help-0

[keen plinth]

LOL

[dark trail]

Yup

[keen plinth]

this is right

[dark trail]

Finally!!!

I think I have to go over 6iii again

And ii cuz

Both ii and iv are the same

Ohhhhh nvm

We haven’t done iv yet

@keen plinth how would u propose i do 6iv

[keen plinth]

NO wrong SNOW

[dark trail]

ohh

[keen plinth]

you just pinged some random person

[dark trail]

edited it my bad

[keen plinth]

https://tenor.com/bjGy8.gif

[dark trail]

hopefully it werent a mod

[keen plinth]

i propose you fill out this table

do you know what it means

[dark trail]

not really

i know its kinda like a multiplication tble

AXB = a and b

but outermost row and column idk

uh oh

[keen plinth]

  |    A      |     A'     |
--+-----------+------------+-------
B | Pr(A, B)  | Pr(A', B)  | Pr(B)
--+-----------+------------+-------
B'| Pr(A, B') | Pr(A', B') | Pr(B')
--+-----------+------------+-------
  |   Pr(A)   |   Pr(A')   |   1

LOL

[dark trail]

my bad snow didnt mean to ping

[gilded crypt]

i would love to help you but it seems another fellow snow has it covered xD

[keen plinth]

sNOw is here

[dark trail]

i thought i edited it

and deleted the ping

[keen plinth]

pings cant be edited out

[gilded crypt]

it’s alright dont worry about it!

[keen plinth]

this is the full table where ive used commas to denote intersection

basically

the middle 4 are all intersections of the corresponding headers

and the outer cells

are the sum of the inner cells

so Pr(A) = Pr(A, B) + Pr(A, B')

since A happening can be split into A happening with B and A happening without B

across the rows too

Pr(A, B') + Pr(A', B') = Pr(B')

  |  A  |  A' |
--+-----+-----+----
B | 0.4 |  ?  | 0.5
--+-----+-----+----
B'|  ?  |  ?  |  ? 
--+-----+-----+----
  | 0.6 |  ?  |  ?

so given something like this

with unknown entries

your job is to just fill in the entries using the fact that rows/columns sum

[dark trail]

@keen plinth thanks bro imma do this now

[dark trail]

so

if the total of column 1 where AB is 0.4 and total is0.6

will the B' = 0.2?

@keen plinth

[keen plinth]

that isnt B'

thats Pr(A, B')

[dark trail]

oh yes

[keen plinth]

its both A and B'

but yes

you are right

[dark trail]

awesome

BA' is .1 then

[keen plinth]

yes

[keen plinth]

the lower right corner is 1

because thats everything added up

[dark trail]

yes

just added that

[keen plinth]

so everything is filled

now you can calculate anything you want

[dark trail]

unbelievable

to think its so simple

so q6iv is just 0.2

[keen plinth]

yep

[dark trail]

thank you

nah wait

ty madam

i remembered looool

how do i calculate this sort of probability

7% of 6 apples?

ive stated A = bad apple (0.7) B = good apple(0.93)

@keen plinth what methd should i use

i feel like the probability of picking 1 apple every time resets

[keen plinth]

bad Anne

this is a binomial distribution

[dark trail]

so maybe i have to stack probability

[keen plinth]

do you know about that

[dark trail]

oh

no but i think i can figure it out

wait

thats the thing where

you have power C o

POWER outside bracket) then C then power of x

for whatever range they specify like fourth expansion

oh no thats expansion not distrivution

[dark trail]

what does the x curly bracket mean?

reading on it now

i think i can answer the first part on my own the questions seem similar in my book

[keen plinth]

which curly bracket?

[dark trail]

ooof

curly symbol i mean in blue

[keen plinth]

oh

thats saying X is a following a binomial distribution

[dark trail]

my book is saying normal dist to binomial dist i dont think i can use it to solve this

[keen plinth]

here X means the number of bad apples you get

[dark trail]

ahhh

[keen plinth]

hmmm

i dont think youd use normal here

[dark trail]

yeah the q doesnt ask tht

but ok k is equal to the 6 apples she gets

or for every instance of getting an apple

now whats n

[keen plinth]

n = how many apples you buy in total

[dark trail]

ok

[keen plinth]

p = probability of getting a bad apple

k = number of bad apples you get

[dark trail]

how do i find k?

[dark trail]

OHHH

[keen plinth]

X is number of bad apples remember

at least 1

X >= 1

[dark trail]

ssooo k = x = some value greater than or equal to

1

[keen plinth]

you could add up all the values of k >= 1

so

[dark trail]

damn what are those values

cuz it can go to infinity

[keen plinth]

no its bounded

[keen plinth]

X cant be larger than 6

[dark trail]

ahhh

[keen plinth]

we arent gonna get more than 6 bad apples if you only buy 6

otherwise youve achieved magic

[dark trail]

so 1 + 2 +3...+6

lol

ok so i do this calculation 6 times with x = 1 then 2 then 3

then add those probabilites

[keen plinth]

yeah

if you want to experience pain that is

[dark trail]

oh damn

[keen plinth]

because the complement of getting at least one

is getting none

[dark trail]

bruh so how do i calcuate this?

