvalues lie in the interval (-1,1]

idk thought that was pretty neat

[astral garnet]

icic thanks!

for this proof type of q

we gotta start with definiton right

[astral sinew]

yeah.

i'll give you the following hints for part a:
*nilpotent matrices are not invertible
*a matrix is diagonalizable if and only if its eigenvectors span R^n

[astral garnet]

so we have $A^k=PD^kP^-{1}=0$

[glossy valve]

ミョーみ

[astral sinew]

yep

[astral garnet]

wait cant we just show that D=0

thats easy

[astral sinew]

exactly!

take the eqn:

P D^k P^-1 = 0
solve for D^k

[astral garnet]

$A=P^{1/k}D\left(P^{-1}\right)^{1/k}=0^{1/k}$

[glossy valve]

ミョーみ

[astral garnet]

and solve for D

right

[astral sinew]

well not quite

we can't really take roots of matrices like that

[astral garnet]

oh cant we

[astral sinew]

we sort of can but it's complicated and there's no need

beyond the scope of what we're doing right now

if we know that D^k = 0, then that implies that D is singular.

when can a diagonal matrix be singular?

[astral garnet]

never

cuz we cant have row of zeros

thats big brain

wait so the 0 matrix

[astral sinew]

yes

a diagonal matrix must be the 0 matrix to be nilpotent

[astral garnet]

cheers!

that was straightforward than i anticipated

[astral sinew]

and that pretty much answers part b as well

[astral garnet]

wait because

0 is an eigenvalue -> nullspace is non-trivial -> matrix is singular -> only matrix that is singular and diagonalisable is nilponent

[astral sinew]

yep

[astral garnet]

which is the 0 matrix

so B=0

@astral sinew is this enough?

[astral sinew]

yes!

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@astral garnet Has your question been resolved?

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[onyx glen]

are you required to use l'hop?

put the limit in a form where l'hop is applicable.

in this case, you can first factor out the $3x$ to get $3x(e^{1/x} - 1)$, then rewrite $x$ as $\frac{1}{x^{-1}}$ and thus get $\frac{3(e^{1/x} - 1)}{x^{-1}}$

[glossy valve]

Ann

[onyx glen]

this will also make your limit into a 0/0 indeterminate form, which is one of the two forms in which l'hop is applicable.

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[severe basin]

When a number p is divided by 5, the remainder is 3 and when pq is divided by 5, the remainder is 2. What is the remainder when q is divided by 5?

[severe basin]

p=5a+3
pq=5b+2

[onyx glen]

is this where you're stuck or is there more coming?

[severe basin]

I'm stuck here

q=(5b+2)/(5a+3)

Maybe this helps

[onyx glen]

gonna be tricky to progress from this point.

$p \equiv 3 \pmod{5} \ pq \equiv 2 \pmod{5}$

[glossy valve]

Ann

[onyx glen]

this implies $3q \equiv 2 \pmod{5}$

[glossy valve]

Ann

[severe basin]

Yes

[onyx glen]

3 times what gives something that's congruent to 2 mod 5?

[severe basin]

q=9

[onyx glen]

it is not true that q is specifically the number 9

you cannot know q itself, but you CAN know its remainder mod 5.

[severe basin]

Ok

[onyx glen]

and its remainder mod 5 is indeed 4 as you claim.

this is a solution that avoids uttering the i-word

[severe basin]

I don't see any pattern in the numbers.

q may be 4 or 9 or some other number

[pure sandal]

and what is the remainder when divided by 5 of all of these numbers

[onyx glen]

i can repeat this

[severe basin]

Hmm

So the answer is four

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[torn jolt]

How to choose a parameter that's adequate for searching if the function is continuous.

[torn jolt]

with a parameter a for example, a in ]0, infinity[

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@torn jolt Has your question been resolved?

[torn jolt]

<@&286206848099549185>

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@torn jolt Has your question been resolved?

[torn jolt]

<@&286206848099549185>

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[torn jolt]

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[runic igloo]

where did t(3) come from

like the number 3

in t(3)=2+6(3-1)

is it the 3rd term

[vast fossil]

It's been chosen

[vast fossil]

Of course

[runic igloo]

@vast fossil

why are they dividing 36/3

i dont get that

is it bc of the constant multiplying of + or - 3

12 36

etc

[vast fossil]

t(2) = r * t(1), right?

