math-help

[left knoll]

for part d it is saying we need to make it so all 9 people were children

i understand it but the wording of the question is strange

unless 9th child age is negligable

i shall take the L nonetheless

[ember kernel]

From what I see in the solution, I think you teacher meant to ask what the means and standard devations would have been if all 9 people were children. Therefore, you rely on the data from the 8 children to get these (since the means and s.d. would remain approximately the same)

But I agree with you the wording was awkward and unclear

Although it makes sense

[solemn sphinx]

https://youtu.be/cUzklzVXJwo
Love this video on how imaginary videos were invented by Veritasium, if anyone else like such videos as well✨

[maiden prism]

Hi bros, what is the dimension of null of tranpose matrix in this question? will it be the same with dim(null(A))?

[left knoll]

True I need to sharpen up on ps questions

[cinder yacht]

r(A) + dim(null(A)) = 7
r(At) + dim(null(At)) = 4
r(A) = r(At)

those are the equations u need

idk what row(A) means though

but I'd assume that dim(row(A)) = r(A)

anyway dim(null(A transpose)) = 4 - r(A transpose) = 4-r(A)

[maiden prism]

so, r(At) is just dim(row(A)) right?

[cinder yacht]

I guess so yea

[left knoll]

Can someone simplify this?
9x - 7i > 3(3x - 7u)

[ember kernel]

From

9X - 7i > 3(3x -7u)

You can multiply the right side by the 3 outside the brackets. That gives 9x - 7i > 9x - 21u

From there, you can substract 9x from both sides to make the "9x" terms disappear. That gives 9x - 7i - 9x > 9x - 21u - 9x
which is equivalent to -7i > -21u

Now, you see that both terms are multiples of 7 and negative. Let's simplify them by dividing both by -7

But be careful ! Multiplying or dividing by negative numbers in inequalities result in a change of the sign's orientation

(That's only natural : we know that 2 > 1. If we happened to multiply both sides by -1 without changing the sign's direction, we'd end up with -2 > -1, which is wrong)

Therefore, let's divide both sides of "-7i > -21u" by -7, and we get i **<**3u

And that's it :)

[maiden prism]

Thanks bro

[left knoll]

Thanks you so much for the explanation

[glacial plaza]

hello, can anyone give me some tips on how to change z to be the subject of the equation? (z=...)

[cinder yacht]

I dont think you can

is this equation for every (x, y, z) in R3?

[glacial plaza]

yes...

maybe i'll find another way to do it then, thanks anyway :3

[cinder yacht]

if you do d/dz(..) 3 times

it gives that -19y(x-y) exp(z) = 0 so y=0 or x=y, so you can forget thzt term -19y(x-y) exp(z) the whole time AND keep in mind that either y=0 or x=y for every solution x, y, z

im not sure this argument works cause I've never solved stuff like this before... for example idk if we're allowed to even derivate ^^ maybe a solution would be an isolated point in space (and u can't derivate on an isolated point, u can only do so in a ball)

anyway sorry I gave more questions than answers xD

[glacial plaza]

hooo okok

thanks a lot!!

actually the concepts in pure maths are still a bit hard for me to grasp as i've never done this before this year... so im still trying to understand the idea of proofs and stuff haha

[atomic dagger]

Hey! The expression circled in red is equal to h. I’m trying to isolate v_0 but I can’t find the final result 😭 could someone help me develop the calculation ?

[ember kernel]

Start off by simplfying the two v0's in the last term, you get :
h = A/v0² + B

With A and B two expressions that don't depend on v0 (identify them in the original equation, it'd take too long to express them here)

Then, multiply all terms by v0², and you get h v0² = A + B v0² which is equivalent to (h-B) v0² = A

Now, just divide both sides by (h-b) and take their square roots to end up with

v0 = + sqrt(A / (h-B))
or v0 = - sqrt(A / (h-B))

[late axle]

hey i need your help!

how to decompose this : x^4 + 1

in simple elements (idk how to say it in english, it's the literal traduction of french)

i m trying with (x2-i)(x2+i)

that's not satisfying

[ember kernel]

Factorise each of the parenthesis once more using De Moivre

x² = i if x = (1+i) / sqrt(2) or if x = -(1+i) / sqrt(2)

[ember kernel]

Do the same with x² = -i

And combine everything to get the answer

[cinder yacht]

looking for the 4 solutions of x^4+1=0
let's think of x as : x = r * exp(ia) with r>=0 and a€[0,2pi[
then the equation is:
r^4 * exp(4ia) = - 1
<=> r^4 * exp(4ia) = 1 * exp(i pi)
<=> r=1 and 4a = pi + 2 * k * pi with k€Z
<=> r = 1 and a = (2k+1) pi/4 with k€Z
we wanted a in [0,2pi[ so the only values of k that work are k=0, k=1, k=2, k=3
so
<=> r=1 and (a = pi/4 or a=3pi/4 or a=5pi/4 or a=7pi/4)
so there are 4 solutions :
x=exp(i pi/4) = (1+i)/sqrt(2),
x =..
...
...

maybe longer than the other method, but works with any x^n + 1 = 0

like u could solve x^1000000 + 1 = 0 with this

[cinder yacht]

do u have an illustration for this?

are u forced to move each turn?

or can u stay in place

okay so grid should be colored in white and black like in chess

if the starting position is in black, then u know
-if T is even, then final position is black
-if T is odd, final position is white

cause each turn u change color

and other than that, if we imagine the grid doesn't have borders, the cells where u can be after T turns are :
the cells of the right color (depending on color of initial position and parity of T) such that |X-X0|+|Y-Y0| <= T

btw what's the exact question

what does the program exactly do

point out how many cells are possible?

I can find a cute formula when X0+T<W, X0-T>=0, Y0+T<H, Y0-T>=0

but other than that (AKA when it's possible to reach the borders after T turns) it would be real ugly

I think u can code something in O(1) still

but

a bit of a hassle

like gotta check which of the 4 borders can be reached

it's not so fast that I'd do it in msgs, I think it'd take me like 30 mins (or maybe 1h idk), sorry :(

if u do care abt finding it, the only quick tip I can give is the parity of T is important (to see if u should count white cells, or black cells)

[left knoll]

very embarrassing to ask this but....

should I find the surface area of the cone then add to area of sphere?

[cinder yacht]

I dont understand the question either ngl, normally the 3 poles (or whatever) don't have much surface area because they're thin (they don't even give u the radius anyway), so I'd assume the total surfzce area is just surface of the sphère ?

[glacial dust]

Is the base a 3d shape? cuz if not then ya it is very thin to a have a surface area

I solved it but I assumed the “3 legs” are like forming a tripod position if thats the case then theres a viable solution to it

[cinder ibex]

Hey, im studying logarithm properties and im having some trouble understanding why they are like this basically im looking for the demonstrations of this properties but i cant seem to find them on the internet maybe it is obvious but anyway if anyone can help i would be grateful, the properties are a^x=e^(x*ln a) and a^(log a (x) )=x i have tried some examples and i know these work but i want to know why

[left knoll]

both are quite obvious just by looking at definition of logarithm: log a (b) = c and a^c=b.Therefore log a (x) =c and a^c = x so a^(log a (x) )=(a^c)=x. e^(x*ln a)= e^(ln a^x) =a^x because ln (a^x) = log e (a^x) = c such that e^c=a^x

[cinder ibex]

Thank you!!

[cinder ibex]

im still kinda confused on the last one tho, maybe its just me complicating this but like i got what u wrote after "because ln (a^x)..." but getting here "e^(x*ln a)= e^(ln a^x) =a^x" is still not making sense in my mind for some reason

[left knoll]

do you understand part: x*ln a = ln a^x

[cinder ibex]

yes, let me just take a photo of what i wrote on my notebook that im not understanding

[left knoll]

k

[cinder ibex]

this is what i got but i need to get to c= a^x right?

[left knoll]

i don't get it. Definition of logarithm states that log g (b) = c and g^c=b. In your example a^x = e^(log e a^x), so now define e=g, b=a^x and substitute it: log e a^x = log g (b) = c and g^c=b => e^c=a^x and c is log e (a^x), this is the same situation as in example a^(log a (x) )=x

[cinder ibex]

that does make sense, substituting makes it easier to understand, thank you!

[cinder yacht]

u mean <=>, not "and", right?