[dark trail]

like 0 ?

cant happen right

since x is either 1 - 6

[keen plinth]

mmm?

if you go to the casino

[dark trail]

yeah

[keen plinth]

are you always guaranteed to win 1

is that how real life works

[dark trail]

no but does x represent a gain here?

good apple probability is 93% , bad = .7

@keen plinth what vid should i watch to answer this q

[keen plinth]

not really sure on vid recommendations

this is just the binomial distribution

id recommend you review that

X represents the number of bad apples you get

in this case

X could be 0, 1, 2, 3, 4, 5, 6

you could get any number of those bad apples

[keen plinth]

the probability of getting k many bad apples is given by this

where n = 6, p = 7% = 0.07

[dark trail]

ok do i start my x count from 0

0,1,2,3,4,5,6

[keen plinth]

what do you mean by x count

[dark trail]

x is a random variable right

ok this page here says

https://a-levelmathstutor.com/stat-binomial-distribution01.php

the possible number of 5s taken in 3 trials is 0,1,2,3

ill do this tmrw thanks for ur help so far

@keen plinth gtg

.close

[lone heart]

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[keen plinth]

np

[lone heart]

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[alpine sable]

when factoring 4x-2xy

[alpine sable]

which is correct -2x(-2+y) or 2(2-y)

not sure if you're supposed to also pull the negative from the -2

[keen plinth]

well for one 2(2-y) is missing an x

[alpine sable]

oh yea sorry I meant to put an x

2x(2-y)

[keen plinth]

okay well

both are correct

[alpine sable]

oh ok

tysm

[keen plinth]

you can write -2x(-2+y) as -2x(y - 2)

pick your favourite

doesnt matter

[alpine sable]

ah ok

ty

.close

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[slate barn]

Yo what is the derivative of 2/x

[slate barn]

Is it a negative sumn or just 2

[alpine sable]

take derivative of $2x^{-1}$

[ocean seal]

rbit ✨

[alpine sable]

is a negative sumn

[keen plinth]

LOL

negative sumn

[slate barn]

-2/x^2

[keen plinth]

[slate barn]

Right?

[keen plinth]

indeed

[slate barn]

Hahaha

I’ve no idea dude

[keen plinth]

you're right my dude

[keen plinth]

it is this

[slate barn]

Eh I got one more question

[alpine sable]

but only one more

[slate barn]

If the given x=2 then slope of the tangent line of y = x/x^2+1

Then it will be 2/5 Isnit?

Just plug the x in

[keen plinth]

rbit wheres your take derivative card

its time to use it again

[alpine sable]

take derivative and plug in 2, but idk where you getting the 2/5 from

[keen plinth]

probably 2/(2^2 + 1)

[slate barn]

So the numerator will be zero right?

[alpine sable]

but doesnt he have to take derivative first

[slate barn]

If I take the derivative

[keen plinth]

yeah

so its not correct

[slate barn]

But fam

[keen plinth]

use the quotient rule

[slate barn]

Lemme show you the options Igor

Ight

[slate barn]

Can u show me

[alpine sable]

whos igor

[keen plinth]

who's igor 💀

LOL

[slate barn]

Lmao this keyboard

[alpine sable]

you know what (f(x) / g(x))' is?

[slate barn]

Yo the derivative of - root of x is -1/2x isn’t it

[slate barn]

Yea I know but ion know how to apply it

It’s for next week’s class but I have to understand it first

[alpine sable]

$\frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{g(x)²}$

[ocean seal]

rbit ✨

[slate barn]

Or -1/2 root of x

[slate barn]

Nvm it’s the right answer

[alpine sable]

$(-\sqrt{x})' = \frac{-1}{2 \sqrt{x}}$

[ocean seal]

rbit ✨

[slate barn]

Yup

Oh yea I’m a bit confused about this wait

Graphically, the derivative of f(x) at r = c calculate
the slope of tangent line of graph y = f(x) at x = c

Is that true?

[alpine sable]

yes f'(c) gives you the slope of tangent of f(x) at x=c

[slate barn]

So it’s true?

