help-42

[amber bolt]

thanks scerball, that's how

it's like suppose you draw 3 aces, and then you draw 2 not aces, neatly ordered

then this would count hands like that
well not hands, it would straight count the probability

and then suppose you draw 2 non aces, and then 3 aces

what would the formula become

@fading blaze

[fading blaze]

im not sure

[amber bolt]

you don't understand that formula in the first place?

[fading blaze]

the formula for the combinations?

[amber bolt]

the formula for the probability of drawing 3 aces at the start ,and then not drawing the 4th one

my picture

[fading blaze]

oh the 48/49

[amber bolt]

no?

yes

maybe

[fading blaze]

you multiply the probabilities of the 3 aces per draw 4/52, 3/52, 2/50 and then the probability of not getting the 4th one which is 48/49

?

[amber bolt]

that's 4 draws

there's 5th draw, you still need to not draw an ace
48/49 is you don't get the 4th one, and then 47/48 is you don't get one again

yes

that would be part of the final answer, one of the ways to have 3 aces is to do that

the final asnwer is the sum of probabilities of all ways to do that

if you do it in reverse, 2 non aces, and then 3 aces in a row, what's the probability of that

[fading blaze]

48/52 etc then 4/50 etc

[amber bolt]

correct

that's actually the same number if you calculate it

the numerators go in a different order

the product is the same

can you see it?

[fading blaze]

yeah

i tried this formula too and divided it with the total amount of hands overall from a 52 deck

and its the same

well around the same

[amber bolt]

so that's 2 ways, but how many more

5 choose 3 is how many total, so you multiply by 10

5th row is actually 6th row i assume

[fading blaze]

wouldnt the 5th row be 5!

[amber bolt]

the 1 5 10 10 5 1 is the row for 5!

it's 6th row

it's shifted like that

[fading blaze]

ah

wait 5 choose 3 is the total of what

hands with 3 aces?

[amber bolt]

it's like orders

how exactly you draw 3 aces, 2 at the start then 1 at the end?

3 at the start?

3 in the middle?

honestly that's also pretty hard to get

i have no f-ing clue, how this question makes sense

[fading blaze]

HAHA mb

but also even if in a diff order the probability would still stay the same

[amber bolt]

it's too advanced to say "use pascal's triangle buddy"

guess im dumb

[fading blaze]

its okay im dumber

i think ill just find a way to bs that i used pascals triangle maybe my prof will accept it

[amber bolt]

there's one use of pascal's triangle to find 5 choose 3

that's reasonable

the solution makes sense

but yeah

the professor is doing parkour

[fading blaze]

very confusing

im not even in like a math focused course so i completely forget these things

thank u for the help i think ill close this now

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[rugged monolith]

How do they integrate that expression

I haven't encountered integrating a function over a function

[amber hearth]

I would try integration by substitution

[rugged monolith]

could you give me an example

[amber hearth]

sure (I'll do the one above and explain)

[rugged monolith]

Thanks

[amber hearth]

A lot of the time to try and get the correct method you might need to try smething and see if it works

however with this we are sort of forced to use substituion as integration by parts doesn't apply

after the bottom line we can integrate with respect to u and then after substitute it back in:

[amber hearth]

(last line is to power of -1/2)

hang on

I've made a mistake smewhere one sec

got it (i made a mistake early on)

[rugged monolith]

yeah my solution didn't get that

Here is the full

Do they use the quotient rule?

[amber hearth]

I'm not confident with that method so idk

I can just never see it

[rugged monolith]

I don't know what happened in the middle part

du/dx = -2x

du = -2x dx

how does that got to -1/2 integrate upper bound 2 lower bound 0 1/u^1/2 du

[amber hearth]

the -1/2 is to cancel out the fact that du is -2x. so that means that we are just left with du = x hence why the x is removed from the numerator (as we are effectivly multiplying by x in the du)

[rugged monolith]

oh you make dx = du/-2x

[amber hearth]

yes, sorry its not very clear

and then sub it in an re-arrange

[rugged monolith]

nah it's good

I just didn't realise that dx is actually a part of the equation

I thought that was just there to show what you are differentiating in respect to

I've never had to do anything with the dx I don't get it

Howcome you can do stuff with the dx?

I thought it was just notation

[amber hearth]

dx means "delta x" or change in x

have you done differentiation by first prinicples?

(also know as the delta method)

[rugged monolith]

no

uh yes

actually

[amber hearth]

cool

so you know how you do that has h-->0 so that the increment of x gets infintesimally small

dx is exactly that

it just means a really small change in x

and so when we add different variables (for example du) the "effect of the increment" may be different. Sorry I'm struggling to explain it

https://blog.cambridgecoaching.com/but-what-is-dx-really-calculus-terms-explained

this may be a better explanation

[rugged monolith]

thank you I will look at that

btw for the above question, was there a simpler way to do it

like quotient rule I think you might use for it

because it's a function over a function

or is that substitution method the only way

[amber hearth]

honestly idk, I'll have a go with the quotent rule and see. It is probably possible but whether it is easier or not can start to get subjective about what bits you find easy

[rugged monolith]

it's just because it was an example in the textbook about an unrelated topic

and they just breezed through it without having any explanation which is very unlike the textbook

so I was wondering if it was because there is a easy way to do it

[amber hearth]

is this edexcel fmaths?

or even maths cuz I'll have the textbook and might be able to see what they wanted

[amber hearth]

also I'm gonna say no, its harder

Imo it's very tricky to integrate sqrt(4-x^2)

[rugged monolith]

OH

The dx is there because it is multiplying the height of the integral sign with the function inside it

by infitesimally small widths of delta x

so dx is not notation and is actually part of the equation that makes sense

thank you for the help, that article was really good

[amber hearth]

np :).

