help-0

[summer goblet]

Good evening

[fallow nebula]

both

[fair osprey]

Which are vertical lines? Blue ones? Also what don’t you know how to do?

[fallow nebula]

i dont know the solution to solve a and b

[fair osprey]

What part of them?

So you know how these equations work? Can you plot y=1x? X=1?

[fallow nebula]

no ;

;-;

[fair osprey]

Did your teacher give notes or anything like that?

Wait a second I’m finding a vid that explains it

Here’s a vid that might help https://youtu.be/Z65mz__8DQ0

https://youtu.be/lz8zVJxRFX8

https://youtu.be/IL3UCuXrUzE

@fallow nebula try one of those?

[fallow nebula]

wait

[fair osprey]

clueless.file on wait mode

[alpine sable]

how am I supposed to solve this with system of equations

[buoyant kayak]

set up two equations using those pictures

[fair osprey]

Make an equation

@fallow nebula do any of those vids help?

[fallow nebula]

i am watching them

[alpine sable]

what two equations

help

[buoyant kayak]

for the first picture, how many widths and how many heights make up that tower?

[alpine sable]

it says the height is 30 cm

for the first brick

and 23 for the second

[buoyant kayak]

not what i asked

that's a stack of bricks

look at each brick

[alpine sable]

2 widths and 2 heights

per prick

[buoyant kayak]

no?

look at the towers

how many bricks are laying horizontally and how many are vertical

in the first tower

[fair osprey]

How many kinds of bricks if that helps

[alpine sable]

only horizontal bricks?

vertical

sorry

[buoyant kayak]

what

[alpine sable]

only vertical bricks

[buoyant kayak]

i'm asking you how many of each there are

[alpine sable]

2

[buoyant kayak]

your answer should be two numbers

[alpine sable]

what

[fair osprey]

You can make vertical v and horizontal h(for variable names, or just use x and y)

[buoyant kayak]

this picture right here

how many bricks are laying on their side

[alpine sable]

oh

[fair osprey]

Count how many bricks are vertical in the 30 cm tower

[alpine sable]

5

[buoyant kayak]

and how many are standing up

[alpine sable]

the right tower?

[buoyant kayak]

left tower. how many are standing up

[alpine sable]

4

[buoyant kayak]

so you have 5 on their side and 4 standing up

now look at the dimensions of the bricks

[alpine sable]

5x + 4y = 30

[buoyant kayak]

there ya go

now do the same for the other tower

[alpine sable]

reallly?

[buoyant kayak]

yep

[alpine sable]

is that it

[buoyant kayak]

that's one of them

this channel is in use

[alpine sable]

ok thanks

[buoyant kayak]

there's still another equation you need to solve it. do the same thing for the right tower.

[alpine sable]

ok let me try

[buoyant kayak]

you used x as the width and y as the height of the brick

[alpine sable]

4x + 3y=23

[buoyant kayak]

there ya go

now solve one of the equations for one of the variables

[alpine sable]

ok

ill try using elimination

wait I think I did somehting wrong

so I multiplied 5x + 4y = 30 by 4

and multiplied 4x + 3y=23 by 5

[buoyant kayak]

you'd multiply one by a negative factor

[alpine sable]

I got 20x+16y=90 and 20x+15y=115

[buoyant kayak]

if you're using elimination

[alpine sable]

(20x+16y=90 )-(20x+15y=115)

why?

20x-20x cancels still right?

[buoyant kayak]

that's not how elimination works

you just put a negative there randomly

[alpine sable]

oh wait

I figured mu mistake

30 * 4 != 90

ok I got the answer

thanks my dud

[buoyant kayak]

just curious, what are your answers?

[alpine sable]

I got 5

for y

now I can solve for x

[marsh shuttle]

Anybody free to help

[harsh belfry]

This is probably a stupid question

But I was wondering what occurred between these two steps algebraically.

to this

I know it has something to do with the bottom part of the fraction obviously right

but for some reason Im having a hard time visualizing it

[raw shard]

@harsh belfry no you’re correct

as long as there’s still a 4 on the very left

[harsh belfry]

ah ok

so are the two expression equivalent, just one is more simplified?

[raw shard]

yes

[harsh belfry]

Ah thanks for the clarification.

[raw shard]

no problem

[marsh shuttle]

Hey quantum do you know anything about linear programming for algebra 2

[raw shard]

no

[alpine sable]

@wispy olive oo i've done that problem before

i dont remember exactly how to solve

but i can pm u the solution if u want?

[wispy olive]

Shhh it is IMO Problem 6, 1996, something I am far far from to understand. I do not even have this question. Just that this is a seriously hard problems. So if anyone solves this I want them in my friends list and respect +100 trust +100 for them.

This problem took some mathematicians like 1 years to solve.

And I am doing Grade 9 Math .

[haughty falcon]

One message removed from a suspended account.

One message removed from a suspended account.

[alpine sable]

Anyone active?

[alpine sable]

speak

I also have a question

say

it

"kitten"

9/20/2021 how many days

???/

Is that in our present

Time

what

[wispy olive]

Did you solve it correctly?!!?

[alpine sable]

If it was 9/20/2021 how many days do I have to add to make it into our present time

[alpine sable]

well no...

[wispy olive]

Are you an IMO winner?

[alpine sable]

...................

[wispy olive]

Still xD.

[alpine sable]

Can anyone answer ;w;

If the (m+n)th term of a G.P is p and the
(m-n)th term is q , show that the mth and nth
terms are √pq and {p/q}m/2n

Can anyone help me with this?

Hewwo?