Can I simplify the words

You*

[alpine sable]

idk whats there to simplify

[slate barn]

Yo dude if the numerator is 0 after taking the derivative it ok be zero right don’t care what the denominators are

So what should i do

[alpine sable]

what is your derivative

[slate barn]

0/2x

X=0 cuz constant yet 1 gone

[alpine sable]

thats not how to take derivatives of such fractions

[slate barn]

Ok tell me

[alpine sable]

$(\frac{f(x)}{g(x)})'=\frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{g(x)²}$

[ocean seal]

rbit ✨

[alpine sable]

what is f and what is g in this case

[slate barn]

Ok wait lemme write it down and show it to you

[runic trench]

differentiate using quotient rule

[slate barn]

I’ll give u my answer

2 minutes

So fx = x right and gx is the denominator

[alpine sable]

yes

[slate barn]

So x^3+x-x^3+x will be 2x

So final answer is 2x/x^4+2x^2+1

@alpine sable

[alpine sable]

,w derivative x/(x²+1)

[ocean seal]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by rbit ✨#4860

[slate barn]

Wait what

[alpine sable]

ok lets see again

[slate barn]

This is wrong ?

[alpine sable]

f(x) = x
g(x) = x² + 1

what is f'(x) and g'(x)?

[slate barn]

ohh yea I see

So the final answer is -3/25

Right

Eh 25?

[alpine sable]

what is f'(x) and g'(x)?

[slate barn]

Wym? Hmmm

Prime / derivative

[alpine sable]

yeah f'(x) means derivative of f(x)

[slate barn]

Yeah oke

So the numerator is 25 right

So it will be -3/25 since the x is at 2

Isn’t it?

[alpine sable]

i mean yes, but you still gotta know how to get there

[slate barn]

Yeah just take derivative and put them into the formula

But if you wanna take them derivatives you can’t do it one by one or you’ll end up having the numerator becomes zero

So how to do it?

[alpine sable]

right

i guess youll still learn it from your teacher?

[slate barn]

It’s for next week’s material but I wanna learn it from now

[alpine sable]

$(\frac{f(x)}{g(x)})'=\frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{g(x)²}$

[ocean seal]

rbit ✨

[slate barn]

This is correct right

[alpine sable]

we had
f(x) = x
g(x) = x² + 1
then
f'(x) = 1
g'(x) = 2x

putting it into the formula

[slate barn]

Yeah keep going

I’ll note them down

[alpine sable]

$\frac{1 \cdot (x² + 1) - x \cdot 2x}{(x² + 1)²}$

[ocean seal]

rbit ✨

[slate barn]

Yeah right

Oh the derivative of 0 is 1? I got it’s zero

Thought *

[alpine sable]

the derivative of x is 1 yeah

[slate barn]

Cuz I seen an article that says that it’s zero because it’s a constant etc etc

Okok das why I got confused at first

[alpine sable]

x isnt constant

the derivative of 1 is 0

[slate barn]

Remember I said the numerator equals to 0

Das why hahahaha

Bro I got questions again will u help me out?

Just like teach me to do it

[alpine sable]

if i know the answer

[slate barn]

The gradient of tangent line of y =(2x-1)^5 at x=2

Of course you can’t just substitute the x right

Well do I have to take the derivative or apply the formula of gradient or wha

[alpine sable]

gradient is slope right?

[slate barn]

Yeah

It stands for m

Y=mx+b that s the formula but It will not be applied in this case

Ion know

[alpine sable]

then whats the derivative of (2x-1)^5?

[slate barn]

5(2x)^4

Nah?

[alpine sable]

na

these questions are hard if you havent learnt the rules yey

you know the chain rule?

[slate barn]

Yeah cuz I just took a look at a few examples so das why I came up with that answer

[slate barn]

Nah

Wait so the question above should be used with the chain rule?

What’s the formula

[alpine sable]

$(f(g(x)))' = g'(x) \cdot f'(g(x))$

[ocean seal]

rbit ✨

[slate barn]

Ion know which one is g or f of x

[alpine sable]

in this case,
f(x) is x^5
g(x) is 2x-1

[slate barn]

Hmmm how do I put them into the f(g(x))’

[alpine sable]

so 2x-1 is the "inner" function and x^5 the "outer" function

[slate barn]

Mhm okay okay i see

[alpine sable]

what if you plug in 2x-1 into x^5?

[slate barn]

2x-1^5

??

[alpine sable]

(2x-1)^5 yeah

[slate barn]

Okay okay fam lets keep going

So that’s for the left side

[alpine sable]

then
f'(x) = 5*x^4
g'(x) = 2

[slate barn]

This is (2x-1)^5 then

[alpine sable]

$g'(x) \cdot f'(g(x)) = 2 \cdot 5(2x-1)^4$

[ocean seal]

rbit ✨

[alpine sable]

you see how?

