help-36

[shell mountain]

You want 1/x² dx

How can you turn -1/x² dx into 1/x² dx?

[random scaffold]

o

h

so it's that easy

[shell mountain]

Optionally you can write the integral as - ∫ -1/x² e^(1/x) dx

[tropic trellis]

That depends

[shell mountain]

The two negatives would cancel giving the same integral

[random scaffold]

hmm

[tropic trellis]

You can rewrite it like this too

[random scaffold]

yeah i think that's what he was getting at, that -du = 1/x^2dx

but one last question

how do you intuitively know to use 1/x over like x^2 for the u substitution

[tropic trellis]

Transfer all the things that cause the trouble over to dx

That solves all of the problem

[shell mountain]

It's a mixture of experience, or trial and error if you don't have that

If you tried u = x², you'll see that the substitution doesn't really work

[tropic trellis]

1. which thing looks most out of place
2. take outs it's deravite and see if it exists in the integral
3. repeat If integral not found

¯_(ツ)_/¯

[random scaffold]

ty for the help, things are clearer now

.close

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[hasty heron]

Show that the inequality is satisfied for all x ∈ R.

[hasty heron]

Sen is sin

I think thats a mistype

But everything else is correct

[vital crag]

Did you try using sum formulas?

[hasty heron]

Mmm no

How is that?

[wraith crater]

$\sin(a+b) = \sin a \cos b + \sin b \cos a$

[soft zealot]

Pure

[wraith crater]

$\cos(a+b) = \cos a \cos b - \sin a \sin b$

[soft zealot]

Pure

[hasty heron]

But its not the same right

Because in my problem we have a but not b

Or im wrong?

[wraith crater]

a b can be anything

[hasty heron]

Yes but

5 sen(x + 37◦) + √2 cos(x − 45◦) <=
√41

a = x

b = ?

[wraith crater]

You have two cases here

[vital crag]

You use it twice for each sine and cosine

[wraith crater]

I just used a and b arbitrarily

[hasty heron]

Oh ok

so

5sin(x + 37◦) = sin(x) * cos(37◦) + sin(37◦) *cos(x)

[wraith crater]

Yeah don’t forget the 5

[hasty heron]

5sin(x + 37◦) = 5(sin(x) * cos(37◦) + sin(37◦) *cos(x))

√2 cos(x − 45◦) = √2(cos(x) * cos(-45◦) - sin(x) * sin(-45))

Ok, but how do I continue?

[wraith crater]

sin(-45) is -sin(45) and cos(-45) is cos(45)

Then do you know the values for some of these trigs

[hasty heron]

cos(-45) = -cos(45)

?

[wraith crater]

No cos is an even function

So cos(-a) = cos(a)

While sin is an odd function

So sin(-a) = -sin(a)

[hasty heron]

Oh bro I didnt know that hahaha

But cos(45) = sin(45)

?

[wraith crater]

Yes

[hasty heron]

how is this called?

Like this method

[wraith crater]

It’s just sin/cos addition formula

[hasty heron]

So √2 cos(x − 45◦) = √2(cos(x) * cos(45◦) - sin(x) * -sin(45))

=

√2 cos(x − 45◦) = √2(cos(x) - sin(x))

?

[wraith crater]

Sorry bro I gotta go

[hasty heron]

Oh thats fine

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@hasty heron Has your question been resolved?

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[final saddle]

@brisk cipher Has your question been resolved?

[void valley]

Can you determine domain

it would help

because it's specific

[brisk cipher]

nothing specified

[void valley]

I meant if you can find it on your own

[brisk cipher]

we can do by common domain

1 and - 1

[void valley]

yes

so now range

just plug x = -1 and x = 1

and that will give you possible values of y

[brisk cipher]

yes

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[quiet forge]

Hoping for a bit of quick insight

[quiet forge]

The questions, and the answers. The math all makes perfect sense to me

But how do I know if a solution set is a line or a plane? (Be gentle with me, its very early)

[fresh relic]

line if 1 parameter, plane if 2

Technically it depends on the dimension of the solution set

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[hollow turret]

A line has slope -2 and passes through the point (r, -3). A second line, K, is perpendicular to the first lline at (a, b) and passes through the point (6, r). Find a in terms of r.

[wraith crater]

What have you tried

[hollow turret]

so I've tried to substitute r, a, and b with some slope values

[final saddle]

@hollow turret Has your question been resolved?

[final saddle]

@hollow turret Has your question been resolved?

