help-4

[vale dock]

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[north scarab]

Sorry

[north scarab]

I was doing something

[marble arch]

no problem

[north scarab]

oh yes

I forgot the x^2

[valid marsh]

I know it's a stupid question but can someone tell me how to study the slope From the beginning because I am lost

[north scarab]

!occupied

[vale dock]

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).

[marble arch]

start a different channel please

[north scarab]

but please be more specific in what you mean also

[marble arch]

Exactly, add that and you should get the answer

[stark wedge]

what's the original question

[marble arch]

.

[valid marsh]

Oh sorry

[north scarab]

Here

[marble arch]

And now double differentiate it, if it's negative, then the value was maximum, and vice versa

[valid marsh]

Can you tell me where I can ask for slope 😅

[marble arch]

#❓how-to-get-help

[stark wedge]

#study-discussion tbh since you are asking about an entire topic

[valid marsh]

Thank you

[marble arch]

You got it @north scarab ?

[north scarab]

x=5sqrt(2)

[marble arch]

Yup

and for c

?

[vale dock]

@north scarab Has your question been resolved?

[north scarab]

hmm

for b I think I still got it wrong

the final volume

[north scarab]

Anyway, this is my justification for C

Correct?

[vale dock]

@north scarab Has your question been resolved?

[marble arch]

Yeah that's correct, as long as it's negative, the value was maximum

[marble arch]

This one however is wrong

sqrt(2) cubed gives 2*sqrt(2), not 4

Just correct that mistake and the remaining answer is correct

[vale dock]

@north scarab Has your question been resolved?

[north scarab]

oh ok thank you

.close

[vale dock]

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[wooden jungle]

how does the help thing work

[tawny moon]

.close (OP used another channel)

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[obtuse osprey]

is 10N the resultant force from force F and 12N?

[tawny moon]

not according to the question

10N is just another force

[lyric sundial]

No, if the ball Is in equilibrium the resultant force must be ...

[obtuse osprey]

Wait if the ball is stationery does it mean its equilbrium?

[tawny moon]

yes

otherwise it would move refer to the explanation below

[short solar]

a state in which all external forces acting on a body are balanced , then we can say its in equilibrium ,
here yes stationary will simply mean that , but it can also be asked in a case where an object is moving at a constant velocity ( net force=0 , a=0)

[obtuse osprey]

actually what does it mean when forces are equilbrium?

[short solar]

its asking for F , so find F such that the net force acting on the block in any direction is zero

[tawny moon]

when two or more forces are in equilibrium, the resultant force is 0

if it's moving, it will stay moving at that same velocity

if it's stationary, it will stay stationary

[obtuse osprey]

to find F I assume we need to x and y components of the 12N force right?

[tawny moon]

that would make it easier, yes

[short solar]

you can do that , but as i said any direction , so you can pic any direction , i.e. along F and perpendicular to F ( would be an easy pick)

your method is a go to too , try with both so you see whats the difference

[obtuse osprey]

I honestly dont get what your method means😅

[short solar]

try with what you thought first , lets get the answer , then we can discuss alternatives

also is the question bychance asked in context of vectors , like is this in a math book or a physics book , maybe the question wants you to apply the vector method ( lami or triangle law of addition)

[obtuse osprey]

its just a practice paper

[tawny moon]

in what subject

[obtuse osprey]

by the way is the x component of force F = x component of 12N?

[obtuse osprey]

Physics

[tawny moon]

their magnitudes are equal, but the sign should be opposite

[short solar]

can you figure out why you did that?

[obtuse osprey]

cus its equilbrium?

[short solar]

yep did you get the asnwer

[obtuse osprey]

idk why but for some reason I got 30 degrees as the angle

[vale dock]

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

[vale dock]

@obtuse osprey Has your question been resolved?

[obtuse osprey]

<@&286206848099549185>

[lyric harness]

yeas

[obtuse osprey]

is my working so far correct or?

[lyric harness]

ah wait

there is also the 10N force

to consider in the uy component

we know F+[12Ncos(30), 12Nsin(30)]+[0, -10N] =[0, 0]

try solving THAT

@obtuse osprey ?

Ok I fixed it now

[obtuse osprey]

so for the y resultant force, I take 6-10?

[lyric harness]

yeah

and negative of that

because we want Fy + 6 -10 = 0

so Fy = 10-6

Fy = 4

k?

[obtuse osprey]

alright I got 21.1 degrees?

[lyric harness]

umm

lemme chack

Fx + 12Ncos30 = 0

Fx = -12N*0.866

Fx = 10.582N

ok no ur right

theta is about 21degrees and magnitude is about 11.1N

[obtuse osprey]

thanks man

how we do this?

[civic umbra]

calculate moments 👍

anticlockwise moments = clockwise moments

use the tension as ur pivot

[obtuse osprey]

how would I know the distance between mass M from the pivot?

[civic umbra]

coz it's a metre rule

lol

[obtuse osprey]

the distance must be in metres right?

[civic umbra]

not necessarily

just make sure it balances

u can have ur moment in units of Newton cm

[obtuse osprey]

is it 20 x 40?

[civic umbra]

nah coz

ok calculate the anticlockwise moments first

[obtuse osprey]

60 x 20?

[civic umbra]

nah coz 60 is the distance from the very end

use the line of tension as ur pivot

the anticlockwise moment is 20N x 10cm = 200Ncm

[obtuse osprey]

how you get 10?

[civic umbra]

look at the 20N

it's a metre rule right

i'm assuming it's uniform

so the 20N will be at the point 50cm

so it's 10cm away from the line of tension (at 60cm)

[obtuse osprey]

so we dont take the whole length?