[left knoll]

how do i factor 4x^2 - 25x -21

[ember kernel]

Quadratic formula

[left knoll]

oh i dont think ive learned that yet but ill go learn it rn

thanks

[ember kernel]

Then that's probably not how you're supposed to solve it

The thing is, unless it's actually +21 (not -21), there's no direct way to find the roots

[left knoll]

the question asked that if it was completely factored what would 1 of the factors be, and the answer was 4x + 3 for that 1 factor, im not really sure how to get there hmm

[ember kernel]

Looks like you have to go through trial and error

One way that generally works is trying plugging in the dividers of the independent term

Eg, here, you could try to plug in 1, 3, 7 and 21

[left knoll]

i dont think its possible to solve through decomposition so yea maybe ill just do trial and error..

[ember kernel]

There you could realise that 7 is a root

And calculate that the other root is -3/4

Oh I think it's that method you're supposed to use

[ember kernel]

I'd forgotten about that tip sry :D

[left knoll]

what does root mean again...mb

[left knoll]

nono tysm for taking ur time to help me

[ember kernel]

A root is a value for x such that the polynomial is 0

Example with your polynomial 4x² - 25x - 21

If you replace x by 7, you get 4 * 7² - 25*7 - 21 = 196 - 175 - 21 = 0. Therefore, x = 7 is said to be a root of your polynomial

And when you factorise a polynomial, the different terms are all associated with one root each

Say you have an arbitrary polynomial ax² + bx + c (don't pay too much attention to the values of a,b,c) which has x=m and x=n as roots,
then you can factorise that polynomial that way : ax² + bx + c = (x-m)*(x-n)

If you want to convince yourself it's true, then you could prove that (x+3)(x-2) = x² + x - 6 does have -3 and 2 as its only roots

[queen thistle]

Hello. Sorry for taking your time but can someone help me with this question, please. I do not understand what the second part of the question is asking... (thank you in advance)

[ember kernel]

It would help us if you could also send screenshots of the Table of Random Digits as well, if you can

Or at least the region between lines 100 and approx 130

[queen thistle]

my apologies, here you go

[ember kernel]

Oh wow

I'll be honest, I'm a bit confused as to how I'm supposed to read them

[queen thistle]

for this you have to read the number from left to right by digits of 3. so for example starting from line 116 it would go like 843|070|501 and so on

I already completed the first part (a) that way but it seems like part 2 (b) asks for something else. I am not sure what kind of proportion it is talking about

for part (a) I have 6 numbers for both categories (total: 12)

but here it says that the owner wants "the number of stores in the sample to be proportional to the number of stores in each category." <--- I am not sure what they mean by that...

[ember kernel]

Honestly I'm intrigued by this question

I don't know what your teacher meant either

[queen thistle]

haha so am I

[ember kernel]

I'm sorry :/

[queen thistle]

no no, thank you for trying to help. I truly appreciate it :3

[left knoll]

for this statement

just want to see

if u say

wait

nvm

can someone explain the logic to part c

dw i think i get it

null hypothesis accepted if there is no reduction

[left knoll]

however, shouldn't this say if it is =>7?

[left knoll]

yo guys i have returned

can someone help me find the critical values

i keep getting 18 as my upper value

[glossy latch]

If you look at the individual numbers, you see that 3/4 of the numbers are in category 1 whereas the last 4th is in category 2

if you were to take a sample of 12 stores, 9 would be in category 1 and the other three in category 2

[queen thistle]

May you show me how u found this please (if you dont mind)

[glossy latch]

oh it's based on the numbers they gave

they have a total set of 200 numbers

any random number chosen between 1 to 200 can be in two categroies

if the chosen number is below 151, its in category one. If it's greater than that, it's category 2

[queen thistle]

Which 200 numbers 😭

Sorry for asking

[glossy latch]

001-151 = jewelry. 151-200 is food. Overall, 200 numbers

[glossy latch]

np

[queen thistle]

Oohhhh okay

Starting to make sense now 😭

[glossy latch]

since jewelry takes 150/200 of those numbers, if you were to choose a random store, you'd probably get a jewelry store 3/4 times

[queen thistle]

Ohhh okay got it 👌

[glossy latch]

so if we apply that to 12 stores *3/4 means about 9 stores will have jewelry

[queen thistle]

So i just need to find the first 9 numbers for cat 1 and 3 for cat2

[glossy latch]

tbh, the 116 line is a bit weird. are u sure it's every three numbers?

[queen thistle]

I believe so because i need to get numbers with three digits from 001-150 and 151-200

I was also confused about that too tbh

[glossy latch]

tru but you end up with remainders and can only go for 36 of those numbers

they give you 40 total tho

[queen thistle]

Oh no no they are continuous

[glossy latch]

??

[queen thistle]

So you keep counting digits of three

Its does not have breaks

[glossy latch]

ohh, but then why create separations every 5 numbers?

[queen thistle]

Or stops

Ots jist how it is 🥲

[glossy latch]

and then there's also the issue of some of those three numbers going beyond 200

[queen thistle]

I was also confused when i first completed my lessons

[queen thistle]

Yes so you skip those

[glossy latch]

oh, so we only accept numbers below 200?

[queen thistle]

Yes

[glossy latch]

so instead, you'd get a sequence like 70,131, etc?

[queen thistle]

So that means i need to get the first 9 3-digit numbers for category 1 and 3 3-digit numbers

[queen thistle]

Yess

[glossy latch]

basically half of the sequence is skipped then. Man ur hw is confusing lol

u have to do this kind of stuff everyday?

[queen thistle]

Yes 😭 and i dont even have a book to help me

Its just reading, you are given 2 examples and then do it on your own typa thing

[glossy latch]

i still think there's something that we've misunderstood bc the format of the thing is a bit too strange

[queen thistle]

So for example if you were to start counting from line 116 it would look like this 843, 070,

And so on

[glossy latch]

right, but 843 is greater than 200 so we discount that

[queen thistle]

Correct

[glossy latch]

if we go like that, we end up skipping a lot of numbers

[queen thistle]

Exactly 😭 counting and double checking is important

Its super time consuming

Thank you for helping me understand how i should start with the 3/4 and 1/4

I cant thank you enough 🙏

[glossy latch]

but alr then, let's see. the following 12 valid sequences would be 070,192,143,043,048, and i lost my patience for the rest lol. would u be able to fact check the current numbers?

[glossy latch]

oh yeah np. i have to go now but gl

[queen thistle]

Thank youuu i will update you on the grade i get 😭🙏

[queen thistle]

BAHAHAHA I FELT THAT 💀

[queen thistle]

Byeee thank youuu

[atomic dagger]

thank you!

is there a faster way to check how many solutions a system of equations has than to transform the matrix into an echelon form?

[cinder yacht]

if it's n equations n unknown variables, then u could compute the determinant and if it's not 0 then u know there's only 1 solution

if it's 0 then either 0 solution or infinite solutions, but can't know which from determinant

that's the only "other way" I can remember of checking that

[atomic dagger]

thank you!

[crisp lodge]

please help!

[cyan night]

theres this concept called the conjugate and its really useful especially in rationalizing this with square roots

[ember kernel]

The conjugate of a + b sqrt(c)
is a - b sqrt(c)

[cyan night]

yeah it leverages the algebraic formula (a-b)(a+b) = a^2 - b^2, where the square root is "get rid of" or "rationalized" because it is squared

[ember kernel]

And as @cyan night said, multiplying a + b sqrt(c) to its conjugate gives you a rational result

[cinder yacht]

idk the conjugate of a-b-c

[left knoll]

yeah i took it as a tripod shape too, what answer did u get?

