help-38

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[errant forge]

help.

[errant forge]

ahhhhhhh math is so hard 💀💀💀

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[night zodiac]

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[idle plinth]

M(6,-2)
C(3,4)
A(2,-4)
B(10,0)

M is the midpoint of the perpendicular bisector [AB]

❓ Q: Check if C belongs to the perpendicular bisector [AB].

[idle plinth]

I assume you don't need the points A and B since I've already given M

[knotty oriole]

You can find the line through A and B using algebra. If you have that line you can find the slope line of the perpendicular.

If you have the slope of the perpendicular you can find the line perpendicular to A and B that passes through M

Once you have the equation of the line perpendicular to A and B thru M you can just plug in C to verify it satisfies the equation. If it does C is on the line, if not then C is not on the line.

[idle plinth]

y = mx + b
m[MC] = -(1/2)

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[silk helm]

No clue how that works out

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[dusty stratus]

I am trying to find the domain and range of z=arccosine(1/y-sqrt(x)). So far for domain i have -1 < 1/y-sqrt(x) < 1. I also need it in setbuilder notation

[dusty stratus]

i guess what im trying to do right now is simplify the inequality in order to obtain the domain

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[knotty locust]

Can someone check my solution for this?

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[knotty locust]

nvm

[marble wharf]

bye

[zinc ginkgo]

bye

[amber python]

bye

[nova spire]

hello

oops sorry came in late

bye

[trail ingot]

hi

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[woven drum]

Could you explain how to do this

[solid kiln]

QuasiStar 超新星

[woven drum]

I’m still trying to learn this have you ever heard of a method called foil

[solid kiln]

QuasiStar 超新星

[woven drum]

Wouldn’t it cancel the squares or would it be (sqrt3r x sqrt 3r) and so on

[tiny nexus]

Solve 3x+1

[solid kiln]

QuasiStar 超新星

[woven drum]

Ok

Yeah I did that

[tiny nexus]

Whats 3x+1

[woven drum]

But I think I did the next step wrong

[tiny nexus]

???

[woven drum]

Yeah

[solid kiln]

QuasiStar 超新星

QuasiStar 超新星

[woven drum]

So it does cancel the radical only if if it’s inside a sqrt

[solid kiln]

QuasiStar 超新星

[woven drum]

So two square root equal 1 hole

[solid kiln]

QuasiStar 超新星

[solid kiln]

QuasiStar 超新星

QuasiStar 超新星

[woven drum]

Yes

[woven drum]

I thought this

Or do I not mutilply the 8

[solid kiln]

QuasiStar 超新星

[woven drum]

No

This is probably to advanced for me

Never learned this in school

[solid kiln]

QuasiStar 超新星

QuasiStar 超新星

[woven drum]

Ohhhhh

Yeah

[solid kiln]

QuasiStar 超新星

QuasiStar 超新星

[woven drum]

Yeah

That makes a lot of sense

So only if the terms are all to the second power inside the radicall they cancel out

Or Is that a bad way to think of it

[solid kiln]

QuasiStar 超新星

[woven drum]

Wait

Ignore the radicals

[solid kiln]

QuasiStar 超新星

QuasiStar 超新星

[woven drum]

That’s the final anwser

By distributing sqrt 5r

[solid kiln]

QuasiStar 超新星

[woven drum]

Ok

Ty

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[frosty mist]

How do I solve the missing values? I got the answers in class, which is why the blank spots r filled in, but I don’t exactly understand how to solve it.

[trim jolt]

@frosty mist Has your question been resolved?

[frosty mist]

Anyoneeee

Plz helppp

[broken kiln]

Well, could you erase the answers? So we can work from the ground up.

[frosty mist]

Ok one moment

[broken kiln]

Perfect.

Now, we have a expression for the g(x).

It just depends on the f(x) right?

[frosty mist]

Yes

[broken kiln]

Let's plug x = 3 at g(x) because we need to fill the row.

$g(3) = 4 f(\frac{3 - 1}{2}) + 3$

Everything simple until now.

$g(3) = 4 f(1) + 3$

[solid kiln]

M.

M.

[broken kiln]

If we simplify the fraction, then we get 1. Right?

3 - 1 = 2. 2 / 2 = 1.

[frosty mist]

Right

[broken kiln]

We're left with this expression.

What is the value of f(1)?

[frosty mist]

-3

[broken kiln]

In the table.

What is f(1)?

Right.

So now we just have to replace f(1) by -3.

$g(3) = -4 (-3) + 3$

What would be the result?

[frosty mist]

15

It’s -4

[broken kiln]

Oh, sorry.

I am monkey sometimes.

[frosty mist]

Loll ur good

[solid kiln]

M.

[broken kiln]

Great.

We discover one value.

g(3).

[frosty mist]

Riiight so g(3) = 15

[broken kiln]

Exactly.

Do you think you can do g(5) by yourself?

[frosty mist]

Hmmm I’ll try it

[broken kiln]

$g(5) = -4 f(\frac{5 - 1}{2}) + 3$

[solid kiln]

M.

[frosty mist]

I got 3

[broken kiln]

Because f(2) = 0. Right?