[civic umbra]

nah coz centre of mass is at 50cm

remember ur formula for moments

[obtuse osprey]

oh so it like this 20x10 = 40 x 20 x M?

[civic umbra]

wot's the 20 on the rhs from

so the anticlockwise moments are 20 x 10, clockwise moment should be 40 x g x m

coz remember u gotta convert ur mass to weight

[obtuse osprey]

arent they asking for mass?

[civic umbra]

yuh

but when u calculate ur moment

it's force x distance innit

so the clockwise moment is mg x 40

that's why you got
20 x 10 = 40 x mg

[obtuse osprey]

I've gotten 0.5kg

[civic umbra]

nice

did you use g=10N/kg

[obtuse osprey]

yes

[civic umbra]

nice

gj

[obtuse osprey]

does this method also apply to the tension

[civic umbra]

and for part ii you need help asw?

[obtuse osprey]

thanks man

[civic umbra]

part 2 you can do in two ways

coz you can do balance moments

or you could balance vertical forces

wait mb

just balance vertical forces

lol

[obtuse osprey]

vertical forces is upward force = downward force?

[civic umbra]

yuh

[obtuse osprey]

so tension force = 20N + the mass x 10?

[civic umbra]

yeh

[craggy rock]

torque

there's a resnick and haliday Question just like this, but instead of weight professor walker replaced them with electrodynamic forces. but the fundamentals were the same. you had to calculate torque using T=r cross F and then simply get it

[craggy rock]

pretty much. No need to calculate torque for that

[craggy rock]

Lami's theorem or sine rule. whichever works

both would because u can interprove sine role from lami's theorem and vice versa

[vale dock]

@obtuse osprey Has your question been resolved?

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[rancid dagger]

can someone help me solve this

[crimson yoke]

!status

[vale dock]

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

[rancid dagger]

1

[crimson yoke]

me too lol, let's try together

[rancid dagger]

lol oke

i think we gotta use summation of definite intergrals?

[crimson yoke]

i was thinking of writing the denominator as $n!(x^2+n^2)\frac{(2x)^2+n^2}{4}\frac{(3x)^2+n^2}{9}...\frac{(nx)^2+n^2}{n^2}$

[rocky lotus]

Suika

[crimson yoke]

and a similar thing for the numerator

and all the n! will cancel out

[rancid dagger]

yeah and we end up with n/n summation graph?

[crimson yoke]

what is that?

[rancid dagger]

we need convert the limit to intergral ig

if we take k/n as x

i give up

ts js too hard

[distant pulsar]

u’d want to consider log f

[rancid dagger]

idk how to that

[vale dock]

@rancid dagger Has your question been resolved?

[vale dock]

@rancid dagger Has your question been resolved?

[fallen mesa]

How were you taught at school to solve this?

[vale dock]

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[spark burrow]

\hr \{\bf Problem.} Let $f(x) = e^x$ for $x \in [0, \pi]$.

\begin{enumerate}[\bf a)]
\item Calculate the Fourier series of the even $2\pi$-periodic extension of $f$.
\item Does this Fourier series converge in the quadratic mean?
\item Does this Fourier series converge uniformly?
\item Where does this Fourier series converge pointwise?
\end{enumerate}

\textit{Hint (a):} The even $2\pi$-periodic extension of $f$ is the $2\pi$-periodic extension of the function $x \mapsto e^{|x|}$, $x \in [-\pi, \pi]$. Calculate the real Fourier coefficients.
\ \hr \ Regarding {\bf c)}; I got that the Fourier series is [\frac{e^\pi - 1}{\pi} + 2 \sum_{k = 1}^\infty \frac{(-1)^k e^{\pi} - 1}{\pi(k^2 + 1)} \cos(kx),] because of $C^1$ and that we can continuously continue the derivative to $-\pi$ and $\pi$, it should be uniformly convergent on $[-\pi, 0)$ and $(0, \pi]$. $x = 0$ ruins this, right? How would we prove this though?

[vale dock]

@spark burrow Has your question been resolved?

[vale dock]

@spark burrow Has your question been resolved?

[vale dock]

@spark burrow Has your question been resolved?

[spark burrow]

.close

[vale dock]

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[vale dock]

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[acoustic laurel]

what is everything i need to know with logarithms ln and e in calc

[thin valley]

😳

[acoustic laurel]

ive recently passed calc 1 but just started calc 2 and on the frst day we were reviewing and there were ln stuff and in my calc 1 we only mentioned it once and did nothing with it

[thin valley]

Your question is too general + the more you know the better

[acoustic laurel]

only thing i know now is intx^-1 = ln abs(x)

[thin valley]

But definition of logarithm, arithmetic properties, logb(x) as a function and its graph, assymptotes, base change, etc.

[thin valley]

That is a good start bc that is the definition of natural logarithm

Now I'd learn the characterization as the inverse of the exponential e^x

[acoustic laurel]

any way i could learn this in an hour?

[thin valley]

e^(ln(x))=x
ln(e^x))=x

[thin valley]

Yes, that is what "int" means.

