help-10

[obtuse pebble]

@uncut fulcrum Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• 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.

[gleaming rose]

@drowsy drum im confused on how to get a rainbow to pass through this

[drowsy drum]

what do you mean rainbow pass through it?

[gleaming rose]

it tells me to make a rainbow pass through the prism

and explain why

I GOT IT

AYO

[drowsy drum]

if you slow it down enough, the speeds of all the light separate

[gleaming rose]

thats what occuring?

[drowsy drum]

yes

[gleaming rose]

could you help me with this?

6. Show that you can use Snell’s Law ( n1sinΘ1 = n2sinΘ2) to predict the angle of reflection and angle of refraction for several scenarios. Show your work. After you have completed the calculations, use simulation to check your work. For incident angle of 30 degrees light shining

[drowsy drum]

is that it

for angle 30*

[gleaming rose]

im confused is this two questions lol

[drowsy drum]

it is 1 question with 5 parts

[gleaming rose]

okay how do i do it once

so i can replicate it 5 more times?

[drowsy drum]

alright do you know the refractive index for air and water

[gleaming rose]

yes

air is 1.00

water is 1.33

[drowsy drum]

ok so n1 is the air

and n2 is water

[gleaming rose]

what unit?

[drowsy drum]

doesnt have a unit

so 1 x sin(30) = 1.33 sin (θ)

[gleaming rose]

yes am i solving for theta 2?

[drowsy drum]

the angle of reflection is always the same as angle of incidence

yes

[gleaming rose]

so should i just divide by 1.33

[drowsy drum]

yes then do inverse sine

[gleaming rose]

arcsin

so just repeat for all the n's?

[gleaming rose]

?

[drowsy drum]

,tex $\sin^{-1}$

[warm shale]

suds

[drowsy drum]

because when you divide it by 1.33 it is = sin(θ) not the actual angle

[gleaming rose]

o

so for all the different one just change n right?

for n2?

[obtuse pebble]

@gleaming rose Has your question been resolved?

[drowsy drum]

yes

[obtuse pebble]

Channel closed

Closed due to timeout

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>.
• Type the command .close to free the channel when you're done.
• 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.

[obtuse pebble]

@gusty marsh Has your question been resolved?

[spice citrus]

I think you should extend AB and use the theorem on A @gusty marsh

Add x^2 to both sides yeah

Nice

[spice citrus]

After the third line you can add r^2 to both sides and factor , that makes a bit quicker

The proof is valid though

[obtuse pebble]

@gusty marsh Has your question been resolved?

Channel closed

Closed by @gusty marsh

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• 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.

[long hedge]

how did they get from 1 to 2

[tardy epoch]

Expand (x-1)^2

[long hedge]

?

[unreal musk]

E.g. make (yet another!) substitution $u = \sin(\theta)$ in the indefinite integral $\int \sqrt{1 - u^2} \dd u$

[warm shale]

@unreal musk

[long hedge]

can u show work

[unreal musk]

No u (too lazy to typeset it!)

[long hedge]

then idk

[unreal musk]

Short version is that when you do, du/dt is cos(t) and the integral becomes of cos^2(t) dt, use double angle to rewrite that and you get integral of (cos(2t) + 1)/2

Integrating that gives (sin(2t)/2 * t)/2; double angle on first one and making use of cos(t) = sqrt{1 - u^2} should get it for you

[obtuse pebble]

@long hedge Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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>.
• Type the command .close to free the channel when you're done.
• 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.

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

@timid silo Has your question been resolved?

[remote skiff]

You are just writing the solution to the system in parametric form there, should be easy to google

You can clearly see your first and second columns are not orthogonal. You should review your notes on what to do in that case

You have two eigenvectors v_1 and v_2, they are not orthogonal. Any two linearly independent vectors form a plane

You found two linearly independent eigenvectors corresponding to a single eigenvalue. They are not orthogonal, you need to change them into an orthogonal set using that above procedure

Once you have that set, you normalize that

It doesn't matter what v_1 is, as long as it's a vector in the plane

yes

[obtuse pebble]

@timid silo Has your question been resolved?

[burnt garnet]

hello guys can u help me this?

[obtuse pebble]

@timid silo Has your question been resolved?

Channel closed

Closed by @shell briar

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• 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.

[indigo arrow]

Help.

What do I do with the one?

Do I just ignore it?

Can I just take the x+2 as u?

[rigid shard]

let sqrt[x+2] be t

or smthing

[rigid shard]

i have a feeling if you do that , you would have to do double substitution

[indigo arrow]

I do NOT want to do double substitution. T can be u, right?

[rigid shard]

yes

take sqrt[x+2]

i think i got it

it becomes fairly simple

$2\ln\left(\sqrt{x+2}+1\right)+\dfrac{2\left(x-1\right)\sqrt{x+2}}{3}-x$

[warm shale]

Singularity

[rigid shard]

and + c ofc

[indigo arrow]

Yeah, your answer's right. But how did you do it?

[rigid shard]

$taking \left(\sqrt{x+2}\right as u$

$taking \left(\sqrt{x+2}\right) as u$

[warm shale]

Singularity

[rigid shard]

wait lemme latex full

nah imma just send a pic

ok so i took 1 + sqrt[x+2] as u nd it becomes simpler

@indigo arrow

i left the expression there cause i think it become fairly linear after that

[indigo arrow]

OH okay so 1+sqrt(x+2) is u!

I got it!

Thank youuu

.close

[obtuse pebble]

Channel closed

Closed by @indigo arrow

Use .reopen if this was a mistake.

[rigid shard]

glad to help

[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>.
• Type the command .close to free the channel when you're done.
• 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.

