help-10

[hidden compass]

Big means nothing in maths actually

[novel pewter]

Yea true

[hidden compass]

You should revise the theoretical concepts about functions, otherwise you will find dificult to solve these basic exercises

Anyway, the domain is the set of possible x values that you can give in input to the function

[novel pewter]

alright

[novel pewter]

I see

[hidden compass]

Ok, so based on that definition of domain, what would you answer for the domain of your function?

[novel pewter]

Well you said set of possible x values correct?

would that make it infinite

[hidden compass]

Yeah, so which of those 4 is the correct answer?

[novel pewter]

the first one?

[hidden compass]

$$\left(-\infty, +\infty\right)$$

[warm shale]

Alberto Z.

[hidden compass]

Exact

[novel pewter]

yea that one

[hidden compass]

Now, the range is the same but in the vertical direction (in other words for the y values)

[novel pewter]

and for the range wouldnt it also be infinite because it goes endlessly?

[novel pewter]

yes

[hidden compass]

Does y = 3 have a corresponding x-value?

In other words, if you draw the horizontal line y = 3, does this intersect the red graph anywhere?

[novel pewter]

yes

[hidden compass]

Where? 👀 For which value(s) of x?

[novel pewter]

at x = 3 if it were to be horizontal

right?

[hidden compass]

Noo, the y axis is the vertical axis, whereas the x axis is the horizontal axis

[novel pewter]

yes

[hidden compass]

Then why did you say this?

[novel pewter]

I meant like

like wouldnt it be at the middle of the graph

[hidden compass]

Nope

I mean, draw a point on the plane that has 3 as its y-coord @novel pewter

[novel pewter]

oh

[hidden compass]

No, don't do this, follow me at each step first

[novel pewter]

btw I have to go in 15 min

[novel pewter]

alright

[hidden compass]

Ok, what is the x-coord of that red point? @novel pewter

[novel pewter]

wdym

[hidden compass]

? do you know what x and y coordinates of a point are?

[novel pewter]

yes

[hidden compass]

So what's the problem in my question?

[novel pewter]

sorry

the x coordinate would b 0

be

[hidden compass]

Ahn ok alright

Now, draw the point (4 , 3) onto the previous graph (don't delete the point you did before)

[hidden compass]

Ok, now draw again on THIS graph the point you had before, (0 , 3)

[novel pewter]

the point I had before 0,3??

what was the point before that

[hidden compass]

(0, 3)

[novel pewter]

oh

[hidden compass]

Perfect, now connect these two points (with a horizontal line)

[hidden compass]

Line, not segment

Lines go infinitely long, whereas segments connect two points

[novel pewter]

like this

?

[hidden compass]

Yes

So, what is the property of all points belonging to that red line? They all share something in common, which is...

[novel pewter]

I actually dont know.

[hidden compass]

If I pick any point in that line, what can I say for sure about its coordinates??

[novel pewter]

y = 3?

[hidden compass]

Of course!!

So that's the line y = 3 @novel pewter

[novel pewter]

right

I have to go man

[hidden compass]

Alright

The answer for the range anyway was D, i.e. (-inf, 2] @novel pewter

Because horizontal lines abovbe y = 2 never intersect your function

https://caps.unm.edu/mathrefresh/assets/DomainsandRanges.pdf

[hidden compass]

And also this file can be very helpful, I suggest you to read it very carefully @novel pewter

[obtuse pebble]

@novel pewter Has your question been resolved?

[obtuse pebble]

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[keen moat]

how do i solve for b

[quaint glen]

can you make the picture more blurry somehow please

i can almost read it

[keen moat]

bud

a+b/c=5

$$a+b/c=5$$

[warm shale]

wildercat

[keen moat]

how do i solve for b

i multiplied c on both sides

[brisk hamlet]

$\frac{a + b}{c} = 5$

[warm shale]

Jangler

[keen moat]

then do i just subtract a

to get b=5/c-a

[brisk hamlet]

So you are done

[keen moat]

is that correct

or would it be

b=5c-a

?

[jagged cargo]

what did you get before moving the a

sounds like needless question, just making sure you get why to do said operation instead of just guessing just because

[keen moat]

nah i think i got it

a+b=5/c

then subtract a

[jagged cargo]

that is not correct

[keen moat]

to get b=5/c-a

it isn’t?

[jagged cargo]

i agree with the a operation. i dont agree with what you did with c

[keen moat]

what should i do then

i multiply both sides

so would it be 5c or 5/c

[jagged cargo]

yes thats correct. what is the right side multiplied by c

[keen moat]

wdym

5c

[jagged cargo]

yes

5c =/= 5/c

[keen moat]

so what’s the answer

[jagged cargo]

5c. make sure whatever you do to one side, you do the exact same to the other. otherwise the sides do not equal

[keen moat]

5c-a=b

?

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[faint kestrel]

im stuck there

[faint kestrel]

this is the question

[wet moss]

what can you do with the sinx/cosx?

[faint kestrel]

i was thinking maybe multiply it to 2 sin x cos x?

but i dont know if thats a thing

[wet moss]

do you know this rule?

[faint kestrel]

ooh i didnt

but thats what i thought

let me try

i can cancel out the cosines no?

leaving me with 2 sin x * sin x

[wet moss]

not quite

imagine b/c is sinx/cosx

[faint kestrel]

bc cos is multiplied by 2?

[wet moss]

so if you put it into that rule what do you have?

[wet moss]

the 2sinxcosx we can put as the "a" in this case

[faint kestrel]

2 sin x cos x * (sin x/cosx)

[wet moss]

remember b/c = sinx/cosx

[faint kestrel]

@warm shale $2 sin x cos x * (sin x/cosx)$

[warm shale]

suka

[wet moss]

you are then multiplying by c/b

[faint kestrel]

oh right

ohhh

so i divide evrything by sin x

[wet moss]

yeah

[faint kestrel]

which is 2 cos xcos x

and that is equal to 2*cos x2

[wet moss]

exactly

[faint kestrel]

squared

greate

thanks

!!!