[acoustic laurel]

like most of log, ln stuff

[thin valley]

It depends how much of a genius you are

[lyric sundial]

Probably, but probably not as well 😅
Maybe just the formulas and rules, but you would need to practice

[acoustic laurel]

ik

[lyric sundial]

And how well you master your algebra

[acoustic laurel]

i need to know most if not all in a few weeks

[acoustic laurel]

not sure

like what

algebra is pretty good

[lyric sundial]

Computations with fractions, powers, and so on and so forth

[acoustic laurel]

its not mastered

but really good

i think

[lyric sundial]

And also solving equations and inequalities

[thin valley]

y=e^x ←→ x = ln(y)
ln(xy) = ln(x)+ln(y)
ln(xⁿ) = n·ln(x)
ln(1)=0
Dom(ln(x))=(0,+∞)
lim x→0+ ln(x)=-∞

[acoustic laurel]

this everything?

[thin valley]

ln(x/y) = ln(x)-ln(y)

ln(x1·...·xn) = Σ ln(xi)

[thin valley]

The basics you find in the first page on logarithms in any book

[acoustic laurel]

how do i do calc with this

like diritive and anti dirivative

[lyric sundial]

You only need to know that the derivative of lnx is 1/x

[lyric sundial]

And this for the integration

(remember the + C)

[acoustic laurel]

what if its the int of e^x or something

[thin valley]

$$\int e^x dx = e^x+C$$

[rocky lotus]

Categorist

[thin valley]

You ar every lost if you don't know this ↑

[acoustic laurel]

how does that work

anti dirivative of e^x is e^x

+c

[lyric sundial]

Also change of base formula:
$$\log_a b = \frac{\log_c b}{\log_c a}$$

[rocky lotus]

Alberto Z.

[lyric sundial]

??

[acoustic laurel]

i remember the change base from pre calc i think

[lyric sundial]

Good

[acoustic laurel]

i forgot what the c would be

is it irrelivent

[thin valley]

$$\lim_{h\longrightarrow 0} \frac{e^{x+h}-e^x}{h}=\lim_{h\longrightarrow 0}\frac{e^x(e^h-1)}{e^h}=\lim_{h\longrightarrow 0}e^x\cdot 1 = e^x$$

[rocky lotus]

Categorist

[acoustic laurel]

what is that

[lyric sundial]

Whatever number (positive and not equal to 1)

[lyric sundial]

The limit definition of the derivative of e^x

[acoustic laurel]

so .6 would be allowed

[lyric sundial]

Sure

[acoustic laurel]

how does log get to ln

need a refresher

[lyric sundial]

$$\ln x = \log_e x$$

[rocky lotus]

Alberto Z.

[acoustic laurel]

what does e mean

[lyric sundial]

...

[dire cloud]

its a irrational constant

[acoustic laurel]

what constant

[dire cloud]

2.71828

[lyric sundial]

THAT constant

[acoustic laurel]

so not relevnt

[lyric sundial]

Lol

[dire cloud]

no extremely relevant

[acoustic laurel]

Eulers number

i google

[thin valley]

Knowing its decimal value by heart is not that relevant but as a concept it is bc it is very common

[lyric sundial]

Yes, that's it

Also called Neper's number

[dire cloud]

ive seen it in binomial expansions, complex numbers, limits, logarithm, calculus and there are probably way more

like WAY more

[red tulip]

so youve already done some calculus right? so youve done derivatives and stuff

[acoustic laurel]

Calc 1 yes

Just started 2

[lyric sundial]

And you have never encountered e^x??

[red tulip]

the special thing about e^x is that the derivative is itself

[acoustic laurel]

It was mentioned once and we never did anything with it

[lyric sundial]

Mmh weird

[acoustic laurel]

We had to cram everything in 6 weeks

2 classes per week

[thin valley]

Any exponential can be based-changed to e^x so it appears anytime you multiply repeatedly so it is very common it is like 0 or 1. I'd say it's importanter than π which is a shit number

[red tulip]

e always appears in functions formed from differential equations because with rates of 1/x it directly ties to ln

and other stuff

[acoustic laurel]

Could I get easy examples in calc that would show me the rules

[thin valley]

Ask chatgpt

[red tulip]

$\frac{d}{dx}e^x=e^x$

[rocky lotus]

ImOakley

[thin valley]

Already done

[acoustic laurel]

That's it?

d/dx ln=?

[red tulip]

the derivative of ln x is 1/x

[acoustic laurel]

Oh

[lyric sundial]

Calculate the derivative of ln(3x) in two ways

[acoustic laurel]

u sub

[lyric sundial]

This could be an application of both calculus and log rules

[lyric sundial]

That's for integration...

[acoustic laurel]

oh

u right

been awhile

1/3x?

maybe

[lyric sundial]

Nope

[acoustic laurel]

Just 1/x?

[lyric sundial]

Yes

[acoustic laurel]

3 dont matter?

[lyric sundial]

How did you reach this (correct) answer?

[lyric sundial]

Indeed, but I recommend you do not learn it as a rule

[acoustic laurel]

Oh

d/dx ln 3x = 1/3x ×3?

Chain rule

[lyric sundial]

Awesome 👍

Another way would be using the log properties

I.e. first write ln(3x) as ln3 + lnx

And then differentiate

[acoustic laurel]

Ok ty I do have tj go to class now

Appreciate the help

.close

[vale dock]

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• Show your work, and if possible, explain where you are stuck.
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[twin geyser]

Can someone identify my mistake please

[twin geyser]

I'm getting 2sqrroot5 right

But not the 53/7

[quasi valve]

what happened to the sqrt(3) at the end of this line

[twin geyser]

Oh damn it

Thank you

That's the issue

[quasi valve]

and how did 16sqrt(3) become 16sqrt(9)

(the first term)

[twin geyser]

Wait

[twin geyser]

This is the issue

[twin geyser]

Wait I think I multiplied that square root 3 with 16 square root 3

Getting 16sqrt(9)

[quasi valve]

why would you be able to do that?

