help-36

[fathom walrus]

$\langle 2(x-4) , 2(y-2) \rangle = \lambda \langle 2, -1 \rangle \newline g(x,y) = 2x - y + 3$

[fathom walrus]

ok so now u got

$\lambda = x - 4 \newline \lambda = 4 - 2y \newline g(x,y) = 2x - y + 3$

[fathom walrus]

correct? @gilded flame

[gilded flame]

Yes

[fathom walrus]

and then u set the lambdas equal and u got

$x = 8 - 2y \newline g(x,y) = 2x - y + 3$

[fathom walrus]

remember this

set g to 0 now

[gilded flame]

Yes, so I set g to 0 and the plug in that value of x, I get x= 2/3

Which is what I got before

Oh wait no its y

Yeah y=19/5

[fathom walrus]

2x - y + 3 = 8 - 2y - x

[gilded flame]

Because it simplified to -5y+19

[gilded flame]

Huh?

[fathom walrus]

?

[gilded flame]

How did you get 8-2y-x

Oh wait

I see

So you don't substitute it in

[fathom walrus]

,w solve x = 8- 2y, y = 2x + 3

[soft zealot]

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Requested by northsteve

[fathom walrus]

there u go

[gilded flame]

Oh so I just got 2/3 instead of 2/5 for some reason?

[fathom walrus]

well that means u did the algebra wrong

[gilded flame]

Oof

[fathom walrus]

x = 8 - 2y
y = 2x + 3

[gilded flame]

Yeah

[fathom walrus]

y = 2(8-2y) + 3

y = 16 - 4y + 3

5y = 19

y = 19/5

[gilded flame]

y=-x+4

[fathom walrus]

x = 8 - 2y

x = 8 - 2(19/5)

x = 8 - 38/5

x = 40/5 - 38/5

x = 2/5

this means u did all the multivariable calculus correct, and the algebra 1 wrong

[gilded flame]

Ouch

[fathom walrus]

XD

[gilded flame]

Ok but wait one second

I still get x = 2/3 if I solve for y instead of x from the lamda equation

Because y=-1/2x+4

So then 2x-y+3=3/2x-1=0

Which means 3/2x = 1 or x = 2/3

Oh would that mean there are two possible values for lamda and now I need to solve for the other point?

[fathom walrus]

no

u made an algebra mistake somewhere, let me find it

ok so

y = -1/2 x + 4

2x - y + 3 = 0

plug in the first equation into the second

2x - (-1/2x + 4) + 3 = 0

2x + 1/2 x -4 + 3 = 0

5/2 x - 1 = 0

[gilded flame]

Omfg I am so dumb

[fathom walrus]

5/2 x = 1

[gilded flame]

Lmao

Thank you so much

[fathom walrus]

dw, it happens

np

[gilded flame]

Seriously I don't know if I would have gotten it, I've spent so long on this question 🥲

Really appreciate it

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[fathom walrus]

sometimes it helps to take a break and come back to redo the question without looking at ur previous work for the question

[gilded flame]

Very true

I've also been working on this since 6 am 🫡

[fathom walrus]

this 1 problem ?!

[gilded flame]

No no

The pset

Well I had an exam at 6

And this is the pset for next weeks exam

Oh god one problem for 13 hours 🥶

[fathom walrus]

ah i see

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[arctic swallow]

1.08*1000+1.08*1000+960.9+802.6+1.02*1000+693.3+604.9+678.8+518.6+429.3+388.3+351.6+341+408.4+290.4+255.4+235.8+226.6+214.8+205.6+207+204.4*2+187.9+186.3+184.6+173.1+173+171.1+171.7+171.4+169.5+167.7+167.2+163.2+159.2+157*2+154.4+150.2+149+143.8+142.5+140.1+138.6+136.2+130.6+128.2+119.3+117.5+116.9+115.2 + 110.2+106.2+98.2+94.2+94+92*3+91.9+90.6+86.4+79.3+79+77.5+77.3+73.7+72.9+72+69.9+69.8+69.3+63.2+60.6+59.5+50.1+47.6+44.4+42.9+42.8+41.3+41.2

[arctic swallow]

Can someone calculate this?

My calculator won't do it and i need it for some computer storage stuff 😭

[formal trail]

,calc 1.081000+1.081000+960.9+802.6+1.021000+693.3+604.9+678.8+518.6+429.3+388.3+351.6+341+408.4+290.4+255.4+235.8+226.6+214.8+205.6+207+204.42+187.9+186.3+184.6+173.1+173+171.1+171.7+171.4+169.5+167.7+167.2+163.2+159.2+1572+154.4+150.2+149+143.8+142.5+140.1+138.6+136.2+130.6+128.2+119.3+117.5+116.9+115.2 + 110.2+106.2+98.2+94.2+94+923+91.9+90.6+86.4+79.3+79+77.5+77.3+73.7+72.9+72+69.9+69.8+69.3+63.2+60.6+59.5+50.1+47.6+44.4+42.9+42.8+41.3+41.2

[soft zealot]

Result:

18126.5

[wary egret]

Got the same

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[arctic swallow]

TYSM

UR THE BES

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[tranquil pine]

i need some clarification

[tranquil pine]

why is this? and how?

