help-10

[long hedge]

$\frac{(x+h)^2-6-(x^2-6)}{h}$

$\frac{x^2+2xh+h^2-6-x^2-6}{h}$

[warm shale]

putridplanet

[long hedge]

$\frac{2xh+h^2-6-6}{h}$

[warm shale]

putridplanet

[long hedge]

im guessing one of the 6's was supposed to be positive but idk how

[coarse ibex]

You dropped parens here

[warm shale]

putridplanet

[long hedge]

this better

[coarse ibex]

Yep

[long hedge]

ok

abuse parenthesis

[obtuse pebble]

@long hedge Has your question been resolved?

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[reef jasper]

Sorry to ask another question, but how would you do this? I can't find the ratio

[high lily]

did you make any attempt to find the ratio

[reef jasper]

yeah I tried but I couldn't find it, I tried using x and listing out the arcs. i did 686, 686x, 686x^2 and then 128

i wasnt sure if i was correct in my approach though

[high lily]

what's the fourth arc in terms of x

[reef jasper]

686x^3 oh omg

wait im sorry

lmao

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[serene swift]

Suppose that point is on a circle with radius $10$ cm, and ray $OP$ is rotating with angular velocity $\frac{\pi}{8} \frac{\text{radian}}{\text{second}}$.

[warm shale]

mXd

[serene swift]

b) Find the distance traveled by P along the circle in $6$ seconds.

[warm shale]

mXd

[formal finch]

I would first calculate the radians the point travels and convert that into cm

[serene swift]

so (pi/ 8 sec)(10 cm)=5pi cm / 4 sec

thats the linear velocity

[formal finch]

Not quite

First you have to multiply it by the seconds it travels for

[serene swift]

pi/8 * 6

3pi/4

[formal finch]

And then calculate the circumference

[serene swift]

20pi cm

x/20pi=(3pi/4)/2pi

[formal finch]

And divide that by 2pi bc that's how many rad you have in a circle

And then multiply by the number of rad you have

[serene swift]

(3/8)(20pi)=15pi/2

wait

isnt linear velocity equal to radius times angular velocity

[formal finch]

This should be correct

[formal finch]

Not sure

[serene swift]

ok i did (pi/8 rad/sec)(10 cm)

[formal finch]

Yes i think

[serene swift]

and got linear velocity=5pi/4 cm/sec

[formal finch]

Why is linear velocity relevant-

[serene swift]

you can use it to find the distance

[serene swift]

so (5pi/4 cm/sec)(6 sec)=30pi/4 cm

=15pi/2 cm

[formal finch]

Ohhh i just used the circumference/rad to solve

Same solution tho

[serene swift]

ye thx

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[obtuse pebble]

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[formal finch]

Np

[obtuse pebble]

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[silent jolt]

what does this question mean

[native inlet]

find $\theta$ such that $\cot\theta=0$ where $\theta$ is in between 0 and 720 degrees

[warm shale]

XxMrFancyu2xX

[silent jolt]

is it just the x and y cooridnates of tan theta but reciprocal

[native inlet]

wait that doesn't make sense

[silent jolt]

like tan is y/x

so cot = x/y

[native inlet]

nvm im just not awake

[silent jolt]

lol

[native inlet]

find a $\theta$ such that $\frac{\cos\theta}{\sin\theta}=0$ or in other words $\cos\theta=0$

[warm shale]

XxMrFancyu2xX

[silent jolt]

isnt cos the x coordinate

[native inlet]

yep :))

[silent jolt]

so like 90 , 270 , 450 etc

how come when i type 1/tan(90) in my calucalotr it says error since tan 180is 0

[native inlet]

is undefined because division by 0 is mega cring

[silent jolt]

how do u do cot in calcualtor then

since its 1/tan

is it not 1/ tan(theta)

[native inlet]

but think about what tan is

in terms of sine and cosine

[silent jolt]

sin / cos so cot is cos/sin

[native inlet]

yep, now put cos(90)/sin(90) into your calculator :))

[silent jolt]

why dpes it say error

[native inlet]

uhhh

[silent jolt]

isnt it 0/1

so 1

0

[native inlet]

yes

so idk ur calculator is bad at life

[silent jolt]

my brothers calcualtor is just ass then

[native inlet]

yea maybe consider using a different calculator

[silent jolt]

it should say 0 tho rught instead of error

[native inlet]

yep

[silent jolt]

alright thanks

cya man

.close

[obtuse pebble]

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[fair whale]

The chapter is explaining how to show a limit doesn't exist by negating the delta-epsilon definition of a limit. Where did it get |f(x)| = 1 from? And why is x0 = d/2?