[slate barn]

Yo yo

The power was 5 and you took the derivative and it became 4 I got it got it

But the thing is where did the 2 come from and why didn’t the -1 disappear

[alpine sable]

so basically, you first take the derivative of the outer function, so
(2x-1)^5 you get
5*(2x-1)^4
and then you multiply that by the derivstive of the inner functiom 2x-1, which is 2

[slate barn]

Also 2x supposed to be just 2 isn’t it

[slate barn]

How did u come up with 2 fam? I’m sorry for asking many questions

[alpine sable]

,w derivative (2x-1)^5

[ocean seal]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by rbit ✨#4860

[alpine sable]

the 2 is the derivative of 2x-1

[slate barn]

Oh yea the inner right

[alpine sable]

you know sin(x)?

[slate barn]

Now it’s understandable

[slate barn]

Sin? Mhm yeah

Sinus right

[alpine sable]

and its derivative is cos(x) yeah?

[slate barn]

I haven’t learned that far

Hehehe

[alpine sable]

well what is the derivative of
sin(x²-2x+3) then, using the chain rule?

[slate barn]

Cos 2x-2

[alpine sable]

not quite, what is the outer and what is the inner function?

[slate barn]

Outer 1 inner is the quadratic

[alpine sable]

outer is sin(x)

[slate barn]

Yeah oke

Mhm

[alpine sable]

then first only take the derivative of the outer and keep the inside:
cos(x²-2x+3)

[slate barn]

Okay show me more fam

Noted noted

[alpine sable]

and multiply that by the derivative of the inside:
(2x-2) * cos(x²-2x+3)

thats the chain rule

[slate barn]

So from the question I have I took the 5 and multiply it by the derivative of the inner

Correct?

So the inner is 2 outer is 5

[alpine sable]

(2x-1)^5

[slate barn]

I got 10

[alpine sable]

we get 5*(2x-1)^4

[slate barn]

Yea das it fam lesgoo

[alpine sable]

then multiply by 2 being the inner derivative

[slate barn]

So what now

[slate barn]

Just plug in the x =2 ??

[alpine sable]

yes

[slate barn]

Yessssss

It’s 810 fam right

[alpine sable]

i think so

[slate barn]

Well I’ll be right back and ask u some questions alright

Ima have to eat first

[lone heart]

@slate barn Has your question been resolved?

[lone heart]

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[trim oracle]

Hi

[trim oracle]

I have this working from my teacher

My issue is this line:

How is this true?

I’ve tried inserting values for x and y and it seems false to me

[worn fox]

The "same as" part?

[trim oracle]

Yes

[worn fox]

What x and y values made you think it was false?

[trim oracle]

Ngl I just did the numerator and compared answer

Just realised it’s the numerator and denominator together that make them equal

I get the same value from both now

[worn fox]

Indeed

[trim oracle]

I see now

[worn fox]

The trick to go from one to the other is to write x-y as -(y-x)

[trim oracle]

Little confused

From what I understand if you have an equation with xs and ys on numerator and denominator

Then as long as the equations are values of x minus values of y then you can just flip them to be ys minus xs

Because the negatives for the latter will cancel on numerator and denominator ?

[worn fox]

Basically $\frac{a}{b}=\frac{-a}{-b}$

[ocean seal]

ΣAC

[worn fox]

It's got nothing to do with x-y

[trim oracle]

Oh

Ohhh I see now

🙏 thanks

[trim oracle]

This makes sense now

.close

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[alpine sable]

question

[worn fox]

answer

[alpine sable]

lets say you dont know how to graph 3 given functions. but, you have to find the area between them. they are also supposed to be put into terms of y

how do you proceed?

[mortal trellis]

integrate

[alpine sable]

as in, how do you know which is on the right and left

(given 3 curves)

[mortal trellis]

I don't know what you mean by right and left

if anything upper and lower

start by calculating where they intersect I guess

[alpine sable]

yes, but in terms of y

[mortal trellis]

do you have a picture of what you mean

[alpine sable]

hmm hold on let me get one

okay if you are given these two equations but are unable to graph it, how do i know which curve is on the top/bottom? or in y-terms, right/left?

[mortal trellis]

what exactly do you mean with unable to graph it

you could still make a sketch

[alpine sable]

yeah but lets say i didnt know how

like i had no idea what shape the equations would make

[mortal trellis]

well you could calculate a few points and try to draw some line between them

to get a feel

I don't really know why this has to be in terms of y

doesn't really change much

[alpine sable]

.clos

.close

[lone heart]

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[mortal trellis]

assuming you are only interested in positive area, you could just integrate as usual

and if you get something negative, just ignore the sign

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[proper lake]

Can anyone help me with this please? I tried plugging it into the first equation but it doesn't seem to be right.

[half epoch]

What is the slope of the tangent line in terms of f_x and f_y?

[proper lake]

The derivative of x and the derivative of y?

[half epoch]

f_x is the derivative w.r.t x, f_y is the derivative w.r.t y

There is a formula for the tangent line in terms of f_x and f_y you should have learned

[proper lake]

I dont think I learned that in class

[half epoch]

Does "gradient" ring a bell maybe?