[twin geyser]

Cus same square roots can be multiplied

Right

[quasi valve]

only if they're in the same term

the sqrt(3) in the last term can't just be moved to the first term

[twin geyser]

Wait why, they're on the same line though

They're in the numerator

[quasi valve]

the sqrt(3) at the end of the first line is only multiplying the sqrt(25), not the rest

you can't just move it somewhere else

[twin geyser]

Hm

[twin geyser]

I see

But then

You see

Sqrt(25) became 5

So where would sqrt(3) go

[quasi valve]

so then you would have:
$$16\sqrt{3} - 8\sqrt{5} - 6\sqrt{5} + 5\sqrt{3}$$

[rocky lotus]

Bungo

[twin geyser]

It wouldn't just start getting multiplied with 14 sqrt 5?

[twin geyser]

Oh

Ohhhh

[quasi valve]

now you can combine terms that have the same square root, using distributivity

for example:

[twin geyser]

Oh yeah since they were getting multiplied

You're right, sorry.

[quasi valve]

$$16\sqrt{3} + 5\sqrt{3} = 21\sqrt{3}$$

[rocky lotus]

Bungo

[twin geyser]

Thank you!

[quasi valve]

yw

[twin geyser]

Yes

.close

[vale dock]

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• Show your work, and if possible, explain where you are stuck.
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[ivory quiver]

Helppo

[weary pilot]

,rccw

[weary pilot]

uh

what

[ivory quiver]

What do you mean what

[gloomy harness]

what’s the question

[ivory quiver]

Number 18

[gloomy harness]

what

sorry we’re gonna need a little more than that

could u be a little more explicit?

[ivory quiver]

How do i do it im confused

[lyric sundial]

Do what??

[ivory quiver]

I would type it but i don’t know how

[gloomy harness]

is ur question
$24 - 25 \div 5^2$

[rocky lotus]

Flatus

[gloomy harness]

do u want us to help u evaluate this?

[ivory quiver]

Yes

[weary pilot]

oh

[lyric sundial]

Next time, put a space after the 18, 19, 20

[gloomy harness]

put a period

[lyric sundial]

Otherwise it seems part of the exercise

[weary pilot]

well

i mean

!status

[vale dock]

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

[ivory quiver]

1

I was told it’s this thing called bit math maybe and I’m confused

[weary pilot]

'bit math'?

[ivory quiver]

I don’t know that’s why I’m confused

I was told the answer is 23

[weary pilot]

do you know what $5^2$ is

[rocky lotus]

Percy

[vale dock]

@ivory quiver Has your question been resolved?

[ivory quiver]

25 is it not

[stable parcel]

Yes.

[ivory quiver]

mhm

[stable parcel]

So, do you know PEMDAS or BODMAS?

[ivory quiver]

nope

[stable parcel]

OK, so PEMDAS is like this:

• Parentheses
• Exponents
• Multiplication or Division (left to right)
• Addition or Subtraction (left to right)

So we start from the top.

Do you have any Parentheses in your expression?

[ivory quiver]

nope

[stable parcel]

Do you have any Exponents?

[ivory quiver]

remins me what those are

[stable parcel]

That's where you have a number, then a number that's raised up to the right of it.

[rocky lotus]

Chai T. Rex

[stable parcel]

Do you have something like that in your problem?

[ivory quiver]

yes at the end

[stable parcel]

OK, so we do that next.

[ivory quiver]

i forgot to hit enter

[stable parcel]

(24 - 25 \div 5^2) becomes (24 - 25 \div 25).

[rocky lotus]

Chai T. Rex

[weary pilot]

....

divergence lol

hilarious

[stable parcel]

$\div$

[rocky lotus]

Chai T. Rex

[stable parcel]

(24 - 25 \div 5^2) becomes (24 - 25 \div 25).

[rocky lotus]

Chai T. Rex

[weary pilot]

did you go change your preamble lol

[stable parcel]

Yes.

[weary pilot]

hah

[stable parcel]

@ivory quiver Do you see how we replaced (5^2) with (25)?

[rocky lotus]

Chai T. Rex

[ivory quiver]

mhmm

[vale dock]

@ivory quiver Has your question been resolved?

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[vale dock]

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[coarse granite]

Hello, I have a proofs issue and it seems there's a discrepancy between the question and the book.

Let p, q, and r be the propositions
p: You get an A on the final exam.
q: You do every exercise in this book.
r: You get an A in this class.

Identify the expression for the proposition "To get an A in this class, it is necessary for you to get an A on the final" using p, q, and r and logical connectives (including negations).

It would seem to me that the question is written as " R is necessary for P", and a SS from our book shows that for p -> r, it can be written as "r is necessary for p."

But when I select p -> r, it tells me I'm wrong and that it's r -> p. I'll send SS below

[thin valley]

It's because that natural English is like doing a roundabout.

It is r → p

It is necessary for you to "get A on the final" to "get A in this class"

It's because of the comma

"For A, it is necessary B" = "B is necessary for A" =

I think it is that

And because that is how exams and final califications work

Maximal calification implies perfect exam

[coarse granite]

what an asinine way to write this question I guess

[thin valley]

Yes 🙁

I don't like these questions confusingly posed

[coarse granite]

alright, thanks a lot

.close

[vale dock]

Channel closed

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[vale dock]

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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.