[royal basin]

nobody is claiming f(x) itself is 1.

it is |f(x)| that is 1

And why is x0 = d/2?
it'll be d/3 next time you ask

[fair whale]

What do you mean?

[polar fossil]

it's arbitrary, we just need to divide by something to make it smaller than d

could be d/3, could be d/7, we choose d/2 because it's simple and works for us

[fair whale]

Ahh, makes sense

And how did they arrive at |f(x)| = 1?

[royal basin]

f(x) = x/|x|

[fair whale]

I got |f(x) - L| = |x/|x| - 0| < e (which is 1/2)

[royal basin]

this is 1 for positive x and -1 for negative x

[fair whale]

I thought we couldn't take 1 as a bound because then f(x) will be inside the epsilon range

Wait wait, is it because x/|x| = x/x = 1?

Was I too blind to notice that

[polar fossil]

when x>0, x/|x| = x/x = 1
when x<0, x/|x| = x/-x = -1

[fair whale]

That was so obvious, I just realized haha

Thanks for the help!!

.close

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[royal basin]

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[balmy sequoia]

1

[royal basin]

do you know how to solve linear equations in general?

[balmy sequoia]

ye

[gleaming ridge]

Then what's the issue here?

[balmy sequoia]

somehow the answer is a fraction

[gleaming ridge]

Okay, show what you've done

And we'll help correct

[balmy sequoia]

$-8x+24=10x$

[warm shale]

nicholas

[balmy sequoia]

$24=18x$

[warm shale]

nicholas

[balmy sequoia]

$x=0.75$

[warm shale]

nicholas

[high lily]

how are you getting x=0.75

[balmy sequoia]

18 divided by 24

[high lily]

why are you dividing 18 by 24

[gleaming ridge]

Shouldn't this be the other way around

[balmy sequoia]

i dont know

wait

1.333

into fraction is 4/3

oh my goddd

4/3 is the answer

[thick oracle]

https://tenor.com/view/boom-mind-blown-mind-blowing-eyeglasses-gif-15569009

[balmy sequoia]

thx

i also need help with this

[brisk matrix]

what methods have you been taught to solve something like that?

[balmy sequoia]

elimination

i guess

[brisk matrix]

have you attempted applying that to the problem?

[balmy sequoia]

yes

i think i end up with

$7x-y=28$

[warm shale]

nicholas

[balmy sequoia]

not sure if its right or not

[brisk matrix]

show how you got there

[balmy sequoia]

if its the same you would subtract but since its different i would add

so 3x+4x is 7x

wait

there woulnt be a y

because they cancel out

y would be gone

so $7x=28$

[warm shale]

nicholas

[brisk matrix]

that looks good

[balmy sequoia]

$x=0.25 or 1/4$

[warm shale]

nicholas

[brisk matrix]

no

try that again

[balmy sequoia]

im looking at my calculator right now

[high lily]

calc won't give you what you need if you don't enter the correct thing

[balmy sequoia]

oh

whoops

$x=4$

[warm shale]

nicholas

[high lily]

consider what you could do to 7x that'll give you just x
and perform that action on both sides

[balmy sequoia]

now we apply to equationm

[high lily]

don't divide stuff in a random order

[balmy sequoia]

i know

i got confused

with the other one since that was in a different order than this one

[brisk matrix]

now that you have x, do you think you can find y?

[balmy sequoia]

$4(4)+y=10$

[warm shale]

nicholas

[balmy sequoia]

$16+y=10$

[warm shale]

nicholas

[balmy sequoia]

$y=6$

[warm shale]

nicholas

[balmy sequoia]

final answer

[high lily]

no

[balmy sequoia]

(4,6)

[high lily]

how are you getting
y=6

[balmy sequoia]

minus 16 from both sides

[high lily]

10-16 isn't 6

[balmy sequoia]

wait

oh

ok makes sense

$y=-6$

[warm shale]

nicholas

[balmy sequoia]

(4,-6)

[high lily]

yes

[balmy sequoia]

how do i find 10.4% of something

[high lily]

1% = 1/100 = 0.01
use whichever depending on personal preference

[balmy sequoia]

could i do 30500 x 10.4/100

[obtuse pebble]

@balmy sequoia Has your question been resolved?

[high lily]

yes

[balmy sequoia]

yes

[obtuse pebble]

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[balmy sequoia]

which is correct

[balmy sequoia]

same websites btw

they jst coinfusing me tbh

[wooden cipher]

Theyre both right

Just different numbers

[dapper silo]

the first picture has a mistake, since you're supposed to divide by P(B)

so it should be 2/5

[wooden cipher]

Oh wait

[dapper silo]

second picture is fine

[wooden cipher]

Oh i misread

Yeah it should be the second

[balmy sequoia]

ok

the website must be on drugs

thanks for the clear up

.close

[obtuse pebble]

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[balmy sequoia]

how to put division in text pot

[balmy sequoia]

bot

[brisk matrix]

$\div$

[warm shale]

maximofs

[brisk matrix]

like that?