[runic igloo]

yeah

[vast fossil]

So t(1) = t(2)/r

[runic igloo]

yeah

so

how did they know it was 2 r

is it because theres 2 spots we need to find?

the 1st and 3rd

and since its constant

[vast fossil]

Looks like they chose positive r for some reason

[runic igloo]

the d

its squared?

[vast fossil]

Yeah there should be 2 answers

[vast fossil]

Wdym?

Oh

Why * r twice?

[runic igloo]

yes

why

is that the case

[vast fossil]

t(3) = r * t(2) and we know that t(2) = r * t(1)

So t(3) = r * r * t(1) = r^2 * t(1)

[thick minnow]

You can also solve (a) as 8-2 = 3y+5 - 8

As the successive difference remains constant

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[ocean helm]

I know one solution that is: (1,1) and another I have though of was that Y = 1/x^2, so than x^2/x^2 = 1, but I am unsure about that

[nova basin]

y = 1/x^2 and then unless x = 1 or -1 the powers of x must be equal in the first equation

[ocean helm]

Oh yeah, I had done that then got a wrong answer, but I just miscalculated something, ty

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[slow gulch]

hi

[slow gulch]

ive been trying a question for hours but nothing worked

part (c) of the question

i tried replacing x with 1/x

but im stuck after that

here’s a clearer pic of what i tried to do

i need to show the form they want me to in (c)

<@&286206848099549185>

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[slow gulch]

<@&286206848099549185>

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[lime jolt]

this is 12.6 right?

[willow wagon]

Why do you say so

[lime jolt]

i solved it earlier and this is what i got with some friends but we entered it in and it says its wrong so im wondering if the answer overall is wrong

[vast fossil]

Approximately, yes

[lime jolt]

well i did 12.69 at first but remembered it said to the tenths so i did 12.6 but now idk it said its wrong

[willow wagon]

Maybe you guys have rounded it off wrong

[brittle steeple]

Round to the nearest

[willow wagon]

Yep

Round of mistake

[lime jolt]

so 12.7?

[willow wagon]

Try it

And notify us

[lime jolt]

is it right tho we only get 2 trys

[willow wagon]

I got no idea

Link me up

Maybe I could try and tell you the answer

[lime jolt]

wdym link u up

[willow wagon]

Send me the link

To the test

[lime jolt]

it wont allow you lol

its an online school

[willow wagon]

Alright

[lime jolt]

you would need to sign in my stf/account 💀

[willow wagon]

Nah

I wouldn't go that far

[lime jolt]

ye

[willow wagon]

I think the answer is 10

[lime jolt]

im not that desperate for an answer either

[lime jolt]

straight up?

[willow wagon]

If it does mean round off to the nearest tens, yes

[static bramble]

It doesn't say that

[lime jolt]

it said tenTH

so is 12.7 right

[static bramble]

12.7 is correct

[willow wagon]

Yep

Definitely

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[lime jolt]

it was correct

.close

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[thin prism]

.close

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[crisp zodiac]

Can someone please explain the last 2 dots. It has to be solved for G

[merry swallow]

is there more context to what part of math this is? I have no recognition of whats going on

[crisp zodiac]

They want the first dot to be rewritten to this The little dots are supposed to be a number

[brittle steeple]

Seems they evaluated the fraction

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[scenic prism]

wHy is lim x --> 0 x/sinx =1

[empty sapphire]

so you know the fact that that lim x -> 0 sin(x)/x = 1?

[scenic prism]

no I don't understand that too

[stiff musk]

what kind of proof do you want?

there's a reasonably persuasive geometric argument here: https://math.stackexchange.com/a/75151

[scenic prism]

Oh isnt it a rule?

mb I totally forgot

[stiff musk]

it's a provable fact, but the way a rigorous proof would proceed depends on how you have defined sin

[scenic prism]

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[latent spoke]

can someone explain the reasoning behind this

[empty sapphire]

because the domain of f(x) is all real numbers except for x = 1 and the domain of g(x) is all real number except for x= 0

so when you add them together you need to account forthat

[latent spoke]

is that the rule for it? like domain of f(x) is always real numbers except for x = 1?

how did you come to that conclusion if you dont mind me asking

[empty sapphire]

sure thing

so the domain is all of the x inputs that are possible

so for f(x), the only numbers that don't work are where the denominator is 0

because we can't divide by 0

[latent spoke]

ohh ok i get it

so in those equations you need a value that would make it impossible, so in this case the f(x) equation if you input 1 for x it would be 1/1-1 which is 1/0 and g(x) is 0 because its impossible to have 1/0?