[distant imp]

Hey

[distant imp]

So like does a Rn->Rm transformation only work on 1xn vectors? Or

.close

[vale dock]

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[ivory quiver]

I know that this is a math server however nobody in the biology server answers, so would anyone wanna call and help me with my biology it’s simple work. I’m just doing a mass amount of it if you’re just here to text don’t answer

[normal raft]

Post your questions if you have specific questions

[ivory quiver]

It’s not specific tho i just said a mass amount

When I say a mass amount I mean from like chapter 3 to chapter 10

Even better if I can get to chapter 16

[thin valley]

😳

[normal raft]

Okay, good luck! Ask again if you have any specific questions

[ivory quiver]

Thank you

[cloud coral]

Yeah, these channels are best for one-time questions. Chapter 3 to 10 sounds quite long and probably take at least an hour. I don't think it's likely people would be likely to stay all the way through.

[ivory quiver]

In question one would classification mean like the name

[cloud coral]

on what... [properties] i guess?

[ivory quiver]

What do you mean

[cloud coral]

So to determine that two organisms are of the same species (or different species), what properties do biologists use

(btw if you want people to work with you, you can just do each topic/question one at a time, with no expectation that people would be around for the entire thing.)

[ivory quiver]

The reason I worded it like that is because I don’t like texting at all. It’s fine if you’re not around for the whole thing, it’s just I would prefer a call.

Texting is gonna make this go from an hour to two hours

[ivory quiver]

The reason I don’t think this is what this means is because this whole page is about names

[ivory quiver]

Would you be willing to help with bio or math?

[cloud coral]

well, if you have an answer for every way of understanding the question, then you are pretty much done for that question - the rest would come from understanding what sort of questions the teacher/professor poses - which you may be able to see from the homework/previous tests.

[vale dock]

@ivory quiver Has your question been resolved?

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[rotund flower]

@cloud coral

[rotund flower]

can u help me

@here

@stark wedge

@next vortex

[charred burrow]

what is your question

[rotund flower]

i send rn

[charred burrow]

and i dont think you should ping individual people

[next vortex]

could you also please send an image instead of a file?

[rotund flower]

uh

[rotund flower]

like this

[final adder]

yep, thats ok!

[rotund flower]

so can u help me solve

[final adder]

hmm, where's your data?

[rotund flower]

bro what data

i need help solbing it

[final adder]

uhh your data points for your graph

im assuming your doing 1?

[rotund flower]

yeah i need help making a table

idk how

[final adder]

ok but do you have numbers

[bitter carbon]

wait he does have some data points written on the side

[rotund flower]

idk is the table correct

[final adder]

are those given?

[rotund flower]

brop

NO

[bitter carbon]

was there a question before this?

[rotund flower]

all i need is someone to teach me how to make a table and equation

[rotund flower]

theres 2 pages

[rotund flower]

one is the word problem

[bitter carbon]

oh wait

you need to figure out the equation from the word problem

[rotund flower]

yes and a table

[bitter carbon]

this is necessary to make the table

[rotund flower]

ok

can u help

[bitter carbon]

what is the area of the rectangle (asking you to find it so i can guide you through)

[rotund flower]

l x w

[rotund flower]

a= l x w

the permiter is 120

[bitter carbon]

yes in this case A = x*y cuz that's the variables we're given

[rotund flower]

ok

[bitter carbon]

can you find y in terms of x given the perimeter?

[rotund flower]

yes

we can do comnbinataiuons

for example

60+60

59+49

then we can multiply

to see what has the greatest area

and when we graph

its at top of prabola

[bitter carbon]

yes but we want a general formula

[rotund flower]

ik how w get

[bitter carbon]

then you're good to go because you can then plug in values of x to graph

[rotund flower]

how we get the formula

[bitter carbon]

do you know the perimeter formula?

[rotund flower]

yeah

p= 2(x times y)

[bitter carbon]

plus

[rotund flower]

i mean plus

[bitter carbon]

byt yes

[rotund flower]

yeah mb

[bitter carbon]

it's alright

[rotund flower]

miss click

[bitter carbon]

p here is 120

so 120=2(x+y)

now you can solve for y here

[rotund flower]

yeah

sorryu my wifi went out

@bitter carbon

[bitter carbon]

it's okay

[rotund flower]

now whats next

[bitter carbon]

as a note, we want y because we want to input y into A=xy

[rotund flower]

ok

so we do x all the way 1-120

[bitter carbon]

we could do that but theres a better way

120 = 2(x+y)
120 = 2x + 2y (expand)
120 - 2x = 2y (subtract 2y from both sides)
60 - x = y (divide by 2 for both sides)

so we have y = 60 - x

do you remember what's next?

[vale dock]

@rotund flower Has your question been resolved?

[rotund flower]

@bitter carbon

fixed my wifi

sorry

it went out for some reason

.close

[vale dock]

Channel closed

Closed by @rotund flower

Use .reopen if this was a mistake.

[bitter carbon]

oh

[rotund flower]

@bitter carbon

.reopen

[vale dock]

✅

[rotund flower]

U CAN HELP

@bitter carbon

i need help with equation

i fxied the rest'

PLEASE HELP

IM STRESSING I HAVE A TEST TMRW

[wraith gull]

I eat tree

[rotund flower]

bro what

[wraith gull]

Yum

[rotund flower]

fuck u

[wraith gull]

Aw man

Ok

Gl on ur test

[rotund flower]

ty

[vale dock]

@rotund flower Has your question been resolved?