[high lily]

use superior fraction notation

if you're going to use the bot

\frac{numerator}{denominator}

[balmy sequoia]

what abt multiply

[high lily]

use \cdot

[balmy sequoia]

.

[high lily]

try avoid \times

[balmy sequoia]

ok

does it confuse?

[brisk matrix]

it’s more cumbersome notation

$a\times b$ versus $a\cdot b$

[warm shale]

maximofs

[balmy sequoia]

also guys

if i have

$(y)^3$

[warm shale]

nicholas

[balmy sequoia]

do i just times it by 3?

and it becomes

and it becomes $y^3$?

[warm shale]

nicholas

[high lily]

(y)^3 is y^3,
but that isn't timesing anything by 3

[royal basin]

also you should not be using "times" as a verb

[balmy sequoia]

for like index law

[royal basin]

(y)^3 is y^3

these are not two different expressions

[balmy sequoia]

like

for example

[warm shale]

nicholas

[balmy sequoia]

would it become $2^3x^6y^3$

[warm shale]

nicholas

[royal basin]

sure would

[balmy sequoia]

ok

thx

.close

[obtuse pebble]

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[vestal trout]

having a problem with c)

[vestal trout]

what can i say?

[sage geode]

Tried making the denominators common?

[vestal trout]

can i say there is an axis of symmetry?

on zero?

[sage geode]

1. There is no axis of symmetry on zero
2. Irrelevant to the equation

Just make the denominators common in $\frac1{e^x + 1} + \frac1{e^{-x} + 1}$

[warm shale]

A Lonely Bean

[vestal trout]

i got zero

[sage geode]

Show your work

[vestal trout]

that means f(x) + f(-x) = 1 isn't it?

but what should i conclude?

[sage geode]

e^x * e^-x is 1

[sage geode]

How come e^x * e^-x cancels out with another +1?

Oh wait

[vestal trout]

there is 2 already in the other side

[sage geode]

You got 0 = 0 as an equation

Yeah it's right

It finished the proof

[vestal trout]

what should i conclude bro

that what i get doomed

[sage geode]

Yes, it does

That's the conclusion

f(x) + f(-x) = 1 is equivalent to 0 = 0 and 0 = 0 is true

So f(x) + f(-x) = 1 is true as well

[thick chasm]

Just write e^-x = 1/e^x
1/(e^x+1)+1/(1/e^x +1)
1/e^x+1 +e^x/e^x+1
e^x+1/e^x+1 = 1

[vestal trout]

it's not that simple xD

wait a minute it's f(x) + f(2a - x) = 1 where a = 0 isn't it?

[sage geode]

It is

[vestal trout]

it's center symetry as well

[sage geode]

Don't use chatgpt

[vestal trout]

why?

[sage geode]

"Do not post responses written by ChatGPT or a similar AI tool." - #rules

[vestal trout]

ok

so it's axes of symmetry?

because f(x) = -f(2a - x) + 2b

[sage geode]

Yeah, but you are gonna end up proving that equation that way or another

There is no point in mentioning that

[vestal trout]

that what my teacher used to say

maybe because our system is different

thanks for help

.close

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[gloomy trail]

make

wait

try make X the same

@ruby elm

[ruby elm]

So we multiply the second equation by 1

Or no

[gloomy trail]

nah

[ruby elm]

Wait

[gloomy trail]

so

for the first equation

you want to times everything by 5

which will be 20x + 15y = 45

[ruby elm]

And second equation we multiply by 4 or we just multiply the first

[gloomy trail]

yes you have to times 4 for the second equation

to make the X the same

[ruby elm]

20x+8y=52

?

Or wait

I forgot to multiply 13

[gloomy trail]

all goods

[ruby elm]

there

[gloomy trail]

there we go

now

u know what to do next?

[ruby elm]

Nope

[gloomy trail]

alright

lemme grab a note

[ruby elm]

alr

[gloomy trail]

lets solve y

in order to do that

[ruby elm]

k

[gloomy trail]

we minus

to take out the x

[ruby elm]

WAOT

lemme guess what happens next

[gloomy trail]

?

sure

[ruby elm]

So now

we do 15+8?

Which is 23

[gloomy trail]

it would be

minus

as well

not plus

[ruby elm]

only minus?

