help-23

[lean otter]

yeah

[frozen forge]

You need to find c where the point (c, 0) is in L

When you apply the equation to (c, 0)

You got : 0 = -c +10

so c=10

[lean otter]

huh

god damn

thanks dude you helped a lot

[frozen forge]

dw

[lean otter]

.close

[safe radish]

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[lean otter]

P is a power set here

[plucky elk]

do you know the definition of injective

also what is the function

[lean otter]

every element in one set can be mapped onto a unique element of the other set

[lean otter]

sorry my goal is to construct a function that is injective onto T

[plucky elk]

[lean otter]

or like

[plucky elk]

oh i guess binary encoding might work

[lean otter]

any uncountable set would work. All I need to do is to find a function f that is an injection from P(S) onto an uncountable set

so I can show that P(S) is uncountable

does this make sense? @plucky elk

[plucky elk]

maybe? hard to judge without an actual funciton

[lean otter]

well how would you prove that P(S) is uncountable?

if I can prove this I basically solved my whole assignment

[plucky elk]

power set of any countably infinite set is uncountable

[lean otter]

yea intuitively this is of course true

but how do you actually prove it

[plucky elk]

https://en.wikipedia.org/wiki/Cantor's_theorem#When_A_is_countably_infinite

[lean otter]

okay so this is a proof to why P(N) is uncounable

S is bijective onto N (I already proved it). so can we imply from that fact that P(S) is also uncountable?

@plucky elk

[plucky elk]

yes

[lean otter]

okay cool thx

another question:

[lean otter]

and if an uncountable set is a subset of another set, then this set is also uncountable right?

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@lean otter Has your question been resolved?

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[silk flicker]

Can anyone give a pair of x,y,z that show that this is not a linear transformation?

[silk flicker]

Every pair I have tried pass by both scalar multiplication and matrix addition. I dont understand how this can not be a linear transformation

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@silk flicker Has your question been resolved?

[silk flicker]

<@&286206848099549185>

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@silk flicker Has your question been resolved?

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@silk flicker Has your question been resolved?

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@silk flicker Has your question been resolved?

[scarlet perch]

Just test (0,0) and if it moves it’s not linear

[sleek narwhal]

.close

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[sleek narwhal]

Due to inactiveness.

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[true nimbus]

why is arctan(1) = pi/4?

[true nimbus]

i am confused on how to solve for this answer

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[lusty widget]

aye aye

wassup

uh

do you mean

[lean otter]

write that as a multiplication, by taking the reciprocal

it is much easier for you

[lusty widget]

$b^{\frac12} \cdot{b^{\frac45}}$

[flat frigate]

Springsskateboard

[lusty widget]

this?

ah okay

subtract exponents

[lean otter]

oh nevermind 💀 i thought those were full fractions

[lusty widget]

yeah

take the reciprocal

and exponent becomes positive

yep

nice

ggwp

mhm

in ur mom’s house

[lean otter]

💀

that took a turn

[lusty widget]

HAHAHHA

ikr

[lean otter]

tell me where you live NOW @lusty widget

[lusty widget]

korea ❤️

but rn I’m in NZ

Vacation

[lean otter]

Ayee sounds good

[lusty widget]

yeye

lucky I wasn’t in korea

after what happened at Seoul 💀

[lean otter]

yeah..all over the news :(

[lusty widget]

Yeah

heart goes out to em

yeah I studied in Singapore

[lean otter]

honestly, i doubt anything can beat what they have in india

[lusty widget]

didn’t study in Korea’s curriculum

[lusty widget]

bro India built diff

respect to them

[lean otter]

india's school system is literally just a grindstone

literally hell's entrance

[lusty widget]

HAHAH

90% also failiure

[lean otter]

you are going to fight Cthulhu

[lusty widget]

I’m going to fight the belt

If I catch the belt then I’ll get a slipper thrown at my face

[lean otter]

💀

ouch

[lusty widget]

oh damn

[lean otter]

calculus isn't thaaaaaat bad

[lusty widget]

calculus is love

[lean otter]

until you get to integrals, then things kinda get harder

[lusty widget]

except for when it’s not

[lean otter]

exactly

[lusty widget]

let me quote what my very insightful teacher said

“>50% of students fail this class”

fking hell

I found out he was lying

Everyone passed

HAHAHHA

here we don’t have calculus as a separate class

it’s all combined into one subject

math ✨

[safe radish]

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[lean otter]

from my end

to quote my inspirational electromagnetism teacher, "you are screwed"

[lusty widget]

you too my bro

[lusty widget]

HAHAHAH FUCK

I dropped physics 😎

[lean otter]

based

[safe radish]

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[lusty widget]

Hopped on to the good ol biology/ chemistry side

[lusty widget]

shit

.close

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[lean otter]

okay nvm u went to worse

[lusty widget]

HAHAHAH FK OFF

[lean otter]

😵‍💫

[lusty widget]

bio>phy

[lean otter]

tbh by this point...maybe

[halcyon carbon]

Wrong

[lusty widget]

yes, I’m shedding light on the truth

[lusty widget]

okay yea

LITERALLY EVERYTHING ELSE > phy

[halcyon carbon]

Bruh

[lusty widget]

HAHAHHAH jokes jokes

luv to all phy students

<33

[halcyon carbon]

[solar hazel]

🙌

[lusty widget]

YAS

[safe radish]

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[white jay]

Need help with some math. Lol. I want to trade in a phone with T-Mobile. T-Mobile has two promotions. Trade-ins are worth either 1000 or 800 credit depending on model.

I want to purchase a used phone for 100-300 and trade that one in for credit towards the iPhone 14 pro max (1100). Can someone help write this equation out? I have an excel spreadsheet that gives me an idea. For my budget, phones worth 1000 credit I shouldn't spend more than 300. For phones worth 800, I shouldn't spend more than 100 (maximum I can stomach coming out of pocket is 400)

(cost of iPhone 14 pro max) minus- (trade-in value of used phone) plus+ (cost of purchasing trade phone) plus+ (whatever owed/remainder to get to 1100) = (total out of pocket cost)

(1100-800=300)+X==Y
(1100-1000=100)+X==Y

X is the variability in used phone prices. I figure "Y" is my budget. Is there some sort of equation or desmos graph that will tell me at what price point the deals come out even?

