help-26

[deep forum]

just simplified the problem

[plain abyss]

先吃晚餐先

[hoary burrow]

好

[plain abyss]

我等下try 看
如果不能还找你们

[hoary burrow]

Don't ask me cuh

Bagelguy started wilding

[sinful flint]

they both speak chinese, so they decided to talk in chinese

nothing crazy

[hoary burrow]

Nah I mean the other doggo picture

The meme one

I was the one talking in Chinese 😭

[sinful flint]

ah okay

i didnt pay much attention lol

[hoary burrow]

Bro pulled out the emojis😭

[merry hill]

is ur question solved @plain abyss

[hoary burrow]

He has more questions to try later

But he's eating dinner first

And he wants to try them himself first too

[merry hill]

then close channel and open a new one later when he has any doubt

.close

[topaz sinew]

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[hoary burrow]

Open another channel when you come back

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[static rune]

Can anyone tell me how to solve these I have 0 idea how

[sinful flint]

!status

[topaz sinew]

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[smoky sparrow]

2nd pic, use the angle bisector theorem
BC/BA = CD/DA

now note triangle DBA is isosceles
can you proceed from here?

(and triangle DCB is also isosceles)

there's a lot of similarity going on as well

for the 1st pic I would draw AE such that AEC = ACD = y and hence angle EAB = x

isosceles triangles again

[static rune]

Thanks

[smoky sparrow]

np!

[static rune]

I'll close it now

Thanks alot

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[plain abyss]

.reopen

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[neon iron]

yo

[neon iron]

4c

i dont get what to d

for part b

ur boundaries are 150,180

so

lower boundary for heavy

180

we find z score of 100%

and find boundary

which is 305

so heavy

is

180<W<305

we find z score of 200

200-165/35

then find probability

which is

0.8413

is probability weighs less than 200

and since it must be freater 180

0.8413-2/3

then u must find probaility of it being heavy

so then

(0.8413-2/3)/(1/3)

and done

ffs i answered the question

but what i dont get also

is D

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@neon iron Has your question been resolved?

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[obtuse robin]

Help please

[acoustic pecan]

what have you tried

[obtuse robin]

Idk where to start

Well

I know we have 12

[acoustic pecan]

perhaps try rationalising the denominator

[obtuse robin]

But how do I do square root 2 -1

[obtuse robin]

How do I do that-

[acoustic pecan]

multiply the fraction only by (sqrt(2)+1)/(sqrt(2)+1)

(ie, 1)

[obtuse robin]

Ohh

wait the whole fraction or?

[acoustic pecan]

just the 12/(sqrt(2) -1)

[obtuse robin]

oh ok

This?

[topaz sinew]

@obtuse robin Has your question been resolved?

[obtuse robin]

Eid I do it right

Did&

Did*

[acoustic pecan]

Yeah

[obtuse robin]

so what do I do now?

[acoustic pecan]

simplify it

[obtuse robin]

?

[acoustic pecan]

what does your denominator say?

[obtuse robin]

Square root 2-1 x Square root 2+1

[acoustic pecan]

ah

expand it

[obtuse robin]

ok

[obtuse robin]

This?

[acoustic pecan]

im unsure how you came to that

[obtuse robin]

Wait I made a mistake

Idk how to expand it

Maybe I'll watch a video on surds and come back here later-

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[barren arrow]

can someone help me with finding f'(x) because i genuinely don't know how to even approach this

[shut obsidian]

Do you know the product rule?

$\Big(f(x)g(x)\Big)' = f'(x)g(x) + f(x)g'(x)$

[thorny flame]

Kepe

[barren arrow]

what would i take as u and v?

[shut obsidian]

$e^{3 \sin(x)}$ and $\cos(x)$

[barren arrow]

so e^3sin(x) would be u and cos(x) would be v?

[shut obsidian]

Yes

[barren arrow]

okay wait i'll try it with this

one sec

[shut obsidian]

(Since multiplication is commutative in R you can also exchange u and v, it doesn't matter)

You should get $f'(x) = 2\Big(\big(e^{3\sin(x)} \cdot 3\cos(x)\big)\cos(x) + e^{3\sin(x)} \cdot (-\sin(x))\Big)$

[topaz sinew]

@barren arrow Has your question been resolved?

[barren arrow]

ah i forgot to get back on here

[barren arrow]

yeah i got the same thing thanks man

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[blazing lotus]

can someone give me an expression that i can use to solve it. I dont want to solve it by listing out all possible combinations

[pseudo bear]

OK, so there are two major ways of making a square.

You can pick two of two different colors.

You can pick two of one color and one each from two other colors.

Let's do the first first.

Let's say you have two reds and two blues.

How many different (according to the rules here) squares are there with two red sides and two blue sides?

[blazing lotus]

2 right?

[pseudo bear]

Right.

How many ways are there to choose two colors?

[blazing lotus]

so would you do that for
RB
RG
GB

so 2x3 is 6

[pseudo bear]

Right, so 6 'different' squares on the first part.

Let's do the second part.

Let's say we have two red sides, one green side, and one blue side.

How many 'different' squares are there?

[blazing lotus]

2?

is there one where reds are connecting

and one where reds arent

becasue if you reflect you can haveb the green ands blue swapped

[pseudo bear]

Right.

RGRB can flip to RBRG.

[blazing lotus]

so do you do that for 2 greens and 2 blues asw?

so its also 2x3

[pseudo bear]

Yes, you do the same thing.