[hollow turret]

<@&286206848099549185>

[loud sundial]

Becuase the lines are perpendicular, the slope of line K is ||1/2||

this allows you to find the equations of both lines in terms of r

and then you can find their intersection in terms of r, the x coordinate is a

[hollow turret]

hm ok I'll try that

ty

[zealous root]

close the channel bro

[wary helm]

.close

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[vestal hamlet]

is y=sqrt(x+1)
in vertex form?

[grim badger]

Quadratic*

I was literally just correcting your spelling

[vestal hamlet]

so when I have like

a parent function of h(x) = sqrt(x)

g(x) = -1/2h(2x-2)-3

I get g(x)=-1/2sqrt(2(x-1))-3

why is the hortizantal shift +1

when its -1

[spark kettle]

Bc $x - 1 \geq 0$ for $x \geq 1$

[soft zealot]

Pluton

[spark kettle]

And bc its in denominator the domain doesnt include 1 so the domain is $x > 1$

[soft zealot]

Pluton

[vestal hamlet]

huh? wdym?

[spark kettle]

What?

[vestal hamlet]

I dont get why the horizantal shift would be +1

when its x-1

#help-36 message

[spark kettle]

because at x = 1 -> x - 1 = 0

Which means it shifted the 0 of the new function is at x = 1 now

[vestal hamlet]

How could it shift if it equals to sqrt(0)?

[spark kettle]

Exactly

In sqrt(x) at x = 0 you have the "start" and at sqrt(x - 1) at x = 1 you have the "start"

The 0 shifted from x = 0 to x = 1

And so did every other value with it

,w plot (x - 1)^2

[soft zealot]

Results provided by WolframAlpha

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

Requested by Pluton#6301

[spark kettle]

(x-1)^2 shifts for 1 to the right

Because at x = 1 its now a new 0

[vestal hamlet]

then what about the x

the x still exists

[spark kettle]

Who said it didnt?

[vestal hamlet]

but then how can we ignore the x to see the shift?

[spark kettle]

Replace any equations x with x - a it will shift a places to the right and replace it with x + a and it will shift a places to the left

[vestal hamlet]

If we take a look at vertex from its a(x-h)^2

the h is positive

do we ahve a "form" for this also?

[spark kettle]

Umm sure
$$y = a\sqrt{x - b} + c$$

[soft zealot]

Pluton

[vestal hamlet]

so if my parent function would be

x^3

then it would be

y=a(k(x-b))^3+c

[spark kettle]

What is that k doing there?

[vestal hamlet]

the horizantal stretch

[spark kettle]

a is the horizontal stretch

[vestal hamlet]

sry had to unload groceries.

[vestal hamlet]

a is the vertical stretch, since its outside the x

[spark kettle]

Ngl idk the difference between horizontal and vertical but ik k is just a with extra steps

[vestal hamlet]

k is like 1/k

when calculating

thats my second problem

i dont rly get that either.

[spark kettle]

This just equals
$$ak^3(x - b)^3 + c$$

[soft zealot]

Pluton

[spark kettle]

And bc ak^3 is constant replace it with a and you get

$$a(x-b)^3 + c$$

[soft zealot]

Pluton

[vestal hamlet]

I thought the K was with the X

to show the horizxantal stretch

[spark kettle]

You put k where you put it

Idk whats its supposed to be

k is "vertical stretch"

Just cubed

Basically its all a 1 constant called a

[vestal hamlet]

I sorta get it

Give me a while to soak this in

For now, thank you Pluton.

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[cosmic idol]

Hi,

I’m doing algebra right now and the problems I’ve been doing are finding products.

I then got this problem:

[8y + (7-3x)] [8y - (7-3x)]

What do the [] brackets mean? I feel like this should be obvious but I just have no clue

[alpine oriole]

im not sure how to explain this preoperly

[void crest]

#❓how-to-get-help

[void crest]

its just brackets as per normal

its just to differentiate from the () brackets

especially as they use both

[cosmic idol]

oh thank you

[void crest]

u can interchange them but like

if u start with [ u shld end with ]

likewise for ()

[cosmic idol]

ok

[void crest]

[steel vortex]

Identify the transformation that maps ∆RST onto ∆R’S’T’, and
∆R’S’T’ onto ∆R”S”T

[void crest]

#❓how-to-get-help

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[flat flame]

help pls

[wraith crater]

