help-10

[median dome]

oh

um

if it's random then you will have to use some vector shenanigans

[obsidian wagon]

its not exactly a random, its a random in a range 🤔

but lets say for instance that number doubles every time

but its closer to the log speed

.close

[obtuse pebble]

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[tender stratus]

Given a set of three positive integers (X, Y, Z), and an operation that allows you to subtract one from two of these and add one two to the third, what is the minimum number of operations required to reduce two of them to zero? example: (1, 4, 6) -> (0, 5, 5) -> (1, 4, 4) .... -> (5, 0, 0): 6 operations

[tender stratus]

this was actually an algorithm question(coding) but I think its solution should be more mathematical in nature. for the obvious, I know that if two numbers are equal the minimum operations required would be their value. I'm not sure how to go about from there though

[royal basin]

are our integers allowed to be negative in the interim?

[drifting wraith]

i think you can't make numbers have the same parity if they had it different

[royal basin]

also ^

[drifting wraith]

so you always know which one you're going for

unless they are all the same parity

[royal basin]

that's true but the three integers will always be either all of the same parity or a two-to-one split

[drifting wraith]

sure

so that's the algorithm

if there's a split, reduce the ones with the same parity until you get one zero

[tender stratus]

the original 'flavor' text called the numbers as currencies(removed for brevity), so I'm inclined think it shouldn't be

[drifting wraith]

then bounce that between 1 and 0
if there's no split do the same with the smallest numbers

[royal basin]

can you perhaps share the original flavor text

[tender stratus]

sure give me a moment

Problem Statement

We have x rupees, y dollars, and z euros. In this wonderland, in a single operation, any single unit of two different currencies can be exchanged with the remaining one. We want to have all of our money in a single type of currency only. Return the minimum number of operations required to do this or else return -1 if it is not possible to do so.

[drifting wraith]

the answer is the second smallest number of the majority parity

neat

[tender stratus]

shouldn't (1, 4, 6) be 4 then?

I could be wrong about 6 operations being required tbh

this was the only example in the question

oh wait

I realized I had the question misunderstood

its +2 not +1

[drifting wraith]

6 is second smallest out of 2

[errant oriole]

Numbers can be negative during process?

[drifting wraith]

so subtract 1 from two, and add 2 to one?

that doesn't even change the algorithm, same answer

[tender stratus]

seems to be

[drifting wraith]

wait nvm

[tender stratus]

the problem statement doesnt mention anything about it but I think it's a no

[drifting wraith]

then i suppose it's not possible if they are all different terity

not a real word surprisingly

[tender stratus]

seems to be true, at least I haven't been able to think of a counterexample

[drifting wraith]

but it's the same answer though, second lowest of the majority terity

[tender stratus]

can we somehow use the fact that the ending number must be the sum of all 3?

[tender stratus]

after trying this out for a while this looks to be correct

unfortunately I don't have access to the test cases so I cant verify it 100%

thanks for the help

.close

[obtuse pebble]

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[obtuse pebble]

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[worthy ferry]

Hi, my question is relating to 8-bit subtraction.

[quasi current]

,w compute 11001010 base 2-10011010 base 2

[worthy ferry]

I am unsure as to how to subtract a binary number.

[warm shale]

Results provided by WolframAlpha

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

Requested by ether#2229

[quasi current]

Do you know the table of base 2s?

[worthy ferry]

Ah yes, most certainly.

[quasi current]

okay wait it's not that one

give me a moment

[worthy ferry]

Sure.

[quasi current]

I just woke up so lemme get my laptop

[trim portal]

https://www.youtube.com/watch?v=C5EkxfNEMjE

[worthy ferry]

That's alright man.

I have seen this video, it is great.

I would like to add four binary numbers. I am unsure as to how that is done

[trim portal]

Show your work

[worthy ferry]

I will show you my work right now.

[trim portal]

you can then verify your calculation by converting everything to base 10

[frosty spoke]

no

you're asked to do two's complement

[trim portal]

Oh I see

[frosty spoke]

change the second number to two's complement (I really mean negate the second number and represent it in two's complement)

i.e. flip all of the bits and add 1

[worthy ferry]

I have done this a couple of weeks prior, so i have forgotten as to how I did this.

My work is correct, but I do not know to apply this to adding or subtracting 3 or more binary numbers.

[trim portal]

1a is adding 3 or more binary numbers

[worthy ferry]

I did not know that.

Oh

..Yes

Of course.

[trim portal]

Also in this particular task you are supposed to do it by two's complement

[worthy ferry]

What is a two's complement?

[trim portal]

https://www.youtube.com/watch?v=L_m7jBvtzpQ

[worthy ferry]

Thank you

[trim portal]

To make the complement do this

[frosty spoke]

what's the equivalent in base 10

tens complement? lol

[worthy ferry]

I want to create a practice problem, is it okay if I can try to do it and you can tell me if I did it correctly?

[trim portal]

That would be a bit harder to make

[trim portal]

Yeah

[worthy ferry]

Or maybe you can make a problem and I can do it.

Okay, I will make the problem.

I will try to add four binary numbers.

[frosty spoke]

-11001
-1011
10100
100000
10011
-10111
-11110
-11101
10
1110
-100100
-11101
-11111
10110
-110
1111
-110
-1
10111
101000
-1001
-1100
10000
100101
1011
10100
11111
1
0
-110
-100100
-1101
-100110
-11001
-100100
1
11000
100011
1001
11011
-100
11010
100110
10111
-11101
11000
1011
10000
11111
101
11011
-100111
1
-100001
1
-101000
-101
101000
100001
101000

here is a list of random binary numbers for you

there are all of your addition problems

[worthy ferry]

I can't even read all of that, but your efforts are much appreciated.

I am almost finished with my problem.