[frosty mist]

Yep

So g(5) = -4(0) + 3

And then g(5) = 3

[broken kiln]

Perfect.

You don't seem to need help.

Now let's fill f(x) table.

[frosty mist]

Yeahh I was just about to say how to do f(x)

[broken kiln]

We don't have a explict formula for f(x).

But maybe we can rearenge the expression to get one.

Do you know about function inverse?

[frosty mist]

I don’t think so

[broken kiln]

Fine.

Are you confortable with algebra?

[frosty mist]

Yeah

[broken kiln]

So, it will be easy.

Imagine the following.

$f(x) = 3x + 4$

[solid kiln]

M.

[broken kiln]

$g(x) = \frac{x - 4}{3}$

[solid kiln]

M.

[broken kiln]

Plug x = 1 at f(x) and we get...

[frosty mist]

-1

[broken kiln]

That f(x).

f(1) = 3 × 1 + 4

[frosty mist]

Oh 7

[broken kiln]

Right.

Now plug that 7 in g(x).

[frosty mist]

1

[broken kiln]

The value we started.

Try f(2).

And then put the result of f(2) into g(x).

[frosty mist]

So f(2) = 10

[broken kiln]

Right.

[frosty mist]

So g(x) = 2

[broken kiln]

Well.

You start to see something.

It appears that the g(x) is doing the exact oposite of f(x).

If you put something in f(x) and the result in g(x), then will get the something you started.

g( f( x ) ) = x
| |
| | the value
| |__result of f(x)
|__applying g(x) to result of f(x)

Why is this usefull?

$g(x) = -4 f(\frac{x - 1}{2}) + 3$

[solid kiln]

M.

[broken kiln]

We see a little function inside the f. Right?

[frosty mist]

Yeah

[broken kiln]

If we find the reverse of that little function, then we would get a explict formula for f(x).

$f(\frac{x - 1}{2}) \mapsto f(x)$

[solid kiln]

M.

[broken kiln]

Does it make sense?

[frosty mist]

Yes

[broken kiln]

Now we just have to find the inverse of $\frac{x -1}{2}$.

[solid kiln]

M.

[broken kiln]

Can you find it?

Do i write too much? I am trying to be as obvious as possible.

[frosty mist]

No it’s good

2x + 1

[broken kiln]

How did you find it?

It's right.

I just want to know your reasoning.

[frosty mist]

Tbh I looked at the one u previously did and realized the relationship between them

Like how the /2 part is 2x in the inverse and the - turns into a +

[broken kiln]

Ok.

Ok, we can enter into how to find the inverse function later if you want.

[frosty mist]

Okk

[broken kiln]

Let's apply to that function 2x + 1.

In other words.

$g(2x +1)$

[solid kiln]

M.

[broken kiln]

I know that if i pass x trough 2x + 1 and then trough (x - 1)/2 i will get x back.

This will leave us with f(x).

[broken kiln]

How would it be?

The right hand side.

[frosty mist]

Wdym

?

[broken kiln]

If we put 2x + 1 at g(x), how would it simplify?

$g(3) = -4 f(\frac{3 - 1}{2}) + 3$

[solid kiln]

M.

[broken kiln]

$g(13) = -4 f(\frac{13 - 1}{2}) + 3$

[solid kiln]

M.

[broken kiln]

$g(2x + 1) = -4 f(\frac{(2x + 1) - 1}{2}) + 3$

[solid kiln]

M.

[broken kiln]

Do you get it?

[frosty mist]

Kind of

I get that u plug the equation in but idk what x value to put

[broken kiln]

We are not going to put a x value yet.

We're gonna simplify first.

[frosty mist]

Oh

So how would u simplify it?

[broken kiln]

We know, that the expression in the f() would result in x.

Like: we built that 2x + 1 just to that.

$g(2x + 1) = -4 f(x) + 3$

[solid kiln]

M.

[broken kiln]

Makes sense?

[frosty mist]

Yeah

[broken kiln]

Then i would isolate f(x).

You just move 3 to the other side and then move the -4.

[frosty mist]

Is it f(x) = g(2x + 1) - 3 / -4

[broken kiln]

Exaclty.

Now we have a explicit expression for f(x).

Now we just plug the missing values in the table and use the g(x) table.

[frosty mist]

So if I had to plug in 3 for x it would be f(x) = g(7) - 3 / -4

?

[broken kiln]

Exactly.

$f(x) = \frac{g(2x + 1) - 3}{4}$

[solid kiln]

M.

[broken kiln]

g(3) is 15.

15 - 3 is 12.

12 / -4 = -3.

f(3) = -3.

Do you think you can solve all the orthers by yourself?

[frosty mist]

So -10 - 3 / 4

And then -13/4?

[broken kiln]

Exactly.

[broken kiln]

But what value you plugged?

[frosty mist]

It would be g(0) = -10

I get it now tho so tyyy! Srry for taking up ur time

.close

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[broken kiln]

No problem. I love to explain.

[broken kiln]

.close

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[frozen kindle]

what does the plus under the contained symbol mean?

