help-10

[pseudo swift]

just focus on r'

that's all we need

[slow phoenix]

oh well kinda haha

ill just use this acct issok :>

u dun mind can write it out maybe? a bit hard to visualise

[pseudo swift]

like the idea is we don't give a shit where our tangent exactly is

we just need the general direction it points to

and that general direction is given by r'

[slow phoenix]

uh huh so how do i use part a to cont from what i got

[pseudo swift]

just evaluate r'(t), at the times t when r(t) touches the plane

that's it

no complicated expansion around the curve to write

[slow phoenix]

er u dun mind can write it out :">

[pseudo swift]

nah

you should already have computed everything you need

it's just a matter of plugging the right stuff in the right place

you know r'(t)

[slow phoenix]

exactly thats the problem TvT

[slow phoenix]

yes yes

[pseudo swift]

you know at what times the curve r(t) intersects the plane (from the previous question normally)

say t0 and t1

[slow phoenix]

the issue is idk how to sub in wtv im sup to sub in

[pseudo swift]

you need r'(t0) and r'(t1)

[slow phoenix]

uh huh mhm

wait i still feel so lost i understand bttr with the actual workings uk TvT sorry if its troublesome but ye

[pseudo swift]

do you have your working out for the previous question ?

[slow phoenix]

mhm yes yes

[pseudo swift]

ok

this is what I mean with "times at which r(t) intersect the plane"

and this is what I mean by plug those times into r'(t)

[obtuse pebble]

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[undone sentinel]

im trying to isolate y to find the slope so what im getting is
y + ky = kx - x+ 3
y+ ky/k = kx/k - x/k + 3/k
2y = x/2 - x+3/k
y= x/2 - x+3/k

am i doing it right

[bold bane]

I would start by setting x = 0 and solving for possible values of y.

[undone sentinel]

why am i setting x = 0?

[bold bane]

Because the line crosses the y-axis at x = 0.

[undone sentinel]

i still have two variables left

okay i see

if its at the center of the graph is it y = 0 or x = 0

its 0 on both axes

[bold bane]

Well, try solving for x = 0 and y = 0 and see what happens.

[undone sentinel]

I cant

[undone sentinel]

y + 0 = k(0-y) + 3
y = -yk + 3

[bold bane]

And what happens when y = 0?

[undone sentinel]

same thing i guess?

[bold bane]

What do you get? Work it out.

[undone sentinel]

x = kx +3
x- kx = 3
x- kx/k = 3/k
1 = -3/k

[bold bane]

There is something called Proof by Contradiction. It's a way of proving something is false by showing one example where it doesn't work. In this case, by solving with x = 0 and y = 0, you can show whether b is a possible valid answer.

[bold bane]

This is incorrect. You didn't divide the first y by k.

[undone sentinel]

huh

[undone sentinel]

why are you isolating the constant

[bold bane]

So that I can put it into y = mx + b form.

[undone sentinel]

y = -(1 - k)x + 3/(k+1)

[bold bane]

You mathed incorrectly.

[undone sentinel]

divide the other side by (1+k)

subtract (1- k)x and put it before the y intercept

so its mx + b

thats what i did

[undone sentinel]

y = (k-1)x+3/(1+k)

so the slope is k-1/1+k

[bold bane]

Yes.

[undone sentinel]

so how do i put that on a graph

does k not being equal to -1 one mean it cant be negative?

[bold bane]

k not equal to -1 is there because that would result in k + 1 = 0 in the denominator.

[obtuse pebble]

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[viscid zenith]

i need help deriving the single linear regression model using the MLE method

[viscid zenith]

like i understand setting up L(yi|xi;b0,b1,s^2) but like i dont really get the math after that lol

[obtuse pebble]

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[fickle ether]

Okay this might be a weird question, and i really should know how to calculate that, since im studying CS, lol.

i've booked 5 traintickets for 5 people, and the sum of that is 455€.
now one person has an German Train Card, which gives them 25% of the price.
How much does that person pay, and how much does everyone else pay?

[fickle ether]

so my first calculation was,
455€ / 5 = 91 per person
91 *0,25 = 22,75
91 - 22,75 = 68,25 (Thats how much the person with discount has to pay)
22 ,75 / 4 = 5,68 (that adds on, on top of the other peoples price)
91+5,68 = 96,68 price of the other ones

[civic zealot]

25% of the price or off the price?

[fickle ether]

25 discount, so 25% off the price, excuse me :D

[civic zealot]

that works.

... i think...

[fickle ether]

yea alright, i dont even know what happened here, brainfart

[fickle ether]

nono, it makes sense for me now aswell

.close

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[civic zealot]

ah... wait, it's off

[civic zealot]

the 455 already has the discount included, so dividing by 5 is not correct.

[obtuse pebble]

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[tardy epoch]

,rotate

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[peak cargo]

Is this E?

[tame quartz]

I believe so

5^3 = 125. 125 x 4 = 500

[sacred violet]

but the answer aren't even in cubic inches?

what is this lol

[uneven palm]

so it is D..

it would be E if the radius was 10 in

[sacred violet]

yeah of course

lol

[peak cargo]

Oh ok

I have radius and diameter confused

Thank you

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[gusty panther]

how to solve linear equations

[gusty panther]

especialy this

-1 - (-3)y =3

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[charred topaz]

Let X and Y be two random variables representing the heights of a randomly selected man and woman, respectively. X follows a normal distribution with a mean of 175 cm and a standard deviation of 5 cm, and Y follows a normal distribution with a mean of 160 cm and a standard deviation of 4 cm. Calculate the probability that a randomly selected man is taller than a randomly selected woman.

