help-39

[pearl pond]

Please don't occupy multiple help channels.

[high geyser]

I dont know how to set up the matrix

[wicked hull]

make a matrix and edit it to the coornets

[high geyser]

?

[wicked hull]

edit the values till you find what you need?

[high geyser]

so whats the matrix

nvm fatass

.close

[pearl pond]

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[brazen meteor]

How can this be simplified? Not used (a+b)³ formula

[brazen meteor]

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[pearl pond]

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[pearl pond]

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[midnight haven]

I've figured out that AB has to be 8, so does B

And BC is 6

and CD is also 10

So I have enough information to determine the area of both circles

But I don't know what they mean by the area between the two circles

<@&286206848099549185>

[lusty minnow]

probably the concentric circle

the area excluding the inner circle's area

[midnight haven]

ok

so then its

64?

*64pi

?

<@&286206848099549185>

Cans you please confirm the answer

?

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

no

Can somone pls confirm

that the answer is 64pi

bruh

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[rain flame]

This ain’t right

[light helm]

-i * i isn't i^2

(last term in the expansion)

[rain flame]

WAIT

That’s

-i^2

Which becomes 1+1 instead of 1-1

So my answer is 2?

[light helm]

yes

[rain flame]

Iiii gotcha!

[pearl pond]

@rain flame Has your question been resolved?

[pearl pond]

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[cedar trench]

I’m just wondering if I’ve done problem 12 a right

,rotate

[cedar trench]

I don’t know where to go from here

[hybrid sentinel]

can someone help me?

[cedar trench]

.close

[pearl pond]

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[hollow pike]

81 27 9 3

explicit sequence

[west sapphire]

neat

[hollow pike]

hey

[west sapphire]

do you have a question about that sequence?

[hollow pike]

the 0th tterm is 81

so the 1st term is 27

[west sapphire]

ok

[hollow pike]

so what can the explict seuqnce be? 81*1/3^n-1?

but that owuldnt make sense

[west sapphire]

these are all powers of 3 yes?

[hollow pike]

yea, the previous term is multiplyed by 1/3

[west sapphire]

ok so you could make a formula that looks like 3 to the something power

where "something" is a formula involving n

[hollow pike]

so 81*1/3^n?

,calc 81*1/3^0

[jolly parrot]

Result:

81

[hollow pike]

,calc 81*1/3^1

[jolly parrot]

Result:

27

[west sapphire]

yep that'll work

[hollow pike]

i got it

@west sapphire wait

i have a question tho

is n equal then or grater then 0?

[west sapphire]

well you plugged in n=0 and n=1 and got the right answers

[hollow pike]

so n equal then or greater then 1 or 0?

[west sapphire]

assuming your sequence index is starting at n=0

[hollow pike]

so its 1

[west sapphire]

oh

well then you have to adjust

[hollow pike]

@west sapphire n≥1

right?

[west sapphire]

i don't know, you tell me

[hollow pike]

it cant be 0

[west sapphire]

why not

[hollow pike]

or wait

im super confused

bro

[west sapphire]

people define sequences starting at n=0 all the time

but it's your sequence, you tell me what the first n value is

[hollow pike]

but only if the 1st term is considered the 0th term?

[west sapphire]

it's not up to me to decide

[hollow pike]

the 0th term is 81

[west sapphire]

if we call your sequence x(n)

[hollow pike]

mhm

[west sapphire]

does it start like x(0), x(1), x(2),...

[hollow pike]

yer

[west sapphire]

or does it start like x(1), x(2), x(3)...

both are legal

it's up to you, not me

[hollow pike]

not in my case

because the problem spefically says that the guy makes cuts

if i make 0 cuts on my paper

im left with original size

which is 81

[west sapphire]

well you haven't pasted the problem statement

[hollow pike]

sorry, but u understand right?

[west sapphire]

sounds correct based on what you're saying

so yes your formula is fine

[hollow pike]

what if it was 81 27 9 3

81 was first term

it would be n equal then or grater then 2

and 81*1/3^n-1

[west sapphire]

why

[hollow pike]

,calc 81*1/3^1

[jolly parrot]

Result:

27

[hollow pike]

yup

i think im correct

@west sapphire am i

@west sapphire

@west sapphire bro

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[pearl pond]

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[midnight haven]

The question says to find k so that the equation is a perfect square trinomial but how is it 12??

[midnight haven]

Does the 9x^2 have anything to do with it?

[autumn merlin]

b^2 - 4ac = 0

[midnight haven]

Do I just remove 9 by divide

Or do I have to multiply 9 by 12 as well

[autumn merlin]

what

where is 12

tell me what a, b, and c are

for this quadratic

[midnight haven]

9x^2 is a

-kx is b

4 is c

[autumn merlin]

that is not correct

a, b, and c are the coefficients of the x terms

so we dont take the x term with them when we say what a, b, and c are

so it would just be
a = 9
b = -k
c = 4

[midnight haven]

Oh

[autumn merlin]

yeah

so what would b^2 - 4ac = 0

imply

[midnight haven]

-k^2 -4(9)(4)=0

What formula is that?