[keen tinsel]

i have this question but I dont know how to do parts b and c, but i got the answer for part a correct

[obtuse pebble]

@keen tinsel Has your question been resolved?

[keen tinsel]

<@&286206848099549185>

[obtuse pebble]

@keen tinsel Has your question been resolved?

[drifting sun]

First do you understand the concept of what is going on here?

Why the formula is $\int_{a}^{b}\pi [R(x)]^2 dx$

[warm shale]

Blue Guilmon

[drifting sun]

For the disk method

[keen tinsel]

i did cylindrical shell for a

i got 4

[drifting sun]

It should be $2\pi$ if my mental math checks out, what's your work

[warm shale]

Blue Guilmon

[keen tinsel]

$\int_{0}^{2}\x [x^2] dx$

[warm shale]

mr man
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[keen tinsel]

wait

that times x

idk how to use the command

[drifting sun]

Just removed the \

[keen tinsel]

$\int_{0}^{2}x [x^2] dx$

[warm shale]

mr man

[keen tinsel]

yeah that was what i solved

so integral of x^3 from 0 to 2

[drifting sun]

This should have a $2\pi$ if you're doing cylindrical method

[warm shale]

Blue Guilmon

[drifting sun]

First off

[keen tinsel]

oh i forgot

[drifting sun]

Secondly the volume you're computing is incorrect

It's actually everything outside your shape

Because the height dx here is vertical, it doesn't touch the y axis

The disk method is much more straightforward in this case, I would only use the cylindrical method if the integration is too difficult as an alternative or if you're revolving about a different line that isn't the axis

[keen tinsel]

ok

[drifting sun]

But do you see what I meant by it being outside

[keen tinsel]

yeah

because of the gap between the curve and the x axis

[drifting sun]

Essentially the shells you're computing with that formula has a radius of x, a width of dx, and a height underneath the function which when you revolve around y computes the area outside

Or volume outside

If you want the volume inside with the shell method you would need to compute the difference of the line y=2 and x^2

It's easier to just use the disk and integrate dy

Which would be $\int_{0}^{4}\pi(\sqrt{y})^2 dy$ I believe

0 to 4 because 0 to 2 in x is equivalent to 0 to 4 in y

[warm shale]

Blue Guilmon

[drifting sun]

Fixed the typo

[keen tinsel]

that gave me the same answer as the shell method after i added the 2pi

which was 8pi

is that what u got?

[drifting sun]

I fixed the bounds

One sec

First time I did 0 to 2 in y on accident I believe

Yeah 8po

8pi

[keen tinsel]

ok and then my friend helped me with b a couple mins ago and he said to do distance x mass x acceleration

so i did 2 x 8pi x 2 and got 32pi

is that right

[drifting sun]

Work is the integral of force I believe

[keen tinsel]

i thought work was force x distance

[drifting sun]

Force is mass times accelerate

Yeah integrated over distance

Or time

It's been a bit since I did this and depending on where you are in calculus it is a bit different

[keen tinsel]

im in calc 2

[drifting sun]

Yeah it's the integral over the displacement

Or over distance just checked

[keen tinsel]

so should it be $\int_{0}^{2} [16pi] dx$

[drifting sun]

But I think in this case we can assume simple

[warm shale]

mr man

[drifting sun]

What are you taking as the distance and how/why

[keen tinsel]

i just did the original bounds of the function because i wasnt sure

[drifting sun]

Yeah tbh I'm not sure either

It's being dumped out I assume gravity is the thing doing the work, now how far down it's dumped is a big unspecified amount

I would need to review how calc books ask this since I haven't seen a problem written this way for this

But what your friend said is correct

That is work

I assume the mass you can take to be the volume of water times its density

Volume you computed

[keen tinsel]

isnt the density of water 1

[drifting sun]

Approx ye

[keen tinsel]

ok so i should just be able to plug the volume in there right

for the mass part

[drifting sun]

Just wasn't sure how accurate your prof wanted

Some people use 0.997

[keen tinsel]

he used 1 in the lectures im p sure

so i will just go with that

[drifting sun]

Kk

[keen tinsel]

and then for acceleration i just do second deriv of x^2 right

[drifting sun]

I don't think so, the thing accelerating mass on earth is gravity

X^2 simply represents the walls of the container

It has little effect on the acceleration by gravity

[keen tinsel]

oh i guess since it isnt a moving object doing that wouldnt make sense

so 9.8 for acceleration

[drifting sun]

Did you get part x

C*

Yesh

Yeah*

[keen tinsel]

i tried doing part c im not sure if correct tho

iu dont think i did it correct

[drifting sun]

It should be $\int_{0}^{4}2\pi(\sqrt{y})\sqrt{1+\frac{1}{4y}} dx$

If I did that correctly

[warm shale]

Blue Guilmon

[keen tinsel]

ok thats what i wrote

[drifting sun]

Use a numerical method for thus

This*

For sure

[keen tinsel]

also the answer for b would be found with 2 x 8pi x 9.8 right

[drifting sun]

Yes if you assume the distance is 2

[keen tinsel]

alright

ok for c i just put the integral into a calculator because my professor told us to do that

could you help me with another question?

[drifting sun]

Sure ask it, I am actually preparing to give a lecture soon if I have to drop out I'll let you know but others here should be able to as well

[keen tinsel]

alright thanks so much

https://cdn.discordapp.com/attachments/912855332110356480/1176200448366092409/image.png?ex=656e0107&is=655b8c07&hm=4ee47d7f6b7172c223ac538761dedba3bcd8a5204c27b105d8cbaae175580686&

[drifting sun]

What's your question here, simply the whole thing?