[wispy olive]

Meow.

[alpine sable]

Meow

[alpine sable]

Meow so know the answer

[outer scaffold]

7?

[wispy olive]

Ayoo just add with some brains?

Of course 8 days.

[outer scaffold]

no 7

[wispy olive]

Depends on time zone.

Ah yeah.

Depends on time zone.

Area.

[outer scaffold]

its 7

[wispy olive]

In the universe.

What space.

Where do you live.

[outer scaffold]

milk

[wispy olive]

How is the gravity.

[outer scaffold]

strong

[wispy olive]

Are you near Black Holes or Huge Star or Small Star.

[outer scaffold]

opposite of china

[wispy olive]

OH SHIT.

China will dig a hole through Earth then what will happen to you.

[outer scaffold]

cover hole up with big rock

[wispy olive]

You are not Naruto.

Or Goku.

[outer scaffold]

eren yeager

[wispy olive]

Nah.

[outer scaffold]

yes

[wispy olive]

You are Chungus.

Monkey.

[outer scaffold]

Chungus yeager

[wispy olive]

Cats and Monkey Supremacy.

[wispy olive]

OH NO-

[sly mantle]

pls move to #chill

[outer scaffold]

ok

[wispy olive]

Oh sorry sorry.

[outer scaffold]

sorry

[sly mantle]

@strange lion don't shitpost here

[manic torrent]

@marsh shuttle sorry for the long wait OBJECTIVE FUNCTION : 500x+150y

CONSTRAINTS:
x ≥ 0
y ≥ 0
7x + 3y ≤ 16
6x + 5y ≤ 14

where x = number of tables and y = number of chairs

[rigid wind]

using lagrange?

[vital prawn]

Anyone free to help?

[alpine sable]

I am dumb you know

[wispy olive]

Nah, me more.

[alpine sable]

I can't even answer the coordinates of X and Y

[rigid wind]

7 days ? 20 to 27

[alpine sable]

Ohhhh

It's 28 for me and I been banned for 8 days?

Something is not right

I'ma do an appeal

[candid minnow]

for 29 my graph looks something like this but how do I use this to draw my graph in Polar coordinates

[ornate lantern]

is anyone good at solving substitution and elimination

example: 2x+y=3

3x+2y=5

[candid minnow]

multply the first equation by 2 and the subtract that by 3x+2y=5

[tender belfry]

Hi

Log4/log3 = log(4-3)
Is the above equation true?

[vital prawn]

But does 1.5in roundest to the nearest square inch = 1 or 2?

[tender belfry]

2

[dusty quiver]

I am having trouble reasoning through the first 4 options

Since A is compact, it is closed and bounded, so any subset of a closed set is also closed

riht?

Nvm

[0,1] (0,1) is a counter example

specifically the first two

[candid minnow]

any1 know?

[dusty quiver]

option 1: R^n is not compact so since A is compact R^n \ A =/= empty

[winged spindle]

how much vbucks can i buy with 41 billion dollars?

[dusty quiver]

5

[robust flame]

Can anyone help me with b) ?

[minor root]

Is there someone here who's really good at solving three equations three unknowns?

[harsh belfry]

Wasn't sure what I was doing wrong here

f'(c):

Which should equal -0.469

[harsh belfry]

I remember going over this briefly in class, but Fairly certain I may be going about the process incorrectly.

Any advice?

Do I not insert c for x?

[undone dock]

what are you having trouble with

[alpine sable]

just put 5 in for x

[gentle jolt]

ive got no idea where to put this but i got an idea that id like to hammer out that has to do with the lucas numbers pascals triangle and f(x,y,n) = x^n + y^n
if anyone would like to discuss this new idea in mathematics voice, @ me
I'd love to iron it out with a second perspective

my "question" is a bit too complicated to just write it down, im not even sure what id write atm

[fast oak]

idk how to solve it

[candid fog]

u put 5 where x is

[hoary shell]

wouldnt it just be 12(3x3x3x3x3) + 9?

[wispy olive]

Yes, ask.

[candid fog]

yes

[fast oak]

ok

[candid fog]

or u could just use a calculator

[undone dock]

$f(x) =12\cdot 3^x +9 \implies f(5) = 12 \cdot 3^5 +9$

[fast oak]

what the answer?

[ocean seal]

K.

[fast oak]

ok ty u

[ornate lantern]

2x+y=3
3x+2y=5 thats the equation

[ornate lantern]

ok but i need someone to explain me the concept since this is new for me

[wispy olive]

Okay.

[fast oak]

f( 5 ) = answer is 2925?

[wispy olive]

See.

What you do is.

Let us say.

[undone dock]

yes

[wispy olive]

We are gonna use Substitution.

[fast oak]

ok ty u sir 🙂

[ornate lantern]

mhm

[wispy olive]

So for that, first we simplify the equation in terms of 1 variable.

2x+y=3
3x+2y=5

Let us simplify for y here.

With 2x+y=3, we get y = 3 - 2x.

Right?

[ornate lantern]

yes

[wispy olive]

Okay.

So we know that y is 3-2x.

So in the other equation which is 3x+2y=5.

We substitute this value.

So we do -
3x + 2(3-2x) =5.

We solve for this.

3x + 6 - 4x = 5.

-x = -1.

x = 1.

Right?

We get the value of x.

[ornate lantern]

ok i kinda understand it

[wispy olive]

Now we substitute the value of x in the other equation to get y.

2x + y =3.

2(2) + y =3.

4 + y =3.

y = -1.

And hence, we know that x = 1 and y = -1.

So you understood substitution?