Which bit are you stuck with

@flat flame

[flat flame]

the whole thing

[void crest]

um

u know that

the 283^2003

and

(-283)^2003

can u rewrite one of them

to simplify

[wraith crater]

What’s (-1)^odd power

[flat flame]

-1

[wraith crater]

Yeah

So what’s (-283)^2003

[flat flame]

-283

[wraith crater]

You mean -(283)^2003

[flat flame]

ye

[wraith crater]

Then what happens to the first 2 terms

[flat flame]

283^2003-(283)^2003

[wraith crater]

What’s that then

[flat flame]

wdym

[wraith crater]

$283^{2003}-(283)^{2003}$

[soft zealot]

Pure

[wraith crater]

What’s this

[flat flame]

i dont know what u mean by whats this

[vital crag]

What's it equal to?

You should be able to simplify it to a single number

[flat flame]

ohh ok wait

0

[wraith crater]

Yes

So what does the first bracket become?

[flat flame]

0-10=-10

[wraith crater]

Yes

So we have

$-10 \cdot -2 \div -\frac{1}{5} \cdot (-1)^{2002}$

[soft zealot]

Pure

[wraith crater]

Do it from left to right

[flat flame]

so

-100?

coudl someone confirm??

[wraith crater]

Yeah

100

[bold lagoon]

-100

[wraith crater]

Yeah

Mb

-100

[final saddle]

@flat flame Has your question been resolved?

[flat flame]

ok ty

hang on

would someone help me with another question?

[vital crag]

It's easier if you closed this and opened a new one because the bot autopins your first message so helpers can easily find your question

[flat flame]

oh ok tyy

.close

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[short sinew]

What does “as much as” mean in algebra word problems?

[sonic cairn]

can you send the problem

[short sinew]

For example:
The computer cost 2.5 times as much as the television.

How would I convert it as a equation?

[sonic cairn]

c=2.5t

[vital crag]

If the television cost $100, then the computer costs $250

[short sinew]

So it basically means multiplication?

[sonic cairn]

yeah pretty much

[vital crag]

"2.5 times" means multiply by 2.5

[short sinew]

Ok but what if it said no times and instead only said “as much as”

What would I do?

[sonic cairn]

c=t

[short sinew]

Ohh

ok thanks

.close

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[solid elm]

this question is confusing me sm

The choices are:
There is no zero between
There is at least one zero between
it is unknown whether there is a zero between

[celest crane]

This question pertains to the Intermediate Value Theorem. Do you have a solid understanding of what the IVT implies?

[solid elm]

kind of

[celest crane]

What is your understanding of IVT?

[solid elm]

its a continuous function and it has to pass the x axis

[celest crane]

That's a specific example. There is a more general interpretation of IVT.

If a function is continuous on an interval [a,b], then there every y-value exists between f(a) and f(b).

[solid elm]

oh

so how does this apply to the question

[celest crane]

Because the question states that the function is continuous on the interval [-18,20], that implies that all the y-values in the range between the given f(x) values exist.

eg.

x=-18 and x=-8
            f(-18) = 26 and f(-8) = 19

Every y value between 19 and 26 exists on the interval [-18,-8] for this continuous function.

[solid elm]

oh so the numbers that are equal to -18,20 or any number outside the interval dne

[celest crane]

IVT makes not claims on if a y value does not exist, IVT only states that we know the y values on that interval must exist.

[solid elm]

oh

so for the first time the answer that its unknown if there was a zero or not?

[celest crane]

https://www.geogebra.org/classic/qpthscry

Look at this graph to help you visualize what I mean.

Now your question is utilizing a specific case of IVT, if f(a) is negative and f(b) is positive, or vice versa, then there must be some point on the function in which (x,f(x)=0) must exist because -N < 0 < +N.

[solid elm]

Oh

[celest crane]

y = 0 lies in the range of -N and +N.

[solid elm]

what about if they were both positive like the graph u sent me?

[celest crane]

If there are both positive, then it would be indeterminant if a zero lied on the interval meaning you would not be able to prove a zero exists on that interval strictly using IVT as a proof.

You could mathematically prove a zero exists in other ways, but not strictly using IVT.

[solid elm]

oh

ngl im so confused

ig i dont really understand what IVT really is

[celest crane]

It's a convenient method for proving at least one root of a function exists on an interval.

[solid elm]

aha

[celest crane]

Reload that page, I made a slight change.

[solid elm]

ok

done

[celest crane]

I restricted the highlighted area to the interval.

