help-13

[quasi plover]

Ahhhh

[hollow trail]

so what we do is we interview a few random students and find the mean number for that random number. But we don't know if the mean we got is exactly equal to the mean for all students. But we can find a margin of error, which says it's likely the mean for all students is within that margin of error above or below the mean we got

so it might be up to 3 dramas more or it might be up to 3 dramas less than 15 and still be "inside" the margin of error

[quasi plover]

Wait whats margin of error?

Is it like, the dot that are not on the line of best fit?

[hollow trail]

when we use the sample to estimate something about everyone we think the mean for the whole population is pretty close to the one we got, but it may not be exactly equal

we say something has a margin of error of 3 if we think the "real" value is somewhere between 3 points above and 3 points below our estimate

[quasi plover]

Ok

So the answer ia A?

[hollow trail]

the margin of error is just for estimating the average. it doesn't tell us how close any individual student might be to the average

[quasi plover]

But it does tells the number that are not on the line of best fit righth?

C is wrong for sure

But B seems the most correct

[hollow trail]

remember, we're trying to find the average number of dramas watched by all students, without asking every single student. We know it's probably close to 15, since that was the average we calculated from the students that we did ask. But it might not be exactly 15, which is why we have the margin of error

[quasi plover]

So its either A or D? Because it aaid between 12-18

[cedar kiln]

@quasi plover Has your question been resolved?

[cedar kiln]

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[oblique raven]

D maybe C

[oblique raven]

no c is possible nvm

A doesn’t list population so…

is the population 5, 20, 1000?

[sonic thicket]

Question?

[oblique raven]

nah i was just looking at the last one

@quasi plover

im interested

what was the answer here?

because A looks plausible (i dont remember standard deviation rn)

wait A should be like 68% and true

but D should definitely be true

[cedar kiln]

@oblique raven Has your question been resolved?

[quasi plover]

No clue

That dude never answer me

Question A and D looks fairly similar

I dont even know the diff between those 2

[cedar kiln]

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[dim tiger]

hello everyone

[dim tiger]

where does the - sign come from

[cedar kiln]

@dim tiger Has your question been resolved?

[dim tiger]

the angle between $\hat{\phi}(t_1)$ and $\hat{\phi}(t_2)$ is $\Delta\hat{\phi}$

[wraith dagger]

calculus is fun

[limpid ore]

maybe its because ф is reducing?

[dim tiger]

no this is not the reason because $\frac{d\hat{r}}{dt} = \dot{\phi}\hat{\phi}$

[wraith dagger]

calculus is fun

[dim tiger]

do you mean $\Delta\phi$ is reducing ??

[wraith dagger]

calculus is fun

[dim tiger]

because $\phi$ is not necessarily reducing

[wraith dagger]

calculus is fun

[dim tiger]

how is the direction of $\frac{d\hat{\vec\phi}}{dt}$ opposite to the direction of $\hat{\vec{r}}$

[wraith dagger]

calculus is fun

[dim tiger]

.close

[cedar kiln]

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[simple bane]

In a group of 30 people, 10 play football (A), 5 play basketball (B), 2 people play both

[simple bane]

find $p(A) p(B) p(A\capB) p(A\cupB)$

[wraith dagger]

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

[lunar lynx]

\cap?

You want U?

[simple bane]

yes

forgot the command💀

[lunar lynx]

Well, i just checked. \cap seems correct. Lol

For intersection.

$p(A) p(B) p(A\capB) p(A\capB)$

[junior dome]

magneto, whats the command for d/dx?

sorry for interrupting

[lunar lynx]

\frac{\dd {expression}}{\dd {x}}

[tropic oxide]

missing spaces

[lunar lynx]

$\frac{\dd {expression}}{\dd {x}}$

[wraith dagger]

Enemagneto

[junior dome]

oh so frac must be used

[tropic oxide]

$p(A) p(B) p(A\cap B) p(A\cup B)$

[junior dome]

thanks

[wraith dagger]

Ann

[tropic oxide]

@simple bane so you want to find the product of these 4 probabilities?

[simple bane]

yes

[lunar lynx]

At least, i do. Others might know a better way.

[tropic oxide]

or just each one individually?

which you'll have to do anyway

[simple bane]

i know p(A) and p(B) but i don't have any idea how to solve the intersection and union

[tropic oxide]

how did you find p(A)?

(also did not answer my question...)

[simple bane]

if football is A, then it would be 10/30

[simple bane]

what is the difference between the "product" and "each one individually"?

[tropic oxide]

product means you're multiplying them all together and report one number

each one individually means you don't multiply them together, and report four numbers

[simple bane]

like

1/3+1/4+1/5?

[simple bane]

can we do both?

[tropic oxide]

...

i mean sure

if you want

you got that p(A) = 10/30, yes?

[simple bane]

yes

[tropic oxide]

and you also got that p(B) = 5/30

so what's p(A n B)?

you are told how many people play both games

[simple bane]

2

[tender acorn]

TAX

[simple bane]

should i draw Venn diagram?

[tropic oxide]

not yet

you know the number of ppl who play both games is 2

p(A n B) can be figured out the exact same way you did the prev two

[simple bane]

should i use this formula?

[tropic oxide]

no, not yet

again

you had no trouble going from |A| = 10 to P(A) = 10/30

do the same thing here for P(A n B)

you know |A n B| = 2

[simple bane]

2/30?

[simple bane]

so is 1/15 correct?

[tropic oxide]

yes...

and NOW you can use that formula you sent to find P(A u B)

[simple bane]

P(A u B) = 13/30

[tropic oxide]

yeah

do you have any further questions?

[simple bane]

the problem is the answers are different

in my textbook

[tropic oxide]

show your textbook's answers.

also show the problem as it was written in the textbook.