[gloomy trail]

yep

[ruby elm]

Ohh k

so 15-8

[ruby elm]

=7

Why negative ?

[gloomy trail]

positive times negative

[gloomy vector]

45-52=-7

[gloomy trail]

will always be negative

[ruby elm]

OH sorry sorry I thought u mean -7 was for 15-8

mb

[gloomy trail]

its okay

[ruby elm]

Now we divide

[gloomy trail]

yepp

what would y be?

[ruby elm]

=-1

[gloomy trail]

yep

bingo

we found what y was

now we solve X

pretty much we should plug in -1 to either equation 1 or 2

[ruby elm]

hm

uh

1

[gloomy trail]

here

alright

this is what u got

[ruby elm]

so now it’s -15

[gloomy trail]

yes

[ruby elm]

so now we move the

-15

Next to 45

[gloomy vector]

what do you uae to make dat

[ruby elm]

It will be 45+15?

[gloomy trail]

20x = 60

[gloomy trail]

?

[gloomy vector]

what program do you use to write that stuff

[gloomy trail]

oh

[ruby elm]

Notes

[gloomy trail]

google docs lol

[ruby elm]

I think

[gloomy vector]

zam

[gloomy trail]

?

[gloomy vector]

yeah that is what i asked

[ruby elm]

anyways thanks now I understand the whole thing

[gloomy vector]

you didnt undwrstand so i rewrote the question

[ruby elm]

I do have one question tho

[gloomy trail]

no worries

[gloomy trail]

sure what is it

[ruby elm]

Why do we only do minus

[gloomy trail]

since we are trying to solve what y is first

[gloomy vector]

you dont

there are many cases where you do plus but slitghtly rarer

[gloomy trail]

and we see both 20x in the same equation

we minus it to get rid of it to make it easier

[ruby elm]

wait let me check the answe

[gloomy vector]

an even more rarer is multiplication and division

[gloomy trail]

alright

um

whats the problem?

[gloomy vector]

@gloomy trail how do you have 2 full blown active conversation in 2 diffrent channels

[gloomy trail]

tbh i dont even know ☠️

how did u even see

[gloomy vector]

i was writing a long paragraph for a question when you took the channel 😠 so i remembered you were asking and i was scrolling down and i noticed you were asking a question and i scrolled down and you were answering a wuestion soo

[gloomy trail]

ohhh.

sorry for taking ur channel

[ruby elm]

I’ll try do this myself now

[gloomy trail]

sure

remember to make 3x and 5x the same

[ruby elm]

ya

I’ll show u my work

Nvm I don’t have storage

[gloomy trail]

oh unlucky

[ruby elm]

Nvm I do

A bit left

tho

lol

ok I did x

Wait is the firs tone X

one

which is x and which is y

or is the first one y

And second X

@gloomy trail

[gloomy trail]

uh

ur kinda confusing me here

[ruby elm]

sorry lol

I mean

like

Wait I’ll just show u

Is it y

Or X

Y right

?

[gloomy trail]

sorry im back

[gloomy trail]

the answer should be

31y = 62

y = 2

[ruby elm]

The 31 y I fixed after I took the pic

k thanks

I just wanted to check

[gloomy vector]

technixhally it wasnt anyones to begin with
but because of you i got help-1 😎

[ruby elm]

If it’s X or y

[gloomy trail]

oh alright

[gloomy trail]

damnn lucky u

[gloomy trail]

now plug y = 2 either equation 1 or 2

to solve what X is

[ruby elm]

I’m doing 2

[gloomy vector]

you can do enlimination but you can go cool with 3x+8y=5x+3y and subtitute x for 5/2y

[gloomy trail]

sure since its easier

[ruby elm]

Do I need to write 2,5

or it doesn’t matter

Is it correct anyway?

[gloomy trail]

back

sorry

[gloomy trail]

yes its correct way

simultaneous equation is for

linear relationships

[ruby elm]

What

[gloomy trail]

those to numbers 2 and 5 will be a point (2,5)

[ruby elm]

Oh alr

[gloomy trail]

like these

dw abt it too much

[ruby elm]

Oh that’s formula we use

[gloomy trail]

yes

bingo

[ruby elm]

ALR IM TAKING A BREAK FOR 1H AND PLAY FORTNITE

[gloomy trail]

yoo

[ruby elm]

I been studying for 6h

[gloomy trail]

which region u play on?

[ruby elm]

uh

I put auto

[gloomy trail]

whats ur default?