[safe radish]

Please don't occupy multiple help channels.

[trim swan]

Wait, so you essentially want to buy a used phone for the purposes of trading it into t-mobile to make a profit?

[lean otter]

this feels...sketchy

[trim swan]

and unrealistic

[lean otter]

are we doing the maths for a possibly illegal business?

Sounds fun to me

[lean otter]

Yeah I am not sure how something like guarantee, warranty, etc. is going to be considered here

if they are not, then i am not sure who is buying this..

[trim swan]

Yeah I just think t-mobile is gonna offer a pretty low amount for most phones, I don't really see this being a net gain

but @white jay I'm not really sure I understand your question anyway

If you don't want to spend more than 400, then you need the used phone to cover 700. Which means you need to buy it for 700 less than what t-mobile will offer you for it

[white jay]

Profit implies I'm making money; I'm not. I need a new phone and want to get the best value and savings I can.

[trim swan]

Well, what I mean is, essentially you need to have a net gain of 700 from when you buy the phone to when you sell it to t-mobile

[trim swan]

Right. So you want to buy a phone for $100 and sell it to T-mobile for $800

for example

That's what I meant by profit. Obviously when you buy a new phone at the same time, you're gonna walk away with less money

[white jay]

Yes. However. The 14 costs 1100. If I spent less than 300 purchasing a used phone to trade, that would be a better deal than the 800 credit offer.

[white jay]

How did you get 700?

[trim swan]

In both your examples, the difference between "trade-in cost" and "trade-in credit" is 700

800-100=700
and
1000-300=700

I don't understand your "difference" column

[white jay]

Ahh. I see. The 100/300 are arbitrarily put there bc my max budget is 400. Online prices are all over the place and the 100/300 should be variable

[trim swan]

Oh yeah ok

[white jay]

Difference column is difference between trade credit and price of iphone 14 pro max

[trim swan]

Ok

Yeah bottom line is if you want to pay 400 instead of 1100, then you need to profit 700 buying and selling the used phone

400+700=1100

Basically just that

[white jay]

It would be neat to see a desmos graph with an x slider that plugged into both equations (trade credit offers)

[trim swan]

I'm sorry, I'm not sure what you mean. Here x represents what?

[white jay]

The variable cost of purchasing a used phone to trade in.

[trim swan]

And the outputs of the functions would be?

How much you need to be able to sell it for?

It's not really a complicated function, if I understand correctly.

You need to buy the phone for x and trade it in for x+700

[white jay]

You lost me there. I suppose the question is: with a max budget of 400, what method could I use to quickly determine which is offer is the better deal? Keep in mind: later models trade-in for 1000 credit, but cost more; older models cost less but only trade for 800 credit

[trim swan]

Oh, the 1000 and 800 are known for sure?

[white jay]

My best equation for this is to know exactly my budget, and use excel to quickly tell me total out of pocket cost

[white jay]

Yes. With qualifying models of phones

For example: the iphone XR lowest price I've seen is 100, but only trades for 800 credit. An iphone 11 pro trades for 1000, but if I spend 300 purchasing it, I break even with the other deal. The deals are effectively the same after all is said and done with

Because I'm OOP 400

[trim swan]

Gotcha

So if you buy the used phone for x dollars

If you buy an older ($800) phone, your OOP cost will be x+300

If you buy a newer ($1000) phone, your OOP cost will be x+100

[white jay]

That doesn't sound right

[trim swan]

OOP = 1100 - (trade in value) + (what you paid for used phone)

If the trade in value is 800, that's 300 + x

If the trade in value is 1000, that's 100 + x

[white jay]

Older phone worth 800 trade in will cost me 300+X

[white jay]

Bingo

[trim swan]

that's literally what I said lol

[white jay]

Beat me to it

[trim swan]

I mean, it's the same thing you just said sounded wrong

[white jay]

The wording got me

[trim swan]

Okay lol

Yeah, if you buy an older phone for x, your OOP is 300+x

And if you buy a newer phone for x, your OOP is 100+x

[white jay]

Is it really just that simple? Is there a graph that I can put a slider on X for both equations that would quickly tell me my OOP?

[trim swan]

Yeah lol I'll make a desmos graph

But yes it is that simple

https://www.desmos.com/calculator/hs0kdivswc

x-value is the price you pay for the used phone

Green y-value is your OOP if it's an older model.
Blue y-value is your OOP if it's a newer model

You can drag the points back and forth like a slider

@white jay

[safe radish]

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[white jay]

Thanks!

[trim swan]

No problem 👍

[trim swan]

Good luck. I wouldn't have thought it's actually possible to buy phones that cheap to trade in

[safe radish]

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[crude star]

If im tryin to find the equation of a secant line that passes through two points (a, b) and (c, d) is y - b = [(d - b)/(c - a)] (x - a) the valid formula?

[crude star]

where [(d-b)/(c-a)] is to find the slope

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@crude star Has your question been resolved?

[crude star]

Determine a linear mathematical model for the temperature, T (in degrees Celsius), as a function of depth, z (in meters) by finding an equation of the secant line that goes between Nico’s two data points.

Is this right?

Or am I missing something

[safe radish]

@crude star Has your question been resolved?

[safe radish]

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[fierce ore]

dy/dx = x+y

[safe radish]

@fierce ore Has your question been resolved?

[fierce ore]

Idk if it’s right to use partial derivatives here

<@&286206848099549185>

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[safe radish]

@teal bobcat Has your question been resolved?

[teal bobcat]

<@&286206848099549185>

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[magic mantle]

y’’ -4y’ +5y = 21e^(2x)

I got Yp = 21e^(2x)

Which is the same as f(x)

Is this possible?