[wet moss]

np

[faint kestrel]

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[sage dagger]

To differentiate 2xy using product rule so 2x • y

Then f’ = [2x•dy/dx] + 2y?

[wet moss]

yes

[sage dagger]

ok and if my original function was x^2 + 2xy = 8 it would be 2x + [2x•dy/dx] + 2y = 0

[wet moss]

yeah

[sage dagger]

ok nice

So 2x•dy/dx = -2x - 2y

Then dy/dx = (-2x - 2y)/2x right

[wet moss]

yep

[sage dagger]

Alright

Cool

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[noble kindle]

should I use double or triple integral?

Or is both fine.

The projection, I imagine, is just a unit circle on the xy plane

So if I just integrate over how much the x and y changes, that would give me many circles depending on the x and y value... SO I guess that could work.

But does it work if I do triple integrals?

[sullen nova]

try and parametrise the thing you're trying to find the volume of

by that i mean you need to give intervals for x y and z that you wanna integrate over

[noble kindle]

Im not sure what the bounds should be

..

Oh, I already did the double integral

Im just trying to see if the triple integral works?

i got it

[sullen nova]

the double integral will give you the surface area not the volume

of something

[noble kindle]

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[lavish nymph]

how does 1+ make sense

what does it represent

[shadow tangle]

log_7(7) = 1

[lavish nymph]

oh lol

.close

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[sweet lake]

Help

[tall abyss]

so basically what add 100 and 48

[sweet lake]

148

[tall abyss]

then do 360-148 and see

if not 360 then 180

[sweet lake]

212 for 360

32 for 180

[tall abyss]

it is one of those sorry i dont know which😔

[sweet lake]

issok

ty tho

[safe mortar]

no, it is just 148 degrees

since QRS is QRG combined with GRS, its measure is the sum of the measures of QRG and GRS

so 100+48=148

[obtuse pebble]

@sweet lake Has your question been resolved?

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[honest crystal]

is this right?

[obtuse pebble]

@honest crystal Has your question been resolved?

[daring rock]

I'd probably add the function f(x) itself too

or like, a graph of what f(x) could look like, since you don't know what it actually is

[honest crystal]

so y=0?

[daring rock]

No, the inequality says f(x) is between the two functions you drew for all x

[honest crystal]

I think I understand

something like this?

[daring rock]

Yeah, exactly, looks good

I'd maybe label the middle one f(x) too just to be safe but I'm sure that's basically what your teacher is looking for

[honest crystal]

ok thank you

.close

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[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

[wraith pike]

I have been stuck midway

[nocturne minnow]

Where are you stuck at?

What f(-2) means?

[wraith pike]

yeah what does it want me to put

[nocturne minnow]

You use that graph

As an example, f(4) means the function when x = 4

So f(4) just means find the y value when x = 4

So using that graph, you go to x = 4, then find what the y value is

[wraith pike]

would it be 1?

the y value

[nocturne minnow]

No

Using what I said, f(-2) means the y value when x = -2

So go to x = -2 on the axis, move until you reach the graph, then find what that y value is

[wraith pike]

mmmm

idkk

[nocturne minnow]

go to x = -2 on the axis, move until you reach the graph, then find what that y value is

[wraith pike]

what do you mean by reach the graph, sorry

the intercept?

[nocturne minnow]

The curve, that's drawn

[wraith pike]

ok once ive reached the graph what do i do

[nocturne minnow]

Like I'm saying first, start at the origin, move your cursor to x = -2, on the x axis, then move the cursor until you reach the curve, then move the cursor to the y axis, what's that value?

[wraith pike]

?

[nocturne minnow]

You move along the lines

[wraith pike]

started from orgin

[nocturne minnow]

So you reached x = -2

[wraith pike]

yes

[nocturne minnow]

Now you need to move until you reach the graph, the blue line

So travel vertically until you do that

[wraith pike]

so id get y= -3?

[nocturne minnow]

Yes

That's it

[wraith pike]

YAHH

THANK YOU BRODIE

i swear my math teacher is so bad at teaching

[obtuse pebble]

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[limber token]

3/4% of 200

[obtuse pebble]

Please don't occupy multiple help channels.

[tardy epoch]

!onechannel

[obtuse pebble]

Please stick to your channel.

[tardy epoch]

.close

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[timid silo]

I'm not sure how to start proving part b?

[sullen nova]

can you write the definition of v being in Span(S)

[timid silo]

V can be represented by some linear combinations of vectors in S?

[sullen nova]

so from there could you show that there's some linear combination of vectors in S union v that sum to 0

[timid silo]

Yes?
If say, S is as defined for a, and V is represented by say 4x_1+2x_2+8x_k
Then -4x-2x_2-8x_k+v=0?

[polar fossil]

ya

[timid silo]

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[wary seal]

How to do the (b) part?

[obtuse pebble]

Please don't occupy multiple help channels.

[wary seal]

<@&286206848099549185>

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

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[wary seal]

How do u do the (b) part?

[wary seal]

<@&286206848099549185>

<@&286206848099549185>

[timid silo]

!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

[wary seal]

6

Idk how to find it

[royal basin]

this sounds more like a 1 than a 6

anyway, consider that E, D and B are collinear means BE is a scalar multiple of BD

[wary seal]

Oh I didn't notice it

[wary seal]

Yeah I know that B, D and E lie on a straight line

[royal basin]

you may write this as BE = μBD, where mu is another unknown scalar

then express the equality BO + λOA = μBD entirely in terms of the vectors a and b

[wary seal]

Yeah

Next

?

[obtuse pebble]

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.close

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[stark bay]

cant find my mistake, the final value when differentiated isn't giving the same integral. can someone pls help

<@&286206848099549185>

[obtuse pebble]

@stark bay Has your question been resolved?

[stark bay]

<@&286206848099549185>

[obtuse pebble]

@stark bay Has your question been resolved?

[undone quartz]

can i get a more clear image

[obtuse pebble]

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[bright pagoda]

quick question, if i had like (a+b)/x=c, how could i make x the subject, is it possible?