[ruby elm]

Wait what do u play on

[gloomy trail]

oceania

[ruby elm]

I’m in the Middle East

[gloomy trail]

oh man rip

[ruby elm]

ohhh won’t work for me

[gloomy trail]

big ping difference

oh well

[ruby elm]

yup

[gloomy trail]

have fun with ur game

[ruby elm]

Ima play bio

[gloomy trail]

go ahead

[ruby elm]

.close

[obtuse pebble]

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[obtuse pebble]

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[astral aurora]

There is a point p(x,y,z) and it's distance from
X axis is $root(y^2+z^2)$

Y axis $root(x^2+z^2)$

Z axis $root (x^2+y^2)$
So how these three equation are derived? I wanted to know most plane,line formulas derivation which is presented directly in books. Help

[warm shale]

! Arjunn

[royal basin]

\sqrt{}, not root()

the point closest to (x,y,z) on the x-axis is (x,0,0)

[astral aurora]

Thank you ma'am

[astral aurora]

Hmm yes

Then distance formula in 3d?

Ohh got it thanks

One more doubt, most derivations aren't available in my 3d book what should I do?

Like angle between planes, lines

[royal basin]

most angle shit comes down to angles between vectors in one way or another

[astral aurora]

I am just learning these formulas with root method

No proofs only formulas

. close

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[proper kelp]

,rotate

[proper kelp]

Im not sure how to do question ii)

sorry im new to vectors but im not sure what expression to get for ii) and how that expression would show FE is parallel to DC

[supple trail]

Then, what is FE?

[proper kelp]

well i got an expression for FE but its probably wrong

i got FE = -r-kp+q

and it said in terms of k and p so im wrong there too

[supple trail]

No, I mean I don't know which points are E AND F

[proper kelp]

oh

[supple trail]

It isn't mentioned

[proper kelp]

i have a diagram above

[supple trail]

Oh yes I didn't see before

[proper kelp]

yeah all goods

[supple trail]

Let me see and answer you

Right?

[proper kelp]

ahh

that makes alot of sense

since its just double

tysm

[supple trail]

🤗

[proper kelp]

wait quick question

i find that you wrote both of the vectors backwards which i assume doesnt really change anything, but is there a specific reason for that

oh nvm

i just drew my diagram wrong, but the answer is still valid

thanks!

.cloes

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[timid silo]

For part b) am I making the wrong D.E?

[timid silo]

This is my DE

This is the one the solution uses

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[pseudo swift]

@timid silo uh how do you get that ?

why is there that additional v ?

[hidden garnet]

physics thing, and you can derive it using chain rule actually

$v = \dv{x}{t}$

[warm shale]

Riku

[pseudo swift]

ah it's dv dx

ok didn't read lol

[hidden garnet]

yes

[timid silo]

That's not the problem

The problem is the q says 1/10 v

The ms is using 1/10 v^2

[timid silo]

Also that's the mark scheme

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[balmy sequoia]

i dont think his problem was resolved

[obtuse pebble]

@timid silo Has your question been resolved?

[boreal acorn]

nope

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

.

[timid silo]

By comparison with the solution, there is a missing r dot on the left hand side?

[timid silo]

Yeah bcz the part b q says

Resistive forces have mag. 1/10v

But the ms is using 1/10 v^2

I don't understand why it's doing that

Am I reading the q wrong

[timid silo]

Maybe it comes from power?
https://byjus.com/question-answer/derive-relation-between-power-force-and-velocity/

[timid silo]

No that's not the problem

I used the eq of power for the driving force correctly

$$1600\ddot r = \frac{100000}{\dot r} - \frac{1}{10}\dot r$$

[warm shale]

brotherImusttalk1234

[timid silo]

This is the differential equation I get

That's my problem

Ms is using v^2 at the end tho

I used p/v correctly I've got the same second term as the ms... Third term is missing a v term but idk am I reading thenresistive force line wrong

v(dv/dx) = (dx/dt)(d(dx/dt)/dx)
= (dx/dt)(d^2x/dtdx)
= d^2x/dt^2 = x dot dot

[timid silo]

Ok I think ur DE is correct. The mysterious v seems to come from r dot dot = v dv/dr

[timid silo]

Oh that's v easy to get lol

One sec lemme show

[timid silo]

I derived it above

[timid silo]

Huh

I think

Huh wait then shouldn't the P/v be P

Anyway I think ur DE is correct and matches the solution DE. Right?

No

It doesn't

Why not?

That's the problem

This is mine.
$$1600\ddot r = \frac{100000}{\dot r} - \frac{1}{10}\dot r$$
In the ms it's...
$$1600\ddot r = \frac{100000}{\dot r} - \frac{1}{10}\dot r^2$$

[warm shale]

brotherImusttalk1234

[timid silo]

Take this eqn and substitute the r dot dot with v dv/dr. Divide by right hand side and then multiply by dr to get the second to last line?