[keen tinsel]

yeah i dont really understand the hydrostatic force part

[drifting sun]

That's more a matter of physics, lemme see

Fluid pressure times surface area it seems

Pressure is given by $P= \rho g h$ for a depth $h$ under the water

[warm shale]

Blue Guilmon

[drifting sun]

Rho is the density of water so, 1 here

g is the gravitational constant

So the pressure on the plate is changing the lower we go, which makes sense

[keen tinsel]

depth would be arccos x?

[drifting sun]

Depth is just y here

[keen tinsel]

oh ok

[drifting sun]

Depth under the water

Well sort of

It's dependent on y but it isn't aligned with the y axis

X-axis I mean

Try to set it up

You first need to find the surface area of the plate which, we can just assume it is infinitely thin

Hence it's just the area bounded there

[keen tinsel]

surface area is arccos x?

from whatever to whatever

[drifting sun]

Yes

May be better to integrate it with respect to y

It shouldn't matter

Arccos has problems, cos doesn't

So I would do with respect to y

[keen tinsel]

ok

sin 3

[drifting sun]

So the surface area isn't changing

What is changing is the pressure as we go deeper

So your hydrostatic force would he integrating over this depth

In general you can think of

The integral as a glorified infinitesimally added summation

And so it's uses become natural to generalize in this way with handwavey arguments

In this case we would need to add up the pressure at every depth across the whole surface area of that infinitely small segment

Etc.

So we're integrating over depth is tldr

[keen tinsel]

ok

so it would be like

the integral of the pressure?

from 0 to 3

[drifting sun]

Pressure with respect to height times surface area

So you can choose to put the surface area inside or outside up to you

[keen tinsel]

wait so for the depth do i just put y

and then density is 1 and then multiply that by 9.8?

[drifting sun]

One sec won't have internet briefly since on the phone and a dead zone

Brb in like 5 minutes

You're integrating from 0 to pi/2

Because in the problem they said it is submerged to the tip, and we can reason out since the function is cosine, the tip occurs when cosine is 0 for the first time

Hence pi/2

[keen tinsel]

oh u right

so the area would be 1

cuz integral of cos y is sin y and then sin y from 0 to pi/2 is 1-0

[drifting sun]

Yes

Lemme look at the rest now that I can better

Ok so how would you represent your depth starting from pi/2 going down to 0 in y?

[keen tinsel]

would it be pi/2?

[drifting sun]

I would write it as pi/2-y, why? Because at y=pi/2 you are at the surface of the water, so depth 0 and pi/2-pi/2=0

At y=0 you are at depth pi/2 which also adds up here

This is your "h" value representing your depth

[keen tinsel]

got it

[drifting sun]

We have to integrate over this, and h is a function of y

h(y)=pi/2-y

[keen tinsel]

so pressure = (pi/2 - y) x 9.8?

[drifting sun]

Hence your hydrostatic force should be $\int_{0}^{\frac{\pi}{2}}(1)g(\frac{\pi}{2}-y)\cos{(y)}dy$

If I'm correct

The reason why I kept the surface area inside is because the force experienced at each depth is the surface area of that infinitely small rectangle under the curve at that depth times the depth, g, and density of water

So I think it has to be inside because the area of this infinitely small rectangle changes depending on where you are under the curve

If that makes sense to you

[keen tinsel]

yeah

[drifting sun]

This is where I got this formula from

Btw

But I just translated it from averages to the limiting process of this thing which is integrals

[warm shale]

Blue Guilmon

[drifting sun]

I fixed a typo in the bounds

I probably have to go now so I hope that answered all your questions

You will likely need to do integration by parts for this integral btw

[keen tinsel]

yeah i just have to solve that integral and im good right

[drifting sun]

Cause you're undoing the product rule

[keen tinsel]

oh also for part b of the first question the unit is J right

[drifting sun]

Joules? Yes

[keen tinsel]

ok thank you so much for the help

[drifting sun]

Joules is newtons over distance

Hence work yeah

[keen tinsel]

.close

[obtuse pebble]

Channel closed

Closed by @keen tinsel

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• 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.

[timid silo]

hey all. how to solve this integral? i do if all of the denominator would be on top but i cannot solve this.

[fickle oracle]

use trigonometric identities to simplify the expression

[timid silo]

i have come from cot csc sec to this and hope it would be much more easy 😭

[fickle oracle]

hmm

use this i would say sin(2x)=2sin(x)cos(x)

[timid silo]

i tried to use this with sin.cos = u but it terrified me

[fickle oracle]

btw sqrt(sin(theta)) * sqrt(cos(theta)) = sqrt(sin(theta) * cos(theta))

[timid silo]

i could find cos(2x) instead of sin(2x) due to cos diff = -sin

[fickle oracle]

it dosent really matter

after all that maybe think about substituting:
u = sqrt(sin(theta) * cos(theta))
then find du

[timid silo]

trying..

ok thanks

[fickle oracle]

got it?

[timid silo]

no

[fickle oracle]

where u at?

[timid silo]

thanks anyways

[fickle oracle]

talk haha

show where u get

[timid silo]

du = right part

right part comes very hard to rearrange

[fickle oracle]

wait

i will show you where i got

before the substituting

if u follow the steps u should get something like this:

dtheta / [sqrt(2sin(theta)cos(theta))) * sin(theta)cos(theta) ]

then if you substitute :
u = sqrt(sin(theta) * cos(theta))
and find du u will get something that is easy to calc

[obtuse pebble]

@timid silo Has your question been resolved?

Channel closed

Closed by @toxic parcel

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• 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.

[waxen raft]

Can someone help me solve thiss?

[pallid aurora]

maybe you could use mods?