[hollow dove]

you can multiply both sides by x, as long as x isn't zero

then do some rearrangement from there

[bright pagoda]

is that the only way?

[hollow dove]

hmm, yeah basically

[bright pagoda]

ok thanks

[hollow dove]

any other way will basically be this in disguise

[high lily]

you could take a detour, end reuslts going to be the same

[hollow dove]

I guess you could take reciprocals of both sides

but it's the same thing

[kind hawk]

you could start by adding 2 on both sides, then multiplying by 17, cube both sides, rearrange further, at some point undo the three things I listed and get the same result anyway

[bright pagoda]

this is what ive got

i want to make x the subject

[hollow dove]

start by multiplying by x

so you get $ba^2-b^2a=x(-ba-b^2-ca-cb)$

[warm shale]

Desync

[hollow dove]

then you have (stuff) = x * (other stuff)

so you can just divide through

[bright pagoda]

oh ok

gotcha

thanks

[hollow dove]

(again, assuming (other stuff) is non-zero)

[bright pagoda]

yeah

bye

.close

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[versed stratus]

!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

[rough stratus]

1

Like, I tried doing something, but they result nowhere

[versed stratus]

Add both equations, and then use the trig addition/subtraction formulae

[rough stratus]

3sinp+4sinq?

[versed stratus]

add both equations first

[rough stratus]

I did

[versed stratus]

you'll get sinpsinq+cospcosq=1

do you recognize that formula

[rough stratus]

How does adding result in that

[rough stratus]

It's cos p-q

[versed stratus]

yup

so cos(p-q)=0

[rough stratus]

...

[versed stratus]

wdym

you have two eqns , don't you?

add them

[rough stratus]

So, this is wrong?

,rccw

[versed stratus]

it's right, but i don't think it helps you solve the problem in any way

oh, oops, I was multiplying them.

sorry

[rough stratus]

💀

[versed stratus]

yeah, my bad

[rough stratus]

Now what

@versed stratus

[versed stratus]

I'd probably sub sinp=x , cosq=y. But I'm sure there's a more elegant method

[rough stratus]

<@&286206848099549185>

[obtuse pebble]

@rough stratus Has your question been resolved?

[rough stratus]

<@&286206848099549185>

[obtuse pebble]

@rough stratus Has your question been resolved?

[dawn iris]

Little me found a way.

[formal breach]

i think u should drop it

[obtuse pebble]

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[royal basin]

!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

[frosty valve]

4

the ans from my textbook is 1.38m

but i keep getting more than that lol-

@royal basin

[royal basin]

are you sure you definitely meant 1.38 m (just meters) and not 1.38 m**^3** when quoting the book's answer?

your work is very sparse, but it looks like you've confused yourself as to the meaning of the letter x.

in the equation as you wrote it, x is the volume of the statue in cubic centimeters,

so once you get the value of x, you need to also convert it to the right units.

[frosty valve]

yea sorry

[frosty valve]

i did

[royal basin]

can you show your work for the unit conversion?

[frosty valve]

pretty much got the same thing again

[royal basin]

your work is once again sparse,

but your mistake is thinking that 1 m^3 is 100 cm^3.

it is not.

[frosty valve]

lol it's not?💀

[royal basin]

it's not.

[frosty valve]

lemme try again

[frosty valve]

so since it's m³ i'll have to divide x by 100 (3times)?

[royal basin]

sure if you wanna phrase it that way

[frosty valve]

what's the proper way to phrase it?

[royal basin]

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

therefore 1,385,400 cm^3 = 1.3854 m^3

[frosty valve]

ah alr

thx

[obtuse pebble]

@frosty valve Has your question been resolved?

Channel closed

Closed by @frosty valve

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[obtuse pebble]

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[ashen saddle]

I'm confused about solving some differential equations. I understand how to deal with cases where you get an absolute value of y. However, when you end up with absolute values of x, I get confused about how people are getting rid of the absolute value.

[ashen saddle]

So, for example, take the differential equation dy/dx+y*(1/x)=1.

I'm getting an absolute value in my answer. However, other people don't. The reason why is that when finding the integration factor, they change e^(ln(abs(x)) to just x, whereas I change that to abs(x).

I don't see how you can just get rid of the absolute value like that.

Wolfram alpha does the same thing: https://www.wolframalpha.com/input?i=dy%2Fdx%2By*1%2Fx%3D1

[compact shadow]

You need to treat it as two different ODE

One with domain (0, infty)

Another one with domain (-infty, 0)

Though they have the same equation, but these need to be viewed as two different ODEs

Each of them you can take absolute value symbol off

[ruby path]

I mean

[compact shadow]

x and -x respectively

[ruby path]

What OP did was fine tbh

[ashen saddle]

You mean what I did?

[ruby path]

Yes

[ashen saddle]

Why?

[ruby path]

People tend to ignore absolute values for simplicity

[ashen saddle]

Doesn't that not give you all solutions?

[compact shadow]

Because R-{0} is not connected

[ruby path]

yeah you do lose solutions I suppose

[ashen saddle]

But even wolfram alpha does it.

[ruby path]

I mean when was the last time you did $\int \frac{\dd x}{|x|\sqrt{x^2 - 1}} = \arcsec x$

[warm shale]

NEON

[ruby path]

Indeed it does

,w solve y' + y/x = 1

[warm shale]

Results provided by WolframAlpha

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

Requested by neonperseus

[compact shadow]

For (0, infty) you have solutions , each two of them are different by a constant.
For (-infty, 0) you have some solutions each two of them different by a constant of course. Take any solution f for (0, infty) and any solution g for (-infty,0) is a solution on R-{0}

But two pairs (f,g), (f’,g’) no longer have to be different by one constant

[ruby path]

It's one function just not defined at 0

[compact shadow]

You can have f’=f+a, g’=g+b

That’s why I treat it separately, on each of the connected components

I guess you will do the same when you encounter something like tan(x). This time 1/x has only two components, of its domain

[ashen saddle]

I understand what you're saying I think. You would just handle the left and right sides of the coordinate plane seperately, right?