[tropic oxide]

hold on a minute

show me the problem itself.

[simple bane]

Sorry, my first language is not english so i don't think you will be able to understand.

[tropic oxide]

send it anyway.

i speak a few languages other than english.

standard procedure for problems written in foreign languages is to send them anyway, and provide a translation if somebody asks.

[tropic oxide]

ok i thought the language might have been chinese

but i was thankfully wrong about that

[simple bane]

lol

[tropic oxide]

i think the way they did it is a little strange, and also for some reason they're rounding everything to the nearest 0.01 while we had exact answers all along

[simple bane]

i will show you the formulas that the textbook gave

[tropic oxide]

oh also another thing

it looks like you might have lied about the goal of the problem

[simple bane]

how?

[tropic oxide]

well the things they're calculating aren't the same as the ones you mentioned at the start

they don't calculate P(A) and they don't calculate P(B) and they don't calculate P(A u B)

[simple bane]

i just wanted to "warm up"

[tropic oxide]

hjkdsjflh

and then you cry "OH BUT THE ANSWERS ARE DIFFERENT"

when you yourself ask for different things than the textbook

.-.

[simple bane]

wait

[tropic oxide]

also $P(A \cap B) = P(A) P(B|A)$ is true, yes. but imo very excessive here.

[wraith dagger]

Ann

[simple bane]

what is imo?

[tropic oxide]

in my opinion

[simple bane]

so i just memorised a bunch of formulas, and didn't really thoroughly understand them

so 🥲

and how come it's excessive to use that formula

Pleas don't leave me. I am very sorry for everything i have done

[tropic oxide]

it involves more computation and concepts than necessary

like, if you can calculate P(B|A) as 2/5, then why not calculate P(A&B) as 2/30 in the first place?

[simple bane]

what is P(B|A) equal to?

and is P(A|B) the same?

[tropic oxide]

no they aren't the same.

also do you mean "what's the number for P(B|A) here?" or "what is the meaning of P(B|A) anyway?"

[simple bane]

both

the textbook explained P(B|A) really vague

[tropic oxide]

the value of P(B|A) here is 2/5

P(B|A) is the probability that B happens given A

informally, what fraction of scenarios where A happens are also part of B?

[simple bane]

i didn't catch that

[tropic oxide]

P(B) is what fraction of our group are basketball players

P(B|A) is what fraction of the footballers are basketball players

[simple bane]

2/5

ah

makes sense

[simple bane]

Thank you for the help, i will work on this later 🙂

.close

[cedar kiln]

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[keen nova]

$a^x * b^x$

[wraith dagger]

marty

[keen nova]

will the answer be:

$ab^2^x$

[crimson sedge]

No...

[wraith dagger]

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

[crimson sedge]

||a^x × b^x = (ab)^x||

[keen nova]

but isn't the rules of powers saying to add the powers? 1 + 1 ?

[crimson sedge]

It's when the base is same

Meaning

||a^x × a^y = a^(x+y)||

[keen nova]

right okay, so when bases are not same, exponent remains similar

and if both bases and exponents are different, we can do no further simplification, e.g:

$a^x * b^y = a^x * b^y$

[wraith dagger]

marty

[crimson sedge]

Ye...

[keen nova]

okay

thank you

[naive canyon]

||a^x * b^x = (ab)^x (not sure why you said equal or superior)||

[crimson sedge]

Typo

I meant to says implies

There you tl

*go

.close to close the channel @keen nova

If you're done

[keen nova]

.close

[cedar kiln]

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[crimson sedge]

does anyone know what it means when it says to tell the values of 𝜃 for which the graph would be drawn once, and only once? i know someone already talked about it but i am still kind of confused

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[oblique raven]

.reopen

/reopen

[oblique raven]

a or d?

someone posted this earlier

B is straight wrong

and we dont have populations,

so C is wrong

[ancient lodge]

Probably D, since there could be some outliers

[cedar kiln]

@oblique raven Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[keen nova]

i just don't understand

[keen nova]

this simplification step is insane

first term should be 1

[naive canyon]

they didn't multiply by the inverse squareroot

[naive canyon]

they brought it down to the denominator

[keen nova]

😦

so sqrt(x)^2 = x, and this other sqrt(x) is from that inverse sqrt(x)

then the first term is sqrt(x) * 1

i no longer understand how simply multiplication works apparently anymore after starting to do derivations but

how do i start on the second term

[naive canyon]

yeah

second term

imagine the inverse square root just went down and all that's left is (2 sqrt(x) -4) * 1/2

[keen nova]

hold on

[naive canyon]

?

[keen nova]

what is even the second term

is it all after the negative sign

$-(2sqrt(x)+4)\frac{1}{2sqrt(x)}$

[wraith dagger]

marty

[naive canyon]

yes

now the sqrt went down to the denominator

[keen nova]

which one

the inverse right

[naive canyon]

yeah

[keen nova]

okay

[naive canyon]

what's left is (2 sqrt(x) +4) * (1/2)

which is?

[keen nova]

so now the denominator is

$x * sqrt(x) * 2sqrt(x)$

[wraith dagger]

marty

[naive canyon]

noo

the reason why they could bring it to the denominator is because 1/(sqrt(x)) is a common factor of both terms

the denominator now is just $x*sqrt(x)$

[wraith dagger]

lotus man

[keen nova]

lost

it is (sqrt(x))^2 to begin with, which equals x, we bring down one inverse square root, so we have x * sqrt(x). Why would we not add 2sqrt(x) if we bring them down too?

[naive canyon]

alright

look back at the original line

[naive canyon]

.

there are 2 terms in the numerator

we can factor them by the inverse square root

[obsidian coral]

@keen nova Do you understand this part so far?