[charred topaz]

I'm trying to find P(X>Y) and this is equal to P(X-Y>0) so I set up ((x-175)/5) - ((x-160)/4) > 0

Not sure where to go from here

nah

all good 🙂

lol, more math in the future for you

[polar fossil]

doesn't middle school start at like

11 years old

[charred topaz]

Hey Hayley could you help? I see you have a helpful tag

[polar fossil]

yeah i was strategically ignoring this question because i don't remember how to calculate the sum of two normal distributions

i know it'll be a normal dist

and the mean should be clear

[charred topaz]

ahh all good

[polar fossil]

but idk how the stdev works

[charred topaz]

yea sorry for drawing you into this question you were strategically trying to ignore

I'll tag helpful in about 10min if no response comes in, ik yall are busy with your own things to do

[weary delta]

Linear combination of normal distributions isssss.... drumroll also normal 😄

[charred topaz]

ohhh

okay wait

okay nvm

i failed

I thought i had something

but then i didn't

[weary delta]

In particular $X \sim N(\mu, \sigma^2) \implies aX \sim N(a\mu, a^2\sigma^2)$

[warm shale]

learath2

[weary delta]

And $X_1 \sim N(\mu_1, \sigma_1^2), \land , X_2 \sim N(\mu_2, \sigma_2^2) \implies X_1 + X_2 \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$

[charred topaz]

yes agreed

[warm shale]

learath2

[charred topaz]

okayokay

[weary delta]

So you can use the first property to get -Y with a = -1, then the second to get X - Y

[charred topaz]

So X-Y is a new random variable in itself and thus you need to recalculate the new mean and variance for the normal distribution of two random variables in this case

is that correct ish?

[weary delta]

Or you can standardize both at the start and use the fact that sum or difference of two standard norms is N(0, 2)

[weary delta]

Yes Z = X - Y if you will

and Z will be another normal distribution

[charred topaz]

could you elaborate?

[weary delta]

Normal distributions are closed under linear transformations, that's why they are so nice to work with

[charred topaz]

I think i need a better understanding of normal distributions and linear transformations before approaching this problem, i think that is where my issue lies

[weary delta]

$X \sim N(\mu, \sigma^2) , \land , Z = \frac{X - \mu}{\sigma} \sim N(0,1)$

[warm shale]

learath2

[weary delta]

When you subtract the mean and divide by the std. dev you get a "standard normal"

[charred topaz]

yea

[weary delta]

Note the second property I gave, with the 0 means of the standard normal you'll get another norm at 0, and sigma^2 + sigma^2 is trivially 2

So sum of 2 standard normals is N(0, 2)

[charred topaz]

ohh this is when there are two random variables that are normally distributed

[weary delta]

Yes

[charred topaz]

ok N(0,2) makes complete sense now

So when i finish this problem in theory

I should find the new mean and variance for what we labeled as Z to be Z = X-Y

wait then what.... because I need an x to evaluate this problem and then use z-table

to find the probability

[weary delta]

Well when you find Z, you can normalize it, and then you can use your z-table, no?

[charred topaz]

To find Z don't I need a numerical value?

((x-175)/5) - ((y-160)/4) > 0

Wouldn't i need a numerical value? at x and y

then i can normalize it

[weary delta]

I think you may be a little confused. We want P(X > Y) = P(X - Y > 0). Why not just start by figuring out the distribution of X - Y?

[charred topaz]

okay lets do that one step at a time

[weary delta]

It's just a suggestion, what you did looks like you normalized first, which is also completely fine, we can do that too

Actually, no we can't, it's very unfun

Let's do it my way because it's almost 4am here 😄

[charred topaz]

oh my

so sorry

lets make this quick lol

Dude love your help but get some rest

I'll just ask tmmr

Dw about me, get some rest 🙂

[weary delta]

Well let's give it a go, if I fall asleep I fall asleep 😄

[charred topaz]

I'll try to do this again with the info you gave me

if your good with it, but I've been in your spot and lets say I wouldn't be helping a stranger on DC at 4am lol

[weary delta]

Any progress on the distribution for X - Y? It's just applying the two properties I gave in order

[charred topaz]

looks good?

[weary delta]

The variances get added, note the a^2 in the 1st one

[charred topaz]

but we are subtracting?

X-Y > 0 right?

[weary delta]

X - Y = X + (-1)Y is what we are really doing though, no?

[charred topaz]

ohh

linear combination

theres that negative one infront right

[weary delta]

Doing (-1)Y with the first property, you'll see the a^2 kills negative values

[charred topaz]

Hey man, I think im gonna watch a video, because I don't think i am too familiar with the basics

[weary delta]

So it's not exactly a linear combination of the mean and variance. The property of normal distributions is that their linear combinations are also normal. That doesn't say anything about the mean and the variance of the new normal distributions that arise from combining them

[charred topaz]

At this your probably gonna walk me through every step ngl

I'll learn a bit more tonight and ask tmmr

I just couldn't get the problem started because it seems like I don't quite know what I was doing from the start lol

I know the Variacne property you mentioned

But idk how that applied to normal because we are looking for the probability...?

[weary delta]

After you figure out the distribution for X - Y, the next step is creating a standard normal from it, so you subtract the mean from both sides of the inequality and divide both sides by the stddev, that'll give you your standard normal on the left and then you can use your z-table

Watching a video is a good idea, cya tomorrow

[charred topaz]

Ok I follow

Get some rest, thanks for kickstarting this issue for me and helping me find out what the root of my problem was in this

preciate it 🙂

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[zinc agate]

I need to figure out what number put on both sides makes the following equal true:
x^2 + 6x + blank = -2 + blank
How do I do this?