I may have the working for that q is lesson, lemme see what we did then

sorry I don't, I also think that substution is the way to go with those questions where you think that differentiating one of the functions may be tricky otherwise the quotiont rule is fair game

[amber hearth]

https://youtu.be/GzEx5x_ph2g?si=Zjk9u36ywMjGoPFa I recommend you get comfortable with substiution as it is useful in later questions. (I'm still getting to grips with this chapter tbh)

Any more questions?

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[rugged monolith]

sorry I was away from my pc for a bit

[rugged monolith]

no that's all thank you for that

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[versed roost]

Please tell me how to calculate the tasks a and b. I am in 9th Grade in Russia and i got into this discord because its late and thare is no other way of getting help because its to late. a) In the plane, two points A and B are given. Determine all points P in the plane for which the sum of the distances |AP| + |BP| from point P to points A and B is minimal (i.e., as small as possible). Provide the minimum value.

b) In the plane, a square ABCD is given. Determine all points P in the plane for which the sum of the distances |AP| + |BP| + |CP| + |DP| is minimal.

[rugged monolith]

For a) it would just be any point on the line between A and B

and the mininal value is just AB

minimal*

[versed roost]

thanks. And what about b)

[calm coral]

@versed roost Has your question been resolved?

[rugged monolith]

I don't know I don't do the same syllabus as you

so I don't know about the problem with squares

[calm coral]

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[neon pine]

hi i need help understanding the solution to this problem

[neon pine]

im doing a similar question

and i dont wanna just copy the answer

this is the way my teacher solved it

the question is the exact same except is a 9 instead of a 8

what I dont understand is how she got y''

i get how she got y'

but after that I'm lost

or is it that its wrong?

<@&286206848099549185>

<@&286206848099549185>

[midnight knoll]

that first derivative isnt even correct

there shouldnt be any ln(x) terms

(in question 16)

[neon pine]

so how would it be solved?

ill show my question

oh wait

wrong one

[midnight knoll]

what do you think y' is first of all?

[neon pine]

the way she does it confuses me

[midnight knoll]

to get the first derivative

we do the power rule as we see a (...)^4

which gives us 4 ( ... ) ^3

but then we need to remember the chain rule

[neon pine]

bring it to the front and multiply

[midnight knoll]

yes

[neon pine]

then multiply everyhthing by the derivative of the inside function

[midnight knoll]

exactly

[neon pine]

but then

how do i get that

[midnight knoll]

and the derivative of 9 + 3/x is -3/x^2

[neon pine]

fractions r scary

[midnight knoll]

the derivative of 9 is 0

and the derivative of 3/x is the derivative of 3x^-1

so we get (-1) 3x^(-2)

which is -3/x^2

[neon pine]

wait so

its 4(9+3/x)^3 * -3/x^2?

[midnight knoll]

yes

wait

the first bracket should be cubed

[neon pine]

yea sorry just saw that

mb

[midnight knoll]

np

thats correct for the first derivative

[neon pine]

so how do I get y''

[midnight knoll]

$y' = 4\left(9 + \frac{3}{x}\right)^3 \times -3x^{-2}$

thats the first derivative

[neon pine]

yea

[midnight knoll]

so to get the second derivative

we need to use the product rule

[neon pine]

oh

f' x g + g' x f?

[midnight knoll]

yes

and you know f and g

[neon pine]

hm

f is 4(9+3/x)^3 ?

[midnight knoll]

yes

[neon pine]

and g is 3x^-2

[midnight knoll]

yes

[neon pine]

is it that

for f' its chain rule again?

[midnight knoll]

yes

[neon pine]

ok

so f' is 12(9+3/x)^2 * 3x^-2?

[midnight knoll]

mhm

[neon pine]

and g' is -6x^-3?

[potent lotus]

Frog_Man

[midnight knoll]

well ye apart from the fact that i realised i forgot a minus sign earlier

[neon pine]

so its positive 6?

[midnight knoll]

yes

[neon pine]

i got this

[midnight knoll]

yes

and that indeed matches with one of the multiple choice answers

[neon pine]

Does it?

[midnight knoll]

12 x 3 = 36

6 x 4 = 24

[neon pine]

Ah I see

let me put it all together

Like that

Right ?