[compact shadow]

Yeah

[ashen saddle]

Is what wolfram alpha did wrong then?

I was wondering if wolfram alpha assumed x>0

Based on its graphs.

[compact shadow]

(0,infty) probably.
more than 1 connected components, two different solutions are no longer different by a constant necessarily

[ashen saddle]

Yeah that's what I thought. That's why I've been so confused because I thought that just adding a constant isn't going to be able to replace the absolute value.

But that's what everyone seems to think.

[compact shadow]

I guess sometimes people aren’t cautious enough

[ashen saddle]

Hmm

I think they need to fix wolfram alpha.

It should state that it is assuming x>0

I might report that

Hold on

It's not assuming x>0

[compact shadow]

Because on (-infty,0) the solution happens to be

Of the same form

The only issue is, you can have two different constants for each branch

[ashen saddle]

For each branch of the absolute value?

[compact shadow]

Each branch of domain of 1/x

Like f(x)=x/2+5293/x for x>0, =x/2-19374/x for x<0 works

[ashen saddle]

Wait what, I'm confused. But nothing is equal to anything with that equation.

The domains of both of those functions are completely seperate.

So I guess that is true, but what is your point?

[compact shadow]

My point is

There are two components, one constant is not enough

[ashen saddle]

Aaah

I see

[compact shadow]

f(x)=x/2+c/x for x>0, =x/2+d/x for x<0, is sufficient

[ashen saddle]

So wolfram alpha is hiding its constants from us.

Rather deceptive...

[compact shadow]

It either omitted one component, or chose two constants being equal

[ashen saddle]

I see...

So if you were actually doing an initial value problem and using some numerical differential equation solver or something, you would need to negate your constant as you move over the y axis?

[compact shadow]

A given value on a component , can make me fix one solution on that component

If you tell me y(1)=8, sure I now have a fixed y(x) for x>0. I can’t determine y(x) for x<0, unless you give me something like y(-6)=71 too, or simply tell me the domain now is only (0, infty)

You have m components, you will need m values to determine the solution , one value for each component

[ashen saddle]

But can't you just negate the constant?

As you cross over the y axis?

[compact shadow]

Idk, I just can’t determine y(x) for x<0 if (-infty,0) is part of domain and you don’t provide me a value of y(x): x<0.

Any solution on (-infty, 0) works

y(x)=x/2+1/x: x doesn’t equal 0, and z(x)=x/2+1/x: x>0, =x/2+5/x: x<0. There is no one of them being more special than the other one thing

[ashen saddle]

But doesn't my equation work?

I could find C given some y(x) for x>0, and then have a solution for x<0.

I understand your components idea, but can't I just use my equation with the absolute value?

[compact shadow]

Then you only gave solutions whose constants are of the form (C,C)

But solutions whose constants are (C,D) are also solutions

[ashen saddle]

Is that because it's undefined at 0?

[compact shadow]

It is because , again, the domain has two connected components

[ashen saddle]

What is a component?

[compact shadow]

If it’s not 1/x, but 1/x(x-1) then three connected components

For now you can view it as how many parts the domain is divided into

[ashen saddle]

Because it's undefined at two places?

[compact shadow]

You can say that

[ashen saddle]

K

So

[compact shadow]

Because the domain is divided into three parts

(An exact definition is , for a topological space X, a subset A is said to be connected, if A is both open and closed. A connected component of X , is a maximum (in terms of inclusion) connected subset of X. Then X is disjoint union of connected components of it)

[ashen saddle]

I'm a computer programmer, so I often thing about differential equations in the light of numerical approximations. If I was performing an approximation, then I could pass that undefined line fine just because I'm using finite steps.

[compact shadow]

I see. I have no idea. I only know math

[ashen saddle]

Do you know about numerical approximation of differential equation solutions?

[compact shadow]

No

[ashen saddle]

Ooh ok

[compact shadow]

I don’t know approximation of any kind. I am only interested in pure math

[ashen saddle]

Interesting

Do you think approximation is not a part of math?

pure math I mean

[compact shadow]

Maybe it is, just looks very tedious to me personally

By comparison. I am usually interested in category theory kind of things, or algebraic geometry

[ashen saddle]

I would say that the fact that sin(x) approx= x for small x is a great mathematical fact.

I see

[compact shadow]

That’s math

lim sin(x)/x=1

But if someone asks me to approx sin (37 degrees) , things like that

No looks like just calculation

[ashen saddle]

So you care more about general cases?

[compact shadow]

Yeah

I answer analysis problems, but I don’t study them unless I need to. Like if I read Lie algebra , topological group , then analysis is not avoidable , you know, Haar measure , things like that , I guess I will read it.

I never answer PDE, or Statistic, in these channels

Ode I am okay with, PDE no, complicated

[ashen saddle]

Yeah, I respect that

I definitely see the importance of computation and approximation though.

[compact shadow]

Yeah. Of course they are important.

Just problem of my personal taste.

[ashen saddle]

Discrete fourier transform > continuous, because it's more useful.

[compact shadow]

More general a result is, I like it more

[ashen saddle]

Yep, makes sense.

Well, thank you so much for your help. I understand this way better now.

[compact shadow]

Np

[ashen saddle]

I still think wolfram alpha should be more explicit about what it is doing with the constants and components.

[compact shadow]

Yeah, you are very responsible

I like that. They should have been more rigorous

[ashen saddle]

Have a nice day!

.close

[obtuse pebble]

Channel closed

Closed by @ashen saddle

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[compact shadow]

You too☺️

[obtuse pebble]

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[knotty crow]

what have you tried

do you know how does rationalizing the denom work?

[timid silo]

Nope

I gotta do something with the 14

Idk tho

7x2 but neither can be rooted

[mellow wraith]

you know that (a+b)(a-b)= a^2-b^2

[timid silo]

No

[knotty crow]

it's not about taking a square root of 14 in some way

[cloud igloo]

Full question?

Rationalize the denominator

[pulsar quarry]

do you even know what rationalizing means?