[ember kernel]

For clarity let's define
x := sqrt(m)
y := sqrt(n)

We want the conjugate of a + bx + cy
Let's work in two steps :

1. We make lone y's disappear from the initial expression (a+bx+cy) by multiplying it to its conjugate relative to y
2. Then, we make lone x's disappear from the product we get in step 1 by multiplying it to its conjugate relative to x

STEP 1
(a+bx+cy)(a+bx-cy) = ((a+bx)²-c²y²)
= ((a²+b²x²-c²y²) + 2abx)

STEP 2
((a²+b²x²-c²y²) + 2abx) * ((a²+b²x²-c²y²) - 2abx)
= ((a²+b²x²-c²y²)² - 4a²b²x²)

And just like that, all x's and y's are squared (thus rationalised)
That means the conjugate of (a+bx+cy) is the product of the two terms we multiplied it by

In other words, the conjugate of
a + bx + cy
is (a+bx-cy) * ((a²+b²x²-c²y²) - 2abx)

And the final, rationalised expression is
((a²+b²x²-c²y²)² - 4a²b²x²)

@crisp lodge if you still need the answer

Or want the method you'll need to apply next time

(Correct me if I made a mistake somewhere pls, the calculations were done by head)

[cinder yacht]

seems good to me

[ember kernel]

⬆️ There's your answer. The calculations are pretty dense and confusing but dw the approach is easy to pick up ; we just get rid of one square root at a time. Finally, don't forget to multiply the numerator by the same thing you do the denominator, to keep the global expression equal at all time

[atomic dagger]

How do I get to this level 😂

[crisp lodge]

thanks a lot

[ember kernel]

Think n train

I'm familiar with maths and that method only seemed natural 🤷‍♂️

[cinder yacht]

I think understanding what it means to conjugate (which is : raise the square roots we don't like to the power of 2), and doing things step by step calmly and logically (like here, after trying some incomplete conjugation at first, he looked at "what is still problematic here? and what is already solved?" which made him split the quantity into 2 sums : one where all square roots are raised to the power of 2, and one where they're not, then ask himself "ok i only got ONE term left that needs to be raised to the power of 2, how can we fix this? oh wait this is exactly what usual conjugation does! let's do it")

and the step by step calm & logic gets better with time and training

for me when I have smth hard to solve, one of the things that help me most solve it is to clarify exactly what the goal is (instead of letting it be blurry and ambiguous in my head)

and potentially look for some condition that is sufficient to reach the goal (for example if the question is "prove that this f * g is continuous" i could tell myself "ok a sufficient condition for that is that f and g are both continuous, let's try to prove that") shit like this

[ember kernel]

I totally agree with everything above. Solving most maths problem is a clever mix of these :
- Clarify what your goal is (squaring all the squareroots in a given sum of terms)
- List everything relevant you have with you to get started : formulas that could help you reach the goal, concepts, theorems, ... (the concept of conjugates & the formula (a+b)(a-b) = a²-b² which squares all parameters involved, the formula (a+b)² = a²+2ab+b²)
- Break the big goal down into smaller, successive or parallel, tasks that could help you make progress (I squared one square root at a time, because I didn't know any formula that could take care of two square roots at once)

Past highschool, most maths problems will require more than one line of calculations, so being able to mentally map a problem and make links between concepts or formulas are two crucial skills needed to be "good at maths"

And it all comes by :

1. Understanding the stuff you learn in maths classes
2. Training to get used to working with it

[karmic whale]

What is the correct value for line FG? The figure is not drawn to scale. Using sine law and cosine law provides you with 2 possible angles: supplementary angles. Idk which one is correct when both angles satisfy the condition wherein the longest side in a triangle is always opposite the largest angle, and the shortest side is opposite the smallest angle.

[pine dust]

consider checking out ambigous case

[ember kernel]

What I would do is solve for x and y in this system of equations

@karmic whale

Then, just compute FG = sqrt(x²+y²)

[left knoll]

yo

help me out with this

part c

[atomic dagger]

Hi there! I’m a little confused with this expression. I’m trying to prove Pascal’s identity but I’m not sure how we got the underlined part here? I’m also not sure how to work with factorials, like can I substract them , add them up…?

[ember kernel]

The factorial notation n! is just defined as n * (n-1) * (n-2) * ... * 3 * 2 * 1

Example : 4! = 4 * 3 * 2 = 24

You were confused about whether you could add or substract factorials, so let's find out by ourselves !

2! = 2
3! = 3 * 2
Therefore 3! + 2! = 3 * 2 + 2 = 4 * 2 = 8

Whereas 5! = 5 * 4 * 3 * 2 = 5 * 4! (which we computed earlier) = 5 * 24 = 120

Clearly, you see that factorials can't be summed : 3! + 2! is not 5!

However, there's an interesting property linked to division

n! / k! = n * (n-1) * ... * (k+2) * (k+1)

(if n > k)
Just try it with 8! / 4! if you want to convince yourself it's true

[ember kernel]

About the "underlined part here", you haven't sent anything yet so I can't help you :D

[karmic whale]

@ember kernel It's a quadratic equation, therefore it has 2 roots. I've tried Sine Law, Cosine Law, and the one you recommend, but I still got 2 possible answers. Slr, I didn't notice your reply since I was solving it the whole day and my head hurts. Jkkk hahaha

[atomic dagger]

lol sorry about that 😆 the underlined part was supposed to be the expression we obtain at the right end of the screen. we d o have n!/k! but I don't think we are transforming it as n * (n-1) * ... * (k+1) here ?

[hardy kite]

i keep messing up with applying log laws to find x with these types of questions :/

[cyan night]

theres actually also a very underrated logarithmic property that i know

[ember kernel]

On the right end of the screen, we factor out the n! /k!(n+1-k)! which leaves us with a bracket (k + (n+1-k)) which is just (n+1)

[cyan night]

this property is very underrated

[ember kernel]

Therefore we end up with (n+1) * n! / k!(n+1-k)!
Since n! * (n+1) = (n+1)! we simplify it to
(n+1)! / k!(n+1-k)!
And that's how you get to last line :D

(Regarding the n! / k! = n(n-1)...(k+1) thing, you're only meant to use it with actual natural numbers, not parameters. I just said it because you hadn't sent the screenshot yet, so I just gave out a few info in hope one of them would be the one you struggle with. Now that I have access to the screenshot, I realise it wasn't necessary)

[hardy kite]

hmmm thing is i understand the log laws but i'm struggling to apply them correctly to solve for x if that makes sense, currently trying to complete a) again :')

[cyan night]

it also works the other way around fyi

just be cautious about the base

they must have the same base to apply the property

[ember kernel]

log3(32) + log3(2) = log3(32*2)

From 2 log3(x) = log3(64)
get log3(x) = log3(64) / 2
by dividing both sides by 2

log3(64)/2 = log3(64^(1/2)) = log3(sqrt(64)) = log3(8)

[cyan night]

actually i think a better way to do it is 2 log 3(x) = log 3 (x^2)

but yeah

[ember kernel]

Therefore log3(x) = log3(8) and x = 8

[cyan night]

yes

[hardy kite]

OHHH okay I think I see what you're saying I'm gonna try writing it out to see if it still makes sense, i just keep getting confused cause the way I've been taught is to first move all log terms to one side and then use log laws to convert the terms into a single log and go on from there

hope I'm making sense and all :')

[hardy kite]

hmmm gonna try writing this out too

[cyan night]

:^)

[ember kernel]

I don't think so. The reason is that then you'd get the equation x² = 64
And umi not being familiar enough with logs yet, she might say that the answers are x = 8 and x = -8, forgetting that x cannot be negative. That can't happen if you never square x

[ember kernel]

You could also do that, get log(...) = 0, which means ... = 1 and solve for x from there. You'd get the same answer

But by saying that log(x) = log(y) is equivalent to x = y (because logarithms are injective) you're skipping a step

Do note it's only possible because the log is injective tho

[cyan night]

yeah i agree

[hardy kite]

I'm currently on this step and currently got log_3 x = log_3 (64)/2 but just to clarify how did you get from "64^(1/2)" to "sqrt(64)"?

i keep forgetting that one basic rule for square rooting

[ember kernel]

Have you considered the possibility of actually having two solutions ? ^^
Because here that's the case

[hardy kite]

yeahhh that's how I've been taught as in getting it to be log(...) = 0 but I just kept messing up with applying the log laws correctly

[ember kernel]

N^(1/2) is literally defined as sqrt(N)

[hardy kite]

OHHHH RIGHT

oml I keep forgetting that

[ember kernel]

That's kind of important to remember with logs

'cause you'll be dealing with fractional exponents a lot

For obvious reasons that you're seeing right now ^^

[hardy kite]

rightt I just finished solving it and got x = 8 from how you've explained it and I'll definitely keep that in mind

it makes sense with the way you've explained it but I just kept messing up with where to apply log laws when I was solving it as log(...) = 0 :/

[ember kernel]

Both methods are valid and very similar

But mine will more make use of loga + logb, where as your teacher's will more often use loga - logb, and highschool students tend to like multiplication more than division x)

[hardy kite]

lmao I can definitely agree with this but when I tried solving it as log(...) = 0, I first got 2log_3 (x) - log_3(32/2) = 0 and when I tried it again I got log_3(64) - 2log_3(x) = 0 then just gave up cause I didn't know how to continue :/

[ember kernel]

Divide both sides by 2 to get log3(64^(1/2)) - log3(x) = 0
Therefore log3(8/x) = 0
Therefore 8/x = 1 and x = 8

[ember kernel]

And you messed up the first one

[cinder yacht]

i was taught to just write every loga(b) as ln(b) /ln(a)

and forget the log_a stuff

[random flame]

Can someone help me with part iii? The working for the previous parts are there ady. Ping me if you do and thanks in advance!!