[waxen bobcat]

You did it right

Then what happens when you have 3^12/1/Y^10

@tranquil pine

[tranquil pine]

that's what im asking for

🤔

[waxen bobcat]

When you divide a fraction the denominator goes up yes?

If I divide 1 by 1/2 it becomes 2

[tranquil pine]

hmm

lemme think

[molten remnant]

basically dividing 1 by 1/2 is simply multiplying 1 by the reciprocal of 1/2 that is 2

[waxen bobcat]

^

Brother explained it better

[tranquil pine]

yeah ig it does

[waxen bobcat]

Exactly

[frozen sparrow]

Hi what you need dawg

[waxen bobcat]

So apply the same logic

[tranquil pine]

lemme think even further

[waxen bobcat]

It’s exactly the same

[frozen sparrow]

No dude is a law

[waxen bobcat]

💀

[frozen sparrow]

Look

Im seeing derivades and they ask us for this

[tranquil pine]

ONG YOU'RE RIGHT

[waxen bobcat]

Ofc

[tranquil pine]

mind blowns

[frozen sparrow]

Then you cant derivate so you do this

[waxen bobcat]

No questions close

Don’t know what this dude yapping about

!done

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[tranquil pine]

.close

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[lime cloak]

Hi. Gotta need small hint. Do i strike for uh sqrt(3)^2+1 or sqrt(3)^2-1?

[barren hound]

$\sqrt3^2 = 3$

[soft zealot]

open up my e-girl eyes

[tranquil pine]

What does the question even say? 🗿

[lime cloak]

Transform into multiply

[tranquil pine]

Ah..

[lime cloak]

Yeah

[tranquil pine]

Is ur question this?

[lime cloak]

Just that first

[tranquil pine]

Do we need to find the value of x?

[lime cloak]

Sqrt(3)*sinx-cosx

[tranquil pine]

What are we sposed to do?

[lime cloak]

To that equation transform into multiplication

[tranquil pine]

Ah..

I may be dimwitted

[lime cloak]

Um i just need to know

To use sqrt(3)^2+1 or sqrt(3)^2-1?

I can solve

I just need that hint

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@lime cloak Has your question been resolved?

[lime cloak]

Huh

Why

[tranquil pine]

I still cant get ur problem

[lime cloak]

Don't worry. I realised what to do now

[tranquil pine]

So, ur problem fixed?

[lime cloak]

Yeah. Thank you

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[fathom stream]

this is a binomial distribution question right?

[fathom stream]

n = 5
k = 0, 1, 2
p = 0.8
1-p = 0.2
n-k = 5, 4, 3

so it would be

P(X <= 2) = P(X=0) + P(X=1) + P(X=2)

[muted prairie]

Yes

[fathom stream]

ty

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[lime cloak]

Hello. Is this correct? The question is transform into multiply

[warm python]

idts

$sin(45)\neq 1/2$

[soft zealot]

ƒ(Why am. I here)=I don't know

[lime cloak]

oh

wait

.close

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[lone dove]

the closure of a closed set is itself right?

[covert tree]

yes

it is the union of the set with its boundary. and for a closed set the boundery is within the set. It immediately follows that the closure of a closed set is itself

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[lone dove]

based thank you

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[ruby hinge]

Can someone please give me a hint for this problem?

[mellow axle]

how is |F| defined for you

[ruby hinge]

Outer measure

@mellow axle

[trail mango]

lol

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[mellow axle]

hmm, not sure how good of a hint this is, but i would try constructing open sets containing all the rationals in $[0, 1]$, say $X_i$ and then consider what happens with $(\bigcup_i X_i)' \cap [0, 1]$

[soft zealot]

p norm reformed

[mellow axle]

@ruby hinge

sorry for late reply, my fucking cat decided she needed to excrement on me

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[violet rune]

about complexe number, when we're searching the real part is this true:

Re(a/b) = Re(a)/Re(b)

[tranquil pine]

If a and b are complex numbers

No

[violet rune]

hum ok

if a is real and b complex?

[tranquil pine]

Nope

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[lavish spade]

I have to count it in real numbers someone help?

[tranquil pine]

!status

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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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[violet stream]

you need to find the solution to this inequality?

[tranquil pine]

did u check the discriminant of this quadratic?

