help-10

[warm shale]

madmike

[gloomy pine]

Nah just focus for a sec

You got this one

[merry vault]

n -> infty

and this frac becomes -1 ?

[gloomy pine]

Nice

[merry vault]

thanks a lot

I am confused why we can now basically set the 1/n fractions to 0, but not earlier

like here

I guess you first need 1 fraction

[spiral knot]

the denominator will be equal to 0 so

[obtuse pebble]

@merry vault Has your question been resolved?

[merry vault]

yeah what does that mean?

just that it's impossible to get a result right now?

[spiral knot]

at the end you will find it undefined so we need to find another method to find the limit

[merry vault]

ok thanks

.close

[obtuse pebble]

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[merry vault]

.close

.close

[obtuse pebble]

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[polar bramble]

question

[polar bramble]

when do we put the postive/negative sign

is this correct ?

or do we only use the sign when defining a variable

[trim portal]

This doesn't seem correct

[polar bramble]

this ?

[warm shale]

MathIsAlwaysRight

[trim portal]

This is correct, if you define variable X to be + or - square root of 36

[polar bramble]

okay thank you

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[tall arrow]

The kilo price for blackcurrants is 75, while the kilo price for sugar is 15. We are going to mix blackcurrants and sugar to make blackcurrant jam with a kilo price of 51. How much sugar is in the jam?

I have made up a formula for solving this:

b = berries
s = sugar

(75b + 15s) / (b + s) = 51
(75b + 15s) = 51(b + s)
75b + 15s = 51b + 51s
24b = 36s
makes the ratio of amount of berries and sugar in the jam equal to

24:36
this is also equal to

4:6
which means that the amount of sugar is 60% of the amount of the total jam
this is kinda wrong considering the equation:

0,4(75) + 0,6(15) = 39 and NOT 51
i kind of do not see what i have done wrong here, either, so i'm confused

[oak bear]

no first is that it is (kilo of bc)(price of bc)/kilo + (kilo of sugar)(price of sugar)/kilo = (kilo of bc + kilo of sugar)(total price)

@tall arrow

[tall arrow]

do you mind clarifying a bit?

[oak bear]

so the bc is 75$ per kilo or 75$/kg

we need to multiply kg there to cancel it right?

the same can be done for sugar

15$/kg

(kg of sugar )15$/kg + (kg of berry)75$/kg = (kg of both products)(price of jam/kg of jam)

[tall arrow]

yeah i seem to kind of get more confused on this one
i believe it would be great to show the correct way of solving this, so that it possibly makes the explanation above clearer

[oak bear]

how about u try it first

[tall arrow]

this is what i've tried

[oak bear]

Let's say the kilo of ur sugar is 10 kilos

[tall arrow]

sure

[oak bear]

ok lemme show it to u

u can also do it ur own while im doing it

AHHAH we cannot do 10 for sugar, its actually too low

Wait lemme do something

[tall arrow]

for reference, the answer is 40%, but let's assume we don't know that

i've seen the book for the answer, but i would rather like to know how it was achieved

[oak bear]

OK it's not

U actually can

I'm just dumb that I thought 51 is 5

HAHAHAHA

[tall arrow]

well let's then get solving

[oak bear]

15

if sugar is 10 kilos

then the berries is 15

[tall arrow]

okay, but what made you come to that conclusion

[oak bear]

try and check it

[oak bear]

use the one u used here

ah don't forget the kilos for the right side

[tall arrow]

yep that's correct

i'm guessing you've inputted a random kg of sugar amount, which in this case was 10, into the equation and then you've gotten the amount of blackcurrents

[silent zodiac]

Let's assume that we mix x kilograms of blackcurrants with y kilograms of sugar to make a total of 1 kilogram of blackcurrant jam. We know the kilo price of blackcurrants and sugar, and we want the kilo price of the jam to be 51. We can set up the following equation:

75x + 15y = 51

We can simplify this equation by dividing both sides by 3:

25x + 5y = 17

We also know that the total weight of the jam is 1 kilogram, so we have:

x + y = 1

We can solve this system of equations by using substitution. Rearranging the second equation, we get:

y = 1 - x

Substituting this into the first equation, we get:

25x + 5(1 - x) = 17

Simplifying and solving for x, we get:

20x = 12

x = 0.6

So we need 0.6 kilograms of blackcurrants in the jam. We can find the amount of sugar needed by using the equation y = 1 - x:

y = 1 - 0.6 = 0.4

Therefore, we need 0.4 kilograms of sugar in the jam.

[tall arrow]

i believe that's also a correct way of doing it

[tall arrow]

and that method also works

They both give the same answer, so I belive I understood it now

Thank you a lot to @silent zodiac and @oak bear, I really appreciate your time

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[oak bear]

.reopen

[obtuse pebble]

✅

[oak bear]

@tall arrow

I think what ur trying to find is the ratio right?

