help-0

[elder kindle]

What's with people just throwing in questions while I'm getting help -_-

[fading musk]

🙂

we all need help

[elder kindle]

You need to read the rules before you ask questions...

Go to a channel that isn't occupied

[fading musk]

oops

[elder kindle]

It's all good 👍

[gusty gorge]

well

the red line starts at (x, y), right?

[elder kindle]

mhm

[gusty gorge]

and it ends at (2, y), right?

[elder kindle]

Why?

[gusty gorge]

well, if we examine the right end of the line segment

it ends at x = 2 and the line segment is horizontal, so the y-coordinate is the same as that of the left endpoint

[elder kindle]

By line segment you mean (x,y) right?
"The y-coordinate is the same as that of the left endpoint" do you mean that the y value is constant along the segment?
Just seeing if I follow

[alpine sable]

Guys I need some help finding how I got an answer. I got it right one a test a while ago but the teacher gave me a zero on the problem. I found it an unusual way an totally forgot how I did. Now to get credit and stay in honors, I need to show how I got it. Anyone willing to help?

[elder kindle]

https://tenor.com/view/in-there-out-get-out-get-in-door-gif-15810355

[gusty gorge]

No, by line segment, I literally mean the red thing I drew

[fiery turret]

can someone teach me algebra 2 inverse trig 🥺

[elder kindle]

Yes I meant the red line

[gusty gorge]

The left endpoint is (x, y), and the right end point is (2, y)

because the red line is horizontal

[elder kindle]

So is it 2-x in reference to the distance from the right to the left? i.e x=0 covers the entire shape and x=2 covers nothing

[thick shard]

<@&286206848099549185> Sorry to tag. But, I really need your help.

I hope you don't mind.

[elder kindle]

Occupied

I figured it out

[strong furnace]

channel occupied?

I assume no ?

ok so I have this physics problem and I am sure I have the right answer but I have been told I might be missing something , what I want is input on what I might be missing

A coin (treat it as uniform disc) of radius r with negligible thickness and mass M rolls without slipping in a circle. The centre of mass of the coin moves in a horizontal circle of radius R and the axis of the coin is tilted at an angle Δ with respect to the vertical. Find the angular velocity of the centre of mass of the coin?

ok so the COM is rolling in a circle about the axis through our assumed origin

if we know the normal force we should be able to get the component of this normal fore along the radius of rotation

and then get angular velocity

what am I missing /?

[civic crypt]

What does it mean "roll in a circle"

[gusty gorge]

imagine that you roll a coin on its side

and it kinda goes in a circle before falling flat on its side

that's basically what the situation is

[civic crypt]

I dont understand

[civic crypt]

I see

[strong furnace]

also I could be right so if you agree with me you can share that as well

[gusty gorge]

I feel like this is an angular momentum problem

[strong furnace]

but I am leaning on me being incorrect as well due to the (treat it as disc) which sounds dumb

so do I

[civic crypt]

ok so the COM is rolling in a circle about the axis through our assumed origin
COM is not rolling, it is just moving in a circle

[gusty gorge]

you have a net torque on the coin which constantly changes its angular momentum

[strong furnace]

thats what I meant

[civic crypt]

Friction is ignored right?

[gusty gorge]

I think rolling resistance is ignored, but the friction that lets the coin tilt isn't

[civic crypt]

So you think we cannot take the only force acting on the coin to be gravity?

[strong furnace]

normal has to act as well

[civic crypt]

You mean normal friction?

[strong furnace]

normal reaction from the ground

" contact force perpendicular to surface "

[civic crypt]

There are two forces acting on the coin.

Gravitational force, and friction force in the direction of the radius of rolling circle.

[gusty gorge]

no, and there's an upward force on the coin's edge

[gusty gorge]

there's an up arrow too

[civic crypt]

oh

[gusty gorge]

probably doesn't matter, though

[civic crypt]

But can we ignore this red friction.

Can the coin travel like this without this friction?

I guess not

[strong furnace]

I think vrafaeli is correct

[civic crypt]

Something must push it inwards

[strong furnace]

there is friction involved

and that is my mistake

[civic crypt]

That's what Saccharine said

[gusty gorge]

there is the non-slip condition, which demands that there is static friction involved

[strong furnace]

yeah I mean both of you are correct

[strong furnace]

I don't know how this did not cross my mind maybe I should sleep thank you guys

[strong furnace]

I think I'll try solving it again

[civic crypt]

THis is actually pretty strange, I dont know how does the coin not fall 😄

[strong furnace]

conservation of angular momentum

pretty interesting actually

[civic crypt]

Similarly there is that rolling wheel on the rope that rises vertically if you turn it horizontally.

[strong furnace]

yeah these examples are pretty similar iirc

[civic crypt]

Never understood that.
And also that gyroscope or how is it called. That small turning thing.

Well, and the bycicle itself.

[gusty gorge]

basically, there's a net torque, but it's used to change the direction of the angular momentum

similar to how if you have uniform circular motion, you can have a force toward the center, but the speed always stays the same

[civic crypt]

Yea, I understand there must be similar analogy, but I kind of dont see it clearly 😐

---

So there is a torque (blue) with respect to the orange axis.

[native nimbus]

I’m having trouble with some problems like one that at the end the answer to x are fractions like let’s take 3x+4 =6 - x the value is 1/2 and I couldn’t solve it cna some help like explain it

[civic crypt]

And the angular momentum is with respect the cyan axis.

[sage jacinth]

try isolating the x

[strong furnace]

there should be two components of angular momentum actually if I am not wrong

[civic crypt]

Oh yea, one is rotation around purple.

[strong furnace]

that is what makes this problem royally messy and why I wanted to avoid this but there's no way to avoid it 😦

<@&268886789983436800>

[civic crypt]

The torque comming from blue force can be represented by the orange arrow.