I had to put them back as fractions to understand

[midnight knoll]

ye nice job

[neon pine]

Thank you very much

[midnight knoll]

np

[neon pine]

have a good day

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[remote mural]

HI

[remote mural]

I need help with the next

problem

[supple needle]

[swift marsh]

https://dontasktoask.com/

[remote mural]

I have a problem

I thought that the translator would work but it have math signs and it is a mess

I am doing it manual

[supple needle]

Take your time

[remote mural]

I will

close

wouldnt it be fair if you closed this and ask when ready to ask

and open again

yeah

.close

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[remote mural]

already

I will upload my try

I know it is wrong

[calm coral]

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[remote mural]

How do I get a reciprocal of -12?

[remote mural]

Or any whole numbers

[hollow perch]

reciprocal means 1/()

that's all

[remote mural]

So the reciprocal would be 1/12?

[hollow perch]

-1/12

doesn't affect the negative

[remote mural]

Oh alr thank u

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[spare pecan]

I haven't done anything except showing d is a metric @warm warren
Intuitively we can think of an infinite matrix, with entries x_{m,n}
we want to show that if the sequence sequence of rows converge, then every column converges to the elements of the row sequence

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[spare pecan]

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✅

[spare pecan]

(\tilde{A}=A/\sim)

I have worked it out and found proved that every ball in (B_r([a])=\bigcup_{a'\in[a]} B_{r}(a'))

[spare pecan]

thus I can write (A=\bigcup {[a]\in \tilde{A}}B{r(a))
}(a))

[spare pecan]

is this what they meant by union of equivalence classes?

oh and here r is dependent on the elements of \tilde A, similar to when we express open sets as union of open balls

sry I have made intermixed [a] and a in some places

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[remote mural]

Trying to simplify equation above to standard sphere formula. Here’s my work but website says my answer is wrong. Not sure which part

[calm coral]

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[main spade]

given

[main spade]

how do you do

i am very confused.

[calm coral]

@main spade Has your question been resolved?

[main spade]

<@&286206848099549185>

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[hollow perch]

what's the problem

[inland geyser]

im tryna sort these least to greastest

[hollow perch]

yeah so what don't you understand

[ancient vapor]

-3 7/10 -3.45

-3 1/4

-3 7/10= -3.7

-3 1/4 = -3.25

always use a calculator

if youre able to

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[mortal lantern]

hi math guys

[mortal lantern]

i need help with a question on my assignment if anyone knows smth about vectorfunctions

"In a coordinate system, a point P moves such that at time t the position vector r(t) of the point P is given by r(t)=(t^2+3t,t), -5≤t≤5"

Determine the time when the velocity vector is parallel to a line l: x-4y-4=0.

ive tried drawing it but it doesnt make sense

first i find the derivate of the vectorfunction r(t) obviously

then im not sure but i think im supposed to find the n vector for the line l ?

and if theyre parallel r(t) should be a multiple of the vector n

where k is a constant

then im lost on what to do next

[calm coral]

@mortal lantern Has your question been resolved?

[calm coral]

@mortal lantern Has your question been resolved?

[pure kayak]

(x-4)/4 =(y-0)/1=s
x=4s+4
y=s so the line can be said to be of the equation r=(4,0)+s(4,1) with direction vector (4,1) if v is parallel to the line then v is a multiple of (4,1)

@mortal lantern

[mortal lantern]

ayy

ty for pinging me

didnt see

[pure kayak]

np

[mortal lantern]

what

xd

[pure kayak]

its a way of making cartesian equations into vector equations
you have to make them parametric
so i start with x-4y-4=0, then x-4=4y, (x-4)/4=y=s, where s is going to be our parametric variable
then (x-4)/4=s and y=s

from there you just rearrange to get x=4s+4

then you can express (x,y) for a point on the line as (4+4s, s)

=(4,0)+s(4,1)

hence (4,1) is our direction vector for the line l

[mortal lantern]

yes

i am so far

i just didnt really understand how u got to it

but yes thats true

[pure kayak]

once you have that, you know that a vector is parallel to a line if it is a multiple of the lines direction vector

[mortal lantern]

yes

[pure kayak]

so you just apply that, v=(2t+3, 1) so you must then know that 2t+3=4

[mortal lantern]

wait go back

how do i know that k = 4

or the multiple is 4

why is it = 4

[pure kayak]

we have direction vector (4,1), we have v=(2t+3, 1) since the second entry of both is 1, then 4=2t+3 also

they are just equal vectors

[mortal lantern]

ooh ye ok

so t = 0.5 ?

[pure kayak]

should do

[mortal lantern]

ok i think i get it Hmm

"so i start with x-4y-4=0, then x-4=4y, (x-4)/4=y=s, where s is going to be our parametric variable
then (x-4)/4=s and y=s
from there you just rearrange to get x=4s+4
then you can express (x,y) for a point on the line as (4+4s, s)
=(4,0)+s(4,1)
hence (4,1) is our direction vector for the line l"
this is just to find the directional vector right?

[pure kayak]

yup

[mortal lantern]

okay

then i understand it

ty alot bro

[pure kayak]

nw

[mortal lantern]

.close

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[spare pecan]

could someone guide me with these

[spare pecan]

and is there some general strategy for a given quotient

[calm coral]

@spare pecan Has your question been resolved?