Or have I made an oopsie somewhere in my solution?

[crimson field]

,w y’’ -4y’ +5y = 21e^(2x)

[flat frigate]

Results provided by WolframAlpha

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

Requested by stabulo#9301

[magic mantle]

Wow, very useful 👀

[crimson field]

Your general solution is correct. Just a coincidence that y_p(x) = f(x).

[magic mantle]

Uh but wait

My general solution is e^x * (c1cosx + c2sinx)

The bot’s solution has e^(2x) instead?

[crimson field]

That's only a general solution to the related homogeneous DE.

Oh yeah.

[magic mantle]

That means I did the easy part wrong

[crimson field]

= 2 +- i.

[magic mantle]

Omg you’re right

🤩 Thanks a lot dude

[crimson field]

🙂

[magic mantle]

You have taken my breath

.close

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[hidden steppe]

hello, how can I solve this?

[white hollow]

convert the units

how many grams in a kilogram

[hidden steppe]

I tried but I got the wrong answer

1000

[white hollow]

right so what's g in terms of kg

[hidden steppe]

2.5 grams should be 0.0025 kg

but it's wrong

[white hollow]

that's correct, but you still need to deal with the cm

[hidden steppe]

oh

and how can I do that

[white hollow]

how many cm in 1 meter

[hidden steppe]

1000

100

[white hollow]

yeah 100

so how many cm^3 in 1 meter^3

[hidden steppe]

100

[white hollow]

no

100 cm = 1 m

not the same as 100 cm^3 = 1m^3

[hidden steppe]

100 cm^3 = 1 m^3

[white hollow]

no

1m^3 represents a cube of length 1m

what woudl a cube of length 100cm be

[hidden steppe]

but it both has cubic so it should be 100:1

cancels out

[white hollow]

no

you have to apply it to the 100 as well

[hidden steppe]

so how can I do that?

how can I know how many m^3 2.5 cm^3 is

[white hollow]

it's not 2.5

look at the diagram and have a go

v is volume

[hidden steppe]

1 m^3 = 1000m

I don't understand your picture sorry

[white hollow]

the cubes are the same

volume of a cube is just side^3

so volume of the left cube is (100cm)^3 which is 1000000cm^3

and volume of the right cube is (1m)^3 which is 1m^3

so 1,000,000cm^3 = 1m^3

[hidden steppe]

that's what I said earlier

100 cubic cm = 1 cubic meter

[white hollow]

it isn't

1 million cubic cm = 1 cubic meter

[hidden steppe]

but how can that be, cubic just means that it's 3 dimensions

[white hollow]

yeah exactly

it's volume

the volume of a cube with side length 100cm is not 100cm^3

it's 100^3 cm^3

[hidden steppe]

I'm afraid I don't follow you

[white hollow]

ok let me illustrate in 2 dimensions, might help you understand better

these 2 squares are the same

as 100cm = 1m

what's the area of each of them?

[hidden steppe]

1m * 1m

and if you add volume to it you get cube

or depth

[white hollow]

don't worry about tha rn

just tell me the area of left square, in terms of cm^2, and area of right square in terms of m^2

[hidden steppe]

10 000cm^2

[white hollow]

and the right square?

[hidden steppe]

1m^2

[fathom fjord]

Guys on day 0 you send a chain letter to 6 friends each of them sends it to 4 friends the next day and each of those people send it to 4 more friends the next day and so on i need recursive and explicit

[white hollow]

right so since the squares are the same, the areas are the same

so 10 000 cm^2 = 1m^2

in other words ten thousand square centimeters makes 1 square meter

now do you see how the same logic applies to the cube?

[sand minnow]

square*

[white hollow]

yeah ty

[white hollow]

this channel is occupied, check out #❓how-to-get-help

[hidden steppe]

yeah I see that now but how can I know this? is there some formula I can use to figure this out

[white hollow]

what do you mean

you just convert the units

for example lets say we have km and m

1000m = km

and then cube / square

[hidden steppe]

yes but I know that k stands for thousand. How can I know that 10 000 square cm equals 1 square meter

[white hollow]

(100cm)^2 = 1m^2

by doing exactly what I did

[hidden steppe]

but it doesn't make sense to me since they are both using cubic

[sand minnow]

easiest way is probably just knowing the first one. so you have that $km=1000m$ then you square both sides to get $(km)^2=(1000m)^2=1 000 000 m^2$

[flat frigate]

Duh Hello

[sand minnow]

this is how it works

[hidden steppe]

but lets say I had something else like 1 square meter to square mm

how can I calculate that

[sand minnow]

so you know that $m=1000mm$ right? then you have that $m^2=(1000mm)^2=1000000mm^2$

[flat frigate]

Duh Hello

[hidden steppe]

oh so I can just square it

[sand minnow]

yep

or cube it

if thats what you need

[hidden steppe]

what if there is 4 dimensions?

[sand minnow]

then u take it to the 4th power

works for any amount

[hidden steppe]

okay

[sand minnow]

so for the start problem, you have $m=100cm$, how much is $m^3$ in $cm^3$?

[flat frigate]

Duh Hello

[hidden steppe]

so to convert cm3 to m3 I have to divide by 10 000?

[sand minnow]

not quite, if it was cm^2 to m^2 then yes you would divide by 10 000

[hidden steppe]

so now it's 1 000 000

[sand minnow]

since $m^2=(100cm)^2=10000cm^2$ but you have $m^3=(100cm)^3=?$

[flat frigate]

Duh Hello

[sand minnow]

exactly yes

[sand minnow]

this is what bo luo was illustrating here

would recommend looking at it and try to understand why

[hidden steppe]

I couldn't understnad that

but I still can't grasp how I can go about solving this problem

[sand minnow]

its two cubes, one measured in cm and one measured in m

let me just repost it so you dont have to scroll

[hidden steppe]

yeah I can see the cubes but I can't read the symbols

[sand minnow]

the one on the left shows that one side is 100cm, the one on the right shows that the side is 1m

but we know that 1m=100cm so they cubes are equal

[hidden steppe]

yes I already knew that

[sand minnow]

so what is it that you dont understand about the picture?