Any part of the function that is in the highlighted area, you can prove exists using IVT.

[solid elm]

oh

[celest crane]

I should note that the graph doesn't work properly if the point B is below point A. You'll have to use your imagination to fill in the area. 😉

[solid elm]

oh i understand

[celest crane]

Fixed that problem.

So do you understand how to solve those problems?

[solid elm]

i graph them then i see if it fill out the area?

[celest crane]

For a, you are given the interval [-18,-8]. f(-18) = 26 and f(-8) = 19. That means every value of y exists in this range.

19 <= y <= 26.

On the interval [-8, -1], every value of y exists in the range 19 <= y <= 29. That range is found by looking up f(x) in the chart provided.

[solid elm]

so for example on the interval [-1,19], every value of y exists in the range -24 <= y =< 29?

[celest crane]

Correct.

[solid elm]

so the answers are

1. there is no zero between
2. there is no zero between
   3.there is at least 1 zero between
   4.there is at least 1 zero between?

[celest crane]

You cannot prove there is no zero on an interval strictly using IVT. You can only prove the range between f(a) and f(b) exists.

[solid elm]

so the first 2 answers are wrong

?

[celest crane]

They should both be "It is unknown ..."

[solid elm]

oh

[celest crane]

As you can see in the graph I posted, there are times when a zero exists on the interval [a,b], but they do not fall in the range of [f(a), f(b)].

[solid elm]

OH

now i get it

and for the last 2 there is at least 1 zero right?

[celest crane]

Correct.

[solid elm]

thank you so much m8

[celest crane]

yw

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[tacit knot]

i need help with these 2 problems

for #4. i realised that x = 2y - 3 == 2y + 1 - 4, where 2y + 1

is usually the odd integers

idk where to go from there

[tacit knot]

so show that htye are both the subsets of each other

that makes them equivalent?

[pine root]

yep thats how you show set equivalency

[tacit knot]

ok ill try that

[final saddle]

@tacit knot Has your question been resolved?

[tacit knot]

@pine root

I don’t think this is right ?

[pine root]

then i'll leave it to you to do the other way

[tacit knot]

ok

[tacit knot]

ohh 2(y-2)+1

[pine root]

but you don't need to add that

[tacit knot]

I get it was approaching it the wrong way

Thanks

[tacit knot]

I have no idea what I’m doing with this second one

i know its like making sure the intersection of s and t with the intersection of the non negative integers is not the empty set, or that the intersection exists with some members of the set

but i dont know how to go further

i think i found the set for the intersection of s and t ?

any idea? @pine root

[tacit knot]

is that the correct set for the s intersects t though

i just multiplied them

i wasnt too sure on that

I just gave a single case that fits, I think that’s fine

But I’m still not sure it x= cd/2^e*3^f is the intersection

also are my reasoning for these 3 correct

@pine root sorry for pinging so much

<@&286206848099549185>

[pine root]

for 2 you can shorten the proof considerably

you can just assume y is any integer then show xy is even

[tacit knot]

oh

[pine root]

i dont think your definiiton for the intersection is right

[tacit knot]

i thought so

i just multiplied them

[pine root]

if i choose c = d = 1, then i get x = 1/(2^e 3^f) and theres no way to split that

[tacit knot]

hm

but wouldnt that still fufill the intersection

for s and t anyway

like what other way would you find the intersection besides multiplying

[pine root]

honestly finding the intersection explicitly is annoying

i'd just find one element

and show that it is an integer

like a=2,n=1,b=3,m=1

[tacit knot]

oh like find an elemnt of both s and t now show its an interger

[pine root]

yep

[tacit knot]

i get it

[pine root]

that are the same integer

[tacit knot]

ok thanks i got it now

.clsoe

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[stray wasp]

How do i solve this

[lucid marsh]

what

work out what

[stray wasp]

Q5

[lucid marsh]

ok

[tranquil pine]

its sqrt(25)

[lucid marsh]

what is sqrt(25)

[stray wasp]

5

[lucid marsh]

done

[stray wasp]

But what is the fraction for

[lucid marsh]

what is cbrt(8)

[lucid marsh]

wdym

[stray wasp]

Look at the whole q5

Theres a fraction

[lucid marsh]

,r

DUDE

[stray wasp]

Next to the number

What

[lucid marsh]

,r

.r

what was the cmd

ok

just rotate it

[lucid marsh]

wdym

[stray wasp]

Look theres a half

Or one third

[lucid marsh]

ok

do you know what exponents are

[stray wasp]

No

[lucid marsh]

then how did you do q4

[stray wasp]

Oh u mean power

I just searched up what it meant

Its the same as power right?