[obsidian coral]

The sqrt(x) * 1/sqrt(x) cancels to be one

Then I multiplied the 1/(2sqrt(x)) into the parentheses

[keen nova]

hold on, so we solve the first term, i get that, then solve the second term. Okay, understandable. And we didnt put any inverse square roots down to the denominator

[obsidian coral]

No not yet

[keen nova]

that makes more sense to me

[obsidian coral]

Do you see how the parentheses, simplify to this?

[keen nova]

okay from here, first term of right term cancels = 1, second term 4, we dont add it i think?

oh wait

hold on

[obsidian coral]

And (sqrt(x))^2 = x

[keen nova]

okay so we can divide 4 and 2sqrt(x) both by 2

i wasnt sure if we could do this

so thats 2 over sqrt(x)

yes okay i am with you so far

and now, can you omit the parenthesis? i dont see it

what to do

[obsidian coral]

Then this simplifies to

Then what you can do is multiply the numerator and denominator by sqrt(x)

[keen nova]

okay, thats okay, but how do you get -2 over sqrt(x)

[obsidian coral]

From here

That -(1 + 2/sqrt(x)) becomes -1 - 2/sqrt(x)

And 1 - 1 = 0

[keen nova]

i cannot see it

[obsidian coral]

-(a + b) = -a - b

[keen nova]

okay

sure

negative distribution

but

[obsidian coral]

That's what you can do here

Distribute that negative

To get this
-(1 + 2/sqrt(x)) becomes -1 - 2/sqrt(x)

[keen nova]

okay so i see what u did. But hold on a second

lemme just go back to my calculations

my second term i had

[glass sky]

can i ask one question about this

[keen nova]

$(\frac{1}{2sqrt(x)}) * (2*sqrt(x)+4)$

[wraith dagger]

marty

[obsidian coral]

Multiply that 1/(2sqrt(x)) into the stuff in the parentheses

So a(b + c) = ab + ac

[glass sky]

which solver did you use for this btw

or was it given in your assignment solutions

[keen nova]

OMPT-D

[glass sky]

so this?

or is it a solver which solves equations

and operators

[keen nova]

do you mean the first term, or the first factor of this second term

[glass sky]

i was asking this cause the steps are crazy

[obsidian coral]

.

[glass sky]

cant we just do this

[obsidian coral]

You can treat $\frac{1}{2\sqrt{x}}$ as a so then it's $a(2\sqrt{x} + 4) = a \cdot 2\sqrt{x} + a \cdot 4$, plug back in a

[wraith dagger]

dldh06

[glass sky]

ah thats the answer

mb

mb

[keen nova]

okay i see

[glass sky]

wait is it? whats the question

[obsidian coral]

So then it's $\frac{1}{2\sqrt{x}} \cdot 2\sqrt{x} + 4 \cdot \frac{1}{2\sqrt{x}}$

[wraith dagger]

dldh06

[glass sky]

nvm he is helping , and understands i will just go , i dont want to create more confusioon

sorry

[keen nova]

okay i see although i would swap order of factors for the second term

but obv ye i see

[obsidian coral]

Then the 2sqrt(x) cancels in the first part to be 1

The second part becomes 2/sqrt(x)

[obsidian coral]

Like what I showed here

[keen nova]

okay so now the entire second term is split up like that. Hold on a second now. Should the terms be separated by parenthesis each, or in one full parenthesis (or rather no parenthesis)

[obsidian coral]

It doesn't matter because of order of operations

[keen nova]

$(\frac{1}{2sqrt(x)} * 2sqrt(x)) + (4* \frac{1}{2sqrt(x)})$

[obsidian coral]

You forgot the sqrt(x), you edited and left of the x

You have 2sqrt)

[wraith dagger]

marty

[obsidian coral]

.

[keen nova]

so when doing factoring, it will never matter

?

i'm trying to understand when i need to apply parenthesis in general

because i have made a thousand such mistakes today

like let's step back one second. I think I understand where to go from there you showed me

but this is the whole process:

quotient rule: I lay out g(x) and h(x)

i take their derivatives each

[obsidian coral]

Take in consideration of PEMDAS

[keen nova]

now I do g'(x) * h(x) - h'(x) * g(x) / h(x)^2 (quotient rule)

[obsidian coral]

The parentheses are redundant as multiplication would be next

[keen nova]

so i put in these functions, and their derivatives in the formula like this

i put a parenthesis around EACH of these. So 4 parentheses in total

am i thinking correct so far.

[obsidian coral]

If it helps you understand the problem better, you can do that

[keen nova]

$(g'(x)) * (h(x)) - (h'(x)) * (g(x))$

[wraith dagger]

marty

[obsidian coral]

It might make it overwhelming with that amount of parentheses if you do it that way but it works

[keen nova]

okay, then the right term is a good example. Here we end up with this:

[keen nova]

.

[keen nova]

we make it into this

[obsidian coral]

Yes

[keen nova]

woops i mean this

[obsidian coral]

Both this

[obsidian coral]

And this are the same

[keen nova]

right they're the same

i see now

but okay

now to be clear

there are no longer any considerations of parentheses

it is strictly order of operations

across the entire nominator

[obsidian coral]

Pretty much

[keen nova]

i'm struggling to imagine where exactly parentheses are needed

but i know sometimes they matter

in this case i think they dont

if im gonna be able to solve these problems i need to know this

[obsidian coral]

If you had $2+ \frac{1}{2\sqrt{x}} \cdot 2\sqrt{x} + 4 \cdot \frac{1}{2\sqrt{x}}$ then parentheses would need to taken in consideration due to distribution
$$(2+ \frac{1}{2\sqrt{x}}) \cdot 2\sqrt{x} + 4 \cdot \frac{1}{2\sqrt{x}}$$ and what is shown above are two different statements because 2sqrt(x) would get distributed on the second expression but in the first, it does not get distributed

[wraith dagger]

dldh06

[keen nova]

okay yeah. So because 2+1/2sqrt(x) is a term, and we want to distribute it, it needs to be in parenthesis

so relating to the quotient rule

$(g'(x)) * (h(x)) - (h'(x)) * (g(x))$

[wraith dagger]

marty

[keen nova]

we don't want to distribute the negative across the entire second term

is that correct?