[tardy epoch]

did you try induction?

[spice citrus]

I think you can pair up terms

1^k + 500^k

2^k + 499^k

Etc..

[pallid aurora]

yeah that’s what i was thinking but with mods
1^k + (-1)^k
2^k + (-2)^k
etc

[spice citrus]

Yeah, same thing. You can factor as well

[obtuse pebble]

@waxen raft Has your question been resolved?

[waxen raft]

thank youu, got it

[obtuse pebble]

Channel closed

Closed by @waxen raft

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• 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.

[timid silo]

how can you tell if there is only one solution to this problem?

[dark stirrup]

Solving it is one way.

Determinants are another way if you just want to know if exactly one solution exists.

[timid silo]

hmm true that

interesting

ok thank you

.close

[obtuse pebble]

Channel closed

Closed by @timid shadow

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• 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.

[merry wave]

ii did i convert ml to l correctly

[obtuse pebble]

@merry wave Has your question been resolved?

[obtuse pebble]

@merry wave Has your question been resolved?

[merry wave]

.CLOSE

/CLOSE

.close

[obtuse pebble]

Channel closed

Closed by @merry wave

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• 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.

[graceful stirrup]

Can anyone explain why we multiply by the interval by 2 here and how the solutions are obtained? (Apart from looking at the graph of sin x) Also, is there a better way of doing this?

[latent walrus]

because the argument of your cos is 2x, not x

so youre solving the equation for 2x

then they divide by 2 to get x

[graceful stirrup]

ok i understand that

how are we coming to the solutions in this case

[latent walrus]

by solving sin(2x)=0

so 2x must be an integer multiple of pi

[graceful stirrup]

alright just one more thing

here's a similar example

so i solved cos 2x = 0

[latent walrus]

yeah

[graceful stirrup]

and obtained pi/4

and then just adding pi and subtracting pi within my interval

would that be an efficient method to solve something like this

[latent walrus]

i guess that works here
but more generally its knowing how the trig functions work, mainly things like symmetry
if you were solving cos(2x)=0.5 for example, you couldnt just add and subtract pi and get your solutions

[graceful stirrup]

right

if you wouldnt mind

could you take me through how you would approach cos(2x)=0.5

i just cant really grasp this method

nevermind i think i got it

.close

[obtuse pebble]

Channel closed

Closed by @graceful stirrup

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• 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.

[viscid cosmos]

Hi, can anyone help me with this hw question? We just started this topic and I’m not rlly sure how to solve. Thanks.

[viscid cosmos]

This was my attempt on solving it. Not sure if it’s correct

[remote skiff]

Your step 3 looks wrong at least

it is -1/(2ysqrt(y)) * dy/dx

I don't know how you ended up at 1/2sqrty there, but that's the derivative of sqrt(y), not 1/sqrt(y)

[obtuse pebble]

@viscid cosmos Has your question been resolved?

[viscid cosmos]

Final answer should be negative ysqrt(y)/xsqrt(x) correct?

went and resolved step 3 because yea I missed that one bit

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• 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.

[fading bough]

Is there such thing as being a help hog?

[fading bough]

5/6 x 2/15 x 20/40 simplifys down to 1/18 right?

[candid rock]

there's such thing as not wanting to put in the work yourself and only wanting an answer

but no, if you're putting in the effort and do need help explaining, then there is no such thing as a help hog

[fading bough]

My book is saying its 5/54 and I've done it 3 different ways and it comes out 1/18 every time.

[candid rock]

so you can multiply all the numerators and multiply all the denominators

[subtle sinew]

,calc 5/6 * 2/15 * 20/40

[warm shale]

Result:

0.055555555555556

[subtle sinew]

Looks like 1/18

So you did it right

[fading bough]

I'M NOT CRAZY

No, but I did write the problem down wrong in the first place............

[candid rock]

20/40 is 1/2
and 5*2 = 10
6*15 = 90
so those first two are 10/90 which is 1/9
halve that and you get 1/18

[fading bough]

UGH

[fading bough]

Right thats what I was doing. At least I know Im doing it correctly now

Imma take a break. Thanks guys!

.close

[obtuse pebble]

Channel closed

Closed by @fading bough

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>.
• Type the command .close to free the channel when you're done.
• 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.

[amber edge]

Can somebody please help me solving this:
Let ∼ be an equivalence relation on a set M, and let R ⊆ M. We call R a complete representative system with respect to ∼ if for every equivalence class [m] ∈ M/∼, there exists exactly one r ∈ R such that [r] = [m].

Decide whether the given sets form complete representative systems with respect to the equivalence relation a ∼ b :⇔ 7 | (a − b). Justify.
(a) {−6, 9, 10, 11, 12, 20}
(b) {0,7,14,21,28,35,42}
(c) {0,1,2,3,4,5,6}
(d) {−3,−1,1,4,5,7,9}
(e) {−9,−4,0,1,4,13,16}"

[obtuse pebble]

Please don't occupy multiple help channels.

[waxen raft]

I really want to know the answer to this question

[obtuse pebble]

@amber edge Has your question been resolved?

[waxen raft]

<@&286206848099549185>

[obtuse pebble]

@amber edge Has your question been resolved?

[lone dirge]

@waxen raft you just need all the remainders mod 7

@waxen raft because 7|(a-b) is equivalent to a = b(mod 7)

(a) is an anwser, (b) isn't since 0=7 mod 7, (c) is an answer, (d) isn't since -3 = 4 (mod 7), (e) is an answer

@amber edge

[amber edge]

Thank you. You really safed us the evening!

.close

[obtuse pebble]

Channel closed

Closed by @amber edge

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• 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.