Oh no... Its v cubed...

Now u see it

[timid silo]

The link https://byjus.com/question-answer/derive-relation-between-power-force-and-velocity/
says P = Fv
If F = 1/10v Newtons then
P = 1/10v^2? =/= P/v

The d.e is for force

[timid silo]

Man

P = Fv

Ik that

But the de is for F=ma

Not for P

That's why it's ma=P/v - R

R being resistive force

I think the de in the ms is wrong leave it be

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[timid silo]

I think the resistive force is 1/10v N not 1/10v^2 N

[timid silo]

That's what I said too

But the differential equation is not for power

It's for force

Power doesn't necessarily have superpositions like that

Force does bcz it's a linear quantity

And then the P/V term and the ma term deffo don't make any sense

[timid silo]

I was commenting that the first equation in the solution picture doesn't make sense because it says P/v = 1/10v^2 but F = 1/10v and Fv = 1/10v^2 which is power not P/v

Ah ok

Oki oki

But just saying power doesn't necessarily add up like that

So making a de for power anyways won't be correct

You can close the channel honestly

[obtuse pebble]

@timid silo Has your question been resolved?

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[opaque galleon]

I will type my solution here, I got stuck somewhere

[opaque galleon]

oh ncm

nvm

lol i forgot to take the derivative

i will still type regardless

$
\vec{R}'(u) = \langle -12 \cos(3u), 5, 12 \sin(3u) \rangle
\\t_0 = 0
\\s = \int_{t_0}^{t} \|R'(u)\|
\\s = \int_{0}^{t} \sqrt{144 + 25} du
\\s=13t
\\t = \frac{s}{13}
$

[warm shale]

steamhahasteamhaha

[opaque galleon]

then

eh not gonna type it

you just sub t xD

am I wrong

can someone tell me pls

also why do we write it as $\vec{R}^*(s)$

[warm shale]

steamhahasteamhaha

[spice basalt]

looks good to me. I've not seen this star notation for labeling the arclength parameterization but I guess it's just a way of distinguishing it from the original function.

[opaque galleon]

nice

[obtuse pebble]

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[unreal vessel]

I dont understand how step 2-3 works

[wooden cipher]

Complete the square

[unreal vessel]

wdym

[grizzled shore]

Have you heard of completing the square

[unreal vessel]

no

[grizzled shore]

Have you heard of quadratics

[unreal vessel]

I dont speak english so like I dont know any of the english terminology

[grizzled shore]

What do you speak

[unreal vessel]

dutch

[grizzled shore]

Is deutsch dutch

[unreal vessel]

no

[grizzled shore]

What’s Dutch in Dutch

[unreal vessel]

I can speak english fluently I just dont know the mathematics terminologies because I dont get taught in english

[unreal vessel]

nederlands

[grizzled shore]

https://nl.m.wikipedia.org/wiki/Kwadraatafsplitsen

Have you seen that word before?

[unreal vessel]

ye

[grizzled shore]

That’s what they are doing

[grizzled shore]

In here

[unreal vessel]

I dont get how

• 1 1/2 becomes - 1 /16th tho

[grizzled shore]

Ok

So

Did you write that

(I hope not)

[unreal vessel]

wdym

the formula?

no

[grizzled shore]

I have no idea why they have elected to use mixed fractions

But it is disgusting

[unreal vessel]

so what you normally do is just use x right instead of 2

and do like x (2x + 5) + 3

or smth

[grizzled shore]

Can you show me how you would complete the square on x² + 2x + 3

[unreal vessel]

uh

[grizzled shore]

Complete the square is whatever this word is

[unreal vessel]

x(x + 2) + 3

[grizzled shore]

Where’s the square

[unreal vessel]

oh

uhhh

nah im lost

wait can I

[grizzled shore]

Expand (x+1)²

[unreal vessel]

x^2 + 1 + 2x

[grizzled shore]

Is that close to this

[unreal vessel]

yeah

oooooh right

I forgot you could do that

that would be

id know if it was -3 but idk tbh

yeah im lost idk

<@&286206848099549185>

@grizzled shore dunno if you mind but would you help me again?

[grizzled shore]

Um

So (x+1)² gets you the x² + 2x

But it also comes with an annoying +1 that we don’t want

So let’s just do (x+1)² - 1

That way when we expand we get just x² + 2x

Do you follow?