[alpine sable]

I am afraid to ask a question

[civic crypt]

Now this is perpendicular to both the cyan and purple angular momentums

[alpine sable]

Can help me

[strong furnace]

try a different channel please?

[civic crypt]

@alpine sable Read #❓how-to-get-help

[alpine sable]

I know I read that already

I am afraid to show my work

And you might laugh at me if you think it's easy

[strong furnace]

no1's gonna laugh that is against the rule too but just use a different channel

the second rule

[civic crypt]

Are you trolling?

Read #❓how-to-get-help

[alpine sable]

No

[civic crypt]

Then read #❓how-to-get-help

[alpine sable]

I know I just need to send my work right

[strong furnace]

this guy is trolling -.-

[alpine sable]

And ask if it's right

[alpine sable]

Eh?

[true torrent]

anyways sincde he's trolling

[civic crypt]

If you dont want to read #❓how-to-get-help ill have to ping moderator

[strong furnace]

man we were doing something

please use a different channel

[true torrent]

oh my bad

[strong furnace]

read #❓how-to-get-help

[alpine sable]

I did read it

[civic crypt]

@strong furnace Is there only one angular momentum, as the sum of all momentums?

[alpine sable]

It says don't ask to ask

[strong furnace]

I guess we could write it in vector form

[alpine sable]

Just ask

[strong furnace]

since the directions don't match

[alpine sable]

It says

[strong furnace]

but the part that is changing is only one part of this

[civic crypt]

We can sum cyan and purple anuglar momentums to get the resulting momentum

And orange is added to that

Or not 😐

Can we like sum angular momentums of different axis

[strong furnace]

the thing is

the torque is perpendicular

to both our angular momentum components

which means its only changing the direction

[civic crypt]

Yes.

[strong furnace]

which makes sense

[civic crypt]

yes

But changes direction of which angular momentum of the two, and how

[brittle oxide]

For 18 how do I find y?

I’m confused

[civic crypt]

@brittle oxide #❓how-to-get-help

read

[brittle oxide]

Sorry, didn’t make sure this channel was used

[strong furnace]

not the purple one for sure so it has to be cyan

[civic crypt]

How do you know it is not the purple

[strong furnace]

I am visualizing it

com is rotating about an axis

that is what the cause of purple is

no matter what happens the direction wouldn't change

but maybe a better argument is needed

more mathematical

[civic crypt]

You need to listen to 90min of Richard Gere: https://youtu.be/ZTNip78TUvA?t=1186

[strong furnace]

but this is addition of angular momentum in quantum mechanics

[civic crypt]

I saw that something is fishy 😄

[sacred sleet]

so basically someone bought a two products the total amount is 134.8euro and how much he bought was 28.4kg the products prices are 4.1euro/kg and the other one 5.7euro/kg. Then I need to know how much he bought each of the items.

[civic crypt]

OK, if you check this image, and if we immagine that you can sum angular momentums, then adding torque (orange) will rotate the angular momentum sum around purple.

[strong furnace]

this is actually first year physics maybe walter lewin covers it but his lectures got removed

[civic crypt]

cause he sexual molester 😐

[strong furnace]

gimme a sec to think about this

[sacred sleet]

?

[strong furnace]

orange is coming out of the screen right? if yes then this all makes sense to me

[gusty gorge]

angular momentum is a vector quantity, and it's related to angular velocity by the inertia tensor / moments of inertia

but I forgot almost everything about this a long time ago

[strong furnace]

we will have to individually add the angular velocity product with moment along axis for every component

to gt the angular momentum

[lethal vortex]

thanks!!!

[civic crypt]

yes, but can you sum the two angular momentums from the different axis or rotation. I guess, why not?

[gusty gorge]

I don't think this gets you the angular momentum, because IIRC, you have cross-moments between the axes

and these also go in

[civic crypt]

You can likely sum angular momentums that are repective to the COM.

[strong furnace]

this is something I agree with and what I meant

[civic crypt]

But here we have that axis and purple axis. These two also intersect.

Just not at the COM.

[sacred sleet]

so basically someone bought a two products the total amount is 134.8euro and how much he bought was 28.4kg the products prices are 4.1euro/kg and the other one 5.7euro/kg. Then I need to know how much he bought each of the items.
Can someone help me out literally confused a f.

[civic crypt]

But if we can sum them, then we are done 🙂

@sacred sleet Have you read #❓how-to-get-help

[strong furnace]

I was thinking of going through a basic approach

taking an instant t and then t+dt

finding the vector differencce

and equating that to the torque

that acted in this instant

[sacred sleet]

my bad

I thought u were done

[gusty gorge]

that seems like a workable approach

[strong furnace]

and whatever I get from there

using that to understand how to approach this

using whatever we already have here

[civic crypt]

https://www.youtube.com/watch?v=8pSLffliCk0

[strong furnace]

the video won't load for me for some reason but euler disk are something else I think

[civic crypt]

Yea 😄

4AM here, gtg sleep

[strong furnace]

gn thank you for the help

[civic crypt]

cannot think any more

[strong furnace]

I will go through what I thought and see where that takes me

[sacred sleet]

ok is this like

[civic crypt]

( @strong furnace also try ask on Phisics discord )

[sacred sleet]

done now?

[civic crypt]

@sacred sleet yes

[sacred sleet]

mk

so basically someone bought a two products the total amount is 134.8euro and how much he bought was 28.4kg the products prices are 4.1euro/kg and the other one 5.7euro/kg. Then I need to know how much he bought each of the items.

[civic crypt]

(btw you have 9 other channels :D, why you like this one so much)

[sacred sleet]

because ppl go here first

[steady rune]

is there anyone on here who can help with intro to stats?