[ornate lantern]

ya i have an understanding of it

[wispy olive]

Cool.

Next we can do Elimination.

And solve 2 more of each kind for you to understand better, if you want.

[ornate lantern]

thanks btw!

[ornate lantern]

ok sure

[vital prawn]

Is 15.4919 rounded to the nearest cubic inch = 15.5?

[wispy olive]

Elimination is simply subtracting one equation from the other.

[wispy olive]

Welcome!

So now we do Elimination for the same equation.

2x+y=3
3x+2y=5.

We subtract or add the first or second equation such that we get only 1 variable's value in the answer.

So.

We do.

3x + 2y = 5.
-2x + y = 3.

But see.

Here, if we solve this, x + 3y = 8 does not give us the value of 1 variables.

Hence, we use the basic laws of arithmetic we know.

If we multiply both the equation by some number which gets us the value of 1 variable, we get it right.

So we multiply 2x+y=3 by 3.
And 3x+2y=5 by 2.

[ornate lantern]

like a common multiple?

[wispy olive]

Well, not really.

See.

We need 2 variable values of the top part and bottom part to completely cancel each other.

So the answers/values should be the same.

[wispy olive]

We get -

1. 6x + 3y = 9.
2. 6x + 4y = 10.

We can subtract any of the 2 from 1 another.

For the sake of simplicity, let us do it simply.

6x + 3y = 9.
-(6x + 4y = 10).

Which gets us -

6x + 3y = 9.
-6x -4y = -10.

We get that -y = -1.

[ornate lantern]

ya

[wispy olive]

Now we know the value of y is 1.

Right?

[ornate lantern]

yes'

[wispy olive]

So we can simply substitute for y = 1 to solve for x.

[ornate lantern]

so we need to find value of x

[wispy olive]

6x + 3y = 9.
6x + 3(1) = 9.
6x = 9 - 3.
6x = 6.
x = 1.

And we are done.

With both the methods.

[ornate lantern]

ok so thats all i would right?

would i put 1,1

(1,1)*

[wispy olive]

1 more method which is much faster and easier is graphing, but dunno if you are allowed to use a graphing Calculator.

[ornate lantern]

we are aloud to use calculators

[wispy olive]

Yes.

The answer is (1,1).

[ornate lantern]

ok thanks for the help

[wispy olive]

Welcome.

You understood how to do these 2?

[ornate lantern]

yes i think i get the concept i just need to practice more

[wispy olive]

Cool.

I can send lots of practise.

But those will probably be harder than the kind of problem you sent.

[ornate lantern]

what level grade?

im in tenth academic

[wispy olive]

Grade 9-10 only.

But.

They are essentially more challenging.

[ornate lantern]

could you send them to me? id like to see because i think on my test it would be much harder

[wispy olive]

And for further use in life if you would, you can just graph the 2 equations in Desmos, since your main problem will merely require you to solve for 2+ variables as a tool.

For bigger problems.

[wispy olive]

Sure.

DM or here?

[ornate lantern]

u can send them in dms and thanks again

[wispy olive]

Aight.

Most/many of those I could not solve.

[fast oak]

sir

can i ask

[wanton spoke]

https://dontasktoask.com/

[fast oak]

oh ok

[tough coral]

statistics

is stupid

i hate it

bru i studied for like 5 hours for that test and i still got an 87

[crisp grove]

huh

[fast oak]

What is right A or B

[crisp grove]

#help-1 ????

[fast oak]

Use the following function rule to find f( 40 )

[crisp grove]

are you posting your entire question paper here?

[alpine sable]

No

[crisp grove]

definitely looks like it

[alpine sable]

Ya it looks like that

[lime moth]

One message removed from a suspended account.

[light tusk]

1440 is equal to all 5 books

1440 = 360 degrees

Find how much 1 degrees is

[light tusk]

Then multiply it 1 by 1 with the degrees of each book category

You should have 4 answer, each answer for different book

[lime moth]

One message removed from a suspended account.

[echo spire]

Hello! Can Someone help with A7 B)? I need to express that logarithm with the help of a

[somber osprey]

Uhh

[sour echo]

anyone knows how to do 4b and 5b

[echo spire]

This is what I managed to solve

[verbal ermine]

just square f

[sour echo]

yea but it’ll be a squared fraction

im stuck there

[manic torrent]

your answer can't be a squared fraction?

[sour echo]

lemme try

[verbal ermine]

do you know how $\left(\frac{a}{b}\right)^2$ simplifies

[ocean seal]

Aerials

[manic torrent]

i mean

[verbal ermine]

it’s just $\frac{a^2}{b^2}$

[ocean seal]

Aerials

[verbal ermine]

so just square the top and bottom

[manic torrent]

i asked if you can't have an answer in fraction

like did your teacher say you cant have that answer?

[verbal ermine]

nah it says how u should express ur answer

[sour echo]

yea

[verbal ermine]

$(2x)^2 \neq 2x^2$

[ocean seal]

Aerials

[manic torrent]

@sour echo

[sour echo]

that’s what i did

[manic torrent]

no

look at it again

[sour echo]

oh

[verbal ermine]

that means not equal to

[sour echo]

it doesn’t equal?

o omg

[verbal ermine]

you have to square both terms

[sour echo]

it’s 4x^2?

[verbal ermine]

yes

[manic torrent]

yup

[tidal thistle]

what's the difference of rational equation and rational function

[manic torrent]

they're both rational meaning they have fractions

then they just differ on the equation and function part

[daring pebble]

One is an equation, other is a function

[sour echo]

this is what i got for 5b most likely wrong tho!!