[hidden steppe]

well I don't understand the right side

right of the V=

[sand minnow]

right side is a cube measured in meters

there it says $V=(1m)^3$

[flat frigate]

Duh Hello

[hidden steppe]

ah okay

so to convert 2.5cm3 to m3 I have to divide by 1 000 000

[sand minnow]

so the left one has volume $$V=l\cdot w\cdot h=100 cm\cdot 100 cm\cdot 100cm$$ and the one on the right has volume $$V=l\cdot w\cdot h=1m \cdot 1m\cdot 1m$$

[flat frigate]

Duh Hello

[sand minnow]

yes

[hidden steppe]

but I still got the wrong answer

8800

[sand minnow]

could you show your work?

[sand minnow]

there should not be a 2.5 in the denominator

[hidden steppe]

2.5 / 1 000 000

[sand minnow]

you have that $kg=1000 g$ so $g =\frac{1}{1000}kg$, you then have that $$m^3=10^6cm^3$$ so you then have that $$cm^3=10^{-6}m^3$$you then have that$$2.5\frac{g}{cm^3}=2.5\frac{\frac{1}{1000}kg}{10^{-6}m^3}=2500\frac{kg}{m^3}$$

[flat frigate]

Duh Hello

[hidden steppe]

didn't understand any of that

[sand minnow]

ok ill remove scientific notation

you have that $kg=1000 g$ so $g =\frac{1}{1000}kg$, you then have that $$m^3=1000000cm^3$$ so you then have that $$cm^3=\frac{1}{1000000}m^3$$you then have that$$2.5\frac{g}{cm^3}=2.5\frac{\frac{1}{1000}kg}{\frac{1}{1000000}m^3}=2500\frac{kg}{m^3}$$

[hidden steppe]

I understand scientific notation

[flat frigate]

Duh Hello

[hidden steppe]

but I have never seen the format you are using

[sand minnow]

i find an expression for $g$ and for $cm^3$ in terms of $kg$ and $m^3$ and then i just put them into the original expression which is $2.5\frac{g}{cm^3}$

[flat frigate]

Duh Hello

[sand minnow]

which is what you are doing when you convert units

[hidden steppe]

what step did I get wrong?

[sand minnow]

well you only showed the answers both times and both were wrong so i dont know

you gotta show all the steps for me to see where you went wrong

[hidden steppe]

well first I converted 2.5g to kg by dividing by 1000

then I did the same with the cm3 to m3

but dividing by 1000 000

and that got me the wrong answer

[sand minnow]

but leave the 2.5 alone

you just put that on afterwards

what is 1 gram in kg?

[hidden steppe]

0.001

[sand minnow]

ok. what is 1 cm^3 in m^3?

[hidden steppe]

I don't know how I can calculate that

[sand minnow]

well you know that $m^3=10^6cm^3$ right? what happens when you solve for $cm^3$?

[flat frigate]

Duh Hello

[sand minnow]

going to scientific notation since you said you understand it, then i dont have to count zeroes anymore

[hidden steppe]

yeah

but I still don't know the formula for converting cm to cm3

or m to m3

[sand minnow]

but you cant convert those, they arent something you convert

they are different dimensions

its like trying to convert seconds to meters

they arent the same measurement so you cant convert them

but look at this equation $$m^3=10^6cm^3$$

[flat frigate]

Duh Hello

[sand minnow]

what do you have to do to get $cm^3$ alone?

[flat frigate]

Duh Hello

[hidden steppe]

no idea

[sand minnow]

well you divide by $10^6$

[flat frigate]

Duh Hello

[hidden steppe]

is there a formula I can use to calculate that?

or to know that

[sand minnow]

this is just basic algebra

its the same as saying $y=3x$, what do you do to get $x$ alone?

[flat frigate]

Duh Hello

[sand minnow]

maths isnt just about remembering equations, its about using logic

[hidden steppe]

well you don't know x so that isnt possible

[sand minnow]

$x=\frac{y}{3}$

[flat frigate]

Duh Hello

[hidden steppe]

of course but it's important to know the formulas

[sand minnow]

thats solving for x

[hidden steppe]

for example calculating distance by doing rate * time

so it's vital to know

[sand minnow]

but you can use logic to know that

[hidden steppe]

actually math is pretty much built on that, pemdas, distributive property etc

but anyways I just need to know the method for calculating this problem

[sand minnow]

yes im trying to show you

get cm^3 alone

[hidden steppe]

maybe you can do cubic root to get rid of the exponent

[sand minnow]

but you want the exponent

you just have to divide by $10^6$

[flat frigate]

Duh Hello

[sand minnow]

that is what i am trying to show you

so $cm^3=\frac{1}{10^6}m^3$

[flat frigate]

Duh Hello

[hidden steppe]

okay I don't understand the logic of it

[sand minnow]

the logic is that you want $g$ in terms of $kg$ and $cm^3$ in terms of $m^3$ so you can substitute them into the equation $$2.5\frac{g}{cm^3}$$to get the right answer in $$\frac{kg}{m^3}$$

[flat frigate]

Duh Hello

[hidden steppe]

but how do you get cm3 to m3?

[sand minnow]

this is what i am showing

[hidden steppe]

where

[sand minnow]

here

[hidden steppe]

where do you get the 1 and million from?

[sand minnow]

but to get there i used the equation $$m^3=10^6cm^3$$ and solved for $cm^3$ to get $cm^3$ in terms of $m^3$

[flat frigate]

Duh Hello

[hidden steppe]

and where did you get the 10^6 from?

[sand minnow]

from that $1 m^3$ is 1 million $cm^3$

[flat frigate]

Duh Hello

[sand minnow]

maybe i should switch away from scientiffic notation

seems you are getting confused by it

[hidden steppe]

10^6 is 1 000 000 no?