[lucid marsh]

yes

exponent is for powers in general

[stray wasp]

Ok

[lucid marsh]

power is used more for "a to the power of b"

anyway

you know exponents

[stray wasp]

Yh

[lucid marsh]

those fractions are exponents

[stray wasp]

I dont get how to solve it tho

I just square the number

[lucid marsh]

what

what is $x^1/2$

[soft zealot]

hired

[lucid marsh]

i meant what is x to the (1/2)

but latex sucks

anyway

what is it

[stray wasp]

X squared?

[lucid marsh]

no

it's sqrt(x)

[stray wasp]

How

[lucid marsh]

in general, a^b/c = cth root of (a^b)

[stray wasp]

Whats cth

[final saddle]

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[primal root]

does anyone know ho to do this question if it werte to be 2x-1?

[quiet crag]

yo guys i need helo understanding functions

[oak arrow]

whoops sorry

wrong channel

[primal root]

mb @oak arrow

[ancient thistle]

#❓how-to-get-help

[primal root]

could someone help me with this?

[quiet crag]

please i dont understand and i have couple days left till a test

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[zenith python]

i got a already

just need help wit b.

i have no clue how to start

lol..

help 😦

[zenith python]

<@&286206848099549185> heeeellp

.close

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[grave moss]

Hi

[grave moss]

I was wondering if someone could help me with some of my delta math

If i learn how to do one or two i can do the rest

@fierce ridge

Do u think u can help me with this

[fierce ridge]

uhh yeah

btw whyd u ping me lmao

do you know the slope equation of a line

@grave moss

[grave moss]

Oh ive never seen that before

This new teacher just assigns us things

[fierce ridge]

well then u dont need to use it

[grave moss]

He doesnt rlly teach us

[fierce ridge]

do u know y=mx+b?

[grave moss]

Yeah

Slope intercept form

[fierce ridge]

ok lets use the 2nd one then

ok

[grave moss]

Okay wait

[fierce ridge]

now how would u find m here

[grave moss]

I have a new problem, because i got the other one wrong

[fierce ridge]

alright send that

[fierce ridge]

alright

how would u find m here

[grave moss]

Midpoint formula right

Y1 - y2 / x1 - x2?

Or wait

Id graph it

And use rise over run

[fierce ridge]

yeah

no do the first one

it will be easier

[grave moss]

Okay

[fierce ridge]

than making a whole graph

[grave moss]

I was just gonna desmos it

But il do the other way

[fierce ridge]

did you get a value for m? or do you need help

[grave moss]

I got -1/4

Im bot sure if thats right

Not*

[fierce ridge]

yeah it is

perfect

so now

$y=mx+b$

[soft zealot]

imagine

[fierce ridge]

you know m

so u can put

[grave moss]

Y=-1/4x+b

[fierce ridge]

$y=-\frac{1}{4}x+b$

yeah

[soft zealot]

imagine

[fierce ridge]

now u have 2 points

take any random one

[grave moss]

OHHH

[fierce ridge]

lets say -1, 3

[grave moss]

I GET IT

PLUG IT IN

[fierce ridge]

and plug in the values in y and x

and u get b

yep!

[grave moss]

Thank you

[fierce ridge]

np

[grave moss]

Sm

Once i finish this ima sleep

[grave moss]

I got y= -1/4x +11/4

[fierce ridge]

yeah that seems right

[grave moss]

I hate desmos so much

[fierce ridge]

uhhh

maybe you have to give it in decimals?

because it is correct

[grave moss]

Okay

It said the one next to simplify negatives was right

Thag makes absolutely no sense to me

[fierce ridge]

oh

u had to use point slope form

[fierce ridge]

this one

so here

basically after u found m

u would get

$y-y1:=:m\left(x-x1\right)$

[soft zealot]

imagine

[fierce ridge]

and then u would put the -1, 3 value

-1 = x1

and 3 = y1

[grave moss]

So i did it too far

[fierce ridge]

basically yeah, once u simplify, you wouldve gotten the equation you had

it said to use point slope equation

i didnt see that, mb, sorry

[fierce ridge]

this is the point slope equation

[grave moss]

Ima just do this in the morning

Im suffering

[fierce ridge]

fair enough 💀

[grave moss]

But think you for your help

I get it now

[fierce ridge]

np

[grave moss]

Thank*

[final saddle]

@grave moss Has your question been resolved?