[barren shuttle]

Hello, is it possible for this system to be inconsistent when all the variables are on the right?

[daring pebble]

One that isn’t a function

And one that isn’t an expression*

[verbal ermine]

if u solved that quadratic correctly, it’s right

[sour echo]

o ok

[sick torrent]

How come the vertical asymptote is only x=2 and not x=-5

[jagged imp]

f has a hole at x=-5, instead of an asymptote, since the numerator and denominator both have a factor of x+5

[eternal osprey]

How does one determine the vertical and horizontal asymptotes of this function?

[sick torrent]

Ahhhhhh

Thx boo

[tender belfry]

Technically yes

[tender belfry]

Non technically also yes

[sick torrent]

Why isn’t the answer DNE?

[edgy sand]

There is a trick you can do with fractions that are approaching inf. Subtract highest factor of variable in the denominator from numerator

[sick torrent]

Huh?

[rigid wind]

divide both numerator & denominator by x

[sick torrent]

Why

[rigid wind]

this limit approaches -inf

[sick torrent]

Ik the answer I just don’t know how

[sick torrent]

Is it cuz it’s the highest factor?

[rigid wind]

yeah, and you can evaluate individual terms as x goes to +inf

[sick torrent]

And to know if it’s a negative or positive infinity it depends on the answer?

[rigid wind]

a simple example: lim x/2 as x -> inf

or lim x/(-2) as x -> inf

1 approach +inf, the other one -inf

[sick torrent]

But how would u do it on a complex question like the one I sent

[edgy sand]

Its the greatest common factor in the denom. Therefore you divide all of you numerator and denominator by x. See if you can set it up

So divide everything on top by x, and everything on the bottom by x

[tight locust]

for infinite limits of rational functions all you need to do is compare the leading terms

[sick torrent]

Like so?

[tight locust]

everything else becomes negligible as you go to infinity

[edgy sand]

.

[tight locust]

like in that example you can ignore the lower degree terms

and turn it into x^3/(-8x)

[edgy sand]

No don't cancel them yet. Just set up

[tight locust]

= x^2/-8

(infinity)^2/-8 = -infinity

[sick torrent]

@tight locust I’m mad confused

With what u just said

[tight locust]

it's simple.

[sick torrent]

Then how?

[tight locust]

what is bigger: 100^2 or 100?

[sick torrent]

The first one

[tight locust]

yep

so the terms of higher degree grow A LOT faster than the terms of smaller degree.

you can say that x^2 is a lot "bigger" than x

[sick torrent]

Yeah

[tight locust]

so when you take an infinite limit, you can ignore all the lower degree terms

because they become very small, relative to the higher degree term

[sick torrent]

How do u wrote that?

Write*

[tight locust]

wdym

[sick torrent]

How to u use that information to find out if it’s a -inf + inf or DNE

[tight locust]

i just showed you. in your example you ignore all the terms but the biggest one. so you get the limit as x goes to infinity of x^3/(-8x)

[sick torrent]

But

How do u know if it’s -Inf with that information

[tight locust]

well you can just simplify

what is x^3/(-8x) ?

[sick torrent]

Uhhhh

Is it already simplified?

[tight locust]

do you see any common factors on the top and bottom?

[sick torrent]

Oh yeah

-x^2/8

[tight locust]

yes

[edgy sand]

anything divided by infinity is 0 as its getting very close to 0

[tight locust]

and now you can just evaluate

what is (infinity)^2

[sick torrent]

Infinity?

[tight locust]

yes

[edgy sand]

Any pos int atleast

[sick torrent]

But

[tight locust]

and what is -1/8 * infinity?

[sick torrent]

Um

-infinity?

[tight locust]

yes

[sick torrent]

But y?

Is it cuz there’s a negative?

[tight locust]

yes

[sick torrent]

Oh

[tight locust]

1/8 * infinity = infinity

-1 * infinity = -infinity

[sick torrent]

The whole process of getting to -1/8 is confusing

[tight locust]

how?

[sick torrent]

I didn’t understand

[tight locust]

what don't you understand?

[sick torrent]

Everything.

Ion understand this infinity shi

[tight locust]

all you have to do is take the ratio of the biggest terms and evaluate

[sick torrent]

How do u do that

I don’t understand words can you please write a step by step

[tight locust]

ok. let's take that limit you have there

the limit as x approaches infinity of (x^3-6x^2-5x-2)/(7-8x)

now i want you to look at the top and bottom.

what is the biggest term on top?

what is the biggest term on the bottom?

[sick torrent]

X^3 for top and -8x for bottom

[tight locust]

yes.

now since it's an infinite limit, those other terms don't matter. so we can take the limit of this much simpler fraction

the limit as x approaches infinity of x^3/(-8x)

[sick torrent]

Uh huh

Then it becomes -x^2/8?

[tight locust]

now you have a power of x on top and the power of x on the bottom. they cancel. you get:
x^2/(-8)

yep

[sick torrent]

So then what?

[tight locust]

now this is a function you can directly evaluate.

essentially, we have: (infinity)^2/-8

[sick torrent]

Oh yeah

And then?

That just equals -infinity?

[tight locust]

yes. infinity^2 is just infinity. infinity times 1/8 is also just infinity. then the negative changes the sign so you have -infinity.

[sick torrent]

Oh my gosh I understand

Thank you so much

[tight locust]

😎

[unique tiger]

is this channel open?

can someone help me with this?

[faint hill]

Who can help with that

[unique tiger]

<@&286206848099549185>

[unique tiger]

so 8 log x e right?