[autumn merlin]

b^2 - 4ac is the determinant

it determines how many roots a quadratic has

since your questions asks for a perfect trinomial

that means it has only 1 root

so b^2 - 4ac = 0

[midnight haven]

I don’t think I learned about determinants in my class yet

[autumn merlin]

oh

alright

another way to do this question then

is to look at the x^2 term and the constant term

which have been given

9x^2 = (3x)^2

correct?

and 4 = (2)^2

[midnight haven]

I don’t see 3x or 2 in the equation

[autumn merlin]

well no but im just simplifying the terms

[midnight haven]

Oh yeah

Sorry I thought you were doing something to the problem

[autumn merlin]

if we want $9x^2 - kx + 4$ to be a perfect square, then it has to be of the form $(ax+b)^2$

[jolly parrot]

nalin

[midnight haven]

Yes

[autumn merlin]

and $(ax + b)^2 = a^2x^2 + 2abx + b^2$

[jolly parrot]

nalin

[midnight haven]

Ok now you lost me

[autumn merlin]

i just expanded (ax+b)^2

[midnight haven]

Ok

[autumn merlin]

understand what's going on?

[midnight haven]

Sort of but the only issue is the 9 on the x^2

I just need to get rid of it

[autumn merlin]

im coming to that

you dont have to get rid of it

so $a^2x^2 + 2abx + b^2 = 9x^2 - kx + 4$

[jolly parrot]

nalin

[autumn merlin]

do you understand this ^

[midnight haven]

Yea

[autumn merlin]

ok so now we just have to compare the coefficients

$a^2 = 9$; $2ab = -k$; $b^2 = 4$

[jolly parrot]

nalin

[autumn merlin]

alright?

[midnight haven]

It’s weird because their were 2 other problems very similar to this right before and they were easy because a=1

[autumn merlin]

yeah but now they've taken it to the next step

do you understand this though?

[midnight haven]

Yes

[autumn merlin]

perfect

so $a = \pm 3$; $b = \pm 2$

[jolly parrot]

nalin

[autumn merlin]

okay?

[midnight haven]

I got the answer already but again I’m not sure if we learned the 2ab part in class 😅

[autumn merlin]

which 2ab?

[midnight haven]

You plugged in 2ab to get 12

[autumn merlin]

oh

yeah

i think you would've learnt it but you must have learnt it only for a = 1 so you would've seen it as 2b instead of 2ab

but idk how else you're supposed to find k than to find 2ab

[midnight haven]

Next time I’m at math I will ask if there was a specific way to do it because I don’t remember doing these kind of questions tbh

.close

[pearl pond]

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[rocky bluff]

how do i solve this

using logarithms

[pearl pond]

@rocky bluff Has your question been resolved?

[rocky bluff]

.close

[pearl pond]

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[gloomy aurora]

this question is stressing me out i dont know how to partition a segment or do this at all

[charred ocean]

Can you figure out the length of pq?

[light helm]

consider midpoint formula

length of PQ not needed

[pearl pond]

@gloomy aurora Has your question been resolved?

[gloomy aurora]

what confuses me is how to apply the midpoint formula to my question but more specifically the steps

[hollow condor]

wdym?

[gloomy aurora]

i understand the concept but i dont know what to do to find R

[pearl pond]

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✅

[light helm]

.reopen

well what's the midpoint formula

[gloomy aurora]

add both x1 x2 and y1 y2 then divide both of them by 2 to find the coordinates

ive done that with PQ but im not sure if im right

[light helm]

show what you've done

[light helm]

yeh, its correct

[gloomy aurora]

okay but as for RQ i dont understand what to do…

i think my issue is im not understanding it the way i should

[pearl pond]

@gloomy aurora Has your question been resolved?

[light helm]

wdym

you have the midpoint
i.e. the coordinates of R
you're done

[gloomy aurora]

oh really??

[pearl pond]

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[gloomy aurora]

dont i have to use the ratio for something though

[pearl pond]

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[true skiff]

Yo

[true skiff]

How would i solve this?

https://i.imgur.com/J5QJLUA.png

is there a actual process or do i just keep guessing numbers?

[hearty parcel]

Are both 'F's the same?

[true skiff]

yeah

[hearty parcel]

I'll take ur word for it

It's already said that

f( T ) = 0

[hearty parcel]

So sub this into the the line above

[true skiff]

So i put 0 for the Ts?

and solve?

[hearty parcel]

No

You said
$$f(T) = T^{2} - 9T + 18$$
$$f(T) = 0$$

[jolly parrot]

Kai Funaba

[hearty parcel]

Notice anything similar?

[true skiff]

the F(T)

[hearty parcel]

Good

And the value of f(T) is..?

[true skiff]

0

[hearty parcel]

Now sub 0 into the first line

[true skiff]

sub it for what tho in the firts line?

[hearty parcel]

Which is "f(T)" ?

[odd dirge]

So you can solve it with factorising or quadratic equation

[hearty parcel]

Step by step
They can't see y they need to sub 0

[odd dirge]

Oh ok

[true skiff]

So 0=T^2-9T+18?