[civic socket]

does the question asking to factor it?

[zinc agate]

I guess I have to factor it

But it’s confusing because idk what number I would need to try out

[civic socket]

you put the numbers like this $x^2+6x+2=0$

[warm shale]

Akira

[civic socket]

if it doesn't factor then you gotta use quadratic formula

[zinc agate]

Yea ik how to factor but I don’t know what numbers to put in

Because it says blank and it needs to match both sides

[teal turret]

Looks like a completing the square exercise

Any number works, but one specific number allows u to factor

[zinc agate]

I thought so, but that means I have to keep trying over and over until I get the right number?

There isn’t a way to find out which number it is without randomly putting numbers?

[obtuse pebble]

@zinc agate Has your question been resolved?

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[analog fjord]

Calc 2

[analog fjord]

This is what I have so far

The formula

nvm I figured it out

.close

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[river marsh]

This is what I have so far for trying to find the arc length but don’t know where to go from here or if this is right so far

[obtuse pebble]

@river marsh Has your question been resolved?

[river marsh]

<@&286206848099549185>

[polar fossil]

what happened in this step?

seems like you lost a few powers of y

[river marsh]

wait should it be 1/4 * y^-6 not y^-1?

[polar fossil]

correct

[river marsh]

ok thank you very much

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[timid silo]

just a quick question what should my coordinate plane number range be like?

[timid silo]

so that i can graph using 2x +3y = -1 and 20y+30y=-10

to find their intersecting point

i know they're already coincident lines

but what should my coordinate plane be like?

1-10?

1-100?

10-100 sorry

it is 20x + 30y = -10

yes, but what range of number should i use for the coordinate plane?

like those number

but i have to put up a graph about my points for both equation

i have to provide a x and y

or have an x and solve for y using the equation provided

ive already checked it out and it is coincident plane just that i need and x and y value

oh okok thanks

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[autumn spade]

if i have two discrete Random Variables with their PMF's how do i find E[XY] when they are not independent?

[autumn spade]

do i just multiply their PMFs?

lets say that i have two PMFs

i need to find their covarience

the formula for that is

E[XY] -E[X]E[Y]

i need to find E[XY]

[desert sinew]

You basically have to calculate on some double summations. 🙂

[autumn spade]

its a probability MASS function

so discrete

[autumn spade]

f(x,y) being their joint pmf?

[warm shale]

antichronos

[autumn spade]

alr imma try this

umm sooo

An urn contains 8 white balls and 6 red balls. 2 balls are selected at random. Let X denote the number of red balls drawn and let Y denote the number of white balls drawn

in this case

my join pmf

comes to be

how to do double summitions?

imma just watch a YT vid on this or smth

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[sacred root]

Hi

[sacred root]

Ik, I can just assume some value and get the ans

But how do i do it manually

[candid yarrow]

this is probably the easiest and fastest way if you’re facing this on an exam

[sacred root]

I'm not, it a assignment

But It might be that the same question appears and says show you work

Then I'm done for

[candid yarrow]

other than that, there is the form $$ \frac ab vs. \frac{a + c}{b + d} $$ in the first statement

[warm shale]

even order group => solvable

[candid yarrow]

substituting numbers in is a valid form of work

“Substitute x = …, y = …, z = … into the equation. We see that it does not hold. Therefore, this inequality is false” if you can find counterexamples

[sacred root]

Alr alr, so I'll not brother solving finding out then

Thanks

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[timid silo]

Can you help me solve 20/0.15 on paper

[ruby path]

what part is confusing? The decimal being in the denominator?

[timid silo]

I never practiced division and I forgot the process

Been using a calculaor

for like 20 years

[whole dock]

20 years

[ruby path]

Do you know that multiplying the numerator and denominator of a fraction keeps it unchanged?

$\frac 12 = \frac 24 = \frac 48$

[warm shale]

neonperseus

[silent jolt]

U can make denomistor whole number

[timid silo]

No I want to divide by a decimal

[silent jolt]

Why

[ruby path]

well that's complicating it unnecessarily

[candid yarrow]

$$ \frac {100}1 = \frac{15}{0.15} $$

[warm shale]

even order group => solvable

[timid silo]

I don't want to discuss fractions at this time. I am asking for help dividing by a decimal

With the process they taught in school, on paper

[ruby path]

I don't think long division can be done with the decimal in the denominator

[peak tinsel]

the way I was taught to divide by decimal was to not divide by decimal and instead multiply both parts until they are both integers

[ruby path]

Or maybe it can be, and I just don't remember the process

$\frac{0.15}{3}$

[warm shale]

neonperseus

[ruby path]

Something like this, sure

[timid silo]

Is this possible?

[ruby path]

Yes but that's 0.15/2

Not 2/0.15

[timid silo]

I felt rushed, let me repost it

[silent jolt]

??

U have to add zeroes to it anyways

[ruby path]

just copy the decimal point directly above where it is and do normal long division in this case

[timid silo]

I don't understand, can you show me with paint

[ruby path]

I'm nowhere near a PC now

Do you know how normal long division works?

Like 20/3?

[timid silo]

,calc 20/0.15

[warm shale]

Result:

133.33333333333

[candid yarrow]

you’re supposed to put 0.15 on the left

[timid silo]

0.15/20

,calc 0.15/20

[warm shale]

Result:

0.0075

[timid silo]

I figured out what I was trying to figure out

But I'm just going to forget it again soon, but thanks

[candid yarrow]

you can refer to this if you forget

[timid silo]

Take care

,solved

[candid yarrow]

do you mean .coose

[timid silo]

,close

.close

[obtuse pebble]

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[long hedge]

what is the implication if a point you plug in is neg or pos

[long hedge]

where did he get the -3

[royal basin]

zeros: -3, 1, 4
this one?