[vale dock]

Channel closed

Closed due to timeout

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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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[north scarab]

whats the difference between 2x vs x^2

[north scarab]

dont they give the same value

[junior jay]

2 * 3 = 6
3^2 = 3 * 3 = 9

[frozen ledge]

only if x = 0 or 2

2x is 2 * x whereas x^2 is x * x

[north scarab]

because I was thinking If I was able to cancel one of cos^2x and one of 2cosx

[frozen ledge]

!xy

[vale dock]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[frozen ledge]

just factor out cos from the denominator

[tawny moon]

i don't understand your working

[north scarab]

did I do something wrong with the question?

[frozen ledge]

he meant cosxsinx- cosx presumably

[tawny moon]

if that's the case he probably should have shown it, because there's now a missing sin(x)

[frozen ledge]

well yea of course

he was wrong

[tawny moon]

specifically, the orange term's numerator

[frozen ledge]

i just meant he should have

[tawny moon]

yea i get what you mean

[frozen ledge]

whether he was aware that it was wrong or not

[copper stump]

its nice that you so thoughtfully decided to write the wrong stuff in orange colour, so we can easily point ti out

[frozen ledge]

[north scarab]

wait ill just redo the q

[tawny moon]

we should have an "always check your work" sticker here

then again i would be the primary victim so

[frozen ledge]

if you factor correctly you avoid this issue

[tawny moon]

when redoing, pay close attention to your fractions

doesn't hurt to equalize the denominators explicitly

[north scarab]

wdym?

[frozen ledge]

common denominator

so that addition can be carried out

[tawny moon]

finding the common denominator and figuring out what to multiply each numerator by

otherwise you're gonna skip them like you did here

[north scarab]

could you check my other q

the one circled in red

I''m not really sure of

but check if the one without circled are correct

[frozen ledge]

for 2 just multiply top and bottom by cos

don’t do this combining fractions nonsense

you’re just confusing yourself

[frozen ledge]

for 6 why are you writing it in terms of sin and cos

cos(-x) = cos(x)

tan(-x) = -tan(x)

[frozen ledge]

which follows from cos being even and sin being odd

[north scarab]

could you show me visual

I forgot about this

[frozen ledge]

sin(-x) = -sinx

cos(-x) = cosx

tan(-x) = sin(-x)/cos(-x) = -sinx/cosx

= -tan(x)

[north scarab]

how does cos (-x) = cosx

[frozen ledge]

for 8 you combined wrong

[frozen ledge]

what’s your definition of cos?

you don’t have one i’m sure

which means you should take it as given

[north scarab]

cos is the x axis

[frozen ledge]

nope

that’s y = 0

[north scarab]

i mean cosx is the x coordinates

[frozen ledge]

of what?

[north scarab]

this

of a point

[vale dock]

@north scarab Has your question been resolved?

[north scarab]

.close

[vale dock]

Channel closed

Closed by @north scarab

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[vale dock]

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

[gray patrol]

is my answer is correct ?

[vale dock]

Please don't occupy multiple help channels.

[gray patrol]

this is solving by complete the square

i want to get the vertex

[tawny moon]

!1c

[vale dock]

Please stick to your channel.

[gray patrol]

sorry

@tawny moon i just closed it

[stiff lily]

you shouldn't be closing and opening a new channel with the same question (within close duration)
but cts has been performed correctly
you should now be able to identify the vertex from that directly

[gray patrol]

yes

thx

<3

[vale dock]

@gray patrol Has your question been resolved?

Channel closed

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• 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.

[torpid ocean]

how do i solve this

[stark wedge]

!status

[vale dock]

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

[torpid ocean]

3

[stark wedge]

ok show work and answer

[torpid ocean]

alr so firstly i started with the inner limit

so for g(x) as x->0 its obv 0

[stark wedge]

that is so

[torpid ocean]

but then i plugged it into the rest of the limit at it became 0^2+1

so that means that im finding the limit of f(1) x->0

but since there is discontinuity

[stark wedge]

uh nope

[torpid ocean]

i think its dne

ok

so what did i screw up

[sharp walrus]

is it 0?

[stark wedge]

you missed an important detail

namely that [g(x)]^2 + 1 will approach 1 from above

so your limit becomes a one sided limit: $$\lim_{t\to 1^+} f(t)$$

[rocky lotus]

Ann

[torpid ocean]

above meaning from the right?

[stark wedge]

yes

[torpid ocean]

is that because the original limit is x->0

which implies the - and +

[stark wedge]

no it's because g(x) is squared

so the square is always positive

[torpid ocean]

do i just have to memorize that

or is there a reason

[stark wedge]

memorize what

that squares are positive?

[torpid ocean]

ik that

but that also means that we find it from above?

ohhh

i see

[stark wedge]

well yes that does mean our limit is one sided

[torpid ocean]

because you approach

[stark wedge]

and like, specifically because f is discontinuous at the relevant point, that's why it matters we keep track of the approach direction

[torpid ocean]

thanks!

.solved

[vale dock]

Channel closed

Closed by @torpid ocean

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[vale dock]

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Send your question here to claim the channel.

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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[lean umbra]

Can someone please explain to me what is going on here and what the use of this is?

[supple breach]

Basically

They are calculating the integration by breaking it into two or more than two integrations

This is useful

When dealing with a piecewise function such as modulus

Or gif

[woeful trench]

Seems like they're proving the more generalised case of "integrals splitting" from the mention that a is between b and c

[viscid spade]

(Why would they do this order also kek)

[lean umbra]

I don’t get the first step tho

[woeful trench]

Guessing cause they showed it for the case of b to c but didn't explain why for the "other way round"

[jade ivy]

I think it just makes a lot of sense if you had access to the previous slides

[viscid spade]

Do you know the relation for vectors sum ?

[lean umbra]

?