U actually almost got it in ur answer

24b=36 s

The problem is that u didn't equate it

so it would be 24b/36 = s

or 2/3b

2b/3 + b = 1

b = 3/5

and we know that 1-3/5 = 2/5

So s is 2/5

That is the ratio

Ah thats it

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[boreal glade]

Hi all, I am looking for someone very good at math and radio astronomy ideally. It is about horn antenna(s) , calculating angles and ideally estimating power of a radio signal source.Here is the link to the website with all details: thewowsignal.epizy.com

[boreal glade]

What is known:

1. a distant static radio signal source about 1420MHz
2. two locations on Earth, both visible to the distant signal source

What needs calculating:

1. angles in two locations (between North and a distant radio signal source)
2. horn antenna calculations (angles of reflections, gain, etc.)

[dark stirrup]

You may have better luck finding the physics server in #old-network

[boreal glade]

ok, thanks

[dark stirrup]

Not to discourage you here, but you may not get many people with a physics background

Heck, my degree is in physics and we never covered radio signals more than mentioning they existed

(Or I just forgot it all)

[boreal glade]

I know, I hope to find a real radio astronomer here on this website

[kind hawk]

good luck with that

[dark stirrup]

You can always email professors that research radio astronomy. They're giant nerds and love to talk shop

I was infrared astronomy, so can't help. :/

[boreal glade]

I emailed a few people already, but no response

[dark stirrup]

Aw

That sucks

Sorry man

[boreal glade]

so, is there any physics forum here?

[sweet edge]

not specifically but check #old-network

[obtuse pebble]

@boreal glade Has your question been resolved?

[boreal glade]

not yet

there is someone on Astropedia interested in it

[obtuse pebble]

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[merry vault]

How do you start here?

[sage geode]

Have you tried multiplying by the conjugrate?

[merry vault]

I don't know what that is

multiply the entire term by sqrt(n^2-8n-10) / sqrt(n^2-8n-10) ?

[sage geode]

In this case, it's sqrt(n^2 + 8n + 10) + n + 1

You should get the idea of what a conjugate is from this example

[merry vault]

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[crude ice]

for this qn how should i differentiate wrt

,rotate

[glossy yacht]

wrt

[crude ice]

i tried to differentiate this and i think it didnt really work out

[glossy yacht]

write it like 10u(-100+100)

[crude ice]

wdym

[glossy yacht]

$10(u-10) + 100 = 10u$

[warm shale]

Xetrov

[glossy yacht]

so you have a constant and a fraction

it'll be nicer to differentiate

[crude ice]

you got that from this equation?

[glossy yacht]

i just rewrote 10u

trust me

write it

[crude ice]

no no im trying to understand how you got it cuz im still a little confused

oh

wait u multiplied both sides by u-10?

idk bro idk how you got that

[glossy yacht]

i just replaced the top

the bottom starts 3

stays

$\frac{10(u-10) + 100}{u-10}$

[warm shale]

Xetrov

[glossy yacht]

then you'll differentiating a constant and a nice frac

[crude ice]

the thing is i dont understand where you got that u-10 +100 from im sorry

[glossy yacht]

ah ok

[crude ice]

i thnk im just very tired you might have to make it more obvious

[glossy yacht]

I added a 100 then took away a 100

the reason being I want to factorise the terms 10u and -100 into a 10(u-10)

[crude ice]

oh

yeah i dont think i'll be able to do that if i was to be in an exam

if i just wanna differentiate it without doing the method you did

[acoustic mortar]

why don't you just simply differentiate the equation using quotient rule

[crude ice]

do i have to use quotient rule

well its a very long step

[glossy yacht]

I can't find where she went wrong using product

[crude ice]

and im differentiating with respect to time too so its gonna look very messy

[acoustic mortar]

you can always write it as dv/dt = dv/du * du/dt

you already have du/dt as -0.15

and using quotient rule you can find dv/du

which would come out to be -100 / (u - 10)^2

[crude ice]

how did you do this did you just apply quotient rule

[acoustic mortar]

yes

[spiral knot]

$\frac{\dd u}{\dd t} = -0.15 cm/s$

$\frac{\dd v}{\dd t}= \frac{\dd}{\dd t} \qty(\frac{10u}{u-10})$

[warm shale]

Mehdi_Moulati

Mehdi_Moulati

[acoustic mortar]

differentiating wrt time would make this question more complex

[spiral knot]

not that much

[acoustic mortar]

but why do we even need to differentiate with time when the question is giving a direct hint for a direct approach

[crude ice]

wdym by direct approach

[spiral knot]

i assumed that he didn't see dv/dt = dv/du * du/dt this yet

[crude ice]

could you guys also help me with this qn, like what each order of derivative means

[crude ice]

like the maximum and minimum thing

im kinda confused

[acoustic mortar]

let me solve it for you

[spiral knot]

A is maximum when dA/dx = 0

[spiral knot]

something isn't right

[crude ice]

could you represent this on a graph so i can visualize it

[crude ice]

i mean it kinda make sense tho ? 4 square on each side

or should i write (24-2x)(36-2x)

[spiral knot]

they want the volume

[crude ice]

oh

so (36-2x)(24-2x)(x)

yeah

i think thats it

[crude ice]

yo did u solve it i wanna see how you do it

[acoustic mortar]

yes

[crude ice]

oh wow

this i really neat

[crude ice]

could u explain the dv/du and du/dt part

[acoustic mortar]

see i just added the du part in dv/dt by

multiplying and dividing dv/dt at the same time with du

[acoustic mortar]

did you get it?