[empty sapphire]

yep exactly.

let's take another example

what's the domain of f(x) = 2x + 3?

[latent spoke]

so you would want 2x to equal -3

to cancel out to equal 0

[empty sapphire]

no not quite

remember that domain is "all valid values of x"

are there any values of x here that won't work?

[latent spoke]

i have no clue, all i understand is if x is 0 then its considered invalid, so 2x would need to equal to -3 in order to make 0 with the +3?

[empty sapphire]

no x = 0 it's valid

all real numbers for x are valie

*valid

here I think that a video will help you

one moment

https://www.youtube.com/watch?v=djT6-YamHaA&ab_channel=TheOrganicChemistryTutor

I would recommend giving this a watch

[latent spoke]

alright ill get back to you

[empty sapphire]

it'll help solidify the domain for ya

yep pn

[latent spoke]

@empty sapphire so

there is no number that can make it impossible

so its just any number?

[empty sapphire]

yep exactly!

for f(x) = 3x + 2, all real number is the domain!

you might've seen (-inf, inf) as the notation

or a big giant capital R

[latent spoke]

ya

[empty sapphire]

okay

[latent spoke]

with the lines

[empty sapphire]

now what if I were to say g(x) = sqrt(x)

what's the domain?

can you have negative numbers inside the square root?

[latent spoke]

no

[empty sapphire]

ok so it's fair to say the domain is x >= 0 right?

[latent spoke]

mhm

[empty sapphire]

so what about f(x) + g(x) here?

both are taking the same x input

[latent spoke]

1+0 = 1

and x > 0?

[empty sapphire]

nah not quite

the domain for f(x) is (-inf, inf)

the domain for g(x) is [0, inf)

so the domain for f(x) + g(x) will be [0, inf)

[latent spoke]

Ohhhhh ok i get that

[empty sapphire]

so it's the same with your problem

f(x) is all real number except for 1, g(x) is all real number except for 0

so the domain will be (-inf, 0)U(0,1)U(1, inf)

[latent spoke]

@empty sapphire im reviewing the problems and i got to this one where i dont know how they simplified these equations even after my friend explained it

do you have a video for these problems? It's on the topic of radicals

[willow sedge]

You need to take common

[empty sapphire]

is there anything in $x^3 + x^2$ that can be factored out? What about $4 + 4x$?

[glossy valve]

MellowDramaLlama

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[gaunt bobcat]

Where did I go wrong?

[viral jasper]

Where did the extra $+\cos^2{x}$ come from?

[glossy valve]

SWR

[viral jasper]

Oh that's divide symbol

Where did that come from?

[gaunt bobcat]

I'm dividing it to simplify it

Is that not allowed? Would dividing it by cos alter the value?

[viral jasper]

Yes

Dividing by anything but 1 will lead to a different value

[gaunt bobcat]

Then where do I go from here to get 1-2sin²x?

[vast fossil]

Rewrite it as cos^2 + sin^2 - sin^2 - sin^2

[vivid wren]

You can also just use the trig id

cos^2 = 1-sin^2

[gaunt bobcat]

Where did the extra sin come from?

[vast fossil]

We added and subtracted it

[vast fossil]

But using this is slightly more straightforward

[gaunt bobcat]

Thanks! School didn't teach me this for some reason, it isn't in my notes

[vast fossil]

Is that a SEELE pfp

[gaunt bobcat]

Yep

[latent spoke]

i have no clue

i watched videos but cant find problems similar to that

[vivid wren]

For the first one you can factor out an x^2 to be left with x^2 *(x+1)

[latent spoke]

how would you do that

[vivid wren]

You just pull out an x^2 term?