[proper lake]

Nope. No idea sorry

[half epoch]

Ok another idea is that you can find the slope (dy/dx) by using the partial derivatives through the "chain rule", maybe that rings a bell?

[proper lake]

I think I remember that sorta? But ive never been taught to make an direction vector

[half epoch]

Well the direction vector is just the gradient, honestly idk what else to say about that

$\nabla f(x,y) = (f_x(x,y), f_y(x,y))$ which is the direction vector of the tangent at (x,y)

[ocean seal]

Learath2

[proper lake]

then the f(x,y) = (-4x , -2y)

Then what am I supposed to do with the trace and x=-1?

[half epoch]

Ok, forgive me for I seem to have misread this question. It is asking the tangent vector for the trace of specific lines. I thought it was looking for a tangent vector to the plane

[proper lake]

Totally fine lol I dont even understand the question

[half epoch]

So we want the specific curve f(x, -3). You can just plug in -3 for y, then find dy/dx normally as you always would

[proper lake]

So I plug in -3 to y like to the first equation?

11-2x^2-(-3)^2?

[half epoch]

Yes, then you can take the derivative to figure out the slope of the tangent vector

[proper lake]

I got -4x

if i did it right

Then how do I find the vector? I think it should be <?,?>

[half epoch]

at x = -1 we get 4 for the derivative, so (1,4) is a tangent vector. These questions are usually looking for a unit tangent, so we need to divide it by it's length

so (sqrt(17)/17, 4sqrt(17)/17) would be my answer

[proper lake]

Im not allowed to put the sqrt so i tried calculating it and put (0,2425, 0.9701) but it says incorrect

Am I doing something wrong?

[half epoch]

Mh, maybe try (1,4)

It looks good to me tbh

[proper lake]

Nope. Says its wrong.

[half epoch]

Then you'll just have to wait for someone else to come along, I'm fairly confident this is the tangent vector. Maybe there is some other way to read this question I'm just not getting

[proper lake]

Alright. Thank you for your timee

<@&286206848099549185>

[lone heart]

@proper lake Has your question been resolved?

[lone heart]

@proper lake Has your question been resolved?

[lone heart]

@proper lake Has your question been resolved?

[proper lake]

<@&286206848099549185>

[gusty gorge]

so because you're talking about a 3d space, the direction vector is probably 3-dimensional

so try (1, 0, 4)

also, fwiw, you can compute this directly without substituting in the value for y

you're basically asked the directional derivative in the direction [1, 0]

[gusty gorge]

does that make sense?

and if not (1, 0, 4), try (1/sqrt(17), 0, 4/sqrt(17))

[lone heart]

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[alpine sable]

Could someone explain why the probability doesn't change when calculating E(X^2)?

[ornate condor]

i think its just substituition

so like they could use anything else

[alpine sable]

Wdym

[keen plinth]

because there's a 1-1 correspondence between x and x^2 here

if you know x^2 = 9

you know x = 3

[alpine sable]

What does 1-1 correspondence mean?

[keen plinth]

one to one correspondence

[alpine sable]

Yeah, what does that mean?

[keen plinth]

it means

one x matches with one x^2

and vice versa

[alpine sable]

OH

that makes a lot of sense

Thank you so much

.close

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[ornate condor]

feng jiu ge is truly the genius of a generation

able to battle rank 8s with a rank 7 cultivation

crossing the gap between realms

[keen plinth]

wtf are you speaking

[alpine sable]

[keen plinth]

LOL

[ornate condor]

yes

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[ocean seal]

osmium

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[bold zephyr]

hey all how would i do this, as i dont understand what the comma does.

[long axle]

It’s like an if then statement ig

If x<3, then x^2 < 9

Disprove that

Show a case in which that’s not true

[bold zephyr]

ahh

2<3,2²<9

2<3,4<9

[abstract nova]

just find any value of x that is less than 3 that is also greater than 9 when squared

think about negative numbers

[long axle]

Yes that’s true, but ur trying to disprove the statement

[bold zephyr]

Ohh i cant read....

[long axle]

Xd

[bold zephyr]

4 will work

[abstract nova]

well 4 is greater than 3

[long axle]

Nah, so the first statement must be true

It’s the second one that’s not always true

[long axle]

use this @bold zephyr

[bold zephyr]

-4?

-4<3,16<9

[long axle]

Yes!

Nice job

[bold zephyr]

thanks alot 👍

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[candid compass]

How can this answer be further simplified?

[abstract fractal]

It can't, though you can write ⁶√x as an exponent

[candid compass]

Im not sure what you mean

[abstract fractal]

Like how you can write √x as x^(1/2)

[candid compass]

so i'd put something where a is

[abstract fractal]

Yea

[candid compass]

what would i put

[abstract fractal]

What's ⁶√x as an exponent?