[obtuse pebble]

@crude ice Has your question been resolved?

[obtuse pebble]

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[warm frigate]

hello there i was wondering how the magnitude of this is root 20

shouldn't it be 2 root 5?

[short spruce]

they are equivalent

[warm frigate]

would do you mean by equivalent?

what is equal to what

sorry just a little confuised

[timid silo]

root(20) = 2root(5)

[short spruce]

$\sqrt{20}=2\sqrt{5}$

[warm shale]

a disappointing son

[warm frigate]

oh i see

alright no worries , thank you for your help both of you

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[tepid urchin]

how would i go about using an online calculator or wolfram to solve this

[timid silo]

I think you would still have to write it all out, but just have wolfram evaluate it

[tepid urchin]

how do I do that?

[timid silo]

do u know how to use midpoint rule ?

[tepid urchin]

ish

[timid silo]

what do u have so far

[tepid urchin]

nothing

[timid silo]

ok so find delta x

[tepid urchin]

I just want the answer then go to bed

how

[timid silo]

(b-a)/n

[tepid urchin]

right

2?

[timid silo]

yea

then just delta x[f(1) + f(3) + f(5) + f(7) + f(9)]

[tepid urchin]

ok

do I just add up all the valuses then add 2

[timid silo]

huh

[tepid urchin]

no

[timid silo]

no what

[tepid urchin]

to what I said

[timid silo]

I dont understand why you would add all the values and add 2

[tepid urchin]

multiply sorry

but thats wrong too

so add the values for 1,3,5,7,and9?

[timid silo]

its literally just midpoint rule formula

yeah

[tepid urchin]

ok

one sec

[timid silo]

https://www.youtube.com/watch?v=5XreKMJDJsg

this helps

[tepid urchin]

tysm

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[stone folio]

Need help with my algebra homework please and thanks

[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

[stone folio]

2

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

[royal basin]

so you have |3x-6| = 4 now

[stone folio]

Idk what to do with the problem after that part

[royal basin]

have you solved absolute value equations before?

[stone folio]

Yeah

[royal basin]

at this point you would typically write down two linear equations: 3x-6 = 4 and 3x-6 = -4, and solve them...

and take the union of their solution sets

[stone folio]

So one positive and a negative problem?

[royal basin]

... if you wish to phrase it that way sure

[stone folio]

10/3 & 2/3

[royal basin]

yes

be mindful of when the equation says an absolute value is negative though

those have no solutions

[stone folio]

Like #8

I got 5/2 and -5/2

[royal basin]

exactly what i warned you not to do...

[obtuse pebble]

@stone folio Has your question been resolved?

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[smoky zenith]

Where did the x come from?

[elder tusk]

That may have been a typo (someone correct me if I’m wrong) but the final result is correct, I also got -1/(3 + sqrtx)

[smoky zenith]

I got this

[elder tusk]

Yep, but you can simplify it further if you multiply numerator and denominator by -1

Since you have (+9 -x) on the top and (-9 + x) on the bottom

[smoky zenith]

Where would the 1 come from?

Are we just "rewording" the problem?

[elder tusk]

Well you can multiply your fraction by -1/-1 and it would still be the same value, since multiplying by -1/-1 is the same as multiplying by 1
Yeah you could say you’re kinda rewording or rearranging things

[smoky zenith]

Oh I get it

[elder tusk]

I think your answer would be fine too btw

Just letting you know how they got to that final answer

[smoky zenith]

I really don't think so, we would be multiplying by zero in the denominator

[elder tusk]

Oh oops you’re right

Forgot it’s the limit

[smoky zenith]

Am I doing it right?

[elder tusk]

That works, you’d end up with 1 / -(3 + sqrtx) but it’s the same result as their answer

Just with the negative in the denominator instead of numerator

[smoky zenith]

K thank you 👍

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[timid silo]

how do i put that into simplest radical form?

[smoky zenith]

Have you ever heard about the Tree method?

Its a pretty dope trick

[timid silo]

prime factorise 60

pretty sure my teacher taught it but I wasn't there that day

[smoky zenith]

you essentially divide by prime numbers
heres a 4 min video
https://www.google.com/search?q=radical+tree+method&rlz=1C1VDKB_enUS1009US1009&oq=Radical+tree+&aqs=chrome.1.69i57j0i512l2j46i175i199i512j0i22i30i625l3j0i390l3.3185j0j7&sourceid=chrome&ie=UTF-8#kpvalbx=_0HHtY_LTLsmt5NoPkJWAqAU_29

[timid silo]

thanks

.close

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[fathom flicker]

,, y''+4y=3cos(2x)

[warm shale]

AustinU

[fathom flicker]

I am trying to solve this using the method of undetermined coefficients

I went with guessing

[warm shale]

AustinU

[fathom flicker]

because there is a cosine term in the solution

After differentiating this twice and plugging all the values back into the ODE I end up with

0=3cos(2x)

so I think it means my guess was wrong

what do I need to do instead?

[dark stirrup]

[fathom flicker]

I am hmming aswell

[dark stirrup]

This feels off

oh wait..

[fathom flicker]

?