How does this look?

yes

[frosty spoke]

im too lazy to actually add any of that up

you should try using a calculator

[worthy ferry]

As wonderful of a suggestion that was, I do not think I would be able to use a calculator to do such problem in certain situations.

[high lily]

1111+1111 isn't 1110

1+1+1 isn't 1

[worthy ferry]

You're right

[frosty spoke]

you can certainly use a calculator when you practice to check your work

[worthy ferry]

I messed up.

[frosty spoke]

what's the difference between asking someone else to calculate it and using a machine built to do it?

one's wasting someone else's time with something that a machine can do

[worthy ferry]

I did not use common sense, you are right.

[worthy ferry]

It is actually 1 1 1 1 0

I will fix the problem now.

The answer is 1 1 0 0 1 1

[trim portal]

what method did you use btw? I dont get the steps

Ah I see, you added them one by one

[worthy ferry]

Yes.

[trim portal]

I was confused with the upper lines, those are the carries

[worthy ferry]

Yes. I should have made it more clear.

[trim portal]

as long as you understand it it's okay

[worthy ferry]

Thank you

I will now have to learn about adding boolean equations. Very difficult subject for me.

I used an online source to show me the answer and the work, but i still do not understand.

Perhaps I will return when I research the topic more.

Thank you Math

.close

[obtuse pebble]

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[trim portal]

actually ramonov was the one who pointed out your mistake

[worthy ferry]

You are right

[obtuse pebble]

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[velvet cypress]

.open

[frosty spoke]

can you type that out or make it larger

[trim portal]

A wizard having powers of mystic in candations and magical medicines seeing a cock fight going on, spoke privately to both the owners of cocks. To one he said: If your bird wins, then you give me your stake-money, but if you do not win, I shall give you two third of that. Going to the other, he promised in the same way to give three fourth. From both of them his gain would be only 12 gold coins. Find the stake of money each of the cock-owners have.

[hasty swan]

-2/3x+y=12
x-3/4y=12

solve

[velvet cypress]

k thank u very much

[hasty swan]

you get why its these eqns?

[velvet cypress]

yes

i'm 15

[hasty swan]

<33

[velvet cypress]

thanks again

.close

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[austere holly]

how do i determine wheter the first surface is only on one side of the second surface?

i cnat figure out how to determine that, and i dont get much help from the book i have

[main rose]

Determine if there are any intersections. If there are none one surface must be on one side of the other

The there are intersections however it doesnt garantue that they pass through each other

[austere holly]

so can i look at both gradients and see if they are linear?

and if they are then they intersect?

[main rose]

Well you know that the gradient of the first function is not linear

So that approach will not work

[austere holly]

i meant linearly independent sorry

i dont know all terms in english sorry

[main rose]

Its alright, I am not sure if that will work

I think it might be easier to try and solve the system of equations

[austere holly]

hmm but i have only two equations

[main rose]

Usually that means there isnt one specific solution

but a range of solutions

y - x + sqrt(8-2x^2 - y^2) = x + 2y - 7
2x + y - 7 = sqrt(8 - 2x^2 - y^2)
(2x + y - 7)^2 = 8 - 2x^2 - y^2
It doesnt look like its going to be very easy

Ooh, the domain of the first function is limited

[austere holly]

yee the aquareroot

[main rose]

since (2x^2 + 2y^2) <= 8

[austere holly]

is that a constraint?

[main rose]

So were looking at the unit circle with radius 2

Yes because otherwise the square root has somthing negative inside of it

and thats bad

but sadly I need to catch a train

I am sorry I can help you further

Hopefully you can get further with that extra constraint

[austere holly]

yes thank you!

when i calculated x from the equations i got, my x became an imaginary number, does that mean there is no intersection of the two functions?

<@&286206848099549185>

[obtuse pebble]

@austere holly Has your question been resolved?

[austere holly]

someone?

[obtuse pebble]

@austere holly Has your question been resolved?

[fallow cove]

You can't pull the y^2 out here

[austere holly]

why not?🤔

[fallow cove]

Take an example sqrt(5-4)

If we take out the sqrt4 here it becomes 2

And we get sqrt4 - 2

But 1 is the correct answer

Cuz we have to simplify the stuff inside the sqrt first then we can take it out

[austere holly]

but i have expressed y as x and the things under the sqrt becomes 0

[fallow cove]

Huh

Ohh

I see

[austere holly]

do you know if i have solved the problem correctly?

[fallow cove]

Yes

Are you grade 8-ish?

[austere holly]

first year at uni

[fallow cove]

Oh ok

[austere holly]

how would you determine wheter the surface and the flat plane intersect?

[fallow cove]

I think that the 2 function did not touch the x axis?

[austere holly]

they do

[austere holly]

.close

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[obtuse pebble]

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[shut carbon]

How do I calculate this thing

[obtuse pebble]

Please don't occupy multiple help channels.

[shut carbon]

Sorry sir

[obtuse pebble]

@shut carbon Has your question been resolved?

[shut carbon]

Yes sir

[obtuse pebble]

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[obtuse pebble]

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[edgy needle]

hello

[edgy needle]

what did i do wrong?

for BH

(part ii)

oh i get it now

i brought over the squared from 80

whoops

thx

.close

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[jaunty dew]

Air is escaping from a spherical balloon at a constant rate of 15 cm³/s. Find the rate at which the radius of the balloon is decreasing when its volume is 972π cm³

[jaunty dew]

How exactly do we solve this question?

Assuming that we have to use the chain rule

We can't really do dV/dr (4/3πr³), can we? I've tried and the answer seems wrong

[plain stag]

You use dV/dt instead

Treating V and r as functions of t

[jaunty dew]

I did

It's part of the chain rule

[plain stag]

So what do you get after you differentiate?