[hearty parcel]

Yep

[true skiff]

gotcha

[hearty parcel]

$$T^{2} - 9T + 18 = 0$$

[jolly parrot]

Kai Funaba

[true skiff]

so i slide it over?

[hearty parcel]

U mean the 0?

Either side is fine

[true skiff]

oh okay

[hearty parcel]

I just prefer it on the right

[true skiff]

I see

[hearty parcel]

Then just factorise the rest

[true skiff]

So would the answer be 6 and 3?

[odd dirge]

You can substitute back to the equation to verify the answer if it's equal to 0, but yes, that's correct

[true skiff]

So id write T=6,3?

[odd dirge]

T = 6 or T = 3

[true skiff]

Cool ty both very much for helping me

.close

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

[opal forge]

. close

[last summit]

Perchance

. Close

[toxic lichen]

.close

[pearl pond]

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[last summit]

. Reopen

[opal forge]

.close

[fast zealot]

Try .end

[pearl pond]

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[ornate tundra]

Cos(x+π/3) is = sin(π/6-x) right?

[muted void]

use cosθ = sin(π/2 - θ)

θ = x + π/3

@ornate tundra

[pearl pond]

@ornate tundra Has your question been resolved?

[ornate tundra]

Ooo

Ty

[pearl pond]

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[opal forge]

hints

[toxic lichen]

you will have to explain what this notation means.

[buoyant ridge]

divides?

[toxic lichen]

otherwise it is impossible to give hints.

[pearl pond]

@opal forge Has your question been resolved?

[opal forge]

I don't know what they meant

[toxic lichen]

then neither do we.

[opal forge]

I thought maybe it is a famous notation and you guys know

Answer is given B

Let me show you the solution

654

[toxic lichen]

formally this does not give us enough info to decipher

but MAYBE (big maybe) this fraction-looking slash symbol means "divides"

which i personally have seen written with a vertical bar

in which case yeah this is about remembering properties of divisibility

no tips can be given you just need to know this

it is almost like vocabulary knowledge

[opal forge]

The solution is okay?

I meant correct

What if i apply divides is this true

[toxic lichen]

yes the solution is correct

[opal forge]

I am taking random numbers for checking this

I checked by taking random numbers

.close

[pearl pond]

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[bright sentinel]

need help on this

[spiral birch]

to remove the log on the left side, raise x to the power of a on the right side. then to isolate x, raise the power of the a on the left side to 1/a

[brazen vector]

you mean multiply by 1/a?

[spiral birch]

no

[brazen vector]

oh wait i got what you mean

[bright sentinel]

i dont get it

can u elaborate pls

[brazen vector]

$log_{x}{a}=a\implies x^{log_{x}{a}}=x^{a}$

[jolly parrot]

calculus is fun

[bright sentinel]

oh i get it

[brazen vector]

from here you get $a=x^a\implies a^{\frac{1}{a}}=(x^a)^{\frac{1}{a}}\implies a^{\frac{1}{a}}=x$

[jolly parrot]

calculus is fun

[bright sentinel]

.close

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[bright sentinel]

for this

[bright sentinel]

If i write that in root does that make sens?

[pearl pond]

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[oblique reef]

At a catering establishment, Crissie bought some trays of food for $31 each and some side dish for $9 each. If she spent a total of $305, how much did she spend on side dishes

[oblique reef]

Is there a quick way to do this?

Like to restrict solutions to whole numbers instead of guesswork or diophantine (for integers)

[pearl pond]

@oblique reef Has your question been resolved?

[pearl pond]

@oblique reef Has your question been resolved?

[pearl pond]

@oblique reef Has your question been resolved?

[pearl pond]

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[pearl pond]

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[scenic rock]

some basic trigonometry help, what's CB length ?

[old geyser]

Is PDA a right angle

Or CDA

(Whatever's on the end of that lever)DA

[scenic rock]

it's supposed to be a right triangle ye

[old geyser]

You can use the geometric mean theorem

[scenic rock]

and the P is the force applied which will make a little more angle

[scenic rock]

awww tysm !!

life saver

[pearl pond]

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[pearl pond]

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[stuck widget]

what is the thought process of doing this

[stuck widget]

this is a multiple choice question so I can't spend more than like 2 minutes on it

that is if I were in an actual test but I'm not

[blissful cloak]

Concave up happens when $f''(x)>0$

[jolly parrot]

PajamaMamaLlama

[stuck widget]

but its the graph of f(x) so wouldn't that change things around?

[regal herald]

the graph of f is concave up when:

[blissful cloak]

yes this, bad wording on my part

[stuck widget]

because I only know using the 2nd deriviative to find the nature of the first derivative

or inflection points

so I don't really know how f(x) and f"(x) relates

[regal herald]

f'(x) is the gradients, f''(x) is the 'gradient of the gradient function' i guess
when f''(x)>0, f'(x) is increasing, so f is concave up

[blissful cloak]

and hence f(x) is concave up as a result

[stuck widget]

ah

I need to remember that 😅

thanks for the help

.close

[pearl pond]

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[fickle finch]

for number 11

[fickle finch]

how do they find the length of the integral from t=0 to t=2?

how do the point corresponds to the t values?

heres the full solution

is it because we are dealing with trig function so it has to start at 0 and end with 2 pi as a full rotation?