[candid compass]

yeah

1/8?

[abstract fractal]

How is it 1/8?

[candid compass]

if 1/2 is one square root

then triple that

for 6

im home sick with covid right now and my teacher isnt very responsive

[abstract fractal]

No

[candid compass]

i have no idea what im doing

[abstract fractal]

How did you get your answer in the first place?

[candid compass]

used this as reference^

so are you telling me its 1/6

cause theres a six

ok this is being accepted thank you for your help gamer dio

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[wise thistle]

Hello

[wise thistle]

How

[candid compass]

isolate x on first equation

simplify

until found x

then put that number as x in the second equation

[wise thistle]

What

[candid compass]

isolate x

[wise thistle]

I cant solve for x

[candid compass]

how come

[wise thistle]

Its in the quesrion

It said “without solving for x”

[candid compass]

nevermind then

i am of no use

[wise thistle]

Thx for trying tho bro

[candid compass]

no problem at all

[wanton tusk]

hihii

for such qns,, u wld have to express the first eqn in a form which u can substitute in ur second eqn

not rly easy to see

but for u can express ur first eqn 2x^2 + x - 8 as 2x^2 + x = 8

[wise thistle]

How?

[wanton tusk]

hmm

manipulate ur second eqn 2x^3 + 17x^2 +123 to x(2x^2 +17x) + 123

then express it as x(2x^2 + x + 16x) + 123

from our first eqn,, we know 2x^2 + x = 8

so sub it in

then u can go from there (:

not a very easy qn I think

[wise thistle]

Oh shit

OH SHIR

SHIT I SEE IT

Wdf how am i suppose to know im suppose to do that

tf

Thx im gonna try

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[lusty wasp]

Im having trouble inserting the numbers in order to isolate x and prove theyre inverses.

I got f(g(x)) and g(f(x)) but idk what to do now

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[mystic sinew]

any idea why the last 2 answers are wrong

[mystic sinew]

its ays they are wrong

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[mystic sinew]

<@&286206848099549185>

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[tender magnet]

Im new to the differentiation topic, would anyone be free enough to guide me through that?

[tender magnet]

this is a note my lecturer sent me, but i still dont understand it

[vale wigeon]

this note explains how to take the derivatives of constant functions and power functions

[lone heart]

@tender magnet Has your question been resolved?

[tender magnet]

what does it mean by that?

sorry for late reply

[vale wigeon]

this is your first time learning calculus, yes?

[tender magnet]

yes

[vale wigeon]

so your professor must have told you about the derivative, which is one of the two main objects of study in your course...

[tender magnet]

this is physics actually, we were learning the kinematics

[vale wigeon]

well you're gonna have to know your way around basic calculus.

[tender magnet]

oh isee

thanks

ill try google through that

[tender magnet]

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[desert tusk]

Struggling on 10b, why is it not 1024? The answer shows 484.

[desert tusk]

This is my working:

Never mind. I included that the old cells that are still alive will produce 3 more new cells on each day, it only states each new cells produced another 3 cells, not all cells.

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[snow pawn]

hello

[snow pawn]

I am working on the epsilon-delta proof of this limit

I have factorized into

and if I choose delta = 1, I get

which in turn

does it make sense in this case to let delta either be 1 or negative epsilon?

im not sure how to proceed from now

[alpine sable]

both delta and epsilon need to be positive

[snow pawn]

that was my concern

[alpine sable]

you cannot assume this. you are trying to show this

in general, delta depends on epsilon

https://www.milefoot.com/math/calculus/limits/DeltaEpsilonProofs03.htm

[snow pawn]

thanks for the link!

I will read up.
would you say my mistake in this case is to let the factor be -1? should it just be 1?

if I let delta be 1 can

[alpine sable]

you want to show |x-1||x-3|<epsilon, not use it to show |x-1|<epsilon

[snow pawn]

mkay. I will read up some more to get my head around this. Thanks!

[lone heart]

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[fresh flame]

hi its me again xD, i figured how to use the online calculator to check my answers. Im supposed to get 6x+1 but i some how ended with just 6x. what did i do wrong?

[solemn juniper]

very first f'(x), your numerator has +x when it should have +x+h

[fresh flame]

Ohhh

okie thank sorry haha

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[marble prawn]

question, is this question written wrong? or is it correct, because from my understanding, should ab = ac ?

[wanton nova]

Yeah it should be ab = ac

I measured ab and bc and they arent equal

[marble prawn]

im helping out a friend on their maths paper and they asked me to check this, its meant to be an official mock, i dont know what to tell them

[wanton nova]

I'm telling you i used a ruler and ab and bc arent equal

You can check for yourself

[vale wigeon]

@wanton nova who said the diagram was to scale?