[lavish spade]

Yes

[tranquil pine]

what was it

[lavish spade]

D= -11

[tranquil pine]

do you know waht that means?

-ve discriminant?

[lavish spade]

It’s in Iracional numbers?

[tranquil pine]

the roots are not real

ie the curve doesnt intersect the axis

[west berry]

Imaginary, you mean?

[lavish spade]

If it was

[lavish spade]

Maybe

[tranquil pine]

what is this

[lavish spade]

Idk 🤷🏽

[tranquil pine]

ok so lets do it again

[lavish spade]

My teacher said it has solution

I know D can be -11

[tranquil pine]

so you know the curve of this quadratic will never intersect the axis

[lavish spade]

But she said it has solution that’s why I am confused

[tranquil pine]

right?

yea it has

[tranquil pine]

do you know this?

[lavish spade]

But in R?

[tranquil pine]

yea

[lavish spade]

So it doesn’t have solution in real member

[tranquil pine]

there will be real soutions to this inequality but non real solutions of the quadratic

[lavish spade]

Alright

[tranquil pine]

do you know how a graph of a quadratic equation looks like

[lavish spade]

Yes

[tranquil pine]

so if i ask you, if we have a quadratic equation ax^2+bx+c=0, and a is positive

what direction will the graph move

or in what direction will the mouth of the curve be

[lavish spade]

What is ^

[tranquil pine]

what do you mean

[vocal forge]

\text{ax^2} = $ax^2$

[tranquil pine]

put the lhs in $$ too

and what is this

[lavish spade]

Oh yea

[lavish spade]

The parabola will start on 0 and then it will go up in both sides

[tranquil pine]

it will not start from 0

but yes it will go up

[lavish spade]

I mean it that way 😭

So thank you for help

[tranquil pine]

yea

so there will be infinite solutions to this inequality

np

[lavish spade]

Okii

[tranquil pine]

6. ?

!done

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[stone flume]

What do dots mean on lines?

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[jagged flare]

#help-32 message

[jagged flare]

ive discussed it up to getting the equality $$b=r+\frac{1+\sqrt{8r+1}}{2}$$ and 8r+1 will produce all perfect odd squares

[soft zealot]

Skill_Issue

[jagged flare]

8r+1=(2k-1)^2=4k^2-4k+1
8r=4k^2-4k
8r=4k(k-1)
r=k(k-1)/2

wait this seems wrong

where did i go wrong?

take k=1, r=0 but its not like that here

wait

brb

so $$b=\frac{(k)(k+1)}{2}+k$$

[soft zealot]

Skill_Issue

[jagged flare]

establishing some bounds, $$4\leq k \leq 63$$

uh oh

i went wrong somewhere

hold on

ah

[jagged flare]

this is suppoaed to be a -

[soft zealot]

Skill_Issue

[jagged flare]

so 60

yay

integers x,y,z follows this
$$x^3-y^3-z^3=3xyz$$
$$x^2=6(y+z)$$
$$x^2+y^2+z^2 \neq yz-x(y+z)$$
how many positive factors of x are there

[soft zealot]

Skill_Issue

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[tranquil pine]

that thing looks factorable

x³-y³-z³ - 3xyz

changes to (x-y-z)(x²+y²+z²+xy-yz+xz)

ye so x = y+z

[jagged flare]

how did you get this

oh wait nvm

x=6

so 4?

in a circle with center O, there are 2 congruent rectangles ABCD and EFGH where all the verticies lie in the edge of the circle, AB//EF, AB=EH=2 and AD=EF=4, there is also square IJKL where the verticies are on the edge of the triangle and AB//LJ, what is the area of the shaded area? (shaded area is red)

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[lapis wedge]

came across this in my book and never seen this done before. I was wondering what rule is this, or is it a composition of rules that allows it, does it work or all real numbers and how could we prove that?

[lapis wedge]

sorry nvm

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.reopen

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✅

[lapis wedge]

does a^n = (a/x)^(n*x) where a and n and x is all real numbers?

[marsh temple]

2^2 = (2/2)^(2*2)?

[marsh temple]

here they just rewrote 4 as 2^2 and invoked an exponent law

[lapis wedge]

oh ok that makes sense

thank you

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[amber light]

I have no clue if my answers are correct:
a) y = 3x
b) k = 3
c) 12
d) 5

[onyx peak]

Are you sure your formula works?

y = 3x

try plugging in x = 12

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[amber light]

apparently it doesn't

oop

||Okay but the truth is, it's that these answers were on the answer page||

[onyx peak]

that's strange

[amber light]

yeah

[onyx peak]

that's task 1 and 2

[onyx peak]

this is task 3

[amber light]

wth

so they switched it around?