[spare pecan]

<@&286206848099549185>

seems like I can reduce
(x^2-3,2x+4)=(4x+6,2x+4)=(2,2x+4)=(2)
So (a) is Z[x]/(2)?

nvm what I did here was wrong

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[spare pecan]

reopening

[spare pecan]

.close

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[feral urchin]

I have no idea how to solve this problem

[feral urchin]

Am I suppose to do like 4x=36?

[quiet swan]

x is the length of the corner piece being cut off

Use the formula: volume=length*width*height to figure out the volume of the box

[feral urchin]

so 36x36x36?

46656?

[quiet swan]

try and imagine folding the cardboard up into a box

[feral urchin]

alright

(thank you for helping btw)

[quiet swan]

no worries

[quiet swan]

some examples

[feral urchin]

I got the whole concept of me trying to find volume and the algorithm to get it but I dont understand how to incorporate using x or what function the question is askiing about

[quiet swan]

x is the length of the corner piece we cut off, Lets say x=1cm, what would be the volume of the box?

[dull wagon]

first get an expression for the side length of the base

[feral urchin]

39304? 34x34x34?

[quiet swan]

is the height 34?

[feral urchin]

because it was a box i assumed it was even on all sides though im realizing i dont know the height of the box so that's x/

*?

[quiet swan]

oh, look at the picture i sent, what do you think is the relation between the corner piece that was cutoff and the height?

[feral urchin]

okokok so itd be something like 36x36x(4x)

[quiet swan]

wait wait wait, we are still assuming that x=1 and 1cm was cut off

[quiet swan]

*inch

[quiet swan]

the length and width are 34 inch and you are correct on that

[feral urchin]

okay cool

[quiet swan]

just that the height is not 34 inch

[feral urchin]

the height is a inch

?

[quiet swan]

yay

u r right

[feral urchin]

mind was doing backflips

[quiet swan]

so the volume in this case is 34*34*1 inch^3

So now lets move on to the more general case: what would be the volume if x inch is cutoff?

We define V(x) to be a function that outputs the volume if x inch is cutoff

[feral urchin]

V(x)=1156x?

[quiet swan]

thats what the question means by "volume" with respect to x

[quiet swan]

well, the length and witdth also changes if x changes

[feral urchin]

V(x)=34x34x(x)

wait

[quiet swan]

if we have 2 inch cut off the volume would be 32*32*2

[quiet swan]

imma let u cook for a bit

[feral urchin]

ok

V(x)=(36-x)(36-x)(x)?

[quiet swan]

plug in x=1 see if u r correct

[feral urchin]

oh

2x?

replace those x's with 2x

with the 36?

[quiet swan]

(36-2x)(36-2x)x right?

[feral urchin]

yes

[quiet swan]

correct!

[feral urchin]

lordt

thank you

[quiet swan]

also, side note

usually it is best to draw some diagrams

writing down steps clearly would help figuring out where the mistake is

is that all you need help with?

[feral urchin]

yeah see I was trynna do that with my brain

[feral urchin]

To find the maximum value

[quiet swan]

probably need more practice before brain starts braining

[feral urchin]

Yeahhh I've taken this whole class back in my junior year for pre calc but freshman college i guess is too far of a gap

[quiet swan]

Hint: Figure out the domain (possible x where the box could be made). and try to find the maximum with the methods u were taught

[feral urchin]

gotcha

[quiet swan]

once you build up some intuition (for math, not this question), it will click.

Depending on the prof though

[feral urchin]

Yeah majority of this class is looking self taught to hell due to the professor but aye we gonna get this done

[quiet swan]

sadge and fire, you got an idea of how to solve part 2 tho?

[feral urchin]

keep it a buck no I was thinking it has to be in the dimensions of 36

I checked using desmos thinking the x's btwn infinity would give me a clue but

[quiet swan]

remember the domain

imagine cutting off 37inch out of a 36 inch box

💀

[feral urchin]

lmao gotcha on that but for values x wouldn't equal 36 at its max

[quiet swan]

we have (36-2x)(36-2x)x

so we need 36-2x>0 and x>0

the bound is a bit tighter

[feral urchin]

OKOKOK

SO 18

or 17

wait maximize the volume

[quiet swan]

no because x could be 17.2

so it is just 0<x<18

[feral urchin]

gotchaaa and at this point the last input can just plug in the numbers giving the biggest which would be 6?

[quiet swan]

yeah!

depending on how much stuff you were taught u can prove it is 6

[feral urchin]

dope dope dope

prove it was 6 how exactly? see I took ap stats and pre calc and have all the notes so after this im going on a review bing

e

[quiet swan]

have you done calc?

[feral urchin]

ap calc terrified me after pre calc

[quiet swan]

personally found it better to learn the stuff rigorously, AP calc kinda hand wavey

so with calc, the idea is as follows:

the function is differentiable everywhere, so local max/min is when derivative is 0

global maximum in [0,18] must be achieved either at the end points or at the local max

[feral urchin]

gotchaaaaa okay and graphing seeing 6 on the graph its the local max

[quiet swan]

it is indeed the local max, the derivative may look ugly but it is ok

[feral urchin]

Im glad my prof lets me use my calculator but the whole graphing shit on ti 84s is atrocious

[quiet swan]

calculator in general is overrated

charging over hundreds for a graphing calc

[feral urchin]

its ridiculous

IT SAID IM WRONG I HATE LUMEN

https://tenor.com/view/rage-gif-24341837

[quiet swan]

bro second question is the x that maximize VOLUME

third is maximum volume

[feral urchin]

...