[strong furnace]

not true I always prefer channel 8 or 9 over any other channel

[civic crypt]

Me too.

[sacred sleet]

ah

idk I have a test in like 3h

[bold token]

channel 8 and 9 gang

[civic crypt]

134.8 [euro] = 4.1 [euro/kg]*mass_x + 5.7 [euro/kg]*mass_y

mass_x + mass_y = 28.4 [kg]

[sacred sleet]

mhm

[civic crypt]

These are two equations with two unknowns. So you solve it as a system.

For example isolate mass_x from the second equation.
mass_x = 28.4[kg] - mass_y
and put this into first equation.

In brackets I put the units, which are nice to preserve, but you can ignore them

[sacred sleet]

Mk tysm

[civic crypt]

You can use wolfram alpha to verify your result

It can solve these equations in no time

[sacred sleet]

I thought I needed a value which left me confused

[civic crypt]

gn

[sacred sleet]

gn

[native nimbus]

What does that mean

[bold token]

finding x, leaving x to one side

so you can find x, which is x = 1/2

[lilac sparrow]

hey

what is A gonna be

[wary stream]

Do you know how to do end behavior with limits?

[lilac sparrow]

Yes i think so

Figured it out: A=4/7

[native nimbus]

I still don’t understand I’m stupid bye I’m in 7th grade and these are 5th grade questions

[bold token]

so,

you see x is unknown right?

you gotta know x

since x is in both sides of the equation,

you gotta try isolating x

so it comes out like x=blah blah

so in your question, 3x+4 =6 - x

you move x to a side you desire, in this case left,

it will become 3x+x=6-4

note when you want to move a number to another side of the equals, it turns negative to positive, and vice vers

then find x

[pale terrace]

is this channel in use

[thorn kindle]

Hey buddy

[pale terrace]

hi

log(0.1(y-2)^2)=1

I know I have to use log rules for this but I'm stuck near the end

I got as far as finding that sqrt(y-2) = 10

[ocean seal]

waler

[rigid smelt]

if so i dont see why there would be a square root

[pale terrace]

my bad

[rigid smelt]

btw, the log is base 10 right?

[pale terrace]

it's supposed to be to the power of 2 after y-2

yeah

not multiplied by 2

[rigid smelt]

hmm dont take the square root on both sides

i believe you got to here (y-2)^2 = 100 already?

[pale terrace]

yeah

[rigid smelt]

alright, so dont take the square root, that only makes it more complicated, just expand the square

now you got a quadratic equation

[pale terrace]

yes

[rigid smelt]

yep, only need to solve for it

technically you can take square root on both sides, but you also forgot to split them into two cases

[alpine sable]

With the equation (y-2)^2=100 the easiest way is to square root both sides

Just make sure you cover the positive and negative case

[rigid smelt]

yeah, which is what i said in the above

[pale terrace]

thank you

[pale terrace]

thank you too

should I get y = 12 and y=-8 as my answer?

[rigid smelt]

yes

[restive orchid]

Can someone help out on his question please

[gusty gorge]

nice pencil

[ionic jewel]

$$u^2r^2 = \frac{sr}{s+r}$$\
$$(s+r)u^2r^2 -sr = 0$$\
$$su^2r^2 + u^2r^3-sr = 0$$\
$$s(u^2r^2-r) + u^2r^3 = 0$$\
$$s(u^2r^2-r) = -u^2r^3$$\
$$s = -\frac{u^2r^3}{u^2r^2-r}$$\

@restive orchid

[ocean seal]

bunny

[severe badger]

Need help, its a trapezoid rule problem

cos(x)/sqrt(x), integral from a=0, b=2. How am I supposed to apply the trapezoid rule if f(0) is undefined?

I know I can split it into two integrals, but I still with have "x" on the denominator

[gusty gorge]

I think you can just use the limit

,w graph cos(x) / sqrt(x)

[ocean seal]

Results provided by WolframAlpha

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

Requested by Saccharine#3780

[gusty gorge]

oh

nevermind then

well, I was dumb enough to think there was a limit

[severe badger]

I just checked mail and the prof said I was supposed to use trapezoid on the Latter Integral, I got it now, you split cos(x) to (1-x^2/2)+cos(x)-(1-x^2/2). First integral is solvable and second can be solved using trap rule

[alpine sable]

Linear Algebra problem that im struggling to understand

if anyone has taken classes higher than Calculus would be greatly appreciated

[placid zinc]

Consider differentiating the first line, and working with that

[rare ivy]

Got an impossible question anyone good at advanced calculus?

A circular disc of radius 2 units is centred at the origin in the x − y plane. The areal
density of the disc is given by σ(y
2 + 1), where σ is a constant.
(a) Find the mass of the disc in terms of σ. Explain each step of your working. [10]
(b) How does the answer to part (a) change if the disc is centred at the point (0,1),
instead of at the origin? [5]

[alpine sable]

so x''(t) = ax'(t) + by'(t)

and then do I substitute y' into the equation ?

[placid zinc]

Yes. Then note you can rearrange the first equation for y, and use that to elim y completely

Differentiating basically brings a third equation in, and you can use the three equations to eliminate two unknowns

[alpine sable]

ok so when I substitute and distribute I get x'' = ax' + bcx + bdy

how do I get a characteristic equation from that

[placid zinc]

Should have no y after that last step

[alpine sable]

but I plugged in cx +dy into y'

[placid zinc]

Yeah! Then y = (x' - ax) / b

[alpine sable]

oh

oh and then the characteristic equation is the 0 = x'' + x' + x

and your eigenvalues are the solution to the characterstic equation

how would I got about B now

[placid zinc]

Solve for x(t). Sub that in for y = ...