[sand minnow]

yes

[hidden steppe]

how am I getting confused

[sand minnow]

so why is it hard for you to get $cm^3$ alone?

[hidden steppe]

I just don't understand the logic behind the formulas you are using

[flat frigate]

Duh Hello

[hidden steppe]

because I don't know the method

[sand minnow]

its just dividing both sides by 1 million

in words, 1/(1 millionth) of a m^3 is 1 cm^3

[hidden steppe]

I feel like you are jumping over steps. I have cm3 and I want to get to m3

how can I do that?

I don't know where you get your numbers from

[sand minnow]

you find an expression for cm^3 from m^3

and since you know the relation $$m^3=(100cm)^3=10^6cm^3$$ you can then divide both sides by $10^6$ to get $$cm^3=\frac{1}{10^6}m^3$$

[flat frigate]

Duh Hello

[hidden steppe]

I don't know that relation

[sand minnow]

thats what we were going through

in the start

[hidden steppe]

the other guy told me that m3 was not equal 100cm3

[sand minnow]

exactly. and you found that 1 m^3 is 1 million cm^3

after some time

and now i have been using that

[hidden steppe]

how can 1m3 be 1 000 000 cm3

[sand minnow]

i thought you understood it at this point

[hidden steppe]

you haven't explained it. You just stated it was the case but I don't know the method for calculating it

[sand minnow]

follow this

i was explaining it here

but you seemed to understand it

since you had the right answer while i was typing it

either way i think you have the wrong mindset when it comes to math. you just want to know how to do things with zero understanding of how it works. this will lead you down a bad path. i think im gonna have to call it quits here, maybe someone else can help you

[hidden steppe]

I'm not sure what you mean by that. I'm trying to figure out a method I can use for converting this

anyways thanks for taking the time

I appreciate the help

[sand minnow]

would probably ping helpers

[hidden steppe]

ok

<@&286206848099549185> could someone help me with this please

[safe radish]

@hidden steppe Has your question been resolved?

[lean otter]

convert each unit using powers of 10
show ur workings
where did u go wrong?

[hidden steppe]

I don't know how I can convert cm3 to m3

[lean otter]

ok

how many cm in 1 m?

@hidden steppe

@hidden steppe ??

[hidden steppe]

100

sorry was getting some water

@lean otter

[lean otter]

ok

what is 100 in scientific form

@hidden steppe

[hidden steppe]

10^2

[lean otter]

ok

so

1 cm^3 = 1 cm * 1 cm * 1 cm

makes sense?

@hidden steppe

[hidden steppe]

yes makes sense

[lean otter]

how many m is 1 cm?

@hidden steppe

[hidden steppe]

0.01

[lean otter]

1 cm^3 = 1 cm * 1 cm * 1 cm
= 0.01 m * 0.01 m * 0.01m

makes sense?

[hidden steppe]

what

[lean otter]

converted each of the 1 cm to meters

[hidden steppe]

I don't understand the 2nd line

[lean otter]

^

1 cm how many meters?

[hidden steppe]

0.01

[lean otter]

okay so thats what I did there

?

[hidden steppe]

oh to get m3

?

[lean otter]

1 cm^3 = 1 cm * 1 cm * 1 cm
= 0.01 m * 0.01 m * 0.01m
= 10^{-6} m^3

[lean otter]

yes

sensible?

@hidden steppe

[hidden steppe]

what is last line

[lean otter]

10^[-6]

m^3

I just multiplied

[hidden steppe]

I don't understand it

[lean otter]

0.01 x 0.01 x 0.01

= 0.00001

= 10^-6

?

[hidden steppe]

10^-6 = 1/10^6

[lean otter]

yes

@hidden steppe

z????

[hidden steppe]

yes

no

[lean otter]

what??

[hidden steppe]

no z

[lean otter]

wat?

thats a typo

bruh

10^-6 m^3

got it???

are u done?

[hidden steppe]

no I don't understand

[lean otter]

what??

[hidden steppe]

the last line

[lean otter]

dude

u know how to multiply??

[hidden steppe]

yes

[lean otter]

what grade r u in?

[hidden steppe]

I am not in school

[lean otter]

so i just multiplied

home schooled?

[hidden steppe]

no

I graduated

[lean otter]

doesnt mean that you dont know basics at all costs

[lean otter]

r u srs?

[hidden steppe]

yes

[lean otter]

graduated with what

[hidden steppe]

but I forgot pretty much all math I did

[lean otter]

okay

anyways

[hidden steppe]

so I had to redo everything

[lean otter]

if u dont know how to multiply

I cant help much

[hidden steppe]

I know how to multiply

[lean otter]

so then tell me

what do you get

[hidden steppe]

also I don't even need to know I got calculator

so I can just put it in there

[lean otter]

when you multuply 0.01 by 0.01 by 0.01

[lean otter]

idrc how u get the answer

so do u get the answer?

yes or no?

[hidden steppe]

yes

[lean otter]

ok

[hidden steppe]

0.000001

[lean otter]

yes

m^3

so thats it

ima go then

[hidden steppe]

I have no idea what you mean

[lean otter]

this is in meter cubed

and is = 1 cm^3

and that was ur q

so done

gl

[hidden steppe]

I need to convert 2.5 cm3

to m3

but I'm not sure how

[lean otter]

multiply this by 2.5 then

[hidden steppe]

I don't understand why that number is relevant

[lean otter]

because it shows how many m^3 in 1 cm^3

[hidden steppe]

I still have no idea how to solve my problem

that doesn't help me

[lean otter]

ah srry cant help i gtg

[hidden steppe]

alright thanks anyways for taking the time

<@&286206848099549185> could someone help me with this problem please

[safe radish]

@hidden steppe Has your question been resolved?

[hidden steppe]

<@&286206848099549185>

[lean otter]

i got you bro

[safe radish]

@hidden steppe Has your question been resolved?