[vale wigeon]

8 log_x(e)

type logs properly

[unique tiger]

okay ....then

= (1/e)* log _x(e)*d/dx ---- is this right?

so 8 In (e) i think

so = (In e^8)/(In x) ???

okiee thanks

[narrow meadow]

wait what we can just move logs outside of limits?

can someone explain to me why

[harsh ocean]

Does there exist an integral with lower bound only?

[crisp grove]

u definitely can

[narrow meadow]

any kind of logs?

[harsh ocean]

like a symbol

[crisp grove]

reason being log(x) is continuous function

[narrow meadow]

🤔

[crisp grove]

u mean like Riemann lower sum?

[narrow meadow]

so any sort of continuous function I can just shift in and out of limits?

[harsh ocean]

to represent the lower value given to the constant

[placid zinc]

You can put the limit in, if the new limit you are creating exists

[narrow meadow]

wdym

[placid zinc]

Which is the same as "is continuous there"

[crisp grove]

I don't there is

[alpine sable]

im having some trouble with this problem

especially with the -2

how do I take that into account

[placid zinc]

Have you found the homogeneous solution yet?

Or the complementary solution or whatever you call it

[alpine sable]

yes

$y_h (x) = c_1 + c_2e^{5x}$

[ocean seal]

bee_ryan

[placid zinc]

Okay! So you're left with the particular solution now

We can do this with undetermined coefficients

[alpine sable]

im not sure what the general form is tho

[placid zinc]

For sin(x), the guess is Asin(x) + Bcos(x)

[alpine sable]

yes

[placid zinc]

For 2, the guess would normally be A

There's a problem with that however. We can't just put a constant in, because our homogeneous solution tells us that constants are eliminated

(Don't even need the homogeneous solution, obviously the left will bring a constant to zero since there's no y, only y' and y")

Whenever this happens, just multiply your guess by x.

Correct guess is Ax

[alpine sable]

ohh i see

ok ill try that

so the overall guess would be $y_p(x) = Acos(x) + Bsin(x) + Cx$?

[ocean seal]

bee_ryan

[harsh ocean]

Better seen with a ss

[crisp grove]

[harsh ocean]

Theres an upper bound one too

Both combined form an integral

Maybe

[crisp grove]

[harsh ocean]

Its better to do a normal integral with upper bound 0 and lower bound what you want

[opaque oxide]

im dying inside

[vale wigeon]

this is a fraction addition problem.

do you know how to add fractions?

[opaque oxide]

Yes

[vale wigeon]

so what is troubling you here?

[opaque oxide]

im just supposed to add it then put a negative sign right?

[harsh ocean]

-144/85

Multiply by 5/5 and 17/17

[opaque oxide]

thats what i answered

[harsh ocean]

Yea

[crisp grove]

u mean you are in a test rn?

[opaque oxide]

no

[harsh ocean]

Lol

[crisp grove]

you wouldn't say yes, would you?

[opaque oxide]

Its practice

[crisp grove]

then why the sweat?

'dying inside'?

[opaque oxide]

cuz its hard

[crisp grove]

ah

[opaque oxide]

U know what tatsu bot is?

Cuz idk that

[crisp grove]

a bot

[spice mortar]

hello

may i ask for help here? 😛

i am kinda stuck with something that i think it's fairly simple

it's probability

i have to determine a.) and b.) but i keep getting weird results

[crisp grove]

It's just inverse Bayesian

[spice mortar]

yeah

i have the formula

but i legit don't know what values to put instead of A/B

respectfully

[crisp grove]

what is the probability that you'll pick a black ball if you were to pick from the first urn?

P(B| K1) =?

[fast oak]

Is this correct

[crisp grove]

@fast oakchannel busy

[fast oak]

Ohh ok sorry

[spice mortar]

@crisp grove ok, can you please tell me - i know that B stands for black ball chosen, doesn't matter the box

what exactly would i do

i am legit loosing my mind because i have been trying to solve this since last night and i don't get it right

first it's

we look at the number of all the balls

and the black ones

and we get probability for that right?

[crisp grove]

So you happened to pick from the first urn, you have thee prob P(B|K1) = (number of black balls in K1)/(total number of balls in K1)

so what's P(B|K1)

[spice mortar]

sec

10/5

2

[crisp grove]

?

[spice mortar]

oh

5/10

0.5 - sry

1/2

[crisp grove]

and what about P(B|K2) then?

[spice mortar]

just a sec

i put it wrong

it's 4 black balls in K2

so it's 4/12

or 1/3

[crisp grove]

you sure you counted them right?

[spice mortar]

yeah yeah

i drew it wrong

it should be 4 black ones in K2

[crisp grove]

ok

[spice mortar]

i am confused because

[crisp grove]

we know that $$P(B) = P(K1)P(B|K1) + P(K2)P(B|K2)$$

[spice mortar]

A1 and A2 stand for box 1 and box 2 being chosen

[ocean seal]

Ryuzaki

[crisp grove]

then what's P(B)?

[spice mortar]

so in that case

both A1 and A2 should be 50%

ie. 1/2?

[crisp grove]

y

[spice mortar]

so

[crisp grove]

so P(B) = ?

[spice mortar]

let me write it down

P(B) = 0.5 x 0.5 + 0.5 x 1/3

or

1/2 x 1/2 + 1/2 + 1/3 ?