[onyx peak]

is there anything else in the answer key?

[amber light]

only these (as in answers that are for questions that are somewhat similar)

[onyx peak]

3a is right

[amber light]

oh

[onyx peak]

this is exercise 3

[amber light]

oh my

wait

I'm actually blind as heck

💀

sorry for wastin' ur time T-T

[onyx peak]

np lol

[amber light]

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[steel kite]

can someone explain to me why the hyperbola is like this?

[grand fossil]

guys please help, the teacher is coming how much is 4*6?

[steel kite]

😟

[grand fossil]

quick she's almost here

[steel kite]

6+6+6+6

quick quick do it

@grand fossil

[grand fossil]

I can;t

[steel kite]

whats 6+6?

[grand fossil]

uhm

12

[steel kite]

ok

[grand fossil]

i think

[steel kite]

what is 12 + 6 + 6?

[grand fossil]

uhh

18?

[steel kite]

12 + 6 = 18

correct

now

add one more 6

[grand fossil]

24?

[steel kite]

see you can do it

dont be silly

[grand fossil]

YES

now I won't fail

you're the best

thanks

[steel kite]

😭 no worries

[final tangle]

not a hyperbola

[steel kite]

hmm

why is it not a hyperbola

because x^2?

[final tangle]

yeh

[steel kite]

hmm

so how does it work then

[final tangle]

just take the reciprocal of each y coord at each x

[ripe wharf]

when x=0 that means the function is 1/0 which is undefined

[final tangle]

and consider limits to infinities and when its undefined for asymptotes

[steel kite]

im not sure what this means

[final tangle]

you know what y = x^2 looks like right?

[steel kite]

yep

just a normal parabola

starting from origin

[final tangle]

and you know the y-coords at each value of x right?

1/x^2 is just 1/that

in simpler words,
you have
y = 1/x^2
just plug in values of x to get your points

[steel kite]

yeah cause if i have x, i can get the y for that coordinate

[steel kite]

ohh ok that makes sense

ok so why would that purple (3rd equation) be drawn like that

it intercepts the y axis at (0, 0.25)

which is 1/4

$\frac{1}{x^2+4}$

[soft zealot]

pixel

[steel kite]

x^2 + 4 = 1/4?

im confused

@final tangle

[final tangle]

when x=0, you get
1/(0^2 + 4) which simplfies to 1/4

[steel kite]

ooo right right

to get y intercept, we let x = 0

and to get x intercept, we let y = 0

which doesnt work because

$0 = \frac{1}{x^2+4} \ 0 = 1$

[soft zealot]

pixel

[steel kite]

which is like

what

but yeah

this is 1/f(0), i understand that bit

and this is 1/f(x) where x is anything but 0?

(purple lines)

@final tangle

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[warm python]

$\int \frac{\sqrt{1+\sqrt{x}}}{x}$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

would a trig sub help here?

or would $\sqrt{x}=u$ be a better sub

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

that would give $\int \sqrt{\frac{1+u}{u}}$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

oh

right

but can this be done with a trig sub

[ivory ingot]

Why trig sub when you might be able to solve this without?

[warm python]

just for fun

[ivory ingot]

Ah well, this could still be awful I guess

[warm python]

I know how to do this without a trig sub

[warm python]

multiply and divide by $\sqrt{1+u}$

[soft zealot]

ƒ(Why am. I here)=I don't know

[ivory ingot]

Maybe since you got 1 + sqrt(x), substitute x = tan^4(t) or stj

[hybrid basalt]

wait substituting u = sqrt(x) does not give that for me

[warm python]

there should be a 2 in the denom

my bad

[ivory ingot]

So that you can use 1 + tan^2(x) = sec^2(x)

[warm python]

hmm, yeah

thanks!

[hybrid basalt]

$\int 2\frac{\sqrt{1+u}}{u}$

i can't use latex lol

[soft zealot]

smygalz

[hybrid basalt]

ok i get this

[warm python]

oo

yeah

my bad

made a mistake

let $u=tan^2(t)$

then?

[soft zealot]

ƒ(Why am. I here)=I don't know

[ivory ingot]

Yeah that seems good

[warm python]

thanks!

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[warm python]

thank you!

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[pallid patrol]

f(x,y) = x^2-4x+y^2+6y, x,y are from 0 to 1, find min of f(x,y)

[pallid patrol]

How to do this with calculus

[spring haven]

with calculus?

[forest ember]

Compute critical points

[spring haven]

you don't really need that here though

ig you can say $f(x+y)=g(x)+h(y)$ where g and h are independent functions of one variable only

and find their minimum values

[soft zealot]

kheerii

[final saddle]

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[rare umbra]

i want to check if i have done this correctly

[rare umbra]

since 0/infinity is not an indeterminate form i can just equate it to 0 right?