[obsidian coral]

Because you're not distributing anything, you don't need to surround each term with parentheses

[obsidian coral]

You can but I did it later

[keen nova]

but hold on here we wouldn't even distribute would we

[obsidian coral]

This step

[obsidian coral]

To this

[obsidian coral]

You do need parentheses surrounded everything though, because of the negative sign outside

[obsidian coral]

If you look at this image

There's still that negative on the outside

[keen nova]

so the second term needs to be respected as a negative term

and that would distribute the negative within

at one point at least

[obsidian coral]

Yes

[keen nova]

and this is true for all formulas or rules or whatever

since this is the definition of the quotient rule

[obsidian coral]

And because the negative would get distributed, that's why you need parentheses

[keen nova]

but i thought the quotient rule wasn't even defined like that

g'(x) * h(x) - h'(x) * g(x)

we're not saying

gh(x) - hx(x)

[obsidian coral]

Notice here how you have the negative outside 2sqrt(x) + 4

[keen nova]

it's not enclosed as a function

[obsidian coral]

You can distribute that negative first into 2sqrt(x) + 4

But I didn't

I multiplied 2sqrt(x) + 4 by 1/(2sqrt(x)) first

Because multiplication is commutative

You could have done -(2sqrt(x) + 4) = -2sqrt(x) - 4 first if you wanted

[keen nova]

take that original image

what if it was

$-(2sqrt(x)+4)\frac{1}{2sqrt(x)}+6$

[wraith dagger]

marty

[keen nova]

6 is now for some reason part of the term

should it be distributed

[obsidian coral]

No

[keen nova]

with negative sign

[obsidian coral]

Because it's not $-\left((2\sqrt{x}+4)\frac{1}{2\sqrt{x}}+6\right)$

[keen nova]

so only the first function of the term

let's say quotient rule is a - b

of term b

[wraith dagger]

dldh06

[keen nova]

you have two functions

only the first function is in contact with the negative sign

only that function should be distributed with negative sign

not the entire term

correct ?

[obsidian coral]

No it depends on the parentheses

[obsidian coral]

Whether it's this, this means all the terms the negative gets distributed to

[keen nova]

i understand that

[obsidian coral]

Or this where only the first term gets the negative

[keen nova]

i'm asking if the parenthesis should be inserted in this example

because we're inserting two functions into the formula, the quotient rule

should we treat the term as having parenthesis

[obsidian coral]

This example meaning the original problem?

[keen nova]

i mean in general for the quotient rule, but also any formula that might be used, i guess

i think the parenthesis should only be considered around the functions each

so 4 paranthesis in total for the quotient rule

[obsidian coral]

No because as I mentioned it's based on distribution

[keen nova]

so then two parenthesis

one for each term

[obsidian coral]

This is quotient rule here

[keen nova]

yes

there i see 4 parenthesis

well not really

but

imagine one around each function

i mean

surely it has to be either that we put parentheses around the terms, OR around the functions inserted

or else it should mean we don't put parenthese at all?

either

$(a) - (b)$

or

[wraith dagger]

marty

[keen nova]

$(f1) * (f2) - (g1) * (g2)$

[wraith dagger]

marty

[obsidian coral]

I think you are just confusing yourself even more

[keen nova]

i don't think so, cus i've been getting almost exclusively wrong answers using the quotient rule so far

i'm obv confused lol but

i def don't get it yet

[obsidian coral]

https://www.youtube.com/watch?v=8jVDEcQ0wXk&ab_channel=TheOrganicChemistryTutor
Trying watching this and doing the examples with the video

[keen nova]

I like this channel v much btw, been using it for everything so far

3 min in, and it seems like he puts parantheses arounds the functions

this helps distribute correctly

and also means that only the first function of the second term has the negative sign

[obsidian coral]

Yeah, like I said parentheses are used to based on what gets distributed

[keen nova]

my mind is v slow rn but... isn't it also opposite?

what gets distributed is based on the parantheses?

we put parantheses around the functions thus we distribute this way

[obsidian coral]

Yeah like that

English is wild at times

[keen nova]

i should have just watched this video to begin with

and so

after this distribution, between : f1g1 and f2g2

we no longer have to consider any parentheses

i think we clarified this at the beginning

[obsidian coral]

Yes

After you distribute, you won't need to use parentheses

[keen nova]

so sometimes we need to distribute. That is based on what parantheses are set up. Functions are naturally put in parantheses. After distribution, order of operations rule

[obsidian coral]

Yes

If that is all you have, feel free to close this channel and come back if you need more help, otherwise good luck

[keen nova]

okay, i followed these principles, redid the exercise, and got it right

i think i get it now

thank you so much

.close

[cedar kiln]

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[wintry galleon]

i've gotten so lost while doing my integral

[wintry galleon]

i have been using this formula to solve it

so i get $ln(|1+sin(x)|)$

[wraith dagger]

stiroy

[slate lintel]

ok yes keep going

[wintry galleon]

then i write $ln(|1+sin(pi/6)|)+ln(|1+sin(-pi/6)|) = ln(3/2) + ln(|1/2|)$

[wraith dagger]

stiroy

[slate lintel]

it's F(top) - F(bottom) not plus

[wintry galleon]

so it should be ln(1)?

[slate lintel]

how's that

[wintry galleon]

well if i rewrite it, i get ln(3/2) - ln(|1/2|)

[slate lintel]

yes

,tex .log rules

[wraith dagger]

Hayley

[wintry galleon]

😮

i really need to get better with the basic rules thank you

[versed fulcrum]

I think he's talking about this

$\int_{a}^{b}f\left(x\right)dx=\left|F\left(x\right)\right|_{a}^{b}=F\left(b\right)-F\left(a\right)$

[wraith dagger]

LE SSERAFIM

[wintry galleon]

i messed up the quotient rule

[versed fulcrum]

Oh sry nvm

[wintry galleon]

dw, i did mess that one up to after playing around with it soo much

[cedar kiln]

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[latent smelt]

Determine all solutions to the equation sin x = 0.6 in the interval 0° ≤ x ≤ 450°

[latent smelt]

Feels all wrong

Is my solution correct?

[stiff totem]

I mean, it wants x to be in the interval [0,450]

and one of your solutions is 756.87

so that doesn't quite seem right

you've found additional solutions by adding 360, which is fine

[latent smelt]

K = 0,1,2

[stiff totem]

but you're missing a few from symmetries

[latent smelt]

How do I go on from there?

[stiff totem]

so, the two solutions 36 and 396 are fine

but recall that sin has horizontal symmetry at x=90

try drawing a diagram

that might be helpful

[latent smelt]

Already done that, but not sure

Can it have only x1 and x2?

Just ditch x3

[stiff totem]

there's a third solution you're missing though

you have the first and last intersection

but notice that sin cuts the line y=0.6 again between those two points

the black line is a line of symmetry

so we can reflect the x=36.87 solution across

[latent smelt]

the third one that is ≈ 143
(90 - 36,87) + 0 * 360 = 53,13
But that's still not correct

[stiff totem]

it's not 90 - 36.87

but you've almost got it

so,

this distance here

we want to add it to 90

to get to x3

but this distance isn't 36.87

the distance on the left from 0 to x1 is 36.87

probably an easier way to think about it is to look at the line x=180

these distances are equal, because it's symmetric across x=90

[latent smelt]

wait so does it become 180

[stiff totem]

the black line on the right is x=180

[latent smelt]

so 180 - 2 * 36,87

[stiff totem]

not 2*

just 180-36.87

x1 is 36.87 to the right of 0, so x3 is 36.87 to the left of 180

[latent smelt]

but don't we need to substract from the right (the blue right arrow)

Ok ok

I got it

[stiff totem]

yeah, the two blue arrows are equal in length

[latent smelt]

Thanks a bunch!!!!

(180 - 36,87) + 0 * 360 = 143,13

[stiff totem]

that looks right, yep

[latent smelt]

tysm

.close

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[cedar kiln]

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[keen nova]

why

[keen nova]

we distribute -cos(x) to 10x, and get -10x * cos(x)? And aren't they the same?

maximo

[royal loom]

@short blade

[royal loom]

they are the same yes

[keen nova]

so why....

[royal loom]

did it say you were wrong?

[upper ruin]

Commutative property

[keen nova]

noo... i guess not.. but i suspect that i got wrong answers on my quotient rule exercises when distributing

so i checked what was the real answer to the distribution

but surely its the same

[royal loom]

yes those are the same

you could show your work for your exercise though and we can double check it

[keen nova]

this is why they're the same, not why it's that's the result on wolfram alpha, right

if you got 1 min let me re'do it and i will show you, especially if im wrong

first i did this and then i put the 2 in front of the x, and that was wrong, idk why i did that, but now it's right

i guess i was just sloppy

[short blade]

did someone call me

[upper ruin]

The steps look good

[keen nova]

could i further simply another way

[upper ruin]

I don't think so

[keen nova]

another q

derivative of sin(4x)

chain rule because of 4x?

in other words is it cos(4x) * 4?

nvm i just checked it is

ok i have a proper question

[keen nova]

regarding these steps. Was the division of 5x possible only because I divided both terms on the nominator as well as the denominator?

[cedar kiln]

@keen nova Has your question been resolved?

[cedar kiln]

@keen nova Has your question been resolved?

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[keen nova]

I've done the product rule, as well as chain rule for right term of f(x). How can I simplify this

[tropic oxide]

could factor out x^2 * 4^(x^5)

btw your notation is bad

you wrote g(x^3) when you meant g(x):=x^3, likewise you should have h(x):=4^(x^5) and not h(4^(x^5))

[keen nova]

oh yeah i know

[tropic oxide]

and f'(x^3 * 4^(x^5)) should instead be [x^3 * 4^(x^5)]', xor f(x) = x^3 * 4^(x^5); f'(x) = ?

[keen nova]

im a lil tired today

[tropic oxide]

if you know then don't write bad notation tired or not

[keen nova]

i was going to omit parentheses

for those

its not meant to be the actual function definition

i just enclosed it in parentheses dont worry bout it

[jaunty mural]

🤨

it just lit dont make sense

f(x^2) does not mean f(x) = x^2

[cedar kiln]

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[keen nova]

@inner quartz

[inner quartz]

yo

[keen nova]

ayo

[inner quartz]

who was it

[dire geode]

Help channels are for math questions.

.close

[cedar kiln]

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[inner quartz]

Yes, we know.

[dire geode]

.reopen

[cedar kiln]

✅

[dire geode]

Ohok continue

[inner quartz]

thanks!

[inner quartz]

nw

@keen nova what was your question?

[keen nova]

its ok its answered

thank u very much sir

[inner quartz]

Okay! ill be around if you need math help

[keen nova]

.close

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[twilit obsidian]

Someone check this please?

[slate lintel]

the graph seems to match the words

[twilit obsidian]

The graph is already set by the assignment btw

The question (image 1) is based on the pre made graph

[slate lintel]

i guess my main thing is pay close attention to the graph spacing

[twilit obsidian]

are you saying I did it correctly?

[slate lintel]

i'd say almost? look at where the vertex is vertically

[twilit obsidian]

Would that not be 24.24?

[slate lintel]

what does each gridline represent? lok at the axis

(this is probably not that big of a deal because you're just estimating anyway)

[twilit obsidian]

It represents throwing something in the air and how long it takes to fall(?) (the graph)

[slate lintel]

yeah but like

the horizontal lines

how much is each one worth?

[twilit obsidian]

Well it looks like it goes by 2s

Wait

On the x it goes by 1s

Y goes by 6s

[slate lintel]

okay, i'd have said 3s

for y

[twilit obsidian]

Mm ok

[slate lintel]

the point is that to me the vertex looks like it's at 25 rather than 24.25

but it's nbd

[twilit obsidian]

Ahh I see

I’ve changed it to 25 I stead of 24.25

[twilit obsidian]

I think the x of 1.25 would still stand or no?

I’m just gonna submit it

.close

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[random heart]

Having trouble to solve part D, why does the E(x) has to be equals to -0.2P?

[upper abyss]

Answerer is drunk

1.2E(x) is the amount Ravi should charge.

[random heart]

I see, thanks so much

[upper abyss]

They somehow got E(x) = 0.8P which is approximately right, but still not exact

$7.33 is the correct answer

[random heart]

For this question, the expected value is for the player, so in the organiser’s perspective, the expected amount of will be negative ?

For example, expected value for the player is $6.11
For the organiser, they will expect to lose $6.11. (Without any charges)?

[upper abyss]

Wait, no I am lying to you

Let P be what Ravi charges.

Then, 0.8P = E(x)

That's their third line

That is, 80% of what Ravi charges, he should expect to give back. That way, he gets to keep 20% of it

I don't know why they did the first line the way they did, since -0.2P is not equal to E(x)

[random heart]

I see, but I think by understanding the third line would be enough to solve the question

[cedar kiln]

@random heart Has your question been resolved?

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[sharp viper]

can some1 explain

ill show the answer too

i understand the summation part

but how does that equal

the summation under them

this one

[drifting marlin]

it's just 1+1+1+... a total of i times

which is i

[cedar kiln]

@sharp viper Has your question been resolved?

[cedar kiln]

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[bronze pivot]

why is dS equal to that

?

<@&286206848099549185>

??????????????????????

<@&268886789983436800> he is trolling

[wanton sail]

@molten magnet stop

[bronze pivot]

why you troll, now people can't see my question

😦

[jade creek]

?

[bronze pivot]

.close

[cedar kiln]

Channel closed

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[jade creek]

i can help

[bronze pivot]

.reopen

[cedar kiln]

✅

[jade creek]

yoo

what you need help

[bronze pivot]

why is dS

equal to that?

[vestal bear]

is this a surface integral

[bronze pivot]

yeah

[vestal bear]

I bet there's some derivation of the Jacobian for this

just look online

[bronze pivot]

we're not there yet

[jade creek]

cal 3

[dense hornet]

it's just how it is

[jade creek]

right?

[vestal bear]

wdym

[dense hornet]

that is called the surface element

[jade creek]

vecot projectioons

[bronze pivot]

yeah

[jade creek]

and orthogonal compontents

ok

[jade creek]

so whats the question?

[dense hornet]

I believe there are some derivative for it online? But that's just how it's defined. The surface element for a given parameterics of r(u,v) is the magnitude of the cross product of it's two gradients

[jade creek]

THAS CORRECT

sorry

caps

mc

fb

[bronze pivot]

why so many trolls

[dense hornet]

so in a similar way, if you have a surface represented by g(x,y), that could be expressed as the parametrics r(x,y)=(x,y,g(x,y)). To which the surface element is the one given in your picutre

[jade creek]

wdym

[vestal bear]

this is the same thing as the Jacobian right waler

[jade creek]

what in cal 3 you need help

a part of a problem

explanied

[dense hornet]

well no, the jacobian bares a different meaning

[jade creek]

a meaning

a topic

[dense hornet]

<@&268886789983436800>

[vestal bear]

(x,y,g(x,y)) is familiar though

[bronze pivot]

okay i guess I just gotta remember dS = the magnitude of the cross product of the surface's two gradients

[wanton sail]

please stop posting in this channel if you have nothing to contribute

[jade creek]

sorry ma'mm

i was ttying to ask

[wanton sail]

remember that the cross product just gives the area of a parallelogram spanned by two vectors, so this should make geometric sense

[jade creek]

what was he talking about

since im learning it

it was helpful for me

[wanton sail]

the two derivatives just give two vectors tangent to the surface

[vestal bear]

[wanton sail]

This is not the place to ask for help. If you want help, then open another channel. This is someone else's.

[jade creek]

no thats not what im saying

im trying to aid him

sense i have some knowlegde on the topic

i wanted to see

what he was doing

[wanton sail]

You haven't said a single word to aid them yet, so please just stop posting in the channel

[jade creek]

and try it myself

[bronze pivot]

I block him

[jade creek]

im looking at the problem ma'mm

sorry

ill leave

[wanton sail]

Okay anyways, LeGM does this geometric explanation make sense

[bronze pivot]

yeah

[wanton sail]

cool :)

[bronze pivot]

wait where did the +1 come from

[wanton sail]

basically you can parameterize the surface as <x, y, z(x,y)> and then the derivative in the x direction is <1, 0, dz/dx> and the derivative in the y direction is <0, 1, dz/dy>

taking the cross product of these gives you

<-dz/dx, -dz/dy, 1>

and then the magnitude of that is

[bronze pivot]

ohh

[wanton sail]

what they have :)

dS

Do you have any more questions btw @bronze pivot

[bronze pivot]

nah

thanks

.close

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[wanton sail]

you're welcome!

[cedar kiln]

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[junior coral]

can anyone explain why 0.5x still negative and and where's the 1x.?

[flint plinth]

1x - 0.5x = ?

[junior coral]

oh HAHAHA thanks

.close

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[crimson sedge]

Why is multiplication for formal power series defined by convolution

[dreamy sleet]

because you can’t foil

(1 + x + x^2 + x^3 ….)(1 + x + x^2 + x^3 …)

You can’t just say that the result is 1 + x + x^2 + x^3 … + x + x^2 + x^3 … + x^2 + x^3 + x^4 + …

[muted bear]

that would be convoluted

bad joke, carry on

[dreamy sleet]

instead you take all the terms of a certain degree and put those first

[dreamy sleet]

I laughed

[cedar kiln]

@crimson sedge Has your question been resolved?

[slate lintel]

this is a good way to understand convolutions tbh

[crimson sedge]

Well okay Ty boys

. close

.close

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[dreamy sleet]

.coose

[cedar kiln]

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[azure marsh]

Does it matter what variable i use for the bounds of an integral as long as i get the same answer?

[royal loom]

in what context

please show the question

[azure marsh]

I dont have the question but i did a trial CSSA 2023 and i got my marks back and i lost a mark for using x as a bound value instead of t but i got the same answer as the criteria

[stiff totem]

if it's an integral dx

[azure marsh]

i havent been given the paper due to securityy release dates

[stiff totem]

then you can't use x (in the limits)

[azure marsh]

its notdx

its dt

the equation required to integrate was in terms of t and i used x as my bound but my teacher argues that i lost a mark due as i should have apparently used t instead of x

[slate lintel]

if it's an integral dt then you shouldn't use t in the bounds

(with one exception which is the Leibniz multiple integration formula but I'm sure you weren't doing that)

[azure marsh]

for some reason my teacher thinks i needed to use t not x

[stiff totem]

I think we're confusing bound as in bound variable and integral bound

[azure marsh]

yyeah it was 2u

i can write an example if u want

[stiff totem]

the bound variable can be whatever as long as it isn't used before

[slate lintel]

yes please

[azure marsh]

ok

one sec

[stiff totem]

if you just replaced all t's with x's, that should be fine

[azure marsh]

as i integrated in the exam i subbed in x and continued to solve for x and get a value in minutes which i also wrote with units

[stiff totem]

the right side isn't right

you can't use the bound variable also in the limits

[azure marsh]

can u explain why so i can say it to my teacher

[stiff totem]

I mean, it's kind of a matter of scoping: the t only exists inside the integral

it's the same problem as using any other variable twice but with different meanings

[azure marsh]

i see

also

u know how the intergral isnt from 0

and if the question asked i.e the time required to accumlate a specific area you would do the upper bound - lower bound

and say that is the net time from the lower bound

[stiff totem]

yes, that's how you would evaluate that integral

[azure marsh]

alr ty

.close

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[misty kelp]

how do you answer questions like this? A knife has a blade that is 31.3 mm wide and 2.02 mm thick. If it takes 1.39 lb of force to cut into a ribeye steak and you apply 0.237 lb of force on the knife, what is the efficiency of the blade? Answer in percent, and round your answer to 2 decimal places.

[dreamy sleet]

to answer questions like this, start from what they’re asking for

then find the relevant formulas for that quantity

[misty kelp]

which in this case is efficency

[dreamy sleet]

and do you know how to calculate efficiency?

[misty kelp]

output/input

[dreamy sleet]

Can you use that here?

[misty kelp]

yes but it gives the incorrect asnwer

[dreamy sleet]

What is the correct answer?

[misty kelp]

here is an example question i have the answer for

it is similar

A knife has a blade that is 33.7 mm wide and 2.40 mm thick. If it takes 1.25 lb of force to cut into a ribeye steak and you apply 0.239 lb of force on the knife, what is the efficiency of the blade? Answer in percent, and round your answer to 2 decimal places.

and

the answer is 37.25

so i am a bit lost

[dreamy sleet]

does the question give any explanation

[misty kelp]

no further explanation than that unfortuantely

[dreamy sleet]

has your teacher explained why

[misty kelp]

no he has not

not that i asked him

because we have open tutorials and i hadnt come across this question

until now

we have no lectures

i am not majoring in math its just a course I am required to take

i am actually quite bad at it lol

its definitely a weird question but there is plenty along these lines on this worksheet

The wheel of a wheelbarrow has a 10-in diameter with a 2.00-in wide axle. If you apply 78 lb of force to move a load of 243 lb, what is the efficiency of the wheel-and-axle system? Answer in percent, and round your answer to 2 decimal places.

such as this

as well

but lets focus on the first question for now

[dreamy sleet]

I don’t know how you’re supposed to get that answer either

[dreamy sleet]

I thought it would just be this

[misty kelp]

yes me too

but i am very lost now

well

thank you for trying

ill see if anyone else has an idea

thank u

.close

[cedar kiln]

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[teal magnet]

Why for negative #s dividing by √ of 2X the leading term odd exponent (of the numerator) result in a negative #? See the attached; for both screenshots a positive # was 1st submitted. I don't understand why the formulas in hints make the result in either screenshot make the result negative.

Thank you kindly for your help

[tropic oxide]

√ of 2X the leading term odd exponent (of the numerator) result in a negative #?
bad place to use the letter X as a substitute for "times", typing abbreviations notwithstanding.

@teal magnet are you asking why, for these limits at negative infinity, $\sqrt{x^2} = -x$ rather than $+x$ as one might have expected?

oh actually no that's not quite right.

[wraith dagger]

Ann

[teal magnet]

I'm asking why a=odd integer; x^a/sqrt(x^(2a)) isn'T always positive.

[tropic oxide]

right...

so you know how every real number has two square roots, one positive and one negative, and the radical symbol always refers specifically to the positive root, right?

[teal magnet]

I know of it technically but barely ever used that.

Like 99.9% of my uses were +

[tropic oxide]

wym "barely ever used that"

do you or do you not know that that's the meaning of the radical symbol and also the meaning one puts into the phrase "the square root of ___"?

[teal magnet]

I know that technically a square root can be negative, but I've barely ever touch that subject.

[tropic oxide]

you've... successfully managed to avoid confirming that you do in fact know what i'm talking about.

i don't appreciate that.

[teal magnet]

let'S say I don'T know it, wouldn't that be better?

[tropic oxide]

it would not be all that much better.

i would still need you to acknowledge it as said by me.

$\sqrt{x}$ means the \textbf{positive} square root of $x$, i.e. the one of the two that's greater than zero.

agree or disagree?

[wraith dagger]

Ann

[teal magnet]

agree

[tropic oxide]

ok

a consequence of this is that $\sqrt{x^2} = |x|$ rather than just $x$.

[wraith dagger]

Ann

[tropic oxide]

the number $x^2$ has two square roots: $x$ and $-x$.

if $x$ is positive, then the positive root among those two is $x$, so $\sqrt{x^2} = x$.

but if $x$ is negative, then the positive root among those two is $-x$ (since $x$ itself is negative), so $\sqrt{x^2} = -x$.

[wraith dagger]

Ann

[tropic oxide]

agree or disagree?

[teal magnet]

if you mean the real #, the original is -|x| I understand

[tropic oxide]

... ??

[teal magnet]

or sqrt((real #)^2)=x

[tropic oxide]

i don't know what you mean by "the original".

i would appreciate if you did not strip my variables of their names.

otherwise you will force me to resort to convoluted and ambiguous variable-free phrasings.

which i don't want to do.

[teal magnet]

let's go with and example, is that ok with you?

an*

[tropic oxide]

i mean... sure

it'll take two examples to illustrate my point ig

$\sqrt{7^2} = 7$, but $\sqrt{(-7)^2}$ is also $7$ and not $-7$.

[wraith dagger]

Ann

[teal magnet]

I was to make the example....

[tropic oxide]

ok then do it

[teal magnet]

(-2)^2=a a is 2, if I want the original #, I need to refer to -a

[tropic oxide]

(-2)^2=a
a is 2
these two contradict each other...

[teal magnet]

sorry

[tropic oxide]

you sound like you are first declaring a to be 4 and then in the same breath saying it is 2

which i don't think is what you wanted to say.

[teal magnet]

sqrt((-2)^2)=a a is 2

[tropic oxide]

... ok

[teal magnet]

my example is over

please either @ or use the reply feature if you decide to try to help more.

[tropic oxide]

@teal magnet so... did your original question get answered to your satisfaction? Y/N

[teal magnet]

N. if you could point to where in the hint formula the result becomes negative it would be nice.

[tropic oxide]

$\sqrt{x^2} = |x|$, so for negative $x$ you have $\sqrt{x^2} = -x$.

[wraith dagger]

Ann

[tropic oxide]

in the first image, at the end of step 1/3, it says: ``In the denominator, let's divide by $-\sqrt{x^{10}}$, since for negative values, $x^5=-\sqrt{x^{10}}$.''

[wraith dagger]

Ann

[tropic oxide]

in the second image in the exact same location it says the exact same thing but with x^2 instead of x^10.

@teal magnet

[teal magnet]

where's the 1st part of any formula of that image spot making the formula output a negative?

[tropic oxide]

... i am struggling to understand what you want

here??

[teal magnet]

I gotta do something 5-10 mins

[cedar kiln]

@teal magnet Has your question been resolved?

[teal magnet]

I guess I'll have to memorize this. 😩 It doesn'T make sense to me why 2 negatives dividing one another don't become positive.

That sqrt(x^10) doesn't output the same as x^5, which AFAIK if it did both x^5 or negatives would cancel each negatives out.

I hate memorizing when there'S probably logic behind it that I just don'T get.

[upper abyss]

Are you asking why √[x^10] = -x⁵, for negative x?

Because x¹⁰ is positive, but x⁵ is negative, so you'd need a negative sign to make them equal

[tropic oxide]

you are confusing "is negative" for "has a negative sign in front of it".

[teal magnet]

In my head sqrt(x^10) is the same as x^5. To me x^5/x^5 should always get y>=0

[upper abyss]

Well, try it.
What's (-1)¹⁰?
What's the square root of that?
Is that equal to (-1)⁵?

[tropic oxide]

In my head sqrt(x^10) is the same as x^5.

this is what i spent a good amount of time trying to explain you is NOT the case.

but i see now it was all for naught.

[teal magnet]

I think they will output something different.

then why does the denominator introduce a negative?

[tropic oxide]

sqrt(x^10) = |x^5|, and for x<0 this is -x^5 !!!

[teal magnet]

So the negative sign of "-x^5" makes it positive?

[tropic oxide]

yes

[teal magnet]

That's a nifty trick!

I don't understand how it workd yes, but it sounds ingenious.

yet*

[cedar kiln]

@teal magnet Has your question been resolved?