[long hedge]

the domain/solution

[royal basin]

so you're asking why -3 belongs to the solution

[long hedge]

yes

[royal basin]

what's f(-3)

[long hedge]

and the other

0

[royal basin]

does that answer your question of "why does -3 belong to the solution set?"

[long hedge]

yes but why did he have to do all the interval stuff figure that out

[candid yarrow]

he wasn’t doing that to figure it out

[long hedge]

im entirely lost at the purpose

[royal basin]

what is the implication if a point you plug in is neg or pos
you're gonna have to try to phrase your question clearer btw, i don't really understand what you're saying here

[long hedge]

so he listed the test points as f(x)= pos or neg

what does that tell us

[candid yarrow]

he was doing it to figure out if the points in between were solutions

[long hedge]

how does it show if they are solutions

if they are neg ?

[royal basin]

ok so

heres what we are doing here

f is a continuous function, which means [informally] that in order to change sign, it must pass through zero

so to find the points where f changes sign, we find its zeros

and then assert that on any interval between two adjacent zeros, f does not change sign

do you follow thus far Y/N

[long hedge]

so if it does change sign theres a zero

[royal basin]

yes that is what i said

[long hedge]

and change sign from the point you plugged in

so -4 and 2 changed sign

[royal basin]

weh.

wording.

[long hedge]

where is the original sign

[candid yarrow]

what do you mean by that

[long hedge]

SO if he plugged in NEGATIVE 4 and f(-4)= a pos number the sign changed

[candid yarrow]

the sign changed from what to what?

[long hedge]

neg to pos

can you just tell me whart this is called so i can look up a video

[candid yarrow]

try something like “solving polynomial inequalities”

[long hedge]

so if you evaluate f(-4) and the result is positive it changed sign and thats means what ?

[royal basin]

it changed sign
remove this

[long hedge]

[royal basin]

f(-4) is positive, which means that f is positive on the entire interval that contains -4

in this case (-∞, -3)

[long hedge]

no clue

[royal basin]

"what you just said was as if you spoke Chinese to me"

okay

look

we found the zeros of the function: -3, 1, 4

these points break the number line into four intervals:
(-∞, -3), or x < -3
(-3, 1), or -3 < x < 1
(1, 4), or 1 < x < 4
(4, +∞), or x > 4

WITHIN each of these intervals, f(x) is sign-cosntant

do you understand this yes or no

[long hedge]

yes

[royal basin]

ok

f(-4) is positive

-4 lies in (-∞, -3)

therefore, within (-∞, -3), f(x) is positive

do you understand this yes or no

[long hedge]

what does within mean

[royal basin]

same as "in" in this case

[long hedge]

oh i get it

[royal basin]

or on

or inside of

you couldve asked the first time i said "within"

why didnt you

[long hedge]

if you plug numbers in that range into f(x)=pos

ok

and

[royal basin]

therefore, for all x in (-∞, -3), f(x) is positive

[long hedge]

ok

[royal basin]

we can find the same info for the other 3 intervals

and in doing so we will have found out the sign behavior of f everywhere

and all that remains is to pick out only the parts we want

[long hedge]

which are

[royal basin]

f(x) ≤ 0

≤ 0 means negative or zero

[long hedge]

so thats why he picked 1<x<4

why did it change to greater than or equal to ?

because he found out theres a zero ?

ok but where did the x=-3 come from in the final solution
so that was just left over ?

[royal basin]

≤ 0 means negative or zero

it means you want to include all those points at which your function was 0

i.e. the endpoints of the interval (1, 4), turning it into the closed interval [1, 4]

and the point -3

[long hedge]

what does this solution look like in interval notation ?

[long hedge]

so thats just what was left over ?

[royal basin]

idk what you mean by "left over"

[royal basin]

{-3} ∪ [1,4]

[long hedge]

no

(-inf,-3)v[1,4]

[south oracle]

(-inf,-3) means the set of numbers that are smaller than -3

[royal basin]

set not group

[royal basin]

i mean you have some balls

[long hedge]

https://tenor.com/view/hog-boar-emoji-mindblown-mind-blown-slideshow-meme-gif-26390794

i get it

-4 is inside -inf,-3

why did you leave the -inf out

[south oracle]

Is -4 included in your answer

[long hedge]

no

[south oracle]

(-inf, -3) does include -4

[long hedge]

so he just threw x=-3 into the domain because it was in the original set ?

[royal basin]

no

f is positive on (-∞, -3), which is not what we want, so that interval does not go into our solution.

[long hedge]

ive never seen one term alone

[royal basin]

only its endpoint -3 goes in, because at that point the value of f is 0.

[long hedge]

where is this confirmed

[royal basin]

in the written solution, you mean?

[long hedge]

from the original set ?

yes

[royal basin]

zeros: -3, 1, 4

[long hedge]

ok so it was left over

[royal basin]

if you insist on putting it that way, yes.

[long hedge]

they turned into that because plugging 2 in made f(x) neg ?

[royal basin]

no

[obtuse pebble]

@long hedge Has your question been resolved?

[obtuse pebble]

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[primal ferry]

how do I find the power series for 4*ln |x+5|

[candid yarrow]

Do you know the power series for ln (x + 1)

[primal ferry]

yes

.close

[obtuse pebble]

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[timid silo]

3/7 * 14/1

I'm watching a video, the guy divided both the 7 and 14 by 7 to make 3/1 * 2/1.

Can someone explain to me why this works please?because I'm confused why he did to the numorator what he did to the denominator of the 3/7, and also why It's okay to leave the 3 and the 1 as they are..

