help-0

[hard harness]

I know the first row contains 4 options

The second row would be 5 because the 3rd option branches out to 2 options

After that, the first dot would have the value of 3 because 3 of the current choices lead there, and the other contains 2 for the same reason

Now, for the next one, the first dot has 2 options, so we multiply by 3 to account for the 3 previous choices you can make, which can create 3 more possibilities because for every choice of the second option, you also have the previous choices to consider.

And then the other choice automatically gets added to the next choice.

So at that point, 3 + 3 + 2 = 8

So for the next row, you have a dot with the value of 8.

The key strategy here is to multiply from the original value to the possible choices, and split them evenly.

And with extra original values added to a choice, you add those together to calculate the value of that dot.

In the end, you add the values up to get your final answer

Does that make sense?

[slow citrus]

am I understanding it correctly?

[hard harness]

I may be reading 1 as 7

[slow citrus]

ye ye it's 1

[hard harness]

But considering that is a one, you have the concept.

[slow citrus]

I'm still not sure about how divvy it up when it splits

[hard harness]

Basically, this is what the next row should look like

[slow citrus]

ahh

[hard harness]

So the values here would be 16-16-48

@slow citrus You think you can handle the rest of the problem?

[slow citrus]

I'll try my best 😄

[strange plover]

A typical question thanks for giving your valuable time

What is a bi - tensor?

[hard harness]

Good luck! 🙂👍

[slow citrus]

I think I'll figure it out with a bit of logic if I get stuck somewhere

now that I understand the basic principle

[strange plover]

Anyone there?

[hard harness]

Sorry, I'm not able to answer that question. Hopefully, someone will help you. :)

[strange plover]

Ok 👌 no problem

[hard harness]

And you can ping helpers if they don't respond within 15 minutes (10:50 AM EDT)

[strange plover]

Ohh thanks

[hard harness]

Yw! :)

[slow citrus]

I don't know if I figured it out, or if I'm stupid haha

[slow citrus]

Which one is it 😄

[strange plover]

This one

[slow citrus]

Sorry, I was tagging Bammm cause I assume he's not looking at this channel.

[strange plover]

Ohh ok

[hard harness]

Oops

Hmm

[slow citrus]

I don't know the answer to your question 256. Way beyond my pay grade lol

[strange plover]

No problem 👍

[hard harness]

I see what happened

Wait

Nvm

Yeah, I think 1248 is the answer

You did good. 👍

[slow citrus]

Thanks for you help

[hard harness]

Yw! :)

[hard harness]

@golden echo Sorry, but there is already a question

[slow citrus]

@hard harness Seeing the other questions being asked here, I feel a bit out of place hahha. Guess that's what art school does to ya. 😄

@golden echo There is another person waiting beside me

[hard harness]

<@&286206848099549185> Hey, there! Could you help him out?

[strange plover]

Ohhh sorry I was afk thanks for the call-out

[hard harness]

Yw

[strange plover]

<@&286206848099549185> Hey, Could you please help me out?

[pliant estuary]

Hey, probably dumb question, but Im trying to derive $$\vec{r}=cos(\omega t)$$ and I dont see why $$\dv{r}\omega t$$ become $\omega$, can maybe someone show me, how this was achieved?

[ocean seal]

marejak023

[vale wigeon]

d/dt (ωt) = ω.

also omega is a constant

[bronze trench]

Yes please. I know I have to use ratios but that part is messing me up.

[dawn wraith]

What are the properties of a bisector?

[pliant estuary]

the r was typo sorry, but when the omega is constant, shouldnt then be its derivation 0?

[vale wigeon]

but we aren't taking the derivative of omega

[dawn wraith]

You’re taking the derivative wrt t.

[vale wigeon]

we're taking the derivative of omega*t

[bronze trench]

QU bisects PQR

[dawn wraith]

Hold on bud

[bronze trench]

oop wrong message I replied to

k

[pliant estuary]

Okey thank you, I just started and Im following book on classical mechanics, so I was bit confused (bcs it uses less mathematics and just goes straight forward)

[dawn wraith]

Soo, I guess this means the channel is free

Let us continue

Yes, QU does bisect angle PQR

but it also creates a ratio

What could be that ratio

[bronze trench]

15/x=10/21

[dawn wraith]

Could you please send the problem again?

[bronze trench]

https://i.gamings.host/‍⁠‌⁠‌⁠‍​‌‍​‌⁠‌⁠‍‍​‌⁠‍‌‍​‍​‌⁠‌‍‍​‍⁠‌‍⁠‌​⁠‍​‌​⁠‌‍​⁠​​

[dawn wraith]

All right

So

No, that isn’t quite it

You see

[bronze trench]

not sure where I got 10 from 😅

[dawn wraith]

The bisector has a theorem in which is depicted like such:
RQ/PQ = RU/PU

Specifically speaking about this triangle

We’ll keep that in mind for later.

Now

Look at the triangles TSQ and URQ

they have a special connection

What could that be?

[bronze trench]

they are similar?

[dawn wraith]

Well-said

[strange plover]

Hey! could anyone please help me out?

With a tensor calculus problem?

[dawn wraith]

From there you should be fine

[bronze trench]

so I could do RQ/SQ=RU/ST

or 21/11=x/15

[molten terrace]

what is 2+3

[low dove]

5

[remote dagger]

Can someone help me with this question, i know how to do the coefficient with 1 bracket, but 2?

[manic quail]

What do you mean by "do the coefficient"?

[remote dagger]

This is the question

I know how to do this when there is only 1 bracket, but not with 2

[manic quail]

So just expand it, I think.

Are you having troubles expanding it?

[remote dagger]

No

I know how to expand but what then

[manic quail]

I think it wants to know the coefficient of x^2, once you expand and simplify.

[limpid rock]

i think you have problem with the left term..see when you multiply left term to whatever you're getting in right term, once the powers of x will remain same and once they will increase by two so just take x^2 and x^0 term from right and multiply to left, then add

[inland saffron]

can someone please explain the advantages to using quasi-Eucidean distance transform vs City-block?