[hidden steppe]

<@&286206848099549185> could anyone help me

[broken shell]

Whats the problem

@hidden steppe

[broken shell]

Ok so

Im not on pc so i cant use latex but try to understand me

1kg = 1000g right

[hidden steppe]

yes

[broken shell]

1cm³ = 1•10^-6 m³

Right?

[hidden steppe]

I don't follow

[broken shell]

So basically when you are converting "cubes"

If you jump for example, from m³ to dm³

You multiply by 1000

1 step = 1000 units

Ok?

[hidden steppe]

I see, is it the same principle for m2 to dm2? but with 100

[safe radish]

@hidden steppe Has your question been resolved?

[safe radish]

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[worthy hemlock]

Why does that say 25 marks?

I'm not stupid

I'm questioning if that's a test or not

[swift fjord]

look at the numbers man

[cedar rivet]

wtf

..

[worthy hemlock]

<@&268886789983436800> Got a troll

[twilit furnace]

cheers

[safe radish]

@olive mirage Has your question been resolved?

[safe radish]

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[safe radish]

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[prime nacelle]

i have tried to equate

like

set up equation

it didnt work

[deep depot]

Let me try but I am not sure that I could do it

180-(10y+30)=5x+3y+3x+20

[prime nacelle]

yea simplifies to 8x+13y=130

[deep depot]

,w Plot 8x+13y=130

[flat frigate]

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Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[prime nacelle]

ah

so

[deep depot]

No I just did it without any reason

[prime nacelle]

anywhere on the line would satisfy

well positives

[deep depot]

Wait you have to find x+y

[prime nacelle]

ya

[deep depot]

Oh I forgot the integer

Word

Here it will be

[twin lantern]

plotting isnt necessary bro its trigonometry

[deep depot]

13+2=15

[prime nacelle]

how u get 13 and 2

[deep depot]

It is triangular algebra cause I have to find integer I have to find by using graph

It would be easier

[deep depot]

Yeah

Here

[prime nacelle]

looking at graph?

oh right

[deep depot]

8x+13y=130 there are some other possibilities like
x=0 and y= 10
But we know that x and y are positive integer so

Yeah

Bro at first I was confused but when I saw the integer word I got that in my mind

[twin lantern]

no

it should sum up to 180

[prime nacelle]

it does

[deep depot]

Yupp

[prime nacelle]

it was simplified

tyty

[prime nacelle]

how would u propose to do it

[deep depot]

,w plot x^2 + {y -3/4(x^2)^(1/3)}^2 = 1

[flat frigate]

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[prime nacelle]

.close

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[plush juniper]

Hey there, I would appreciate some advice to approach the following question (underlined and circled part).

This is about joint continuity of partial derivatives of the jacobian. Currently seeing this in Calculus.

Thanks!

[safe radish]

@plush juniper Has your question been resolved?

[crimson field]

I feel like you should include in information from the start of the question.

[plush juniper]

yea sorry

[crimson field]

Does your book say what jointly continuous means?

[plush juniper]

We dont get a textbook, just slides which i will post a pic of

[crimson field]

I just get statistics stuff when I search for jointly continuous.

[plush juniper]

Yea true I cant seem to easily find this on the internet as well

[crimson field]

Oh. I think that's uniform continuity.

Well for x > 0 each of the partial derivatives are continuous and continuous functions are also uniformly continuous.

I think showing they are continuous for x > 0 will be enough.

[plush juniper]

so I got these partial derivatives, how do i check if the bottom left arrow is continuous?

[crimson field]

They are all clearly continuous for x > 0. Proving them seems unnecessary though.

I think just saying they are all defined for x > 0 will be sufficient and the rest of the problem will come from showing that the jacobian 2(x^2 + y^2)/x^2 > 0.

[plush juniper]

Ok, the thinking process is the following:

1. Each partial derivative is continuous and such is uniformly continuous
2. If all are uniformly continuous then x and y are also jointly continuous

Is this right?

[crimson field]

I think that jointly continuous is just your courses terminology for uniform continuous.

[plush juniper]

But in this it uses the derivatives of the function at a particular point right? I don't think the definition I got uses that.

[crimson field]

They seem very similar but your definition defined it for a point but has the conditions to be satisfied on a neighbourhood about it.

x', x'', y' and y'' are not derivatives in my image. Just names for points.

[plush juniper]

I see

[crimson field]

I'm just noting the similarity between them so I think they might be equivalent. Which then allows us to only need to explain why they are continuous.

[plush juniper]

Yea, ok I'll stick with what we discussed then. Thank you Stabulo!

.close

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[lean otter]

1/3 + 1/6 = 2/6 + 1/6 = 3/6 = 0.5

1/3.5 + 1/5 = 14/35 + 7/35 = 21/35 ≈ 0.6

Is this correct?

[grizzled fossil]

How is 1/3.5=14/35?

[lean otter]

hmmm not quite

[lean otter]

Oh, should it be 10/35?

[lean otter]

yes, try to be consistent with your numerator and denominator scaling

[lean otter]

yeppie

17/35 ≈ 0.5

Right?

yeah

good job

Thanks

.close

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[raven stream]

Why would the pointwise limit not be integrable? Isn't it just infinity at 0 and 0 elsewhere? Which would be measurable with integral 0, no?

[safe radish]

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[safe radish]

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[dense yacht]

Hi all. [Real Analysis]

I'm proving that the sequence 1/2, 2/5, 5/12, 12/29, ... , x_{n+1} = 1 / (2 + x_{n}) is convergent

I need to show that the odd terms (n starts at 1) form a decreasing sequence and the even terms form an increasing sequence.

I already figured out that x_{n+1} - x_{n} = -2 (x_n^2 + 2x_n - 1) / (5 + 2x_n)... which means that for x_n < sqrt(2) - 1, it is decreasing and for x_n > sqrt(2) - 1 it is increasing.

My question is: how do I tie this information with the subsequences? I mean, how do I guarantee that the odd sequence will stay above sqrt(2) - 1 and the even subsequence will stay bellow sqrt(2)-1?