[quiet swan]

💀

[feral urchin]

im gonna cry

I can do this thing 4x but I will do inexplicable things to the world trade center if I get to that point

[severe dawn]

Online problem sets in math is hell on earth

[quiet swan]

misreading questions happens to all of us lol

[feral urchin]

Alright with this new info i'm gonna see how this other stuff goes thank you qwerty I hope inexplicably good things happen to you

[quiet swan]

@feral urchin btw .close to close the channel

[feral urchin]

.close

[calm coral]

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[feral urchin]

.reopen

[calm coral]

✅

[feral urchin]

I dont know where i went wrong

[dull wagon]

how are you getting 4.5

[feral urchin]

desmos graphing the local minimum being 4.5

[dull wagon]

you're not being asked for min

[feral urchin]

so it's still 1.5?

[dull wagon]

no

[feral urchin]

oh then it'd be 54?

[dull wagon]

yes

V(x) = 54 when x=1.5

[feral urchin]

gotchaaaaa

thanks

.close

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[calm coral]

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[barren nymph]

how are these two angles related?

[dull wagon]

supplementary

[barren nymph]

but wouldn't that cause the answer to change?

[dull wagon]

sin(x) = sin(180° -x)

[barren nymph]

oh

[dull wagon]

supplementary property for sine

[barren nymph]

oh

well could you elaborate a bit why cos and tan become negative when you do the supplementary angle thing on them?

[dull wagon]

unit circle and/or compound angle identitites

[barren nymph]

so is theta always smaller or larger than 180

because assuming lets say...

I have cos(355)

oh

its negative because when you subtract 180 from the angle goes into the other quadrant it becomes a negative right

[velvet osprey]

cos(355) is positive

[barren nymph]

indeed

[velvet osprey]

"the other quadrant" is vague

[barren nymph]

when cos355 is in Quadrant 4 its positive

so when you subtract 180 it goes into quadrant 3

[velvet osprey]

bad wording

also no, 355 - 180 = 175 and that is in quadrant 2 not 3

[barren nymph]

oh yea your right

wait, wouldn't sin(355) become sin(-175)

so it would be in quadrant 3 in the "negative unit circle"

so its technically sin(5)?

so it would be sin 5 or sin175

<@&286206848099549185>

:/

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[ripe trout]

v1,v2...,vm,w are vectors in V
prove that if {v1,...,v1m} is linearly independent
and
{v1-w,v2-w,..,vm-w} is linearly dependent
then w is a linear product of {v1,..,vm}

[ripe trout]

if a1v1+...+aivi+...+amvm = 0 then a1 = ... = ai = .. = an = 0

and there are values b1...bi...bm so bi != 0 and b1(v1-w) + ...+ bi(vi-w) +...+bm(vm-w) = 0

but we can open the brakects and get
b1v1 + ... +bivi+...+bmvm -w(b1+..+bi+...+bm) = 0

move it to the other side
b1v1 + ... +bivi+...+bmvm = w(b1+..+bi+...+bm)

bi != 0 =>b1v1 + ... +bivi+...+bmvm != 0

(b1+..+bi+...+bm) != 0

and now we divide by (b1+..+bi+...+bm) and we get w as a linear prudct of v1,,,,vm

is it ok?

[vagrant bobcat]

Do you hae a picture of the probnlem instead?

or the work

Im confused on what we are trying to prove

[novel turtle]

yes, everything is fine

[ripe trout]

thx

the problem is not in english so I had to translate it

that might have caused mistakes

[ripe trout]

but I dont rly want to re explane it so if its fine by you ill just close the room

.close

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[formal sinew]

$a^2 - 1 = 0$

can u say "using the commutavity of multiplication we get:"

$(a-1)(a+1) = 0$

[potent lotus]

amin

[formal sinew]

Or are there more steps

Field axioms

[vagrant bobcat]

No because this isnt the commutativity of multiplication its not applicable I think

I think this relies solely because we are factoring a quadratic equation

This looks like a factorization rule of

$x^2 - y^2 = (x-y)(x+y)$

[potent lotus]

downhill

[prime skiff]

difference of two squares is name of rule

[calm coral]

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[sudden nebula]

Graphing Calculator X84

[sudden nebula]

I'm using a Graphing Calculator X84 and notied I get a syntax error when I try this:

Anyone have an idea why ? My professor didn't get one

[prisma mango]

How are you typing it into your calculator?