[alpine sable]

can someone help with with this

[ornate ridge]

A scale factor of 2/3 means that 21 inch becomes 14 in and 15 in becomes 10 in. So new area would be 140 sq in.

[alpine sable]

Thank you! I appreciate it !

any ideas on how to get the value of x in terms of y?

@alpine sable let mw know (ping me) if you find the answer. I'm also curious but can't solve it

okay shore

will doo

so i dont understand what it means by this

i get that i have to take a derivative

but how exactly do i find the point

[dense current]

Hello there

[limpid rock]

make the derivative zero

[dense current]

I think u need to equate the derivative with zero

[limpid rock]

youll get the value(s) of x for which derivative is zero

find y for those corresponding values

did you try solving the quadratic inequality in e^x?

[chrome salmon]

You have to show $e^{2x}-3e^x +2 <0$

[ocean seal]

it's sam

[chrome salmon]

Taking e^x =t you can solve quadratic equation

$$t²-3t+2<0$$
$$t²-2t-t+2<0$$
$$t(t-2)-1(t-2)<0$$
$$(t-1)(t-2)<0$$

[ocean seal]

it's sam

[chrome salmon]

This will happen when exactly one of the factor is negative

$e^x + 2e^{-x} <3$
\ multiplying by e^x
$\ e^{2x}+ 2 < 3e^x$

[ocean seal]

it's sam
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[chrome salmon]

And e^x is always positive

So the inequality won't undergo any change

Just by experience

I've done such problems before so

Your welcome sire jealous

Take log on both sides

Its give x>0 so u can use log

[limpid rock]

just keep moving constant to other side and keep squaring

[chrome salmon]

Do as tough guy said

[limpid rock]

[visual fractal]

5^2x-1 + 5^ x-1 = 130

need help! and explanation
thanks :)

[strong furnace]

multiply by 5 on both sides and you have a quadratic in 5^x

x=+-3/4?

solution must be trivial x=1

you take log on both sides

get lnx(1+P(x))=lnx*x

[limpid rock]

x can be 2 as well

you just cancelled out (x-2)^2

[strong furnace]

never cancel terms when finding the roots just take them in bracket :/ that is how you can prove there are no other roots

I was going to finish that

then take them both to LHS we get lnx*(1+P(x)-x)=0

for roots ln(x)=0 and 1+P(x)-x=0 as they are the two factors

you're not supposed to divide by 0 and when you divide by a term which may be 0 at some x you are not considering x to be equal to that number hence ignoring that root

[visual fractal]

@strong furnace can i dm you about my question?

so we don't make a mess here

[stiff tiger]

can some1 help with 17g

[graceful lance]

how to do (iv)

[strong furnace]

4x+3=0 gives us x=-3/4 , we were given x>0

[graceful lance]

how to do (iv)

[strong furnace]

I think what they want you to do is take the answer to (ii) and divide by the answer to (iii)

[graceful lance]

its not that

[wooden saddle]

Why do you need the cosinus rule to calculate R ?

Why is R not simply sqrt(40^2 + 60^2) ?

Oh cause it's not a right triangle

That's what I call a good start to the semester

[harsh swallow]

you figured it out before i could explain 😄

[honest echo]

Is ((x+iy)(x+iy))* the same as (x+iy)* (x+iy)*?

[harsh swallow]

*?

[honest echo]

complex conjugate

[sly mantle]

in general conjugate distributes over products

[harsh swallow]

you can look

[sly mantle]

it also distributes over sums

[harsh swallow]

multiply then conjugate and conjugate then multiple, check if they're the same

[honest echo]

Referring to 36 btw

It's really easy if it does distribute over products

But I can't recall learning that so I feel like I have to somehow prove that

[harsh swallow]

i did it

it is the same

[honest echo]

How?

[harsh swallow]

the answer of ((a + bi)(c + di))* = (a + bi)* (c + di)* = (ac - bd) - (bc + ad)i

[honest echo]

Wonder if I have to show that in the task

Because if I can just assume that to be true it seems suspiciously simple

[harsh swallow]

it's really easy to prove if you do have to

[honest echo]

We haven't learned how to do conjugates of expressions that aren't in the form x+iy

[harsh swallow]

it's always in the form a + bi

you just have to group the a bit together and the b bit together

ac - bd = x

[honest echo]

but if you expand the (a+bi)(a+bi)

[harsh swallow]

bc + ad = y

expand that

i'll explain

[honest echo]

expand what

[harsh swallow]

this

[honest echo]

xˆ2 - xˆ2 yˆ2 -yˆ2

or?

sry im used to using x and y

[harsh swallow]

it's all good

[honest echo]

aˆ2 - aˆ2 bˆ2 - bˆ2

[harsh swallow]

$x^2 + y^2 i^2 + 2xyi$

[ocean seal]

Katharine

[harsh swallow]

$(x + yi)(x + yi)$

[ocean seal]

Katharine

[harsh swallow]

you first multiply the x

first term

then multiply the yi

second term

then multiply the first x with the second yi

and then the second x first yi

now

[honest echo]

yep

[harsh swallow]

i^2

[honest echo]

-1

[harsh swallow]

exactly

so we can rewrite this as

$(a^2 - b^2)+ (2ab)i$

i put the specific bits into brackets

to better show it

that is now in the form

x + yi

[ocean seal]

Katharine

[harsh swallow]

maybe this is less confusing

this is now x + yi

do you see that

[honest echo]

oh you leave the xyi

[harsh swallow]

you can easily get the complex conjugate

[honest echo]

so the real part stays the same and the sign of the Im(z) changes to a negative?

where Im(z) is (2ab)i

[harsh swallow]

if $(a^2 - b^2)+ (2ab)i = z$
then $Im(z) = 2ab$

[honest echo]

yeah

[ocean seal]

Katharine

[harsh swallow]

no i

anyway

if we want to know

[honest echo]

oh i got it

thanks so much

[harsh swallow]