[crisp grove]

1/2*1/3

[spice mortar]

oh

righyt

[crisp grove]

Now you need to find P(K1|B) ie given the ball is black, probability that it's picked from A1

[spice mortar]

wait so i get the value for P(B)

sec

[crisp grove]

DO you know that $P(B|K1)*P(K1) = P(B)*P(K1|B)$

[ocean seal]

Ryuzaki

[crisp grove]

can you figure out P(K1|B) from here?

[spice mortar]

so B is

probability of black ball chosen

so it's number of black balls divided by number or total balls?

[spice mortar]

regardless of the box

i am sorry, i am kinda losing my mind here 😅

[crisp grove]

yeah if you write it out in eng you'll get something like P(K1|B) = (number of black balls in K1)/(total number of black balls in K1, K2 combined)

[crisp grove]

are u not familiar with prob rules/identities? Bayes' rule?

[crisp grove]

you can use this one if you are really confused

[spice mortar]

to be honest, i am doing this last moment and trying to figure out on my own

so yeah

everything i did so far was from internet and my uni docs

but they don't give answers xD

in uni docs

[crisp grove]

ofc they don't

[spice mortar]

so P(K1|B) is 5/9

so in the case of this

[spice mortar]

this

it should be:

[crisp grove]

also check if you get the same answer by doing

P(K1)P(B|K1) / P(B)

[spice mortar]

this is

0.027

2.7% ?

[spice mortar]

2.7%

[crisp grove]

2.7%???

[spice mortar]

xD

so i did

1/2 x 1/2 / 9

[crisp grove]

P(B) is not 9

also if you are using the counting method, then there are 5 balls in K1 not .5*.5

you literally took one method for nominator and another for denominator

[spice mortar]

wait

so

k1 is 0.5

that right?

P(A1)

[crisp grove]

P(k1) = .5 yes

[spice mortar]

ok

next part

P(B|A1) is nm of black in A1 / total nm of balls

so it's 5/10

0.5

[crisp grove]

hm

[spice mortar]

am i right?

[crisp grove]

.

[spice mortar]

so it is

0.5 x 0.5 and then divided by P(B)

[crisp grove]

what's the value of P(B)?

[spice mortar]

and P(B) is probability that you will get the first ball black out of all the balls

both A1 and A2

[crisp grove]

ya

i mean the value

that you get of P(B)

[crisp grove]

this

[spice mortar]

so we do number of black balls / total nm of balls?

wait

[crisp grove]

ya

you should get the same ans anyways

first of all the + sign in the denominator suddenly turned into -

and also 0/0 is not 0

[spice mortar]

P(B) is 9/22?

right

[crisp grove]

bruh

[crisp grove]

could be i didn't calculate

[gloomy perch]

@crisp grove Hi are u good in complex number ?

[spice mortar]

it's 9 total black balls out of total 22 balls

[crisp grove]

then yeah

I trust your counting

[spice mortar]

ok i will do the math

[crisp grove]

,calc 9/22

[ocean seal]

Result:

0.40909090909091

[spice mortar]

xD don't

[crisp grove]

,calc .5*.5 + .5*1/3

[ocean seal]

Result:

0.41666666666667

[crisp grove]

,calc .5.5 + .5*1/3

[ocean seal]

The following error occured while calculating:
Error: Unexpected part ".5" (char 3)

[crisp grove]

,calc .5*.5 + .5*1/3

[ocean seal]

Result:

0.41666666666667

[spice mortar]

so it's 42% that it's going to be black one from box K1

i mean a.)

[crisp grove]

NO that's the value of P(B)

[spice mortar]

oh right

wait

yeah

i get the same value for 9/22

aprox. 0.41

[crisp grove]

now calc P(K1|B) = P(K1)*P(B|K1)/P(B)

[spice mortar]

=0.60975

so that was A

a.)

60%

and to do b.) which is

we take the probabilities of P(A1|B) and P(A2|B)

[alpine sable]

How do I do this?

[spice mortar]

i am calculating

i got values for P(A1|B) and P(A2|B) - 0.6 and 0.4 = 3/5 and 2/5

we consider as P (A1) and P (A2):

Event A1 - the first ball, black, drawn from box K1 - then, after that first draw, a total of x balls remained in box K1, of which x black;

Event A2 - the first ball, black, drawn from the box K2 - then, after that first draw, a total of x balls remained in the box K2, of which x black;

Event B2 - the second drawn ball is black (of course, it must be drawn from the same box as the first ball).

[daring schooner]

Hello, may I ask if the labeling of the vertices are correct, considering the quadrilateral is called ABCD?

[spice mortar]

@crisp grove Okay, i think i got it

Thanks for your patience

🙂

[harsh belfry]

I've been working on this for over two hours

and I just cant figure out how to find the derivative for this.

When I put it into quotient form I can't figure out how the trig functions work with one another.

If anyone could help me out on this I'd really appreciate it.

This is the problem in quotient rule

Any help at all would be appreciated. Im just so worried.

I have a test today and I cant do this.

I feel like I'm about to break down. I haven't slept in over 24 hours now.

[halcyon comet]

can anyone help me in a practice test

[vague coral]

about what ?

[icy trail]

could someone help me simplify this fraction?