Try foiling what I just got

It's the exact same expression

[latent spoke]

Ah ok

I got that

[vivid wren]

As for 2

You can only factor out a 4 to have 4(1+x)

[latent spoke]

Right but why are they in the square root sign

[vivid wren]

I dont understand your question

Thats what the problem gives you no?

[latent spoke]

So if you factored out x^2

Why is it x sqrt(x+1)

[vivid wren]

You would factor the x^2 inside the square root, then perform the appropriate operation to simplify that

sqrt[x^2 *(x+1)]

Remember your root properties, namely sqrt(a*b) = sqrt(a)*sqrt(b)

bruh discord does NOT like asterisks

[latent spoke]

Yeah, im just having trouble understanding how there is an x outside of the sqrt and in the inside its x+1

[vivid wren]

Because of this

What is sqrt[(x^2)*(x+1)] using this rule?

[latent spoke]

sqrt(x^2) sqrt(x+1)

[vivid wren]

Right

[latent spoke]

OHHHH

and sqrt(x^2) is uust x right

[vivid wren]

Yes

[latent spoke]

Since ^2 cancels the sqrt

Now it all makes sense

[vivid wren]

In your case yes but realistically it would be + or - x cause either one squared yields x^2

Just stick to the positives here to avoid confusion

[latent spoke]

Alright

Ao the process would be

Factor the xs inside of the sqrt then use the rule

[vivid wren]

Yes

[latent spoke]

That applies to the second one too?

[vivid wren]

Yoi just gotta make sure to be careful on what the factor is

Yes

[latent spoke]

wdym by this?

[vivid wren]

Cause in some other problems the factor may not be x^2

Just a side note to keep an eye out for

Example: x^5 + x^4

[latent spoke]

you cant factor those?

[vivid wren]

Yes you can

The factor isjust different

x^5 can be rewritten as x^4 *x

and x^4 is just x^4 *1

[latent spoke]

right

[vivid wren]

Thats your facto

R

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[crimson terrace]

Help needed

[full forum]

Please don't occupy multiple help channels.

[crimson terrace]

Mb

[void warren]

?

[crimson terrace]

Idk where to start

<@&286206848099549185>

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@crimson terrace Has your question been resolved?

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@crimson terrace Has your question been resolved?

[gaunt quartz]

He collected 10 each level and there are 5 levels he also collected 30 points per level and there’s 5 levels. The 10 is a variable because we don’t know how many points per gem stone

[crimson terrace]

Dw

I worked it out and got to 100

🗿

Like the chad I am

[minor ravine]

good

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[barren iris]

Problem: I need to decide what should be u for my substitution.
Question: Without telling me which one I should choose, what are some things to consider when deciding which one to choose? Should I just try both and see where either of them lead me? Example: choose u = sec theta or u = tan theta?

[sharp flame]

$$\int f(x) \cdot f'(x) \dd x$$
$$f(x) = u \implies f'(x) \dd x = \dd u$$
$$\int u \dd u$$

[glossy valve]

NEONPerseus

[sharp flame]

This is the general structure you want to look for while substituting

(at a lower level)

[barren iris]

I want to look how to get the indefinite integral to u du?

okay

[sharp flame]

Well not exactly

[barren iris]

I think that's what I've normally been doing

okay

[sharp flame]

Sometimes you'll get an easier to integrate function in u

[terse karma]

Could've used g(f(x)) to make it more general

[sharp flame]

I suppose

[barren iris]

hmm

well this one has me confused

I don't think I can combine the terms in a way to get that

[gaunt quartz]

Try letting u = tan

[barren iris]

Maybe I need to rewrite the trig functions into an trig identity ?

[gaunt quartz]

Because du = sec^2

[barren iris]

OH

cause

d/dtheta tan = sec^2 theta

wow

[gaunt quartz]

Yes

[barren iris]

that was way easier than I thought

[gaunt quartz]

Then it’s just a basic power rule

[barren iris]

I was out here looking up trig identities

like oh man

[gaunt quartz]

Lmaoo I felt that

[barren iris]

what do I do

[gaunt quartz]

You really only need to use Pythagorean’s identities when it’s like sin^2cos^3

Even and odd

But for close powers

[barren iris]

ty for your help as well

[gaunt quartz]

Np

[barren iris]

ty for your help

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[royal sedge]

h prime was constant so thats y we didnt have any input of g prime?

with 2 i mean

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[pale jackal]

How can I find other values of angles using the sin, cos, and tan rules? Forgot the rules

[tardy matrix]

sohcahtoa

[pale jackal]

I think sin is 180 - x but I don't remember the others

[tardy matrix]

wait no

💀

[pale jackal]

xD

[tardy matrix]

it's 2am here my eyes are bleeding

[grave elm]

sin(x) = sin(180-x), is that what you meant?