[frozen forge]

What is {} ?

i don't understand

Oh i got it

I got it

[dense yacht]

Sorry I was writing the full thing and hit enter by mistake

[frozen forge]

x_1 = 0 i guess ?

[dense yacht]

x_1 = 1/2

[compact wraith]

Hi all. [Real Analysis]

I'm proving that the sequence $1/2, 2/5, 5/12, 12/29, ... , x_{n+1} = \frac{1}{ (2 + x_{n})}$is convergent

I need to show that the odd terms (n starts at 1) form a decreasing sequence and the even terms form an increasing sequence.

I already figured out that $x_{n+1} - x_{n} = \frac{-2 (x_n^2 + 2x_n - 1)}{(5 + 2x_n)}$... which means that for $x_n < \sqrt(2) - 1$, it is decreasing and for $x_n > \sqrt(2) - 1$ it is increasing.

My question is: how do I tie this information with the subsequences? I mean, how do I guarantee that the odd sequence will stay above $\sqrt(2) - 1$ and the even subsequence will stay bellow $\sqrt(2)-1$?

[flat frigate]

Max..

[compact wraith]

just so i can actually read it easier lol

[dense yacht]

That's cool, thanks

[compact wraith]

i think for this question you need to show that the sequence is bounded and it's montonically increasing or decreasing

[dense yacht]

Yes exactly.

[compact wraith]

first of all we know $x_n>0$ so $x_{n+1}$ can be bounded from above

[flat frigate]

Max..

[dense yacht]

The $\frac{1}{2 + x_n}$ gives us the 1/2 upper bound, since all $x_n > 0$

[compact wraith]

yeah nice

then you have shown $x_{n+1}-x_n=\frac{-2 (x_n^2 + 2x_n - 1)}{(5 + 2x_n)}$

[flat frigate]

Max..

keepitalphanumeric

[compact wraith]

im not sure where you got that?

$x_{n+1}-x_n=\frac{1}{2+x_n}-x_n=\frac{1-2x_n-x_n^2}{2+x_n}$

[flat frigate]

Max..

[dense yacht]

Oh, I jumped some steps... this comes from the rule for skipping one element... $skip_{n+1} = \frac{2 + skip_{n}}{5+2*skip_{n}}$... since the original sequence is alternating, but the skip elements are all decreasing (for the odd ones) and increasing (for the even ones)... I mean, that's what I have to show...

[flat frigate]

keepitalphanumeric

[compact wraith]

ah ok ic

[dense yacht]

So odd(1) = 1/2, odd(n+1) = skip(n+1); even(1) = 2/5, even(n+1) = skip(n+1)... just to define what I was talking about...

1/2 is above sqrt(2) - 1, I hope to show that all odd(k) stay that way. While 2/5 is bellow sqrt(2) - 1, and I hope to show that all even(k) stay that way...

[compact wraith]

I mean for even it's easy as you can show it's monontically decreasing between even terms by the fact x_2=2/5 < 0.414 (sqrt(2)-1), then for even terms it's monotonically increasing

[dense yacht]

The original sequence is oscilating towards the limit... Both the even and the odd subsequences converge to the same limit.... like a dampening sine wave kind of thing...

The original sequence isn't monotonic, but it's even and odd terms are.

At least that's what i'm going for...

[compact wraith]

my bad yea i got confused with increasing and decreasing lool

yeah i mean you've got it no?

you've found the difference between the odd and even terms

and then as they are monotonically increasing but bounded above and monotonically decreasing but bounded below you have shown both sequences converge

you havent shown what it converges to but you've shown it converges

[dense yacht]

My problem is garanteeing that it doesn't overshoot the limit... why can I say that from one step to the other this subsequences doesn't pierce the sqrt(2) - 1 bound (I mean how do a garantee that all odd terms will stay above, for instance). They are above and decreasing, but they may decrease too fast and pierce the limit ... I have to prove that they can't do that.

I mean, why is odd(k) > sqrt(2) - 1 for all k? Once I establish this, I can say that for all k, odd(k) is decreasing... But, I have just shown that at k = 1, odd is decreasing (since 1/2 is greater than sqrt(2) -1 ).

My question should be rephrased to:

[devout flume]

I think you can create new sequence y_n = x_n - (sqrt(2)-1) for n is odd and (sqrt(2)-1) - x_n when n is even

and prove than y_n > y_(n+1)

[dense yacht]

i'm thinking about it...

[devout flume]

Actually i can prove that ||y_n+1 < y_n/2||

[safe radish]

@dense yacht Has your question been resolved?

[devout flume]

Hint: ||In original recursion, minus both sides by sqrt(2)-1 and use conjugate on fraction side||

[dense yacht]

@devout flume Cool. I think I got it.
$y_{n+1} = \frac{1}{(2+x_n)(sqrt(2)+1)}} * y_{n}$. So {y} is decreasing.

[flat frigate]

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

[dense yacht]

Now why can I say that Y is bounded by sqrt(2) - 1? I have to use a previous knowledge that the original sequences were bounded by that, right? I keep coming back to this statement.

[devout flume]

If limit of y when n->inf is 0, then we can guaranteed that sequence x_n converges

it is the definition of limit

the epsilon thing

[dense yacht]

Once I know that each subsequence converges to that same limit, I can transform the original sequence into Y, and know that lim Y -> 0. Got it! Thank you so much. This problem was destroying by brain! I am finally set free! hahaha THANKS! @devout flume @compact wraith

[devout flume]

It is nice problem tho

[dense yacht]

It is. I learned a lot trying to solve this. Thanks again!

.close

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[safe radish]

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[unique basalt]

in the interval [1, 7] why is 4 a local maximum

[flat frigate]

cookie2

[unique basalt]

And so the only critical points are 0 and -8 right?

but they are not in the interval

[flat frigate]

cookie2

[unique basalt]

Ok this is the derivative

[ornate creek]

you have to find the derivative of (x-1)/x(x+8)

yea

the bottom is positive

[unique basalt]

Then what

Idk where the 4 come from

[ornate creek]

wait a second I'll do it on a piece of paper

[unique basalt]

ok

[ornate creek]

wait a second, are you sure there isn't a mistake in the derivative you've sent?