[lethal anvil]

It just means positive rational numbers

[frozen kindle]

So positive rational numbers are contained in Real numbers, is that what it's saying?

why not just say Q

[lethal anvil]

OH WAIT

lol

my bad

that looks like a plus but what I think it means proper subset

it's crossing out the "equal portion"

[frozen kindle]

ohhh

ok thanks

.close

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[crimson rover]

can someone show me the formula for difference of a quartic for whatever reason i just cant find it anywhere

[crimson rover]

like

a^4-b^4

[crimson rover]

ty

[tidal forge]

there u go buddy

[crimson rover]

@tidal forge do you mind answering another question

|a-b| < |a| - |b| or is it + |-b|

[tidal forge]

let me see...

for this type of question, try to subsitude numbers in it

because I forgot the properties of abs too, lol

[maiden zinc]

Buddy is insane 😭😭😭😭

[crimson rover]

?

[maiden zinc]

.

[tidal forge]

like for a = -2, b = -3

wait

[tidal forge]

Is this an statement or question?

[crimson rover]

it would help a lot rn if it was |a|-|b|

statement

[tidal forge]

umm

[crimson rover]

its either |a-b| <|a|-|b|

or

|a-b|<|a|+|-b|

[tidal forge]

|a-b| < |a| - |b| false

if a and b are pos, they equal

if a and b one of them neg, |a-b| > |a| - |b|

[tidal forge]

that's |a-b| < |a| + |b|

which ys

in logic

or equal

[crimson rover]

ugh

[tidal forge]

so...

ur problem finished?

[crimson rover]

yeah

ty

.close

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[wraith hinge]

Disclaimer: I don't have problem regarded maths till now.

I just want to know the meaning of Highlighted part

[wraith hinge]

Does it mean that 500KG body which can provide 400N of force will only provide 395N force effectively?

Does it mean by friction?

Kind of

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[silent knoll]

Given cot(a) = -3√2 and π/2 < a < π. What's the value of tan(a/2) + cot(a/2)

[cyan zinc]

parantheses please

[silent knoll]

Ah okay okay sorry

It just we aren't taught to write it with () here so sometimes I just forgot to write them

Somehow they still understand it without parantheses

[silent knoll]

So tan(a) = 1/cot(a)

Idk about the (a/2)

[cyan zinc]

what is a?

[silent knoll]

Eh

[cyan zinc]

cot(a) = -3sqrt(2)
that tells you that cos(a)/sin(a) = -3sqrt(2)

and you are also given that it lies in the second quadrant

[silent knoll]

Ah

[silent knoll]

Mhmm

So I'll have to solve for a

[cyan zinc]

wait

are you sure that cot(a) = -3 * sqrt(2)?

that doesn't really check out

[silent knoll]

Yes it is

It was written in Vietnamese so yea I translated it into English

[cyan zinc]

,w cot(x) = -3*sqrt(2), pi/2 < x < pi

[solid kiln]

Results provided by WolframAlpha

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Requested by the.mr.gamer

[cyan zinc]

that doesn't really have a closed form unfortunately

looks like we'll have to try a different way

[silent knoll]

Ow

Mhm

Idk what to try

[acoustic trout]

given cot(alpha), can you find csc(alpha)?

[silent knoll]

What is csc

Haven't learned that yet

[acoustic trout]

alright I forgot, we don't have that in vietnamese curriculumn

[silent knoll]

1/sin(a)?

[acoustic trout]

do you know the identity cot^2(x) + 1 = 1/sin^2(x)?

yes

[silent knoll]

Yes

Ik that

[acoustic trout]

alright, that's all you need. We can express tan(x/2) + cot(x/2) in terms of 1/sin(x)

[silent knoll]

Wahhhh

[acoustic trout]

firstly, since you already knew that cot(x/2)=1/tan(x/2), can you use this to simplify tan(x/2)+cot(x/2)?

[tame solstice]

Assuming that u r allowed to use a calculator,
In order to go through the question, u need to figure out the Angle 'A' so that u can get the ratio of tan(a/2) & cot(a/2)

Firstly, to get the value of angle A,
Accroding to the unit circle, u have to look for the angle that has cotangent ratio of 3√2
Like, in a calculator just press "shift" + "1/" + "tan" and then 3√2

The angle the calc will show u is the angle of first quadrent

Now since it's negative, (-3√2) u have to find out in which quadrents cotangent ratios are negative

Since 1st & 3rd quadrent makes cotangent ratio positive, basically 2nd and 4th will lie in the negative one

However, as mentioned in the question (π/2 < A < π) indicates the angle will only be appreciated in the 2nd quadrent.

Morever, cot(π-A) = -cotA

So the angle would be (π-A)

[silent knoll]

[silent knoll]

What I thought it should be 1/ + shift + tan

[tame solstice]

Whatever u do. Just getting the angle A would help :V

[silent knoll]

I used the calculator and got 0.01301.... it should arccot right?

[silent knoll]

Ehhh

Making it 1/cot(x/2) + cot(x/2)

?