[jaunty dew]

Wait

This is what I did 🥲

[plain stag]

dt/da?

[jaunty dew]

Air escaping per seconds 💀

[jaunty dew]

I'm now sure how we'd work with radius here

[plain stag]

The units used for the air escaping is cm^3/s

[jaunty dew]

Yes

[plain stag]

Volume/time

[jaunty dew]

Meaning volume

[plain stag]

So not "dt/da", but dV/dt

[jaunty dew]

Lemme retry

It correct, tysm

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[novel gale]

What is tan • x

[grizzled shore]

what is •

[novel gale]

Multiplying

[severe reef]

Tan is a function, needs an argument

[novel gale]

Ik but wth is this

[severe reef]

That's just tanx

I guess the • is a misprint

[novel gale]

Ah alr thanks

[obtuse pebble]

@novel gale Has your question been resolved?

[obtuse pebble]

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[agile siren]

AD equation is: y = 1/2 x-3. so far I've worked out: A(6,0) AB equation: y = -2 x +12. B(0,12). D(10,?) is given. what is the y of D if the x is 10?

[night lodge]

D lies on which line

[agile siren]

AD

[night lodge]

u have eqn of AD

and u know X coordinate

[agile siren]

gosh thank you

.close

[obtuse pebble]

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[placid wadi]

i'm working through khan academy and i'm trying to do it by myself.

what i need to know is how to differentiate Y= 2 / 9+x^2

I legit just dont know how to go about solving it.

[slender nebula]

do uk the quotient rule for differentiation?

[placid wadi]

i don't believe i remember that.

[slender nebula]

okay do uk the chain and product rules?

[placid wadi]

i took calc 2, 3 to 4 years ago and i'm in diff. eq. now. i have quite a few holes in my knowledge.

Chain rule and product rule seem familiar. i can look it up.

[slender nebula]

alright if u know the quotient rule, use that (I wd recommend learning it if u're already doin diff eqns)

[placid wadi]

okay i'll refresh myself. thanks!

[slender nebula]

or u can rewrite this as 2(9 + x^2)^(-1)

and use the chain rule

[placid wadi]

is it okay to leave this open and comeback with a response?

[slender nebula]

sure

[obtuse pebble]

@placid wadi Has your question been resolved?

[placid wadi]

Still working on it, But i did have a side question, what would the derivatives of the signs called? like what could i look up to learn or memorize them.

when you derivate a number it just becomes zero right?

[slender nebula]

yes

[placid wadi]

okay so i got 4x / (9+x^2)^2 for y' which is not equal to xy^2

using the quotient rule.

Did i maybe go about it wrong?

@slender nebula (sorry for the ping)

[arctic raptor]

remember y = 2/(9+x^2)

you can substitute that value into your new equation

[placid wadi]

So what i calculated was y' so you mean to say set what i calculated equal to xy^2? orr?

[arctic raptor]

no no no

what is y^2 equal to

[placid wadi]

2y right?

[arctic raptor]

no no

[placid wadi]

OH

[arctic raptor]

y squared

[placid wadi]

like take the entire (2 / 9 + x^2) and square it?

[arctic raptor]

yupo

[placid wadi]

okay gimmie a few.

oh.. [ 4 /( 9 + x^2)^2 ] unless i need to go further with the denom.

[arctic raptor]

yup

and your differential is x[4/(9+x^2)^2]

[placid wadi]

right so its not equal due to the x on the right side?

[arctic raptor]

from what you have done by substituting in y^2

y" = xy^2

because y" = x (y^2)

where y^2 is 4/(9+x^2)^2

[placid wadi]

right, right, so its as you said. x[4/(9+x^2)^2].

so i didnt need to derive the Y?

[arctic raptor]

yes you did to prove it

[placid wadi]

Oh okay, and i proved that it is not equal due to the addition of the X right?

[arctic raptor]

the question asks if y" = xy^2

[placid wadi]

did i maybe derivate wrong?

[arctic raptor]

whats the answer

is the answer no

[placid wadi]

i got y' = [( 4x / 9 + x^2)^2]

i haven't clicked an answer yet.

[arctic raptor]

yes

[placid wadi]

its not gonna kill me if i do i just wanted to learn and be sure.

[arctic raptor]

i can write it out and send a pic of my working real quick

you just need to sub in the value of y again

[placid wadi]

If its quick, sure. i would appreciate it.

I just dont want to waste your time.

[placid wadi]

Awesome okay, thank you!

what other rules should i look up to remember.

[obtuse pebble]

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[timid silo]

I am doing Rational exponents with Paul's notes:

a^n = b <=> a = b^(1/n)

what if a is -5
for example

(-5)² = 25 <=> (-5) = 25^(1/2)

but below, in the examples, only 5 is the solution, not -5.

he says

In other words, when evaluating b^(1/n)
 we are really asking what number (in this case a) did we raise to the n to get b

We can raise both -5 and 5 to the power of 2, to get 25..

The question is, why's -5 not the solution too?

[timid silo]

https://tutorial.math.lamar.edu/classes/alg/RationalExponents.aspx

I am reading about this and it seems that
sqrt(25) = 5 is just a convention..
mathematicians decided to throw away -5..

if so, then isn't it wrong to say that

a^n = b <=> a = b^(1/n)

?

would it be fine if we explicitly say that a is greater or equal to 0

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[tardy epoch]

square root is defined to be the positive solution

That way it's a function

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[long dove]

need help construct the proof

what idea might be useful

[unreal musk]

The definitions(!)

[obtuse pebble]

@long dove Has your question been resolved?