@sudden nebula

[sudden nebula]

exactly the same

oooo

I figured it out

So there's a symbol for a negative sign that I have to click

I tried just putting a minus but I guess it read it differently

[prisma mango]

Yeah there's a separate button for minus and negative

[sudden nebula]

Yeah some other calculators don't have this issue

from what I recalll

[prisma mango]

The ti calculators are designed like that

[sudden nebula]

ohhh

okay

gotcha

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[sudden nebula]

Can someone give me a hint for the total weight

The distribution of the sample is non-normal left skew, the mean is 150lb and SD is 27 with n = 35

Would I just take 5,000 / 35 ?

[calm coral]

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[unborn stone]

How do I do this

[old frigate]

Domain and range?

[unborn stone]

Yeah

[void frost]

Define domain

Also, define range

After that, you should notice how to get them

[old frigate]

Domain is basically all the possible values of x

[unborn stone]

So it’s [-4,4)?

[uneven shell]

Why )?

You've got a full dot

[unborn stone]

What full dot

[steady notch]

on the lower bound of your domain

[unborn stone]

??

[steady notch]

on the graph

at x=-4

[unborn stone]

Yeah

[steady notch]

its shaded

[unborn stone]

Yeah so it’s [

[steady notch]

and at the upper bound

its also shaded

so both have to be

]

[unborn stone]

How come on this they just combined both of them together then

And not (-6,-1]U(-1,5]

[steady notch]

this is also correct

but notice how -1 is defined for one line and not the other

however the domain are the values of x where it is defined for the whole graph

and not just the line

[unborn stone]

-6 is not defined tho

It’s open

[steady notch]

yes and?

thats why its a parentheses

[unborn stone]

So why is -1 not included

[steady notch]

it is for the domain

[unborn stone]

Their answer was (-6,5]

[steady notch]

yeah all the values from -6 to 5

it is included

[unborn stone]

So this should be [4,4) then??

Cus that’s through entire graph

@steady notch

[steady notch]

upper bound should be a bracket

since its not open

[unborn stone]

It is in a bracket

[steady notch]

its in a parentheses

[unborn stone]

@steady notch

How about these

It’s same graph

[steady notch]

same shiz

values of x

where the line would be equal to and under y=0

[unborn stone]

Só c is just 4

D I got 0,-1

[steady notch]

oh i thought you were asking d

[steady notch]

no

[steady notch]

negative 4

[unborn stone]

How

[unborn stone]

What is it

[steady notch]

2nd option

[steady notch]

look at what numbers of x

is y=4

[unborn stone]

I’m confused why was the g on question a count as the x axis but now it counts as y axis

So is it -4,4

[steady notch]

its g(x)

so the values inside the parentheses are x

and your answers are the corresponding y value

[calm coral]

@unborn stone Has your question been resolved?

[calm coral]

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[old pagoda]

<@&286206848099549185> I'll rephrase the question,
I'm trying to to approximate tan(0,1) using the Taylor polynomial around x = 0 with a remainder less than 0,001.
I solved one way of doing it but digitally, how would you show that the remainder of the polynomial you get is less then 0,001?

[old pagoda]

Tried Lagrange form but it's not like sinx or cosx, you cant really know a boundary for the derivatives of tan x right?

[hushed sequoia]

!15m

[calm coral]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[old pagoda]

Oh I should've said, if you check #help-37 I posted the question 3 hours ago.

Only switched chanel cause mine got taken while trying to phrase my question better in desperate need of help.

[calm coral]

@old pagoda Has your question been resolved?

[calm coral]

@old pagoda Has your question been resolved?

[calm coral]

@old pagoda Has your question been resolved?

[old pagoda]

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[glad dock]

$a$ and $b$ are two integers such that:
$$(1+\sqrt{2})^{2020} = a + b\sqrt{2}$$

What is the value of: $a^2 - 2b^2$?

[potent lotus]

gkeocog

[glad dock]

Without using newton's binomials or something related to it

[old frigate]

Isn't this just a linear equation?

[calm coral]

@glad dock Has your question been resolved?

[glad dock]

<@&286206848099549185>

[leaden thunder]

why can't you use this?

"anything related to it" is also vague

[glad dock]

I didn't learn it

[leaden thunder]

so why are you doing this problem

[hollow mural]

try writting it as a difference of two squares, might lead to something

thse olympiad questions always have weird tricks

[hollow mural]

question likely doesnt need it

[calm coral]

@glad dock Has your question been resolved?

[calm coral]

@glad dock Has your question been resolved?

[still marlin]

@glad dock Notice that $a^2 - 2b^2$ is $(a - b\sqrt{2})(a + b\sqrt{2}))$, then try to express $a - b\sqrt{2}$ in terms of known numbers

[potent lotus]

992qqoloy

[still marlin]

You'll have to mentally expand a lil in your mind but you don't need the coefficients, so that's doable without binomial theorem

Well you do need to know something about the coefficients, just not the exact values of them

Wishful thinking is rly helpful herw

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[empty tartan]

i dont get what im meant to do.

[velvet osprey]

find the x^10 and x^11 terms themselves maybe, for a start?