[acoustic trout]

sure, that works

[silent knoll]

Nice

Now I got no clue

[acoustic trout]

show me what you have, I can't really read your mind

[silent knoll]

Basically this

,rotate

[acoustic trout]

alright, now try to do common denominator

you will see the identity cot^2(u) + 1 = csc^2(u) appear

also, I'm going to use u as substitute for x/2 since I'm really lazy

[silent knoll]

ok I'll try

[silent knoll]

Howwww

[acoustic trout]

have you done common denominator for 1/cot(u) + cot(u)?

[silent knoll]

Nope

[acoustic trout]

...

[silent knoll]

I'm doing it

[silent knoll]

Oh wait I get it a little now

So multiply cot(u) with cot(u)

[acoustic trout]

mhm

[silent knoll]

And you get 1 + cot²(u)

[acoustic trout]

rather, you get (1+cot^2(u))/cot(u)

but anyway, you get the gist

[silent knoll]

Yay

[acoustic trout]

alright, so we get that tan(u)+cot(u) = csc^2(u)/cot(u)

do you follow up until this step?

[silent knoll]

Yes sir

[acoustic trout]

great, now since cot(u) = cosu/sinu, that means we can cancel out a 1/sinu in the numerator and denominator. What will we have after this?

[silent knoll]

Eh 1/cosu?

[acoustic trout]

I meant cancelling the 1/sinu in the right hand side of this equation

[silent knoll]

Wahhhh

[silent knoll]

The right hand side of this?

[acoustic trout]

alright, I'm going to skip ahead then, do you agree that after cancelling the 1/sinu in csc^2(u)/cot(u), we will get 1/(sinucosu)?

[silent knoll]

Yea I just got it

[acoustic trout]

alright, and you probably know the double angle identity, that says 2sin(u)cos(u) = sin(2u)?

[silent knoll]

Its basically

(1.sin(u))/(sin(u)².cos(u)) right?

[silent knoll]

Yes

[acoustic trout]

eh no, we are not going that way

[acoustic trout]

alright, good

so that means 1/(sinucosu) = 2/sin(2u)

do you agree with this?

[silent knoll]

Then we subtract it into 1/(sinucosu) if I right

[acoustic trout]

again no, that path is just going to make stuff more complicated

[silent knoll]

Oh

Okay

[silent knoll]

Yes

Make sense

[acoustic trout]

alright, since u=x/2, 2u=x. So we have just proved that tan(x/2) + cot(x/2) = 2/sin(x)

and you know that cot^2(x) + 1 = 1/sin^2(x)

can you find what 2/sin(x) is?

[silent knoll]

Uh

That's confusing

[acoustic trout]

do you agree that cot(x)=-3sqrt(2) is given?

[silent knoll]

Yes

[acoustic trout]

and that x is in (pi/2, pi)

so what can we conclude about the sign of sin(x)?

[silent knoll]

sin(x) is > 0 ?

[acoustic trout]

yes

so can you plug everything into cot^2(x) + 1 = 1/sin^2(x) and solve for 1/sin(x)?

[silent knoll]

± 1/√((-3√(2)²+1)?

[acoustic trout]

I'm not asking you to solve for sin(x), I'm asking you to solve for 1/sin(x)

[silent knoll]

Oh

[acoustic trout]

basically just look at the equation as y^2= 1 + cot^2(x) and solve for y, given that y>0

[silent knoll]

= √((-3√(2))²+1)

[acoustic trout]

ok, I'm just going to assume you wrote $\sqrt{(-3\sqrt2)^2 + 1}$

[solid kiln]

waler

[silent knoll]

Yea exactly that

[acoustic trout]

anyway, this should simplify down to sqrt(19)

[silent knoll]

Yes

[acoustic trout]

alright, so you know that 1/sin(x) = sqrt(19), what is tan(x/2)+cot(x/2)?

[silent knoll]

2.√19 ?

[acoustic trout]

yes

[silent knoll]

Ohhhhh finally

Ty ty

I get it now

Thank you very much

.close

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[vapid lynx]

!status

[trim jolt]

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

[wide charm]

1

I only can say integration will help

[vapid lynx]

yeah thats right

as a broad set of steps:

Find points of intersection between the two functions
Find out which one is "on top"
integrate

[wide charm]

x^2+x-4=0

This will give intersection points

@vapid lynx

Check it

[trim jolt]

@wide charm Has your question been resolved?

[trim jolt]

@wide charm Has your question been resolved?

[delicate belfry]

You've now found the intersection points. Now, two things to be checked: $$\textcircled{1}; \text{is the whole area in the same y-plane?}$$ $$\textcircled{2}; \text{which graph/function is above/under?}$$

[solid kiln]

Good

[delicate belfry]

After confirming, then the area is just the bigger area subtracts the smaller area.

@wide charm

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@wide charm Has your question been resolved?

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[terse portal]

anyone who can help me with matlab

[terse portal]

<@&286206848099549185>

[trim jolt]

@terse portal Has your question been resolved?

[terse portal]

<@&286206848099549185>

can someone plz help me :/

[hallow crater]

Ask.

[terse portal]

an=n-sqrt n^2-4n using matlab without toolbox

i need to evaluate and find the answer in matlab

[hallow crater]

Sorry, but I am not sure what matlab. I may not be able to help with this one.