$\overline{((a+ bi)(c + di))} \stackrel{?}{=} \overline{(a + bi)} \cdot \overline{(c + di)}$

[ocean seal]

Katharine

[harsh swallow]

whatever you know what i mean 😄

[honest echo]

lmao ive never seen that notation somehow but yeah i have the answer now at least

[harsh swallow]

overline notation is another way of showing complex conjugate iirc

[honest echo]

oh I see

[vestal nebula]

Yes $\overline{((a+bi)(c+di))} \stackrel{=} \overline{(a+bi)} \cdot \overline{(c+di)}$

[ocean seal]

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

[harsh swallow]

just write equal

not stackrel

[vestal nebula]

Okay

Same happens when you divide two conjugates

[graceful lance]

How to do this

[wary stream]

You could factor by grouping but because that's not possible for that equation, start plugging and chugging. Plug in 1, and see if that equals 0, if not do -1, and when you find the first root, do long division for the others

[strong furnace]

assume roots a,b,c 2b=a+c plug in a+b+c=6 to get b = 2 , assume c = 2+d a =2-d use abc=-10 to get d =+-3 c and a are interchangable so you get -1,2,5

[wooden saddle]

Bruh why is the hypothenuse x instead of 7

I mean

Why is the hypothenuse 7 instead of x

[strong furnace]

hypotenuse is the side opposite to the right angle(that square thing)

[wooden saddle]

Ohh okay

I love how I just passed diff. eq and now for Physics 1 I go back to the absolute basics of which side is the hypothenuse :' )

[graceful lance]

Whoa

[tawdry plover]

I can also assume the roots to be (a-1), a and (a+1) find a by sum of roots and then use that

[strong furnace]

you can factorize it and break the number line into intervals between its roots and assign signs or you can plot the graph and do the same thing

(x-1)(x-2)<0 now make a numberline mark 1 and 2 these are the points where LHS changes signs , now assign signs to the remaining intervals

[still crest]

Could someone help me with this?

[strong furnace]

write slope equation for AE and EB they should be equal for A,B,E to be colinear replace x1 with y1^2/4a and x2 with y2^2/4a

[solar pebble]

Don't suppose anyone has any exercises for differential equations using Laplace transforms?

[still crest]

ty

[still crest]

How did we come up with y1^2/4a?

[still crest]

@graceful lance

[graceful lance]

yea

I just saw that it was in use

[still crest]

no worries^^

[graceful lance]

It was already posted, I am having slow network

[strong furnace]

the point lies on the parabola so it must satisfy parabola's equation

[still crest]

Oh wait there's an x on the parabola equation, oops

Let me give it a shot now

[flat gust]

how do u do this

[still crest]

ty

[flat gust]

Oh sorfry

[still crest]

no worries^^

This question might sound dumb but when you multiply 2 fractions, the denominators get multiplied too correct?

[strong furnace]

yes e.g. - 1/2*3/4 = 3/8

[still crest]

I had to make sure thanks

This got quite complicated

Let me send you a photo

@alpine sable

[alpine sable]

My bad

[still crest]

no worries^^

@strong furnace Have I done anything wrong? It seems wrong to me:/

[strong furnace]

you messed up the slope equation for RHS in line 2 you have x above y

[still crest]

Ohhh

Correct

I feel that this is still incorrect

[strong furnace]

2nd line has y1-a in LHS should y1-0/x1-a

[still crest]

Correct

ty

[keen wasp]

oh i thought correct meant he was done

[still crest]

Not yet, I'm sorry

I proved the first one

What about x1x2 = a^2

Do I substitute, where y1 = sqrt(4ax1) etc?

[strong furnace]

square y1y2=-4a^2 on both sides replace y1^2 with x1 and y2^2 with x2

[still crest]

Oh I did it opposite

That works

Let me clarify

Perfect

Thanks so much!

[thin berry]

What does this mean

I don't understand the whole thing lol

[strong furnace]

fog(x) = f(g(x))

[thin berry]

Alright thank you

[still crest]

I have done a)

[thick yoke]

try representing point B by x1 (bot its x and y values) and then find line AB and point C (intersection). then calculate the distances

I don't see any easier way

[still crest]

For starting with something like this I made a graph

And I put A two times away from (0,0) than the E(focus)

Something like this

,rotate

[still crest]

Does this seem right?

[rotund trout]

1+1 = ??

[still crest]

2

@worn trail better helper wins

🥲

[alpine sable]

brr

texit hard

Ok, so im stuck on this question, I've got the derivative of the function, but I can't piece together the rest of the question.

[cursive badger]

find the zeroes of the derivative, use sign analysis to see where the derivative is positive (original graph increasing) and negative (original graph decreasing)

[alpine sable]

find the zeroes of the derivative
You lost me there. sorry for being slow, but what??

solve f'(x)=0

[wary stream]

Find the derivative and set it equal to 0, is what they mean

[alpine sable]

ah

[candid yoke]

And then you can check the values of the derivate on each interval. Say you have 2 solutions to f'(x) = 0, x1 = 3, x2 = 5.
If the value of f'(X) is negative on interval -infinity 3, then the function is decreasing on interval, you do the same for interval 3 - 5 and 5 + infinity

[viscid yacht]

Hey is this channel open

I would like assistance with this packet ( I already tried😫

it’s a practice for my regents

[alpine sable]

Even i have we have the same packet

[viscid yacht]

Ye

[alpine sable]

Trying to work together

[viscid yacht]

Question 1 first

[wary stream]

Find the slope at each interval and compare them

[viscid yacht]

Uh

how do I find slope again

Change in y over x?

[wary stream]

Yeah

$slope = (y_2 - y_1)/(x_2 - x_1)$

[ocean seal]

dldh06

[alpine sable]

So could you walk it with us through the first choice?

2002-2004?

[wary stream]

How would you find the slope for that interval?