[timid silo]

Wtf

[pulsar quarry]

did'nt your teacher taught you this?

[timid silo]

Nope

[cloud igloo]

Or you didn’t pay attention in class?

[pulsar quarry]

then i wouldnt recommend you attempting this ques

[pulsar quarry]

lol

[nocturne minnow]

What does "More instruction" do?

[timid silo]

I’m in college now. It’s been since 9th grade since I took algebra

[cloud igloo]

😭

[timid silo]

I took algebra one in 8th and 9th was 2

[timid silo]

Gives me a terrible explanation that I’ve already tried lol

[nocturne minnow]

Show

And
https://www.youtube.com/watch?v=tpA-F-cz6Mw&ab_channel=MartyBrandl

[cloud igloo]

If it ain’t Organic Chemistry Tutor it ain’t right 💯

[nocturne minnow]

I tried tbh but I was too lazy to find one what did conjugates, that was the first one I found that did conjugates

[cloud igloo]

Fair enough

@timid silo do you understand now?

[timid silo]

I got 6 + 10 to square root of 14

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

[cloud igloo]

Ehr

[timid silo]

Right

?

[mellow wraith]

nah

[cloud igloo]

It’s not correct

[timid silo]

bruh

[mellow wraith]

u got the signs wrong

[warm shale]

𝕏

[timid silo]

How’d you do that

[nocturne minnow]

Your simplifying on the last line is wrong

[cloud igloo]

LaTeX

[nocturne minnow]

$\frac{a+b}{a} \neq b$

[warm shale]

dldh06

[nocturne minnow]

Meaning you can't do only -30/-5 = 6

[timid silo]

Oh I gotta do 14/6

-5*

Square root

[nocturne minnow]

You can't just divide one of the terms by 5

[nocturne minnow]

No

The 10

[cloud igloo]

You can apple difference of two scares btw

Squares *

For denominator

[nocturne minnow]

They did that math right, just the longer way

[cloud igloo]

Oh damn I didn’t even look at the pic right

[nocturne minnow]

It was the simplifying of $\frac{-30 +10\sqrt{14}}{-5}$ was wrong

[warm shale]

dldh06

[cloud igloo]

Factor out 10

So you have multiplication

[nocturne minnow]

Factor 5 is better

[cloud igloo]

And proceed

[nocturne minnow]

From the numerator

[cloud igloo]

I guess

[nocturne minnow]

Well -5 from the numerator

[timid silo]

Wtf

[nocturne minnow]

$ab + ac = a(b + c)$

[warm shale]

dldh06

[nocturne minnow]

No

[cloud igloo]

Bloodclat confuse wid di maths😭

[nocturne minnow]

You want to simplify this

You can factor out a -5 from the numerator first

[timid silo]

Ok

[nocturne minnow]

Do you see how the -5's cancel out?

[cloud igloo]

$$
\frac{-5(6-2 \sqrt{14})}{-5}
$$

[warm shale]

𝕏

[timid silo]

Yes

[cloud igloo]

Damn my WiFi slow 😣

[timid silo]

So my answer is 6-2 square root of 14

[nocturne minnow]

Yes

[timid silo]

😭

I’m dropping out dawg

[cloud igloo]

Good luck bredrin

[timid silo]

I’m jk I ain’t dropping out but bruh

.del

[slim cove]

.close

[obtuse pebble]

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[pulsar quarry]

how do i convert this in the form of degree and minutes

actual ques

my work

[slim cove]

I guess I should say first, it's not equal to 11/25 degrees

It's equal to 11/25 radians

How do you convert from radians to degrees?

[pulsar quarry]

we multiply by

180/pi

but wait

[pulsar quarry]

how is that in radians?

[slim cove]

"An angle is x radians" means that "The ratio between the arc and the radius is x"

That's the definition of radian

So since the ratio between the arc and the radius is 11/25

The angle is 11/25 radians

Does that make sense?

[pulsar quarry]

yes

yes it does

okay, didnt know that

[slim cove]

Okay awesome

[pulsar quarry]

now i can do that

ty mam

[slim cove]

So in degrees it's 11/25 * 180/pi

And then you need a calculator to figure out what this is

[pulsar quarry]

well, calculators are restricted

i can figure it out now

ty

.close

[obtuse pebble]

Channel closed

Closed by @pulsar quarry

Use .reopen if this was a mistake.

[pulsar quarry]

.reopen

[obtuse pebble]

✅

[pulsar quarry]

sorry

i still cant do it

[royal basin]

what's troubling you?

converting 11/25 radians into degrees?

[pulsar quarry]

okay so i only

converted radian measure into degrees when it had the pi in it

and then multiply it by

180/pi

pi-pi gets cancelled

idk

[royal basin]

you are ascribing some kind of mysticism to the conversion between degrees and radians.

when in fact it is just like any other unit conversion.

just like 1 meter = 100 cm, you have 1 rad = (180/pi) degrees

the conversion factor has pi in it but that does not really make things any different

[pulsar quarry]

so should it be 11pi/25?

[royal basin]

no

11/25 is 11/25.

if we wanted to write 11pi/25 we would have written 11pi/25.

you will not have a cancellation of the kind you are used to.

[pulsar quarry]

okay so after multiplying 11/25*180/pi ?

[royal basin]

,calc 11/25 * 180/pi

[warm shale]

Result:

25.210142985756

[royal basin]

25 and a bit degrees.

,calc 0.210142985756 * 60

[warm shale]

Result:

12.60857914536

[royal basin]

should be 25°13' if we are rounding to the nearest arc-minute, but only 25°12' is among the options, so that is probably what they want.

[pulsar quarry]

but is there a way i can do this without using calculator?

[royal basin]

no

[pulsar quarry]

like simply putting value of pi and then solving

[royal basin]

well ok yes there kind of is

long division

but it is painful and time-consuming

[pulsar quarry]

like 11/25*180/3.14?

[royal basin]

i think using 3.14 for pi might already lead to too much inaccuracy.