[terse portal]

ok np

thx anyways

[terse portal]

.close

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[fading pebble]

Hi

[fading pebble]

I need to find a solution to this system which holds that x4 = 0. And also one that holds that x4 = 1

Not sure how to begin with this. should I start making elementary operations?

or should I just put 0 and 1 in x4's place in all rows?

[trim jolt]

@fading pebble Has your question been resolved?

[chilly remnant]

x_4=0 means you can cancel the fourth column of the matrix

[fading pebble]

great, thanks!

.close

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[silver sail]

Need help finding another solution

[silver sail]

That’s part 1

[winged hinge]

try switching around real and imaginary parts of one of w and z

[silver sail]

Huh

[inner urchin]

yeah

[silver sail]

Wdym

[inner urchin]

try it

[silver sail]

No but wdym

Like swap

@inner urchin

Like $w=\frac{3+12i}{5}$

[solid kiln]

Sukiie

[inner urchin]

could anyone pls answ my quiz first pls

[nova spire]

stick to your own channel

[inner urchin]

$w=\frac{3+12i}{5}$

[solid kiln]

Finicalfire

[inner urchin]

u mean?

[nova spire]

I mean #help-9

[silver sail]

@winged hinge I don’t understand

[nova spire]

you have w = Re(w) + iIm(w) and same for z

what if you did Im(w) + iRe(w)

swap real and imaginary*

[silver sail]

Owoe

Does that change anything tho?

[nova spire]

actually keep z as it is

but change w as such

[silver sail]

So swap on w

Leave for z

Then do what I done prev?

[nova spire]

so like w' = (4+3i)/5

yep, do the same with your "new" w

[silver sail]

Oki I’ll try that

Just a question why would that work?

Like just wondering

[nova spire]

well we still have |w'| = |z| = 1

[silver sail]

Mm

[nova spire]

so when we compute 13*5*wz, it's module squared will still be 65^2

[silver sail]

Yea

[nova spire]

and we will most likely get different real and imaginary parts, since we swapped them

[silver sail]

Yep

Thx

😊

.close

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[topaz mauve]

how do I solve the rest? I'm stuck

[supple copper]

did you read what the other guy sent before

[topaz mauve]

yes but didn't understand still

[supple copper]

this thing

so if i had a sample of 2

my X bar would be (X_1 + X_2)/2

do you agree?

Let $\overline X = \frac{1}{2}(X_1+X_2)$

What is the distribution of $\overline X$ according to this? ($X_1$ and $X_2$ are independent)

[topaz mauve]

ye

[solid kiln]

Frosst

[nova spire]

Let's not forget the final property : if $X\sim\mathcal{N}(\mu,\sigma^2)$ and $\lambda \neq 0$, then $\lambda X \sim \mathcal{N}(\lambda\mu,\lambda^2\sigma^2)$

[supple copper]

ah yeah

maybe we should ignore that part for now

[solid kiln]

rafilou2003

[supple copper]

maybe if you're feeling overwhelmed we can consider just the distribution of $X_1 + X_2$

[solid kiln]

Frosst

[topaz mauve]

whatever is the simplest way to go about it

I'm not familiar with this much at all

[supple copper]

we'll have to build it in steps then

[nova spire]

so here we're helping you, you saw that $X\sim \mathcal{N}(3,1.44)$

[solid kiln]

rafilou2003

[nova spire]

since it's the same law for all reviews

it's the same for $X_1$, $X_2$ and so on...

[topaz mauve]

how did you get 1.44?

[solid kiln]

rafilou2003

[nova spire]

1.2^2, I told you earlier

[topaz mauve]

ohhh

[nova spire]

So, when we're adding independent normal distributions, we're just adding their mean and their std^2

[supple copper]

std^2 is also called variance

or var for short

var(X), Var(X), σ²

[topaz mauve]

ok

what do I do next?

[supple copper]

what is the distribution of $X_1 + X_2$ given that they are $\mathbf{independent}$ and are individually distributed $X_i \sim N(3, 1.44)$

[solid kiln]

Frosst

[supple copper]

use this to help you

[topaz mauve]

all I'm just seeing is formulas and my mind isn't clicking together any of this to solve the puzzle sorry

[supple copper]

do you know what you're looking at

[topaz mauve]

not really?

fully

[supple copper]

do you know what this is saying

[topaz mauve]

I understood x~n(3,1.2) a bit

[supple copper]

what does it mean

[topaz mauve]

just nothing beyond that

[supple copper]

what does X ~ N(3, 1.44) mean

[topaz mauve]

In my own terms x~n(3,1.2) stays the same because it doesnt have the x-

I'm sure theres more in depth terms

[supple copper]

no i want you to explain what these symbols mean
" X ~ N(3, 1.44)"

[topaz mauve]

But I work at a slow pace regarding math I've had problems for years

[supple copper]

(it's okay if you cant explain)

[topaz mauve]

X is the random variable representing in this problem the time it takes her to complete one review and N means it's normally distributed

[supple copper]

ok that's great

what about the 3 and 1.44

[topaz mauve]

3 is the mean because it's the approx time it will take her to review employees

and 1.44 was found because of the STD being 1.2 and following some sort of formula we did 1.2^2 to get 1.44

[supple copper]

okay pretty good

the 1.44 is the variance of this random variable X

and because we know the variance is the standard deviation ^2 and the std was 1.2

we know the variance is 1.44

yeah?