[grave moss]

A little update

I didnt sleep

[final saddle]

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[vale python]

I’m not sure if this induction proof is good to go or not. It pertains to the Euclidean algorithm

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@vale python Has your question been resolved?

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[vale python]

<@&286206848099549185>

[vale python]

.close

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[unkempt swift]

@grim nebula hello it is me yet again

so for a linear system to be inconsistent i need to show that one row is 0 0 ... 0 | non zero right?

[grim nebula]

#help-44｜stanley-🌲-v2-dans message

[unkempt swift]

wut

[grim nebula]

you wrote S_inf

[unkempt swift]

ooops

lmao

itsok lets not look at that question anymore

[grim nebula]

im on a computer now so i can type algebra

hold on

[unkempt swift]

flex

[grim nebula]

you can get rid of all the negative signs

[wraith tulip]

inconsistent you just need some components of x requiring 'multiple identities'

[unkempt swift]

wait do u actually wanna go back to that question

[grim nebula]

and its pret ez

well do you

im just saying like you can do that

and so its not actually that bad to solve

[wraith tulip]

for example if x_1 is required to be both 1 and 0, the system Ax=b is inconsistent

[unkempt swift]

ill do it later LOL

[grim nebula]

kkk

[unkempt swift]

right

wait so that or like 0 0 0 | 1

either

ok so

[wraith tulip]

this works too. no linear combination of 0 whatever (0 * x_1 + 0*x_2 + 0*x_3 + ...) gives a non-zero value

[unkempt swift]

right

[grim nebula]

when you start out linalg and stuff

the 0 row having a nonzero afterwards is definitely the way its usually taught

[unkempt swift]

ye

i started linear alg last week

still a noobie

[wraith tulip]

for this question I recommend fixing m, n to some value first (lol it's a lot easier to show 1 example), then using algebraic symbols

[unkempt swift]

uhh

ok wait so what im tryna do is to find a matrix such that the value of the last row last column is not 0

wait but this si correct right?

[wraith tulip]

seems fine?

[unkempt swift]

ok so is it safe to say i can ignore the

x1 ... xn

since theyre just vairables

[wraith tulip]

no lol, you want to show inconsistency

[unkempt swift]

heh

wait but how does the x1...xn column help tho

[grim nebula]

you dont need it

the augmented matrix doesnt contain them

theyre in the background

but you dont need to keep track

[unkempt swift]

yes

[unkempt swift]

wait then what does he mean

or did he do an oopsie

[grim nebula]

theoretically theyre in the background

but i mean you dont need to think about them

[unkempt swift]

okay so lemme just rid that

boom

[grim nebula]

oh no

no you cant do that

[unkempt swift]

oh um like its wrong

[grim nebula]

you need them in this equation

you dont need them in the augmented matrix

[unkempt swift]

oh i just wanted to look at it more easy-ish

nice

[grim nebula]

you want k!=0

[unkempt swift]

yes

so i need a matrix

to multiply this entire thing

right?

[grim nebula]

no probably not

you want to show that k can be nonzero

[unkempt swift]

wait but the question asked me for a matrix b

oh wait showing k != 1 also implies that such a matrix exists

wait can u give me clues as to how to solve this
i wanna try to think by mysefl

[grim nebula]

nice

[unkempt swift]

lma

[grim nebula]

so you know when solving with augmented matrices

you can do row ops to get to equivalent matrices

but if the left matrix is inconsistent so is the right one

[unkempt swift]

COLUMN OPS?

[grim nebula]

so what you want to know is "can the k be nonzero?"

NO

[unkempt swift]

o

[grim nebula]

LOL

[unkempt swift]

hmmm

[unkempt swift]

yes

[grim nebula]

so the b vector

in the last column of the RHS

completely determines how you get the last column on the LHS yeah?

so thats

[unkempt swift]

$\begin{pmatrix}
x\y\\rdots\z
\end{pmatrix}$
where z is non zero

[grim nebula]

probably what you want to do

[unkempt swift]

how is there a compile error

[grim nebula]

i havent played around with this problem tho so idk if theres another way to do this

you need math env i think

surround with $

[soft zealot]