[empty tartan]

okay

do i equate them like something x^10=soemthing x^11

[visual nimbus]

yea

[empty tartan]

would the coefficient for x^10 be 15C5 x 5^5 x 2^10

[visual nimbus]

$\binom{15}{10} 2^{10} 5^{5} x^{10}$

[potent lotus]

Herels

[empty tartan]

I think thats what i wrote

right

[visual nimbus]

you wrote 15C5

🤔

[empty tartan]

oh

wait

i got the right answer somehow

[visual nimbus]

yea because 15C5 = 15C10 too

[empty tartan]

oh what

[visual nimbus]

$\binom{n}{k} = \binom{n}{n-k}$

[empty tartan]

then does it matter whit your write for nCr for 'r'

[potent lotus]

Herels

[empty tartan]

is it just better to write 10

[visual nimbus]

you write what you want i guess

15C5 and 15C10 are the same

[empty tartan]

wow thats so weird

thanks anyways, got me questioning this nCr thingy

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[empty tartan]

what does it mean by using the binomial theorem

[empty tartan]

why is image so zoomed in

yeah anyways how do i use binomial theorem

cant i just use nCr formula

[rugged crater]

yes, it means use the formula

[empty tartan]

the nCr or binomial one

[teal drift]

nCr is called binomial coefficient

I dunno if you were asking about this

[empty tartan]

i was just asking what they want me to do

do i just make nC1 = n!/n! and then thats it

[rugged crater]

is there anything else on top is that the whole question?

[empty tartan]

no

[teal drift]

Yeah I would do it that way

[empty tartan]

okay

i think i got the gist of this question where i do nC2 = 28 but it should be nC2 x a^something = 28 right?

and nC3 x a^something = 264 but not sure what that something needs to be

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[valid geyser]

pls help

[prisma mango]

Are you allowed to use a calculator

[valid geyser]

yes

[void frost]

What's the context? Have you learned about Taylor series?

[prisma mango]

Did you use it?

[valid geyser]

?

[prisma mango]

You don't need this

[valid geyser]

bonjour

[calm coral]

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

[prisma mango]

.

[valid geyser]

yes

[prisma mango]

Did you use the calculator to find sin(47)?

What did you get

[void frost]

You don't if this is a calculator exercise..
But I could imagine this coming up after you learn about Taylor series, using a = 45° and estimating the error

[valid geyser]

i got 0.12357312274

[void frost]

I assume this is not about Taylor series then, though

Since OP doesn't mention anything about them

[prisma mango]

Are you in degrees or radian mode?

[valid geyser]

what do u mean?

[prisma mango]

It means what it means

[void frost]

It should be around 0.7, not that

[prisma mango]

Are you in degree or radian mode in the calculator

[valid geyser]

wait lemme try again

[prisma mango]

It is 0.123 if OP did sin(47) in radian mode

[void frost]

Yep

[valid geyser]

0.7313537016

[void frost]

Yeah

Around sqrt(2)/2

[prisma mango]

Now round to 2dp

[valid geyser]

0.73?

[prisma mango]

Yes

[valid geyser]

ty

have a loveley day sir

god bless ya

.close

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[languid sentinel]

can somone check the answer

[rancid geyser]

have you got a clearer image i cant read the question

[calm coral]

@languid sentinel Has your question been resolved?

[languid sentinel]

o gotchu

[rancid geyser]

perfect, im dyslexic as shit sorry about that

[languid sentinel]

lol np

[calm coral]

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[languid sentinel]

.reopen

[calm coral]

✅

[rancid geyser]

looks good to me, i got the same basically the same way you did, but i did my working a lil different, i did most of the differentiating then just went straight to making it equal to 18, but it was just a corner cut no real difference, but yeah looks good to me 👍

[languid sentinel]

could u show what u did to get the answer?

[rancid geyser]

sure gimme a sec to write it all back down

i set it out so horribly im sorry, hopefully you get the gist from that

[languid sentinel]

thanks

also ca u help me out with this question as well

[rancid geyser]

sure, ill be back in a second though

[calm coral]

@languid sentinel Has your question been resolved?

[languid sentinel]

could u check the answer?

@rancid geyser

<@&286206848099549185>

[languid sentinel]

@rancid geyser

.close

[calm coral]

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[tulip herald]

Im having trouble phrasing this better

could anyone help me with this or is it fine

[humble ivy]

<@&268886789983436800>

[clear sapphire]

<@&268886789983436800>

[soft kettle]

If you want to promote, promote somewhere else, this is a math server, not some crypto market promotion area.

[humble ivy]

No, seems fine

[tulip herald]

alright thanks for verifying

.close

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[brittle gate]

im not sure if i got these right, could someone check for me?

[raven pier]

In the expansion of27() (3 2 )+−axx, the coefficient of the 8xis -320. Find the value of a.

[polar jetty]

!occupied

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

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<@&286206848099549185>

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<@&286206848099549185>

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@brittle gate Has your question been resolved?

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_ _

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[median schooner]

Find the point on the surface $z^2 = xy + 1$ which is closest to the point $(4, 2, 0)$

[potent lotus]

o.O

[median schooner]

I started with

$d = \sqrt{(x - 4)^2 + (y - 2)^2 + (\pm \sqrt{xy + 1} - 0)^2}$ so we have the equation $d = \sqrt{x^2 - 8x + y^2 - 4y + xy + 21}$

[potent lotus]

o.O

[median schooner]

Then we know this is minimized when $d^2$ is minimized

[potent lotus]

o.O

[median schooner]

How do I setup the lagrange multiplier using the partial derivative

[marsh valley]

Can't you just minimize d^2 using a horizontal gradient?

Unless you have clear constraints you can't really use lagrange multipliers

[hollow totem]

The surface is the constraint, no?

[sweet briar]

yeah i wouldve thought so too

[hollow totem]

First, rewrite the surface as a level curve of some function

or wait, you got rid of the z parameter

[median schooner]

I think I took the wrong approach

[marsh valley]

I'm still not entirely sure why you'd need LM here.

You can do it with it, but since you've plugged it in alreaady you just get a function of 2 variables which you can simply solve for extrema

[hollow totem]

You might need to be careful that your "distance"^2 doesn't attain a negative value anywhere

[median schooner]

wait so we have $z^2 = xy + 1$ which gives us $z^2 - xy = 1$ we have $f(x, y, z) = z^2 - xy$ and our constraint is $f(x, y, z) = 1$

[potent lotus]

o.O

[hollow totem]

By solving for z you made the false assumption that it was real everywhere

[median schooner]

okay

so using this then, I find the gradient / partials and set them equal to 0

$\triangledown f = (-y, -x, 2z)$

[hollow totem]

You find the gradient of the level curve and the gradient of the distance to (4,2,0) squared

and then you set them to be multiples of each other

isn't it +2z?

[median schooner]

oops

[potent lotus]

o.O

[hollow totem]

So then g(x,y,z) is square of distance to (4,2,0)

[median schooner]

ohh

so the constraint is g(x, y, z) = 1?

[hollow totem]

nah

your constraint is f=0

[median schooner]

oh

[hollow totem]

or f=1 ig

If you leave the constant there

g is the function you're trying to minimize

[median schooner]

oh ok so we have $g(x, y, z) = (x - 4)^2 + (y - 2)^2 + z^2$

[potent lotus]

o.O

[hollow totem]

right

[median schooner]

and then we have $\triangledown g = \lambda \triangledown f$

[potent lotus]

o.O

[median schooner]

to setup the lagrange

[hollow totem]

yeah

[median schooner]

so we want

oh I should've calculated the gradient of g

$g(x, y, z) = x^2 - 8x + 16 + y^2 - 4x + 4 + z^2$ so $\triangledown g = (2x - 8, 2y - 4, 2z)$

[potent lotus]

o.O

[median schooner]

and then we can setup the linear system

wait I've messed something up I think

$$2x - 8 = \lambda(-y)$$
$$2y - 4= \lambda(-x)$$
$$2z = \lambda(2z)$$

[potent lotus]

o.O

[hollow totem]

yeah

[median schooner]

this implies $\lambda = 1$ so $8 - 2x = y$ and $4 - 2y = x$

[potent lotus]

o.O

[hollow totem]

or z=0

[median schooner]

o

how do i carry forward with that

[hollow totem]

solve both

it's quite possible to get multiple solutions

then you have to test them all

start with yours though

[median schooner]

do I solve for which x, y satisfies $8 - 2x = y$ and $4 - 2y = x$ first

[potent lotus]

o.O

[hollow totem]

yeah sure

[median schooner]

x = 4, y = 0

[hollow totem]

yeah

then find z

[median schooner]

how do I do that since I assumed lambda = 1

that has all values of z give 2z = 1(2z)

[hollow totem]

right

so it's a different constraint

other than the lambda thing, what do you know about your point

[median schooner]

we need z^2 = xy + 1

[hollow totem]

Yeah

[median schooner]

so z is 1

if x = 4, y = 0

[hollow totem]

or -1

[median schooner]

oh, true

so we have found (4, 0, -1) and (4, 0, 1)

[hollow totem]

Right

[median schooner]

so these points on the surface are equal in distance to (4, 2, 0)

and that sounds right just thinking about it

[hollow totem]

Yeah the surface is symmetric about the xy-plane

[median schooner]

did you know that prior to finding to us finding the points? i didnt recognize that

[hollow totem]

So if there's a closest point where z≠0, its mirror is also closest

yeah

[median schooner]

is it because we are squaring z

[hollow totem]

yeah

[median schooner]

oh i see then

thank you

[hollow totem]

You're welcome

Then there's the z=0 thing

[median schooner]

do we still use the constraint z^2 = xy + 1

[hollow totem]

Yeah

[median schooner]

$$8 - 2x= \lambda{y}$$
$$4 - 2y = \lambda{x}$$
$$z = 0$$

[potent lotus]

o.O

[hollow totem]

Yeah

[median schooner]

$$\frac{8 - 2x}{y}= \lambda$$
$$\frac{4 - 2y}{x} = \lambda$$
$$\Rightarrow \frac{8 - 2x}{y} = \frac{4 - 2y}{x} = \lambda$$

so this follows from the system

[potent lotus]

o.O

[median schooner]

since we have $z = 0$ how do we resolve the $\lambda$ since it is allowed to take on any value and still satisfy $0= \lambda * 0$

[potent lotus]

o.O