[unreal vessel]

yeah

[grizzled shore]

Ok but we don’t want x² + 2x

We want x² + 2x + 3

[grizzled shore]

Ok we just add +3 on this then

(x+1)² -1 + 3

[unreal vessel]

so +2

[grizzled shore]

Yes

Ok let’s do another one

x² + 6x - 5

Now you try it

[unreal vessel]

this one also isnt possible in that form

[grizzled shore]

Think about what kind of (x+a)² expanded gives us x² + 6x

[unreal vessel]

I dont think any does that

but ig whats close is

[grizzled shore]

Then we’ll deal with the extra bits later

[unreal vessel]

(x+3)^2

[grizzled shore]

Yes

What extra bits do you get from that

[unreal vessel]

+9

[grizzled shore]

And we don’t want that

[grizzled shore]

So let’s just take 9 away from this

Now we have (x+3)² -9

But we don’t just want x² + 6x

We want x² + 6x - 5

So what do we do

[unreal vessel]

-14?

[grizzled shore]

Yes

In full?

x² + 6x - 5 = …?

[unreal vessel]

(x+3)^2 -14

[grizzled shore]

Yes

Now let’s try 2x² + 5x + 3

How would we do that?

[grizzled shore]

Btw notice that a is half of 6

[unreal vessel]

would we do it in the same form as the last one

[grizzled shore]

Because expanding (x+a)² gives us x² + 2ax + a²

[unreal vessel]

ig

[grizzled shore]

And also notice that we always get the a² that we don’t want

So for any x² + bx + c the general form is (x + b/2)² - (b/2)² + c

[unreal vessel]

2(x^2 + 2,5x) + 3

[grizzled shore]

Ok good

Let’s use fractions

5/2 not 2.5

[unreal vessel]

so now you want to uh stick a fraction in there

[grizzled shore]

$2\left(x^2 + \frac{5}{2}x\right) +3$

[warm shale]

frosst

[unreal vessel]

so 2(x + 1 1/4)^2 +3 - 1,5625

idk what fraction the last one is so I just did that

[grizzled shore]

Noooooo

No mixed fraction

Do not use mixed fractions they are disgusting

[unreal vessel]

huh

ok but is it right tho

because when I look at the answer sheet it says it isnt but to me it looks like I did it right

[grizzled shore]

How did you get 1.5625

[unreal vessel]

5/4 ^2

made sense to me idk

[grizzled shore]

But that’s inside the bracket

[unreal vessel]

yeah but wouldnt you have to add a counter thing to make it so you can square it in the first place

[unreal vessel]

because this and 2(x + 1 1/4)^2 +3 both have different outcomes

[grizzled shore]

$2\left(\left(x+\frac{5}{4}x\right)^2- \left( \frac{5}{4}\right)^2\right)+3$

[warm shale]

frosst

[unreal vessel]

yeah thats what I did

also isnt it just 5/4 no x behind it

[grizzled shore]

But it’s inside the bracket with the 2 outside

You didn’t multiply this bit by 2

[unreal vessel]

what?

what 2

[grizzled shore]

$2\left(\left(x+\frac{5}{4}x\right)^2- \left( \frac{5}{4}\right)^2\right)+3 \neq 2\left(\left(x+\frac{5}{4}x\right)^2\right)- \left( \frac{5}{4}\right)^2+3$

[warm shale]

frosst

[grizzled shore]

$2\left(\left(x+\frac{5}{4}x\right)^2- \left( \frac{5}{4}\right)^2\right)+3 = 2\left(\left(x+\frac{5}{4}x\right)^2\right)- 2\left( \frac{5}{4}\right)^2+3$

[warm shale]

frosst

[unreal vessel]

ok that makes sense kinda

[grizzled shore]

You missed multiplying it by 2 before moving it out of the bracket

[unreal vessel]

oooh righ

right

[grizzled shore]

If you do that the answer will be correct

(But please don’t use mixed fraction)

[unreal vessel]

alr I think I understand thanks for the help

.close

[obtuse pebble]

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[vapid pecan]

How can i start

[vapid pecan]

,rotate

[vapid pecan]

,rotate

[vapid pecan]

I mean i see that it's 3

But like how do i get there

[normal oracle]

what is that, I can't read it well

[robust sleet]

8

[royal basin]

that looks like an eight

[vapid pecan]

8 😅

[royal basin]

anyway 4-sqrt(15) and 4+sqrt(15) are reciprocals

that should help

[vapid pecan]

Well i only see that ✓15 will be gone after the root gets ridden of

[royal basin]

let a := (4-sqrt(15))^1/3, then your equation is a^x + (1/a)^x = 8

[vapid pecan]

But how am i supposed to see that they're reciprocal

[vapid pecan]

And can i just delete the third root like that

[obtuse pebble]

@vapid pecan Has your question been resolved?