,calc 11/25 * 180/3.14

[warm shale]

Result:

25.222929936306

[royal basin]

,calc 11/25 * 180/3.142

[warm shale]

Result:

25.206874602164

[royal basin]

,calc (11/25 * 180/3.142 - 25) * 60

[warm shale]

Result:

12.412476129854

[pulsar quarry]

i dont have any option, we cannot use calculators, sorry

[royal basin]

,calc (11/25 * 180/3.14 - 25) * 60

[warm shale]

Result:

13.375796178344

[royal basin]

okay so you would need at least 3 decimal digits of pi to get an accurate result ehre

[royal basin]

in how many seconds are you expected to do this question

[pulsar quarry]

i mean

maximum of 2 mins

[royal basin]

i think you are just fucked.

[pulsar quarry]

lol

okay, ill use three digit of pi

thanks tho

.close

[obtuse pebble]

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[obtuse pebble]

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[surreal carbon]

Hello, I am getting two different derivs and wonder which is the correct one. Cause I need to find the extreme points.

[timid silo]

The first one you have

f(x)

Differentiate like you normally do

Equate to the one below

Should get values of x

(idk how to differentiate 3x / (1+4x⁴) btw)

[surreal carbon]

alright, I'll keep doing that

[timid silo]

Only one of them should be correct as clearly one of the answer you found is 9 times the other

[surreal carbon]

yes exacly but the left is the proffessors answer

[timid silo]

Unless x = ⁴√(1/4) => very unlikely

[surreal carbon]

but It didn't make sense even when she showed it

[timid silo]

Hmm my knowledge too little to help you more correctly

[surreal carbon]

the right one I mean

the left answer is mine

aight

[obtuse pebble]

@surreal carbon Has your question been resolved?

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[dry rain]

Hi, I need help please. I have to graph a cubic function with the following characteristics. I'm not sure as to how to draw it.

f(-2) = 0
f(0)=8
f’(0) = 0
f’’(x) > 0 when x > 0, and vice versa

[forest sinew]

what have you tried? how far did you get?\

[timid silo]

!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

[tired skiff]

toki

[forest sinew]

to ki

[timid silo]

@dry rain

[tired skiff]

he's playing fortnite rn

[timid silo]

Bro got carried away

Shit I was doing math too

Forgot

Bye

[timid silo]

f''(x) tell us there's a minima for positive x and maxima for negative x (local)
f(0) tells us y intersect of graph at x=0 is 8
Ofc f''(0) is 0 when f(0) = 0 but also, slope of graph at x=0 is constant?? (Yeah ofc, it's always tangent constant)
X intersect of graph is at y= 0 when x = -2

[tardy epoch]

can you stop

ask questions in your own channel

[timid silo]

Bro I'm ANSWERING THAT QUESTION WDYM

[tardy epoch]

.

[timid silo]

Happy?

Cuz I can be wrong sometimes and I don't want people all over me 😭

[tardy epoch]

be wrong in your own channel then

you're just spamming joel's channel

[timid silo]

I'm not, thankyou

[tardy epoch]

you definitely are

[timid silo]

I did answer the question tho 👍

[obtuse pebble]

@dry rain Has your question been resolved?

[warm canopy]

you should wait for them so they can work through the problem with guidance rather than listing half accurate information at them

[dry rain]

well, i know that the x intercept is -2, and that the y is 8

but i’m not sure how to draw the graph based on the other characteristics

this is what i drew and i’m not sure if it’s right

like, because of f’(0) =0, shouldn’t the y intercept also be a turning point?
but then i don’t have another x intercept

[obtuse pebble]

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[nocturne minnow]

@zenith raft @tardy epoch Help me with math lmao

[nocturne minnow]

.close

[obtuse pebble]

Channel closed

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[tardy epoch]

(╯°□°)╯︵ ┻━┻

[nocturne minnow]

┬─┬ノ( º _ ºノ)

[obtuse pebble]

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[nocturne minnow]

@tardy epoch @zenith raft @polar fossil How to find det of
$\begin{bmatrix}
1 & 1 & 2 & 4 & 4 & 0 & 3 & 0 & 8 & 3 \
7 & 1 & 8 & 9 & 7 & 3 & 2 & 3 & 4 & 4 \
4 & 2 & 9 & 5 & 7 & 4 & 8 & 9 & 2 & 3 \
1 & 3 & 6 & 6 & 7 & 7 & 1 & 9 & 3 & 1 \
4 & 4 & 1 & 8 & 7 & 2 & 0 & 3 & 2 & 8 \
5 & 8 & 8 & 7 & 9 & 9 & 5 & 5 & 1 & 6 \
8 & 5 & 7 & 0 & 0 & 9 & 8 & 7 & 2 & 1 \
2 & 8 & 0 & 9 & 2 & 8 & 3 & 6 & 1 & 3 \
9 & 5 & 4 & 5 & 9 & 0 & 1 & 1 & 8 & 7 \
2 & 5 & 5 & 2 & 1 & 9 & 2 & 5 & 5 & 6
\end{bmatrix}$? Lmao

[nocturne minnow]

.close

[obtuse pebble]

Channel closed

Closed by @nocturne minnow

Use .reopen if this was a mistake.

[polar fossil]

ti-89

[zenith raft]

☠️

[warm shale]

dldh06

[drifting badger]

!show

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

[nocturne minnow]

Don't have one, how to do by hand?

[polar fossil]

it's randomly generated, it's probably 0

right?

[floral garnet]

U serious?

[floral garnet]

Most probably

[tardy epoch]

he isn't

[tardy epoch]

i will murder you

[floral garnet]

Oh good😂

[nocturne minnow]

Help me riemann how to math

I mean meth

[tardy epoch]

there was balance in the force

[drifting badger]

Just multiply every row by 0 and the solution is trivial

[tardy epoch]

then you interrupted it with your bad Midichlorians

[drifting badger]

Mitochondria

[timid silo]

diagonalize it

[nocturne minnow]

I wasn't serious

I don't care about it

[tardy epoch]

!15m

[obtuse pebble]

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

[nocturne minnow]

!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

[tardy epoch]

jezus google come on

[nocturne minnow]

🔪

[nocturne minnow]

That's a 10 x 10....