[dark stirrup]

Let me actually solve this before potentially misleading you

[fathom flicker]

okay

I just am trying to find a particular solution specifically using the method of undetermined coefficients

I thought when there is a cos(bx) in the right side I should guess y=acos(bx)+Bsin(bx)

and then ended up with a bad

[dark stirrup]

My guess is that trivial particular solution needs to be x(A cos(2x)+B sin(2x))

But this is just instinct with no strategy

[fathom flicker]

Possibly, but I don't even want/need the whole solution

[dark stirrup]

I meant trivial

mb

[fathom flicker]

like for the homogenous equation?

I am looking for the particular

[dark stirrup]

right

I meant trivial

Sorry, the wording again

I know how to solve, but forget all the wording

[dark stirrup]

I think this is the approach you need

[fathom flicker]

I didn't try the x

but yeah

[dark stirrup]

Okay just verified. That's definitely the right track. I don't remember why this strategy works, but I do remember it's what you need to do in situations like these.

[obtuse pebble]

@fathom flicker Has your question been resolved?

[obtuse pebble]

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[ashen drift]

how to formulate an inequality on this problem?

is it 30b + 70 >= 1000 or 30b + 70 > 1000?

because i think the answer would be b >= 33.1 or b > 33.1

just making sure

[wooden cipher]

i would say >=

[ashen drift]

i see thank you so much

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[smoky zenith]

I cant find my error (Work included)

[timid silo]

,RCCW

[timid silo]

you divided the numerator with x^2 when u divided the denominator with sqrt(x^2)

[smoky zenith]

But dont we use the greatest power in the denominator to divide the numerator in a infinity limit?

[timid silo]

yes but x^2 isnt your greatest power

it is sqrt(x^2)

[smoky zenith]

if the X in the denominator was X^3, would my greatest power still be sqrt(x^2)? or sqrt(x^3)

[timid silo]

also your operations are illegal because you are not multiplying the numerator and denominator uniformly

[timid silo]

it would be x^3

[smoky zenith]

Sorry i meant denominator

[timid silo]

it would simply be sqrt(x^3) yes and u can see it would be undefined then

[smoky zenith]

how did we get a negative one in the numerator?

[timid silo]

it is the absolute value function

it is 1/|x| and for x < 0 the absolute function could be represented as 1/(-(x))

[smoky zenith]

ohh yaa, okay to summarize
A square root in an equation of a limit to infinity has a completely different greatest power then the limit out side of the square root.
Depending if were going from negative infinity or positive infinity we use (picture)

[obtuse pebble]

@smoky zenith Has your question been resolved?

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[zealous kindle]

hi

[zealous kindle]

.close

[obtuse pebble]

Channel closed

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[zealous kindle]

.close

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[wise talon]

<@&268886789983436800>

[zealous kindle]

@wise talon snitch

im new to this server

[urban ridge]

why are you making help channels to shitpost?

1 day mute

be glad im not just banning

.close

[wise talon]

coose

NO

[dark stirrup]

!help

[obtuse pebble]

Please read #❓how-to-get-help

[wise talon]

I REFUSE TO REPORT MY ANSWER AS AN EXACT DECIMAL

open a new channel btw

[rose heart]

sry didnt see just joined

[obtuse pebble]

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[teal smelt]

whats the command for differentiating

[teal smelt]

sorry im taking a spot

pls dont ban me

[wise talon]

wdym differentiating

[teal smelt]

im getting ignored in the other chats and i cant find it when i use the help command

[gilded needle]

command in what software/language?

[wise talon]

,w differentiate x^2

?

[warm shale]

Results provided by WolframAlpha

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

Requested by DerpZ#9141

[teal smelt]

ty derp

.close

[obtuse pebble]

Channel closed

Closed by @teal smelt

Use .reopen if this was a mistake.

[wise talon]

lol

[gilded needle]

you must be psychic 😁

[wise talon]

my guess was on point

goddamn my knack for

seeing what users want

[gilded needle]

and delivering the goods haha

[wise talon]

me trying to guess what the customers want be like

well thats what swe experience gets ya

knowing what the customer wants more than they do

haha

[gilded needle]

indeed

[obtuse pebble]

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[steel wolf]

Where am I going wrong with this

[gilded needle]

"with no additional information" means that the coin might be unfair?

[steel wolf]

I mean if this question is right then I guess it doesn’t matter if the coin is fair

[gilded needle]

to calculate the set we don't need to know if it's fair or not

to calculate the probability of the set, we do

[steel wolf]

What’s the difference?

[gilded needle]

the set is just the outcomes, for example a few of the outcomes in the set look like HHHHTTTTTT, HHTTHHTTTT, TTTHTHHTHT

basically any sequence of ten heads or tails where there are exactly four H's

those outcomes are in the set regardless

but their probabilities depend on whether the coin is fair or not

like if the coin is weighted heavily in favor of H, then getting only four heads out of ten would be unlikely

but if it's a fair coin then it's more likely

[steel wolf]

Ohhh

So probabilities don’t matter in the case for the set cuz the subsets are just subsets

[gilded needle]

yep

the possible sequences are the same no matter what their probabilities are

probabilities are numbers that we assign to that set of sequences

[steel wolf]

I see now

So besides the unfair case for question 1 is everything good?