[royal basin]

i did not delete the cube root

i simply incorporated it into the value of a

[vapid pecan]

Then my equation is (³✓a)^x + (³✓1/a)^x = 8

Oh

You did 1/3

That removes the root right

[obtuse pebble]

@vapid pecan Has your question been resolved?

[polar fossil]

$a^{1/3} = \sqrt[3]{a}$

[warm shale]

kitten.in.a.teacup

[vapid pecan]

Am i doing something wrong

[polar fossil]

what the heck is squiggle

anyway it's b^2 - 4ac not just b^2 - 4

[vapid pecan]

Well a and b is 1

[vapid pecan]

.

I just did t and not a

[polar fossil]

and c is 7, but also you can just factor it

[vapid pecan]

Its 1

[polar fossil]

[vapid pecan]

I just write ugly

[polar fossil]

oh okay

[vapid pecan]

What does factor mean

[vapid pecan]

Lets go from here

How do i do this

[polar fossil]

like turning x^2 + 3x + 2 into (x+1)(x+2) but that's not possible here

[polar fossil]

you had the right idea setting m = t^x and then it's a quadratic

[vapid pecan]

But i get root of 60

And the urge to end my life

And then i have t^x = 4 + ✓15

Oh but i do have t

This might be possible actually

[polar fossil]

might be!

[obtuse pebble]

Channel closed

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[vapid pecan]

.reopen

[obtuse pebble]

✅

[vapid pecan]

I get -3

I think it should be 3

[polar fossil]

I haven't run the numbers myself but it's fairly straightforward to check a solution

[vapid pecan]

There's no solution

,rotate

[vapid pecan]

But its obvious that its 3

When you import x

[polar fossil]

why is it obvious that it's 3?

[vapid pecan]

It removes the root

[polar fossil]

you have a 1/ and a negative exponent but I think you want one or the other

[vapid pecan]

Nah it should be without -

But i put it there

So it becomes 1/

Wich is equal to the other side

Bc these are reciprocals

[polar fossil]

oh right

-3 will also remove the root

[vapid pecan]

Yeah but it will be 1/

With 3

I have 4-✓15 + 4 + ✓15 =8

Which is correct

Um

I think both are correct

3 and -3

Oh

Ok

Ill kms

There's also m2

I thought 4-✓15 is under 0

But it isnt

😅😅😅😅😂😂😅😂😅😂😅😂😅😂😅😅😅😅😅😅

. close

.close

[obtuse pebble]

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[ruby path]

cancel the log on both sides

[vapid pecan]

But the base isnt the same

[ruby path]

Oh didn't notice that

Then make the base the same first

[vapid pecan]

How

[gentle current]

you can change the base using the formula

[vapid pecan]

What formula

I forgot this stuff

[ruby path]

$\log_{b^c} a = \frac 1c \log_b a$

[vapid pecan]

Oh yeah there's something like that

[warm shale]

NEONPerseus

[vapid pecan]

Kay got it

[ruby path]

Unnecessary here

[gentle current]

yeah but it s in general

[vapid pecan]

I'll need that too later

I forgot almost everything

So like this?

[timid silo]

4 = 2^2

[gentle current]

you can t do that because of that 1/2

so using the properties of log that 1/2 goes to the power

[vapid pecan]

But i can put it as ^?

[gentle current]

ye

[vapid pecan]

Whats that called btw

[gentle current]

raise to the power

[vapid pecan]

Should be like this then?

[gentle current]

exactly

and now

you must check those values if they give you a true statement if you put them in the original form

[gentle current]

in this one

[vapid pecan]

Yeah

They do

Thanks

[gentle current]

no problem

wait

are you sure that x=1 is a solution?

[vapid pecan]

Ye

Its 1=1 in the end

[gentle current]

oh yeah right

sorry

xd

[vapid pecan]

Am i on the right path for this one?

Ignore the last row ik its incorrect

But am i supposed to go witg this formula

[gentle current]

ye it s good

[vapid pecan]

But what do i do

[gentle current]

ok wait i ll try to solve it and see if you can do it some easier way

[vapid pecan]

Is there?

[gentle current]

well not really

so

use the properties of log for the disivion and multiplication of the arguments

and then to be easier you can renote log 5 of x = t or some letter

and then solve the equation

[vapid pecan]

If log5(x) = t

Is log²5(x) = t²

[gentle current]

yes

[vapid pecan]

They somehow got 3 results

I mean photomath

I got 2

[gentle current]

there are 3 results because it s polynomial equation with a 3rd grade

[vapid pecan]

I got t= 0 t= 1 and t= -2

-2 is impossible

[gentle current]

send me a pic pls

[gentle current]

-2 is possible

because t is log 5 of x

and log can give you negative numbers

[vapid pecan]

How

[gentle current]

only x must be positive