I wanted this one
2 & 3 & 5 & 3 & 8 & 2 & 1 & 4 & 4 & 9 & 0 & 3 & 3 & 4 & 8 & 6 & 3 & 9 & 1 & 3\
9 & 2 & 4 & 8 & 0 & 7 & 0 & 7 & 1 & 3 & 0 & 8 & 0 & 1 & 1 & 2 & 4 & 5 & 7 & 2\
3 & 6 & 4 & 4 & 6 & 8 & 1 & 6 & 9 & 4 & 3 & 1 & 5 & 8 & 3 & 3 & 0 & 6 & 1 & 5\
2 & 8 & 2 & 1 & 4 & 0 & 6 & 3 & 4 & 0 & 4 & 5 & 9 & 1 & 9 & 1 & 0 & 0 & 4 & 1\
2 & 3 & 1 & 2 & 7 & 0 & 2 & 6 & 4 & 6 & 0 & 9 & 5 & 2 & 2 & 8 & 2 & 9 & 7 & 7\
0 & 7 & 8 & 2 & 7 & 6 & 2 & 1 & 6 & 7 & 8 & 1 & 4 & 9 & 0 & 2 & 2 & 3 & 6 & 3\
2 & 1 & 2 & 5 & 9 & 2 & 2 & 3 & 1 & 8 & 2 & 1 & 3 & 7 & 1 & 6 & 8 & 4 & 8 & 2\
3 & 2 & 2 & 7 & 0 & 1 & 5 & 3 & 5 & 3 & 0 & 0 & 9 & 7 & 7 & 0 & 1 & 5 & 0 & 5\
4 & 7 & 6 & 4 & 3 & 0 & 0 & 7 & 6 & 3 & 3 & 9 & 7 & 9 & 9 & 2 & 7 & 7 & 2 & 6\
7 & 7 & 0 & 5 & 5 & 2 & 0 & 4 & 1 & 6 & 7 & 2 & 3 & 1 & 9 & 3 & 0 & 4 & 2 & 5\
2 & 6 & 9 & 9 & 4 & 3 & 7 & 8 & 5 & 6 & 1 & 8 & 2 & 1 & 9 & 1 & 2 & 6 & 1 & 6\
5 & 0 & 3 & 9 & 7 & 0 & 2 & 6 & 3 & 4 & 8 & 2 & 7 & 6 & 8 & 1 & 6 & 4 & 2 & 6\
3 & 8 & 1 & 1 & 4 & 7 & 1 & 0 & 7 & 0 & 7 & 9 & 8 & 7 & 8 & 9 & 6 & 6 & 5 & 2\
1 & 2 & 0 & 2 & 7 & 5 & 5 & 8 & 5 & 0 & 0 & 4 & 7 & 7 & 5 & 4 & 1 & 7 & 5 & 3\
7 & 9 & 9 & 3 & 8 & 1 & 5 & 6 & 8 & 4 & 6 & 6 & 5 & 0 & 4 & 8 & 0 & 3 & 4 & 6\
4 & 2 & 8 & 0 & 5 & 5 & 6 & 5 & 1 & 8 & 9 & 4 & 7 & 7 & 2 & 1 & 2 & 2 & 3 & 9\
2 & 8 & 8 & 0 & 5 & 4 & 1 & 5 & 6 & 2 & 1 & 9 & 5 & 1 & 0 & 3 & 6 & 5 & 9 & 2\
2 & 0 & 5 & 9 & 1 & 1 & 1 & 6 & 5 & 5 & 3 & 6 & 2 & 5 & 2 & 8 & 9 & 8 & 8 & 8\
7 & 8 & 2 & 1 & 3 & 2 & 0 & 4 & 2 & 7 & 8 & 4 & 4 & 1 & 6 & 9 & 2 & 2 & 7 & 4\
6 & 5 & 0 & 3 & 5 & 8 & 5 & 5 & 9 & 5 & 7 & 8 & 8 & 8 & 0 & 0 & 7 & 6 & 9 & 5\
But I couldn't format it

[drifting badger]

[nocturne minnow]

@tardy epoch Help me with this instead

[tardy epoch]

i will double murder you

[nocturne minnow]

Is it W?

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[reef nymph]

how does the area of a polygonal face be generalized to $\frac{nl^2}{4tan(frac{180}{n}}$

[warm shale]

AℤØ

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

[tardy epoch]

what's n and l here

[reef nymph]

n is the number of sides and l is the side length

[brittle topaz]

is this a regular polygon

[reef nymph]

yes

[brittle topaz]

think of it as splitting the polygon into n different triangles

with the vertices being, the center of the polygon and 2 adjecent vertices of the polygon

like this for a regular hexagon

or this for a heptagon

you can calculate the area of 1 triangle and multiply by n

[reef nymph]

so the area of one triangle will be 0.5 s^2 sinX given that angle x is the one where one of the vertices is in the center of the triangle and s is the side(s) adj to angle x

but the formula uses l which is the side opp angle x

[brittle topaz]

so s is the length from the center of the polygon to one of the vertices?

[reef nymph]

yes

[brittle topaz]

l/2=s*sin(x/2)

so s=l/(2sin(x/2))

you can sub that into you equation

[reef nymph]

I found that the base angles of the triangle will be (n-2)*180/n

[brittle topaz]

you have to divide by 2 because thats the angle formed at each vertex

and the base angle is half of that

if you sub l/2=s*sin(x/2) into your equation, we get l^2cos(x/2)/4sin(x/2)

simplifying gives us the formula l^2/4tan(x/2)

but x=360/n so we can sub that into the equation to get l^2/4tan(180/n)

and then multiply by n to get our equation for the area of the whole polygon

[reef nymph]

thanks vv much.... I was forgetting to half the angle.... I was getting some vv weird stuff

[brittle topaz]

no problem

[reef nymph]

.close

[obtuse pebble]

Channel closed

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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>.
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• 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.