[marble prawn]

yeah but i ruler wouldnt matter as its not meant to be used to scale bud

[wanton nova]

Oof

[elder furnace]

supposing AB=BC and you dont have a scatch what can you say about <ACB and <CAB?

[vale wigeon]

anyway

[marble prawn]

not meant to sorry

[vale wigeon]

this problem can be solved as written

you just have to be mindful of the potentially misleading diagram

[marble prawn]

if you went off the diagram its simple

[wanton nova]

Still believe its a typo cuz ab and ac are equal on the drawing

[marble prawn]

you just add the left angles and the top one is the same

[sudden hinge]

redraw the diagram as given in the text, then solve both cases lol

[marble prawn]

np thanks

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[small plover]

how to find the domain of arcsin(x/x+2)

[small plover]

i know the domain of arcsin is -1>=x>=1

[dense sleet]

domain of a inverse function f is range of the function f

Yes

[small plover]

how do I solve the eq x/x+2 <= 1

the answer is negative infinity to 1

[dense sleet]

$$-1 \leq \frac{x}{x + 2} \leq 1$$

[small plover]

yea

i cant solve this inequality

[ocean seal]

Pluton

[dense sleet]

Move 1 to left side

[small plover]

wdym

[dense sleet]

Subtract 1 from both sides

[small plover]

gives x-x+2 <= 0

[dense sleet]

Nope

[small plover]

wait sorry

the question is arcsin(x/x-2)

my bad

i typed wrong

[dense sleet]

Doesnt matter

We do same thing

Move 1 to left side

$$\frac{x}{x - 2} \leq 1$$

[ocean seal]

Pluton

[small plover]

(x-x+2)/x-2 <= 0

[dense sleet]

Yes

[dense sleet]

And do you know why you cannot multiply by x - 2

Like you did up there

[small plover]

with zero?

[dense sleet]

Wdym with zero?

[small plover]

i dont get this question

[dense sleet]

Do you know why
These 2 inequalities are not same

[small plover]

nope

[dense sleet]

$$\frac{x - x + 2}{x - 2} \leq 0$$
$$x - x + 2 \leq 0$$

[ocean seal]

Pluton

[dense sleet]

You cannot do that because when doing that you are basically assuming that x - 2 > 0

Because when you multiply by a negative number you have to change the sign

[small plover]

i see

[dense sleet]

So by multiplying both sides with x - 2 you are basically assuming x - 2 is never negative which is false

Ok so you now have
$\frac{2}{x - 2} \leq 0$

[ocean seal]

Pluton

[small plover]

yes

[dense sleet]

This is now pretty straight forward

The fraction has to be negative

[small plover]

yes

[dense sleet]

Therefore either numerator or denominator have to be negative

Numerator is a constant and positive so denominator has to be negative

[small plover]

yes

[dense sleet]

Ok so now solve for x in this inequality

[small plover]

rationalize?

multiply with x+2/x+2

[dense sleet]

No

[small plover]

ig not

yea

x < 2

[dense sleet]

Yes

Very simple

Just make denominator less than 0
x - 2 < 0

[small plover]

yea

[dense sleet]

Ok now $$\frac{x}{x - 2} \geq -1$$

[ocean seal]

Pluton

[dense sleet]

Absolutely same thing now

[small plover]

lemme try

2(x-1)/(x-2) >= 0

[dense sleet]

Yes

You can divide by 2 so you get

$$\frac{x-1}{x-2} \geq 0$$

[ocean seal]

Pluton

[small plover]

right

[dense sleet]

This seems a little harder right

[small plover]

yea

[dense sleet]

when is x - 1 = 0

[small plover]

x = 1

[dense sleet]

how about x - 2

[small plover]

x = 2

ohh

1 != 2

[dense sleet]

No that doesnt mean anything

[small plover]

ohh ok

[dense sleet]

We will just split this now in 3 ranges

$<-\infty , 1>$ , $<1 , 2>$ , $<2 , +\infty$

[ocean seal]

Pluton

[small plover]

yes

but according to restrictions

we can only select first range

• infinity to 1

[dense sleet]

At first range the numerator is negative and so is denominator

And -/- is +

[small plover]

yes

[dense sleet]

I mean yes the <2 , + infty> doesnt meet the x < 2

But in this case its right

In the full answer it isnt tho

[small plover]

full answer?

[dense sleet]

I mean we are first solving the first inequality , then the second

The answer is the intersection of both

Because both must be true

[small plover]

yes

[dense sleet]

So here we get the answer
$<-\infty , 1] U < 2 , + \infty>$

[ocean seal]

Pluton

[small plover]

why dont we include the (2, +infinity) in the domain

[dense sleet]

Because for $<2 , + infinity>$ the $\frac{x}{x - 2}$ is bigger than -1

But by our other inequality its is less or equal to 1 if its less than 2

[ocean seal]

Pluton

[dense sleet]

So it isnt in the domain because it is bigger than 1

[small plover]

right

i get it

[dense sleet]

So now you just find the intersect of those 2 sets

[small plover]

ok

[dense sleet]

Do you know what that would be?