[long dove]

okay but how the idea

By definition of limit of f at c we have \
$\forall \varepsilon >0, \exists \delta=\delta (\varepsilon)>0 \ni 0<|x-c|<\delta \Longrightarrow |f(x)-L|<\varepsilon$

[warm shale]

Welf

[long dove]

and by definition of g tends to $\infty$ as $x\to c$: \
$\forall \alpha \in \mathbb{R}, \exists \delta=\delta(\alpha)>0 \ni 0<|x-c|<\delta \Longrightarrow g(x)>\alpha.$

[warm shale]

Welf

[long dove]

What I think is I need to show that $f(x)g(x)>\varepsilon$ but idk how to make it happen

[warm shale]

Welf

[long dove]

I found it on internet but I don't know how they achieve it

[long dove]

because the definition of limit d at c I know is this

and I also try it this way, \
$|f(x)|g(x)>\alpha$ but Idk how get rid the absolute.

[warm shale]

Welf

[long dove]

and I'm avoiding the \
$\lim_{x\to c}{f(x)\cdot g(x)}=\lim_{x\to c}{f(x)}\cdot \lim_{x\to c}{g(x)}=L\cdot \infty=\infty (\because L>0)$

[warm shale]

Welf

[long dove]

<@&286206848099549185>

Sad

.close

[obtuse pebble]

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[thick gyro]

!status

[obtuse pebble]

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

[charred escarp]

1

[thick gyro]

is the | |

whats confusing you

[charred escarp]

Yea

[thick gyro]

what you do

is

write it as a piecewise function

and then split your integration over it

as in

6|u-1| is 6(u-1) when u is >1

and 6(1-u) when u is <1

so then you do

$\int_{0}^{2} 6|u-1|du = $\int_{0}^{1} 6(1-u)du + $\int_{1}^{2} 6(u-1)du$

[warm shale]

doctor99268
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[thick gyro]

@charred escarp do you get why we did this

[lucid falcon]

cuz from 0 to 1 it's minus the line and from 1 to 2 it's the positive line

?

[thick gyro]

yh

[charred escarp]

Yea

I got 0. Correct answer is 6

[thick gyro]

your limits

read them again

also not sure why you needed to use subsitution on u-1

lmao

[charred escarp]

Yea I don't know what I'm doing

[thick gyro]

@charred escarp

[charred escarp]

Did I do it right?

[obtuse pebble]

@charred escarp Has your question been resolved?

[thick gyro]

yh

[obtuse pebble]

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[dusky scroll]

2x(16-x^2)

idek how to approach the question

[barren estuary]

Looks good to me

[obtuse pebble]

@dusky scroll Has your question been resolved?

[unreal musk]

How would you find the area of a rectangle? What would the coordinates of N be? Does that allow you to find information you’d need to calculate the area of that rectangle?

[dusky scroll]

thats the solutions idk how they reached it

[dusky scroll]

thats the all the information given

[barren estuary]

Oh...

[dusky scroll]

ye

[barren estuary]

How do you calculate the area of a rectangle

[dusky scroll]

LxW

[barren estuary]

Yep

What would be the width of the rectangle in this problem?

[dusky scroll]

ohhhh

i kinda get it since P is x

so the length would be 2x

[barren estuary]

Yes

What's the height

[dusky scroll]

oh since y =16-x^2 we sub it in

facts facts

[barren estuary]

Solved

Good job

[dusky scroll]

ty ty ty

[barren estuary]

Please type .close if you're finished

[dusky scroll]

.close

[obtuse pebble]

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[graceful apex]

1-0.5=0.95???

[graceful apex]

????

That goes again normal math man 💀

[thick gyro]

what?

it says 1 - 0.05

[fiery birch]

0.5 is 50% , 0.05 is 5%
(Decimal) * 100 gives the percentage

@graceful apex

[obtuse pebble]

@graceful apex Has your question been resolved?

[obtuse pebble]

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[zenith raft]

hard for me to tell what the question is

what's x?

[winged aurora]

An integer

[zenith raft]

so just saying this is even?

[winged aurora]

Yup

[zenith raft]

that's $\frac{a+b}{2} \frac{a-b}{2}$? as in multiplication between them?

[warm shale]

💜𝓁𝒶𝓎𝓁𝒶💜

[winged aurora]

Yes

[zenith raft]

so the question is prove that if $a$ and $b$ are even integers such that $\frac{a+b}{2}\cdot \frac{a-b}{2}$ is even, then $\frac{a+b}{2}$ and $\frac{a-b}{2}$ are even

[warm shale]

💜𝓁𝒶𝓎𝓁𝒶💜

[winged aurora]

Exactly

[zenith raft]

loll ok just making sure

[winged aurora]

No problem

[zenith raft]

i will try it out

ok got it i think

so one of (a+b)/2 and (a-b)/2 is even as you said

if (a+b)/2 is even you can write a+b = 4n for some integer n

then take a look at what a-b and (a-b)/2 equal

[winged aurora]

Got it

[zenith raft]

yay

the case where (a-b)/2 is even should be very similar 🙂

[winged aurora]

Yup

My teacher said that they both will even so quickly i thought it was some simple logic

[civic zealot]

What if neither is even?

[winged aurora]

Then thier product won't be even

[zenith raft]

then their product isn't even

[civic zealot]

You also need to show that

[winged aurora]

It is given in the question

[civic zealot]

The statement is “if product is even, then one of them is even”
Your proof assumes one of them is even

[zenith raft]

i thought of something simpler which is that(a+b)/2 and (a-b)/2 differ by b which is even

so they have the same parity

[winged aurora]

What is parity?

[zenith raft]

whether something is even or odd

[winged aurora]

This work?

[zenith raft]

so i just meant they are both even or both odd

if you know they differ by an even integer

,rotate

[winged aurora]

My main confusion was if a-b or a+b equals 2 then one the term turns out to be odd

[zenith raft]

not sure what you're saying there :c

[zenith raft]

are you on this?

[winged aurora]

?