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

[unkempt swift]

wait isnt it just this

[grim nebula]

yeah but

the row ops could mess b up

so then you might get a 0 where k is

when you dont want it to be a 0

[unkempt swift]

o

hm

wait

[unkempt swift]

wait i honestly dont see how row ops are similar to this]

[grim nebula]

what are elementary matrices

[unkempt swift]

$\begin{pmatrix}
1 & 0 & 0\
0 & 1 & 1\
0 & 0 & 1
\end{pmatrix}$

[soft zealot]

HellO

[unkempt swift]

stuflike tese

the R1 + cR2

cR1

[grim nebula]

and they correspond to the row ops

[unkempt swift]

R1 <--> R2

[grim nebula]

so rewriting the hint

[unkempt swift]

$\begin{pmatrix}
1 & 0 & 0
0 & 1 & 1
0 & 0 & 1
\end{pmatrix}$
R2 + 1(R3)

[soft zealot]

HellO

[grim nebula]

If A is row equivalent to a matrix C, then A can be reached from C using some row ops

[unkempt swift]

yes

wait is that the clue

or am i supposed to be waiting

LOL

[grim nebula]

LOL no like

ive already explained

go think

about it

[unkempt swift]

yessor

ok wait

[unkempt swift]

is it right if i say that

if i make b such that 0,0,...1

this can be wrong sometimes

because i can switch the 1

via row operations?

[grim nebula]

that sounds kinda sus

[unkempt swift]

o

[unkempt swift]

this

why is LINEAR ALGEBRA SO DIFFICULT

[grim nebula]

you could make b just a vector with a single 1 in it

but

where that 1 is placed depends

on the row ops

somehow you need that 1 to land where k is

[unkempt swift]

i still dont get it, why cant i just "fix" it at the bottom

is it because of row ops

[grim nebula]

because the last row of A isnt 0

hmmmmmm wait

no you might be

correct

[unkempt swift]

A HA

[grim nebula]

i think we've solved it

[unkempt swift]

i think we overcomplicated it

[grim nebula]

yeah

[unkempt swift]

a

lmao

lemme write down the answer

.close

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[dawn elk]

Polynomial division

[tawny cedar]

yes

how do u do it

no matter what i do

it doesnt work

[dawn elk]

https://youtu.be/FXgV9ySNusc

[tawny cedar]

i know how to do it

it doesnt work for this specific problem

[dawn elk]

Hmm okay

Show ur work then maybe

[tawny cedar]

5x2 doesnt add with 6x

[dawn elk]

Yes, it doesn’t

So you carry forward

[tawny cedar]

?

[dawn elk]

So in next line you get something like

5x^2 +6x -6x +15

= 5x^2 +15

So you do usual steps from here on

So 5x^2/x^2 = 5 so you write + 5 at the top

And 5*3 =15 so we have 5x^2+15

Which perfectly cancels out to give 0 remainder

[tawny cedar]

sorry

so

u did

1 sec

why did u ddo

+6x

-6x

[dawn elk]

+6x is from here

And -6x is from here

Umm wait wrong img

Here

[tawny cedar]

.close

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[south juniper]

Could anybody help me with c and my imaginary d?

[south juniper]

My hypothesis for c is - 0.3×0.7+0.3×.07+0.3×0.7
Or something like that

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[south juniper]

.close

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[wispy sequoia]

the second one

ignore the first one

[wraith crater]

Think of another way to relate x, y and 300

[wispy sequoia]

Sin law?

[wraith crater]

Using the fact that it’s a right angle triangle

You don’t know the angles

[wispy sequoia]

pythegreom?

i spelt that wrong but

[wraith crater]

Yeah

[wispy sequoia]

ok so

how do i do that with one angle

[wraith crater]

You don’t need angles to use thePythagorean theorem

It’s this

[wispy sequoia]

Oh

Kk so

Now what

Only have one angle

[narrow wyvern]

hello

[wraith crater]

Now you have

[narrow wyvern]

is this occupied?

[wraith crater]

$a^2 + b^2 = 300^2$

[soft zealot]

Pure

[wraith crater]

I mean x and y

And $x+y = 170$

[soft zealot]

Pure

[wraith crater]

Rearrange the second eq for y

Then substitute into the first eq

Then solve for x and y

[wispy sequoia]

Uh

ok but

2 variables one known number

how do i

[wraith crater]

You have to equations

You know y=170-x

Then substitute this into the first equation x^2 + y^2 = 300^2

[wispy sequoia]

Ill try

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[rapid totem]

What am I doing wrong?