[pearl pond]

@fickle finch Has your question been resolved?

[pearl pond]

@fickle finch Has your question been resolved?

[pearl pond]

@fickle finch Has your question been resolved?

[pearl pond]

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[still patrol]

where does the 3/2 in the power come from

[unborn abyss]

,tex .exp rules

[jolly parrot]

hayley.bin

[still patrol]

which rule?

[unborn abyss]

well, you'll need a few

perhaps you might start with the one that has the radical symbol in it

[still patrol]

can you elaborate further

[pearl pond]

@still patrol Has your question been resolved?

[still patrol]

<@&286206848099549185>

[still patrol]

<@&286206848099549185>

Can i get anyones help

[frozen lantern]

aight

so

$\frac{tp_2 tp_3}{\sqrt{tp_1}}$

[jolly parrot]

Percy

[frozen lantern]

@still patrol

be alive sir

[still patrol]

Im here 😂 lemme look

Thx btw

[still patrol]

Yes

[frozen lantern]

what is the power of t in the numerator?

[still patrol]

1

[frozen lantern]

oh jesus

what is t * t * p2 * p3

[still patrol]

t^2 * p2 * p3

[frozen lantern]

yeah

[still patrol]

Yeah and😂

[frozen lantern]

and

how can you write $\sqrt[n]{t}$?

[jolly parrot]

Percy

[still patrol]

Nvm got it ! Thanks for your help 🙂

.close

[pearl pond]

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[frozen lantern]

yeah yay

[pearl pond]

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[brittle osprey]

help

[calm wing]

with?

[frozen lantern]

💀

[meager gorge]

😭

[feral sedge]

💀

[unborn abyss]

[frozen lantern]

i think he needs help

[calm wing]

not enough twinks

[meager gorge]

seems like we got epicly trolled

[brittle osprey]

P (2x-1) = x^2 - 3x + m + 1
and
the sum of the coefficients for P(x+6) is 2
then what is the constant of P(x+3)

i dont get the logic behind this at all

my teacher basically solved it like

ok to find the sum of the coefficient u give x 1

[frozen lantern]

oh he does have a question

yeah

[brittle osprey]

so p(7) equals 2 then

but

the thing i dont get is

ok if someone can vc about this thatd be great too but like

if no one can i’Ll jus type

[meager gorge]

we usually type here

[frozen lantern]

i wonder if we have vcs here

[brittle osprey]

i thought to find the sum of coefficients u do

P(1)

[frozen lantern]

we dont

[frozen lantern]

yeah

[brittle osprey]

hm

[calm wing]

yes

[brittle osprey]

my brain is asoup

[frozen lantern]

not enough twinks 💀

[calm wing]

and for constant, you do ||P(0)||

[brittle osprey]

then by logic, shouldnt we

[frozen lantern]

yeah

[brittle osprey]

instead of saying p(7) equals 2

[frozen lantern]

no we shouldn't

[brittle osprey]

x+6= 1

[meager gorge]

life's tough

[brittle osprey]

then x is -5

then so we put that x value in the polynomial given abve

or some hsit

[frozen lantern]

no it isn't 💀

[brittle osprey]

ok

will you guys help me or not

[frozen lantern]

you set the variable equal to one

in this case x

[brittle osprey]

[frozen lantern]

so p(1+6)=2

[brittle osprey]

x+6 isnt the variable? x is?

[frozen lantern]

wait sorry

yeah

x+6=1

[brittle osprey]

see thats wrong

[frozen lantern]

shouldntttt be-

wait

oh god im tired

[brittle osprey]

is anyone able to help me with this?

[meager gorge]

i'm confused about this whole question 😭

[brittle osprey]

bruh

[frozen lantern]

same 💀

[brittle osprey]

<@&286206848099549185>

[meager gorge]

apologies lmao

[brittle osprey]

.close

[pearl pond]

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[midnight haven]

asked to turn this into standard form

[calm wing]

is it a polynomial?

[midnight haven]

yep

[calm wing]

unknown degree?

[pearl pond]

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[vague sandal]

if i have matricies a and b, is (-a)*(-b)=ab? if so, why?

[nimble cave]

yeah. its because -a = (-1)a, and the scalars commute with matrix products

[vague sandal]

damn i need to sleep thanks

.close

[pearl pond]

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[pearl pond]

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[dark sun]

How do I prove that there is no more functions left in functional equations?

[pearl pond]

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[fickle canyon]

A =
[−12 6
14 −7]

[fickle canyon]

Hello I'm trying to find what B would have to be to get AB = 0

how could i achieve that?

[signal atlas]

Let x be column 1 of B, solve Ax=b

Do the same thing for each column

[fickle canyon]

wdym solve for Ax = b?

[signal atlas]

The strategy is to find each column of b independently

A single column can be though of as a vector

[fickle canyon]

im sorry but im unaware of how to do that

[signal atlas]

You mean solving Ax=b?