[alpine sable]

I just dont know what numbers to plug in where

I know the formula

[viscid yacht]

Would slope be 2.5?

[alpine sable]

Year is in x

But what would the change in x be then

I got 2/12 lolll

[viscid yacht]

No 2.5

[alpine sable]

Ho ww

[viscid yacht]

Slope is 2.6 I think

2.5

[alpine sable]

Could you please walk us through how to fund 1

Find

[wary stream]

How would you get 2.5?

[viscid yacht]

Bruh wait

[alpine sable]

She did to 2003

[viscid yacht]

I Fis that wrong

[alpine sable]

Is my answer correct?

A1/6?

1/6

For the first slope

[wary stream]

Yeah, I got 1/6

[alpine sable]

WAIT

no

[viscid yacht]

No just 6

[alpine sable]

Cus x goes on the botton

Bottom

[viscid yacht]

It’s 6

[alpine sable]

I got 6now

[wary stream]

Wait, I had it flipped too, it's 6

[alpine sable]

Ok

[viscid yacht]

Choice 2

Ugh

[alpine sable]

9.5?

For choice 2?

[viscid yacht]

Yea

[wary stream]

Yeah

[alpine sable]

8.5 for 3?

[wary stream]

Yeah

[alpine sable]

7 for 4?

Which means the answer is choice 2?

[wary stream]

Yeah, looks like it

[alpine sable]

For question 2 is the answer number 2?

[wary stream]

Yeah

[alpine sable]

For question three we have to show work how would i solve it?

[wary stream]

Find the vertex of the parabola

[alpine sable]

Ok then what?

How does the vertex lead to the answer

[wary stream]

The vertex is going to be the lowest/highest part of the parabola

[alpine sable]

Ohh

So how would i do the vertex for that specific thing

Like how do i complete the square

[wary stream]

You can, or use $-b/2a$ and plug that in to find y

[ocean seal]

dldh06

[alpine sable]

I think my teacher wants us to complete the square

How would i do it that method

Or wait

Ill just so that method

So

-2 over 2(-1)

[wary stream]

Another to note, is look at the a term, on if it opens up or down

[alpine sable]

So i got 1 for x is that correct?

[wary stream]

Yeah

[alpine sable]

So what do i do next?

[wary stream]

The a term will determine if it's less or greater than the vertex

[alpine sable]

How do i find the y value now

[wary stream]

Plug that x into the equation to find y

[alpine sable]

Ok

I got (1, -7) as vertex did u get that?

[wary stream]

Yeah

[alpine sable]

Wait the a term was supposed to go in parentheses right?

-(1) squared?

Or -1 squared

[wary stream]

-(1)

[alpine sable]

Ok

So its either 1 or 2

[wary stream]

Determine if it opens up or down

[alpine sable]

It opens down wards

So 1?

[wary stream]

Yeah

[alpine sable]

Thanks so much

[lime crescent]

I need help making a function for video game math. Each point of armor awards 1% damage reduction, but as armor increases, that % value gets smaller and smaller. So some kind of square root formula, I would assume. But the slope of the graph is never greater than 1.

How do I control the shape of this curve?

[candid yoke]

Well, it depends on how fast you want it to get smaller and smaller

[alpine sable]

For question 4 would the answer be 4?

[candid yoke]

Like, if you have 1 armour you would like to receive 1% less damage, but for 100 it should be not 100, right?

[lime crescent]

Correct.

So how would I control the shape of this curve?

I know putting multipliers in different places tweaks the shape.

[candid yoke]

Hmm, I think it depends on different parameters, but I would use something as simple as a log function

[lime crescent]

Yeah, I'm not good with that part of graphing. I was starting with y = x / sqrt(x) and trying to figure out how to alter the shape from there.

How would I tweak a log function to get the desired shape?

[candid yoke]

Well, it depends on what shape you want, here is how it would look for some random parameters

I would play around with the base, 100 is not good, since if X = 100, you would have 501% damage reduction which is not good :D

[modest bramble]

hey can someone help me with this problem?

<@&286206848099549185> ? sorry if i annoyed anyone with this ping

[coral pagoda]

Okay sure. So first, what does the first part tell us about our box and whisker plot?

[modest bramble]

All four of the box plots lie on the same scale and contain the same number of data points.

Identify the box plot that best represents the description below?

[ivory veldt]

I. Middle 50% of data is not evenly distributed about the median. This means that the line marking the median doesn't evenly split the box left and right. It's skewed either to the left or to the right.

[coral pagoda]

I am capable of reading the question ues

[modest bramble]

LOL

[coral pagoda]

I asked them

[ivory veldt]

So that means it can't be number 4

[modest bramble]

so its not 4 or 2

[coral pagoda]

This is correct

[modest bramble]

but isnt it asking which one has all 3

[coral pagoda]

It also says the data distribution is not symmetric

[modest bramble]

the first one doesnt have an outlier right?

[coral pagoda]

There are two w/o outliers

[modest bramble]

yeah

so its the third?

[stable dune]

can someone help me get the answer?

[coral pagoda]

You got it Celia

[modest bramble]

thanks! @coral pagoda

[coral pagoda]

You betcha

[modest bramble]

I have couple more if you would like to help?

i got a little mixed up with the iqr

[hybrid pike]

I think the answer is 3

[modest bramble]

can you explain why?

cause it was 3

[wary stream]

@modest bramble Isn't interquartile range the upper quartile minus lower quartile? So 12.5 - 9.5?

[tiny rampart]

I think the answer is 92.11 in

[stable dune]

thank you!

[modest bramble]

wait yes i need help to figure how to do that

i have a median question

isnt it 17@wary stream

[wary stream]

Nope

The median is the middle value of the total number of items in the set

So for example a set, {1, 2, 3}, the median is 2

[modest bramble]

yeah

theres 7

[wary stream]

There's 7 values, but some have multiple

[modest bramble]

OH

but median is middle

[wary stream]

The set for that is {14,14,14,14,15,15,15,16,16,17,18,20,20}, right?