[slim whale]

if Un = sum from k = 0 to n of (minus 1) to the power K / K! then U2n is ?

[latent walrus]

$U_n=\sum_{k=0}^{n}\frac{(-1)^{k}}{k!}}$

[slim whale]

no

no

[latent walrus]

oh

[slim whale]

minus one power k

[warm shale]

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

[latent walrus]

that?

[slim whale]

yessssss

i thought of replacing n by 2n

but what abt k ?

how are they gonna know if they should move with 0 1 2 till 2n or 0 2 4 till 2n

[latent walrus]

the index will always go up by 1

[slim whale]

so it should b 2k

[latent walrus]

nope

[slim whale]

u said the index goes by 1

[latent walrus]

if it really only asks for an expression for 2n

then just replace n with 2n

[slim whale]

the xo only asks for that

but i wanna understand

[latent walrus]

the only thing 'n' affects is the upper limit, not the expression within the sum itself

[slim whale]

if u give me that new expression and u ask me to calculate first 3 un

[slim whale]

how so?

[latent walrus]

because thats how you defined U_n

if it was U_3, replace n with 3

[slim whale]

but we looking for U2n now

[latent walrus]

if its U_2n replace n with 2n

[slim whale]

and how does the k works in this case?

[latent walrus]

the k will just index from 0 up to the upper limit by 1 at a time

if its n you have k=0,1,2,...,n-1,n

if its 2n you have k=0,1,2,...,n-1,n,n+1,...,2n-1,2n

[slim whale]

not logical at all

ah actually it is

but what abt if we replace n by 2n

then k= 0 2 4 6 .... 2n ?

[latent walrus]

no

[slim whale]

why not

[latent walrus]

because sums index increase by 1 per term

the upper limit has no bearing on that fact

[latent walrus]

youd just have this

[slim whale]

just like the n+1 case?

for xmp

if u ask to write un+1

its the same thing as un + ( minus one to the power of n+1 / (n+1)!

[latent walrus]

it would be, yes

[slim whale]

how abt for u2n?

un + minus 1 to the p 2n ....?

[latent walrus]

this is the same as $U_{n+1}=\sum_{k=0}^{n+1}\frac{(-1)^{k}}{k!}}$

uups

[warm shale]

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

[latent walrus]

you just replace n with the new thing

[latent walrus]

no

[slim whale]

k doesn't change

[latent walrus]

k is an index value

[slim whale]

aaaah + n

+n not 2n

since 2n is n + n

get me?

[latent walrus]

what are you saying u_2n could be?

[slim whale]

yeah

[latent walrus]

im asking what you mean specifically

if youre saying U_2n=2U_n

then no

[slim whale]

no

its the sum of un

from k = to n

this alone

PLUS

minus 1 to the power of n ......

how do u do to write the expression like in here

[slim whale]

.

[latent walrus]

just use this format

ill write it here as a template, one sec

[slim whale]

ok

[latent walrus]

U_{n}=\sum_{k=lower lim}^{upper lim}\frac{(-1)^{k}}{k!}}

just fill in your stuff

put $ $ around it

[slim whale]

$U_{2n}=\sum_{k=0}^{n}\frac{(-1)^{k}}{k!}} + \frac{(-1)^{n}{n!}$

[warm shale]

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

[latent walrus]

$U_{2n}=\sum_{k=0}^{n}\frac{(-1)^{k}}{k!}} + \frac{(-1)^{n}}{n!}$

[warm shale]

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

[slim whale]

exact, thanks

[latent walrus]

that is wrong

you just added the last term of the sum again

[slim whale]

you're right

[latent walrus]

if anything, i can write it like this:

$U{2n}=\sum_{k=0}^{n}\frac{(-1)^{k}}{k!}}+\sum_{k=n+1}^{2n}\frac{(-1)^{k}}{k!}}$

[warm shale]

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

[latent walrus]

which is just$$\sum_{k=0}^{2n}\frac{(-1)^{k}}{k!}}$$

[warm shale]

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

[slim whale]

from n+1 to 2n?

[latent walrus]

yes

[slim whale]

yeah you're right

but

1 2 3 4 ... n-1, n, n+1, n+2,.....n+n (aka 2n)

[latent walrus]

yeah?

[slim whale]

this is the process right?

[latent walrus]

i see no helpful reason to write 2n as n+n, but sure ill go with it

[slim whale]

i just wanna understand from where did it come from

the 2n

[latent walrus]

it was given to you

[slim whale]

since here there's a summation pattern

[latent walrus]

thats about it

[slim whale]

je sais

[latent walrus]

if i had the sum of x, from 0 to n
if n was 3 then its 0+1+2+3
if i do 2n then since n is 3, it becomes upper limit 6, 0+1+2+3+4+5+6

very literally youre just replacing n with 2n

and the terms just index up to 2n instead of n

[latent walrus]

you can split it like this but theres no need to

[latent walrus]

its just a substitution if anything

[slim whale]

i get what you're talking abt its simple

but ok maybe i just have to think abt it

[slim whale]

okay

so if

he says calculate u2n where n is 10 it means i calculate u0 u1 ... u20

talking abt the sum obv

[latent walrus]

no

i disagree only because youre writing u0 u1 ...

[slim whale]

ik i wrote the syntax incorrectly

[slim whale]

yeah

i know

[latent walrus]

but it would index up to 20, yeah

[slim whale]

okay got it

i have another question

[latent walrus]

shoot

[slim whale]

i have another un

its a fraction

1.3.5....(2n-1) sur 2 4 6 .... 2n

[latent walrus]

im not too sure what you mean

[slim whale]

Un=(1 * 3 * 5 .....*(2n-1)) /( 2 4 ....*2n)

• = x

[latent walrus]

is the upper limit supposed to be 2n-1

[slim whale]

yes

[latent walrus]

it wouldnt be indexing by 2's

and youre writing products

so im confused

[slim whale]

better?