[gilded needle]

yea, i would exclude answer 4 because "with no additional information" probably means "if we don't assume fairness"

and in that case you don't know the probabilities

the rest look fine

[steel wolf]

Alright thank you so much

[obtuse pebble]

@steel wolf Has your question been resolved?

[obtuse pebble]

Channel closed

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[obtuse pebble]

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[haughty oxide]

https://media.discordapp.net/attachments/269573202018041856/1075623627191496754/IMG_0104_2.jpg?width=536&height=650

[obtuse pebble]

Please don't occupy multiple help channels.

[haughty oxide]

how to do C ?

[fierce lagoon]

You have your own channel already

Go back there

.close

[obtuse pebble]

Channel closed

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[obtuse pebble]

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[humble umbra]

I have tried expanding this several times now but I can't seem to get it

I've gotten to root x - 9 / x - 81

I tried multiplying by the conjugates but that doesn't do anything here

[supple granite]

You need to take the derivative of that function to find the slope

[humble umbra]

I'm pretty sure I need to use this formula

to get the slope

then once I have that, then I can get the equation

[humble umbra]

is this the formula for a derivative ?

[supple granite]

You can subb in x=81 and f(x)=sqrtx

Did you do that?

[humble umbra]

yes

I eventually got root x - 9 / x - 81

and found that multiplying the conjugate doesn't help

[supple granite]

This one, right?

[humble umbra]

my goal is to try and remove the denominator.

Yes

[supple granite]

For this ur goal would be to remove the numerator

Can you factor the denominator using difference of two squares?

x is the square of sqrt x, and 81 is the square of 9

[humble umbra]

wait

but the numerator is x - 81 not x^2 - 81

[supple granite]

Can you factor it now?

[humble umbra]

oh wow

yeah

that solves it

[supple granite]

Now you have the slope, and a point that goes through the linear function, so you can set up a linear equation

[humble umbra]

oh

wait

So, when I factored, I got 1 / root x - 9

does that look correct?

If yes, we have a problem

ah

[supple granite]

Can you show ur work?

[humble umbra]

oh sure

lemme take a picture

[supple granite]

You factored it wrong

Do you know how to factor difference of two squares?

$a^2-b^2=(a-b)(a+b)$

[warm shale]

jay.

[humble umbra]

wait

I did do it correctly, but I accidently left ( root x + 9) as (root x - 9)

ok ok

I see

it works now

I have the slope now for the tangent line

[supple granite]

Which is...?

[humble umbra]

1/18

[supple granite]

Great

So now you have a line whose slope is 1/18 and goes through (81,9)

[humble umbra]

yes

now I use a linear equation

y-y_1 = m ( x - x_1 ) -> y-9 = 1/18 ( x - 81)

y = 1/18 + 9/2

ty

.close

[obtuse pebble]

Channel closed

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[thorny stump]

x^2 + y^2 + 4x + 2y -20 = 0, The question wants me to find the center (h,k) and radius, of circle

[warm shale]

AustinU

[thorny stump]

yes thats the equation

i think because x,y are square that means there must be 2 solutions for both x and y

[fathom flicker]

Can I stop you for a moment

[thorny stump]

yes

[fathom flicker]

We aren't really looking to "solve" for x and y

That equation

will be true for all points of whatever circle it is

that's how the function works

so solving for x and y, won't give us the center

Let me ask you this

Do you know the center and radius of this circle:

[warm shale]

AustinU

[thorny stump]

This is the unit square correct?

[fathom flicker]

Unit circle, but I assume that is what you mean yes

[thorny stump]

unit circle*

[fathom flicker]

mhm

[thorny stump]

it starts at (0,0) origin and then extends 1 in every direction

[fathom flicker]

yes

So for a more standard circle, a less simple one, like yours

we can write it like this

,, (x-h)^2+(y-k)^2=r^2

[warm shale]

AustinU

[fathom flicker]

where h and k are the x and y coordinates of the center, and r is the radius

See how it applies even to the unit circle?

h=0, k=0, and r=1

[thorny stump]

i see

[fathom flicker]

So what we need to do with your equation, is get it into that form

Do you have an idea how we might do that?

[thorny stump]

the first think im thinking of is that -20 is going to turn into our radius because its is value independent of a variable

i think im going to have to simplify

[fathom flicker]

maybe, but I'll lead you in a different direction

obviously we are going to have to simplify, but we need to do so in a specific way

do you know how to complete the square?

[thorny stump]

No I dont recall how to do that

this is my first time touching this material

[fathom flicker]

Okay I'll try my best to help you learn it

first we want to group the x and y terms together, so lets rearrange our equation to look like this

[warm shale]

AustinU

[fathom flicker]

Stop me in the chat if you ever have a question

[thorny stump]

okay so far we moved 20 to the other side and re-arranged the variables

[fathom flicker]

yea

So now let's focus one at a time on x and y

if we have

,, x^2+4x

[warm shale]

AustinU

[fathom flicker]

and we want to rewrite it so it looks like this

[warm shale]

AustinU

[thorny stump]

okay

[fathom flicker]

We need to complete the square to do that

the method is to choose half of the value attached to the x

so in this case 2

[thorny stump]

so we do (x+2)^2 , (x-2)^2 ?

[fathom flicker]

Not quite

If we try something like this

[warm shale]

AustinU

[fathom flicker]

and multiply it out

we will find that it equals

[warm shale]

AustinU

[fathom flicker]

and we only wanted the first two terms of that ^

so what might we be able to do

to make them equivalent

[thorny stump]

make (x+2)^2 equivalent to x^2+4x+4?

[fathom flicker]

no

what do we need to adjust about (x+2)^2

so that it is equal to x^2+4x

^ without the plus 4

because remember in our equation for our circle, we just have x^2+4x

[thorny stump]

can we subtract 4 from both sides of the equation?

[fathom flicker]

Well we do want to subtract 4

but we don't have an equation

they aren't equal

so it isn't from both sides

but yes that is the point

if we have

(x+2)^2 - 4

then it is equal to x^2+4x

do you see how that works?

[thorny stump]

so you removed it from the parentheses and changed the sign?

[fathom flicker]

I'll summarize the steps again to make it clearer for you

So just hold on one second

We have x^2+4x In our equation

** We want it to look like (x-h)^2**

We try (x+2)^2

[warm shale]

AustinU

[fathom flicker]

** This isn't what we want, but it is close**

If we subtract 4, we get

[warm shale]

AustinU

[fathom flicker]

That is what we want ^

Do you follow those steps?

[thorny stump]

im trying really hard to understand

[fathom flicker]

feel free to ask me about any part that confuses you

[thorny stump]

okay so we want to make x^2+4x = (x+2)^2 - 4

[fathom flicker]

yes

[thorny stump]

i dont understand how we can create that -4 value

[fathom flicker]

Here is how

I'll demonstrate

Do you have paper?

[thorny stump]

if i solve (x+2)^2 it equals x^2+4

we are now missing the x from 4

[fathom flicker]

^ you are multiplying incorrectly then

it equals x^2+4x+4

[thorny stump]

okay i see

i have paper now

[fathom flicker]

So follow along with me, I'll tell you what to write

Ready?

[thorny stump]

👍

[fathom flicker]

Write (x+2)^2 = x^2+4x

Multiply out the (x+2)^2 into x^2+4x+4

and now write below it

x^2+4x+4 = x^2+4x

^ and then you can see that they are not equal

so for all your equal signs

put a cross through it

Now what can you do to the left side, to make it actually equal to the right?

subtract 4

So now write

x^2+4x+4 - 4 =x^2+4x

and instead of erasing both the 4s, because they do cancel eachother out

rewrite what we know x^2+4x+4 is

it is (x+2)^2

and you get

(x+2)^2-4=x^2+4x

and ^ that is actually equal

[thorny stump]

Okay i think i understand better now

[fathom flicker]

Perfect, because I'm gonna ask you what you think we should try for the y's in a second here

we have in our equation

[warm shale]

AustinU

[fathom flicker]

what should we try?

[thorny stump]

(y+1)^2 -2

[fathom flicker]

Very close actually

double check to see if you need -2

or some other number

the (y+1)^2 is great

-2 is just slightly off

[thorny stump]

(y+1)^2 -1

[fathom flicker]

yup perfect

okay now we know

[warm shale]

AustinU

AustinU

[fathom flicker]

and our equation to begin with was

[warm shale]

AustinU

[fathom flicker]

So what do you think we should do

[thorny stump]

I think the next step is to square root the parentheses?

[fathom flicker]

no

[thorny stump]

or i guess both sides of the equation

[fathom flicker]

no need

the next step is simpler than you think

all we want to do now, is replace what we just found out

so use what we now know we can replace the x's and y's with

and rewrite that equation

the equation with the crown emoji

[thorny stump]

oh ok

(x+2)^2 -4 + (y+1)^2 -1 = 20

[fathom flicker]

yes perfect

[thorny stump]

then combine like terms?

[fathom flicker]

and then move the constants over with the 20

and rewrite it one more time

[thorny stump]

oh okay

(x+2)^2+(y+1)^2=25

[fathom flicker]

mhm

and I'll remind you now of the equation I gave earlier

[warm shale]

AustinU

[fathom flicker]

For a circle with ^ that equation, the center is (h, k) and the radius is r

[thorny stump]

25 = 5^2

[fathom flicker]

yup

so radius is 5

and be careful with those ^ negative signs when you are finding out the center of your circle

[thorny stump]

yeah thats what i was going to ask

how do i make this equation have those negative signs

[fathom flicker]

Well

[thorny stump]

because i need them to plug this into a graph right?

[fathom flicker]

+2 is the same as -(-2)

No, no need to plug into a graph

Unless your teacher wants you to

[thorny stump]

oh the question wants me to graph it

[fathom flicker]

ok we can do that at the end

but tell me what the center is first on your own

[thorny stump]

(2,1)

(h,k)

[fathom flicker]

no, almost

if it was (x-h) the center is at h

you don't have (x-2)

you have the opposite -> (x+2)

so where is the center

hint: (x+2)=(x - (-2))

[thorny stump]

if it is the opposite is it (-2,-1)

[fathom flicker]

yes

that is the center

and you have the radius

so all that's left is to graph

[thorny stump]

okay so i just rewrite the equation

[fathom flicker]

https://www.desmos.com/calculator/dqfvtgygmf

Here is two ways you can graph it ^

[thorny stump]

(x-(-2))^2+(x-(-1))^2=5^2

[fathom flicker]

Not minus signs

they were plus signs

[thorny stump]

can i write (x+2)^2 same as (x-(-2))^2

[fathom flicker]

Yes you can

[thorny stump]

perfect

[fathom flicker]

so that's correct

[thorny stump]

https://tenor.com/view/love-thank-you-thanks-crypto-nft-gif-26347302

[crude ice]

not really…

[fathom flicker]

No problem @thorny stump

[fathom flicker]

This channel got closed due to inactivity since your question was in it, if you still need help you should open up a new one because it is occupied until @thorny stump closes it

[thorny stump]

.close

[obtuse pebble]

Channel closed

Closed by @thorny stump

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vivid edge]

can some one help me with (ii)

[vivid edge]

i know im supposed to use tan

[lost tree]

Okay how do u think u must use tan

Try drawing a pic

[vivid edge]

tan24.8=55/294-x

[lost tree]

Yes

There u have it

[vivid edge]

so then

i multiply by x?

[lost tree]

It's now just an equation in x

[vivid edge]

wdym

[lost tree]

How wud u solve something like 3 = 2/(1-x)

The x is in the denominator, and it's kinda hard to work with x when it's in the denominator

What can u do to bring it to the numerator

[vivid edge]

inverse

[lost tree]

How

[lost tree]

Show me how u wud do it with this

[vivid edge]

take x to other side

3+x=2/1

3+x=2

[lost tree]

U can't just take x to the other side

[vivid edge]

so multiply by x?

[lost tree]

You have $3=\frac{2}{1-x}$

[warm shale]

SilverSoldier

[vivid edge]

oh thats what you mean

[lost tree]

If u had $3=2-x$, then you can take $x$ to the other side like you did

[warm shale]

SilverSoldier

[vivid edge]

sorry i thought you had that one lol

[lost tree]

Well wasn't it something like this u meant when u wrote this

[vivid edge]

yeah

ok so do i multiply by -X

[lost tree]

Multiplying by -x will give u

$3\times-x=\frac{2}{1-x}\times-x$

[warm shale]

SilverSoldier

[vivid edge]

so then its -3x=2/1

[lost tree]

No

U can't cancel the x in the denominator with the x in the numerator

Coz there's a 1-x

Only if -x was a factor of the denominator can you cancel it like that

The denominator doesn't look like (-x) × (something else)

[lost tree]

It must look like this for you to be able to cancel out the -x from the top and bottom

[vivid edge]

oh i see

[lost tree]

For example, say you had $\frac{10}{3+2}=2$. You can't multiply both sides by $3$ and say $\frac{10}{3+2}\times3=\frac{10}{2}=2\times3=6$

[warm shale]

SilverSoldier

[lost tree]

Clearly 10/2 is not 6, it is 5

[vivid edge]

yeah i see now

[lost tree]

If u want, you could have multiplied both sides by (3+2)

The whole thing

$\frac{10}{3+2}\times(3+2)=\frac{10}{\cancel{3+2}}\times\cancel{3+2}=10=2\times(3+2)=2\times5=10$

[warm shale]

SilverSoldier

[vivid edge]

so can i multiply by (294-x)

[lost tree]

$\frac{10}{3+\cancel{2}}\times\cancel{2}$ and $\frac{10}{\cancel{3}+2}\times\cancel{3}$ are NOT valid

[warm shale]

SilverSoldier

[lost tree]

Yes

[vivid edge]

so

(294-x)tan24.8=55

[lost tree]

Yes

[vivid edge]

can i now move x to the other side

so

[lost tree]

No

Everything inside those brackets must be multiplied by tan24.8

Before you can move them about

$2(5+2)=2\times7=14$.
This does not mean
$2(5)=14-2=12$

[warm shale]

SilverSoldier

[lost tree]

2×5 is clearly not 12

[vivid edge]

i see

[lost tree]

But $2(5+2)=2\times5+2\times2=14$

[warm shale]

SilverSoldier

[lost tree]

And $2\times5=14-2\times2=14-4=10$

[warm shale]

SilverSoldier

[lost tree]

Is valid

You must expand the brackets before u can move terms inside brackets to the other side

[vivid edge]

how do i expand with tan tho

so 294tan24.8

[lost tree]

Well tan24.8 is just some number

[lost tree]

Yes this is one term in the expansion

[vivid edge]

xtan24.8

[lost tree]

That's another term

[vivid edge]

so

[lost tree]

Tan24.8(294-x)

[vivid edge]

294tan24.8-xtan24.8

[lost tree]

Exactly 👍

[vivid edge]

but what about the 55?

[lost tree]

That'll still be there on yje right hand side

So far we only have been simplifying the left hand side

We had tan24.8(294-x) = 55

[vivid edge]

294tan24.8-xtan24.8=55

[lost tree]

Yes

[vivid edge]

so can i now move xtan24.8

to the other side?