[fickle canyon]

yah

[signal atlas]

Have you not done anything like gaussian elimination?

[fickle canyon]

maybe? idk the name of all the things they want me to do

[signal atlas]

Alright I think I got a simpler way

[woven matrix]

you can also notice that -12 = -2*6
and 14 = -2*(-7)
so if you take
B = [1
2]
it will give you 0, and any column of B has to be of this form

the 2nd line of B has to be twice the first line

[fickle canyon]

i wish that made sense to me

but im lost

[signal atlas]

Lmao same

[signal atlas]

I was gonna say use definition of matrix multiplication

[woven matrix]

Haruno -> if you're familiar with linear algebra, B has, as columns, vectors of the kernel of A since AB = 0
but the kernel of A is obv of dim 1 and we guessed (1, 2) was in it, so it's span{(1, 2)}

so 2nd line of B has to be twice 1st line

[signal atlas]

Oh I see

That sounds a bit advanced to understand though if they havent learned how to solve Ax=b

[woven matrix]

yeah but if he can see how it has to be true for each column individually, which you explained, then there's no need to talk about the advanced stuff

[signal atlas]

I see

[woven matrix]

once he does it for one column of B, each column is the same calculation

[fickle canyon]

im not sure what calculation u mean

[woven matrix]

you take a matrix
B = [a1 b1 ...
a2 b2 ...]

and you calc the product AB

and you equate it to 0

you'll see that the only condition is a2 = 2a1, b2 = 2b1 etc

you'll conclude that AB = 0 iff the 2nd line of B is twice the first line

because it's the same calculation for each column of B again and again

it always only gives the condition 2nd line = twice 1st

[pearl pond]

@fickle canyon Has your question been resolved?

[pearl pond]

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[wary fulcrum]

i dont understand this

<@&286206848099549185>

[pearl pond]

@wary fulcrum Has your question been resolved?

[wary fulcrum]

.close

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[languid quest]

Hi, how would I do this?

[strange quiver]

finding g(3) is trivial

for g'(3) (slope at x=3) slope is same as slope of tangnet line

[languid quest]

Oh

Idk what integrate means and idk the equation

[languid quest]

Alr

Yep

I get that

g(3) = 3

[languid quest]

Sorry I'm still unsure

So g(3) = 3

Obviously

What about g'(3)

[strange quiver]

as I said the slope of a function is same as the slope of the tangent line

so you need to find the slope of tangnet line at x=3

the question gave you two points on the line so you can calculate its slope

[languid quest]

Ohhh -1

Tysm

[modest dragon]

you have (3,3), and (1,5). find its slope

[languid quest]

-1

[pearl pond]

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[pearl pond]

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[graceful nexus]

if f(n) = 1/n^2 and g(n) = n what is f = ?(g), answer can be big O or theta and the smaller counterparts

[pearl pond]

Please don't occupy multiple help channels.

[graceful nexus]

this should be O right because when n approaches infinity it gets smaller

or am i missing something

[pearl pond]

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@graceful nexus Has your question been resolved?

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[torn willow]

can someone pls help me with this

[torn willow]

pls pls pls

[pearl pond]

@torn willow Has your question been resolved?

[torn willow]

<@&286206848099549185>

[pearl pond]

@torn willow Has your question been resolved?

[pearl pond]

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[undone oasis]

Would someone explain to me what happened between 1:07 and 1:21 on https://youtu.be/765X_PAxhAw ? Transcript on https://www.khanacademy.org/transcript/x03a2d28c

[tropic saddle]

how do you differentiate $e^{y(x)}$

[jolly parrot]

Denascite

[undone oasis]

I'm not sure does the (x) next to the y mean.

does it mean you input x and it gives you y?

[tropic saddle]

yes

y(x) could for example be x^2 or sin(x)

how would you differentiate those

[undone oasis]

I know the power rule for the former and from repetition know the 2nd.

[tropic saddle]

do you know the chain rule

[undone oasis]

I've seen it but know know in which circumstant to apply it.

[tropic saddle]

you use the chain rule whenever you have a composition of functions

e^sinx is a composition of the functions f(x)=e^x and g(x)=sinx.

e^sinx = f(g(x))

[undone oasis]

oh nice

so e^y*dy/dx mean "we're interested in derivating y in e^y" ?

if not, can you phrase it in common language?

[tropic saddle]

its an application of chain rule

outer derivative times inner derivative

f'(g(x))*g'(x)

if you have seen that before

for us f(x)=e^x and g(x)=y(x)

so we get f'(y(x))*y'(x)

f'(x) is still e^x

so f'(y(x))=e^y

and y'(x) is the same as dy/dx

[undone oasis]

y'(x)=1 right?

[tropic saddle]

in general no

[undone oasis]

y'(x)=1 for this case though?

[tropic saddle]

no

[pearl pond]

@undone oasis Has your question been resolved?

[pearl pond]

@undone oasis Has your question been resolved?

[pearl pond]

@undone oasis Has your question been resolved?

[pearl pond]

@undone oasis Has your question been resolved?