[latent walrus]

not really no one sec let me try write what i believe youre saying

correct me if im wrong

[slim whale]

okay

[latent walrus]

$\sum\frac{1\cdot3\cdot5\cdot...\cdot(2n-1)}{2\cdot4\cdot...\cdot2n}$

this?

[warm shale]

AℤØ

[slim whale]

exact

they want u2n

[latent walrus]

wait, so should i remove the sum? and just say this:

$U_n=\frac{1\cdot3\cdot5\cdot...\cdot(2n-1)}{2\cdot4\cdot...\cdot2n}$

[slim whale]

ah wait there's no sum yes

[warm shale]

AℤØ

[slim whale]

exact

[latent walrus]

yeah that makes more sense

[slim whale]

yeah so we can see that first part of the fraction is the inverse of the second

my explanation is bad

[latent walrus]

i wouldnt say inverse

[slim whale]

im not native

[latent walrus]

but one is the product of odds, the other is evens yeah

[slim whale]

yeah

was there 2n - 1 before 2n?

in the second line

second part of the fraction

[latent walrus]

no

there would have been a 2n-2

[slim whale]

bec its not a sum

?

[latent walrus]

i can extend it slightly if it helps, one minute

[slim whale]

ohh makes sendse

since we can do 2( 1 2 3 ... n)

[latent walrus]

$U_n=\frac{1\cdot3\cdot5\cdot...\cdot(2n-3)\cdot(2n-1)}{2\cdot4\cdot...\cdot(2n-2)\cdot2n}$

[warm shale]

AℤØ

[latent walrus]

yeah that works

[slim whale]

should i just replace n by 2n

i can't siplify it after doing that?

[latent walrus]

pretty much

not really no

you could make it longer but theres no real need to

[slim whale]

exact

it appeared weird to me why our teacher would ask such questions

thats why i thought maybe there were a piege

we're studying real sequences whats could the point be

[latent walrus]

$U_n=\frac{1\cdot3\cdot5\cdot...\cdot(2n-3)\cdot(2n-1)\cdot(2n+1)\cdot...\cdot(4n-3)\cdot(4n-1)}{2\cdot4\cdot...\cdot(2n-2)\cdot 2n\cdot (2n+2) \cdot...\cdot(4n-2)\cdot (4n)}$

[warm shale]

AℤØ

[latent walrus]

is a dragged out version

[slim whale]

of finding expressio of u2n

[latent walrus]

$U_{2n}=\frac{1\cdot3\cdot5\cdot...\cdot(4n-1)}{2\cdot4\cdot...\cdot4n}$

[warm shale]

AℤØ

[latent walrus]

but this would be enough as you said

[slim whale]

yes, so i can't simplify it or use the product notation?

[latent walrus]

you can use product notation if you want

[slim whale]

but for what purpose

[latent walrus]

just to make it look neater

[slim whale]

hum okay

oh wait

i have another question

can a sequence be defined where n belong to N minus 2?

like u0 u 1 exists u2 doesn't then u3 u 4 ... un exists

[latent walrus]

i guess thats possible
if the expression was U_n= 1/(n-2), all terms except U_2 exist ig

[slim whale]

yeah

thats what i mean

so we can

we don't have to say n>2

[latent walrus]

i think so, im not 100% but i think its alright

[slim whale]

okay

[latent walrus]

hmm

actually it may not be valid

one min just doing some research

[slim whale]

its up to def of a sequence

to decide if its valid or not

[slim whale]

okay

[latent walrus]

i think its invalid

sequences require the natural numbers to be their domain

if you exclude one or more its no longer valid

at least thats the general idea im getting

maybe saying n>2 would allow for its validity

im not so certain, but i would think having undefined terms is a nono

[slim whale]

same

[slim whale]

yeah makes more sense

im just comfy with this more

but idk why

a sequence is an application right

an appli of a mapping

[latent walrus]

yeah it maps natural numbers to some other set

[slim whale]

R

right

[latent walrus]

could be R, could be C, Q, etc

[slim whale]

ah true

ok one last q

[slim whale]

remember this

[latent walrus]

yeah

[slim whale]

u0 is undefined

right

[latent walrus]

i can argue 0 isnt a natural number

[slim whale]

its a whole number?

in the french system they accept it as a natural number

u know

or nvm

[latent walrus]

fair enough, if you did u_0 would be 1/2 since thats the first term of the product though

[slim whale]

can we continue tmrw, i have to sleep

[latent walrus]

sure np

[slim whale]

arent we supposed to put 0 in the place of n

[slim whale]

how can i find u tmrw?

[latent walrus]

just ping me, i might appear, our time zones are similar i believe anyway

[slim whale]

its 2 12 am her

e

[latent walrus]

im an hour behind

,ti

[warm shale]

The current time for aldrnari_ is 01:13 AM (GMT) on Tue, 21/11/2023.

[slim whale]

cool then

did i pin u ?

g

[latent walrus]

sure

i wont be free until around 5pm in your time, then ill be busy from 8 onwards ish

but if i see it on my phone while im out, ill try reply

[slim whale]

this is so sweet of u

that would be much appreciated

thanks again

[latent walrus]

no worries

[slim whale]

goodnight

[latent walrus]

gn

[obtuse pebble]

@slim whale Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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>.
• Type the command .close to free the channel when you're done.
• 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.

[astral ivy]

How do i find an area of a triangle?

[astral ivy]

I can not show my work for this question i do not think-.