[cyan night]

sorry do u mind taking a closer picture

its ok if u cant capture ur working i can just see the worksheet

[random flame]

Ah aight

This better?

[cyan night]

yes yes thank you

wait but

how is pqx and srx congruent

they look similar

@random flame in the triangle SRX, how is its area calculated?

or actually, in any triangle?

[random flame]

It's given that the area is 20cm^2

It's in the question

[cyan night]

yes

i know

but what formula is used

[random flame]

I think I got the congruent thing wrong

[cyan night]

to calculate the area of the triangle?

[cyan night]

they are similar because its a cyclic quadrilateral (a quadrilateral inscribed in a circle)

[random flame]

Ah okay

[random flame]

Is it smtg abt cosine and sine rule thing?

[cyan night]

yes but for iii just use the formula for triangle area

nono

think simple

its base * height/2 right?

[random flame]

This is the thing that I can't seem to do 💀

[random flame]

Yeah

[cyan night]

area = base*heigh/2

and you know the base and the area..

so the height is..

[random flame]

... Okay how did I get so complex

Lmao thankiew

[cyan night]

😄

o np

[glacial dust]

can someone help me with this the answer is 6.77 but am getting 5.72

i dontget this 55 v

is this the voltage drop of the unkown resistor?

oh ok i got the answer now but still this 55v doesent make sense to me

[grim goblet]

the answer is 69

🤓

[glacial dust]

uhm its not

thanks for trying to help but i got it now 😄

[maiden prism]

Hi bros, how can I deal with the range |x| < 0.1 in this question? usually it's an estimation at a specific point and how can I extend to a range like this quest?

[karmic whale]

I was confused at first because my instructor got 2 values but he doubted and he thought that the correct answer was the one with the smallest value based on the figure, but I can already see that there are two possible solutions because when I tried to draw it on scale, the one triangle is actually part of the big triangle. Thank you for the explanation and your effort. I appreciate it. ^__^

[ember kernel]

You should teach your instructor to not solve from figures x)

(Or at least that a figure only represents one specific scenario)

[ember kernel]

Worst case scenario : x = 0.1 (since the error grows with x). Calculate the error for x = 0.1, and you'll be assured that the error for any x such that |x|<0.1 will be inferior to it

[cinder yacht]

if he wants an estimation that is a function of x, better use the integral rest of Taylor formula (and i guess approximate e^t as 1 in it)

[spring summit]

hey while reading an important concept and you have a lot to cover which is the most suitable method of learning?

[atomic dagger]

hi, i'm trying to prove this inequation but i'm not sure how to do it, could i get some help?

[limber aspen]

You might wanna compare (√1+X)² and (1+X/2)²

[atomic dagger]

i will do that, but why?

[limber aspen]

As far as I know, if a²<b² and a>0,b>0 then a<b

[ember kernel]

@limber aspen is right, square everything (the signs won't change)

That way, instead of comparing a polynomial to a square root, you'll be able to compare two polynomials, which is waaay easier

That way, from
1 <= sqrt(1+x) <= 1 + x/2
you get
1 <= 1 + x <= 1 + x + x²/4

[atomic dagger]

oh i see

[ember kernel]

The first inequality 1 <= 1 + x is obvious since x >= 0

[atomic dagger]

it does make sense! the square root felt too confusing to work with

[ember kernel]

And the second is also easy to find if you substract 1 + x from both sides and use the fact that x²/4 >= 0 is always true

[atomic dagger]

thank you! i understand now

[limber aspen]

How come

[atomic dagger]

thank you also @limber aspen !

[limber aspen]

Anytime !!

[ember kernel]

Oops, I got confused and mixed that up with something else. Thanks for pointing the mistake !

[limber aspen]

No problem, we all make mistakes

[glacial dust]

Area = 20 cm× 40 cm = 800 cm2 = 800 × 10^−4 m2

can someone explain to me how did this 800 cm became 800 x 10^-4

it should be 8 m2 (because changing it for the si units so divided by a hundred)

idk why there is a 10^-4

[tight storm]

it change to 800 10^-4 because it change a units from cm2 to m2
and cm is 10^-2 m like you already know divided by a hundred but it's cm2
so cm2 is 10 ^-2^2 = 10 ^-4 m

and if you change a unit before you multiply you will see 0.2*0.4= 0.08 which is equal 800 x 10^-4

[glacial dust]

ohhhhh

ok thank youu

[tight storm]

No problem .

[glacial dust]

wait so if that's the case then for mm squared its x 10^-6? m squared

[ember kernel]

Yes

The best way to not get confused is converting the units right away

You know that 1mm = 10^-3 m

Therefore 1mm² = (10^-3 m)² = 10^-6 m²

[random flame]

Hey can someone help me with part b? It’s the very top question, I’ll send in the other part of the question gimme a bit

Refer to Q13

Pls ping me when y’all do help, thanks in advance!!

[ember kernel]

First rise : the ball travels 10m up and then travels 10m down (total : 20m)

Second rise, travels 10 * 4/5 m up, travels 10 * 4/5 m down
(Total : 20 + 20 * (4/5))

Third rise : travels (10 * 4/5) * 4/5 m up, travels the same distance down
(Total : 20 + 20 * (4/5) + 20 * (4/5) * (4/5))

You might recognize a pattern there

[random flame]

Ah yeah

Ik the common ratio is (4/5)

But part b confused me a bunch

[ember kernel]

It's a geometric series

And each term is 20 * (4/5)^n (where n is the number of times the ball has already bounced on the ground)

And there is a specific formula to compute the total value of this series from n = 0 to n = p (p is up to you to decide)

[random flame]

Ah okay

[ember kernel]

I'm 90% sure you've learnt it in class, because that question is clearly meant for you to use the formula

[random flame]

You mean the Sn = a(1-r^n)/1-r ?

[ember kernel]

Exactly

[random flame]

I tried using that but I didn't get the ans

[ember kernel]

(Yeah your n is my p)

What n did you use ?

[random flame]

Erm

5

Cuz it said 5th impact

So I assumed n = 5

[ember kernel]

Ok so finding the actual n requires a little bit of thinking

If your input is n = 5, what does the last (= 6th) term of the series represent ?

[random flame]

The 6th impact with the ground-?

[ember kernel]

Hint : the answer is contained here

[random flame]

Hmm

20*(4/5)^6?

[ember kernel]

Ok you're confused

[cyan night]

isnt that after the 5th impact

[random flame]

Yeah -^-"

[ember kernel]

n=5 involves 5 total impacts since the beginning

[random flame]

The beginning as in before the first impact is counted?

[ember kernel]

When the ball is thrown upwards for the first time (0 impacts with the ground)

When n = 0, the ball has had 0 impacts with the ground yet (because the first launch is not considered an impact)

[random flame]

I see

[ember kernel]

When n = 1, the ball has has 1 impact with the ground

(Same reasoning for n=2, n=3,...)

[random flame]

-?

You gud? 😂

[ember kernel]

So if you use the input n=5, that means you're computing the total distance covered by the ball while it has had 5 or less impacts with the ground

[cyan night]

is n = 4 correct

o

[random flame]

Ohhh

Okay

So does that mean I have to use 6?

[ember kernel]

If you use n=6, that means the ball will have had 6 impacts total, and the final impact (the one at which point we stop incrementing the total distance covered) would be the 7th impact

We want to stop the incrementing at the very moment of the 5th impact

[random flame]

Ah okay

[ember kernel]

Thus, just a few milliseconds before that 5th impact, what is n (the last one we use for our sum) ?

[random flame]

Erm

n-1?

[ember kernel]

Not in terms of n, because n = n-1 makes no sense

Use actual numbers

[random flame]

Hmm

n=4

Right?

[ember kernel]

Just before the 5th impact, how many impacts has the ball had ?

Exactly

[random flame]

4?

Ah

I see now

[cyan night]

xd

[ember kernel]

That's just it

You want the total distance before the 5th impact. That means the sum of all terms of the series from n=0 to n=4 included

[random flame]

Wait lmao I rmb you, you helped me with Maths a while back 😂

[random flame]

Ohhhh

[cyan night]

ye

[ember kernel]

If you use n=5 as well, that means your ball hits he ground 5th times, which is not what the question asks for

[random flame]

Wait it says "from the 1st throw to the 5th impact"

That means it's asking for before the 5th impact?

[ember kernel]

The first throw is the bounce n°0

[random flame]

OHHH

Okay thankss

[ember kernel]

Yeah

You stop measuring the total distance as soon as the 5th impact happens

[random flame]

Bruh I think my textbook ans is having a stroke

[ember kernel]

So you don't use n=5

[random flame]

It's telling me a different ans

[ember kernel]

What formula did you use (with numbers, not n)

[random flame]

10(1-[4/5]^4)/1-[4/5]

[cyan night]

that is n = 4

[random flame]

Yeah

Isn't it-?

[ember kernel]

Welll on the first throw

The ball travels 10m up, and 10m down, doesn't it ?

[random flame]

Uhuh

[ember kernel]

That's a total of 20

[random flame]

Yeah

[ember kernel]

With n = 1, the ball travels 10(4/5) m up, and 10(4/5) m down

[random flame]

Wait a minute-

[ember kernel]

That's a total of 20(4/5)

[random flame]

So that means the first term is 20 and not 10?

[ember kernel]

It's been said since the beginning ^^

[random flame]

💀

[ember kernel]

Here to be precise

[random flame]

That's why my paper is full of careless mistakes

Thanks again

[ember kernel]

Np

[glacial dust]

can some one helop with c

what does it mean by t at 0?

[random flame]

Sorry- one thing- my textbook is telling me the ans is 67.232, I got 59.04-

[ember kernel]

Maths are ambiguous. Before the question involves a 5, doesn't mean you have to use n=5, the same goes for the 10m thing. You'll need to learn to translate from English (which uses the number 5 and 10m) to actual maths (which uses n=4 and 20m). Learn to think, not to copy

[random flame]

Gotchu

[ember kernel]

Ok, I computed it too and apparently the question asks for n=5, which means that the distance travelled after the 5th impact should also be included

If find it nonsensical, but alright

[ember kernel]

#science-help

[random flame]

Ah I see

Thanks a lot!! You're rly patient-

[ember kernel]

np

[random flame]

Heyo can someone help me with Q2b? Ping me if ya do. Any help is appreciated!!

[cyan night]

hi

the formula for arithmetic series is n*(a1+an/2)

n being the no. of terms, a1 is first term and an is last term

actually no

for this one use the different formula

S_n = n*(2a + (n-1)d)/2

we have to find n

substitute and solve

@random flame where are you struggling

[random flame]

I probably had a stroke somewhere substituting the formula with the values

But it’s alright, my teacher will help when I get back to school tmr

Thank you tho!!

[torpid root]

idk where I'm making a mistake

[cinder yacht]

what a cancerous function to derive lol

oh it's not that bad actually

[torpid root]

It's fine if you use the table

[cinder yacht]

I think ur mistake is that

u forgot the 2 in

(sqrt(u)) ' = u' / [2 * sqrt(u)]

[torpid root]

I did it like that

[cinder yacht]

it cancels out with the 2 of - 2 sin(2x)

[torpid root]

wait isn't it 1/(2*sqrt(u))

[cinder yacht]

by u I mean a function

not a variable

anyway the formula u used was fine then I guess

but how is there a 2 in the nominator of what u found

if at the start there was a 2 in the denominator (from the sqrt(u) ' formula)

[cinder yacht]

that should've just gotten canceled out by the 2 that comes out when u derive cos(2x)

wait

why the 2

in 2 * e^cos(2x)

[torpid root]

the number in front of x goes behind the e no?

[cinder yacht]

no

that's if u derive e^(2x) then sure, the derivative is 2 * e^(2x)

[torpid root]

(e^ax)'=ae^ax I though

[cinder yacht]

but u're deriving e^[cos(2x)]
and the derivative is [cos(2x)]' * e^[cos(2x)]

there is no e^ax here

[torpid root]

oh you're right

so when it's not a constant*x

we take the entire power?

[cinder yacht]

in general, (e^something)' = (something)' * e^something

so for example

e^(ax)' = (ax)' * e^(ax) = a * e^(ax)

and for example

e^[cos(2x)] ' = [cos(2x)]' * e^[cos(2x)]
= -2 sin(2x) * e^[cos(2x)]

[torpid root]

ah I see now, we derive the power itself and put it at the front. I thought it was just the constants

[cinder yacht]

constant is just a very particular example of power

[torpid root]

yeah

[cinder yacht]

like when the power is ax

but u can use the formula for any power

[torpid root]

that was how the teacher had explained in the pdf without any other notes

it was literally e^x' = e^x

[cinder yacht]

that's correct

[torpid root]

yes, but it didn't tell me much xd

[cinder yacht]

and when u use the chain rule or whatever u call the formula

then u deduce that thing I wrote

[cinder yacht]

this

the chain rule is : f(g(x)) ' = [g(x)]' * f' (g(x))
so by replacing "f" by "exp"
" f' " by "exp" as well
and "g(x)" by "something"
you get the formula I gave you

[torpid root]

the pdf didn't fail to explain the chain rule thank god

but the e^x' part was pain

I have my final exam in two days and it's just log derivatives and complex numbers

tysm

[cinder yacht]

good luck

[left knoll]

hey guys i have a doubt from complex numbers

soo there is this identity for complex numbers that is (a-ib)(a+ib)=a^2+b^2

and i was doing a questions in which i gotta do (1-2isintheta)(1+2isintheta)

and my answer is coming 1-4sin^2theta

but the real answer is 1+4sin^2theta

soo if anyone knows where i am doing wrong, just ping me here, thanks

[ember kernel]

As you said
(a+ib)(a-ib) = a² + b²

Using a = 1 and b = 2sin(t)
You get
(1 + 2i sint)(1 - 2i sint) = 1 + 4 sin²t

If you struggle to convince yourself it's true, compute the product using distribution, you'll see that you get the same result

[left knoll]

dude i am soo dumb really, i was taking b as 2isintheta instead of just 2sintheta

thanks a lot

[ember kernel]

No worries. Something that could help you not get confused next time is the fact that, in the cartesian representation a+ib of complex numbers, a and b are real numbers ^^

[glacial dust]

its saying that when flux density "B"is at 1.4 then H is 1650 I dont get it cuz in my eyes H is at 1000

how is it 1650 -_-

can someone? help this is bothering me the whole day

I think am blind lol

[ember kernel]

At B = 1.4, the silicon iron curve is a bit more than 3 square units away from the vertical axis
Since 2 square are 1000 H, I let you do the crossed product

You're indeed blind ^^
Maybe using a ruler (even against your screen) might help you analyse thay graph

[glacial dust]

3 square units?

wdym

how do u know the precise coordinates of that

I mean yah from looking at the graph the intersection of the line seem to pass the half value so I would just estimate it at a value of 1600

[glacial dust]

Nvm

I got it

btw I had answer that had a prefix of 10^-4 if so can I say that its melee?

[ember kernel]

Just in case

@glacial dust

[ember kernel]

Ok you have an answer, but I can't check it unless you send me a screenshot of the question

[glacial dust]

This is a diff one but I ll send a similar question wit the graph then Ill let u know if i finished it

[charred charm]

any help?

[cyan night]

maybe trying tan (3x) = tan (x+2x) and use tan (a+b)?

[charred charm]

thats what i did

but seem to get the wrong answer

and the mark scheme isnt very helpful

[cyan night]

hm

[cyan night]

wait is that sign the congruent sign?

[charred charm]

yes

an identity

[ember kernel]

tan3x = tan(x+2x) = (tanx + tan2x)/(1-tanx tan2x)

tan2x = 2tanx / (1-tan²x)

Therefore

[ember kernel]

Simplify the expression to cancel out the 1-tan²x denominators and you get that result

[charred charm]

ty

[ember kernel]

np

[charred charm]

i made a silly a very silly mistake

i forgot the 1

ty so much @ember kernel

[ember kernel]

np

[severe radish]

hey yall got a quick question

the part after (4.40)

i think what theyre implying is that the derivative as in 4.40 is always greater than or equal to zero

but i dont see how the fact that the rows add to zero implies that or even what they mean by it

do they mean adding the rows of a single column results in zero?

[left knoll]

hi guys

im not sure how u do part b

dw i got it

[supple island]

i believe it means the that the sum on j of Wij (for any given i row)=0

[severe radish]

yeah, youre probably right

[somber reef]

help me :'(

[rustic shuttle]

with the ratio you can get the first term

[rustic shuttle]

.

[weary arrow]

<@&717391911132069929> here as well

[trail swift]

thanks, it looks like the guy is already banned, will look for more messages to delete

[balmy star]

you're too fast wtf ⚡

[true steeple]

anyone here have linear algebra by kenneth hoffman? I have a doubt in an exercise

[true steeple]

dw got it

[hidden fable]

Can someone explain why that this is the answer please? I thought that it kinda related how if the speed was a fraction but I don’t understand conceptually

[left knoll]

working out the mean of 2 variables means you have to sum and average them

[ember kernel]

Janie says that the average speeed from Barnsley to York is 70km/h (the mean of 60 and 80)

But actually, the average speed's formula is obtained like this :
Total distance = Speed1 * Duration1 + Speed2 * Duration2

Average speed = total distance/(duration1 + duration2)

As you can see, the formula of the average speed is a weighted mean (the weights are the duration of each section of the drive)

Since the we know that the weighted mean and the classic arithmetic mean are the same (because Janie says the average speed = weighted mean can be calculated using a classic arithmetic mean), that can only say one thing :
Both weights must have the same value

Because if that's the case, using duration1 = duration2 allows us to simplify things in the actual formula of the average speed :

The total distance becomes (Speed1+Speed2)* Duration1

And average speed = total distance / (2* Duration1)

Plugging the first expression into the second, we get
Avg speed = (Speed1+Speed2)/2

=> With duration1 = duration2, we get the exact expression of an arithmetic mean. This proves our earlier suppositions that the weights (durations) were equal

@hidden fable Hope this helps ☺️

[glossy latch]

Great explanation! I had a quick picture of the direct algebra to show if it helps

[crisp heron]

Hello, I have this problem in which I have to determine the line parallel to the bisector of the second and fourth quadrants that is supported by the parable of equation y = x ^ 2-x + 3 making a chord of measure 2 roots of 6. The answer is 6, but it gives me 9; I don't know what I'm doing wrong, I made a system between the equation of the parabola y = -x + q (-x because it is parallel to the bisector) and the distance between the two points gives me the 2 (roots of q-3 ) but when I equate it to 2 root of 6 it gives me nine.Pleaseeee help

[hollow mauve]

Pls solve "B"

[left knoll]

is there a correlation between lnt and arctant that i can use or is the answer supposed to be a long function

[ember kernel]

Calculate each term by hand and sum them all

r= 0 and r=10 are included

[ember kernel]

Denote I(t) as the antiderivative of ln(t) arctan(t)

G(x) = I(x³+1) - I(x²-1)

(that's the Fundamental Theorem of Calculus if I'm not wrong)

Therefore, via chain rule :

G(x) = I'(x³+1) * (x³+1)' - I'(x²-1) * (x²-1)'

Good thing for you, you know by the definition of I(t) that I'(t) = ln(t) arcan(t)

[ember kernel]

And tadaaa, you're not just a few lines of calculations from the result

[left knoll]

so the answer is a long function

yeah i found the same result

thank you!

[ember kernel]

Calculus in a nutshell

Np :D

[left knoll]

if the answer is long or short no matter the case it's always suspicious lol

[ember kernel]

Exactly

Too long is sus

Too short is sus

And right in the middle is too short or too long anyway

[left knoll]

there's no right in the middle it's always extremes lol

[ember kernel]

Acutally, it's not that long. There are only two terms of 3 factors each

[left knoll]

yeah i know i'm just being intentionally dramatic

[ember kernel]

As long as an actual dramatic answer doesn't take place during your exam, you'll be just fine

[random flame]

Heyo can someone help me with Q8b? Ping me if ya do and big thank!!

[atomic dagger]

hi there, i need help to calculate this sequence, with the knowledge that sin(x) < or equal to x as long as I x I < 1

my teacher says that sin(1/n) is superior or equal to 0 from n > or equal to 2 but i don't know why or how

[safe ember]

sin(1/n)≥0, as long as n≥1/π since sin(X)≥0 for values of X between 0 and π

Since sin(X)≤X, sin(1/N)≤1/N. Therefore, the limit for the function substituting 1/n for sin(1/n) must be greater or equal to the limit of the function. Solving that limit you should arrive at a conclusion for the limit of the original function

[safe ember]

According to the Binomial theorem, the coefficient for the term with x^3 is n!/[3!(n-3)!] * 1^(n-3) * (1/2)^3. That equals to 5n/12, so you just gotta do some algebra to get to the answer

[left knoll]

Can you find which one is correct and why?

[atomic dagger]

thank you!

[safe ember]

you should add the +C after integration

[left knoll]

can someone help me please

[safe ember]

First consider the function that describes the line, then calculate the distance between PQ. Then calculate the point with positive coordinates in the line which satisfies PQ:QR = 2:3

Distance between points (a,b) and (c,d) is sqrt[(a-c)^2+(b-d)^2] btw

[left knoll]

ive found the distance between pq, how do i calculate coordinates that satisfy 2:3 after?

[safe ember]

What is the line that goes through p and q?

[left knoll]

6.7

[safe ember]

The line, not the distance

[left knoll]

line L?? 😵‍💫

[safe ember]

yeah in the form y=mx+c

[left knoll]

oh

you want me to find the gradient?

[safe ember]

The whole equation for the line

[left knoll]

okay

y = 1/2x + 3

[safe ember]

therefore you must find a point of coordinates (x, x/2 + 3) which is at a distance of 3|PQ|/2 of Q

so yeah, gotta do some algebra

You can also do it with vectors, it's much easier now that I think of it :v

[ember kernel]

I've been staring at that picture for 20 minutes now and can't find where it went wrong :(

[safe ember]

it should be = t + C, not just t

and each case has a different C

[crisp pagoda]

Consider the function f (x, y) = x^3 + y^3 + xy
(a) Determine the gradient of f and the directional derivative ∂f
∂v of f, where v = (1, −3), at the point P of
coordinate xP = 2, yP = 1.
(b) Write the equation of the line tangent to the level curve of the function f passing through the point
A with coordinate xA = 1, yA = −1.

Can someone kindly tell me the result of this question?

[ember kernel]

Yeah you're right

[safe ember]

the directional derivative of f(x,y) in the direction v is the partial derivative of f(x,y) with respect to x times the x-component of v + the partial derivative of f(x,y) with respect to y times the y-component of v. Then you substitute in the coordiantes of P

[left knoll]

helloooo

[safe ember]

the gradient is the same but without multiplying by the coordinates of v

[crisp pagoda]

thank you!

[safe ember]

And for b, you set the derivative (not partial) of f(x,y) = 0, solve for dy/dx and substitute in the coordinates of A

[lucid hollow]

Can someone explain me this?

can't understand the meaning

[safe ember]

It's just an identity :v, for any value of x taking the square root of its square gives its absolute value

[lucid hollow]

But why can't it be a normal x

[safe ember]

sqrt(2^2)=2
sqrt(5^2)=5
sqrt[(-1)^2]=1

[glossy latch]

It should be the second one. They kind of disregarded one of their coefficients in the second step of the first derivation

[lucid hollow]

if we are cutting the 2 with the square root

[safe ember]

not exactly, that's only true for positive values of x

for x=-1, applying the square and the square root gives 1

[glossy latch]

for example, what is (-2)^2

[lucid hollow]

hmmm i see

thanks @safe ember

[safe ember]

no need to thank me

[glossy latch]

could you give a graphical representation? I kind of don't get what the problem is saying.

[thick pine]

Is anybody familiar with singular value decomposition and Gram-Schmidt process?

[ember kernel]

Another way of seeing it is that square roots are positive by definition, so if you defined sqrt(x²) as x, you could get in a situation where a real square root (sqrt(x²) > 0) is equal to a negative value (x < 0) which is impossible

[glossy latch]

I have limited experience with it. I prob don't know enough but sure, ask ahead

[thick pine]

Okay so I have a to do it for an 4x2 matrix. I got two eigen vectors which I can now use for my matrix U, but I'm still missing two to get a 4x4 matrix. My script tells me something like "we add vectors u_{p+1},...,u_m to get an orthonormal basis" but I don't understand where they come from

[glossy latch]

Huh, I don't get the added part because it was always multiplying the two other matrices that produced an orthogonal matrix that could be used.

But as to the part of the orthonormal basis. Are you wondering why it's there or how you can get there by adding vectors?

If it's the former, then essentially, the other two matrices help greatly with data compression. When multiplied, you get a function similar to a taylor series that approximates the actual value. By cutting out parts of those two matrices, you'e able to get rid of a lot of data while nearly preserving all the values

[thick pine]

The latter, because I need them for calculating the resulting matrix

[glossy latch]

In that case, I wouldn't know. Didn't get what they meant by adding to produce an orthonormal basis.

sry

[thick pine]

I mean I have two eigen vectors but need a basis of R^4, so 2 are missing

[thick pine]

Sure no problem

AHH

I use the eigen vectors of AA' instead of the ones from A'A! Then I get 4 instead of 2

[glossy latch]

didn't get what you meant but it looks like you figured it out lol.

[thick pine]

Hahaha yeah I was dumb

[true steeple]

I dont understand (c). According to what they have given, zero vector is V={f:S->{0}} [ik {0} is not a field, here V is just a set,not a vector space]. How is it possible? Zero vector is v+0=v right?

[lucid hollow]

Thanks 😁

[thick pine]

I'm not sure if I understand your question but the function f: S -> {0} is the zero vector that makes V a vector space (not sure where c is from, probably from F? making V an F vector space)

yes, v+0=v. But you have to take into account that v and 0 are functions from S to F in this case. Meaning, you'll get something like (v + 0)(s) = v(s) + 0(s) = v(s). (v and 0 are bad names for those functions here)

[atomic dagger]

hey everyone ! i was wondering how I can go from :

to:

[glossy latch]

as n goes to infinite, everything except the most influential variable is removed. Here, the most influential variable for the numeroator is -9n and the most influential for the denominator is 2sqrt2 x n. Cancel out n and you get that value

[safe ember]

Yeah

[thick pine]

Reason for the part with the "most influential variable": expand the fraction with 1/n

[thick pine]

333333333333333333333333w656eddddddddddddddddddddddddd5wqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq5w5w55555qqqqqqqqqqqqqqqqqqqqqqqq77

[thick pine]

Is there any easy way to find 2 linear independent vectors to 2 given ones in R^4?

meaning, I want a generator given v1, v2 in R^4

[supple island]

you want a really general solution or something more practical?

[thick pine]

whatever is faster for you

[supple island]

ok sorry i completely forgot to answer, usually i just use elements of the standard basis

like if i have (1,0,2,4),(2,0,3,0) i just use (0,1,0,0),(0,0,1,0)

[true steeple]

Ahhh, I wrote my doubt when I was sleepy lol. Turned out my question was confusing. I got it now. Thanks

[atomic dagger]

i have a question! how can I find the superior or inferior limit of a sequence?

[thick pine]

You need to take a look at all subsequences, calculate their limits and then take the maximal/minimal value of those

[crisp heron]

there is no graphic,it is just text and it's in italian :C

could you please specify the parts that are not understandable.The exercise is everything till the first point.I edited a little bit,but if you could please tell me what parts are not understandable that would help me very much c:

[atomic dagger]

i'm not sure how I can find the subsequences ?

[thick pine]

Yeah that's actually kinda hard I think. I guess trying to plot the values is an option

[silent copper]

hey i need help with part A

[thick pine]

What do you need help with?

[atomic dagger]

okay i see thank you!

hey! i'm a little confused as to how to prove that a sequence is a Cauchy one. How could I do that ?

[glacial dust]

guys can some help me out

this is the equation for the indiced emf

but look at this

why wasnt the inductance (L) not negative?

[glacial dust]

I dont get it

plezzz help 🙂

[thick pine]

Sorry I’m not good at physics, maybe ask in #science-help ?

[thick pine]

One option is to show that |a_{n+1} - a_n| approaches 0 as n -> infinity

[atomic dagger]

does proving that suffice in showing that it is a cauchy sequence ?

[thick pine]

Yes, because then for every epsilon, you'll find an n_0 so that |a_n - a_m| < epsilon for all n,m > n_0

But it's best to explicitely look for the epsilon

[cinder yacht]

so sqrt(n) is a Cauchy sequence according to what ur saying

[cinder yacht]

depends on the I and E conventions used in the circuit diagram (but L is always not negative)

[glacial dust]

wdym it

says on the equation by default its -L

[cinder yacht]

that doesn't mean L is negative

that means L is multiplied by (-1)

[glacial dust]

So when do I know which kne is which

[cinder yacht]

for example if L=1H then L is positive, and yet -L=-1H

[cinder yacht]

kne

?

[glacial dust]

one**

one*

[cinder yacht]

which what is which

u mean when to put a - and when not to?

it depends on the circuit diagram

[glacial dust]

Wait ill send you a question using diff + and - L

[cinder yacht]

the orientation of the E and I arrows

if E and I have opposite directions on the diagram then do +L in the formula, if E and I have the same directions on the diagram then do -L in the firmula

[glacial dust]

and this one

induced emf has to equations

one with +- N

[cinder yacht]

they teach u this stuff in a shitty way, lol

i don't know then

[glacial dust]

I think ill just cover my eyes and pick which sign to use lamao

[cinder yacht]

maybe generally prefer -

and when u see "magnitude" maybe means +

lol

and when u have a circuit diagram with I and E arrows on it, use the rule I said

[glacial dust]

I dont think thats the case

magnitude was not mentioned but L is positive 🙂

same goes for N

[cinder yacht]

LI= N Phi is always correct btw

[cinder yacht]

ur teacher got it wrong lol

[glacial dust]

nah its the book

[cinder yacht]

E = - L dI/dt = - 12.5 [ (0-2)/ (20*10^..)] = 1.25 kV

[glacial dust]

I got it from the book

[cinder yacht]

the book got it wrong

in dI/dt it would be 0 - 2

not 2 - 0

because it was 2 and it dropped later to 0

[cinder yacht]

same here

should be the E=-L dI/dt formula, and dI/dt should have 0-4 not 4-0

makes me sad that physics is taught like this xd

[glacial dust]

so its not the L inductance or Turns the problem?

[cinder yacht]

no L is always positive, and the formula is generally E=-L dI/dt

and u just have to pay attention to what's in the dI/dt

like

[glacial dust]

yah you mean -(L)

but eqaution makes it -

[cinder yacht]

here the formula should be E = - N dPhi/dt

and since Phi went from 35mWb to - 35mWb,
then dPhi/dt formula should have [ (-0.35 * 10^-1) - (0.35 * 10^-1)]

ur book sucks at teaching this xD

they (wrongfully) reversed the formula (to E=N dPhi/dt) and they also (wrongfully) reversed the expression of dPhi/dt
so at the end it gave a correct answer

[glacial dust]

what about N?

always positive?

so simply n?

[cinder yacht]

u don't understand what positive means I think

saying "N is positive" means "N>0" aka "if u write N in numbers, it won't have a - sign next to it. for example : N=300"

"N is negative" means "N = -..."
it DOESN'T mean "N is written next to a - sign, like in -N dPhi/dt"

N is always positive, and it's generally next to a - sign

[glacial dust]

But why does the equation tell emf=-N flux/time

[cinder yacht]

N is positive, but it's next to a - sign

(-N) is negative

[glacial dust]

yah ik now that

but some times they dont include this -

when really equation tells that -(N)

[cinder yacht]

yeah from the 3 examples u showed me,they were wrong

to forget this -

but they were also wrong in the computation of dI/dt or of flux/time

so their final result turned out correct by luck

[glacial dust]

so what is the actual equation?

[cinder yacht]

the actual equation has - in it : E = - N dFlux/dt or E = - L dI/dt

u just have to be careful when computing dI/dt or dFlux/dt

like dI/dt = [ I(of the future) - I(of the past)] / time

and dFlux/dt = [Flux(of the future) - Flux(of the past)] / time

so when they tell u "I was 3A,then it dropped to 0 in 5 seconds" that means that :

• I(of the future) = 0
• I(of the past) = 3A
• time = 5 seconds
  ==> and u use the formula I just gave

[glacial dust]

uhm wheres the formula? xd

[glacial dust]

is it this?

also can u tell me why its 0-2

the current of 2 when to 0 so is it 2-2?

[cinder yacht]

im done sorry

i said everything already, in details and with multiple examples, read what i wrote

[glacial dust]

Yo I got it

thanks alot

[spice basin]

I need someone to explain to me how to find indefinite integrals such as this one 🥺

the division is what causes all the problems for me

[true steeple]

Ig you can take x+1=t, then x-1 becomes t-2. Then try solving it

[spice basin]

ok thanks ill try

[willow basin]

hey. got this question wrong on my assignment and idk why. can anyone see my mistake?

[safe ember]

Profit isn't the same as revenue

[true steeple]

One mistake i can see by just going through is that total profit during september is 780-565 not 780

[atomic dagger]

could someone help me study the convergence of this series ? I tried to use Alembert criterion but i'm finding that it converges when it's not supposed to...

[cinder yacht]

always start looking for an equivalent i guess

anyway here u write it as :
n * sqrt(1+7/n^2) - n
= n * [ (1+7/n^2) ^ (0.5) - 1]
and that is equivalent to :
n * [ 0.5 * 7 / n^2]
which is = 7/(2n)

so :

• Un ~ 7/(2n)
• Un and 7/(2n) are positive sequences
• the series of 7/(2n) diverges
  => therefore the series of Un diverges

[willow basin]

thanks

[atomic dagger]

where is the ^(0.5) - 1 coming from ?

[cinder yacht]

sqrt(y) = y^0.5

the - 1 comes from the " -n" except that we factored by n outside the [] so it became "-1"

[atomic dagger]

okay thank you, now I'm just wondering how did we obtain that, I mean where did the ^0.5 go? n * [ 0.5 * 7 / n^2]

sorry sorry

[cinder yacht]

oh

it's an equivalent

like

(1+x)^a - 1 ~ ax

when x->0

or if you saw it as développement limité : (1+x)^a = 1 + ax + o(x)

if you didn't study this stuff then

you have to do a longer way to prove that Un is equivalent to that stuff

u have to do :
[sqrt(n^2+7)- n]
= [sqrt(n^2+7) - n] * [ sqrt(n^2+7) + n] / [ sqrt(n^2+7) + n]
= [ n^2 + 7 - n^2] / [ sqrt(n^2+7) + n]
= 7 / [ sqrt(n^2+7) + n]
= 7 / ( n * [sqrt(1+7/n^2) + 1])
which is equivalent to 7/(n * 2)

[atomic dagger]

i'm fine up till this point, i'm just confused as to why it's equivalent to 7/(n*2) i'm sorry i'm slow T.T

[signal arch]

does anyone know how to do ii and iii? please im so desperate i dont understand 😭

[ember kernel]

The cubic polynomial having a+1, b+1 and c+1 as roots means that the equation is (x-(a+1)) * (x-(b+1)) * (x-(c+1)) = 0

I let you calculate the products to get a sum of terms. That way you'll be able to compare both notations of the equation, and express p and q in terms of a,b,c

[signal arch]

this is how i wrote it and then i expanded it all (I tried it twice) but im still not sure what to do once ive got the whole thing written out, even when I got an answer for P, which is probably wrong I can't figure out a way to get q, maybe my calculations were wrong

[ember kernel]

Expand the product, it is equal to the cubic function given in the question
The coefficient you get for x² is p
The independant term you get is q

And then you'll be done

[signal arch]

ive been using the formulas a+b+y= -b/a, ab+by+ya= c/a and aby= -d/a, which got me an answer for p, but once ive expanded it all the last term for q actually involves me using aby= -d/a, and like q is d from my understanding so I just don't know what I'm misunderstanding

[ember kernel]

I have ... never heard about those formulas 🤔

You didn't expand the product by hand, right ? That's what I would've done I think (however long this method seems at first)

[signal arch]

I'll show you

[signal arch]

this is what ive been taught

[hardy kite]

so I created an exponential growth problem and I got 17577, if anyone's free could someone lmk if they got the same answer? A city's population is growing at a rate of 12% every 3 years and its current population is 9248. What will the population be in 20 years?

[ember kernel]

Thanks

[ember kernel]

Yeah that makes sense. It would have been a weird exercise asking students to expand a 27 total terms product

Well

Use that formula

Replace alpha by alpha+1 (because the root is alpha+1 in this case, not alpha), dot the same for alpha and lambda

Replace a by 1 (because the coefficient of x³ is 1)

And replace b by p (same)

Now just isolate p and you get its values in terms of alpha, beta and lambda

And the logic for finding q is similar : replace the paremeters in the second formula by relevant values and isolate q

[junior blaze]

Hi guys,
I've been stuck on these 3 problems for over an hour. Could anyone please help? I have the work for 2 of those, I'll send it too, maybe I'm doing it right, but missing a step or something like that. Anyways, sorry for my bad handwriting.

[hardy kite]

nvm I did it again and I now got 19686 but if anyone's free to double check if it's correct or not lmk asap :')

[left knoll]

@signal arch are your doubts clear or you still need help?
This is my level i can send you a solution if needed...

[signal arch]

please send me a solution if you can \😭 🙏

ill probably understand it better once i actually see it put together

[cyan night]

that looks like vieta's formula

[signal arch]

sorry to bother you, just wondering if you're still available :)

[left knoll]

Ok I'll just send it...

[left knoll]

Just confirm this is the ques right?

[signal arch]

yep

[left knoll]

@signal arch
Just see if you understand this-

[signal arch]

oh i understand now i think, did u manage to get a value for q as well? it's fine if not, thank youuuu so muchh anywayy <333

[left knoll]

Q = product of roots

[signal arch]

thank u so much ure a blessing ❤️ i hope you have a great day

[left knoll]

No problem
It was interesting to solve
You too have a good day 🥰

[cinder yacht]

u mean the last line is what u don't understand?

7 / ( n * [sqrt(1+7/n^2) + 1]) is equivalent to 7/(n * 2)
because sqrt(1+7/n^2)+1 -> 2 when n->infinity

did u study equivalents?

[atomic dagger]

no i haven't...

[cinder yacht]

oh lol

[cinder yacht]

@atomic dagger then say [sqrt(..)+1] <= 3 when n is big enough (like when n>=5)
so 7/(n * [sqrt(..)+1]) >= 7/(n*3)
so Un>= 7/3 * (1/n) >0
and the series of 7/3 * (1/n) diverges
so the series of Un diverges

[barren verge]

can someone help me understand this, I've gone through it many times, watched tutorials but i just can't understand it
can any one walk me through the steps

[untold dune]

<@&717391911132069929>

[cinder yacht]

(i)
PS=PQ
so it's an isosceles triangle at P
so PQS=PSQ=70°
and we know that in the triangle PSQ : QPS+PQS+PSQ = 180°
so QPS+70°+70°=180°
so QPS = 40°

[barren verge]

Thank you))

[sudden maple]

can someone help me with this

question 10

[delicate cave]

There is an identity: two lines are perpendicular if and only if the product of their slopes is -1

using that the question is pretty straightforward

Also, when the equation of a line is in the form ax + by + c = 0 the slope is given by m = -a/b

Proof: ax + by + c = 0
==> by = -ax - c
==> y = -ax / b - c / b

comparing this with y = mx + k gives m = -a/b

[hexed zealot]

Any tips before learning calculus and vectors?

[steel mist]

Solve lot of questions

[ember kernel]

Don't just rely on formulas. You need to understand the concepts and be able to visualise what you're doing when solving exercises if you really want to excel in calculus and vectors.

If you just want to scrape through the exam, understanding the topics and exercising a lot should be enough

[left knoll]

yo

i need help

how do u do part b

i dont understand the solution

[sharp delta]

@left knoll can you show me how u did the first part?

[charred charm]

hey

guys

had a qs

why do you multiply by 60

i understand the rest

[safe ember]

Velocity is in metres per second, but time is given in minutes - you must multiply by 60 to turn it into seconds

[charred charm]

ahhhhhhhh ty

bro

@safe ember could you help me

with a stats qs

pls

[safe ember]

Sure?

[charred charm]

bit confused on

hoe to work out

c

thank you @safe ember

[safe ember]

Well, first consider the condition for being an outlier t>Q3+1.5IQR. Calculate this value first

[charred charm]

sorry not c

d

my bad

[safe ember]

Well, we consider 5% of 120 applicants = 6 applicants. This 6 applicants must have times between 5 and 7 minutes. Assuming the 10 in this period are distributed randomly not following any sort of normal distribution, we divide the 2 minutes in 10 (each 12 seconds). Since we get the 6 best 6*12=72s= 1 minute 12s --> Times below 6 minutes 12 seconds

[charred charm]

Thank you so much

[left knoll]

why are these two graphs not equivalent?

[supple island]

why would they be?

they just happen to have the same solutions for y=0

[left knoll]

you see the equation for red line

multipy the whole thing by 4

then devide by three

divide*

its now the equation of the blue line

i realised my mistake

my apologies for wasting ur time

[tardy bone]

hey

can someone tell me what the p stands for?

is it the distribution of all variables or one variable?

[nocturne island]

i'd assume its de PDF of x

based on the second constraint

[coarse cypress]

elp pls

[thick pine]

Rule #5: No help with graded tests allowed