[primal robin]

In the video, this is what they are trying to find y'(x).
They start with y = lnx, then change it to exponential form
e^y = x
Take derivative with respect to x on both sides:
d(e^y)/dx = d(x)/dx
d(e^y)/dx = 1

As Denascite explained, for the left side you need the chain rule because y is a function of x: y(x)

Then they solve for y'(x), use the fact that e^y = x, and you get
d/dx (ln x) = 1/x

[undone oasis]

because y is a function of x: y(x)
does that mean the function y=1x+0 ?

[primal robin]

no, y(x) = ln(x)

a(b) is a way to say that a is a function of b

same as a is a function with variable b

[undone oasis]

so

for us f(x)=e^x and g(x)=y(x)
should have been
for us f(x)=e^x and g(x)=y(x)=ln(x)
?

[primal robin]

Yes g(x)=y(x)=ln(x) is a correct equation, but setting g(x)=ln(x) would be useless in this case because we're trying to find y'(x)

so just say g(x)=y(x)

otherwise, y(x) "disappears" and we won't get any info about y(x)

[undone oasis]

could you expand?

[primal robin]

let's say you use this

for us f(x)=e^x and g(x)=y(x)=ln(x)

then f(g(x)) = e^(lnx) = x

The function y is gone as it was simplified, and we can't get it's derivative

Mmm still that's kind of arbitrary, you can do it in any way you like as long as you know what you're doing 🤔

Did you follow the procedure until this part?
d(e^y)/dx = 1

[undone oasis]

I stopped at the line right above the 1st dy

[primal robin]

They start with y = lnx, then change it to exponential form
e^y = x

Do you get this?

[undone oasis]

I understand it as you give ln x and it gives the result of e^x, and results are y

[primal robin]

mmm ok

[undone oasis]

oh, and the x that was fed to ln was e^x

wait

stop

[primal robin]

y = lnx <-> e^y = x
these are equivalent

[undone oasis]

yes

[primal robin]

ok

Now we use this: e^y = x
Take the derivative with respect to x of both sides

$\frac{d}{dx}\left(e^y\right) = \frac{d}{dx} x$

[jolly parrot]

ELeonardo

[undone oasis]

please @ me when you get back to this

[primal robin]

I was expecting questions haha, ok

d/dx of x is just 1

$\frac{d}{dx}\left(e^y\right) = 1$

[jolly parrot]

ELeonardo

[primal robin]

e^y is a composition of functions:
f(x) = e^x and g(x) = y
When they are composed:
f(g(x)) = e^g(x) = e^y

The chain rule works for compositions

Chain rule:

$\frac{d}{dx} f(g(x)) = f'(g(x)), g'(x)$

[jolly parrot]

ELeonardo

[undone oasis]

g(x) = ln(x) = y right?

[primal robin]

yes

we're working with e^y, not e^lnx

of course in this case they're the same because y=lnx

but we're using y for now

[undone oasis]

using y while not including ln(x) seems obvious to you but not me, I have to keep seeing it.

[primal robin]

I mean is just a way to write it

And it's better for now, so you can learn

after you learn this totally, you can write it as you like

Remember that we want to find the derivative of ln(x)
In order to do that, we start from
y = ln x

whenever you see y, it's lnx

[undone oasis]

ok, but at the end I'll just copy-paste all y's by ln(x)

[primal robin]

ok ok I think I get your point. Sure we can use lnx as long as you don't simplify it

in that case, we have

$\frac{d}{dx}\left(e^{\ln x} \right) = 1$

[jolly parrot]

ELeonardo

[undone oasis]

cool

[primal robin]

Terms we need for chain rule:
f(x) = e^x and g(x) = lnx

f'(x) = e^x and g'(x) = (lnx)'

f'( g(x) ) = e^lnx

$e^{\ln x} \left(\ln x \right)' = 1$

[jolly parrot]

ELeonardo

[primal robin]

ok so far?

[undone oasis]

yes

[primal robin]

ok now, (lnx)' is another way of writing d (lnx) / dx

$\frac{d}{dx}\ln x = \frac{1}{e^{\ln x}}$

[jolly parrot]

ELeonardo

[undone oasis]

I'm not sure if I'd get a mistake if submitting it to a teacher, but for me it's A ok.

[primal robin]

e^lnx = x, so

$\frac{d}{dx}\ln x = \frac{1}{x}$

[jolly parrot]

ELeonardo

[undone oasis]

+thank @primal robin

+close 300

.thank @primal robin

!thank @primal robin

D:

[primal robin]

just .close haha, there are no commands for that here

dw

[undone oasis]

.close 300

[pearl pond]

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[tender geyser]

the answer is 2 + root 5 but how do i get to there

[tender geyser]

x = 2 + root 5 btw

[toxic lichen]

uncrop

[tender geyser]

okay

[pearl pond]

@tender geyser Has your question been resolved?

[tender geyser]

uh no

[pearl pond]

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[torn willow]

can someone pls help me with this?