[proud hatch]

does anyone know how to do this

[alpine sable]

nikoo its a triangular pyramid ontop of a triangular prism

find the area of the bottom triangle and multiply it by 10

then to find the volume of the pyramid you multiply the area of the triangle by 14 and divide by 3

then add the two volumes together

[proud hatch]

bet thank u for explaining

[alpine sable]

np

[alpine sable]

how do I find x(t) ?

[severe badger]

$[ \int_{1}^{infinity} (1)/(1+x^6) ,dx ]$

How do i change this to a definite integral through change of variables? I tried t=1/x, but that doesn't give an equivalent value

[placid zinc]

@alpine sable
That's just a regular second order DE at this point. Solve as per usual

Hah it's been a while since I've solved one, I'll look it up

[ocean seal]

PandaMan-AMB
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[severe badger]

There we go, had to edit it

[placid zinc]

I'm guessing that λ1 and λ2 can be assumed to be real

[alpine sable]

yes

[chilly walrus]

could I grab a hand with some differenciation

[alpine sable]

need to do quotient rule

[chilly walrus]

how would I go about doing so?

[woeful pulsar]

apply quotient rule

identify what f(x) and g(x) are

yeah, that's all

[alpine sable]

yup

f(x) is 2x^2 and g(x) is 4x-1

[chilly walrus]

yeah I get that part

[alpine sable]

f(x)' would be 4x and g(x)' would be 4

then just plug everything in

[chilly walrus]

seems to get very complicatied after that

[alpine sable]

all u need to do is find f g f' and g'

then plug into the equation

and youll get your answer

[chilly walrus]

ok thanks, imma try make sense of this xD

[mellow dome]

https://media.discordapp.net/attachments/490557019623915520/844238677689630761/IMG_20210518_1549322.jpg?width=1440&height=357

how do I do the last part?

[alpine sable]

idk

[mellow dome]

<@&286206848099549185>

[chilly walrus]

what about this?

[alpine sable]

same thing lol

[chilly walrus]

quotient again?

[alpine sable]

yep

[chilly walrus]

THOUGHT IT COULDNT BE THE SAME THING TIWCE

caps

[alpine sable]

derivative of 3e is 3e

and the derivative of (x+1)^2 is 2x+2

then use this again

match everything toghether

[chilly walrus]

yep 🙂

[alpine sable]

i would write down everything and keep track of it

it gets confusing if you try to do it in your head

[chilly walrus]

yep sweet

whats deriv of g(x)

3x??

@alpine sable

[alpine sable]

huh

oh

[chilly walrus]

i guess that makes sense

[alpine sable]

it says

3e

oh

nbvm

nvm

it says 3x

LOL

the derivative of 3x is 3

[chilly walrus]

lolo

right

because the power -1 thing

[alpine sable]

mhm

[chilly walrus]

epicsu

[restive orchid]

thx

[ionic jewel]

didnt really need the ping but yw

[wooden anchor]

Hi! So when plotting a graph, I've taken every little box on Y axis to be 0.25. Should I just write "every box represents 0.25 unit" or is there any proper way to indicate it?

[remote heron]

just mark the thick lines @wooden anchor

as 2, i guess?

[wooden anchor]

I did. Marked them as 2.5. But do I need to indicate how every small box is 0.25 unit?

[remote heron]

nah, as long as theyre evenly graduated

you can be nice and mark the midway point if it fits with your data

[wooden anchor]

Got it, thank you so much!

[alpine sable]

hey do you use the divergence theorem here?

doesn't seem right because it'll give an answer of 0

or should i use this formula?

<@&286206848099549185> thanks!

[brittle wren]

This open?

<@&286206848099549185>

[alpine sable]

this passes through the points wdym

[remote heron]

you just need to find the equation, right?

[alpine sable]

What is that fraction which when multiplied by itself gives 227.798649?

<@&286206848099549185>

[remote heron]

whyd you instantly ping helpers

[alpine sable]

.

help

[remote heron]

anyways how do you undo repeated multiplication

[alpine sable]

what?

[remote heron]

like theyre asking $(\frac{p}{q})^2 = 227.798649$

[ocean seal]

jan Niku

[remote heron]

haha that looks so shitty

[alpine sable]

oh

[ocean seal]

Chai T. Rex

$\qty(\frac pq)$

[alpine sable]

oh

k

then?

[remote heron]

$\overset{!}{\qty(\frac pq)}$

[ocean seal]

jan Niku

[remote heron]

thats the whole process pretty much

how do get rid of that square

[alpine sable]

bruh what r u doing??

[remote heron]

youll probably end up needing a calculator, idk if theres a good way to do this by hand

like if i asked

what number times itself equals 9

we'd set up an equation

call that unknown number x

[alpine sable]

any easy way to do this?

[remote heron]

x*x=9

so x^2=9

[alpine sable]

i am in 8th grade

please

[remote heron]

right

[oak chasm]

@alpine sable Yes, @remote heron is explaining how.

What's the opposite of squaring?

[alpine sable]

cube?

@oak chasm

[oak chasm]

Nope, it's square root.

[alpine sable]

ok

[ocean seal]

Chai T. Rex

[oak chasm]

So, use a calculator to take square root.

[alpine sable]

damn bruh

we have not studied this

can you explain an easier way

my teacher will say this is complicated method

[oak chasm]

You can skip some steps then.

[alpine sable]

such as?

[oak chasm]

Like, is it obvious to you that a number times itself is that number squared?

[alpine sable]

??

[oak chasm]

If it's not obvious, you can't skip the first step.

That's the problem, really. An easier way is for people who understand the math.

[livid jolt]

i think he is supposed to go like 15x15 is 225 so we check something just bigger than 15
like 15.1x15.1 = 228.01
so a lil less and so on

i think

[alpine sable]

can we do the long division method

[oak chasm]

To find a square root, yeah, something like that, but a calculator is easier.