And then there are 3 ways of choosing the color for the two sides.

So, you have 6.

[blazing lotus]

is it just 6+6 then?

[pseudo bear]

Right.

[blazing lotus]

theres no other combinations right

tysm

[pseudo bear]

You're welcome.

[blazing lotus]

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[pseudo bear]

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[opal vault]

class 1 and class 3 are the same according to the rules since you can turn over the square and rotate it

[opal vault]

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[pseudo bear]

Ahh, OK. Thanks.

[cedar wagon]

Need help rafilou?

[opal vault]

I am deciding whether to write "1+1" or not

[cedar wagon]

Thats 1

[sweet shard]

have you tried learning Z_2, + and its group structure first

[cedar wagon]

Just learn bool algebra

Its 1

[opal vault]

in (Boole, * , +) yes

we had the same thought

[pseudo bear]

Fixed my test program. Now it shows two classes: [["BGRR","BRRG","GBRR","GRRB","RBGR","RGBR","RRBG","RRGB"],["BRGR","GRBR","RBRG","RGRB"]].

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[vestal badger]

Hey everyone, I argued with my math teacher and need some help figuring out who’s correct here.

Here’s the situation: We were solving a problem where I had to multiply three ranges of numbers together. Two of them were already negative (e.g., something like -3 ≤ x ≤ -1), but the third was positive (e.g., 1 ≤ y ≤ 3 ). Instead of making everything positive before multiplying (as my teacher wanted), I decided to "give the minus" to the positive range, flipping it to be between negative numbers (-3≤ - y ≤ -1) . Then, I multiplied all the ranges and flipped the result at the end to return to the original signs.

My result was correct. But my teacher said I wasn’t allowed to do it this way. She insisted that I make all the ranges positive before multiplying them because that’s the rule she teaches us. She also said it wasn’t valid to just multiply negative ranges like that. We argued, and she couldn’t give me an actual counterexample to prove me wrong, so I’m stuck wondering: Am I right?

For context, I study in the Common Core in Morocco. I understand that the method I used isn’t the "standard" one, and I’m not saying it’s easier or better. But as far as I can tell, the math works out.

So is my method valid, or did I miss something? Let me know what you think, especially if you’ve dealt with similar arguments before!

Thanks in advance!

PS: She tried showing me a counterexample but couldn't. Also, I write in French, and so does my teacher. The text in red in the copy means you can't multiply a negative by a negative.

-'' on a '' means we got

• P(x) equals (x+2)(x-1)(2x-3)

[glacial sandal]

How do I solve for x with the exponent being a fraction?

[cedar wagon]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[vestal badger]

<@&286206848099549185>

Please <@&286206848099549185>

[light nebula]

Just do square

[rigid ivy]

I'll take a look in a bit. At a glance, what you did is fine, but let me double check

[vestal badger]

Alright

[rigid ivy]

@vestal badger tbh it's too hard to read

[cedar wagon]

Tu ne peux pas multiplier membre à membre sur des inegalites négatives, c'est une règle

[rigid ivy]

First, what is the problem you are trying to solve?

[clear lodge]

i'm not following the image either tbh

[rigid ivy]

[vestal badger]

I want to know where is located P(x)

Between what and what

[rigid ivy]

what is P(x)?

[vestal badger]

P(x) equals (x+2)(x-1)(2x-3)

[rigid ivy]

equals? Or inquality?

[vestal badger]

and we have -1<x<1

[clear lodge]

so you want to know the image of P(x) for x between -1 and 1?

[vestal badger]

Yesss

[clear lodge]

and you're asked to do this without differentiation?

[vestal badger]

P(x) = (x+2)(x-1)(2x-3) and we have to give the image of it for x between -1 and 1 , like knowing between what and what

[vestal badger]

i didn't understand what do you mean by differentiation

[clear lodge]

dont worry about it then

it's just that i find weird being asked what you are being asked, because with the tools that you have your bounds are gonna be significantly larger than the actual answer

okay, so from what i see, you're computing the range of each of the three terms

[rigid ivy]

Oh okay, so you want the range of $P(x)=(x+2)(x-1)(2x-3)$ restricted to $x\in[-1, 1]$?

[thorny flame]

SWR

[vestal badger]

Exaclty

[clear lodge]

so for -1<x<1 you have:
1<x+2<3
-2<x-1<0
-5<2x-3<-1

[vestal badger]

Correct

[cedar wagon]

-1*

[vestal badger]

and i gave a minus to x+2 to make it negative

[clear lodge]

why?

[vestal badger]

Why not

because i found it too long to give the minus to x-1 and also 2x-3

[clear lodge]

okay, so technically, you can say that
-3<-(x+2)<-1

[vestal badger]

Correct

[clear lodge]

however, your initial polynomial does not have that

so while your interval is correct, your usage of it is not

basically, if your polynomial is P=A*B*C, you're computing the bounds of P'=(-A)*B*C

[vestal badger]

Wait wait , when I gave it the minus i multiplied all of them together and at the end i multiplied by -1

[vestal badger]

Exactly and i multiplied everything by -1 to make it ABC

[clear lodge]

no, what you want is on the three exact factors that you have, the product that gives the maximum value, and the product that gives the minimum value

you have 8 possible products, since each factor can have two values

so in theory, you'd neet to compute the 8 products, and get the minimum and maximum values

this would give you bounds of -30<=P<=10

however, there has to be something wrong here, since P can have values over 10 in that range