[pearl pond]

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[pearl pond]

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[fickle canyon]

x^2 + y^2 = 82
-x^2 + 2y = -19

[fickle canyon]

Can i get some help solving the system of linear equations

i have absolutely no idea what to do with it

[sharp vigil]

a good first step would be to add the two together

[fickle canyon]

oh

i was trying to do substitution

[restive briar]

Elimination should be easier here

[dim tangle]

you can do it either way. sometimes its easier to do one way over the other.

[fickle canyon]

so y should be
y = -9
y = 7 ?

[dim tangle]

almost

[fickle canyon]

oh? what did i miss

[dim tangle]

no nvm you got it

[fickle canyon]

oh lol

so im getting

(1, -9)
(sqrt(33), 7)

i believe this is correct

[dim tangle]

looks right to me

[fickle canyon]

thank u so much

.close

[pearl pond]

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[dim tangle]

remember though its +- 1 and +- sqrt(33). since it was a square root.

[fickle canyon]

.reopen

[pearl pond]

✅

[fickle canyon]

oh lord

idk if i'll ever rememebr that

do i put +- in the solution or

just put 4 solutions

prob in the solution that i look at it

[dim tangle]

it's technically 4 solutions, since you are finding where both equations are equal (where they intersect)

and they intersect at 4 points

[fickle canyon]

from the looks of it it'll accept 2 with +- in front of them

i hope

[fickle canyon]

do u know if using a graphing calcualtor (ti-84 plus cs) is there a way to accurately view the points they intersect at

or is that a lost cause

[dim tangle]

you could, i think. let me see

[fickle canyon]

im going to murder this site

[dim tangle]

you forgot an apostrophe

comma

between 2nd and 3rd point

[fickle canyon]

ugh

[dim tangle]

you can graph it but it looks a little wonky

[fickle canyon]

yah but do u know if theres a way to view the points accurately?

where they intersect

i feel like there should be or graphing this stuff on a calculator would be pointless but idk

[dim tangle]

you can find intersections

[fickle canyon]

which button is that?

[dim tangle]

[2nd][Trace]

you have to pick two curves, then look for a point close to the intersection

and it'll find it

you trace with <- -> and you pick curves with ^ and v

when graphing these curves though, since they are relations you have to solve for y to put it into the calculator to graph

thats why i had to use two curves to fully graph the circle

[fickle canyon]

ahh alright

[dim tangle]

graph looks better with ZoomSquare

[fickle canyon]

is that a calculator thing or?

[dim tangle]

yeah

you press [Zoom][5] for the square zoom

[fickle canyon]

ohh okay

tyvm for all ur help

[dim tangle]

np

[fickle canyon]

.closeq

.close

[pearl pond]

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[left crane]

In one model, the biceps muscle of a particular person can be described by the body of revolution that appears when the graph of f is rotated 360° around the prime axis in the interval [0;40].
f(x)=2sin⁡(0,05πx-0,5π)+2

Strength of the biceps muscle is proportional to the maximum cross-sectional area of the muscle. Determine the maximum cross-sectional area of the biceps muscle

[pearl pond]

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[left crane]

nope

[pearl pond]

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[left crane]

nope

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[trim wing]

An arch has the shape of semi-ellipse that is 10 ft. long and 3 ft. high at the center. How high is the arch at a point 3 ft from one endpoint of the base of the arch?

[trim wing]

Anyone have any ideas about this?

[pearl pond]

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.close

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[gaunt ocean]

hi guys I need help on how to find side lengths of a triangle if I only have angles and the triangle isn't a right triangle so I can't use sohcahtoa

[gaunt ocean]

I'm supposed to use sine law and cosine law but I'm not sure which one to use since both of them use side lengths

[woeful fern]

I don’t think it’s possible if you only have angles. No side lengths at all?

[vestal tapir]

yeah, i mean it's not

you can't fide any side from 3 angles

[woeful fern]

There are infinite amount of solutions

[pearl pond]

@gaunt ocean Has your question been resolved?

[gaunt ocean]

wait nvm there is 1

can I show you the question?

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[stuck gorge]

help

[stuck gorge]

can someone explain how to prove k ≠ 0

its not listed as an answer bu

is it part of the answer?

this is a graph, and y = 0 only intercepts once

cuz C = 0 only has one solution

and when you sub k = 0 into the equation

this one

you only get one solution, x = 4

but when you solve discriminant > 0 you get k = 0 within the range

discriminant = 1 when k = 0

is it because the equation becomes linear, so discriminant doesnt aplly when k = 0?

<@&286206848099549185>

[pearl pond]

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[pearl pond]

@stuck gorge Has your question been resolved?

[cloud zephyr]

this may or maybe not be one of the reasons, but I didn't prove or disprove it so i can't tell.
But this is the case causing trouble in some of the "original" quadratic equations.

But another thing is that, when you have kx²+.....=0
you need to divided by k to get the roots, but the point is k cannot be zero in order to be divided, hope this helps!

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[opal forge]

Minimum will be 6

Am I right?

[pearl pond]

@opal forge Has your question been resolved?

[strong glade]

the minimum of what?