[lost tree]

Yes u can move that whole thing

Yes, along with the x

[vivid edge]

so

294tan28.4-55=xtan24.8

[lost tree]

Yes

And now you just have to get rid of tan24.8 from the right hand side

And isolate x

[vivid edge]

how?

divide?

[lost tree]

yes

[vivid edge]

ahhh so

[lost tree]

If u had 2x = 15 or something, how wud u isolate x

The same thing here

[vivid edge]

divide by 2

so

[lost tree]

Yes

In ur case its not 2

[vivid edge]

294tan24.8-55/tan24.8=x

[lost tree]

Yes

[lost tree]

(294tan24.8-55)/tan24.8=x

I'd put parentheses to make it clearer

$\frac{294\cdot\tan24.8^\circ-55}{\tan24.8^\circ}=x$

[warm shale]

SilverSoldier

[vivid edge]

i got 174.9690

so 174.96

97

my bad

[lost tree]

,calc (294*tan(24.8deg) - 55)/(tan(24.8deg))

[warm shale]

Result:

174.96909418559

[lost tree]

Yes

[vivid edge]

ayyyyy

thx for the help

[lost tree]

😄👍

[vivid edge]

.close

[obtuse pebble]

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[latent knoll]

Stuck with part B

[latent knoll]

,rotate

[frigid prism]

use the quotient rule

[gleaming ridge]

quotient rule as suggested above

[latent knoll]

aight hang on

havent learnt yet i think

[gleaming ridge]

hmm... have you learned product rule yet?

[latent knoll]

dont think so

[gleaming ridge]

ah, I see you have to use the above question to get the answer

so, write (x^2 - x + 1) as suggested above

then divide to get a simplified equation, then differentiate

@latent knoll

[latent knoll]

oh wait

hang on i think i got an idea

when u simplify before differentiate, u get x+ 1/(x-1)

[gleaming ridge]

weird grammar but yeah you need to differentiate what you wrote

[latent knoll]

oh oops

wait

isnt that just 1/(x-1)^2

oh wait

im dumb

aight thx i got it

[gleaming ridge]

ah, cool

[latent knoll]

.close

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[royal basin]

the equation f(x)=g(x) needs to have exactly one solution

[obtuse pebble]

@timid silo Has your question been resolved?

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[wild orbit]

Can someone help me with this solution

[dark stirrup]

,rcw

[obtuse pebble]

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@wild orbit Has your question been resolved?

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[tiny isle]

hi can someone teach me part c

[thick gyro]

well everything froom the pan to the two masses must be accelerating at the same rate

[tiny isle]

yes

in part b you simply find the reaction of mass A to get the answer

i wonder if it is the same for c

[thick gyro]

part bs answer isnt just the reaction force though

im pretty sure

[grizzled shore]

What forces are on B

[tiny isle]

ok the answer for part b is just the reaction force of A

[grizzled shore]

Is b moving, if so, find the net force on b

Use the sum of all forces = net force to find the component by the scale-pan

[tiny isle]

this is the solution according to the mark scheme

as for part c the answer is equal to the overall tension

which is roughly 10 newtons

[thick gyro]

lets see, cuz im pretty sure i remember this question

AS edexcel mechanics

[tiny isle]

yep

[obtuse pebble]

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[cold ore]

can someone help me with this question: I'm unsure how to algebraically resolve this. I keep getting to a step where å^2 - b^2 =3/2 ( ab ). Using vector methods to find the equations of the lines passing through ( 2,-1) which are at an angle of 45 degrees, to x +3y + 1 = 0

[timid silo]

What are a and b? Also are you interpreting 3/2(ab) as 3/(2(ab)) or (3/2)(ab)? (there isn't an officially correct way to interpret this as far as I know)

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[vital topaz]

https://media.discordapp.net/attachments/905285721886187580/1075607601200504843/image.png?width=430&height=189

[vital topaz]

tried integrating both equations from intervals 0 to 30 then subtracted their results but i got it incorrect
even adding 125k doesnt help cause it just adds up to 365k
13t + (t^2)/2 was the first integral for birth rate and
5t + (t^2)/4 was the second integral for death rate

12th

[thick gyro]

how so

i litearlly get b

[thick gyro]

lmao

before you add 125k

its 465k

for the integral you literally have

[vital topaz]

I get 240k

Hmm

Error on my part then

Thanks for the help doc

.close

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[pulsar mason]

I’ve tried 10/3 (8+4(7)+0) and a few other variations

[pulsar mason]

Simpsons rule

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[craggy ferry]

At the very bottom did they find the inverse of every single one?

[craggy ferry]

And then multiply them to get the original matrix

Better picture

[kind hawk]

inverses of elementary matrices are easy

but yes that's what they did

[craggy ferry]

Ok

thanks

I get it i think

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[feral prawn]

Hi, I'm not able to find the rule to find the nth term in this sequence. I tried to find the difference and then do it, but it's still not working

[high granite]

1,2,3

Tn=a+(n-1)d

[feral prawn]

No, like if the question asks what is the 6th shape, how can I find that?

[high granite]

to find the neth term

[high granite]

oh ok

[feral prawn]

So I need a rule for that, so what is that rule, since the sequence is non-linear

?

[high granite]

^

n=6