(and is always positive)

[vestal badger]

why would it be always positive

[clear lodge]

because that's the polynomial, and for x between -1 and 1 is positive

[vestal badger]

damn bro this is too advanced for me

[clear lodge]

see, the thing is, usually with differentiation you'd compute where those two maximum and minimum points are, and you'd obtain the exact values for the range

[vestal badger]

Ohhh i see

but we didnt do that , we just did order rules

[clear lodge]

oh, wait, i see the issue

while as you can see on the image, the actual value is between 0 and (approx) 10.05

[vestal badger]

But the correct answer (according to the teachers correction ) is 0<P(x)<30

[vestal badger]

Oooh

[clear lodge]

two mistakes in the table actually

there you go your 0 to +30

im gonna delete the other 2

that's what happens when i try to math at bedtime

[vestal badger]

Yea it arrives to everyone dw

But it doesn’t answer my qst

Because I had the same result but the teacher didn’t count it because I can’t multiply when they are negative

[clear lodge]

well, you can, but you're just doing redundant unnecessary steps

it makes no sense bounding -P when you can bound P directly

[vestal badger]

Yea

But as you said it is still correct

And there is no logical or mathematical issue there

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[vestal badger]

Thank youuuu

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[upbeat minnow]

What am I doing wrong here?

[upbeat minnow]

I solved for a_2 by plugging a_1 in, then used the geometric sequence equation, plugged everything in and its wrong

this is the correct answer and im not sure why

[pseudo horizon]

it's the same

[upbeat minnow]

?

[pseudo horizon]

you need to use a dot rather than a decimal for multiplication

*

[upbeat minnow]

oh, i thought that was a decimal in the textbook?

certainly looked like one, thats weird

[pseudo horizon]

yeah that's strange...

[upbeat minnow]

ok well thank you then

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[jade glade]

If a quadratic function only has positive value for -1 < x < 3 and has a maximum value of 3, then the equation for that quadratic function is?

[pseudo bear]

!status

[topaz sinew]

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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

[ivory sorrel]

many such functions

[pseudo bear]

OK, but how far have you made it in solving it?

[golden wigeon]

This seems a little weird; assuming the function is continuous, arent there a lot more positive values than 2?

[pseudo bear]

They didn't say it was positive values for integers in -1 < x < 3.

[golden wigeon]

Thats exactly my point

Isnt it fully possible for f(x) = 1.99999999999

[pseudo bear]

Is that within -1 < x < 3?

[golden wigeon]

Oh wait i didnt read the question properly

[topaz sinew]

@jade glade Has your question been resolved?

[jade glade]

I can't solve it

Eh to be accurate, I don't know where to begin

[pseudo bear]

OK, so they're saying it's a quadratic.

Which means the graph of it will be a parabola.

Like:

,w plot x^2 - 5

@jade glade

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[pseudo bear]

Does that make sense so far?

[jade glade]

Yes

[pseudo bear]

OK, now notice where it crosses the x axis.

In between those two points, it's all negative.

And it has a minimum value that's halfway between where it crosses the x axis.

[jade glade]

The y value is all negative?

[pseudo bear]

In between the two places where it crosses the x axis, yes.

[jade glade]

Yes that is what I mean

[pseudo bear]

OK, a parabola can also open downwards.

,w plot -(x + 1)(x - 3)

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[pseudo bear]

Notice how that opens downwards.

You can also see where it crosses the x axis.

Between the two places it crosses the x axis, everything is positive.

That's kind of like in your problem.

Everything between x = -1 and x = 3 is positive.

Do you see that in the graph?

[jade glade]

Yes, it is because before x intercept all of the y value is positive

I cant.....

[pseudo bear]

You can't what?

[jade glade]

Connect the dot

Bruh

[pseudo bear]

Oh, right now, we're just noticing what's in the graph.

The parabola curve crosses the x axis at x = -1 and x = 3.

Between those two xs, y is always positive.

[jade glade]

Waitt? So -1+3/2 = 2 so vertex is (3,2)

[pseudo bear]

Nope.

The places where it crosses the x axis (x intercepts) are called the roots.

Roots are where the graph is at y = 0.

Where it crosses the x axis.

The x intercepts.

Now there's a special form for roots called the factored form.

[jade glade]

Yes

[pseudo bear]

y = a(x - (-1))(x - 3)

Notice the x intercepts are at -1 and 3.

And I wrote those subtracted from x.

y = a(x - (-1))(x - 3).

[jade glade]

Ahhhhh

When y = 0 there is our roots

[pseudo bear]

Right.

They're also called zeros because y is zero there.

[jade glade]

I seeee

Waittt

[pseudo bear]

Subtracting a negative number is the same as adding, so:
y = a(x + 1)(x - 3)

Now, you have a maximum of 3.

The maximum or minimum will appear right in the middle of the two roots.

So, like our roots are at x = -1 and x = 3. What's halfway between -1 and 3?

[jade glade]

-1+3/2 is 1?

[pseudo bear]

Right.

So, our maximum will be at x = 1, y = 3.

So, now we can figure out that a value.

y = a(x + 1)(x - 3)

[jade glade]

3 = a(1 + 1)(1 - 3) like this?

[pseudo bear]

Right.

[jade glade]

Okayyyy

I get the general idea

Thank you sir @pseudo bear

[pseudo bear]

3 = a(1 + 1)(1 - 3)
3 = a(2)(-2)
3 = -4a
a = -3/4

No problem.