[small plover]

(-infinity, 1]

[dense sleet]

Yes

Thats the range

I mean domain*

[small plover]

yea

[dense sleet]

And thats done now

[small plover]

Thank you Pluton

[dense sleet]

No problem

[small plover]

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[stable grotto]

guys i need help to find max points of this function

[stable grotto]

we can use calculus

[keen socket]

f’(x)=0

Or use sin(a-b)

[stable grotto]

how can i solve this equation ?

[vale wigeon]

your f'(x) is wrong

the cos(x) should've been cos(x/2)

also sqrt(3)/2 = sin(pi/3) and 1/2 = cos(pi/3). you can put that to use.

[stable grotto]

is this correct ?

because the answer is

this

[lone heart]

@stable grotto Has your question been resolved?

[runic trench]

ok holon

[runic trench]

this one rite

[stable grotto]

I don't understand

<@&286206848099549185>

.close

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[lone heart]

@indigo crown Has your question been resolved?

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[languid river]

That’s not possible because 2y-3x can’t be equal to both 10 and -2

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[spiral furnace]

Hey, can somebody help me complete the following:

Identify and label pairs of congruent triangles that show 1x translation, 1x reflection, and 1x rotation of congruent triangles, and explain each of the 3 different rigid transformations

[south mantle]

i have no idea how to solve this but uh ;-;

mind if I ask what field of math this is from o.o
im very intrigued and ive never seen a question like this before

[spiral furnace]

Congruence, particularly triangles

[south mantle]

hmm i do know what both of those are but never learnt about transformations and rotations i suppose

[spiral furnace]

Oh, okay

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@spiral furnace Has your question been resolved?

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@spiral furnace Has your question been resolved?

[lone heart]

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[orchid chasm]

I need help with using an fx-9750GIII calculator

[ornate condor]

that a first

but can u elab a bit on what exactly u need help with

[orchid chasm]

Basically I needed to know how to do this, teacher only gave me the function for ti 84 while I have a different calculator

Heres what my calculator looks like

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[blazing rose]

How am i supposed to write AB and AD as 3D vectors

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<@&286206848099549185>

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[ancient saddle]

Hello, maybe assume that the third coordinate is 0

[real gazelle]

maybe it means 2D

must be a typo

ask your instructor

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[long kraken]

I need help with answering the second part!

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[rustic coral]

This is a line with slope 5/4

so we can think of this as a right triangle where the opposite is 5x and adjacent is 4x

you can use that to find the hyp, then apply sin=opp/hyp

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[south jasper]

How do u find the answer

Like why is that the answer

I saw it on the board but I don’t understand why it’s that…

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[edgy flare]

2 please

I have this part to show bounded:

Need someone to look at what I have so far and help me with the portion about monotoned.

[alpine sable]

you should also show that x_{n+1} is always greater than 1 (using a quick induction or something)

also, what you've done might be notation abuse, but the idea is right

monotone would probably also be an induction argument

[edgy flare]

Where at?

[alpine sable]

I feel like x_n to infty is unclear

but that might just be me, idk

[edgy flare]

Maybe as n approaches infinity?

[alpine sable]

definitely not

I would just say suppose 1<x_n<2. Then show by induction, 1<x_{n+1}<2

[edgy flare]

Also,

I got this but I’m not sure how to transition into showing x_n is greater than or less than x_n+1 to get the monotoned part

As they are both greater than 1

Ehh

I found some work online and it makes somewhat sense

Take a look

[alpine sable]

ah, thats neat

[edgy flare]

Then they use induction

I’ll add it too.

Yay, nay?

@alpine sable

[alpine sable]

yes, (you used x_k>0 too there)

[edgy flare]

As in I need to add this in the suppose to section? “Also, suppose x_k > 0”

I hate proofs.

This crap really is an art, and I suck at drawing art.

[alpine sable]

yeah

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[edgy flare]

Thanks so much.

.close

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[kind jasper]

question 4

[abstract fractal]

#❓how-to-get-help

[wise thistle]

idc

[abstract fractal]

<@&268886789983436800> this guy is spamming the help channels

[junior blade]

thanks

[abstract fractal]

What have you tried

[kind jasper]

I tried question 3 using below:

2x - 3y = 6

-3y = -2x + 6

y = 1/3x - 2

[abstract fractal]

y = 1/3 - 2?

[kind jasper]

to get y you have to divide both sides with -3 right?

[abstract fractal]

Yeah

How do you get 1/3 - 2 from that?

Where'd the x go?

[kind jasper]

i forgot to put the x

[abstract fractal]

Still, you shouldn't get x/3 - 2