[livid jolt]

he is 8th grader

[alpine sable]

Long division method??

Can we do that?

[oak chasm]

Yes, something like that.

[alpine sable]

yeah

ok

u dont have to make it complicated

[oak chasm]

But again, a calculator is easier.

[alpine sable]

@oak chasm how do i do kn calculator?

please tell

[oak chasm]

Type in 227.798649 and hit the square root button.

,w sqrt(227.798649)

[ocean seal]

Results provided by WolframAlpha

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

Requested by Chai T. Rex#7323

[oak chasm]

,calc 15.093^2

[ocean seal]

Result:

227.798649

[oak chasm]

So, you got 15.093, right?

[alpine sable]

yes

calculator did

[oak chasm]

OK, what is that in a fraction?

[alpine sable]

now what do i do next?

fractio?n

[oak chasm]

Yes, it wants a fraction as the answer.

[alpine sable]

i just square root 227.798649

and 15.093 came

[oak chasm]

Right, that gave you 15.093.

[plush ice]

15093/100

[oak chasm]

No.

[plush ice]

hehe

[oak chasm]

See how many digits are after the decimal place?

[alpine sable]

ik

3

15093/1000

?

[oak chasm]

Yes.

Now, can that be reduced?

[alpine sable]

yes

thx

[oak chasm]

OK, reduce it if it can be.

[alpine sable]

Wait

[oak chasm]

If not, that's the answer.

[alpine sable]

it cant be

thats tha answer

[oak chasm]

OK, then that's the answer.

[alpine sable]

yes

thanx

[oak chasm]

No problem.

[alpine sable]

How to do the first

If (2^3/6x9)^x

9 becomes 3^2

What happens to 6

[ocean seal]

Chai T. Rex

[oak chasm]

So, one power of 2 is cancelled in top and bottom.

Does that make sense?

[alpine sable]

i

Why 3^3?

[oak chasm]

Factor the 6 into primes.

[alpine sable]

Find the square root of each of the following correct to three places of decimal.
(i) 5 (ii) 7 (iii) 23.1

[ocean seal]

Chai T. Rex

[alpine sable]

Ohh

[oak chasm]

@alpine sable Sorry, channel is now busy, but you can enter the number into the calculator, hit the square root, and then round that number to three digits past the decimal point.

[alpine sable]

What how? didnt understood

[oak chasm]

Remember how you took the square root a few minutes ago?

[alpine sable]

yeah

[oak chasm]

Type in the number and hit the square root key.

[alpine sable]

square root of 5??

BRUH

Is this the final ans?

[oak chasm]

@alpine sable Close, but no.

When you take the fraction to the xth power, the top and bottom are taken to the xth power, right?

So, the 2 in the bottom is taken to the xth power, right?

[alpine sable]

Oh

p^3 / pq^3

?

[oak chasm]

Right, now reduce to lowest terms.

[alpine sable]

Find the square root of each of the following correct to three places of decimal.
(i) 5 (ii) 7 (iii) 23.1

<@&286206848099549185>

[woeful pulsar]

yeah, just put into calculator

[alpine sable]

How??

[woeful pulsar]

show your calculator?

[alpine sable]

What do I put in the calcultaor??

Square root 5?

[woeful pulsar]

does your calculator have a square root button?

[alpine sable]

im using google

calcutaor

[woeful pulsar]

then use sqrt(...)

[alpine sable]

calculator*

[alpine sable]

p^2/q^3?

.?

[ocean seal]

Chai T. Rex

[oak chasm]

@alpine sable Yes ^

[alpine sable]

oh cool

I need help for b n c 😅

@woeful pulsar u r not even explaining, do i square root 5?

[woeful pulsar]

yeah

there's a square root button

[alpine sable]

thats the answer

i did

[woeful pulsar]

you can do the same with the other numbers too

[alpine sable]

2.2360679775

nice

[oak chasm]

@alpine sable Let's switch to #help-1.

[alpine sable]

@woeful pulsar so just square rooting them and 2.2360679775 is the answer?

the final answer comes 2.2360679775

[woeful pulsar]

and you said to 3 decimal places

so ?

[alpine sable]

so what do i have to do?

thats the answer?

[woeful pulsar]

round to 3 decimal places

[alpine sable]

how

divide?

100?

???

@woeful pulsar

[woeful pulsar]

erm

[woeful pulsar]

here right?

yeah it says round to 3 decimal places?

[torpid oak]

bruh.

[alpine sable]

so i will divide by 100

right?

[torpid oak]

no

2.2360679775

3 decimal points

[woeful pulsar]

dividing by 100 doesn't round a number

like 1.234 is a number with 3 decimal places

[alpine sable]

ok

22360679.775?

@woeful pulsar

[torpid oak]

no.

[woeful pulsar]

22360679.775 is not anywhere close to 2.2360679775

[torpid oak]

rounding to 3 decimal placing means looking at the 4th decimal point and seeing if its above or below 5

for exmaple 2.3456

the fourth decimal point which is 6 is higher than 5

which means we got to add 1 to the number before it

which is 5

2.346

thats trig right

first u gotta label the sides

side h is adjacent to the angle 60

and 4 is opposite to 60

have u heard about SohCahToa

soh is sine = opposite/hypotenuse

cah is cos = adjacent/hypotenuse

toa is tan = opposite/adjacent

which one do u think we should use here

well there are two angles we can work from

60 and 30

but i chose 60 for sum reason

we would use tan for using the angle 60

yeah teacher is a bit mad sendingu this if u dont know how to do it

anyways

[alpine sable]

wait give me a minute

[torpid oak]

tan60 = 5/h

[alpine sable]

it's trigonometry

[torpid oak]

yeah ik

[alpine sable]

not geom

[torpid oak]

trig is geom

[alpine sable]

yh obv

[torpid oak]

yeah so whats ur point

?