[viscid spade]

AB + BC = AC

[safe fulcrum]

if you want to find the area under f(x) between x = 1 and x = 3 for example

that's just (area under f(x) from x = 1 to x = 2) + (area under f(x) from x = 2 to x = 3)

you're just splitting the region like so

[lean umbra]

but here it’d become the area from x = 1 to x = 2 + the area from x=2 to x = 1 + area of x=1 to x=3? In the first step

[safe fulcrum]

well it's confusing cause b <= a <= c

but yeah, if b = 1, a = 2, c = 3

you actually get (x = 2 to x = 1) + (x = 1 to x = 2) + (x = 2 to x = 3)

the important bit is that (x = 2 to x = 1) + (x = 1 to x = 2) cancel and equal zero

why? because when you are finding the area, you are finding the area of the rectangles from left to right

so if you go from right to left over the same region, it cancels out

[lean umbra]

I think I understand

I got the letters abc a bit mixed up cuz of the order

[safe fulcrum]

yeah I had to check quite carefully too

[lean umbra]

But theyre also asking me to prove for 4 different cases

a<c<b, b<c<a, c<a<b and c<b<a

I just gotta do the same thing but then in the correct order?

[safe fulcrum]

yeah, so you have to keep applying these 2 properties

I'll use numbers to make it easier but you should use variables

$1) \int_1^3 f(x) \ dx = \int_1^2 f(x) \ dx + \int_2^3 f(x) \ dx$

[rocky lotus]

south

[safe fulcrum]

$2) \int_4^6 f(x) \ dx + \int_6^4 f(x) \ dx = 0$

[rocky lotus]

south

[safe fulcrum]

1. is the splitting rule and 2) is the reverse direction rule if that helps

[lean umbra]

Let me try the first case rq

Is it like this?

[safe fulcrum]

let me see

[lean umbra]

I think i messed up no?

[safe fulcrum]

yeah, so in the 2nd line

split (a to b) up into (a to c) + (c to b)

so (a to c) + (c to b) + (b to c) in total, and now which parts cancel?

[lean umbra]

C to b and b to c

[safe fulcrum]

yep!

[lean umbra]

Can I also split like this, for b < c < a: a to b = a to c + c to b?

Like this

[safe fulcrum]

yeah!

so this is an interesting case, cause now you're going backwards from a to b

but then you're not switching direction

so a is more than c, and c is more than b, so you can do that

this may be a bit pedantic but formally, you should do $\int_a^b f(x) \ dx = -\int_b^a f(x) \ dx$ first

[rocky lotus]

south

[safe fulcrum]

then you have $-\left( \int_b^c f(x) \ dx + \int_c^a f(x) \ dx \right)$, so that now the direction is left to right

[rocky lotus]

south

[safe fulcrum]

and then after cancelling you get $-\int_c^a f(x) \ dx = \int_a^c f(x) \ dx$ as claimed

[rocky lotus]

south

[safe fulcrum]

so this would be a better way to write it

[safe fulcrum]

but yes, you've made a very good observation here that allows you to save on working

[lean umbra]

Alr alr

So like this for c < a < b

[safe fulcrum]

yep! that's all correct!

[lean umbra]

And this for the last case

Thanks for the explanation i think i get it now

[late badge]

https://tenor.com/view/sigma-gif-5970294187009681545

[safe fulcrum]

yep! I'm not going to nitpick the 2nd line but yes

you could have written $-\left( \int_c^b f(x) \ dx + \int_b^a f(x) \ dx \right)$ in between

[rocky lotus]

south

[lean umbra]

I was thinking about it but didnt deem it necessary

Or is it

.close

[vale dock]

Channel closed

Closed by @lean umbra

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[vale dock]

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Channel closed

Channel closed due to the original message being deleted.
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as this action is irreversible, and this channel may abruptly lock.

[stark wedge]

welp you messed up

now you have to go make a new channel

[heavy dew]

Oh

My bad

[vale dock]

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[paper herald]

just a quick question,

integral of:

cos u du = sin u + c
sin u du = - cos u du + c
sec^2 u du = tan u + c
csc^2 u du = - cot u + c
sec u tan u du = sec u + c
csc u cot u du = -csc u + c
e^u du = e^u + c
a^u = a^u / ln a + c, a does not equal 1

correct?

[late badge]

yea

[paper herald]

alr thanks zhongli

.close

[vale dock]

Channel closed

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[vale dock]

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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.

[obsidian cloak]

im reviewing last year's work n hopefully pass math for once and I do remember how to do these equations the "long" way (just what I put on the sticky note on the bottom) but I don't understand what the numbers all mean together

[obsidian cloak]

ig this is more of a teach than help / is there a way to shorten the process and graph the equation without having to do the work I have on the bottom

(also why did i put the "y int" on problem 3 when that number is just not consistently the y interecept)

[ebon glade]

drawing |x-a| is just drawing |x| but moving it around to the left or right

and then drawing |x-a|+b is drawing |x-a| but moving it up or down

[obsidian cloak]

got it so just flip the sign and move however many units?

[ebon glade]

basically, yeah

[obsidian cloak]

except for b, the sign remains the same

[ebon glade]

yes

[obsidian cloak]

perfect tysm

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[wide path]

I can't believe I'm having to ask this but can you simplify this further?