[topaz mauve]

ye

[supple copper]

ok so normally we have say variables like x and y

we can do x + y

new variable cool

[topaz mauve]

ye im familiar with x and y

[supple copper]

but in probability we can also add random variables

so we can say X ~ N(1, 2) and Y ~ N(3, 5)

then we can go X + Y is a new random variable!

[topaz mauve]

are those numbers random?

or did u find those

[supple copper]

but just knowing it's a random variable is pretty useless, we are always interested in the distribution of this random variable

i just made them up

[topaz mauve]

ok

making sure I didnt miss somethin

[supple copper]

so we want to consider the distribution of X + Y

yep that's all good

ok so there's some cool theorems that you might see later (if you're interested) that tells us

if we add 2 normal random variables together

we get another normal random variable (albeit with new mean and variances)

[supple copper]

so that means since X and Y are both normally distributed from here

then X + Y ~ N(?, ?)

so it's normal but we haven't discussed what its mean and variance are

do you follow so far?

[topaz mauve]

ye

[supple copper]

so we figured out it's still normal

now we need to find the mean and variance

[topaz mauve]

which is 3

[supple copper]

oh wouldn't you look at that

if i turn these into the example we have

[topaz mauve]

and vari is 1.44

[supple copper]

If $X\sim N(1, 2)$ and $Y\sim N(3, 5)$ are independent, then $X + Y\sim N(1 + 3, 2 + 5)$

[solid kiln]

Frosst

[topaz mauve]

is 5 random?

[supple copper]

this is still from my example (numbers i made up)

[topaz mauve]

ohh

ok

[supple copper]

we'll go back to your question shortly

we just need to understand how adding stuff together works

[supple copper]

do you follow this?

the means add together and the variance adds together

[topaz mauve]

gotcha

[supple copper]

ok now

let's put your question's values into here

If $X_1\sim N(3, 1.44)$ and $X_2\sim N(3, 1.44)$ are independent, then what is the distribution of $X + Y$?

[solid kiln]

Frosst

[supple copper]

i've just changed the numbers to fit your quesiton

and renamed X to X_1, Y to X_2

(otherwise we'll run out of letters soon)

[topaz mauve]

gotcha

[supple copper]

(this is a question i want you to try and answer)

[topaz mauve]

ok

uh

is the mean, std, and sample involved?

or what do I need to put together on my problem instead of the example

[supple copper]

there is nothing more than is written here that you need

[supple copper]

try and follow what was done here

i just changed the variable names and the numbers but the process is the same

[topaz mauve]

I'm not following to a degree since theres like 4 different numbers in the example and I'm left with only 2 for mine

[supple copper]

ok let's try something simpler then

[topaz mauve]

ok

[supple copper]

If $X\sim N(1, 2)$ and $Y\sim N(3, 5)$ are independent, then $X + Y\sim N(1 + 3, 2 + 5)$

[solid kiln]

Frosst

[supple copper]

If $X\sim N(1, 2)$ and $Y\sim N(2, 5)$ are independent, then $X + Y\sim N(1 + 2, 2 + 5)$

[solid kiln]

Frosst

[supple copper]

If $X\sim N(3, 2)$ and $Y\sim N(3, 5)$ are independent, then $X + Y\sim N(3 + 3, 2 + 5)$

[solid kiln]

Frosst

[supple copper]

If $X\sim N(0, 1)$ and $Y\sim N(0, 2)$ are independent, then $X + Y\sim N(0 + 0, 1 + 2)$

[solid kiln]

Frosst

[supple copper]

do these examples help?

In general, if $X\sim N(a, b)$ and $Y\sim N(c, d)$ are independent, then $X + Y\sim N(a + c, b + d)$

[solid kiln]

Frosst

[topaz mauve]

but in my problem what is c,d

[supple copper]

if you look closely that's literally what this says, which new symbols

oh there's a typo in my quesiton

[solid kiln]

Frosst

Frosst

[supple copper]

(that's the same question with different variable names)

[topaz mauve]

X+Y~N (3+3,1.44+1.44)

?

[supple copper]

yes

which would be?

(do the sums)

[topaz mauve]

do we do 3+3 then 1.44+1.44?

[supple copper]

yes

[topaz mauve]

like 6,2.88?

[supple copper]

yep

[topaz mauve]

ok

[supple copper]

ok what if i asked $X_1\sim N(3, 1.44)$ and $X_2\sim N(3, 1.44)$ and $X_3\sim N(3, 1.44)$ are independent, then what is the distribution of $X_1 + X_2 + X_3$?

[solid kiln]

Frosst

[topaz mauve]

(3,1.44)+(3,1.44)+(3,1.44)?