[brisk matrix]

when you multiply fractions, you should think of multiplying the top of each together, and the bottom of each together

so $$\frac{3}{7} \cdot \frac{14}{1} = \frac{3 \cdot 14}{7\cdot 1}$$

[warm shale]

maximofs

[brisk matrix]

does that make sense so far?

[timid silo]

That makes sense yeah, I was confused why he made the 7 a 1 and 14 a 2 by dividing by 7 though and not touching the other numbers of each fraction

[brisk matrix]

i will get to that

[timid silo]

ah okay thanks

[brisk matrix]

now have you heard how multiplication is commutative

so x * y = y * x

[timid silo]

yeah

[brisk matrix]

so we can swap the numbers on the bottom

$$\frac{3 \cdot 14}{7\cdot 1} = \frac{3\cdot 14}{1\cdot 7}$$

[warm shale]

maximofs

[brisk matrix]

now we can split it again for the sake of clarity

so we can write

$$\frac{3\cdot 14}{1\cdot 7} = \frac{3}{1}\cdot \frac{14}{7}$$

[warm shale]

maximofs

[brisk matrix]

what is 14/7?

[timid silo]

2

[brisk matrix]

yes

so this becomes 3/1 * 2/1

and as you may know, a number over 1 is just the number itself

so 3 * 2

what they did is basically the same thing

[timid silo]

I understand perfectly now it just clicked, thank you!

[brisk matrix]

this is just a more elaborate way to do it and should explain why we can do it

no problem

[timid silo]

.close

[obtuse pebble]

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[gleaming rune]

can u help me solve this with working

[royal basin]

,rccw

[royal basin]

what's troubling you with this problem?

[gleaming rune]

like i dont understand what is the first step

[royal basin]

well, since x^2 - 3 is a factor of p(x), what can we say about the zeros of p(x) based on this?

[gleaming rune]

can we say that x^2 - 3 should be multiplied to x^2 - (root(2) - root(5)) - root(10)

[royal basin]

no, bad. also doesn't answer my question.

[gleaming rune]

or root(-3) is zero of the p(x)

[royal basin]

sqrt(-3) isn't even real, but no, it is not a zero of p(x).

[neon blade]

can i get help on determinants ?

[royal basin]

@neon blade please open your own channel. #❓how-to-get-help

[gleaming rune]

can we tell root(3) and root(- 3) are also zeros of p(x)

ma'am

??

ok

.close

[obtuse pebble]

Channel closed

Closed by @gleaming rune

Use .reopen if this was a mistake.

[warm canopy]

.reopen

[obtuse pebble]

✅

[warm canopy]

you can keep it open, no idea why integral said to close

[gleaming rune]

ok

.reopen

[warm canopy]

reposting for people

[royal basin]

no, are you sure you definitely mean sqrt(-3) there?

[gleaming rune]

if we x^2 - [sqrt(-3)]^2 = 0

then we get root 3 and - root3

[royal basin]

1. do you understand that sqrt(-3) isn't a real number
2. even ignoring this, x^2 - [sqrt(-3)]^2 is x^2 + 3, not what you want
3. sqrt(-3) isn't the same thing as -sqrt(3)

[gleaming rune]

oh

i mean -sqrt(3)

[royal basin]

i gave you the chance to correct yourself earlier

[gleaming rune]

i did not notice

[royal basin]

anyway, ok

we now know p(x) must have the following zeros:

-sqrt(5), -sqrt(3), sqrt(2), sqrt(3)

and that p(x) has degree 4

this determines p(x) almost uniquely

[gleaming rune]

ok i got it

i forgot about the root 2 and - 5

[polar fossil]

by the way, we didn't need to calculate those roots, we could have just left (x^2 - 3) as it was

[royal basin]

again -sqrt(5) isn't the same as sqrt(-5)

[royal basin]

i want to emphasize that there are four known roots and therefore none left unfixed

[gleaming rune]

thank you for your help ma'am

.close

[obtuse pebble]

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[dry crow]

how do i draw the djikstra algorithm im confused

[dry crow]

is that the right step?

[polar fossil]

those are indeed the first two nodes Dijkstra's would visit starting at A

[dry crow]

im confused on the next step

i was wondering should i do the A-B-F

but like the distance is longer than if we just go from A-H-G-F

A is the starting point btw

[polar fossil]

make a list of all the nodes the algorithm can see and their current shortest path costs

here I think the lowest cost would be AH at 421 (compare with ABH at 503 and ABCH at 567)

as you visit nodes you can cross them off the list and add new ones or update costs if the new node uncovers a better path

[nocturne minnow]

<@&268886789983436800> Spammed in all the channels

[obtuse pebble]

@dry crow Has your question been resolved?

[dry crow]

is after this F?

[polar fossil]

I think so

you'll want to keep track of the distance to each blue node

[dry crow]

are the steps right?

[dry crow]

i did take notes on the distance

im just unsure of the steps, which one i should make it blue first

[digital pasture]

Wow

[polar fossil]

well, whichever one has the lowest distance to A when taking the blue highway and one additional step

[dry crow]

im confused should i color vertex D blue first before i do the other

[dry crow]

i feel like D should be colored before C

oh wait

[polar fossil]

no, because C is much closer - it has a distance of 65+distance(B)

[dry crow]

ooh

so my steps correct?

[polar fossil]

they seem to be

I usually mark the distance next to each node so I can easily see what comes next

[dry crow]

thankss so much

[obtuse pebble]

@dry crow Has your question been resolved?

[obtuse pebble]

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[sage geode]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[sage geode]

Could you show your work?