[jade glade]

[pseudo bear]

Now we know the whole thing in factored form.

y = -3/4(x + 1)(x - 3)

Then, we multiply out everything on the right side.

And that'll give us one of the answers.

[jade glade]

I see

So that was very simple to begin with

Alright

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[solar ferry]

True or false

1. If the Taylor expansion of ( f(x, y) ) is of the form ( 2 + 3x + 4xy + o ), the gradient of the function is ( (3, 0) ).
2. When interpolating an odd function ( f(x) = -f(-x) ) with 9 points, we obtain a polynomial of degree 7.
3. The Newton method for finding the roots of an equation has quadratic convergence, which means that the number of correct digits doubles from iteration to iteration.
4. Numerical approximation of integrals is more precise when partitions of the integration interval are approximated.
5. An approximation of ( \int_0^1 x^2 dx ) using the trapezoidal rule is exact.
6. Given the function ( \ln x ), we define Newton iterations starting from ( x_0 = 3 ); in this case, the iteration ( x_1 = 3(1 + \ln 3) ).

[thorny flame]

jvrt.

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[serene acorn]

Hi everyone! This is more of a physics question, but I hope someone can still help me. I am trying to solve question 10.

It says:
On the diagram below, results for an experiment which attempts to calculate the heat capacity and heat of vaporization for water are shown. 490 grams of water was warmed until only 395 grams remained. What is the result of heat capacity and heat of vaporization of water?

I am currently trying to calculate what the heat capacity is, but im getting the wrong answer. The correct answer is c = 4.21

[serene acorn]

And here is my calculation:

[topaz sinew]

@serene acorn Has your question been resolved?

[smoky sparrow]

you haven't considered the heat of vaporisation of the water

2 250 000 J/kg

(I got this value online)

adding the two together you have $E = mc \Delta T + mL$

$c = \frac{E - mL}{m \Delta T}$

[thorny flame]

south

[smoky sparrow]

ah and the standard units are J and g (or kJ and kg)

so you should convert this to L = 2250 J / g

[topaz sinew]

@serene acorn Has your question been resolved?

[serene acorn]

I cant find any sources that show heat if vaporization being used in calculating heat capacity

[smoky sparrow]

yeah I think I made a mistake

[serene acorn]

Also not sure if I did it wrong, but if i try doing E - mL, i get:
160 - 2250 = -2090

[smoky sparrow]

,calc 160000/(75 * 490)

[thorny flame]

Result:

4.3537414965986

[serene acorn]

Btw i made a mistake in my notebook, it should be 0.49 kg and not 490 g

[smoky sparrow]

yeah the heat capacity is given by the slope

and then the enthalpy of vaporisation is given by the (change in energy) / (mass lost), Q = mL, so two separate calculations that's correct

[smoky sparrow]

honestly that just sounds like a textbook error then

[serene acorn]

Could be, but the real heat capacity for water is 4.2, which isnt exactly what the textbook says but its close enough, while mine isnt

Which is what worries me that I might have missed something

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[lament fractal]

the question as to find the components of vector q

[lament fractal]

is this correct

[dense lily]

You swapped the values of q1 and q2

[lament fractal]

what do y mean ?

[dense lily]

your value for q1 is actually supposed to be the value of q2

@lament fractal

It's in the 4th quadrant so the x component is positive and the y component is negative

And are you told that || q ||=1

[lament fractal]

no the magnitue of q is 1

the question says it is a unit vector

i confused

nvm

[dense lily]

Oh okay

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[cyan basalt]

How do I find the angle of BAC and the length of BC? Advance thank you :3(I'm bad at geometry)

[junior nimbus]

Can u find the angle CAK?

[cyan basalt]

Wait nvm I know how now ty anyways

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[limber summit]

hello, i have two linearly independent continuous functions u and v from [0,1] to R and i need to prove that for all (a,b) in R^2, there exists a continuous function f from [0,1] to R such that

[limber summit]

$\int_{0}^{1} u(x)f(x) dx = a$ and $\int_{0}^{1} v(x)f(x) dx =b$

[thorny flame]

macilak

[limber ridge]

Well if the functions u and v are linearly independent, think of a way to get two orthogonoal functions from u and v

[limber summit]

what's an orthogonal function ?

is it a function such that the integral of the product of the two is zero ?

[limber ridge]

exactly yes

[limber summit]

so i need to construct f ?

[limber ridge]

if you have a pair of orthogonal functions, f is very straightforward to construct

say you orthogonal functions are g and h

then $\int_0^1g(x)g(x) dx=k$ and $\int_0^1h(x)h(x) dx=l$

[thorny flame]

The د

[limber summit]

mmm why did you take the integral of the square of the functions ?

[limber ridge]

you can now replace the second $g(x)$ with $c_1\cdot g(x)$ so that the inner product is equal to $c_1k$

[thorny flame]

The د

[limber ridge]

and you can choose $c_1$ such that $c_1k = a$

[thorny flame]

The د

[limber ridge]

now you can also replace $c_1g(x)$ with $c_1g(x)+c_2h(x)$ without changing the value of the integral at all since $h$ is orthogonal to $g$

[thorny flame]

The د

[limber ridge]

also choose $c_2$ such that $\int_0^1 h(x) [c_1g(x)+c_2h(x)]dx=b$

[thorny flame]

The د

[limber summit]

are two linearly independent functions orthogonal ?

ah no

you said to get two orthogonal functions from u and v

[limber ridge]

yes exactly

do you know how to do that?