[modest bramble]

yuh

[wary stream]

What's the middle term of that set

[modest bramble]

15

[stable dune]

i need help getting this answer now

[modest bramble]

is it a test?

[wary stream]

Exactly

[modest bramble]

but i have to click the 3

[stable dune]

no its supposed to be homework

[modest bramble]

can i skreen?

[stable dune]

wdym?

wait i got the answer nvm

[flat gust]

How do u do this

So do you put that in vertex form and just find the max height?

[gloomy lintel]

Or you could differentiate and set =0

Up to you, I’d probs complete the square like you said

[flat gust]

the answer is this but i dont get how

we're looking at the y right? y doesnt equal to 2069

wait no

we're looking x

[gloomy lintel]

Yeah, y is pop number but x will tell us how many years after 2000 that pop number occurs

So we care about x rn

[modest bramble]

can someone help with a math question

[flat gust]

oh ok got it

[coral pagoda]

In terms of the box and whisker plot, the 25%-75% quartiles is the box itself. So to find the range, just evaluate the difference between the endpoints of the box.

[modest bramble]

to make it easier
interquartile range 1. mean 2. mean absolute deviation 3. range. 4. mean 5.
and for the second variability or center

[gloomy lintel]

That’s the standard deviation isn’t it

[coral pagoda]

No

Standard deviation does not play a role here

Actually, it's somewhat close. I take it back

[gloomy lintel]

Sorry the variance

[coral pagoda]

Even then, we aren't squaring the differences here

For this you take the average of the absolute value of the differences

[modest bramble]

?

mean is average

._. im so braindead

[coral pagoda]

So you had 5 points ${x_1,x_2,x_3,x_4,x_5}$ and the mean is given be $\mu$. Then the average referred in this problem is given by [ \frac{|x_1-\mu |+|x_2-\mu |+|x_3-\mu |+|x_4-\mu |+|x_5-\mu|}{5}]

[modest bramble]

i got it

mean absolute deviation, and variability

[coral pagoda]

Yes

[ocean seal]

dackid (jump king +)

[modest bramble]

this is my very last question

[coral pagoda]

For variance, you square the differences instead of take the absolute values

Celia... is this a test

[modest bramble]

no

its a quiz to help study

i finished my test last week

[coral pagoda]

Okay

[modest bramble]

i can skreen share it

[coral pagoda]

It's fine. I'll take your word for it.

[modest bramble]

._.

whats interquartile range?

i kind of get mixed up with it

[coral pagoda]

That's what the rectangle was in the box and whisker plot

It is the middle 50% of the data

[muted prairie]

what is the difference between discrete and continuous variable

[wary stream]

Discrete is finite while continuous is infinite

[round nova]

discrete is capped at certain values. I.e, X is 1, 2, 3, or 4.
Continuous is like saying, X can be any value between 0.1 and 0.2

[muted prairie]

oh

thanks guys

[wary stream]

Use pythagorean theorem

[zenith beacon]

how

Isn’t that for if you have two sides

[wary stream]

It's an isosceles right triangle

[west harness]

The two sides identcal ,see the marking

[zenith beacon]

Oh

Ok

[wary stream]

Hence, isosceles right triangle

[zenith beacon]

Is it 3?

[left shore]

Nah it's D

@zenith beacon

[zenith beacon]

oh

I was multiplying

[wary stream]

Do give out answers like that, let people solve it and if they get it wrong, see where they made the mistake and help from there. People don't learn if you spit out answers

[zenith beacon]

I missed class all last week so I don’t have a clue how to do this one

[woeful pulsar]

okay you heard of SOHCAHTOA?

[zenith beacon]

no

[south path]

im kinda stupid so i might need to ask a lot of querstions

[left shore]

Ok sry broo m new here so I don't have any idea about how it works here
I will take care of it

[woeful pulsar]

There are three sides of the triangle, the Adjacent, Opposite and Hypotenuse

Hypotenuse is always opposite the right angle

[zenith beacon]

yes

[woeful pulsar]

Opposite means opposite the angle you have

Adjacent means the leg of the triangle next to the angle

SOHCAHTOA is a mnemonic meant to remember sin, cos and tan as ratios of the sides

[zenith beacon]

ok

so a to c is the hypotenuse

[woeful pulsar]

yeah, AC is the hypotenuse

[zenith beacon]

and b to c is adjacent

A b opposite?

[wary stream]

I don't think it's a rule to spit out answers, but it helps to teach the person the method instead of spit out answers

[zenith beacon]

So how I use that to find sin a

[woeful pulsar]

we can use the SOH part of the mnemonic, which says sin(A)=Opposite/Hypotenuse

can you find those side lengths?

[zenith beacon]

5/12?

[left shore]

What is hypotenuse here bro@zenith beacon

[woeful pulsar]

remember which side is the opposite and which side is the Hypotenuse?

[zenith beacon]

A c

Wait

[left shore]

Yeh nice

[zenith beacon]

so wouldn’t that be 5/AC

[left shore]

It is oppo side of angle / hypo@zenith beacon

[woeful pulsar]

the side opposite A is not of length 5

[zenith beacon]

12

[left shore]

Yes

[zenith beacon]

12/AC

or 12/hypotenuse

[left shore]

Now find out hypotenuse by Pythagoras

[zenith beacon]

Ok

12/13?

[left shore]

Yupp nice

[zenith beacon]

ok so for this one I have the hypotenuse and the opposite

But I’m confused on what tan f even is

[wary stream]

Are these homework or test questions?

[zenith beacon]

this is a homework review for the test Monday

is that ok to do

cause it’s practice

[wary stream]

Practice/homework help is fine

[south path]

could someone help me?

[zenith beacon]

so what exactly is tan f

[left shore]

As you know sinF = oppo/hypo
Just like that tanF = oppo / adjecent

[zenith beacon]

ok

I’m screenshot ing that

Is this for everything

Like is that a rule

[left shore]

Yes yes

[zenith beacon]

Ok

How do I find the adjacent with the hypotenuse and opposite

[earnest pendant]

Can you use a squared plus b squared equals c squared to find the adjacent? And the. Solve for tan that way?