[zenith raft]

what i was saying earlier

so one of (a+b)/2 and (a-b)/2 is even as you said
if (a+b)/2 is even you can write a+b = 4n for some integer n
then take a look at what a-b and (a-b)/2 equal

[winged aurora]

Yeah i solved that and got that it is even

[zenith raft]

and is that what you were trying to do here?

[winged aurora]

I gotta and catch a bus

[winged aurora]

I expressed a and b in terms of 2 as both are even .

And solved further and got the required result

Sorry i gotta go

Thanks for the help

[zenith raft]

ok np ^^

[zenith raft]

i'm still not sure abt this tho

[winged aurora]

I will dm the reason later if you want

But about 6-7 hours later

.close

[obtuse pebble]

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[lime summit]

how do i find out if 2 vectors are linearly independent using matrices

[frosty orchid]

organize the vectors into a matrix as columns then solve

$A\vec{x}=\vec{0}$

[warm shale]

benevo

[frosty orchid]

non-zero solutions should indicate linear independence

this works generally for any number of vectors

if you have more vectors than numbers in your vectors then you can infer non-linear independence of the set of vectors

[lime summit]

what of the opposite

[frosty orchid]

meaning?

[lime summit]

more numbers in vectors than number of vectors

[frosty orchid]

ah if you have the same or fewer you still need to solve for non-zero solutions in the equation above.

[lime summit]

but how though
i cant do row 1 x vector etc

[frosty orchid]

form an augment matrix with the 0 vector and your matrix and use row reduction to form an equation.

[lime summit]

0 vector and my matrix?

[lime summit]

wheres the equation come from then if the x vector isnt there

[frosty orchid]

I am not sure what you mean

[lime summit]

heres what id do if it there were 3 vectors

but i cant do that with 2 vectors that are 3 dimensional

[lime summit]

to have an equation you need an unknown
if i have my 2 vectors and a 0 vector and i form an "equation" out of those (and idek how youd do that) i would just have numbers

[frosty orchid]

are you familiar with row reduction?

[lime summit]

yes

[frosty orchid]

for your example

preform row reduction to solve for your <x, y, z> vector in this case you are guaranteed one free variable (all 0 row) as the there is one fewer vector than the height of the matrix.

that last column of 0 is your 0 vector

[lime summit]

thats not an augmented matrix but a regular one
now if i augment that with the x y z vector and then row reduce
that would make sense

[frosty orchid]

what qualification is the matrix missing to be augmented?

[lime summit]

the vertical line that goes between the matrix and the vector its augmented with

[frosty orchid]

not quite correct, but I assume you get the point while I fiddle with latex.

[frosty orchid]

seems correct enough; I suppose the 0 vector is behaving as an unnecessary component in this case.

[lime summit]

indeed. what this all means, i no longer know. i need sleep

.cloe

.lcose

damn

.close

[obtuse pebble]

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[clever idol]

Hello!

[clever idol]

can you help me with this? i just need the answer

[obtuse pebble]

@clever idol Has your question been resolved?

[neon eagle]

!status

[obtuse pebble]

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

[trail cloak]

@clever idol

[clever idol]

Sorry I'm back

I'm just really confused and i don't really need the steps \

just the answer

[fathom flicker]

Well no one is here to give you the answer or do your work for you. Please read #rules

[clever idol]

ahhh okay

thanks

[obtuse pebble]

@clever idol Has your question been resolved?

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[past peak]

so my question is that i'm not able to understand a concept in functions.

[median dome]

,rccw

[bold nest]

Sorry, I had to

[past peak]

i'm able to understand what f(n) means when only one condition is given on the rhs such x or |x| or something
but what does it mean when 2 conditions are given

[bold nest]

Define |x| for me

In the format above

[past peak]

modulus

[bold nest]

No, define it as above

[past peak]

f(x) = |x|
you mean this

[bold nest]

like this

yes

define what it means to take the absolute value of a number

in the format like this image

f(x)=x if x is non-negative

f(x)=-x if x is negative

that's what the absolute value does

[quasi current]

I was just looking that 😭

[past peak]

wait what?

modulus changes the negative value to positive that's what i've read

[bold nest]

Yes......

that's exactly what it does

[past peak]

i didn't get this

[bold nest]

f(-3)=-(-3)

Do you see now

[past peak]

no

[past peak]

you're a saying a function f(x) = -x ?

[median dome]

$|x| = (x$ if $x \ge 0, -x$ if $x < 0)$

[warm shale]

kheerii

[median dome]

imagine this was written in the same format as on your paper

[bold nest]

In your format, it's this MrBadAttitude

[median dome]

ye

[grand thunder]

You can define a function in all sorts of ways. I can define a function in natural language that maps all real numbers to the letter H

As long as the conditions of the piecewise function cover the entire domain (and are mutually exclusive) then it's allowed

Idk if you were confused why it was possible or just what it meant, sorry

[past peak]

as i understood you just explained the conditions of a function to be in relation

am i right?

[grand thunder]

Not an answer but a function doesn't characteristically have conditions

Us stating conditions is just an approach to defining the function

[past peak]

i mean a relation to be called a function does have conditions that it should follow

only then it can be considered as a function

[grand thunder]

Are you using relation in the sense of a set of ordered pairs

[past peak]

yes exactly

[grand thunder]

Or are u using the term casually

[past peak]

yes

[grand thunder]

Then yeah I agree

I was talking about the conditions like "if x > 0" in our notation

[grand thunder]

Not sure what this means if you're still using relation as a set of ordered pairs

[past peak]

the thing i'm not able to understand what those things mean

like when you have 2 different conditions on the right

[grand thunder]

Yeah

Well

Let me write up soem latex hold on

[past peak]

in f(x) = x
here i understood is that if I input the of x in f(x) it'll be domain or a value of set A
and value of x in right hand side will be the range or value of set B

but when there are 2 conditions i'm not able to understand what they mean/ what they're trying to say

[grand thunder]

[
f \colon { a, b } \to { 1, 2 }; \
f(x) = \begin{cases}
1 & x = a \
2 & x = b
\end{cases}
]

[warm shale]

futsal gkn1

[grand thunder]

@past peak

do you understand this example

we're defining the rule of f using the notation after the semicolon

For our input, x, output 1 if x = a and 2 if x = b

[past peak]

what does this mean? never learnt it before

[grand thunder]

$f \colon A \to B$ is a function that maps from set A to set B

[warm shale]

futsal gkn1

[grand thunder]

A is the domain, B is the codomain

[past peak]

the function is just defining all the values of the set using different conditions. am i right?