[limber summit]

mmm nop

[limber ridge]

I encourage you to look up the Gram-Schmidt orthogonalization process

its not so bad at all

[limber summit]

alright

and what do h and g look like ?

[limber ridge]

i wouldn't know much about how they look other than that $\int_0^1h(x)g(x)dx=0$

[thorny flame]

The د

[limber summit]

but aren't their expressions supposed to have v and u ?

[limber ridge]

yes they would depend on u and v for sure

they would be linear combinatoins of u and v

[limber summit]

is it possible to get an explicit expression of h and g ?

[limber ridge]

yes thats what the Gram-Schmidt orthogonalization process is all about

[limber summit]

okay when you said wouldn't know much about how they look i thought it wasn't possible

[limber ridge]

mb i misunderstood what you meant by how they looked

[limber summit]

so you do know their explicit expression right ?

[limber ridge]

yes

a short summary of how it works is as follows

let $g(x)=u(x)$

[thorny flame]

The د

[limber ridge]

$h(x)=v(x)-\int_0^1u(x)v(x)dx\cdot u(x)$

[thorny flame]

The د

[limber ridge]

this removes all the component that is parallel to u from v

enusruing that v is orthogonal to u

this video explains it well: https://youtu.be/rHonltF77zI

[limber summit]

you mean g orthogonal to h ?

[limber ridge]

its in terms of vectors but same same

[limber ridge]

sorry yes

g is orthogonal to u which is equal to h

[limber summit]

so what would be the expression of f ?

[limber summit]

this f

[limber ridge]

$f=c_1g+c_2h=c_1u+c_2(v-\int_0^1uv \cdot u)$

[thorny flame]

The د

[limber ridge]

where $c_1 ||u||^2=a$ and $c_2 ||v||^2=b$

[thorny flame]

The د

[limber summit]

alright i'll check this, thank you!

also if i have a question can i ping you ?

[limber ridge]

sure no worries

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[final thunder]

yo guys, i may have found a new formula for factorials, it works for every z in C except (1, -1/2 and every other negative integer). the last step to complete it would be to solve the integral but i couldnt do it, i would appriciate it if somebody here has time to waste trying to solve it. (also lemme know eventually if the formula is arleady known or not thanks)

[final thunder]

if you try putting z=1/2 it works so i assumed it works for every other number in the field of existence.

[gilded plume]

i solved it but it is too small to write in the margins of this discord message

[final thunder]

i wrote the formula wrong im sorry here the good one

[final thunder]

lmao

not the fermat ref haha

[gilded plume]

regardless please close this channel for actual problems

[broken niche]

???

[final thunder]

yes

[broken niche]

Ur using a factorial in your formula for a factorial

[final thunder]

yes why not lol

it was just to show off it useless ik but it was kinda fun proving it :)

[cursive thorn]

Disproven by desmos

[final thunder]

try with 1/2

[cursive thorn]

That does work, but that's just one number though

[final thunder]

i dont have clue lmao

guess ill scrap it

[cursive thorn]

Unfortunate

[lethal anchor]

how did you even get to this point

[final thunder]

do you want the short or the long explanation?

[lethal anchor]

there is a long answer?

[final thunder]

i mean the complete proof with long answer

i can screenshot it from notebook

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[wispy burrow]

What does the ± but with the minus on top mean?

[loud oasis]

it means that it takes the opposite sign of the ±

[wispy burrow]

Ah okay

That makes sense

Thank you

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[odd pagoda]

if you write it in a series then that means that the next term would have a minus

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[stable jacinth]

does anyone understand 1b

[golden blade]

trivial solution means L = {(0,0,0)}

[thorny flame]

anti-algebraist 𝔸dωn𝓲²s

[stable jacinth]

ah

and why do we ignore the second formula you just mentioned

[golden blade]

where did I say we ignore something?

[stable jacinth]

the answersheet didnt mention it was what i meant mb

so a isnt equal to minus 1

and how can we follow this up

a isnt equal to -1 and beta is a ?

[golden blade]

2(beta) - 2(alpha) = 0
alpha = beta
so whatever alpha is, is also beta

[stable jacinth]

alright

so we indeed have x3 = 0

but looking at the 2nd row

isnt x2 = 2 then?

[golden blade]

hmm I just noticed too

[cedar wagon]

but its lookin for Ax = 0

not Ax = b

[golden blade]

oh right

that solves it

[cedar wagon]

take the last row reduced matrix

4th column should be fulled of 0

and then yeah

1+a^3 != 0

else the only trivial sol wouldn't be 000

for (c) its quite the same mind but little bit different

same for (d)

[stable jacinth]

omg

thanks

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[upbeat minnow]

Determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.
$a_n=\frac{(n!)^2}{(2n)!}$

[thorny flame]

dingypine

[upbeat minnow]

Ok so I don't really know where to start with this problem. The thing that's really throwing me off is the factorial.

I know to determine whether the sequence converges or diverges I should take the limit of a_n as n -> heads to infinity

[crisp raptor]

first thing that comes to mind is the Stirling formula, are you allowed to use that?

[upbeat minnow]

I've never heard of that, and I don't believe we went over that in class ever

[crisp raptor]

its a approximation for the factorial function, that is actually equal in the limit

[upbeat minnow]

I guess are there any other methods to go about this besides Stirling's formula?