[frosty peak]

A=1/2(bh)

[astral ivy]

Ok thank you!

[frosty peak]

no worries 🙂

[astral ivy]

How do i find the height of a triangle?

[unreal musk]

what triangle are you dealing with? do you have a particular problem you're working on?

[astral ivy]

The triangle I am dealing with is an isoceles triangle. Yes, I do and it is to find the area of the triangle!

:>

@unreal musk

[unreal musk]

Well if it's isosceles, you can "fold it in half" and then you have a right angle triangle

For which one of the sides forms the height

[astral ivy]

How do i find the height of a right angle triangle?

oh

ok

[astral ivy]

The left side!

<--

@unreal musk

[unreal musk]

Pythagoras!

[astral ivy]

Ohhh ok! Tyy

:0

But it has coordinates.

I'm used to numbers.

[unreal musk]

depeneds on how you've "folded" it ofc - not the hypotenuse

[astral ivy]

0-0

[astral ivy]

Oh I folded it in my mind!!

lol

[unreal musk]

From coordinates, you can find numbers

[astral ivy]

OHHH

ok ty

I solved it and I got 8281, but it is none of the answer choices and I doubled checked my work.

@unreal musk

[unreal musk]

Can you show the question and your work please

[astral ivy]

Idk how to take a picture and send it to you because i do not have Discord on my phone but instead my computer.

[unreal musk]

are you able to like send pictures from the phone to the computer?

[astral ivy]

i lost braincells.

0-0

Well idk how.

Do you know how?

@unreal musk

[unreal musk]

Depends on what phone and computer you have I think

Are you not able to get Discord on your phone?

[astral ivy]

I can but i forgot my password-.

;-;

[unreal musk]

Oh, pain ermmm then, maybe email the pictures to yourself from the phone or something?

Just that seeing the question and what you've done make it easier to see how to go about it and all

[astral ivy]

I don't think I can email myself. I rlly wanna show you so bad.

@unreal musk

[unreal musk]

maybe try typing out as much of the question as you can then, and then briefly explain the steps you did?

[astral ivy]

Ohh ok!

The question is "What is the area of the triangle?".

The steps I took were:

1st: I wrote the formula A= 1/2 (bh).

2nd: I counted the base of the triangle.

3rd: I did the Pythagorean Theorem and found my height which was 26.

4th: I multipled seven and twenty-six.

5th: I multiplied 1/2 and 182.

[unreal musk]

sorry to interrupt you but what's the triangle? You mentioned there were coordinates, what were they?

[astral ivy]

Then that's how i got my answer, 8281!

[unreal musk]

Also 1/2 * 182 isn't 8281

,calc 182/2

[warm shale]

Result:

91

[astral ivy]

The triangle is an Isoceles triangle. The coordinates are for A (1,4) B (4,1) and lastly C (9-3,0).

Oh wait sorry i got 91.

not 8281.

AAAAAAAAAAAAAA

I'm internally screaming inside.

My inner child is crying.

[unreal musk]

Awww

[astral ivy]

lol

[unreal musk]

Wait did they give the last one as 9 - 3, or?

[astral ivy]

Oops I meant (9, -3,0)

0-0

[unreal musk]

is the triangle in 3d or 2d?

[astral ivy]

What is 3d?

and 2d?

@unreal musk

[unreal musk]

The first two ones you gave are like two coordinates, (1, 4) and (4, 1), whereas the last one you gave has three entries, (9, -3, 0)

Are there entries missing from the first two?

[astral ivy]

Oops I meant for B it is (-3, 0).

I typed so fast because I was panicking!!

@unreal musk

[unreal musk]

there's no need to panic here

[astral ivy]

ty

: )

I'm more calm! c:

[unreal musk]

But could you double check the points again and write them exactly as they appear in the question? Take your time, it's important we get them correctly as then I can double check what you have

[astral ivy]

ok!

The A coordinate is (1, 4). the B coordinate is (4, 1), and lastly the C coordinate is (-3, 0)!

I double checked it this time!! c:

@unreal musk

[unreal musk]

nice, thank you

[astral ivy]

np!!

[unreal musk]

Give me a second

[astral ivy]

ok

https://tenor.com/view/fknbnuy-funny-dog-waiting-patient-gif-21698298

[unreal musk]

are you sure the triangle they mentioned was supposed to be isosceles? Cause I'm not getting that it is

[astral ivy]

I am not sure actually.

I think it is a normal triangle.

@unreal musk

[unreal musk]

\catthink if it's a normal triangle, then there's a way you can find the area of a triangle given two sides and the angle between them, $A = \frac12 a b \sin(C)$

[warm shale]

@unreal musk

[unreal musk]

If you're familiar with that?

[astral ivy]

Nope, never seen it in my life!!

@unreal musk

[unreal musk]

Hmmm, I don't know then

[astral ivy]

D:

[unreal musk]

I'm assuming that it was supposed to actually be isosceles but idk

[astral ivy]

Oh, what do i do then?

T-T

[unreal musk]

It's really hard to know what to suggest without a picture of the question and what they've said

[astral ivy]

:c

[unreal musk]

do you think you'd be able to reset your password or something so that you could then get Discord on your phone?

[astral ivy]

Sure

[unreal musk]

as long as you have access to the email that you made this account with!

[astral ivy]

It's downloading rn!!!

Yep I do have access to the email!! < 3

Ok downloaded it!

@unreal musk

[unreal musk]

cool cool, try and see if you can get a picture of it now

Let me know when you have (or if you struggle!)

[obtuse pebble]

@astral ivy Has your question been resolved?

[astral ivy]

Omg finally, I'm on my phone right now and it took forever!!