[rapid totem]

3/5 * 7 = 21/5 = 4 2/5

but calculator says 4 1/5

[limpid orchid]

When you divide 21 by 5 whats the remainder?

[rapid totem]

.2

[limpid orchid]

...

5*4=20

[rapid totem]

21/5 = 4.2

So wouldnt it be 4 2/5?

[limpid orchid]

No

You are trying to write it as a mixed fraction right?

[rapid totem]

Yeah

,rotate

[limpid orchid]

21/5 = 4.2 but 4.2 is in the decimal form
And 0.2 is the fractional part

[rapid totem]

bottom one

[limpid orchid]

You need the remainder

Do you know how to find the remainder?

[rapid totem]

no

[limpid orchid]

Do you know how to use long division?

[rapid totem]

yeah

[limpid orchid]

So use that to find the remainder

Do you know what the remainder is?

[rapid totem]

so 5/2 = 2 x 2 = 4 5-4 = 1?

[limpid orchid]

Yes

Now like that find the remainder for 21/5

[rapid totem]

4.1

21/5 = 5 4x5 = 20, 21 - 20 = 1

so I do that for every single question?

[limpid orchid]

Dont write it as 4.1

Decimal notation is a different thing

You write it as the quotient is 4 and remainder is 1

[rapid totem]

Wrote it like that

[limpid orchid]

Basically yea

[rapid totem]

well actually that

[limpid orchid]

So can you now write the mixed fraction for 84/9??

[limpid orchid]

Why 4.1 in the quotient?

It would be only 4

[rapid totem]

ah I remember

[limpid orchid]

.

[rapid totem]

I will do it

[limpid orchid]

K 👍

[rapid totem]

9 3/9?

[limpid orchid]

Yes

[rapid totem]

Great, thank you

[limpid orchid]

Np

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[keen depot]

Anyone know how to go from 57.6kg CO2 to (g CO2)/kWh?

[keen depot]

Gram CO2/kWh

So for example 1000g CO2/kWh

[wise pond]

Careful here because you have two different units which may not convert

you can: kg CO2/kWh -> g CO2/kWh
^ to do this conversation, multiply by 1000

but you cannot: kg CO2 -> g CO2/kWh

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[regal cloak]

w server

yo whoever is helping afk rq

[keen depot]

Well I have a task that asks me to do it

[final saddle]

@regal cloak Has your question been resolved?

[regal cloak]

that question

yooooo

yoo

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[turbid lotus]

"when the 3 and 12 get reduced to 4 and 1. The 12 gets reduced to 4 and 3 gets reduced to 1

so the 12 is being downsized to 4 because thats what you would need to multiply the denominator by to get the common denominator

But why is the denominator (3) being reduced to 1 ?
what is causing that and does that make the fraction disappear ?

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[tranquil pine]

how do you know when you have a trivial or non trivial solution

[tranquil pine]

halp

[deft ravine]

A trivial solution is trivial

[tranquil pine]

so would a trivial solution

be a zero row

and a non trivial solution

would be a zero solution with a non zero LHS

[deft ravine]

Like for the equation y^x = x^y the line y=x are the trivial solutions

[tranquil pine]

so like here

for the bottom row

would this be a trivial or non trivial

@deft ravine++

[deft ravine]

Haven't heard it in context of a matrix yet, sorry

[tranquil pine]

<@&286206848099549185>

[final saddle]

@tranquil pine Has your question been resolved?

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[soft grove]

hello does anyone know how to do this question

[soft grove]

oh ok thanks

.close

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[charred vigil]

How did they factor out this number?

[charred vigil]

im factoring i keep getting this

[shell mountain]

Where's + 12 coming from

[charred vigil]

by completing the square?
am i not suppose to do that?

[shell mountain]

That's not completing the square

[charred vigil]

wait nvm its nine

[shell mountain]

For one, you can't complete the square because it's not a quadratic, it's a cubic

And second, you can't just add numbers to an equation and expect it to be the same equation

[charred vigil]

okay i understand but how do i get from here to there?

[shell mountain]

You have x³ - 6x²

What do both terms have in common

[charred vigil]

x

[shell mountain]

Just x?

They only share a single x?

[charred vigil]

i mean, they cant share the same exponents or actual numbers so x?

[shell mountain]

x * x * x
6 * x * x

They only share 1 x?

[charred vigil]

if they put it that way, 2x?

[shell mountain]

Not 2x