[supple copper]

(3+3+3, 1.44+1.44+1.44)

the addition goes inside

[topaz mauve]

ohh

[supple copper]

so actually i notice i'm typing X ~ N(3, 1.44) a lot

maybe i want to say they all are distributed like that

then i would say $X_i \sim N(3, 1.44)$ using the $i$ to denote every random variable $X_1, X_2, X_3, ...$

[solid kiln]

Frosst

[topaz mauve]

No I'm ok with what we were doing

I was starting to understand it

[supple copper]

on top of that i can also say $X_i \overset{\text{i.i.d.}}\sim N(3, 1.44)$

[solid kiln]

Frosst

[supple copper]

the i.i.d. stands for "Identically and Independently Distributed"

which just means they all have this distribution

and they are all independent

so i can turn $X_1\sim N(3, 1.44)$ and $X_2\sim N(3, 1.44)$ and $X_3\sim N(3, 1.44)$ are independent, then what is the distribution of $X_1 + X_2 + X_3$?
\newline\newline
into $X_1, X_2, X_3 \overset{\text{i.i.d.}}\sim N(3, 1.44)$ then what is the distribution of $X_1 + X_2 + X_3$?

[solid kiln]

Frosst

[supple copper]

do you follow?

i've just contracted the "are independent" and repeating myself into "i.i.d."

[topaz mauve]

are we still on this? if so its 9,4.32

[supple copper]

yep

ok now

new thing

suppose $X\sim N(2, 5)$ then $3*X\sim N(3\cdot 2, 3^2\cdot 5)$

[solid kiln]

Frosst

Frosst

[supple copper]

to be completely honest this isn't really that new

[supple copper]

you can see that's exactly what we're doing here with c = 3

[topaz mauve]

for a,b,d are we adding the new values?

like 9,4.32 or 6,2.88

or old

[supple copper]

um this is not the other example anymore

okay i guess "exactly" isn't quite the right word

[topaz mauve]

ok?

[supple copper]

have you seen expectation before

[topaz mauve]

not sure

[supple copper]

ok anyway then

[supple copper]

try and remember this

[topaz mauve]

im honestly getting overwhelmed do we have to do this much stuff just to figure out one thing for the problem? or is there a more easier way

[supple copper]

this is the same as this

well, these are the 2 steps you need

we've just done the first step

first step is learning to add random variables

[topaz mauve]

ok

[supple copper]

2nd step is the scale the random variable

so if $X \sim N(2, 5)$ then what is the distribution of $3\cdot X$?

[solid kiln]

Frosst

[topaz mauve]

not sure

because for the example I'm not following and I'm understanding more with my problem than the example

[supple copper]

ok then

if $X\sim N(3, 1.44)$ then what is the distribution of $2\cdot X$?

[solid kiln]

Frosst

[topaz mauve]

not sure

[nova spire]

remember what the distribution of $c X$ is

apply to $c = ...$

[solid kiln]

rafilou2003

rafilou2003

[nova spire]

this

[nova spire]

here you can identify that we take a = ... and b =...

[trim jolt]

@topaz mauve Has your question been resolved?

[nova spire]

I'll write it again

if $X\sim\mathcal{N}(a,b)$ and $c \neq 0$, then $c X \sim \mathcal{N}(ca,c^2 b)$

[solid kiln]

rafilou2003

[nova spire]

So if $X\sim\mathcal{N}(3,1.44)$, then $2 X \sim \mathcal{N}(?,?)$

[solid kiln]

rafilou2003

[topaz mauve]

im lost still

[nova spire]

this is always true, no matter a b and c

[nova spire]

now tell me, what did we pick for a,b and c here ?

[trim jolt]

@topaz mauve Has your question been resolved?

[topaz mauve]

I'm still very lost ima just do another help area rq for another problem

.close

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[trim jolt]

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[humble prawn]

Can anyone check and see if its correct?

[trim jolt]

Please don't occupy multiple help channels.

Please stick to your channel.

@humble prawn Has your question been resolved?

[zinc ginkgo]

closed your other channel

can't read your handwriting. what does this say?

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[sly trench]

the exponent ^3 part im not sure whether its for the entire fraction
or just the numerator

for the answer below

[humble prawn]

f(h,rh)/h

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[carmine rain]

is this the correct way to solve inequalitys in set notation?

[carmine rain]

and is there a faster way do solve it?

[stark bison]

(x - 3)(x - 1) < 0 does not imply that x < 3 and x < 1

[cosmic forum]

hi plizz help

[stark bison]

This channel is taken, open your own one
#❓how-to-get-help

[trim jolt]

@carmine rain Has your question been resolved?

[carmine rain]

oh ok

@stark bison would it be this?

[stark bison]

Not sure where you have seen that kind of a step but that's totally invalid

Anyway ab < 0 rather implies that a < 0 and b > 0 or a > 0 and b < 0

So, consider two cases

x - 3 < 0 and x - 1 > 0
or
x - 3 > 0 and x - 1 < 0

[carmine rain]

oh sorry i forgot how to do that one and now remeber

@stark bison

[stark bison]

I hope you wamt to write 1 < x and x < 3 there, so that the answer is 1 < x < 3

And either method works

[carmine rain]

yes 😭

i keep making mistakes

@stark bison is there a way to know which method to use to solve inequlaitys? or is it just if it has x^2 do it this way and if not do it anouther?