[pearl pond]

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[ruby heath]

can you explain this

[pearl pond]

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[opal forge]

.reopen

[pearl pond]

✅

[opal forge]

I only assumed for it

[brisk pike]

z is complex? and |z| = a^2 + b^2 then? (if z = a+bi)

[ruby heath]

try plugging in each number and see if the number you get ends up being less than or equal to 6

you can find the answer that way

or, you can combine the two absolutes

|z + 2| + |z - 2| is the same as |z + 2 + z - 2| or |2z|

so you can also check |2z| <= 6

[woven matrix]

that's so false lmao

[brisk pike]

is || abs value for integer or complex?

like is z integeral or complex

[woven matrix]

both cases lead to spoutshield saying something false, even in the complex you don't have |z+2| + |z-2| = |z+2+z-2| (the triangular inequality implies <=, with them being different for most numbers, like 2i)

|| is comprehensible as a distance, z being complex or real

[brisk pike]

i was confused for the complex case, thanks for confirming my suspicion

[brisk pike]

okay, thx!

[opal forge]

Of course complex

Wow

[brisk pike]

well have u tried substituting? it would be painful but shouldnt be too bad

$\sqrt{(a+2)^2 + b^2} + \sqrt{(a-2)^2 + b^2} \leq 6$ \\
Since square root cant be negative, \\
$0 \leq \sqrt{(a+2)^2 + b^2} + \sqrt{(a-2)^2 + b^2} \leq 6$ \\

[jolly parrot]

ItzKraken

[brisk pike]

$(a+2)^2 + b^2 + (a-2)^2 + b^2 + 2\sqrt{((a+2)^2 + b^2)((a-2)^2 + b^2)} \leq 36$

[jolly parrot]

ItzKraken

[brisk pike]

$2a^2 + 8 + 2b^2 + 2\sqrt{b^4 + ((a+2)(a-2)b)^2 + ((a+2)(a-2))^2} \leq 36$

[jolly parrot]

ItzKraken

[brisk pike]

$2\sqrt{b^4 + ((a+2)(a-2)b)^2 + ((a+2)(a-2))^2} \leq 28 - 2(a^2 + b^2)$

[jolly parrot]

ItzKraken

[brisk pike]

$\sqrt{b^4 + ((a+2)(a-2)b)^2 + ((a+2)(a-2))^2} \leq 14 - (a^2 + b^2)$

[jolly parrot]

ItzKraken

[brisk pike]

isnt that with absolute value?

and equality only and only if a=b?

$b^4 + ((a+2)(a-2)b)^2 + ((a+2)(a-2))^2 \leq 196 + a^4 + b^4 - 28a^2 - 28b^2 + 2a^2b^2$

[jolly parrot]

ItzKraken

[brisk pike]

$b^4 + (a^2b - 4b)^2 + ((a+2)(a-2))^2 \leq 196 + a^4 + b^4 - 28a^2 - 28b^2 + 2a^2b^2$

[jolly parrot]

ItzKraken

[brazen vector]

triangle inequality solves it directly

[brisk pike]

huh..i never applied the inequality outside of geometry

[brazen vector]

i never applied triangle inequality in my life

[brisk pike]

,w simplify b^4 + (a^2b - 4b)^2 + ((a+2)(a-2))^2 \leq 196 + a^4 + b^4 - 28a^2 - 28b^2 + 2a^2b^2

[brisk pike]

how do u know it works then

[brazen vector]

i just applied it when helping others sometimes when needed

[jolly parrot]

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Requested by itzkraken.

[brazen vector]

but i never had to use it

[brisk pike]

oh i see

[brisk pike]

I see

well i gtg

i am gonna continue this algebra sometime later tho

cya

[opal forge]

Sure

Take care

[brazen vector]

hey arjunn

use triangle inequality it gives the answer faster

$|a+b| \leq |a|+|b|$

[jolly parrot]

calculus is fun

[brazen vector]

you have |z+2|+|z-2|=<6

so what are a and b from here

@opal forge

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[pearl pond]

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[opal forge]

.reopen

[pearl pond]

✅

[pearl pond]

@opal forge Has your question been resolved?

[pearl pond]

@opal forge Has your question been resolved?

[regal flax]

Ok I have epic solution -
|Z-2| is distance of z from point (2,0)
|Z+2| is distance of z from point (-2,0)

Now what is the locus of a variable point whose distance from 2 fixed points is constant

It's an ellipse

So you have an ellipse with foci (±2,0) and major axis 6

|z| is distance from origin which is the centre of ellips

So max distance from centre will be semi major axis =3

@opal forge

[opal forge]

Yes I'm reading

[brazen vector]

what are a and b here

[pearl pond]

@opal forge Has your question been resolved?

[last summit]

@opal forge

[pearl pond]

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[formal tiger]

How do solve this

[rigid mist]

square each side

2 times

[formal tiger]

Wdym

[rigid mist]

squaring both side?

[formal tiger]

Ya

How do u do it twice

[rigid mist]

like when you will do it the first time, and simplify it a bit, you will still end up with a root, so you have to square both side again

in order to remove that root

[formal tiger]

Idk whar im doing wrong