$\frac{b+a}{\sqrt{b^2-a^2}}$

[ocean seal]

querty

[icy trail]

totally not from #help-1

yess

$\frac{b+a}{\sqrt{(b-a)(b+a)}}$

[ocean seal]

querty

[icy trail]

copy paste 😎

but im not sure where to go from here?

rationalise?

dosent the bottom have a factor of (b+a)^(0.5)

oh wait can i do $\frac{b+a}{\sqrt{b+a}} = \sqrt{b+a}$

[ocean seal]

querty

[icy trail]

ohh and thats the same as multiplying by (b+a)^0.5

or would lead to the same thing

because the b+a would cancel

$\frac{(b+a)^2}{(b-a)(b+a)}$

[ocean seal]

querty

[icy trail]

thankyou : )

[alpine sable]

I’m not sure how to prove that AC is the diameter

Please ping with response

There are two ways, one of them is a cheeky way that I go by quite often:

1. Show that cord AB and cord CB are perpendicular, thus ACB is a triangle in a semicircle, making AC the diameter.
2. Get the equation of the circle/ the radius, get half of the distance of AC and show that it equals the radius.

@alpine sable

As for method 1, here's a graphic demonstration

This proves that if angle B (which is on the circumference) is a right angle, the line between the two other points is the diameter.

How do you get that AB and BC are perpendicular:
$\frac{yb - ya}{xb - xa}$

[ocean seal]

Samy

[alpine sable]

You multiply this by $\frac{yb - yc}{xb - xc}$ , if your answer is -1, they're perpendicular.

[ocean seal]

Samy

[barren river]

Quick question, what does y=ax^2+bx+c A B and C stand for

[glass lichen]

the co-efficients...?

[alpine sable]

C is the y intercept, a is the coefficient of x^2 which indicates whether the parabola's mouth will be facing up or down

[barren river]

what does that mean for C

[alpine sable]

C is the value of y at x = 0, the place at which the curve cuts the y axis

[barren river]

So like I put the value of Y into C when I am finding the axis X?

[alpine sable]

Oh, no

[barren river]

kinda confused

[alpine sable]

Let me show you

[barren river]

So it's from the picture

[alpine sable]

Okay, but how do you know that the fact that it’s a right triangle makes AC be a diameter? Also, isn’t there a possibility where it’s not a right triangle and AC still passes through the center?

[alpine sable]

As for how do you know that, if all the points lie on the circumference, it must be a diameter.

But let me demonstrate your doubt

If this is the case, AC can't be a diameter because one of the points is not on the circle's circumference

[icy trail]

but they still lie on some circle?

just not the one you drew

[alpine sable]

Yeah that makes AC a diameter for another circle but please no

[icy trail]

please no?

[alpine sable]

My question is how do you actually prove that the fact that it’s a right triangle makes AC the diameter of the circle

Because you assumed that to answer the initial question

But I want to know where that assumption is coming from

[alpine sable]

Do you know the rule that says that an angle in a semi-circle is 90 degrees?

[barren river]

Sorry but, how do I know if a parabola mouth will be facing up or down if no picture are provided

[alpine sable]

a coefficient

[alpine sable]

No

[alpine sable]

If a is negative, it's down. Otherwise it's up. Also you'd be better in luck if you ask in another channel

[alpine sable]

Okay

That's the rule needed here, I'll explain it briefly.

[barren river]

Well there's no A, but X

[alpine sable]

That means a is 1 (if there's an x^2)

[barren river]

alright

[coarse kernel]

is the channel free?

[alpine sable]

https://www.youtube.com/watch?v=oT7arIHd0D8

If you accept this rule (you should lol) then reversing it gives you the answer to your question

[oak jewel]

Is this open?

Find all complex numbers making this true

how would i go about solving that?

[supple tundra]

Do you know how to find the magnitude of a complex number?

[oak jewel]

maybe, can you further explain? i don't do this in english*

[supple tundra]

If $ z = x+iy $ then $|z| = |x+iy| = \sqrt{x^2+y^2}$

[ocean seal]

azeem321

[oak jewel]

Yep, but i learned it today and dont fully understand it

or how to implement it

|z|=sqrt(x^2+y^2)

[hushed pasture]

no wait

[oak jewel]

when its absolute Z, the "i" goes away?

[hushed pasture]

you have that z = x + iy

[oak jewel]

yep

[alpine sable]

Calculate the range of values of x which satisfy the inequality

2x − 3 > 6x + 1

[hushed pasture]

so $|z -5| = |x-5 + iy| = \sqrt{(x-5)^2+y^2}$

[ocean seal]

Salah

[hushed pasture]

@oak jewel clear?

[oak jewel]

the complex led gets cleared like that... ok yep clear

i thought it might be like that, thats great

so what i would do is put that in for all.... let me just quickly do that

|x+1|=|x|+|1| right?

just to be absolutely clear

[alpine sable]

No

[oak jewel]

lol forget that. yep no it isnt

[supple tundra]

If I asked you to find the magnitude of $3+4i$ could you do it?

[ocean seal]

azeem321

[oak jewel]

would that be sqrt(3^2+4^2)=5?

[distant gate]

Good day! Can u help me solved this num? Thanks a lot for help ❤️

[alpine sable]

The chan is occupied

[supple tundra]

Correct. The question you have presented, they simply want you to find all complex numbers $z$ such that when you calculate $z+3+4i$ and find it's magnitude, you get the same magnitude when you calculate the magnitude of $z-5$

[ocean seal]

azeem321

[oak jewel]

so, this is where i get a bit confused

i get x and y, two variables

[alpine sable]

^2

[supple tundra]

This is sufficient to define all the coordinates which satisfy your original equation. To clean it up though, you can square both sides

[oak jewel]

yep i did that

aah

y^2 cancels

[errant tangle]

could someone help me prove this

[alpine sable]

Yeah, you should be able to express x with y now

||
x=F(y)
Your solutions are F(y) + yi, with y in R
||

[oak jewel]

yep, thanks a lot!! @alpine sable @supple tundra i will try to solve it myself, i may come back here if im unsucessfull

[supple tundra]

The equation you get will be in the form $y = mx + c$ all the points on that line $(x,y)$ satisfy your orginal complex magnitude equations. So if the point $(2,3)$ was on your line, then $2+3i$ would be one the solutions

[ocean seal]

azeem321

[oak jewel]

i got a line y=-2

[elder glacier]

is this occupied?