[final tangle]

if you actually have 0/inf, yes
however,
lim as x→0 of e^(2/x) isn't quite inf

you should be considering one sided limits here

the above is what you'd get for lim as x→0^+

[rare umbra]

i see

how do i solve the lim as x→0-

because that gives 0/0 form and differentiating it for l'hospital doesn't help

[final tangle]

doing a substitution like u = -2/x would be helpful

[rare umbra]

i see

so for x→0-, the limit would be equal to infinity right?

[final saddle]

@rare umbra Has your question been resolved?

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[tranquil pine]

Help pls

[humble swift]

what did u try to do?

[final tangle]

continuing on

so
1 : sqrt(2)
and you want
3sqrt(2) : a
so to get what you want, you can multiply both sides of 1 : sqrt(2)
by the appropriate amount

before those jerks invaded

[tranquil pine]

So multiply sqrt2 * sqrt2 ?

[final tangle]

no

you're starting with 1 on the left side of the ratio
and you want to end up with 3sqrt(2) on the left side

[tranquil pine]

Yea

[final tangle]

1 * [what?] = 3sqrt(2)
don't overthink this

[tranquil pine]

Ohhh

Do u’d have to get 3sqrt2 by 1*x

So use the pythagorean

[final tangle]

if you're going this route, you don't need pythag

[tranquil pine]

Oh

[final tangle]

so just multiply both sides of

1 : sqrt(2)
by 3sqrt(2)

[tranquil pine]

Ok wait

I got 6

[final tangle]

yes

good

[tranquil pine]

But how do u know if that was sqrt2

[final tangle]

wdym

[tranquil pine]

Cuz of the ratio?

Oh wait

Nvrm

[final tangle]

yes

[tranquil pine]

I understood

Thanks!!

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[blazing apex]

(vector functions)

[blazing apex]

why is both domain and range not [0,4]

-4,4

oh

pi

no

?

2pi
2t ?

0 clue

yea

2pi/n

so t no ?

2pi/2t

so pi/t

idk

2t

ok so both periods are t

but what does the period have to do with domain and range, i thought only amplitude matters

?

why

yea

the domain is the range of x

and the range is the range of y

so in theory since we have a restriction of [0,pi/2]

both domain and range should be [0,4]

but the domain si [-4,4] and range is [0,4]

sure

wouldnt cos go from 1 to 0 to 1 again

ok

wat

that still doesnt explain why domain is -4,4 and range is 0,4

[blazing apex]

anyway for whomever else who sees the channel

[final saddle]

@blazing apex Has your question been resolved?

[blazing apex]

ahh

makes sense

thanks

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[warm python]

so I'm trying to verify that the space of polynomials is a vector space

[warm python]

to start let

$v=\sum_{i=0}^na_ix^i$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

$u=\sum_{i=0}^nb_iy^i$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

$u+v=\sum_{i=0}^nb_iy^i+a_ix^i$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

$v+u=\sum_{i=0}^na_ix^i+b_iy^i$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

now let's evaluvate the function at an arbitarary point $a$ ; $a\in \R$

[soft zealot]

ƒ(Why am. I here)=I don't know

[hearty zephyr]

your polynomials would be in the same variable(s)

[warm python]

oh, okay

$v+u=\sum_{i=0}^na_ix^i+\sum_{i=0}^mb_ix^i$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

and$u+v=\sum_{i=0}^mb_ix^i+\sum_{i=0}^na_ix^i$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

evaluvating these at a and applying the properties of reals, it;s apparent they are both equal

is that right?

[hearty zephyr]

you don't need to evaluate it

take $k = max{m,n}$\
then $u+v = \sum_{i=0}^k b_ix^i+a_ix^i = \sum_{i=0}^k (b_i+a_i)x^i$ by introducing some 0 coefficients\
Then apply properties of reals to $b_i+a_i$

[warm python]

ah

[soft zealot]

Zybikron

[warm python]

so I use the fact that $a_i+b_i =b_i+a_i$

[soft zealot]

ƒ(Why am. I here)=I don't know

[hearty zephyr]

yep

[warm python]

ok

thanks

[hearty zephyr]

its the same argument of 'addition of reals is commutative' but without going through evaluation

[warm python]

thanks

next to prove associativity

$\left(v+u\right)+r\ =\ \sum_{i=0}^m\left(a_ix^i+b_ix^i\right)+c_ix^i$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

hmm

I take out the $x^i$ by the law of assocation over $\C$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

which gives me

$\left(v+u\right)+r\ =\ \sum_{i=0}^m\left(\left(a_i+b_i\right)+c_i\right)x^i$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

now this is the same as

$\left(v+u\right)+r\ =\ \sum_{i=0}^m\left(a_i+\left(b_i+c_i\right)\right)x^i$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

by the law of associativity over $\C$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

is my proof right so far?