[alpine sable]

d = √3
h = 7√2

should b correct

okay

[vale wigeon]

what diffeq are you looking at

also uh

@lethal otter we might wanna move to a free channel

[lethal otter]

thanks

lets join math

[vale wigeon]

im not gonna go in voice

[lethal otter]

np

[vale wigeon]

i have pinged you in #help-9 let's move there

[lethal otter]

thx for the help

😄

[wise sigil]

Well, regardless of which side you need to solve for, you know that on a 30-60-90 triangle the bigger leg is sqrt(3) times the length of the smaller leg, right? And the hypotenuse is twice the length of the smaller leg

[twin fulcrum]

Can someone please help me solve number 9

[vale wigeon]

go to sleep lmao

@twin fulcrum what's troubling you here?

[twin fulcrum]

My grades are due this week 😭

I’m confused how to set up an equation for it

[vale wigeon]

introduce a variable

say, let x be the number of hours worked by samantha

then write down how long amy worked, in terms of x

then write down "together they worked 19 hours" as an equation

[twin fulcrum]

Fr

Lol

Cant wait for school to be over

[twin fulcrum]

I don’t get how to do that?

[vale wigeon]

okay so

[vale wigeon]

refer to this again

[twin fulcrum]

Mhm

[vale wigeon]

samantha worked x hours

how many hours did amy work?

[twin fulcrum]

5 hours less

[vale wigeon]

write this as an expression in terms of x

@twin fulcrum ?

[twin fulcrum]

I still don’t get it sorry I’m stupid

[vale wigeon]

if you don't get it then say so instead of leaving me on read.

what does it mean to have "5 less than something"?

if i have $5 less than you and you have $30, how do you figure out how much money i have?

[twin fulcrum]

My tab was on open idk why it said read I’m on mobile

[ocean seal]

Ann

[twin fulcrum]

IDK

[vale wigeon]

think about it

[twin fulcrum]

Oh

$25?

[vale wigeon]

you have 30 dollars, i have 5 dollars less than you, how much do i have?

[twin fulcrum]

$25

[vale wigeon]

and how did you find this?

[twin fulcrum]

Because I know I can subtract $5 from what I have

[vale wigeon]

that's right, you subtracted 5

great

now

samantha worked x hours

amy worked 5 hours less than samantha

so wouldn't you agree that amy must have worked x - 5 hours?

[twin fulcrum]

Yes

[vale wigeon]

okay great

are you able to write down the equation for "together they worked 19 hours" now

[twin fulcrum]

Gn fishy

Uh

I get what I need to write I just don’t get like how to write the subtracting part

[vale wigeon]

wym

does it or does it not make sense to you that samantha worked x hours (since that's how we defined x) and amy worked x - 5 hours (since that's what "5 hours less" means)?

[twin fulcrum]

I get that I just don’t get how to write that in an equation

[vale wigeon]

if you have two people who worked for some amount of time, and you know how long each person worked, how do you find how much they worked combined?

for example, if jack worked 9 hours and jill worked 11 hours, how much did they work together?

[twin fulcrum]

20 hours

[vale wigeon]

and how did you find that?

[twin fulcrum]

Because I added

[vale wigeon]

the "because" does not belong in your answer, but yes. you added the work times.

so

[honest umbra]

hi

[vale wigeon]

@honest umbra channel busy please move

[honest umbra]

lol

[vale wigeon]

samantha worked x hours and amy worked x-5 hours. together, they must have worked x + x - 5 hours, or simplifying that, 2x - 5 hours.

@twin fulcrum does this make sense to you?

[twin fulcrum]

Yes

[vale wigeon]

great

and you know that their combined work time actually equals 19 hours

[twin fulcrum]

Mhm

[vale wigeon]

can you write down the equation now?

[twin fulcrum]

Yea

I got 12 I think I did it wrong 🤦‍♀️

2x-5=19 I guess it’s wrong tho

[vale wigeon]

you mean you got x = 12?

[twin fulcrum]

Ya

[vale wigeon]

what reason do you have to believe that this is wrong?

other than your self-perception of incompetence?

[twin fulcrum]

😦

[vale wigeon]

@twin fulcrum if you show your work, i may point at a place where you went wrong, or tell you you did not go wrong anywhere

[twin fulcrum]

I got it

[vale wigeon]

in any case, x = 12 is the correct solution to 2x - 5 = 19

you may have made an even number of mistakes that cancelled each other out, for all i know.

[twin fulcrum]

I just don’t get which hours who worked tho

I know it’s 12 and 7

[vale wigeon]

if you recall the beginning of our convo

i made the suggestion of letting x be the number of hours worked by samantha

this sort of detail should be written down clearly on a piece of paper at the beginning of your work, so that you may consult it after doing the algebra in order to translate your numbers back into the context of the problem.

[twin fulcrum]

BITCH I DID WRITE IT I UNDERSTAND NOW YOU ARE BEING SO RUDE

[vale wigeon]

???

[twin fulcrum]

Like I get I’m kinda stupid but like I’m trying

[vale wigeon]

was i rude just now?

[twin fulcrum]

Yes very

[vale wigeon]

all i did was remind you, in a manner i thought was polite, of the thing you forgot.

[twin fulcrum]

Well I didn’t forget

I wrote it on the paper

At the beginning

[vale wigeon]

did you write "x = number of hours worked by samantha"?

[twin fulcrum]

Yes

[vale wigeon]

and you solved the equation and got x = 12?

[twin fulcrum]

Yes

[vale wigeon]

and it did not click that 12 was samantha's work time?

since that's what we defined x as?