[alpine sable]

rewrite 7pi/18 as pi/3 + pi/18 and 5pi/18 as pi/3 - pi/18 and several trigonometric formula will help

[errant tangle]

thank you!!

[elder glacier]

can someone help me with a and b, im having difficulty wrapping my head around this, and apologies if this in occupation, thanks.

[oak jewel]

i think i messed up. can you help me verify, is the line y=-2? and so all x values and only y=-2 is on the line-?

[supple tundra]

If you got it right, then any point on that y=-2 line will satisfy

[oak jewel]

brilliant thanks a lot!!

[supple tundra]

You can test a point on that line to make sure. (0,-2). Let z = -2 and sub it into equation

[eternal osprey]

Can anyone help me find asymptotes with limits?

[oak jewel]

if z was a point, lets say (-1,4) how would you verify it?

[hushed pasture]

find the domain first

[alpine sable]

for the point (-1,4), z = -1+4i

[supple tundra]

Remember you let $z = x+iy$

[ocean seal]

azeem321

[oak jewel]

so you plug it into on both sides

and just verify?

so -1+4i

[supple tundra]

right

[oak jewel]

great

we have not went through this in class reee

[supple tundra]

is this complex loci stuff? i struggled with it when i was learning it

[oak jewel]

anything

the professor is lost i think

he spend 20 minutes integrating 1 today

and he got the integral of 1= x^2/2

[elder glacier]

ur prof is hitting bongs before class

[oak jewel]

he is drunk before every class

no joke, that guy is a mess

[elder glacier]

prob a bit of both tbh

maybe wife divorce?

[oak jewel]

he looks like walter white

maybe meth, dead people in the basement, los pollos hermanos, mike ehrmantraut

[elder glacier]

LOL

[silver current]

wow

[alpine sable]

nvm i cant type it

[twin pine]

how can I find this without L'Hopital?

\lim _{x\to +\infty }\left(x\cdot \:sin\left(\frac{1}{x}\right)\right)

[rocky atlas]

Would be great if you can write it on copy and show then

[rocky atlas]

.

[crisp grove]

$\lim_{x\to +\infty }\left(x\cdot \sin\left(\frac{1}{x}\right)\right)$

[ocean seal]

Ryuzaki

[glass lichen]

h=1/x then it becomes a known limit

[ocean seal]

marc

[rose prism]

is the domain 0 less than or equal to x greater than or equal to 6

[ancient saddle]

Hello, the set is Z^2, so your drawing should be some points that satisfy the condition

If that's what you've done, it should be ok

[ocean seal]

marc

[ancient saddle]

I guess they are just asking you to draw it to prove you understand the condition

[cold spindle]

guys i ahve exam tommorow can any1 help me study

[harsh belfry]

I'm a little confused since this wasnt covered much in class.

I wasnt sure where to start exactly.

Would I find the derivative of the function?

[glass lichen]

You need slope of the line and a point on the line

[alpine sable]

Yeah that gives you the slope that Mosh is talking about

[harsh belfry]

Got it got it.

I only know these equations mathematically. I'm not good with knowing what they actually do.

[glass lichen]

The definition of derivative us the slope function...

[pliant spruce]

How would you solve this?

[harsh belfry]

If thats the case, once I fiind the derivative and insert the given x value. Would this work out to be y=?

[glass lichen]

If you didn't know that, you can't say you know it mathematically

[glass lichen]

No

[cold spindle]

help

[glass lichen]

If you plug something into a derivative you don't get a function value

You get the slope of the tangent at that point...

[harsh belfry]

Which would be my M value right.

I think I got it now thanks.

[glass lichen]

Yes...

[alpine sable]

To get an equation of a line using just one point and a gradient, you equate the gradient to a fraction of a ("y" - the y value of the point) over ("x" - the x value of the point) and you'll get the equation

[harsh belfry]

I just memorize what equations are to be put where. I really wish I has the opportunity to use more graphs and see what the values im changing and what Im doing has on the graph. Sorry if I'm annoying you.

I'm very new to this.

Thanks samy. Will do.

❤️

[quartz oxide]

Diving one equation by the other should allow you solve for y directly

[alpine sable]

2(26) = 52 I think that's what they mean.

So 2[x + xy + x(y^2)] = x(y^5) - x

One other way I can think of is to get x in terms of y (from the second equation) and remove every x in the first equation with that value you get

np

[hazy lynx]

having a hard time with this one: find the limit as x approaches 1 of x^-1 - 1 / x - 1

i looked it up online and its saying you can simplify it to ((1 - x) / x)) / x -1 but i dont get how you get there? how do you get the -1 in the numerator on top?

[noble sinew]

Is this x^(-1)-1/(x-1)?

[hazy lynx]

uhh its (x^-1) - 1 / (x-1)

[noble sinew]

So like I wrote?

[hazy lynx]

yeah

i was jsut confused by the parenthesis you had lol

[noble sinew]

I had proper parenthesis

You didn’t in your original

[hazy lynx]

oh ok

[noble sinew]

I mean can just use linearity of limit so this is equal to limit as x->1 of 1/x + limit as x->1 of 1/(x-1)

[hazy lynx]

ah ok ty

[queen raven]

i need help with trinomial factorisation