[zenith beacon]

Is the adjacent 12

9/12

?

simplify to 3/4?

@left shore

[wary stream]

Yes

[left shore]

Yes correct

@zenith beacon

[zenith beacon]

what is the rule for cos

[hasty seal]

SOH CAH TOA

[left shore]

For cos it's adjecent/hypo

[zenith beacon]

ok lemme screenshot that

[wary stream]

https://www.khanacademy.org/math/geometry-home/right-triangles-topic/intro-to-the-trig-ratios-geo/v/basic-trigonometry
Watch that to understand the trig ratios

[zenith beacon]

so sin C would be 5/12

[hasty seal]

5/13*

[zenith beacon]

oops

that’s what I meant

[hasty seal]

yeah, you can basically decorate that phrase

[zenith beacon]

Would cos A be 12/13?

[left shore]

No

[zenith beacon]

I just realized that

So it would be cos C

[left shore]

Yes

[zenith beacon]

I’m bad with degrees

so one is 90 right

but how would I find the other two

[woeful pulsar]

angle x

so you'll need to use inverse trigonometric functions

[zenith beacon]

pls explain that

[woeful pulsar]

so, you have trigonometric functions

they allow you to calculate ratios from angles

but now you can calculate the ratios

so you can find the angle from the ratio

[zenith beacon]

so what function would I do

like what side / angle do I put into something

is there a rule for this

[glass lichen]

SOH CAH TOA if you cant remember. .

[zenith beacon]

ohhhh

wait

if I want to find X

I’m confused

[glass lichen]

Yes, so pick your favourite trig ratio since you know all 3 sides, and determine which sides you need relative to X

[zenith beacon]

5 and 12?

[glass lichen]

sure, what trig ratio is that?

[zenith beacon]

I’m not sure this is all new to me I’m returning to school tomorrow for the first time in a while I was in the hospital a bit ago

just really a lot to take in

[glass lichen]

Ok but you said 2 random sides, so Im asking what trig ratio uses 5 and 12

[woeful pulsar]

hmm, a good chance to talk to teachers to figure out

[wary stream]

You should talk to your teachers and watch some Khan academy videos

[zenith beacon]

yea hopefully they push the test back for me

So does the rules like opposite/hypotenuse work for this or

[hasty seal]

they do

but you would find something like cos(x)=5/13

and to solve to x, you would have to use the inverse trig. functions

[zenith beacon]

Is the trig ratio for X Tan X

[vivid brook]

what do you mean by that

[zenith beacon]

I’m confused on how to find what trig ratio to use

[vivid brook]

youre given all the sides, you can use the inverse of any of the different functions

you just want to use the correct set of sides for the trig function you pick

so if you use inverse sin, then use the side opposite of X divided by the hypotenouse

[zenith beacon]

so can you pick any trig function?

[alpine sable]

Hello, I need help on this question. I did the question by myself but i got stuck mid way.

[hybrid pike]

@vivid brook, can I dm u?

[vivid brook]

for what

[hybrid pike]

Math help

[vivid brook]

uh sure?

[zenith beacon]

12/13

for the side opposite of X divide by hypotenuse

? @vivid brook

[vivid brook]

for inverse sin yeah thatd work

[zenith beacon]

So how do I use 12/13 to get X

do I do 12/13=C^2

[vivid brook]

inverse sin of 12/13

[zenith beacon]

I’m confused

How do I use that to get the angle

[vivid brook]

its -25 since the second term is negative

[vivid brook]

inverse sin of a value results in an angle measurement

do you know what the inverse sin function is?

[zenith beacon]

no

[vivid brook]

its like the reverse sin

[zenith beacon]

that’s why I’m confused

[vivid brook]

you put in a proportion of sides and it gives an angle

its the trig function that looks like it has an exponent of -1

[normal mortar]

Soh, cah, toa = sho, cha, tao

[woeful pulsar]

might want to use a different channel

[vivid brook]

o y?

is it cuz this channel is occupied? i tried to answer the questions above me before posting

[alpine sable]

which -25?, There are two answers that have -25

[zenith beacon]

it’s ok I’ll try and figure it out

[woeful pulsar]

yeah but you should make sure that the questions are done tho

[vivid brook]

A since the second term has 27 as a result of 9*3

[woeful pulsar]

otherwise if the person asking still finds it unclear we have two questions in a channel which is tedious

[vivid brook]

okie

[dusky quarry]

how do i show a probability function is well defined ?

like this ones

[woeful pulsar]

probability density?

firstly the area under the entire curve must be 1

and it must be always nonnegative

[dusky quarry]

yep, but i still don't get how am i suppose to show that it is indeed well defined

[woeful pulsar]

hmm looks like you just need to check nonnegative everywhere

and total area 1

[gentle ocean]

Can someone help mee

[glass lichen]

Integrate

[gentle ocean]

Yeh ik

But

[glass lichen]

then you know x(0)=5

[gentle ocean]

I’m lost

[glass lichen]

lost with what?

[gentle ocean]

Huh?