[unique basalt]

im not sure

Let me use symbolab

Here

[ornate creek]

yea

that's the derivative

and now you try to tell where is the top fucntiuon positive and when is it negative

you can then tell where the local maximum is

[unique basalt]

do i just plugin random numbers?

[ornate creek]

nope try to find P(x)=0

P(x) being the thing on the top

you can use the quadradic formula or idk hiow it's called

[unique basalt]

so i just use the derivative

and add an equals 0 to it

and then solve

[ornate creek]

yea now that you'vefound the derivative you find the solutions of the polynomial on the top so you can tell when it's positive and when it's negative

then you find that 4 is the number at which the derivative goes from positive to negative

thus it's a local maximum

@unique basalt did you solve ?

[safe radish]

@unique basalt Has your question been resolved?

[unique basalt]

ive been trying, idk how to solve it

i dont know how i get a 4 anywhere

[unique basalt]

I think its just my algebra

[ornate creek]

you fiond the solutions of the polynomial on the top

[unique basalt]

i needed to simplify the parentheses

[ornate creek]

oh maybe idk but the thing is that you have to find the roots of the polynomial then factor the polynomial out thanks to the roots

and then you can easily find the sing of the polynomial

[unique basalt]

ok got it

.close

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[quick leaf]

L1 = {w ∈ I⋆ | w is divisible by 2 or 3 when converted to its equivalent decimal number}

[quick leaf]

I'm supposed to design a NFA for this

But I'm a little confused by the "when converted to its equivalent decimal number" part

[safe radish]

@quick leaf Has your question been resolved?

[safe radish]

@quick leaf Has your question been resolved?

[quick leaf]

<@&286206848099549185>

[safe radish]

@quick leaf Has your question been resolved?

[quick leaf]

does it look answered 😭

[safe radish]

@quick leaf Has your question been resolved?

[quick leaf]

fine omg

jesus

.close

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[quick leaf]

get me out of this hell

.close

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[feral remnant]

@glass carbon from yesterday if the answer was 2.708

If u convert to a mixed fraction

There is no answer given

[glass carbon]

because it's not 2.708

you've rounded it

,calc 5*6.5/12

[flat frigate]

Result:

2.7083333333333

[glass carbon]

it's not 2.708 !!!

[feral remnant]

Then why did the calc say it

[feral remnant]

Wdym

I’m confused

[glass carbon]

???

calc says 2.708(3)

not 2.708

[feral remnant]

Oooh

So if u make that into a mixed franc

[flat frigate]

Modus

[feral remnant]

Frac

Got it

So what would the mixed number be

[glass carbon]

,calc 2 + 17/24

[flat frigate]

Result:

2.7083333333333

[feral remnant]

Oh

But how do u convert a decimal like that to mixed frac

[neat kiln]

There should be an option in your calculator for that

[feral remnant]

Ye

But I would need to know how to do it irl in case the teacher asks how I solved it

[neat kiln]

,w how to convert decimals into fractions

[flat frigate]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by Diesel#5243

[neat kiln]

This app is very stupid

[glass carbon]

just work on fractions

you'll get that

$$\frac{5 \cdot 6.5}{12}=\frac{5 \cdot 6\frac{1}{2}}{12}=\frac{5 \cdot \frac{13}{2}}{12}=$$
$$=\frac{\frac{65}{2}}{12}=\frac{65}{2 \cdot 12}=\frac{65}{24}=$$
$$=2\frac{17}{24}$$

[flat frigate]

Modus

[feral remnant]

I should pay u for typing this

.close

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[lean otter]

is this correct

[safe radish]

Please don't occupy multiple help channels.

[lean otter]

.close

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[lean otter]

I’m not too sure how else to prove it

[icy crystal]

so how can I solve this problem? I know how to normally determine dy/dx but I'm not sure what to do regarding the logarithmic differentiation

oh this channel is already occupied sorry.

[neat kiln]

So you know induction?

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

Uhhh didn’t learn that yet

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[novel thicket]

Does anyone know how to solve this?

[novel thicket]

i have already tried to let z = c + di and formulated an expression that satisfies the required solution form: d = 3c - 2
thus, z = c + (3c - 2)i
i then expanded the modulus equation using the conjugate property zz* = |z|^2
i am stuck here...

<@&286206848099549185>

[safe radish]

@novel thicket Has your question been resolved?

[novel thicket]

<@&286206848099549185>

[devout flume]

Can you show how much have you done from there

[novel thicket]

hey, thank you so much for responding
i have stated what i have done so far
in the end, i derived an equality in terms of a, b and c

but, i was not sure how to proceed from there

[devout flume]

Ok that equation is true for every c right?

[novel thicket]

yes

but i need to find specific a and b

[devout flume]

I think you can compare coeffcient of 1,c,c^2

[novel thicket]

wait, can u show me how u would approach the derivation
to compare my equality with yours, bcs i am not sure if it is correct?

[devout flume]

Well i didn’t do yet. But it must be able to write in polynomial of c

[novel thicket]

it would be greatly appreciated if u could derive just the equality so i can compare my answer with yours
i can then take it from there and apply ur suggestion

thanks

[novel thicket]

to get the first expression, did u use the conjugate property or the modulus definition |z|^2 = c^2+d^2?

[devout flume]

Yes

Second one

[novel thicket]

so u would then construct a system of equations
but, u would then have two equations for 3 variables which is a deficient system right?

[devout flume]

No, you can see my arrow showing the coefficient of c

It must be equal

[novel thicket]

on no wait, sorry i was being stupid
ur right
thank you so much

🙏

[devout flume]

💯

[safe radish]

@novel thicket Has your question been resolved?

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[lean otter]

Consider the function g(x) = −2x² + 3x + 4. Which inequality correctly orders the values of g(−3), g(3), g(0), and g(2)?