[twin fulcrum]

NO IT DIDNT OK?

I get it now BUT IT WAS CONFUSING

[vale wigeon]

and instead, when i reminded you of it, you chose to yell at me and call me a bitch.

[twin fulcrum]

YOu legit were being so rude

[honest umbra]

tf

[twin fulcrum]

Like sorry I’m not the smartest cookie on the fucking planet

[honest umbra]

she wasn’t rude ur just frustrated @twin fulcrum

[vale wigeon]

i legit do not see how this message was rude

also i do not see what part of "x = hours worked by samantha, x = 12, therefore samantha worked 12 hours" could be confusing

[honest umbra]

Lol

[vale wigeon]

it seems like the most straightforward conclusion possible to me

[honest umbra]

@vale wigeon can u help me

How do I find the phase shift for this

[vale wigeon]

@honest umbra can you maybe move to another channel

[honest umbra]

Ok

[vale wigeon]

oh wait

valgirlchillin ragequit this server ig

[honest umbra]

#help-9 ig

lol

ban her pls

u had so much patience for such a simple question

ban her

[flat turtle]

can anyone explain the difference between the commutative property and the associative property

my teacher doesnt want to respond

[vale wigeon]

commutative is a+b=b+a, associative is (a+b)+c = a+(b+c)

the commutative property says you can switch around the order when adding two things
the associative property says you can switch around which numbers you add first if you have three or more being added

[honest umbra]

can I get help in #help-9

[flat turtle]

oh

thanks

ur a lifesaver

[fallow pagoda]

how would you solve
(a+b)^c

wait no sorry

scratch tha my bad i meant (a/b + x/y)^c

[alpine sable]

There's nothing to solve

[fallow pagoda]

no but like simplify it

[alpine sable]

Its just an expression

[fallow pagoda]

like expand it

[alpine sable]

Oh you meant simplify

[fallow pagoda]

yeh sorry

[alpine sable]

Ok add the fractions a/b and x/y

[fallow pagoda]

okay so ay/by +xb/by

[alpine sable]

Yup

[fallow pagoda]

and then lets say c=2

so (ay/by + xb/by)^2

[alpine sable]

Why are you setting c=2

Does the problem say c=2?

[fallow pagoda]

because perfect squares

[alpine sable]

Can you send a picture of the original problem

[fallow pagoda]

no its just i have a maths exam tomorrow and perfect squares r gonna be in it and idk how to do perfect squares with fractions

uhh i got it from an equation before it was (x+2/x)^2 i think

[alpine sable]

So this was something you made up yourseld?

[fallow pagoda]

yes

u can replace the pro numerals with any number sorry i just wanna understand the process

[alpine sable]

Ok so what do you need help with, squaring fractions, order of operations, something else?

[fallow pagoda]

bc.i know normal perfect square is just a^2+2ab+b^2 but i get confused with fractions

squaring fractions

idk how to do the middle term if the perfect square is fractions

[alpine sable]

If you have (a+b)^2 where a and b are fractions it works exactly the same way

But it would probably be easier to first add a+b

Then square

[fallow pagoda]

wait let me pull up the equation and we can go through it hang on

(x+x/2)^2 was it

so

[alpine sable]

Ok first add x+x/2

[fallow pagoda]

wait why

wouldnt you square the x first

[alpine sable]

Because its easier to do the addition inside first and then square

[fallow pagoda]

x+x/2= 3x/2??

[alpine sable]

Yeah

[fallow pagoda]

ok so

[alpine sable]

Then find (3x/2)^2

[fallow pagoda]

uhhh

6x^2/4???

[alpine sable]

Hiw did you get that

[fallow pagoda]

lol i just squared three, squared x, and squared 2

[alpine sable]

You need to multiply the fraction by itself

[fallow pagoda]

i did that

oHh

[alpine sable]

Ok so you messed up squaring the 3

[fallow pagoda]

wat i times it by two pls

okay so 9x^2/4

[alpine sable]

Correct

[fallow pagoda]

so x^2+9x^2/4

HUH

wait

isnt there supposed to be three terms

oph so now the middle term is 3x/2 times two times x^2?

[alpine sable]

Ok we can do it the ither way and you'll see that you'll get the same answer

(x+x/2)^2=x^2+2(x)(x/2)+(x/2)^2

[fallow pagoda]

okay i see

[alpine sable]

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

Which is what we got before

[fallow pagoda]

wait can you simplify x^2+x^2 to 2x^2

[alpine sable]

Yes

[fallow pagoda]

okay

thank u so much

[night sigil]

I know that A(x) is

how do I find the allowable values

so x/4 = 0 when x = 0

x must be greater then 0

so how could I do 0 = (2a-(6-sqrt3)x)?

[left robin]

u have 2 x's

so when one x is 0 the other one is

probably also cant have a negative area.

[night sigil]

I know

so how do I solve for x in 0 = (2a-(6-sqrt3)x)

<@&286206848099549185>

[vale wigeon]

$2a - (6 - \sqrt{3})x = 0$

[ocean seal]

Ann

[vale wigeon]

is this your equation?

[night sigil]

yes

[vale wigeon]

this is linear in x, is it not?

[night sigil]

I don't know

[vale wigeon]

yes it is

(6 - sqrt(3)) is just a number

[night sigil]

yeah

[night sigil]

so 2a = (6-sqrt3)x

[vale wigeon]

@proven oak channel busy please move

@night sigil yes

[proven oak]

Ok

[night sigil]

and x = 2a/6-sqrt3

I don't know what to do after this

this just isn't correct

[vale wigeon]

2a/(6 - sqrt(3))

and this is correct

unless there is some kind of requirement on the answer that this doesn't fulfill but which you are omitting

[night sigil]

it's the wrong answer

apparently a/3 is the right answer

so this is an equation for area

trying to find the allowable values for x

A(x) > 0 when 0 < x < ?