@unreal musk

[unreal musk]

[astral ivy]

: D

[unreal musk]

respond with ❌ btw

[astral ivy]

mk

c:

[unreal musk]

What did you get as the side lengths of that triangle?

[astral ivy]

Wdym?

@unreal musk

[unreal musk]

The triangle has side lengths - how did you get a base of 7, for example?

[astral ivy]

I got the base of 7 by counting the squares!

:>

@unreal musk

[unreal musk]

I mean, you have to be a tiny bit careful as it isn't entirely accurate

[astral ivy]

ok

How do I come up to my base?

:>

[unreal musk]

I'll give one example, the length of the side between $(-3, 0)$ and $(4, 1)$ is $\sqrt{(4 - (-3))^2 + (1 - 0)^2} = \sqrt{50} = 5\sqrt{2}$

[warm shale]

@unreal musk

[unreal musk]

Have you seen how to work out distances between points like that?

[astral ivy]

Nope, my teachers never taught me if!!

it*

[unreal musk]

Hmm, well if they haven't, then you can't really do this question then

[astral ivy]

.

[unreal musk]

Ask your teacher about the distance between two points formula and if they covered it before

[astral ivy]

DUN DUN dun.

ok

[unreal musk]

Cause you can't find the accurate distance between those points otherwise

[astral ivy]

Oh

I also need help on another problem!!

How do I find how much salad was sold on that day?

@unreal musk

[unreal musk]

Salad?

[astral ivy]

I will take a picture of it! : )

@unreal musk

[unreal musk]

Simultaneous equations

[astral ivy]

-gasp-

[unreal musk]

Let s be the number of salads you sold, and d be the amount of drinks you sold

Can you form two equations from that?

[astral ivy]

Sure!

6.50s+2.00d=836.50

@unreal musk

[unreal musk]

Nice, that's one and the other one is?

[astral ivy]

My brain just sparked another brain cell! :0

836.50-6.50s=2.00d

@unreal musk

[unreal musk]

That's basically the same thing...

[astral ivy]

oop-

[unreal musk]

Remember that 209 salads and drinks were sold...

[astral ivy]

6.50s x 2.00d=836.50

[astral ivy]

: D

oh no

0-0

[unreal musk]

Not quite right

[astral ivy]

mk

6.50s+2.00d=x

: D

[unreal musk]

nope, not that...

[astral ivy]

oop!

0-0

-thinks-

[unreal musk]

See again, 209 salads and drinks

You have s salads and d drinks

[astral ivy]

s+d= 209

or s x d= 209

@unreal musk

[unreal musk]

that's it

[astral ivy]

YESSSS

: D

[unreal musk]

So now you have $s + d = 209$ and $6.5s + 2d = 836.5$

[warm shale]

@unreal musk

[unreal musk]

Can you solve those?

[astral ivy]

Sure! : D

I got 128.15!

;-;

@unreal musk

[unreal musk]

how did you get that?

[astral ivy]

By dividing 8363 by 6.5.

[unreal musk]

Show your working out! You have your phone now, there's no excuse

[astral ivy]

XDD

c:

I worked very hard. UnU

https://tenor.com/view/backpack-kid-backpack-kid-dance-floss-floss-dance-gif-18264451

[unreal musk]

I can barely make the picture out

[astral ivy]

oop

Let me take the picture again then!

[unreal musk]

nice and clear! But this here is very very illegal

First off...

,calc 836.5 - 2

[warm shale]

Result:

834.5

[unreal musk]

Second off, you can't remove that 2d by subtracting just 2, you need to subtract 2d

[astral ivy]

OHHH

[unreal musk]

There's a reason why you have the two equations

[astral ivy]

OH

128.1320 is my answer ms.teacher!

c:

I worked very hard!!

@unreal musk

[unreal musk]

show your work again, I can tell you've done something illegal

[astral ivy]

;-;

[unreal musk]

[unreal musk]

How did you learn how to solve simeultaneous equations?

[astral ivy]

From years ago

from a teacher

:>

in school

in 6th grade

@unreal musk

[unreal musk]

Might be a very good idea to go over them again, espeically if it's been some time, just so you can get used to doing them!

[astral ivy]

ok! c:

Well did I commit anything illegal by my working?

@unreal musk

[unreal musk]

If you've done what you think you have, yes, straight to jail...

[astral ivy]

Wait what-?

[unreal musk]

See here, you didn't get rid of the 2d

[astral ivy]

0-0

[astral ivy]

I did D:

[unreal musk]

Well not legally you didn't, which is why you're wanted

[astral ivy]

I-.

[unreal musk]

Seeing your working out seems like you only corrected the right hand side of this, but not the left hand side

[astral ivy]

I lost brain cells.

: D

[unreal musk]

The left hand side is not 6.5s but 6.5s + 2d - 2

[astral ivy]

i got 192,251!

c:

@unreal musk

[unreal musk]

[astral ivy]

c:

[unreal musk]

Maybe think about the steps that you need to do to solve simeultaneous equations

[astral ivy]

ok!

I checked and I did the steps correctly. c:

[unreal musk]

Show me the steps you did - I'm sure you didn't

[astral ivy]

:0

[unreal musk]

Illegal

[astral ivy]

:0

[unreal musk]

Right hand side becomes 836.5 - 2d

[astral ivy]

oh-

fixed it! c:

[unreal musk]

You need to make use of both equations, both s + d = 209 and 6.5s + 2d = 836.5, to get the correct solution

If you haven't used one of them, you are very likely doing something very wrong

Anyways, I'll have to go soon, I'm gonna get food

[astral ivy]

I got 24.2 for the s+d=209!!

oh ok

c: