help-33

[sinful thistle]

...?

OP ghosted us fr

[sand fable]

u waited like a total of one minute

[static quarry]

sadly not uncommon

[sinful thistle]

i am an impatient man

[sand fable]

valid ig

[sinful thistle]

patience is a virtue i do not possess

and no actually been asking nutka here to expand cos(2x) for approximately 6 minutes

[glossy bobcat]

like that?

[sinful thistle]

jesus chr7st

[glossy bobcat]

XD

[sinful thistle]

what-

[glossy bobcat]

i dont know

[sinful thistle]

cant really understand after line 2

actually half of line 2 is illegible

[glossy bobcat]

I counted x

maybe wrong

[sinful thistle]

okay let's go back for a second

[glossy bobcat]

okay

[sinful thistle]

you've already figured out that sin^2 2x + cos^2 2x = 1

we know that sin 2x = 14/19

[glossy bobcat]

yes

[sinful thistle]

so sin^2 2x = 196/361

all good so far

after that we get sqrt 165 / 19 = cos 2x

,calc 361-196

[elfin berry]

Result:

165

[sinful thistle]

yep

[glossy bobcat]

yes

[sinful thistle]

we know that sin^4 x - cos^4 x = 1(sin^2 x - cos^2 x)

what's the expansion of cos 2x?

[glossy bobcat]

ooo and this is the answer?

cos2x = cos^2x - sin^2x

[sinful thistle]

mhm

good

now what's sin^2 x - cos^2 x

[glossy bobcat]

sin^4x-cos^4x

[sinful thistle]

no no no

in relation to cos (2x)

[glossy bobcat]

thats the same

ooo

no

wait

I see

no i didnt

[sinful thistle]

okay look

[glossy bobcat]

sqrt (165)/19 isnt the answer

?

[sinful thistle]

no

wait ok

what's cos^2 x - sin^2 x = cos 2x IN RELATION TO sin^2 x - cos^2 x

[static quarry]

that's a 165??

you need to work on your writing haha

[glossy bobcat]

hahah

[static quarry]

^ can you justify this answer?

[glossy bobcat]

I have weird "5" in my hand writing

I understand

wait

?

but it dont have any sense

I cant solve it

[static quarry]

i don't know what any of this is

weren't we talking about sin(2x) and cos(2x)?

i don't see either of those in your work

[glossy bobcat]

wait

maybe

but

[sinful thistle]

mb guys

discord crashed

again

okay look

nutka answer thsi

[sinful thistle]

.

[glossy bobcat]

first I had sin2x = 14/19,
then => cos2x= v165/19
<=> cos^x - sin^x = v165/19
and I have to solve |sin^x - cos^x|

[sinful thistle]

yes look

[sinful thistle]

answer this. please

its the literal last step

[glossy bobcat]

and I cang ive the minus befor

so -cos^x + sin^x = -(v165/19)

[sinful thistle]

@static quarry help me out here bro

[static quarry]

how are you getting this answer

wait, lemme try typesetting what you wrote above

[static quarry]

^ this stuff

[sinful thistle]

thanks bungo

[static quarry]

you have established that $\sin(2x) = 14/19$

[elfin berry]

Bungo

[static quarry]

then you used $\sin^2(2x) + \cos^2(2x) = 1$ to get $\cos^2(2x) = 1 - 14^2/19^2 = 165/19^2$

[elfin berry]

Bungo

[static quarry]

so far correct?

[glossy bobcat]

I think this is the answer

[static quarry]

well the problem is i have no idea what cos^x means

[sinful thistle]

^

[static quarry]

and your v165 apparently means square root of 165
this is hard to read
that's why i want to typeset it

[glossy bobcat]

and |-v165/19|

[static quarry]

i think you probably did the right thing you're just not expressing it in a legible way

[glossy bobcat]

yea sorry

Its easier to write that and I ussualy using that way to write

okay so I uderstand it

thank you so much

[static quarry]

i get it
but if you want people to help you have to make a bit of effort to make it more readable, learning a bit of tex will take you a long way

[glossy bobcat]

okay sorry I thought it is understable

[static quarry]

no need to apologize

my handwriting isn't great either haha
i compensate for that by using tex 😆

[sinful thistle]

thanks bungo

was stuck

[glossy bobcat]

okay

[sinful thistle]

couldn't really explain it myself

!done

[marsh citrus]

If you are done with this channel, please mark your problem as solved by typing .close

[glossy bobcat]

Can I dont close that, and ask for more help for few minutes?

[static quarry]

yea sure
although if it's a new question/doubt, you should just open a new channel for that

[glossy bobcat]

okay

[static quarry]

helpers are more likely to reply to new channels

[glossy bobcat]

thanks I dont know how it works

okay thanks

bye

[static quarry]

cheers

[glossy bobcat]

.close

[marsh citrus]

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[marsh citrus]

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[still temple]

I to show that v has a potential, and calculate all the potentials of v.
I kniw that i need to do rot=0 but i don’t understand how i can calculate all if the potentials

[fervent rampart]

since you know that the components of the vector are the partial derivatives of the potential function, you can find it by integration. some examples here: https://tutorial.math.lamar.edu/Classes/CalcIII/ConservativeVectorField.aspx

[still temple]

Is -xsin(xy) the same as -sin(xy)x here?

.close

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[sterile scroll]

So I’m confused the work done by a force F WF = |F| * |deltax| * cosa. If a object’s moving down and a force acts down then the work is positive since the direction of displacement and force is the same cos0 = 1. Why is work by gravity negative when Wmg = |mg| |deltay| cos0(since it’s falling and force is also down). So it’s |mg| |deltay| how can that be negative ?

[hard gull]

it isnt

work done if the object is displaced downwards due to gravity is positive

if the object was going upwards then work is said to be negative cuz cos180 is -1

[sterile scroll]

That’s what I’m thinking as well

[hard gull]

is there any context to this question? like something from ur textbook or smthing

where they took work as negative even though displacement and force is in same direction

[sterile scroll]

The teacher drew some ball that’s falling down and it’s work is mg * (-deltay)

I was confused

[hard gull]

should be positive

are there any other forces acting on the ball?

[sterile scroll]

Nope its under the influence only of gravity a conservative force

[marsh citrus]

@sterile scroll Has your question been resolved?

[sterile scroll]

Ok all good she’s assuming that we plug in for deltay the negative y since it was displaced down. - - is positive

[marsh citrus]

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[sleek topaz]

what is the question

[acoustic vapor]

Prove it

[midnight obsidian]

,rotate

[sleek topaz]

apply $tan^{-1}(a) + tan^{-1}(b)$ formula on lhs

[midnight obsidian]

{-1}

[elfin berry]

swerriee

[marsh citrus]

@acoustic vapor Has your question been resolved?

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[patent vine]

can someone help me read my teacher's hand writing?

just this part

the inner loop (shaded ?????) has ??? graphed ??? 7pi/6 and11pi/6 set r = 0

help me with the question marks 🙏

[paper musk]

maby like the inner loop (shaded margin) has been graphed btw

[patent vine]

oh yeah

that makes more sense

i think btw means between

thanks 👍

.close

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[sinful thistle]

this is pretty simple, just split the summation

[elfin berry]

ren

[sinful thistle]

switch indexes and solve

[knotty ivy]

dang i didnt even think to split them

[crystal lintel]

you shouldn't go splitting things up like this without knowing if they even converge

[bleak ibex]

To make it more rigorous you could put a limit in front

as N goes to infinity

and take the sums from 1 to N

[crystal lintel]

[knotty ivy]

im sorry but im still lost ,how can i find the limit if i dont have an exact function

@helper sorry..

[stark trail]

Did your sir not teach you about uniform convergence

[knotty ivy]

no

i had to go out and learn a bunch of tests today

[stark trail]

The value of the infinite sum if it converges

Is the limit if its partial sums

Do you agree

[knotty ivy]

yes

[stark trail]

Please express that limit

Without computing it

[knotty ivy]

<@&286206848099549185>

[marsh citrus]

@knotty ivy Has your question been resolved?

[marsh citrus]

@knotty ivy Has your question been resolved?

[knotty ivy]

.close

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[dusky viper]

can someone quickly help. I am conducting an experiment on transformers to test the efficiency of various materials in the core, I have the input voltage at 10V and a 1:2 coil ratio so i'm expecting 20V, how would i calculate percentage error? Do i calculate how close each value is to 20V, to me that seems wrong since I should be expecting a different voltage for each material since they have varying efficiencies, but at the same time i am expecting 20V due to the coil ratio

[marsh citrus]

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[fleet sail]

help ive been trying to connect tthem but i cannot find the right sentence and it doesnt sound right, decided to get help

[marsh citrus]

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[marsh citrus]

@fleet sail Has your question been resolved?

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@fleet sail Has your question been resolved?

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[marsh citrus]

@gray maple Has your question been resolved?

[gray maple]

How to start this one?

[marsh citrus]

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[coral lagoon]

I need to use TFD [series] to state if it diverges or converges

This is what i think i need to do:

arctanx = tan^-1x

= 1/1/tanx
= tan(x)

but im unsure where to go after this

[sand fable]

btw this is not correct

arctanx is not 1/tanx

[coral lagoon]

ok

is there any equivalence then i can do to simplify it?

[sand fable]

think abt what happens to arctan(n) as n goes to infinity

[coral lagoon]

it apporoaches pi/2

1/pi/2
= 2/pi

.close

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[lofty sun]

ABCD is a parallelogram. m and n are trisecton points of AD. find the ratios of regions, III:I, II:I, and triangle MDC:AMCB

[lofty sun]

i know the ratio of III:I is 9:1 because similar triangles and areas^2

but if you know how to find II:I or triangle MDC:AMCB plz help because ive been stuck on this question for 5 hrs

<@&286206848099549185>

[marsh citrus]

@lofty sun Has your question been resolved?

[lofty sun]

no help

.close

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[amber viper]

I dont understand how the first claim works

I get the second

[marsh citrus]

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@amber viper Has your question been resolved?

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[tulip fiber]

How do i find the volume? is the volume of a right triangular prism (1/2) Base * Height * length?

If so, how do i also find the length?

[quasi patrol]

They give you the lenght

[tulip fiber]

are both legs 6 and 6?

for the triangle sorry

Is my formula correct for the triangular prism?

[quasi patrol]

Yes

[tulip fiber]

how come the answer comes out to 360?

i got (1/2) * 10 * 6 * 6

[quasi patrol]

Yeah thats right

So 180

[tulip fiber]

For the line 96/12 = (6/x)^3, why did they cube 6/x

[marsh citrus]

@tulip fiber Has your question been resolved?

[tulip fiber]

!

<@&286206848099549185>

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[whole hinge]

Is this right?

[winged parrot]

No it's wrong @whole hinge

If a > c and c > b that doesn't mean a > b

[whole hinge]

so what is the right solution?

[winged parrot]

I haven't done inequalities for a long time, I'll have to think

[brittle shuttle]

First we know by applying AM-GM inequality (AM ≥ GM) that (a⁴ + b⁴)/2 ≥ √(a⁴b⁴), right?

[whole hinge]

how can you get a^4?

[brittle shuttle]

oh, let me show you the process first

[whole hinge]

ok

[brittle shuttle]

so do u get this

[whole hinge]

no i dont

[brittle shuttle]

this is AM-GM inequality

that (x+y)/2 ≥ √(xy)

here x=a^4, and y=b^4

[whole hinge]

oh ya i get, but u got the a^4 by power it right?

[brittle shuttle]

wdym by this?

[whole hinge]

r u indonesian ppl im sorry

i mean like square the equation

idl how can i sa y that in english

[winged parrot]

What he is saying is right

[brittle shuttle]

no, im not, sorry

[brittle shuttle]

what is he trying to say?

sorry

[whole hinge]

i dont know bro, im very confusing rn

[winged parrot]

What you are saying is right @still temple

[brittle shuttle]

no, dont worry

[boreal rose]

do you mind if I take over rq

sorry I don't mean to hijack

[brittle shuttle]

sure

[whole hinge]

no prob

[brittle shuttle]

no worries!

[boreal rose]

okay so basically a good way to solve this question

is to look at the RHS and first come up with a common denominator

we still use AM-GM in my solution

but I think it's a lil more clear

once you find a common denominator and multiply

[brittle shuttle]

yes, you should do this

[whole hinge]

RHS what is that mean

[boreal rose]

the problem becomes $a^4 + b^4 + c^4 \ge a^2bc + ab^2c + abc^2$

[boreal rose]

Right hand side

[elfin berry]

nosqldb

[brittle shuttle]

you can simplify rhs further as abc(a+b+c)

[boreal rose]

yes but I will not bc I want the AM-GM to be clearer

[whole hinge]

oo

i will try

wait guys

[boreal rose]

but basically b4 I go

apply AM-GM to two copies of a^4, one copy of b^4 and one copy of c^4

then also apply AM-GM to two copies of b^4, one copy of a^4 and one copy of c^4

two copies of c^4, one copy of b^4 and one copy of a^4

[boreal rose]

the kinda AM-GM hint is there when u reformulate the q like this

[whole hinge]

a^4+b^4+c^4/3>...

like that u mean

inthe right side apply GM formula

[boreal rose]

just apply it

and you will see it immediately

[brittle shuttle]

@boreal rose , do u know how to hide whatever im saying because i want to share the solution with you but in the hidden form

[boreal rose]

||yes||

the /spoiler

[brittle shuttle]

||thank you||

[whole hinge]

bro i think i found the solution

U guys can read my note?

[whole hinge]

I Apply AM-GM for both sides of equation

N I got it

[brittle shuttle]

||@boreal rose , here is my solution to this problem: by AM-GM inequality, (a⁴+b⁴)/2 ≥ √(a⁴b⁴)=a²b² so a⁴+b⁴ ≥ 2a²b² , similarly b⁴+c⁴ ≥ 2b²c² and a⁴+c⁴ ≥ 2a²c². adding the inequalities we get 2(a⁴+b⁴+c⁴) ≥ 2(a²b²+b²c²+c²a²) and thus a⁴+b⁴+c⁴ ≥ a²b²+b²c²+c²a². on similar lines if we apply AM-GM inequality again, (a²b²+b²c²)/2 ≥ √(a²b²)(b²c²) and thus a²b²+b²c² ≥ 2b²ac, similarly b²c²+c²a² ≥ 2c²ab and c²a²+a²b² ≥ 2a²bc. Adding these three inequalities again, we get 2(a²b²+b²c²+c²a²) ≥ 2(b²ac+c²ab+a²bc) and thus a²b²+b²c²+c²a² ≥ abc(b+c+a). now since a⁴+b⁴+c⁴ ≥ a²b²+b²c²+c²a² and a²b²+b²c²+c²a² ≥ abc(b+c+a), we must have a⁴+b⁴+c⁴ ≥ abc(b+c+a). Now diving by abc on both sides of the inequality, we have a^3/bc + b^3/ac + c^3/ab ≥ a+b+c hence proving the inequality given a, b and c are positive numbers||

[brittle shuttle]

sorry, but im a bit confused. could u explain each line of working?

[whole hinge]

wait

[brittle shuttle]

ok

[whole hinge]

i apply am -gm for the both sides of equation

[whole hinge]

our way is a bit same

[brittle shuttle]

here you proved that both a^4+b^4+c^4 and a^2bc+b^2ac+c^2ab are greater than or equal to (abc)^4/3, but that does not mean that a^4+b^4+c^4 ≥ a^2bc+b^2ac+c^2ab

[brittle shuttle]

do u get this solution?

[whole hinge]

i need time for understand this

but if there is smth i dont understand is it okay if i dm you?

[brittle shuttle]

sure, im sorry i wasnt much help to you. im bad at explaining things. but i shared with you what was my solution to your problem

[brittle shuttle]

sure, feel free to dm me!

[whole hinge]

but i think if u could write it in the note

it will be more easier for me to understand

[marsh citrus]

@whole hinge Has your question been resolved?

[marsh citrus]

@whole hinge Has your question been resolved?

[marsh citrus]

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[proud zealot]

can someone explain me why is 5! needed here

i understand we chose 5 men and 5 womenes and combine them together for every such choice bu using 10C5 12C5

but why 5!

please tag if you answer

[noble oak]

may i?

[worldly yew]

No

You may not

[proud zealot]

?

[worldly yew]

Im just trolling

[noble oak]

[worldly yew]

Because the 5 men can be switched around

[noble oak]

5! is for the differnt combiantions that can be made

[proud zealot]

you mean they can have different partners?

[worldly yew]

Yuh

[proud zealot]

but can't we use combinations instead

10!/5! * 12! * 7!

sorry permutations

the first one was combinations

[worldly yew]

Now im just confused, are we not using permutations in the first place?

[steep storm]

nope

[worldly yew]

5p5?

[steep storm]

its 10 c 5

[proud zealot]

anyway acn i do this instead 10!/5! * 12! * 7!?

[steep storm]

why are you using permutations

[worldly yew]

10c5 is 5! ?

[proud zealot]

is 10!/5!*5!

[steep storm]

yeah

whats the doubt

[steep storm]

why do you thinkl this would work

try explaining your reasoning

@proud zealot

[proud zealot]

becuase it takes care of the order

permutations don't care about the order the cancel the things that repeat

[worldly yew]

WHAT?

[proud zealot]

isn't clear what i said?

[worldly yew]

Permutations does care about the order tho

[steep storm]

yeah

in permutations ABC is different from BCA

[proud zealot]

anyway im confused with the terms

[steep storm]

however we have to choose 5 men from 10, here the order of the men doesnt matter

[proud zealot]

but you got what i men

this n!/(n-r)! cares about the order

[steep storm]

yeah

its n p r

[proud zealot]

thats why i wanted to do this 10!/5! * 12! * 7!?

[steep storm]

could you write it in latex

its pretty confusing

do you mean this

$\frac{10!}{5!} \times 12! \times 7!$

[elfin berry]

flurry

[worldly yew]

He does but that looks wrong

[steep storm]

yeah the 7! is in the denominator

[proud zealot]

sorry my mistake

7 is denominator

i messed when i wrote

thats what i meant

[steep storm]

so when expanding it becomes

$\frac{10!}{5! \times 5!} \times \frac{12!}{7! \times 5!} \times 5!$

[elfin berry]

flurry

[proud zealot]

no bro

[steep storm]

yes bro

[proud zealot]

its (10!/5!) * (12!/7!)

[steep storm]

what

[proud zealot]

never mind

[worldly yew]

He meant 10!/5!

[steep storm]

that still isnt correct

[proud zealot]

this is the answer i get it

my question is why mine is not correct

[steep storm]

could you write yours in latex

its very diff to interpret it in text

or just handwrite it and send a picture

[proud zealot]

i don't know how to use latex

[proud zealot]

is it that hard to understand this

[steep storm]

in your answer

you missed a 5! in the denominator

[proud zealot]

$\frac{10!}{5! } \times \frac{12!}{7!}$

[steep storm]

add a $

[elfin berry]

lordi

[steep storm]

there should be another 5! in either the denominator of 10! or 12!

[proud zealot]

thats wha i don't get

[steep storm]

only one of the 5! gets cancelled

or

[proud zealot]

i dont't understand the idea

not the math

[steep storm]

the idea is to choose 5 from 12, 5 from 10

now rearrange these in 5!

wayus

[proud zealot]

ok thank you very much

[steep storm]

no problem

!done

[marsh citrus]

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[desert prairie]

How do I answer this

[marsh citrus]

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[desert prairie]

.close

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[urban bobcat]

"a )For what values ​​of the constant k form the vectors a base for R^4?"

[urban bobcat]

how does he breakout k from the matrix here?

[remote rampart]

This is just using the definition of a determinant

This is a really good video going over the concept of a determinant
While this one shows you how to compute it

The idea is that for a set of vectors to form a basis, it must have a nonzero determinant. So by finding which values of k give you a zero determinant, you can deduce that the remaining values will form a basis

[urban bobcat]

hmm okay!

[remote rampart]

You can see also in the thumbnail of Professor Dave's video how he is "breaking out" one of the values. This is how your problem is doing it, but for k

[urban bobcat]

ohh yeah i see!

that makes sense!

[remote rampart]

I'm glad 🙂

[urban bobcat]

but why does he take this here can´t i just do k | 0 2 0, K 3 2, 1 1 1 |?

[remote rampart]

You have quite a bit of freedom in choosing how you compute the determinant. The way that it is written is easiest IMO since a lot of the ocmputation is multiplied by 0. You could do it that way and get the same answer, but the computation will be a bit more hairy

[urban bobcat]

oh okay i am not sure how to do it the way he does

but what do i do know that i have -K |0 2 0, K 3 2, 1 1 1| + 3 |0 K 0, K 3 2, 1 1 1| - 1| 0 K 2, K 3 3, 1 1 1|?

[remote rampart]

Now you take the 3x3 matrices and repeat the algorithm to get 2x2 matrices. The algorithm is recursive, and each step reduces the matrix down. Follow along with the professor Dave video. It might be easier to start practicing determinants on a smaller matrix

[urban bobcat]

ohh i see! yes okay i will

i dont quite understand what he does in the last step here

[remote rampart]

Ahh yes, so we \textit{define} the derivative of a 2x2 matrix as follows
$$
\begin{vmatrix}
a & b\
c & d
\end{vmatrix} = ad - bc
$$

So once you get to that step, you just use the definition of a 2x2 determinant to evaluate

[elfin berry]

curry

[urban bobcat]

yes i know that but where does the -2 come from?

2 * 2 - 3 * 0 is 4?

[remote rampart]

that.... that is a good question

hold on, let me check to make sure there isn't a mistake in that

[urban bobcat]

yes i am so confused

yes thank you!

[remote rampart]

Ohhh I see what they did

They combined a few simplification steps together

I'll explain in just a sec

[urban bobcat]

yes okay!

[remote rampart]

wait ...... so I'm not too sure how they did that step either. I'm going to assume that there was a mistake, and the missing parenthesis suggests to me that they were being a bit sloppy. I think for you, what is best is to ignore that last step and chalk it up to them making a mistake

[urban bobcat]

yes okay i agree!

[remote rampart]

yeah Looks like it was a rush job and the professor messed up the algebra a little. The determinant is still correct, however.

[urban bobcat]

yes he has done that a few times before also

okay i will try to calculate it and see if i get the same answer

[remote rampart]

This here shows you some of the steps. You can use it as a guide

But careful not to 100% rely on it as it does things slightly differently than you may have learned.

[urban bobcat]

yes okay i won´t, thank you!

[marsh citrus]

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[still temple]

this problem is literally impossible

$\sum_{k=1}^n \dfrac{k}{(k+1)!}$

[elfin berry]

Derivative

[remote rampart]

Hmm - I have some ideas but I'm not sure either

wait

[still temple]

its in a textbook chapter on combinatorics (chapter on permutations and combinations)

[remote rampart]

yeah that makes sense

So I'm thinking something like this
$$
\sum_{k=1}^n \frac{k}{(k+1)!} = \frac{1}{(n+1)!}\sum_{k=1}^n \frac{k(n + 1)!}{(k+1)!}
$$

[elfin berry]

krotkin

[still temple]

ok but form are you trying to get it in?

try writing it a k + 1 - 1

[still temple]

here

i think

since

its from combinatorics

you should try

turning it

into a sum of binomial

coefficients

$\binom{n}{k} = \dfrac{n!}{k!(n-k)!}$

yes

this form

try to get it into this form

[elfin berry]

Derivative

[still temple]

the general term of the summation

then it can get easier as you can do alot things

once you have the operator nCr

in the summation

ok

let me try

welll thats what ive been trying to do for the past few days 🤣

uh

[still temple]

thiis what i tried rn

$\sum_{k=1}^n \dfrac{n!}{n!(k+1)!}$

[elfin berry]

Derivative

[still temple]

im trying to get it in the form $\binom{n}{k+1}$

[elfin berry]

Derivative

[still temple]

so which means
$\sum_{k=1}^n \dfrac{n!(n-(k+1))!}{n!(k+1)!(n-(k+1))!}$

[elfin berry]

Derivative

[still temple]

now:
$\dfrac{1}{n!} \sum \binom{n}{k+1} \cdot \dfrac{1}{(n-k-1)!}$

[elfin berry]

Derivative

[still temple]

but now i dont know what to do

[cunning fiber]

There’s an easier way

[still temple]

what if you just had sigma 1/n!

[cunning fiber]

Note that $n=(n+1)-1$

[elfin berry]

Civil Service Pigeon

[still temple]

yes

thats what i said

[cunning fiber]

From here, ||the series telescopes||

[still temple]

yup

you know whats funny, the moment i looked at this a few days ago, i said telescoping series 🤣

lmao

i just didnt know what to do

just down write down the terms

[cunning fiber]

Oh I didn’t notice that mb

[still temple]

they cancel

[cunning fiber]

I just saw the barrage of Texit lol

I’ll leave you to it 👋

[still temple]

so i manipulate k?

[still temple]

write k as

(k+1) - 1

now simplify

the expression in the summation

\sum $\dfrac{k+1-1}{(k+1)!}$

[stoic slate]

Before simplifying u need go rewrite denominator

[elfin berry]

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

[still temple]

yes

[stoic slate]

As (k+1) * k!

[still temple]

now

simplify

ok let me see

finally i get

$\sum_{k=1}^n \dfrac{1}{k!} - \dfrac{1}{(k+1)k!}$

[elfin berry]

Derivative

[still temple]

and yes this is telescoping

but, why is this problem in a combinatorics chapter?????

so weird

[still temple]

cause maybe you can turn it into a sum of

binomial coefficients

then solve it using combinatorics

( which i won't recommend you can do it with simple summation or differentiation or integration if some constants are multiplied with the binomial coefficients

[still temple]

yes but i think i should learn it the combinatorics way

so i have the above equation

its telescoping, yes.

but can i convert telescoping into a sum of binomial coefficients?

no

idk

[still temple]

try co7nverting this one into that

ok

[marsh citrus]

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[gray maple]

Is this correct?

[marsh citrus]

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[rocky lark]

if two planes have two points in common, then their intersection is a straight line

[calm harbor]

no, not really

The two place can stick to each other

Their intersection would be a plane under that circumstance

[rocky lark]

wait

would the intersection be that red line?

[calm harbor]

yes

[rocky lark]

which would be a straight line

[calm harbor]

That happens when the two plane are separated

[rocky lark]

also in the book it is written that if it is in more dimensions (space) the intersection could only become a point. But I can't understand how

[rocky lark]

I wrote it wrong, they were two separate plans

if two distinct planes have a point in common, then their intersection is a straight line

if I considered 3 planes their intersection could be a point

[delicate aspen]

This is from this tutorial https://understandinglinearalgebra.org/sec-expect.html

[rocky lark]

What does the one on the right form?

ah.. they have no intersections in common

[delicate aspen]

Three lines, and no common point that is common to all three

[rocky lark]

👍

thank you so much @calm harbor @delicate aspen

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[still temple]

Hello

[still temple]

Can someone explain to me what is arccos and how it works?
Asking because i'm using this formula: $arccos(deltaZ / hypotenuse) * (frac{180 / pi})$ to calculate the Pitch angle..

[elfin berry]

Cut

[upper gust]

Its the Inverse function of the cosine

[still temple]

yea

[upper gust]

just like e and ln are inverses

[still temple]

The thing is, i don't know anything about it, my school isn't teaching it, none of my teachers know this

I'm studying all alone, so all this is new to me

[upper gust]

its like division and multiplication these are also inverse operations for non zero numbers

[nova totem]

arccos takes the ratio from a supposed right triangle and finds the angle that makes that ratio

[upper gust]

can try explaining it

maybe in general what inverse functions are

[still temple]

What is this ratio?

[nova totem]

so let's say cos(θ) = R. Then arccos(R) = θ, or some angle within its proper domain that kicks out

[nova totem]

adjacent/hypotenuse

cos

so arc(whatever) is like inverse(whatever)

arcsin is inverse sin
arccos is inverse cos
etc

[still temple]

So, with cos we trying to get adjacent?

And arccos we get opposite

[still temple]

Is what i said correct?

[nova totem]

No so

If you have a right triangle

Here Gimmie a sec

[delicate aspen]

https://www.math.hkust.edu.hk/~machiang/1013/Notes/arccos.gif

[nova totem]

Well do you even know what sine, cosine, and tangent do

https://youtu.be/Jsiy4TxgIME?si=90luiiETpxrXdber

[still temple]

no idea

[nova totem]

And then the inverse legit does the opposite

Do watch the video first

[still temple]

Okay

[still temple]

Can it be like

Go to the right too?

Ay, @nova totem, thank you very much for the video

May i save your DM so we can chat sometimes?

[nova totem]

Just ask for help in here

But Google literally has resources

[delicate aspen]

I think so, but this is how the unit circle is usually oriented

[still temple]

Okay thanks

[still temple]

Yeah, thanks !

[delicate aspen]

https://www.mathsisfun.com/algebra/trig-interactive-unit-circle.html

https://demoman.net/?a=trig-for-games

[still temple]

THIS IS SOO GOOD

Thanks

[delicate aspen]

You are welcome!

[still temple]

.close

[marsh citrus]

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[cursive jolt]

any suggestions

on how i can approach this

i was thinking of using bayes thereom

but im not quite sure on how

[marsh citrus]

@cursive jolt Has your question been resolved?

[cursive jolt]

😢

[limber imp]

Bayes theorem will probably be useful here. Have you drawn a probability tree yet?

Actually, you can substitute from the word problem itself

In Bayes' theoerm, based on the probelm, what is A and what is B

[cursive jolt]

A could be the result being positive

and B could the aliens having bleep blop condition

[limber imp]

Read the question again, what is the probability that you are asked to find?

[cursive jolt]

the probability that the alien has Bleep Bloop given that the test result is positive?

this is how i denoted my p's thus far

then i believe P(Bleep) would be 1/5

(the probability that an alien has Bleep Bloop)

P(Zp) would be 2/3 i think

[limber imp]

Yeah, so A would be probability of bleep bloop and B conditioned probability of being positive

[limber imp]

I agree yeah

[cursive jolt]

Oh switched, Yea makes sense

[limber imp]

What might be tricky here is finding the denominator

do you know what P(B) stands for in the equation you shared?

[cursive jolt]

the probability that the test result is positive

i think

the probabilities for the two conditions (Bleep Bloop and not Bleep Bloop)

[limber imp]

yeah the total probability of having a positive test

You will have to find this based on wether the alien has had a ZAP or a ZOP

[cursive jolt]

and the numerator?

P(P|B) (P(B)

would that just be

multiplied by P(B) which is 1/5

[limber imp]

P(bleep bloop|positive)P(positive)

[cursive jolt]

right

[cursive jolt]

as in

^?

[limber imp]

what is P?

oh i see

Yes, that looks right

[marsh citrus]

@cursive jolt Has your question been resolved?

[cursive jolt]

for P(P)

which is the probablity the test is positive

how would you approach this

[limber imp]

I haven't reread the question but were you given P(positive |bleep) or were you given P(positive|bleep, zap) and P(positive|bleep, zop)?

if it's the former then yeah, if it's the latter you need to go down each branch of the probability tree to get the total probability

P(positive) @cursive jolt

[cursive jolt]

oh the question is here @limber imp

mb shouldve said that b4

[limber imp]

your approach is correct, just check to use what you are given 👍

[cursive jolt]

thanks

this is what i ended up cooking

sort of just found each part of bayes thereom

and put it all together

[limber imp]

Your approach looks right, i didn't check the number but symbolically it looks correct

you found the marginal of probability of positive which is the long part

and yeah you put it together

[marsh citrus]

@cursive jolt Has your question been resolved?

[marsh citrus]

@cursive jolt Has your question been resolved?

[marsh citrus]

@cursive jolt Has your question been resolved?

[marsh citrus]

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[serene basalt]

I want to know about area under a graph by double integration

[serene basalt]

We do double integration by limits of x and y of a function f(x,y) in 3d to find out the volume between the graph and the limits of x and y right?

For area of a surface we use
dydx straight double integration and limits of surface we use f(x,y)=1 so it is a z=1 plane and we calculating volume between the surface and z=1 plane so why it is equal to its surface area please help

[marsh citrus]

@serene basalt Has your question been resolved?

[serene basalt]

<@&286206848099549185>

[amber birch]

https://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceArea.aspx

[amber birch]

You've seriously misinterpreted the surface area formula

The surface area is the double integral of 1 ds over the surface

You can imagine it as taking the arclength in the x-direction

And then summing all of these arc lengths as you move along the y-direction

[amber birch]

The formula here comes from the distance between (x, y, z) and (x + dx, x + dy, x + dz)

Which is approximately sqrt(dx^2 + dy^2 + dz^2)

Take sqrt(dx)^2 = dx out and you get sqrt(1 + (dy/dx)^2 + (dz/dx)^2)) dx

[amber birch]

https://math.libretexts.org/Bookshelves/Calculus/Calculus_3e_(Apex)/13%3A_Multiple_Integration/13.05%3A_Surface_Area
There's a better derivation here: it should indeed be dA and not dx

[serene basalt]

In my textbook let's say we have to calculate area of Circle x^2+y^2=1, I use limit x 0,1 and y -(√1-x^2) , √(1+x^2) and integrate by double integral

Here I say f(x,y) integrand is 1

[amber birch]

Oh that's not a surface area then

[serene basalt]

So basically it is the volume under circle and its image

[amber birch]

That's just regular integration with respect to dx dy (or dy dx)

[serene basalt]

Ya

[amber birch]

Nope, the circle is already an area

So you don't have to take the area under a multivariable surface

[serene basalt]

No like we use double integral we have to integrate some f(x,y) and we have limits of x and y like I have for circle so we get volume between the f(x,y) map and circle

[amber birch]

There's absolutely no volume in a circle

[serene basalt]

So here f(x,y)=1

[amber birch]

Ah whatever I think it's possible this way

But why would you

[serene basalt]

No we are in 3d because z=f(x,y)

Like volume of 1 height cylinder is same as 1 radius circle surface area

[amber birch]

Fair enough, okay so in the formula

The partial derivative of 1 w.r.t x is 0

The partial derivative of 1 w.r.t y is 0

[serene basalt]

Ok

[amber birch]

So we have sqrt(1 + (fx)^2 + (fy)^2) dA = 1 dA

It's literally the same double integral

Okay yeah so that's kind of nice it simplifies down like that

[serene basalt]

You not getting me

[serene basalt]

In my question f(x,y)=1 so it's a surface z=1, in 3d and by double integration we get volume of a cylinder which is 1 height,1 radius and its same as surface area of circle

[amber birch]

Yes I understand you now

If you take the area under that 3D surface, the integral becomes the same as if you use regular double integration

[serene basalt]

Ya

I not read it anywhere is it true

If again we take a sqaure like length 2 and width 2 so double integral taking f(x,y) =1,is give us a cuboid of height 1 ,now the volume of cuboid is 4, and surface area of square is 4 ,my mind is blown right now

It is like we increase the height of any surface by 1 and calculate volume it is same as curves surface area

@amber birch

[amber birch]

Yeah that's what the formula tells you, it's the same if your surface is a constant

[serene basalt]

Awesome nice talking to you have a good day

. close

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[marsh citrus]

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[amber birch]

npnp

[marsh citrus]

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[upbeat sable]

What have you tried

[zenith sail]

well i have tried to just do the root and then cancel x

[upbeat sable]

What do you mean by "I have tried to just do the root"

You have to look at what happens when you differentiate x^2+1

You get 2x

[zenith sail]

ye already... oh

[upbeat sable]

The numerator is almost 2x

You just need to multiply and divide by 2

[zenith sail]

so I need to do by subtitution and the i will get it?

[upbeat sable]

Yes

[zenith sail]

wait but that would be a chain derivative

[upbeat sable]

I was trying to tell you what substitution you have to make

And in order to see that, differentiate x^2+1

[spice plover]

u=x^2+1
du=2xdx

or

$u=\sqrt{x^2+1}$

$du=\frac{x}{\sqrt{x^2+1}}dx$

So the integral becomes,

$\int 1 du$

$u+C$

$\sqrt{x^2+1}+C$

[elfin berry]

The Prophet Of The Damned

[zenith sail]

wait why 1?

the derivative I already did it

but never understood why the 1?

oh

thx

.close

[marsh citrus]

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[marsh citrus]

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[copper ice]

how do i find the sum of this series? i just dont know what to do with (-1)^k, if we wouldnt have had it, it would be just geometric series which is easy to calculate

[still temple]

separation

even and odd

wait nvk

[twilit arrow]

((-1)^k\left(\frac{1}{2}\right)^k = \left(\frac{-1}{2}\right)^k)

[elfin berry]

maximo¹

[novel juniper]

good, you got the key idea already

try doing that

oh

[stoic slate]

He is not the one with the question

[novel juniper]

just realised

sorry

[copper ice]

🤦‍♂️ i didnt see the obvious

thank you

.close

[marsh citrus]

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[marsh citrus]

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[restive field]

Hi, just wondering does anybody know how to do question 5(a)?

[restive field]

?

Anybody?

[icy scarab]

You can substitute the values of x given, you will get two linear equations

[restive field]

Ok

[icy scarab]

So for f(2) for example, since f(x) = mx + c
f(2) = 2m + c
And it's given that f(2) = 7

Therefore
2m + c = 7

[restive field]

Ya

[icy scarab]

Similarly do for f(4) and solve the two equations

[restive field]

I've done it like 2 equation

But how to solve?

[icy scarab]

What have you got?

As in, what equations have you got?

[icy scarab]

Yes, that is correct so far

[restive field]

Ok

[icy scarab]

So while solving a linear equation with two variables, we try to eliminate one of them.
So that we can get the value of one variable, and then with that we get the other one

[restive field]

Ok

So how do I do it ah?

[icy scarab]

So you can subtract the two equations
I will send a picture, I don't know latex to type it in discord

[restive field]

Ok

I got it

I think so

Ya

I got it

M=-4

Now i need to find c

C=15

I got it already

Thank you soo much ya!

[icy scarab]

No probs

Welcome

[restive field]

Yo bro

May I ask how do I do 5(c)?

I don't really get the c

[icy scarab]

Uhh sorry, I don't know how to do that question either
Maybe another helper can help

[restive field]

Ok

Tks for helping yo

Appreciate it

[icy scarab]

Glad to help

[restive field]

Maybe I'll ask my school teacher

[icy scarab]

Okay

[marsh citrus]

@restive field Has your question been resolved?

[marsh citrus]

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[pastel ridge]

.open

.reopen

.close

[rocky lava]

hello

[rocky lava]

yo so i need some of my work to be checked

i’ll send the work when someone is online cuz i have a feeling most of these are wrong tbh

so lmk when ur online

<@&286206848099549185>

i’ll just send it afchalky

for question 4 would the domain be (-infinity,infinity) and the range (4, infinity), the vertex (-2,4), and the axis of symmetry x=2, and the function increased on the right and decreasing on the left

also would it be as y increases x increases when x < -2

i’ll just post this for now and then when someone responds i’ll post the rest

[past frigate]

Everything seems right except the axis of symmetry

[marsh citrus]

@rocky lava Has your question been resolved?

[rocky lava]

oh shoot sorry for the late reply

r u still there

and also what’s wrong with it

[rocky lava]

wait

sorry i meant x=4

@past frigate

idk why i typed 2

[rocky lava]

this is true also right?

[past frigate]

4?

[rocky lava]

omg wait

-2

[past frigate]

Lol

It's right now

[rocky lava]

i straight up had -2 on my paper lol

idk why i’m saying these i’m getting confused

[rocky lava]

is this true tho?

[past frigate]

As y increases x should decrease, no?

For x<-2

[rocky lava]

oh right

[past frigate]

The y value at x=-3 is big, but the y value at x=-4 is bigger

-4<-3

[rocky lava]

true

and then vice versa it would be when x>-2, y increases as x increases

[past frigate]

Another(better) way to say that would be that y decreases in (-inf,-2)

[past frigate]

Yeah

[rocky lava]

aight ima send the best question

*next

for 8, i put the vertex is the same and the axis of symmetry is the same also, and the transformation is vertical compression by a factor of 0.75

[marsh citrus]

Channel closed

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[rocky lava]

what

.reopen

[marsh citrus]

✅

[rocky lava]

ok look up here to see it

[past frigate]

The same as what again?

[rocky lava]

both graphs have the same vertex and axis of symmetry

vertex is (0,0) and the axis of symmetry is x=0

[past frigate]

y=x² and y=0.75x² you mean?

[rocky lava]

and then i think it’s vertical compression

[rocky lava]

yes

[past frigate]

Kk

Yes that's right then

[past frigate]

One sec

[rocky lava]

alr

ok

[past frigate]

Ok yeah that's right too

Coz 0.75 lies between 0 and 1

[rocky lava]

yeah

lemme send the next question now

[past frigate]

[rocky lava]

ok so for 12 i got the axis of symmetry and vertex to be the same, and it is a vertical compression by -2/3 and reflection in the x axis

[past frigate]

👍

[rocky lava]

14 i got the axis of symmetry and vertex to be the same again, and it’s. a horizontal stretch by -4

[past frigate]

It would be -4 if the function would be -4x²

In this case it's (-4x)²

See the difference?

[rocky lava]

yeah

wait so what would it be then

[past frigate]

What is (-4x)²

[rocky lava]

i forgot to say this is by -2/3 also

[rocky lava]

16x

[past frigate]

16x²**

[rocky lava]

ohh right

so it would horizontal stretch by 16?

[past frigate]

Why horizontal?

The function is 16x²

It would be vertical itself right?

[rocky lava]

oh right

yea

so vertical stretch

right

[past frigate]

It's similar to the last few so I'm sure you'll be able to answer this on your own

[rocky lava]

oh would it be compression then

[past frigate]

No it'll be a stretch itself, you're right

[rocky lava]

oh

[past frigate]

I said similar coz

[rocky lava]

oh ok

so vertical stretch by 16

[past frigate]

In the last few we multiplied with a constant too

[rocky lava]

yea

[past frigate]

That's why

Yeah

[rocky lava]

aight

for graph a would it be a>1 and then graph b it’s |a|<1

[past frigate]

In a, do you think it's a vertical stretch or compression?

[rocky lava]

compression

[past frigate]

Look at the blue curve wrt y=x²

It's more compressed horizontally

[rocky lava]

oh yea true

[past frigate]

So vertical stretch

And when you multiply with a constant>1 then it gets compressed wrt the original one

Try it out on desmos or something

[rocky lava]

ohh

ok

so it should be