This is what I have so far:
,, 9-2\sqrt{2}+2\sqrt{6} -2\sqrt{12}

[wide path]

,,tex 9-2\sqrt{2}+2\sqrt{6} -2\sqrt{12}

[rocky lotus]

UltraSonicSpeed

[wide path]

whoops

[bronze moss]

12=4*3

[wide path]

ah

my mind was being tricked into thinking in 2s and 6s

[bronze moss]

once thats done i think u r good

[wide path]

thanks

so its

,tex 9- 2\sqrt{2} + 2\sqrt{6} - 4\sqrt{3}

[rocky lotus]

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

[stark wedge]

just write it with dollars

[wide path]

bruh

[stark wedge]

$9- 2\sqrt{2} + 2\sqrt{6} - 4\sqrt{3}$

[rocky lotus]

Ann

[wide path]

oh thanks

[stark wedge]

you can put dollar-enclosed formulas in any part of the message like this: $y=x^2$

[rocky lotus]

Ann

[wide path]

first time using it

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[wet tundra]

how do u do question f using integration

[wet tundra]

is it integral of absolute of v(t)

[glass kelp]

|v(t)|

[wet tundra]

and then 3 for upper boundary

and 0 for lower boundary?

[glass kelp]

Yup

[wet tundra]

what if it changes direction?

does that matter

[novel canyon]

hello

[light saddle]

no

that's why you have | |

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[earnest valley]

Just a quick question I have

[earnest valley]

The book says $\ln8-\ln4+\ln\frac{1}{2}\to \ln2\frac{1}{2}=0$

[rocky lotus]

Totalani

[earnest valley]

but couldnt you also do $\ln8-\ln4+\ln\frac{1}{2}\to \ln8-\left( \ln4\cdot\frac{1}{2} \right)$ which gives you a different answer? if not why?

[rocky lotus]

Totalani

[ebon glade]

x-y+z is not the same as x-(y+z)

[earnest valley]

ah you are right

derp

thank you !

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[velvet sequoia]

To solve for x, I believe it's x/2380 = 2640/1320. However, I'm not sure if it's correct. This would mean that the scale factor is 2. But, 1320x2 does not equal 2380.

[stark wedge]

scale factor between what

[velvet sequoia]

I'm not exactly sure what I'm actually saying

[velvet lake]

this is just asking for direct application of Thales'

[stark wedge]

your x value is correct btw

[velvet sequoia]

i see

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[modest hemlock]

can someone explain b

[modest hemlock]

would it just be like

vy/vx * sina/cosa = -1

a it would yeh

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[main blaze]

I'm getting the answer in different form

Is the question wrong or I'm just dumb

Question 4

[stark wedge]

show work?

[main blaze]

1 sec

Was making it understandable

[stark wedge]

hmm\

i cannot find a fault in your working

[main blaze]

I have retired it 3 times

My head hurts

[stark wedge]

it looks like your expression is correct and theirs has the wrong form

[main blaze]

Ohk

[main blaze]

Aww mgod

I found the problem

It's me

I am a dumbass

I interchanged the variables

Cos alpha + beta is a

Sorry for trouble

I'm going to bed ig

[weary pilot]

its me

hi

im the problem

its me

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[wide path]

my first attempt at a STEP question, though I've gotten stuck
I thought I had made a lot of progress and got something that was looking correct, but I didn't know what to do with it so I had to go back and check my workings and review what the question was actually asking for.
However, it turned out I had made an error which set back my progress...
I'm still on the first part of the question to show that the equation has only one real solution when c isn't 1 and if the given equation is true.

[stark wedge]

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[upper sundial]

hey

[upper sundial]

what is x? yeah its easy but im confused im study math nonstop for 8 hours

[thin valley]

Do you know the definition of radical as a power?

So how √3 = 3^?

Square root of three equals 3 to the what

[upper sundial]

1/2=^

damn really i confused

@thin valley

[thin valley]

Yes

So

$$\left(9^x\right)^\frac{1}{2}=\frac{1}{9}$$

[upper sundial]

i found it

[rocky lotus]

Categorist

[thin valley]

Have you solved it?

[upper sundial]

yeah

im gonna explain

one minute

[zinc spear]

this is ok

[upper sundial]

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[thin meadow]

Something feels sketchy about the unit circle.

[stark wedge]

what exactly

& can you show the diagram(s) you're looking at of it

[thin meadow]

The unit circle has multiple has similar coordinate. For example.

Sin(1/2) = 5pi/6 and pi/6

And, If I want to memorize trig ratios, won't it be problematic.

[stark wedge]

you have it backwards...

you mean sin(pi/6) = 1/2 and also sin(5pi/6) = 1/2 ?

[thin meadow]

ye.

[stark wedge]

well i mean that is how the sine function works

like if you doubt me

,w sin(pi/6)

[rocky lotus]

Results provided by WolframAlpha

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Requested by ann_dec

[stark wedge]

,w sin(5pi/6)

[rocky lotus]

Results provided by WolframAlpha

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

Requested by ann_dec

[stark wedge]

these just are both equal to 1/2

[thin meadow]

Oh, Okay.

[stark wedge]

there are some mnemonic/visual tricks if you want to memorize these values better, such as the "trigonometric hand" mnemonic for the first quadrant

[thin meadow]

Oh okay.

[stark wedge]

https://cdn.discordapp.com/attachments/1064643750493691967/1134924411120648213/20230422_100420.jpg

this is the hand in question

labelled in degrees, but still

i can also share a more recent thing that i put into writing for how i personally think about/visualize trig values for the entire unit circle

with... not that much memorization, certainly not just blind memorization of all 32+ of those values

well actually here

[thin meadow]

you did this?

[stark wedge]

wdym

[thin meadow]

I'm asking if you took a pic of your hand and labeled all of those.

[stark wedge]

yes that is my actual hand

[thin meadow]

I see.