[grand thunder]

it defines what each value of the domain maps to using different conditions yeah

[grand thunder]

f(a) = 1 and f(b) = 2

[past peak]

just give me some time. lemme think

[grand thunder]

maybe you're overtaking it or need a perspective shift

the notation is just another way of defining the elements of the domain map to

[obtuse pebble]

@past peak Has your question been resolved?

[past peak]

here the domain being the value of x in f(x)
and
range being the value of the rhs, and its written that the value will be so and so if in the value of x in f(x) follows a certain set of conditons

this is what i've understood

[grand thunder]

the domain is the set of all objects the function maps from

so yes x is in the domain

[grand thunder]

it sounds like you've got it yeah

[past peak]

finally

@grand thunder Thanks.

.close

[obtuse pebble]

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[obtuse pebble]

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[subtle igloo]

I am supposed to prove that the two vectors means the same line

[subtle igloo]

I am guessing that i need to check if they are parallel first. But i have no clue how to do that

[royal shard]

what a nice task

one second

[subtle igloo]

yeah no worries

[royal shard]

when we look at these vectors, we can split them into two parts
one vector with the t's and one with only numbers

for the first one we would get:
(2t,-t,-t) + (1,2,3)

the sum of these two makes up the whole vector again

we can write (2t,-t,-t) as (2,-1,-1)*t

we can now call (2,-1,-1) our direction vector

[subtle igloo]

That is basically parametric vector form right?

[royal shard]

on a different note, if we look at linear function we had:
y=mx+b
here m was our "direction" or slope
here it is the same
v=a*t + b
with a being the "direction"

if we have the vector (2,-1,-1)
the parametric form would be 2x-y-z

[subtle igloo]

ah okay

[royal shard]

two function y=mx+b are parallel when their m are equal

so our vectors are also parallel if their direction vector is equal

so in this case, we get a "direction" of (2,-1,-1)

[subtle igloo]

Okay. so far i understand what you are saying

[royal shard]

now let's get the direction vector of the second one

maybe try it yourself

[subtle igloo]

(-2,1,1) ?

[royal shard]

great

now would you say they are parallel?

[subtle igloo]

well. they would be the same if one of them would be multiplied by (-1). doesnt that mean that they point towards opposite directions?

[royal shard]

exactly

so they are parallel, but point in opposite directions

still parallel though

[subtle igloo]

Ah i see

[royal shard]

now if two lines are parallel and we want to check if they are equal
it is enough to check if there is one point where they intersect
because then since they are parallel, all points will be shared between the two

and thus they will be identical

[subtle igloo]

Okay

[royal shard]

do you have an idea on how to check this?

[subtle igloo]

not really

[royal shard]

we can think of these vectors as functions
so f(x)=ax+b
just that f(x), a and also b are vectors
now if a point P lies on this function, then P=f(x) has a solution

it also works the other way around
so if we find a solution to P=f(x)
then we can know for certain that P lies on f(x)

so let's start by getting this point P
we can choose any point of one of the vectors

so let's just set t=0
the first vector gives us (1,2,3) then

now to check if this point lies on the other line as well, we set them equal

now let's see if we get a solution

do you see, how we would do that?

on the left side we have (1,2,3) and on the right (-2,1,1)t+(3,1,2)

[subtle igloo]

if we set t to a value that makes them become equal?

[royal shard]

yes that works

but guessing can potentially take a long time

we can simplify this equation by subtracting the vector (3,1,2)

so we get
(-2,1,1)=(-2,1,1)t

now we can clearly see that t=1 is a solution

so in total:
-we found that the vectors point in the same, but opposite direction
-therefore they are parallel
-we found a point that both lines go through
-since they are parallel they are equal

[subtle igloo]

You are very good at explaining. I have no clue why my teacher didnt go through this

Thank you for your help

have a good day

[royal shard]

you too ^^

[subtle igloo]

.close

[obtuse pebble]

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[icy yacht]

I have a question. So if an item cost $281 for 50 quantity, how much would I pay if I had 6 of the 50?

[unborn anvil]

uou would do 281/50

to get the price of one

and then multiply by 6

to get the price of 6 of the 50

[icy yacht]

I tried 6 times 50 and its 300 (19 over 281) and doing 5 times 50 is 250 which is 31 under 281

[unborn anvil]

wait what

what was ur answe?

[icy yacht]

Like what I need to spend of the total?

[unborn anvil]

yea

[icy yacht]

The total?

[unborn anvil]

tell me ur answer

[unborn anvil]

the price of 6

[wanton hull]

,w 281/50*6

[warm shale]

Results provided by WolframAlpha

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

Requested by bettim#9852

[wanton hull]

this should be your answer

[icy yacht]

The whole math thing is in a game i'm playing. And the game only goes by cents and non whole dollars

[wanton hull]

1 dollar = 100 cents

just multiply by 100

[obtuse pebble]

@icy yacht Has your question been resolved?