now consider $\sum_{i=0}^na_ix^i+b_i0^i$ ( to prove that teh additive identity is an element)

let $0^0=0$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

this feels very sussy to me

what do I do in its place

[soft zealot]

ƒ(Why am. I here)=I don't know

[hearty zephyr]

the additive identity would be 0x^i

[warm python]

why not $a_i0^i$

[soft zealot]

ƒ(Why am. I here)=I don't know

[hearty zephyr]

you want a polynomial that is = 0, always

yours is (maybe) 0 at x = 0

[warm python]

ah

okay

makes sense

thanks

[hearty zephyr]

The key thing to remember here is that the polynomials are always in variable x, you shouldn't be changing x. The thing you can affect is coefficients.

[warm python]

ah, okie

thanks

next to prove that an additive inverse exists

$\sum_{i=0}^na_ix^i+b_ix^i=0$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

it's evident that $\sum_{i=0}^n\left(a_i+b_i\right)x^i$ is always zero when $a_i=-b_i$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

thus an additive inverse exists

is my proof right so far?

[hearty zephyr]

yep

[warm python]

cool, thanks

next to prove the multiplicative identity

$\sum_{i=0}^n1\cdot b_ix^i=\sum_{i=0}^nb_ix^i$

[soft zealot]

ƒ(Why am. I here)=I don't know

[warm python]

and lastly distributivity can be proven similarly to associativity

is that right?

[hearty zephyr]

yeah, you just need to distribute a constant multiple into a polynomial.

[warm python]

thanks so much !

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[torn prism]

Am I doing something wrong here?

[spring haven]

yeah, you are

you have $N-\text{matrix 1}+\text{matrix 2}$

[soft zealot]

kheerii

[spring haven]

you solved it as $N-(\text{matrix 1+matrix 2})$

[soft zealot]

kheerii

[spring haven]

these two are not equal

[opal plinth]

Are they not?

[spring haven]

no..?

[opal plinth]

Oh I'm dumb

[final saddle]

@torn prism Has your question been resolved?

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[warm python]

here, why does the fact that $a_k=C_k => a_k=c_k=0$

[soft zealot]

ƒ(Why am. I here)=I don't know

[candid hull]

they never say that ak = ck = 0

[hearty zephyr]

you're looking form $\implies$ or $\Rightarrow$

[soft zealot]

Zybikron

[candid hull]

just ak = ck

[warm python]

but they've said each $a_k=c_k=0$

[dense coral]

Axler is showing that linear combinations are unique

[soft zealot]

ƒ(Why am. I here)=I don't know

[candid hull]

no

[dense coral]

no

[candid hull]

where do you see that

[dense coral]

they said that a_k - c_k = 0

i.e. a_k = c_k

[warm python]

yes

[dense coral]

nowhere does Axler say that a_k = c_k = 0

[warm python]

ooh

and how does that imply uniquness?

[candid hull]

they supposed you had 2 ways of writing v

one with the ak's

one with the ck's

[warm python]

right, got it

[candid hull]

but the ak and ck end up being the same cause linear independence

[warm python]

so $a_k=c_k=1$ for instance is valid, right?

[soft zealot]

ƒ(Why am. I here)=I don't know

[candid hull]

it's possible yes

[warm python]

ok, thanks !

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[snow karma]

a|b means which one divides the other to a whole number?

21|3 is True, or is 3|21 (I mean because 21/3=7, whole number)

[clear moon]

we say a|b if there exists some integer c such that b = ac

you cannot find an integer c such that 3 = c * 21

but you can find an integer c such that 21 = c * 3

so 3|21 is true but 21|3 is not

[snow karma]

thank you very much mate

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[shut linden]

Hello all! I have a very low level question, but I am going to ask it anyway.

A bag contains 50 coins. The total value is $1.81. If the bag only contains 5 cent and **2 cent **coins, how many more 5 cent coins than **2 cent ** coins are there?

[warm python]

any ideas?

[lucid marsh]

Try to first turn the scenario into an equation

[shut linden]

a=5 b=2

Total amount of a and b must total to 181.

x5+y2= 181

[rustic sinew]

Yes

And one more piece of information

[small fjord]

yep

[shut linden]

x+y = 50

[small fjord]

exactly!

[small fjord]

now rewrite them both and get this equation in terms of y

[shut linden]

x>y

x>25

y<25

[small fjord]

hmm i like your way of thinking but it wont help you much

you said that
5x+2y= 181

right?