[obtuse pebble]

@fierce horizon Has your question been resolved?

[obtuse pebble]

@fierce horizon Has your question been resolved?

[tardy epoch]

!show

[obtuse pebble]

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

[obtuse pebble]

@fierce horizon Has your question been resolved?

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[obtuse pebble]

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[thin pawn]

Do you remember what polar coordinates are?

[slow fulcrum]

the circle

[thin pawn]

Right, but more than that?

[slow fulcrum]

angle and radius

(r, theta)

don't you just replace x with rcostheta and y with rsintheta

[thin pawn]

yeah exactly

[slow fulcrum]

oh

okay

thx

.clos

.cose

.clos

.closee

.close

[obtuse pebble]

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[vocal cove]

guys help

[vocal cove]

how do you solve this

my option choices are

47 72 69 94

[fierce lagoon]

What are you trying to solve for

Area?

[vocal cove]

perimeter

mb

i calculated 104

but just double checking

[teal turret]

Show how u got 104

I got something else

[vocal cove]

perimeter of 25 and 10

perimeter of 12 and 10 -10

add together

70 +34 = 104

[teal turret]

Draw number labels on image and screenshot

[vocal cove]

?

what do you mean

this is all i got

this is what the question is

[high lily]

it's not clear what you're doing with your numbers

[vocal cove]

?

[high lily]

i.e we have no idea how you're getting
70 and 34

[vocal cove]

perimeter of 25 and 10?

[high lily]

what exactly are you adding up

[vocal cove]

70 and 34

[high lily]

firstly 25 and 10 aren't perimeters
and how does adding 25 and 10 get 70 or 34

[vocal cove]

lmao

[wintry forum]

hii so take what i say with a grain of salt bc i'm not that great at math but this problem doesn't give you that much to work with

[vocal cove]

i just used the google calculator

[vocal cove]

true

i'm trying to get into IB

[high lily]

it gives you enough

[vocal cove]

but i suck at geomatry

like it's the less i know in math

[high lily]

we're trying to get you to mark your diagram so that you actually add up the correct values and not double count or under count

[vocal cove]

yeah i can tell

[vocal cove]

but

25 + 10 is 50% of the thing

x 2 to get final answer

[high lily]

ok, that gets you 70, where's 34 coming from

[vocal cove]

Isn’t that correct?

[vocal cove]

12+10x2-10

Because the side of the rec is 10

You minus 10 to remove it

Right? Maybe idk

[high lily]

we're trying to get you to mark your diagram so that you actually add up the correct values and not double count or under count

[vocal cove]

Yes

[wintry forum]

they're asking you to draw on the picture to show ur calculations i think

[high lily]

(because you're overcouting here)

[wintry forum]

yeah theres an empty space here so you cant do 2(25+10)

[high lily]

its not that you can't do it,
its that you need to be aware of what you're adding

[wintry forum]

o wait i see what you mean

theres an extra 10 cause you counted that side twice

[vocal cove]

Nvm y’all

I solved it

[vocal cove]

Ye

[vocal cove]

Lmao me too

[high lily]

there's a nice trick with these types of perimeter problems

[vocal cove]

Yes

[high lily]

extend lines out, this will have the same perimeter as a 25cm by 22cm rectangle

[obtuse pebble]

@vocal cove Has your question been resolved?

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[vocal cove]

E

[obtuse pebble]

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[chrome horizon]

how do you integrate a composite function f(g(x))

[high lily]

chain rule

[chrome horizon]

chain rule is for differentiating

[high lily]

oh misread

better to use the word integrate

depends on the composition,

[chrome horizon]

yeah ok

I was asking, cuz there's chain rule for any composite function so is there any rule for going the other way

[high lily]

depending on how nice the composition is, u-sub may work
ibp may also work
and if your composition is messed up, then you're pretty much stuffed

[chrome horizon]

sorry what is u-sub

and what is ibp

I don't know what those mean

[south oracle]

U-substitution

[south oracle]

Integration by parts

[chrome horizon]

riiight

i am not familliar with those

[polar fossil]

there is not in general a formula to integrate f(g(x)). Integrals are in some way finicky!

[chrome horizon]

ah ok thanks

[obtuse pebble]

@chrome horizon Has your question been resolved?

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[obtuse pebble]

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[fallow cove]

How do I do this

[fallow cove]

I know I have to find (dR)/(dt)

and I know that (dR_1)/(dt) is 0.5

and (dR_2)/(dt) is -1.1

it's just idk how to find (dR)/(dt)

[unkempt bramble]

R=R1R2/R1+R2

[royal basin]

well you could take the derivative of this equation wrt time as you no doubt have done for the last ten or so problems

[royal basin]

parentheses + unnecessary

[fallow cove]

Yes

[unkempt bramble]

Like why not product quotient rule n stuff u can use ryt

[royal basin]

too long and painful.

differentiating both sides wrt $t$ \textbf{as is}, i.e. \textbf{without} signing yourself up for pain and suffering, you get: $$-\frac{\dot{R}}{R^2} = -\frac{\dot{R}_1}{R_1^2} -\frac{\dot{R}_2}{R_2^2}$$

[warm shale]

Ann

[fallow cove]

what

[royal basin]

or, once you multiply by $-1$ and perhaps also by $R^2$: $$\dot{R} = R^2\paren{\frac{\dot{R}_1}{R_1^2} +\frac{\dot{R}_2}{R_2^2}}$$

[warm shale]

Ann

[royal basin]

overdot means time derivative

[fallow cove]

oh ok

[royal basin]

if you want i can rewrite the same thing w/o the unfamiliar notation, at the cost of it getting a bit more bloated...