I got x=e^3t +C

[glass lichen]

yes

[gentle ocean]

then idk how to find the equation

[glass lichen]

$x(t)=e^{3t}+C$

[ocean seal]

moshill1

[glass lichen]

and you know $x(0)=5$

[ocean seal]

moshill1

[glass lichen]

so $5=e^{3(0)}+C$

[ocean seal]

moshill1

[gentle ocean]

OMG

[alpine sable]

Hey im trying to learn algebra on my own because my school is just too slow so can i ask stuff here (im just starting off)

[gentle ocean]

I’m so dumb

[glass lichen]

so $x(t)=e^{3t}+4$

[ocean seal]

moshill1

[gentle ocean]

it says initial 😭😭

[glass lichen]

yeah

[gentle ocean]

thank you

😭😭😭😭

[alpine sable]

How would you guys solve

6a - 4 = 2a +8

[left sparrow]

carry over like variables, that makes it 4a = 12, divide both sides by four, that gives you a = 3

[alpine sable]

K

I guess i did it in one more step

My other question

5x + 8 = -3x + 40 =
5x = -3x + 32
Can i do this:
8x = 32

Nvm

Lol there are so many different ways to approaching things in math

[glass lichen]

I mean you did it the same way as was suggested

[alpine sable]

I know

Never mind

Its like walking trough 1 door and then seeing there is another door behind you that leads to the same room but then wen you have walked through it you only see 1 door

Its 4:22 am i should probably take some sleep

[woeful pulsar]

depends if you just woke up or stayed up that late

[alpine sable]

Well judging by the context i probably stayed up late

Can i ask why you guys do math

[hearty flower]

How would you go about showing that 5/3 - ln(6) is negative

[cold ingot]

if plotting the graph is an option, that way we can prove it

[strange plover]

Hey, a question here, thanks for spending your time,

What is Stokes theorem on Manifolds and what it means to integrate a one form over a region (a surface integral in formal terms)?

[nocturne gazelle]

is there a channel i can ask about runtime and correctness of an algorithm?

[gloomy remnant]

can someone help me with this please

https://gyazo.com/6f0dcae64de647276aa6ade86192ddd1

<@&286206848099549185>

[west heart]

Hi! Qq, would the node order: A, C, D, B, E, F, G, I, H be a topological order? Thanks!

[strange plover]

<@&286206848099549185> please help me out.

[wary stream]

You can create two equations with the given information

[molten moth]

Work backwards @west heart

[oak chasm]

@west heart Sorry, channel is busy.

@west heart #help-8 is open if you hurry.

[strange plover]

Anyone here?

[oak chasm]

@next pumice My guess would be u substitution.

u = sin(x)
du = cos(x) dx

[molten moth]

https://youtu.be/1lGM5DEdMaw @strange plover

[strange plover]

Thanks

[molten moth]

Can also use taylor series jkjk

[thorn kindle]

I'm good at algebra and calculus but i kinda suck at geometry

Anyone wanna help me take a stab at this?

The goal is to find theta

[ocean seal]

Chai T. Rex

[thorn kindle]

Yes

[onyx sable]

hey i need help with this anyone?

[thorn kindle]

@onyx sable occupied

[oak chasm]

Then the figure is impossible with those measurements, as AB is longer then CD, but the measurements show that that's not true.

[onyx sable]

kk

[thorn kindle]

Definitely not to scale at all

This is actually a physics problem

It should be possible though

The drawing is just for visualization. Itd be impossible to draw it accurately

[oak chasm]

It's not the scale that's the problem. With the measurements and angles given, that figure is impossible independent of the scale of the image.

A would have to be above and to the right of C.

Very close to B.

[thorn kindle]

Dont worry about the actual lengths

I assure you that this is an actual physics problem which does appear in the real world

[oak chasm]

Then you're dealing with some very noneuclidean space.

[thorn kindle]

It's just light lmao

[oak chasm]

So, you can't use Euclidean geometry.

[thorn kindle]

Lmao

Read the units again

AB is 1000 NANOMETERS

[oak chasm]

Right, 1 micrometer and a shorter length is 2 meters.

That's not true.

I don't know any noneuclidean geometry, so I can't really help with this.

[thorn kindle]

I mean what can i say? I really cant draw it to scale.

Its not noneuclidean

[oak chasm]

It's not about the scale.

[thorn kindle]

The lines are not in proportion to each other.

The length of each line does not reflect their actual length

[vale wigeon]

can i see the diagram again

[oak chasm]

You have two right triangles, ACF and FDB. The sum of the hypotenuses of both will be longer than the sums of a leg of both.

[thorn kindle]

A, C, and E really should be right on top of each other

[oak chasm]

@vale wigeon They also said that ACD is a right angle.

[vale wigeon]

so are AE and BD meant to be parallel to each other

[thorn kindle]

Yes

[vale wigeon]

then this is impossible

AB must be at least as long as CD

and yet it's literally one micron

and CD is two whole meters

[oak chasm]

If you're having trouble seeing why, notice that ACF and FDB are right triangles. AB is the sum of the hypotenuses. CD is the sum of two legs.

The hypotenuses must be at least as long as their legs, so the sum of the hypotenuses must be at least the sum of their legs.

This is irrelevant of scaling.

[light kestrel]

@oak chasm once ur done with him dont leave

[vale wigeon]

reddevils123 if you have a question of your own you can and should post it in one of the other nine questions channels

one that's free obviously

[oak chasm]

@light kestrel I might not be able to help, but you can ask in #help-8 if you hurry.

[vale wigeon]

make a coordinate system with C as the origin and the x axis running along CD
then A = (0, -5e-5) and D = (2, 0) and B = (2, y) for some positive y (according to the diagram)

there's no way in hell for AB to be 1e-6

@thorn kindle are you absolutely sure you didn't royally fuck up the lengths

[thorn kindle]

i did fuck it up

it's not 1000 nm

[vale wigeon]

@crude prairie channel busy, please move

[thorn kindle]

it's just gotta be some multiple of 500 nm?