[small fjord]

. i plugged in 5 and 2

[shut linden]

yeah

[small fjord]

and you also said that x + y = 50

[shut linden]

yeah

[small fjord]

when i solve these types of equations

i want only 1 unknown

how can i turn the y in 5x+2y=181 to just something(x) = 181

cuz i dont want 2 unknowns i only want 1 so i can solve for it

[shut linden]

By unknown do you mean like a as in a variable?

[small fjord]

yes

in this equation 5x+ 2y = 181

the x and y are unknowns

[shut linden]

yeah

[small fjord]

or atleast thats what some people call them

how can we make it only 1 unknown

other wise solving with 2 impossible

[shut linden]

Max that y can be is 24, 24x2=48 181-48=133 133/5=26 3 remainder

24 doesn't work

[small fjord]

hm well that would be 1 way of solving it

but what if we had more than 50

it wouldnt be very efficient

lemme give you a little push

re arrange x + y = 50 into terms of y

[shut linden]

What do you mean by in term of y

[small fjord]

like y =

make that equation in terms of y

[shut linden]

well you can get to 180 with 90x2

so y coud be 90

if no limit

and it would be done

[small fjord]

well its not what i really mean

x + y = 50

is just

y = 50 - x

right?

[shut linden]

yeah

[small fjord]

now we know that the original eq 5x + 2y = 181

if we know y, we can replace it with whatever it is

so you can do 5x + 2(50-x) = 181 right?

[shut linden]

oh yeah

[small fjord]

tell me what you get for ur final answer!

[shut linden]

No I've confused myself

Idk how

[small fjord]

hm

how about you break it down into steps

expand 2(50-x)

[shut linden]

100-2x

[small fjord]

now we could say 5x + 100 -2x = 181

right?

[shut linden]

5x + 100-2x

oh yeah

sorry didn't see thems

[small fjord]

but we have alike terms right? the 5x and 2x

so we can just say 3x

[shut linden]

yeah

[small fjord]

try solving it now

[shut linden]

3x+100=181

x=27

y=23

thanks

so the answer is 4

Thank you so much

!

[small fjord]

why is the answer 4?

oh yeah

[shut linden]

27-23=4

[small fjord]

i didnt read the full question

so that solution is called solving by substitution

there is actually another method called solving by elimination

but you might have not taken that yet

[shut linden]

Is elimination basically brute forcing with trial and error?

[small fjord]

not quite

ill show u

this one might seem a little less straight forward

[shut linden]

It does

[small fjord]

so basically we line both equations up

i did a mistake there

actually

its a 2

[small fjord]

5x |+ 2y| = 181
x | + y | = 50

now imagine we could subtract the equations from each other

5x - x then 2y -y and 181-50

[shut linden]

3x+81

I see

[small fjord]

see that gets us 4x + y = 131

but that doesnt help us

[shut linden]

81/3 = 27

[small fjord]

so instead we make one of the terms below equal to the above

so we multiplied it by 2 the whole x + y = 50

[shut linden]

50-27=23

[small fjord]

exactly

[shut linden]

x=27 y=23

27-23=4

[small fjord]

you could try solving a few more examples on the internet or if you have any in a worksheet to get the hang of it

id reccomend that

[shut linden]

Thx

[small fjord]

np

[shut linden]

shall i close this now?

[small fjord]

if you dont have any futher questions then sure

[shut linden]

.close

[final saddle]

Channel closed

Closed by @shut linden

Use .reopen if this was a mistake.

[shut linden]

.reopen

[final saddle]

✅

[shut linden]

sorry

I forgot

[small fjord]

need anything

[shut linden]

Other question

There are some tiles arranged in a square, it is made into a rectangle by making one side 3 tiles shorter, and one 4 tiles longer. Only one more tile is needed. What are the dimensions of the final rectangle?

The square must be at least 4x4

So we can do l (length) and w (width)

sqr(4) = 16

[small fjord]

hm i dont think i understand the question much

[shut linden]

neither

[small fjord]

is that the full question

only 1 more tile is needed

[shut linden]

I'll copy directly

[tranquil pine]

(w-3)(l+4)-1 = wl

[small fjord]

so effectively what we could say is x^2 = (x-3)( x + 4) -1

[tranquil pine]

ywah

that

but x^2 not 2x

[small fjord]

yeah

in a square the l and w must be equal so if we donate them by x and say x * x its the same

[shut linden]

Kevin had some square patio tiles which he arranged to make a square.

He decided to change it into a rectangle by making one of the side 3 tiles shorter, and the other side 4 tiles longer.

To make this change, Kevin only needed to buy one extra tile.

What are the lengths of the sides of the rectangle?