[fallow cove]

uuuh no need

[royal basin]

also the parentheses you put here are redundant

do you have anything else to ask regarding this problem?

[fallow cove]

so how did we get here

[royal basin]

differentiating both sides wrt t

applying chain rule thrice

[unkempt bramble]

Wabbout differentiating both sides of original equation wrt R1

[fallow cove]

so then we just sub right?

[royal basin]

unnecessary, painful, and gets us nowhere near our goal.

[royal basin]

yes, though you will need to find the value of R first.

but that, hopefully, is not hard.

[royal basin]

what?

[unkempt bramble]

Yea ryt

[royal basin]

how did those logarithms happen?

[unkempt bramble]

I differentiated both sides wrt R1

Did I do smthg super dumb

[royal basin]

yes you did. $\dv{R_1} \paren{\frac{1}{R_1}} = -\frac{1}{R_1^2} \neq \ln(R_1)$

[warm shale]

Ann

[unkempt bramble]

Ok ok bye bye

[fallow cove]

got it

thx

.close

[obtuse pebble]

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[humble tangle]

i really am not understanding the difference between these two questions

i cant visualize whats its asking or how to even start

[ivory trout]

do u know what does taking the limit of a function mean?

[humble tangle]

its looking at what values its approaching?

[humble tangle]

^

[ivory trout]

yh it means to see what happens at y when x is near the value of the limit

[humble tangle]

right

[ivory trout]

as t go to 0 from right, y approaches...

[humble tangle]

425

because thats the point its at? (0,425)

[ivory trout]

yup

[humble tangle]

hm okay so the second part is infinity

does that mean i am looking at both sides?

[ivory trout]

well it's not infinity

look at what the graph represents

it says it in the exercise

[humble tangle]

temperature of a pie as it comes out of the oven

[ivory trout]

exactly

clearly the pie temperature won't go to infinity after u take it out the oven

what do u expect to happen IRL

[humble tangle]

eventually it will reach room temp

[ivory trout]

exactly

what's room temp in the graph

[humble tangle]

ooohhhhhhh 72

wow that makes so much sense

[ivory trout]

yup

[humble tangle]

i had someone else try to explain and their explanation was so wordy

thank you so much

.close

[obtuse pebble]

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[timid silo]

Please start explaining by a hint or clue

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

.close

[obtuse pebble]

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[sage dagger]

What is the greatest value of 1 / [ 3^(1-sin(x))]

[sage dagger]

I’m not sure how to approach this

The only thing I can find of is solving 1 - sin(x) separately

And setting it equal to 0 or something

[royal basin]

well you could write your function as 3^(sin(x)-1), for a start...

[sage dagger]

What then

[unkempt bramble]

For 3^(a) to be least 'a' shd be least

[royal basin]

this function's maximums happen at the same values of x as those of sin(x)-1

and also those of sin(x)

and you ought to know when the raw sine is maximized, and what its maximum value is

[sage dagger]

Regular sin(x) is 1-<=x<=1

[royal basin]

typos

[sage dagger]

Oh oops

-1

-1<=x<=1

[royal basin]

not x

[sage dagger]

What?

[royal basin]

what you wrote reads as "sin(x) can only accept inputs between -1 and 1" which is incorrect and irrelevant to your problem

you could have said "the raw sine ranges from -1 to 1" and saved yourself the trouble of forcing me to correct you

[sage dagger]

But isn’t that the correct range of sin(x)

Between -1 and 1

[royal basin]

yes but you wrote it incorrectly

[sage dagger]

breh

[royal basin]

it is not x itself that ranges from -1 to 1

[sage dagger]

So what’s supposed to be in the middle

Instead of x

[royal basin]

it's not x itself that ranges from -1 to 1, but what does?

[sage dagger]

sin(x)

[royal basin]

answered your own question.

[sage dagger]

okay

[royal basin]

in any case, once you know what the maximum value of sin(x) is, you also know (or can easily find) the maximum value of sin(x)-1 and hence also that of 3^(sin(x)-1)

since 3^t is a strictly increasing function

[sage dagger]

I haven’t worked with exponentials that have a trig term in it

What does that look like?

[royal basin]

you want a graph?

it isn't necessary here

but i can give you one anyway if you want

[sage dagger]

Yeah I don’t think I’ve ever seen an exponential with like sin or cos as it’s exponent before

Helps me better visualise what I’m dealing with anyway

[sage dagger]

oh that’s strange

The peaks and trough aren’t symmetrical like sin(x)

So if the maximum of sin(x) - 1 is 0

How can I use that to find the max of 3^(sin(x) - 1)

[royal basin]

this function's [3^(sin(x)-1)] maximums happen at the same values of x as those of sin(x)-1

[royal basin]

sure aren't. 3^t is nonlinear.

[sage dagger]

By Nonlinear you talking about curves?

because sin(x) also doesn’t have straight lines

[royal basin]

by nonlinear i mean nonlinear

i mean a function that is not linear

[sage dagger]

[royal basin]

composing a nonlinear function with a sinusoid gives something that might also look like a sinusoid, but its peaks and troughs generally need not be symmetrical.

[sage dagger]

I see

Anyways,

[sage dagger]

Why is this the case?

[royal basin]

3^t is a strictly increasing function

3^a < 3^b <=> a < b

[sage dagger]

But if something has the form a^x wouldn’t a change the maximum and minimum of the function

[royal basin]

bad notation, but

it changes the max and min value themselves, yes. but it doesn't change the inputs at which those max and min values happen.

[sage dagger]

I see

So what to do next after establishing max of sin(x) is = 0

[royal basin]

the max of sin(x)-1 is 0.