[rancid quest]

i need help

[polar fossil]

what about it?

[rancid quest]

i need help

and i don't understand anything-

[tardy epoch]

<@&268886789983436800>

[rancid quest]

nvm...

allgd ill try again

.close

[obtuse pebble]

Channel closed

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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>.
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• 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.

[pulsar grail]

I have this sprite and I want to find where the green dots are. The players origin us half the width and 9/13 of the height (going from top to bottom and left to right.) How do I find the x and y position of the green dots if the player spins around the origin. right green dot is 11 / 13 of the width and 3 / 13 for the height. left green dot is 3/13 width and height. Never done trig or geometry before because I am a sophmore what formula or whatever do I need?

[pulsar grail]

My player is on a large screen and I have the player x and y which is located at half the width and half the height

[timid silo]

Oooo, a CS-related question! We rarely get those

[pulsar grail]

yea most cs guys are good at math lol

[timid silo]

What information do you have? The player's angle?

[pulsar grail]

yes

[timid silo]

In the picture you sent, is the angle 90° or 0°?

[pulsar grail]

0

[timid silo]

You could precompute the distance from the origin and angle of the 2 green dots when it is at 0°

Then you can calculate their current angle at any time by adding the player's angle to the precomputed angles

After which you can use cos of angle times magnitude and sin of angle times magnitude to get their offset from the origin

Then just add that to the origin's coordinates

[pulsar grail]

ok thanks a ton I'll get back to you when I program all of that 🙂

probably not today it's pretty late

[timid silo]

I'm honestly not sure if that's the most elegant way to do it

It's probably the most efficient though

[pulsar grail]

eh I have no idea how this stuff works so as long as it works thanks

[timid silo]

CS grindset

More seriously though try to understand it

[pulsar grail]

ok thanks

[timid silo]

@pulsar grail By the way, if you're planning on messaging me back in DMs you can .close this channel

[pulsar grail]

.close

[obtuse pebble]

Channel closed

Closed by @pulsar grail

Use .reopen if this was a mistake.

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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.

[timid silo]

What would be the best way to write this $$\abs{\vec{v_2}}$$ term?

[warm shale]

One person

[polar gulch]

huh, you have a text channel before this

[timid silo]

Basically it would be the square root of the vertical velocity of the foot ball at t_0 and the horizontal at t_0... i should've made t_0 t_1

[timid silo]

No i don't?

[polar gulch]

you have help 13 before in this time,

you created more

[timid silo]

Yeah, because Different question

[polar gulch]

different not all you can cut it into another channel

[timid silo]

Green leaf

[polar gulch]

wait why im not active?

[timid silo]

Hmmm

[polar gulch]

you can wait the other, im lazy

[timid silo]

The plan

I'm confused about $$\theta_2$$ tho

[warm shale]

One person

[timid silo]

It only gives one value

But i need two from it

how do i get the 2nd one

the

highest/minimum angle to reach the distance $$R- \Delta x$$

[warm shale]

One person

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

:(

.close

[obtuse pebble]

Channel closed

Closed by @west valve

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Available help channel!

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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.

[haughty lance]

Lets say I have some function that is $g(x) = \frac{-1}{f(x)}$

[warm shale]

SubsonicSpraak

[haughty lance]

Is the global/local minimum of f(x) also a local/global min of g(x)?

[grizzled shore]

doesn't look like it

,w graph x^2

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by frosst

[grizzled shore]

,w graph -1/x^2

[warm shale]

Results provided by WolframAlpha

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

Requested by frosst

[grizzled shore]

doesn't even look like it's defined

doesn't look like it for sinx either

[haughty lance]

Hmm, yeah, fair point. but a local minima for sin(x) is a local minima for -1/sin(x) (as per your graph)?

[slim cove]

First of all,
g'(x) = f'(x)/(f(x))^2, so g' has the same critical points as f' whenever f is not zero
g''(x) = [f(x)^2 f''(x) - 2 f'(x)^2 f(x)]/(f(x))^4, so at any critical point where f is not zero, g'' has the same sign as f''
From these we can basically conclude the property about local minima/maxima of f transferring to local minima/maxima of g, but global minima/maxima are not similarly well behaved, as demonstrated by Frosst

If we assume that f is smooth and nowhere zero though, then I think you do get that g and f share the same local/global minima/maxima

[grizzled shore]

ah there's the 2 parts, x^2 shows the f is 0 situation and sinx shows the global situation

[slim cove]

Since -1/x is a strictly increasing function if you restrict to x>0 or x<0

In fact you don't even need the criterion that f is smooth tbh

Just that it stays either above 0 or below 0 the entire time

[haughty lance]

To be honest this is to do with the breaking time for PDEs (the first time (t) a shock forms), and I'm trying to word how I'm analytically exploring that above function to reason when we get the minimum positive value of:

$\frac{-1}{f(x)}$

[warm shale]

SubsonicSpraak

[grizzled shore]

those are definitely words i've seen in my pdes class but i dont understand them

[elfin burrow]

https://xyproblem.info

[slim cove]

You're going to have to post more context

If you want help with it

Unless you're satisfied with what you already got

[haughty lance]

I just needed to grapple with the initial problem I posted. I have values and everything working, I just wanted to explore some properties of my initial function

to make my logic make sense

So thanks everyone!

.close

[obtuse pebble]

Channel closed

Closed by @haughty lance

Use .reopen if this was a mistake.

[slim cove]

no problem!

[obtuse pebble]

Available help channel!

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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>.
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• 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.

[zealous stag]

for question 8

[zealous stag]

,rotate

[proven zephyr]

!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

[zealous stag]

are my equations correct?

[proven zephyr]

is this number 8?

[zealous stag]

yeah

[proven zephyr]

should be correct

[zealous stag]

aight thanksss

.close

[obtuse pebble]

Channel closed

Closed by @zealous stag

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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• 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.

[oblique nebula]

Hi. How to you solve this type of systems?

[oblique nebula]

I managed to get this:

[warm shale]

themathboi #2137

[oblique nebula]

How am I supposed to proceed?