[obtuse pebble]

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[timid silo]

Hello I need help covering some rather easy topics

I was neglected to be taught as a child because I have autism and my working memory is ass so people got fed up with me rather quickly

I'd appreciate someone super patient willing to eli5 because I'm not the best at even simple things, I still can't divide in my head.

So, that being said

I need to cover real numbers and Algebraic expressions

Properties of exponents and radicals .
Polynomials and factoring

And rational expressions

[timid silo]

also my English is 6/10?

my teacher currently (I go to a technical college for nursing) kind of talks at me rather than to me

which is a common phenomenon with math teachers in my location

[trim portal]

What do you mean by real numbers?

[timid silo]

uhhhh

[royal basin]

those sound like kind of broad topics that would be hard to cover in a discord chat

you might do good to go to http://khanacademy.org/ perhaps

[trim portal]

https://www.khanacademy.org/math/algebra

[timid silo]

this is true but I'll give some example problems

Khan academy makes me freak out ngl

[royal basin]

oof

[timid silo]

genuinely I'm stupid

[trail cloak]

You're not

No one is stupid

[timid silo]

there is almost no hope in me really

this is kind of a last stitch effort before I fail and try to figure things out

like what even is an integer

[trail cloak]

Okay

You know the regular numbers right

[trim portal]

Integers: -1, -4, 1, 0, 10, 100, 532 and everything similar

[trail cloak]

Those ones you use and count with

[royal basin]

an integer is a positive whole number, a negative whole number or zero.

[timid silo]

I just don't see how those are similar

[royal basin]

..., -4, -3, -2, -1, 0, 1, 2, 3, 4, ...

[trim portal]

None of them have decimal part

[royal basin]

do you know natural numbers

[timid silo]

not quite

Like 1 2 3

....

[royal basin]

natural numbers are what you count with

1, 2, 3, 4, ...

[trail cloak]

Yep

[royal basin]

these, their negatives, and zero

those are the integers

[trail cloak]

Now integers is like whole numbers, but they can be plus or minus

[timid silo]

one sec I will pull up an image I need explained

[royal basin]

yeah, do that

[trail cloak]

Okay so like

Natural numbers are the numbers you use for everyday stuff

[timid silo]

so 1 2 3 4 5... Positive ONLY is natural numbers

[trail cloak]

Yea

[royal basin]

oh god the "whole as natural with zero" shit

[timid silo]

Okay so

whole is like

What is whole specifically

[royal basin]

in this picture, whole numbers means natural numbers including zero

[timid silo]

so like

0, 1, 2, 3

is whole numbers

[trail cloak]

$\mathbb{N}_0$

[warm shale]

VulcanOne

[royal basin]

0, 1, 2, 3, ...

yes

[trail cloak]

I think

[timid silo]

what's the point of having whole numbers then if they can be represented by integers as well

or is it just for technicalities

[royal basin]

integers are a wider set

[trail cloak]

They are different sets for different uses

[timid silo]

integers contain negatives as well?

[royal basin]

-3 is an integer but not a whole number, according to that pic...

[trail cloak]

Yep

[timid silo]

So integers would be like -3 -2 -1 and 0 1 2 3

[trail cloak]

Yeppo

[royal basin]

sure

[timid silo]

are there any exceptions to this

[royal basin]

exceptions to what

[timid silo]

I just feel like there's something missing like you said sure

what else is included

[royal basin]

okay so like

the reason why i said a somewhat noncommittal "sure"

is that the way i heard it, you have glossed over the fact that the natural numbers go on forever

they don't stop at 3 !

[timid silo]

oh wait so do rational numbers include fractions in between each number as well

[royal basin]

and the integers go on forever but in two functions

[timid silo]

like negative and positively?

[royal basin]

rational numbers are fractions with an integer on the top and bottom

this includes both positive and negative rationals

[timid silo]

and yeah I know they go on forever I see no reason to stop at 3 I just forgot the ellipsis

[royal basin]

i caught onto your forgetting the ellipsis yes

[timid silo]

Why do they have to have an integer on top and bottom

oh wait nevermind I answered my own question

that's like everything else, it HAS to be

[royal basin]

it's a definition.

[timid silo]

oh so an integer would be defined as any number that is not a fraction

[royal basin]

eh.

i think that's a bad definition.

[timid silo]

what would be a better definition

[royal basin]

rational numbers are fractions with an integer on the top and bottom

[timid silo]

sorry for being obnoxious as well

[royal basin]

idk maybe you don't want me to repeat myself

maybe you want me to write things in a different way somehow

[timid silo]

I don't know what I want I'm just trying to cover fundamentals of things I don't understand before I tackle bigger things

[wanton hull]

quick question, we dont need to apply quotient rule while using l'opital's rule?

[timid silo]

the definition of square root confuses me

[median dome]

open your own help channel, this channel is taken

[wanton hull]

it was just a y/n question thats why i just asked it here sry

[timid silo]

So the definition says a number which produces a specified quantity when multiplied by itself.

like

7 is the square root of 49

my questions are

1. How would that be expressed (7 as the square root of 49)

2. For example would 2 be the square root of 4 since 2x2 is 4 and that's a set quantity

man I wish I could save this chat

I'm gonna take some screenshots but my late class is over so I can't see the math I'm working with anymore

I'll try to be more descriptive tomorrow and this afternoon

this helped though

got my screenshots

so why is Pi so significant

like how the hell does it do the things it does

like finding measurements that it doesn't even concern itself with

it just seems so random

I've got some specific subjects now

Field properties

how do I use them?

And what are they used for

I think tomorrow I'll go through all of the lessons I've completed and send pictures and stuff here for better explanations

with more concise questions

thanks guys

[obtuse pebble]

@timid silo Has your question been resolved?

[trail cloak]

π is really important because it appears everywhere

[obtuse pebble]

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[main goblet]

anybody know this one?