[sage dagger]

Oh yeah

[royal basin]

the max of 3^(sin(x)-1) is 3 to the power of that.

[sage dagger]

3^0 so it’s 1?

[royal basin]

sure is.

[sage dagger]

How do we know this is the case?

[royal basin]

let $f$ and $g$ be functions with $g$ strictly increasing, and let $x_0$ be a point at which $g$ attains its maximum. then $f \circ g$ also attains its maximum at $x_0$, and its value there (as it would be anyway) is $f(g(x_0))$.

[warm shale]

Ann

[sage dagger]

So essentially composite functions have the same max as it’s parent function?

[royal basin]

a composite function has its maximum at the same point as the inner function, provided that the outer function is strictly-increasing.

no such thing as a "parent" function

[sage dagger]

Really, isn’t that a thing?

[royal basin]

at best it's a vague thing that sometimes helps with graphing

"parent function" generally describes a simple-to-write-down or simple-to-graph function whose graph may be transformed into that of the one you want

[leaden ginkgo]

parent function would imply that the child function is dependant on the parent function which is not the case, i think?

in the case of composite functions^

[sage dagger]

Oh

[royal basin]

and these transformations may involve composing stuff on either side of the "parent" function, in or out

[sage dagger]

Alright

So after finding that 3^(sin(x) - 1) has a max of 1, what would be the next step

[sage dagger]

Also is this like a general rule or does it only apply in this context

[royal basin]

i gave you the general rule

[royal basin]

well what did the problem ask you to do?

[sage dagger]

Find the greatest value of that function

But it’s being divided by 1 isn’t it

1/3^(sin(x) - 1)

[royal basin]

no, your function was 1/3^(1-sin(x))

and the first thing we did was write it as 3^(sin(x)-1)

[sage dagger]

Oh

Right

[royal basin]

"divide by 1" would mean 1 is in the denom btw

[sage dagger]

How would I say it

[obtuse pebble]

@sage dagger Has your question been resolved?

[royal basin]

"but we were looking at its reciprocal, weren't we"

or simply "but we had it in the denominator" or something

idk

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

Is this right?

[timid silo]

For i?

Because the solutions say something different

These are the solutions

Why is my i different to the solutions I

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[candid yarrow]

Which one is the answer for i?

Could you write it on a separate piece of paper so we can read and check it more easily?

[obtuse pebble]

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[timid silo]

.reopen

[obtuse pebble]

✅

[timid silo]

This is my answer @candid yarrow

[timid silo]

These are the solutions

[candid yarrow]

Could you write your answer to the question on a separate piece of paper than the question, so we can read and check your answer more easily?

[timid silo]

It was fine to read before but anyways

[candid yarrow]

This expression looks right, maybe the answer key isn’t the right one for this question paper?

or there was a typo

[timid silo]

Oh okay

Thank you

.close

[obtuse pebble]

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[obtuse pebble]

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[lean tendon]

Hello

[lean tendon]

Why is this a complex number?

$\sqrt[3]{-3}$

[warm shale]

Ziggy

[south oracle]

It's not?

[royal basin]

depending on the context, this symbol may refer to either the real root or the principal complex root (with lowest argument)

also like... your question, interpreted to the letter, admits the interpretation of "complex number" as meaning any element of C, i.e. potentially also referring to real numbers, rather than specifically "complex but not real".

in which case the answer becomes "because it is the cube root of a complex number. what else could it be if not a complex number itself?"

@lean tendon did you perhaps mean "I plugged this into WA and got a result with i in it, rather than a real number as I was expecting. What gives?"

[south oracle]

Could this symbol indicate the complex roots of x³=-3

[royal basin]

one of, maybe.

having a symbol like this be multivalued opens up a very deep and disgusting can of worms.

[lean tendon]

well im looking at the polynomial

$x^5+3x^2$

[warm shale]

Ziggy

[lean tendon]

and looking at the rots

and solving this in mathematica yes

and it gives me

${0,0,\sqrt[3]{-3},-\sqrt[3]{3},-\sqrt[3]{-1}^2\left(\sqrt[3]{3}\right)}$

[south oracle]

Hmm this can be factored as x²(x + 3^⅓)(x² - 3^⅓ x +3^⅔)

[warm shale]

Ziggy

[royal basin]

ok so it give you the principal complex root

i.e. $\sqrt[3]{3} e^{i\pi/3}$

[lean tendon]

principal complex root?

[warm shale]

Ann

[royal basin]

every complex number save for 0 has exactly 3 cube roots, which are the same in magnitude but differ by 1/3 of a full turn in argument.

[lean tendon]

3 cube roots?

doesn't it just have a conjugate?

[royal basin]

what's "it"?

[lean tendon]

[royal basin]

no need to sadcat me when you're the one who's not expressing your own questions clearly enough.

[lean tendon]

maybe you need to chill

im just saying im not sure what to say

[royal basin]

you're asking me a question which i have trouble interpreting. what else do you expect me to say?

[lean tendon]

so this polynomial has a complex root

[south oracle]

Are you sure that he knows the basic of e^ix= cos x+ isin x, ann?

[royal basin]

this particular polynomial has a conjugate pair of complex roots, yes.

[lean tendon]

yes

[royal basin]

who's he?

[lean tendon]

but why is that conjugate pair cubed?

[south oracle]

Sorry, he refers to Ziggy

Yeah I assumed his gender, sorry

[lean tendon]

i am a he it's fine

[royal basin]

get yourself a he/him role later

anyway

[royal basin]

not sure what you mean by "cubed"

[lean tendon]

to the power of three

[royal basin]

mathematica's output doesn't contain any 3rd powers...

[lean tendon]

alright well i have to run thanks for your help

.close