[tardy matrix]

i'm guessing for cosine and tangent too?

[grave elm]

for cos and tan it doesnt work

[tardy matrix]

yeah ik

[grave elm]

cos(180-x) = -cos(x)

[tardy matrix]

but i think he wants to know more rules

[grave elm]

and tan(180-x) = -tan(x)

[pale jackal]

I mean for a question like this where ACB when found is 44, but given that BAC is acute, ACB should be 136

[tardy matrix]

yuppers

[tardy matrix]

ok why is it in comic sans

[grave elm]

because it's the best font

[pale jackal]

idk, my school just gives practice questions in that font xD

[grave elm]

cool

sum of angles in triangle is 180°, so 23° + 44° + m<BAC = 180°

[pale jackal]

okay, I found the rules i was looking for, idk what they're called but it's

180 - x for sin, 180 + x for tan, 360 - x for cos

[grave elm]

oh, ACB is 136, so it would be 23° + 136° + m<BAC = 180°

[atomic blade]

Slap your teacher for using comic sand

[grave elm]

no

I dont get what do people have against comic sans

[tardy matrix]

we just follow the sheep

[grave elm]

I use it in every document, in every presentation in everything I do

[tardy matrix]

gotta fit in 😔

[pale jackal]

went from maths questions to comic sans hate 😦

xD

well, thank you guys 🙂

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[sharp perch]

How to find angle d,c?

[glass crystal]

hint for c: AO = OB

and then idk how C is built?

what property does it have?

because it seems like it could be any point really

[sharp perch]

I'm only given these theorems

[glass crystal]

oh yeah ok

i forgot about one

this one

[sharp perch]

But there is only one that is touching the segment

The one in the center doesn't count

I got the answers if that's gonna help

c=42

d=44

[glass crystal]

thats because you also need this

but i didnt forget about it

so lets do c first

using the hint i gave you

[sharp perch]

AO=OB

[glass crystal]

yes

what does that mean for the triangle AOB?

[sharp perch]

c must equal the right angle?

[glass crystal]

why?

[sharp perch]

The B

Angle

That's another theorem

[glass crystal]

state it then

the B angle?

[sharp perch]

Yeah

[glass crystal]

can you say what is the theorem?

never heard of it

maybe i got a different name for it

[sharp perch]

AO=OB right?

[glass crystal]

yes

[sharp perch]

Wait nvm it doesn't apply for angles

[glass crystal]

since A and B are on the circle with center O

[sharp perch]

Mb

I mixed them up

[glass crystal]

what i wanted you to say is that AOB is an "......." triangle

isosceles

[sharp perch]

45?

No Yeah

Wait

[glass crystal]

please lets have an interaction about the problem and how to solve it instead of just throwing guesses,

if you have an idea

pls say it

instead of just a result

[sharp perch]

THat's what I meant at the beginning lmao

[glass crystal]

do you know what an isosceles triangle is,

?

[sharp perch]

OB=OA

So they should have the same angles right?

[glass crystal]

yes

thats it

OAB = OBA

so now do you know something about the sum of angles of a triangle?

[sharp perch]

I'm pretty sure I tried to say that at the beginning

I just wasn't familiar with the term

Just started learning math in english so

[glass crystal]

ooooooh

the right angle

its because right angle = 90°

[glass crystal]

whats your native language?

[sharp perch]

Arabic

[glass crystal]

oh nvm then lol

but yeah c is the same as the angle on the right

but is not a right angle

[sharp perch]

But I'm almost at a native level at english, so I'm just trying to learn those really long terms

[glass crystal]

so

yeah

[glass crystal]

now this

[sharp perch]

I'm not sure

What is it

[glass crystal]

the sum equals 180°

or Pi in rad

[sharp perch]

OH

Yeah

I do