[stark bison]

Like I said, either one works, it's up to you

[trim jolt]

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[carmine rain]

you wouldent do it the same way tho for this would you?

[trim jolt]

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[stark bison]

No, it can be solved straight away, directly

[carmine rain]

👍🏻

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[stark bison]

.close

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[stark bison]

Post your question in an available help channel, this one is about to close

[undone bane]

Oh ok sorry

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[sly trench]

where did i go wrong

[sly trench]

<@&286206848099549185>

[tribal gust]

why does it not work sorry ?

where are you bugging

[sly trench]

idk

the answer is

x/(36sqrt(36+x^2)) + C

[sly trench]

and whenever i integrate the last part i dont get that

[tribal gust]

well youll have to sub back after doing the integral right

[sly trench]

yes

but i still dont get that

[tribal gust]

there can be multiple answers if u didnt take the same route as the correction

[sly trench]

i wanna take the same route

as the correction

[tribal gust]

so its hard for me to confirm here like this since ur halfway there

well yours seems legit

u can use online tools to compare

to make sure

[sly trench]

i did

i used the integral calculator

and doesnt work

they used u sub

or not u sub

,rotate

[sly trench]

see its different

[sly trench]

from this

its just the 6 part

in the answer thats wrong

in the answer sheet its 36 instead of 6

[tribal gust]

watch out how u simplify

6sec^2 isnt the same as (6sec())^3

dont u get

6/216

which is 1/36

[sly trench]

wait what

Omgggggg

i get it

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[tribal gust]

makes sens ?

or im wrong

[sly trench]

.reopen

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✅

[sly trench]

yes

[tribal gust]

ok lol

[sly trench]

sec^3 theta

[tribal gust]

have a good one

[sly trench]

times 216

'ty

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[tribal gust]

happens

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[real herald]

Hi, since this is not multiplied why would this work?

[unique minnow]

The polynomial to the left must be exactly the 0 function (0 for all x), so in particular, this means that every coefficient must be 0.

[real herald]

But they are not miltiplied, so you can have the left = -1 and right = 1 no

or some other combination

im just kinda lost on why they have to be 0

[unique minnow]

Yes but this has to hold for any x

If you have a polynomial, you could not have that the right is always -1 and the left always 1

[real herald]

Oh, it would not shift because x^k is the only x in the equation

so the rest are just constants

ah

[unique minnow]

Yes, sort of. Say I want ax^2 + bx + c = 0 for all x.

Then c = 0 if we plug in x = 0

So we're left with ax^2 + bx = x(ax + b) = 0

[real herald]

yeah I think im understanding it now

[unique minnow]

YOu can go on and see that all the coeff must be 0

[real herald]

Yeah, for this equation those have to be 0 for any x to be 0

[unique minnow]

Usually this comes from a confusion between equating 0 as a value (and solving for x) and equating to the zero function

[real herald]

Yeah that makes sense

sorry idk why that got me lmaooo

tyty

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[sly trench]

on the bottom of the page the denominator is 4

[sly trench]

Where did i make a mistake

[sly trench]

im supposed to get this

<@&286206848099549185>

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[sly trench]

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@slim mountain Has your question been resolved?

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[chilly basin]

hi

[mighty pond]

solve

please

[chilly basin]

Question 30

...

[nimble stone]

do some squaring

[chilly basin]

square both sides

[nimble stone]

yeah

[chilly basin]

2wice right?

or just once

[nimble stone]

wdym?

only once

[chilly basin]

oh

i though u had to to it twice cos theres two square roots

[nimble stone]

no matter how many times you square it youll still have square roots, thats not the point of the question

you want to get the lhs into some form a-sqrt(b) so you can then say what p and q are

[chilly basin]

thank you

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[sudden musk]

What is the negation of the uniqueness quantifier?

[vernal robin]

i need help in this question. The question is telling me to find the net force. I find the magnitude which comes out to be 29 which is correct but the direction is wrong

[sudden musk]

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[stark hatch]

how do ☹️

[fresh vector]

@stark hatch what % of total pedestrian deaths were caused by a drunk driver?

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[stark hatch]

41?

[stark hatch]

wait no actually

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[wraith hinge]

How do I find the inverse of the equation in the very first line

[wraith hinge]

those are my steps so far

yes im using desmos cause idk how to use the bot to make the numbers look good 🤣

[last rune]

You have to first make your subject ‘x’

[wraith hinge]

well yes i swapped x and y

x = y+2/3y-6

forgot to include that in there

[last rune]

I see, if you swapped then solve for y

[wraith hinge]

yes

thats what im stuck on

idk how to go from the last step

these are my steps so far

[last rune]

you can bring the y in one side

and factor y from it

[wraith hinge]

im trying to move the y to the right hand side but idk how to get rid of that y on the left side

yk?

@last rune

[last rune]

3xy-y= 6x+2

y(3x-1)=6x+2

[wraith hinge]

Which step did you start on?

this one?

[last rune]

your last step

[last rune]

here

i added -y and 6x both sides

[wraith hinge]

so you subtracted a y and added a 6x