[tranquil pine]

^^ then solve for x

for length of square

[small fjord]

wait is it x^2 or 2x lmfao

[tranquil pine]

x^2

[small fjord]

i said 2x because im talking about the number of tiles

[tranquil pine]

asking abt area

[small fjord]

oh wait alr

[tranquil pine]

so x*x tiles in a squre

square

[small fjord]

x^2 = (x-3)( x + 4) -1

its like this

[shut linden]

x squared tiles in a square

I don't understand bc ur defining x with x

[small fjord]

lets say we have a square

now we know that is has a legnth and a width

and the product of those is the number of tiles

now since its a square

the length and width is the same

[shut linden]

lw= area

[small fjord]

so if we denote either of them as x, we can say x * x is just the aera

[shut linden]

l=w

x squared

= area

yeah

[small fjord]

thats how we got x^2

now for the other side of the equation

[shut linden]

is x arrow 2 squared?

how do you type the arrow thing

[small fjord]

thats how i personally write it

its shift 6

if your on laptop

[shut linden]

thx

x^2

= area

[small fjord]

yep

[tranquil pine]

yes, x raised to the power of 2 or x squared

[small fjord]

now heres the thing we know that the area is the same exact thing as the area of the rectangle except we are buying 1 more tile

so we could say
area of square = area of rectangle - 1

[shut linden]

((w-3)(l+4))-1=final rectangle area

yeah

[small fjord]

yep exactly

[shut linden]

area of square = y

[small fjord]

the only reason why we are swapping w and l with x is because they told us one side is 3 less and 4 more

from x

[shut linden]

Area of rectangle=r

y=r-1

⏹️⏹️⏹️⏹️⏹️
⏹️⏹️⏹️⏹️⏹️
⏹️⏹️⏹️⏹️⏹️ -3
⏹️⏹️⏹️⏹️⏹️
⏹️⏹️⏹️⏹️⏹️
+4

3x4=12

too many

[small fjord]

why dont you solve the equation?

x^2 = (x-3)(x+4)-1

its much easier than trial and error

[shut linden]

x^2 - 13

[small fjord]

do (x-3)(x+4) by itself first

expand

[shut linden]

x^2 -12

[small fjord]

hmm well not quite

what grade are you

[shut linden]

Idk uk

year 8

Idk what that is equivalent to

[small fjord]

so the way that we expand (x-3)(x+4) is we multiplying each of the value in the left parenthesis with both values on the right

[shut linden]

oh yeah

sorry i forgt that it was evrything

not just like terms

[shut linden]

so x^2 + 4x-3x-12

simplifies to

[small fjord]

wait hold on

its positive x

[shut linden]

x^2 + x - 12

[small fjord]

not negative

yep exactly

so now we have x^2 = x^2 + x -12 - 1

[shut linden]

??

[small fjord]

?

[shut linden]

How is x^2 = x^2 + x-12-1

[small fjord]

we went from
x^2 = (x-3)(x+4) - 1
to x^2 = x^2 + x -12 -1

right?

remember we only expanded those 2

[shut linden]

if that is true everything after x^2 is 0

cancels out

[small fjord]

yeah we get 0 = x - 12 -1

which is just 0 = x - 13

[shut linden]

x=13

[small fjord]

exactly

so ur answer to that question must be 13

[shut linden]

13^2=169

[small fjord]

that would be the area of the square

we dont need that

we need the dimensions of the rectangle

[shut linden]

a=169

r=a-1

168

[small fjord]

nono

no need

we know that x is 13

and we know one side is (13-3) and one is (13+4)

so the dimensions would be the result of those

10, 17

[shut linden]

10x17

[small fjord]

exactly

anyways i need to go, goodluck with your studies tho!

[shut linden]

bye

thx for the help!

[small fjord]

np, bye

[shut linden]

.close

[final saddle]

Channel closed

Closed by @shut linden

Use .reopen if this was a mistake.

[final saddle]

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[quartz spade]

I was looking at this homework assignment, and I know the 1st two are true because they're the definitions. Am I supposed to figure out if the other ones are "close enough" or logically equivalent?

[final saddle]

@quartz spade Has your question been resolved?

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

[short ravine]

is this correct

[soft zealot]

..,,

[final saddle]

@short ravine Has your question been resolved?

[final saddle]

Channel closed

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Available help channel!

Send your question here to claim the channel.

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

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

[vernal pulsar]

For 9, I set my integral as $\int_{0}^{\pi} \int_{0}^{2asin(\theta)} 2r^2 dr d\theta$. This worked and I got a positive volume as my answer. For 10, I set my integral as $\int_{0}^{\pi} $\int_{0}^{2acos(\theta)} 8^{0.5} r^{\frac{3}{2}} (cos(\theta))^{0.5} dr d\theta$. However, this just gave me an answer of zero and I'm not sure why this was the case when volume below the z axis was present in both questions.

[soft zealot]

Secret
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