[wanton hull]

wdym know this one

what do we have to do with this random terms

[golden oriole]

What is the question
Send full picture

[timid silo]

We have to factor it

Probably

[main goblet]

oh welp nvm i already found the answer

sorry

[obtuse pebble]

@main goblet Has your question been resolved?

[obtuse pebble]

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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.

[grave flicker]

I got this wrong I was wondering how to solve this problem

[sage geode]

You can start by combining the logarithms

[grave flicker]

Wait in what way>

?

[sage geode]

Recall that $\log{a} - \log{b} = \log{\frac{a}b}$

[warm shale]

A Lonely Bean

[grave flicker]

omg i forgot that

So ln 3x+1/x+3

as x approaches inf

[supple granite]

consider it a composite function

[sage geode]

Yeah, now, since ln is a continuous function, you can interchange the limit and the ln as long as the expression inside has a limit

[supple granite]

of ln(x) and (3x+1)/(x+3)

[grave flicker]

Wait so one second let me look at this for a second

[grave flicker]

Im kind of confused what you mean by this

[sage geode]

$\lim_{x\to\infty}\ln{\frac{3x+1}{x+3}}=\ln{(\lim_{x\to\infty}\frac{3x+1}{x+3})}$ as long as $\lim_{x\to\infty}\frac{3x+1}{x+3}$ exists

[warm shale]

A Lonely Bean

[sage geode]

Now we need to evaluate the limit that I mentioned the last

[grave flicker]

oh so you took out the ln

so if i find the lim of inside function

[supple granite]

let f(x)=ln{x} and g(x)=(3x+1)/(x+3)
the limit you're looking for is lim x->inf
f(g(x))

[grave flicker]

I can take ln of that

[sage geode]

Yes, thanks to the fact that ln(x) is a continuous function on its domain

There are two ways to evaluate the limit of (3x + 1)/(x + 3)

[grave flicker]

so the lim would be 3

or ln(3)

[sage geode]

1. Divide top and bottom by x
2. Rewrite the fraction as 3 - 8/(x + 3)
   (L'hopital's rule is also an option, but a whack one)

The final answer is ln3, yup

[grave flicker]

omg thx guys

you helped me so much

Next time i wont mess that up

i appreciate it

.close

[obtuse pebble]

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Remember:
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• 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.

[bright geyser]

how do i find a normal to a line

[bright geyser]

vector form

so normal to a line of like

(0,0,0) + (1,1,1) * t

[royal basin]

there are many vectors that are perpendicular to a given straight line.

[bright geyser]

how would i find

some

[royal basin]

you want a vector that is perpendicular to (1,1,1)

i.e. the direction vector of your line

i.e. a vector (x,y,z) satisfying (1,1,1) · (x,y,z) = 0

[bright geyser]

yeah

[royal basin]

this is a system of 1 linear equation in 3 variables.

[bright geyser]

ya

i wouldnt know hwo to do that i thinkl

[royal basin]

have you considered that perhaps you are overthinking it

you yourself said you do not need to find all solutions but only one

[bright geyser]

uh oph

just sub

select a x and y

and find the z

or smth

[royal basin]

yes

[bright geyser]

cheers

it is late

JRHJF

.close

[obtuse pebble]

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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!

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[worthy peak]

How do i find x

[silent hatch]

@worthy peak

[glossy basalt]

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[worthy peak]

either theres a miscalculation in my part or something is wrong

my textbook ahve diff answers

[glossy basalt]

yea, it's a miscalculation

[worthy peak]

from my part?

[silent hatch]

Did I do smth wrong?

[glossy basalt]

usually we won't cancel out something with x

[worthy peak]

oh?

[glossy basalt]

anyways, it's good process, but to further simplify it, actually it good to notice that it's a quadratic

[silent hatch]

Btw, it won't equal 0 cause it just can't

So the only answer is 1/2

[glossy basalt]

Let t=x^(-1/3),
let's see what you can get

[worthy peak]

where do i plug t?

[glossy basalt]

very first expression, it will make the whole thing shorter and easier to spot

[worthy peak]

what do i do with the second polynomial

the -1/3 x^ (-2/3)

[glossy basalt]

$x^{-\frac23}=(x^{-\frac13})^2$

hmmm

LaTeX down

what i meant is
(x^(-2/3))=(x^(-1/3))^2

[worthy peak]

ah got it

[glossy basalt]

do tell if you need more help

[worthy peak]

do i just factor out the t?

[glossy basalt]

yep

[worthy peak]

then the inside i combine into one?

actually not rlly sure abt the next step

[glossy basalt]

did you get this:

t(2/3-t/3)

[worthy peak]

yes

[glossy basalt]

then, since we don't have x in 3, we can multiply the 3 on both sides of the equation

we have

t(2-t)=0

[worthy peak]

oh

[glossy basalt]

then
t=0 or 2-t=0

[worthy peak]

t = 2?

[glossy basalt]

t=0 or t=2

then we plug back in the t=x^(-⅓)

x^(-⅓)=0 or x^(-⅓)=2

[worthy peak]

?

[glossy basalt]

for the first one
"since we cannot have any number x such that 1/(x)^⅓=0, therefore we have to reject this case"

[worthy peak]

Oh

If the second?

Where did i misscalculate

[glossy basalt]

for the second one
t=2

1
_ = 2
x^⅓

x^⅓=½

x=(½)³

x=1/8

[worthy peak]

oh

[glossy basalt]

all good?

[worthy peak]

one moment

can u explain this

[glossy basalt]

yea

[worthy peak]

the only way i see this is cross multiplying

and it doesnt equal to the below one

oh

nvm

u removed the 2

so it became half

[glossy basalt]

1 2
_ = _
x^⅓ 1

can you see it now?

[worthy peak]

yeah