[crisp raptor]

[
\boxed{
\lim_{n \to \infty} \frac{\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n}{n!} = 1
}
]

[thorny flame]

tobi

[crisp raptor]

yes there are other methods, like I said that was just the first thing that came to my mind

[upbeat minnow]

aha okay, what else could I do then?

[agile harness]

do you know how the factorial is defined

[upbeat minnow]

I'm gonna assume no

unless you mean like how it works, 5! = 1 * 2 * 3 * 4 * 5

[crisp raptor]

well thats how its defined

[agile harness]

yes that’s fine now just do that for each and expand

[crisp raptor]

maybe try writing it for a general case n and using the product notation

[upbeat minnow]

Do you guys mean like this?

[crisp raptor]

so far so good

you can do the same for (2n)!

[upbeat minnow]

Ok I'll do that rn

i think this works, i had to change it because it was being multiplied somehow since its being multiplied by 2

assuming that works, what can I do next here then? If I try to take the limit it will just be inf/inf I think

[crisp raptor]

for that matter

[upbeat minnow]

whats wrong with it?

[crisp raptor]

nah sorry my bad

its correct

[upbeat minnow]

oh cool

[crisp raptor]

first of all you can cancel some things

multiplying by the number 1 in the way you are doing is not necessary :D

[upbeat minnow]

aha ok good to know

ok so i think ive cancelled what I could, what can I do from here tho?

i think itll still be inf/inf if i try to take the limit as n -> inf

[crisp raptor]

maybe try writing this in product notation now

do you know how?

[upbeat minnow]

I've honestly never worked with product notation

so idk how

[crisp raptor]

do you know how to write n! in product notation?

[upbeat minnow]

honestly no

[crisp raptor]

have you worked with summation notation before?

[upbeat minnow]

yes ive worked with summation notation before

[crisp raptor]

okay then the product notation wont be that different

its just that instead of adding the terms we multiply that

and write that with a capital pie

[upbeat minnow]

Ok let me try then

[crisp raptor]

try with writing n! in product notation

[upbeat minnow]

alright

[crisp raptor]

what would that be equal to?

[upbeat minnow]

would equal to n!?

[crisp raptor]

that would actually be equal to i!

you want the indices to be the other way around

[upbeat minnow]

Ok that makes sense

oh woops

[crisp raptor]

yeah nice

[upbeat minnow]

Ok cool

[upbeat minnow]

So you want this, to be in that form?

[crisp raptor]

yeah

maybe factor it like this

$\frac{1}{n+1} \cdot \frac{2}{n+2} \cdot \frac{3}{n+3} \cdot \dots \cdot \frac{n}{2n}$

[thorny flame]

tobi

[upbeat minnow]

Alright let me try that

so I'm not 100% sure, but this is what i got

[sinful flint]

so close

[crisp raptor]

👏 👏

[upbeat minnow]

ok awesomee

whats next then?

[crisp raptor]

now products are kinda annoying right?

[upbeat minnow]

i guess so

[crisp raptor]

if there would be any way to turn products into sums hm

thats usally what I try when solving these problems

[upbeat minnow]

how would you do that tho?

[crisp raptor]

however im still not sure how to solve the problem from there but we will figure it out

[crisp raptor]

$\ln(a \cdot b) = \ln(a) + \ln(b)$

[thorny flame]

tobi

[upbeat minnow]

Do you want me to put everything around an ln and work on seperating it from there?

[crisp raptor]

I would try that, that's usually a good idea, when working with a limit of a product

[upbeat minnow]

like that?

[crisp raptor]

yeah

[upbeat minnow]

Ok cool

What can be done from here tho?

[crisp raptor]

you can add 0 in the denominator adding + n - n

[upbeat minnow]

like this?

[crisp raptor]

sry at the top I meant

im not quite sure if this idea will work, but it might work

[upbeat minnow]

Alright

[crisp raptor]

well you can do further algebraic manipulation

[upbeat minnow]

oh lol

ok one sec

[crisp raptor]

that's not how you add or cancel fractions

[upbeat minnow]

ok yeah im a dumbass let me try again

[crisp raptor]

take your time, we are all here to learn

[upbeat minnow]

is this what youre looking for?

I'm honestly not sure what else i could do

wait what

it didn’t paste the right image

hold on

ok that’s the one i was looking to paste

[crisp raptor]

$\frac{i+n-n}{i+n} = \frac{i+n}{i+n} - \frac{n}{i+n}$

[upbeat minnow]

but yeah idk what else i could manipulate algebraically

[thorny flame]

tobi

[crisp raptor]

yeah or you can write it in one ln

[upbeat minnow]

Alright

Ok so one fraction cancels out, and then youre just left with the one all the way on the right

[crisp raptor]

have you had taylor series before?

[upbeat minnow]

unfortunately not

i think that’s gonna come after sequences in my class

[crisp raptor]

first of all do you have an intuition for what the solution should be?

its easier to show something, when you have a goal in mind

[upbeat minnow]

Well I'm honestly not sure, I assume it will be divergent just by looking at some graphs but idk

[crisp raptor]

the number in the numerator, will always be smaller then the one in the denominator

[upbeat minnow]

Does that mean it’ll head to 0?

[crisp raptor]

yeah, but now we have to proof that

[upbeat minnow]

How do you go about doing that? The way I usually go about doing these problems is setting up a limit but that hasn’t worked so far here

[crisp raptor]

well with taylor series you would be done as well

what tools can u use?