[vale wigeon]

wait you said this was for a physics problem

[thorn kindle]

maybe i should go to physics server unless you by chance understand light

[vale wigeon]

can you show the original problem

[thorn kindle]

sure

[vale wigeon]

uh huh

let's see

lemme try to draw a diagram for this

[thorn kindle]

yeah it's definitely not 1000 nm as the entire length. just that the two sources of light have to be in phase (in this case off by two wavelengths.) so in my original diagram you could say EB + 1000 nm = AB

[vale wigeon]

,w 500nm wavelength color

[ocean seal]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by Ann#0413

[vale wigeon]

okay nvm

[thorn kindle]

i don't think color should matter though

[vale wigeon]

i just wanted to know what color pen to use in my diagram lol

[thorn kindle]

Here is an actually possible diagram

Not sure if it's possible to solve though

[young notch]

Help

[vale wigeon]

@young notch channel busy please move

[thorn kindle]

yes

i honestly don't think this is possible without using the equation

[vale wigeon]

using what equation

[thorn kindle]

let the distance between slits = d.
let m represent the order of the maximum. (2)
let theta be the angle from the center line to the maximum.
let lambda represent the wavelength

[vale wigeon]

okay so your d is my h

[thorn kindle]

then dsin(theta) = m*lambda

[vale wigeon]

i have successfully reduced $S_2F - S_1F = 2\lambda$ to a quadratic in $y$

[ocean seal]

Ann

[vale wigeon]

@onyx sable channel busy please move

[onyx sable]

k

[thorn kindle]

S_nF represents the hypotenuse of the right triangle right?

whose horizontal length is the same for both triangles

[vale wigeon]

??

well if you insist

i'm just doing everything in terms of coordinates

[thorn kindle]

how did you do it in terms of y?

[vale wigeon]

algebruh

[thorn kindle]

ohh i see.

yeah using coordinates seems like a good idea

now all that's left to do is take the inverse tangent of y/2 meters

but somehow that whole process is also the same as a simple equation?

[vale wigeon]

not quite

$h \sin(θ) = m\lambda$ is only an approximation and it works reasonably well for small values of $m$

[ocean seal]

Ann

[vale wigeon]

or, well, it works well when the angle between the rays hitting the fringe is small

[thorn kindle]

so when sin theta = theta?

[late tundra]

Hi

I'm new

Not sure what todo

[vale wigeon]

adithyan, do you have a math problem that you are struggling on?

[thorn kindle]

alright ann i think we're pretty much done thanks

[late tundra]

Does this work like someone posts a question and I try to answer it ?

[vale wigeon]

someone posts a question and you try to guide them along, without just giving out the answer

read #❓how-to-get-help

[late tundra]

I don't have Math questions that I am struggling on but I do have questions that I have think might be challenging for you

[vale wigeon]

is that a generic you, or does that refer to me specifically

[late tundra]

And I want more math questions that has an elegant solution or something clever about it
That'

[vale wigeon]

in the latter case i am intrigued

[late tundra]

A generic 'you'

Well generic includes you tooo

[vale wigeon]

there are a lot of people here with very different backgrounds

[late tundra]

Well this is my first time in a social media that connects myself to strangers

So what is your background ?
Your age , your accomplishments,
Your country

I'm from India

17

Likes math

Has rediscovered a few math ideas

[vale wigeon]

calling discord social media is a bit of a stretch

[late tundra]

Well it does connect people doesn't it

[vale wigeon]

but if you're so interested, i'm 21, Russian and writing my bachelor's thesis in applied math

[late tundra]

Wow

Great

[thorn kindle]

antisocial media

[late tundra]

I feel like I'm in the right place

[gray gorge]

Welcome

[little whale]

hi

perms and combs

6 questions (numbered 1 to 6), can score 0,1,2,3 on each question, how many ways to score a total of 15

not sure how. to start

thanks in advance

[gray gorge]

https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)?wprov=sfla1

Theorem 2

Although

[little whale]

oh thanks

[gray gorge]

You're restricted in which scores you can get

[little whale]

i didnt think of thinking of it that way

[glad matrix]

find the area bounded by x axis and the curve x²=4by and the ordinate at the point (b,0)

[gray gorge]

So it's not completely the same

[glad matrix]

helpers plz find the area bounded by x axis and the curve x²=4by and the ordinate at the point (b,0)

[gray gorge]

It requires some extra thinking

[little whale]

ah ic

hmm ill c how i go thanks:d

[gray gorge]

Np

[glad matrix]

find the area bounded by x axis and the curve x²=4by and the ordinate at the point (b,0)( via integration)

[gray gorge]

What have you tried?

[late tundra]

What is the shortest distance from P to Q such that the path from P to Q touches the straight line

[strange plover]

Nice to meet you I joined yesterday and I have to be true the people here are very kind and friendly 👍.

[late tundra]

Thanks for the info

Am in the right age group

That's what I wanna know

[late tundra]

How old r u

[strange plover]

19

[late tundra]

And what country r u from

[late tundra]

OK now try this question I know the answer it's more like a friendly challenge

[glad matrix]

find the area bounded by x axis and the curve x²=4by and the ordinate at the point (b,0)(via integration)

[safe sable]

this makes no sense to me help would be appreciated

[strange plover]

U can ping the helpers within 15 mins of uploading the question, I am from UK btw

[late tundra]

Why don't u answer it

[normal mortar]

For 3, find a coordinate point where you can clearly see what one pair of (x,y) is. Then another two points somewhere else on line A. Use the formula (y2-y1)/(x2-x1) to find the slope and plug in the value for "m". Whichever value you found for y1 and x1, plug in those values into the given equation as well, distribute, and simplify

Hint: (0, -2) is one group of pairs I found. Find another one. Or different one.

Number 4 is just what you turn the equation from number 3 into after simplifying

[safe sable]

so for question 3 i have y-10=-3(x-(-4)) and for 4 i have y=-3x-2

is this correct?

[normal mortar]

I don't think the point you used were from line A. Line A ends near 4 on the x axis but it's still not exactly touching 4

Oh wait your talking about the 4 on QII. My bad lol. Hang on

[wary stream]

@normal mortar That person is getting helped in #help-8 already

[normal mortar]

Okay

[alpine sable]

i need help on this question

someone help

[gray isle]

was it necessary to post the question twice?
what have you tried?

[alpine sable]

@gray isle i've tried but it doesn't work

did i post twice?

<@&286206848099549185>

[gray isle]

wtf

why are you pining helpers now too

[alpine sable]

dude i just want help

[gray isle]

there's wanting help and then there's being rude

[alpine sable]

sorry if i'm being rude

[gray isle]

what exactly are you trying that doesn't seem to work