I have no idea

[rigid smelt]

since x/4 is always positive

you need to solve for 2a - (6-sqrt(3))x > 0 for the lower bound

[night sigil]

yes

how?

[rigid smelt]

trust treat this like a normal inequality

try to isolate x

this is a linear inequality so there wont be involving dividing by x and all that complicated stuff

[night sigil]

well 2a/6-sqrt3 > x

[rigid smelt]

are you sure you arent missing any brackets?

those are quite important

[night sigil]

(2a)/(6-sqrt3) > x

[rigid smelt]

you almost got it right

oh wait nvm

sorry was lookign at wrong sign

[night sigil]

this doesn't seem to be the right answer

[rigid smelt]

what was the original question?

and erm it should be 2a/(6-sqrt(3))<x

[rigid smelt]

oh

ok, so you dont need to look at the domain of x for this

even tho that doesnt matter anyway

firstly, expand the brackets

[rigid smelt]

.

you will see that this is a quadratic expression nright?

[night sigil]

yeah

[rigid smelt]

and can you determine if the leading coefficient is positive or not?

[night sigil]

I don't think so

but I don't really know how I can get the equation in a form I know how to work with

[rigid smelt]

ok, before that, can you write out what you have after expanding?

[night sigil]

sorry

[rigid smelt]

great

that could also be written as ax/2 - (6-sqrt(3))x^2/4

right?

[night sigil]

yes

[rigid smelt]

so now could you determine the leading coeffecient?

[night sigil]

is it a/2?

[rigid smelt]

leading coeffecient should be the coefficient that is attached to the highest power term

[night sigil]

oh so (-(6-sqrt3))/4

[rigid smelt]

yup and is it positive or negative?

[night sigil]

negative

[rigid smelt]

so that means our vertex will be maximum or minimum?

[night sigil]

maximum

[rigid smelt]

yes, and do you know the formula for the vertex of a parabola?

given the quadratic expression

[night sigil]

I just usually find the x intercepts and find the half way point

[rigid smelt]

hmmm that works too but usually it is quite lengthy

so we have a formula which is -b/2a

that is the x-coordinate of the vertex

so you just need to find that number, plug it back into your function

and you will get the maximum value

[night sigil]

that still contains a variable though

[rigid smelt]

a isnt a varible

technically here, a is a constant

thats also the reason why the maximum area has "a" in it

[night sigil]

I also need to do part b of the question anyways

[rigid smelt]

oh alright, but anyway, lets finish this first

so b is part of what we did earlier, well come back to that

[night sigil]

so what next?

[rigid smelt]

plug this into your original equation to find the maximum value

tho you might wanna simplify that first

oh and that isnt right

should be (-6 + sqrt(3))

not -sqrt(3)

you forgot to distribute the - into sqrt(3)

[night sigil]

yep

how can I simplify this?

[rigid smelt]

alright

lets deal with our denominator first

what would our denominator become?

[night sigil]

so is the denominator 4?

no wait

[rigid smelt]

no i meant this

[night sigil]

yeah

[rigid smelt]

yep

so 2 and 4 cancels out

leaving 1/2

times that quantity sqrt(3) - 6

[night sigil]

(-6+sqrt3)/2

[rigid smelt]

yep

so now since 1/2 is in the denominator

basically thats like 1/(1/2)

which is?

[night sigil]

2?

[rigid smelt]

yes

so we can bring that 2 in the denominator to the numerator

so now it becomes (-a/2 * 2)/(sqrt(3) - 6)

[night sigil]

alright

[rigid smelt]

so now 2 cancels each other out

which leaves us with

?

[night sigil]

(-a)/(-6+sqrt3)

[rigid smelt]

yep, but lets just multiply with -1 in the numerator and denominator

so we have a/(6-sqrt(3) )

which looks a bit like what we have in the beginning

[night sigil]

so try put substitute that in as x?

[rigid smelt]

yes

[night sigil]

I can't do it

[ocean seal]

waler

$\frac{2a\frac{a}{6-\sqrt(3))} - (6-\sqrt{3})\left(\frac{a}{6-\sqrt(3))}\right)^2}{4}$

[rigid smelt]

here try this

[night sigil]

ok that's not what I was trying

[rigid smelt]

hmm alright, try again

its easier to plug it in this form

[night sigil]

I have no idea how to do this

[rigid smelt]

$2a\frac{a}{6-\sqrt{3}}$ lets try this first

[ocean seal]

waler

[night sigil]

well that's 2a^2)/(6-sqrt3)

[rigid smelt]

yep

[ocean seal]

waler

[rigid smelt]

how about this

[night sigil]

should you try to square it first?

[rigid smelt]

no dont, you can use the fact that (a/b)^2 = a^2/b^2

[night sigil]

or can you just cancel the (6-sqrt3) out?

[rigid smelt]

yes

so after cancelling you can get a^2/(6-sqrt(3))

[night sigil]

ah yeah

[rigid smelt]

so to sum it all up

[ocean seal]

waler

[night sigil]

yep got that

[rigid smelt]

so now its just easy subtracting fractions

and luckily we already got the common denominator

[night sigil]

(4a^2)/(6-sqrt) right?

[rigid smelt]

ok lets slow down a bit

can you do the numerator first?

[night sigil]

you get a^2)/(6-sqrt3)

[rigid smelt]

yes

[ocean seal]

waler

[night sigil]

yeah

[rigid smelt]

and dividing by 4 is basically multiplying by 1/4

so its just the numerator multiplied by 1/4

[ocean seal]

waler

[night sigil]

so you get

yeah that

[rigid smelt]

yup so that c solved