[upbeat minnow]

In this case? I'm honestly not sure.

[crisp raptor]

okay what we can also do is we can show that

1/a_n tends to + inf as n -> inf

this implies a_n goes to 0 as n goes to inf

[upbeat minnow]

Ok

[crisp raptor]

yeah thats probably easier to show

what would the product be in that case?

[upbeat minnow]

I honestly have no idea

[crisp raptor]

If you look at the case

$\frac{(2n)!}{(n!)^2}$

[thorny flame]

tobi

[crisp raptor]

if this is limit is equal to + (or - inf) then your original limit is equal to 0

[crisp raptor]

just like you wrote down this

[upbeat minnow]

Okay I see

Alright man thank you for all the help and the support . I'll try to finish this on my own since I don't wanna take up much more of your time since we've been working on this for more than an hour, but I appreciate it. Thank you ‼️

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[neon iron]

Alright i just wanna make sure i got a right answer will show my work in a second

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[unkempt bane]

question 3 jai pas compris comment la travaille

[topaz sinew]

Please don't occupy multiple help channels.

[rigid ivy]

stay in one channel please
#help-22

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[pliant cypress]

could I just be walked through why it's done this way? I'm having trouble visualizing the question

[pseudo horizon]

can you visualize the disks?

[pliant cypress]

not in three dimensions, the delta y is confusing me

i keep visualizing it with circles instead

[pseudo horizon]

https://www.desmos.com/3d/2dcb3flvoz

oh

i was thinking about it in circles too

https://www.desmos.com/3d/zazykzaon5

https://www.desmos.com/3d/nvcaqutxx5

[topaz sinew]

@pliant cypress Has your question been resolved?

[pliant cypress]

but where do they get figure 8.4 from 😭

[pallid zenith]

like, where is it coming from, or what is it itself?

[pliant cypress]

where is it coming from?

[pallid zenith]

the "slices" here encompass dy and the inside of the hemisphere; in other words, its representing the infinitesimally small slices (which here are circles, if you take cross-sections parallel to the base) (also use your logic of hemi-"sphere") so i suppose its true derivation would be from a question, but that is all the information i can gather from the picture

[pseudo horizon]

figure 8.4 is a side view

the disk is just a horizontal line here. they didn't give it any thickness

[pliant cypress]

oh wait

and then the radius is the only thing changing

which like

okay i see

thank you

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[pearl brook]

,rotatte

,rotate

[pearl brook]

Hi, i need help with 6

[sinful flint]

!status

[topaz sinew]

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

[pearl brook]

2

[sinful flint]

!show

[topaz sinew]

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

[pearl brook]

ok

[agile harness]

your work is hard to read

[pearl brook]

um

what i did

was

[agile harness]

14 + 4/11?

[pearl brook]

for what

[agile harness]

i was just guessing

[pearl brook]

ok

[agile harness]

did you do that in your work

[pearl brook]

no

[agile harness]

do that

[pearl brook]

where and why

yeah

i know

that

[agile harness]

then you can start building a continued fraction

[pearl brook]

158/11

is 14 at 4/11

[pearl brook]

what is thi

[agile harness]

ouu you don’t know

hold on

i’ll send you a video

[pearl brook]

ok

[agile harness]

https://youtu.be/vw4eYXMWYEw?si=UWOhhYAT5vQXdAYe

start with $\frac{158}{11} = 14 + \frac{4}{11}$

[thorny flame]

knief

[agile harness]

then from there

$14 + \frac{1}{\frac{11}{4}}$

[thorny flame]

knief

[pearl brook]

ok

what

i just asked someone

and they said

1/11/4

no

wait

ok

they a

14

[agile harness]

well it was 2a right

[pearl brook]

ye

ss

[agile harness]

2a = 14

[pearl brook]

yes

then

they did 1 over 4/11 is equal

to b-2 over c

and they b as 3

and c as 8

and they did trail and error

[agile harness]

yea well this way doesn’t require trial and error

[pearl brook]

how

[agile harness]

$\frac{11}{4} = 2 + \frac{3}{4}$

[thorny flame]

knief

[agile harness]

which means we have

[pearl brook]

yes

argee

[agile harness]

$14 + \frac{1}{2 + \frac{3}{4}}$

[thorny flame]

knief

[pearl brook]

ok

and

they is the smae format

same

[agile harness]

yea i guess

[pearl brook]

what were you gonna say

thou

[agile harness]

well i was just going to match terms

[pearl brook]

oh ok

thanks

um

wait

one seced

is is 3/ 4 and in the promnom it is 2

2/c

[agile harness]

yea i was going to make $\frac{3}{4}\to \frac{2}{\frac{8}{3}}$

[thorny flame]

knief

[pearl brook]

then what

wait

how

[agile harness]

oh c needs to be an integer yea

[agile harness]

i just multiplied top and bottom by 2/3

[pearl brook]

ok'

[agile harness]

$14 + \frac{1}{3 - \frac{2}{8}}$

[pearl brook]

wait then you

what

i have to go eat

can u just right down how you got the soultion

sorry i have to go

[agile harness]

you can just read the chats back

oh it says -2/c

ok so 2 + 3/4 = 3 - 1/4

then just write 1/4 as 2/8

done

[thorny flame]

knief

[agile harness]

no need for trial and error

@pearl brook

make sense?

a = 7, b = 3, c = 8

[topaz sinew]

@pearl brook Has your question been resolved?

[pearl brook]

wait

ok

thanks

.close

[topaz sinew]

Channel closed

Closed by @pearl brook

Use .reopen if this was a mistake.

[topaz sinew]

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.

[manic yew]

$x^2 +6x + 9 = 1$

[thorny flame]

Simon James B

[manic yew]

can someone explain how to solve this using delta? i am stuck because we have = 1 and always i had = 0 idk what to do with this 1 to we take it to the left hand side and then say = 0 ? or??

[cedar wagon]

don't use delta

[manic yew]

?

[cedar wagon]

whats the lhs

[manic yew]

x^2 +6x + 9

[cedar wagon]

don't you see smth about it ?

[manic yew]

(x+3)^2

[cedar wagon]

so

(x+3)² = 1

proceed

[manic yew]

OH

[cedar wagon]

no need goofy delta

[manic yew]

x + 3 = 1
x = 1 - 3
x = -2

x+3 = -1
x = - 4

[manic yew]

i am on the lesson with delta so i think it's point was to use it

[cedar wagon]

then -1 both sides

so that you have quad = 0

and use delta

[manic yew]

$x^2 + 6x + 9 -1 = 0$

[thorny flame]

Simon James B

[manic yew]

oh and we would also do 9 -1 now

i gett it

$x^2 + 6x + 8 = 0$ right?

[thorny flame]

Simon James B

[manic yew]

so delta here is = 36 - 32 = 4

[cedar wagon]

y

[manic yew]

x1 = (-6 + 2)/2 = -2

x2 = (-6-2)/2 = -4

[cedar wagon]

perfect

all good ?

[manic yew]

easy.

yes all good

thanks

[cedar wagon]

wunderbar

[manic yew]

.close

[topaz sinew]

Channel closed

Closed by @manic yew

Use .reopen if this was a mistake.

[topaz sinew]

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.

@quasi raft Has your question been resolved?

[quasi raft]

.close

[topaz sinew]

Channel closed

Closed by @quasi raft

Use .reopen if this was a mistake.

[topaz sinew]

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.

[forest ibex]

Could someone help me here pls

[agile harness]

distribute

[forest ibex]

Will the second one just be

4x Sec^2 x

[agile harness]

🤔

[coarse herald]

wont the 'x' cancel out?

[agile harness]

what happened to the /x

[forest ibex]

OH

I see wait

This is the final integral thing?

[coarse herald]

yes

[forest ibex]

so 8x + 4tanx + C is the final answer right…

thank you guys

.close

[topaz sinew]

Channel closed

Closed by @forest ibex

Use .reopen if this was a mistake.

[topaz sinew]

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.

[languid tapir]

limx→0 sin3x /x

[languid tapir]

I know that limx→0 sinx /x = 1

is it the same ?

[craggy haven]

no, it's not the same

try to get it in the form of $\lim_{x\to0} \frac{\sin\bsq}{\bsq}$ where $\bsq$ is anything

[thorny flame]

hayley (measurable)

[craggy haven]

like you have 3x on the top, can you multiply or divide by something to get 3x on the bottom as well?

[languid tapir]

3/3

limx→0 3 * sin3x/3x

yeah I get it

thx

[topaz sinew]

@languid tapir Has your question been resolved?

[topaz sinew]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[hybrid pasture]

I am struggling to understand exactly why their solution works, and I don't know how to prove that mine works. So basically I would like to know how I could construct a truth table for my solution to see if it matches the textbooks so I can test my logic

[prisma mesa]

yours works

their says
"If (you are under 4 feet tall and you're not older than 16 years old), then you can't ride a roller coaster"

your says
"Either (if you're under 4 feet tall, you can't ride a roller coaster) or you're older than 16"

[hybrid pasture]

I see

Mine is more convoluted

[prisma mesa]

Yeah, it actually took me a bit longer to understand why yours works

but it's fine

[hybrid pasture]

ok I have one more example to ask

if I get this one

I think I got it

for my solution (give me 1 sec to draw in paint)

[prisma mesa]

okay, now this is not the same

[hybrid pasture]

how come?

[prisma mesa]

yours says that the automated reply cannot be sent exactly when the file system is full

[hybrid pasture]

doesn't that symbol mean iff?

[prisma mesa]

but there could also be kind of malfunction in the system (although it isnt necessarily full) and the reply wouldnt be said either

[prisma mesa]

yeah, it does

[hybrid pasture]

oh I see what you mean

my statement makes it seem like file system = full is the only way a reply cannot be sent

instead of it being just 1 of the cases

[prisma mesa]

"The automated reply cannot be sent when the file system is full" only tells you that if the system is full, then it cant send the reply

[prisma mesa]

Yep

[prisma mesa]

yeah

[hybrid pasture]

so then I would say if the file system is full, then* the* automated reply cannot be sent

[prisma mesa]

yeah, this is fine

[prisma mesa]

and its the same thing as their solution, except they named the constants differently

[hybrid pasture]

so c -> a

[prisma mesa]

their -p is same as your a

[hybrid pasture]

gotcha

yeah I try to solve these without looking at solutions first so I usually come to something different

I think I get it a bit better now thanks

kind of hard to go from language -> logic

and vice versa

[prisma mesa]

Yeah, the point of that is just to make your understand logic on intuitive level

once you get some understanding, no more translations like these will be needed

you'll be able to do coherent reasoning in natural language

[hybrid pasture]

is this how ai works?