help-27

[glossy dew]

just rotated at an angle

use the right hand thumb rule

[river pebble]

k vector is associated w z-axis, j vector y-axis tho?

[pseudo basin]

yes, in fact it applies in any number of dimensions

[pseudo basin]

yes

[glossy dew]

physics axes look weird

but they are correct

[pseudo basin]

i j k are the basis vectors pointing along the x y z axes resp.

[glossy dew]

with reference to physics

[river pebble]

i mean these unit vectors are notated wrong

[glossy dew]

cross product of i and j should give k

and it indeed does in organic chemistry tutors diagram

using right hand thumb rule

[inner ibex]

my blue line is OA

im confused

at where OA actually is.............

[river pebble]

yeah im aware.

i'm saying it's wrong that the unit vector j is in the z direction and k is in the y direction

[inner ibex]

my green is OB

where is OA

[glossy dew]

it's not really wrong

just oriented differently

like i rotate a book

along an axis

but the distance and direction of a point from one of its corners is the same

[inner ibex]

guys where would OA be

[glossy dew]

hmm

[inner ibex]

do u like my arts and craft

[river pebble]

bro this is NOT what im saying 😭

[river pebble]

ofc u can rotate the axis but the j vector is literally the unit vector in the Y direction

[glossy dew]

sigh

just search axes of unit vectors

you'll see

[river pebble]

do you not see what's wrong here

[glossy dew]

because theres nothing wrong

there's no rule that y points up

anything is fine

[meager swift]

i think she means z axis is being represented by j

[pseudo basin]

she

[meager swift]

mb

[glossy dew]

you can even turn it to make y point up and z point into the screen

that's alright too

[river pebble]

i get that you can rotate an axis yes 😭 🙏

[inner ibex]

this is my other one

[inner ibex]

but guys where would OA be

pls 😢 is my bed time

[glossy dew]

ou

oh shit

oh shit

oh shit

did my sleep issues kick in

[inner ibex]

hi guys

can i have some help now 😭😭😢😢😭😭😭😢😢😢😭😭😭

[glossy dew]

sorry

yeah

let me just

[inner ibex]

U CANT FALL ASKEEP NOW

IVE BEEN WAITING

is this OA

[glossy dew]

why is there a cube

[inner ibex]

um sorry

that was to help me

is only supposde

to be the green line

[glossy dew]

oh

[inner ibex]

but i kinda hit it

wait

um it’s kind of hard to see..

but that’s.. it..

[glossy dew]

hm

can you possibly use

OR = (OA + OB)/2?

[inner ibex]

wouldnt the line segmetn AB

be OB - OA

I did this but it’s wrong

[river pebble]

why did u subtract?

[inner ibex]

um..

bcoz i thought line segment AB

would be OB - OA

guys..

[glossy dew]

that's true..

[inner ibex]

but i did that

didnt i

[glossy dew]

but how would you find the midpoint of it

from O

[inner ibex]

find AB and then half it

[glossy dew]

that's not the same

[inner ibex]

idk because i cant draw my diagram

):

[glossy dew]

can you see it?

[inner ibex]

omg what is this

[glossy dew]

im guessing you average OA and OB

[glossy dew]

the vectorw

OR is drawn from O to the midpoint of AB

[river pebble]

this is true but you don't actually have to find AB here

[inner ibex]

im sad

.close

[devout snow]

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[bitter jacinth]

what app is that chat

[inner ibex]

is one note

i think its free if u have student email

[glossy dew]

https://youtu.be/4v1TRst4QCw?si=_3AbqDQmMsgF4im3
before i fade away

[inner ibex]

i keep all my school stuff in folders lol

[bitter jacinth]

Is it available on in pc lol

[inner ibex]

yes

[bitter jacinth]

ty chat

[inner ibex]

no

problem

[inner ibex]

thank u

are u on

🍃🍃

[devout snow]

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[lime plaza]

Uh isn't there formulas

[dim lantern]

Which ones?

[devout snow]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[lime plaza]

But half angle formulas won't give you answer in terms of the options

Or how about componendo and dividendo

[dim lantern]

Ahh i don't remember the formula for c n d

[blissful igloo]

what about rationalise

[misty crest]

no just common denominator

[dim lantern]

A+b/(a-b)=c+d/(c-d)??

[blissful igloo]

yeah just lcm maybe

[dim lantern]

I tried that

[solar goblet]

it's just adding two fractions

[dim lantern]

I'm getting 2/1-sin

Thats not even there in the options

[solar goblet]

show your work?

[river pebble]

can you show ur process

[dim lantern]

K

[misty crest]

$\sqrt{\frac{1 + \sin q}{1 - \sin q}} + \sqrt{\frac{1 - \sin q}{1 + \sin q}} = \frac{\sqrt{1 + \sin q}}{\sqrt{1 - \sin q}} + \frac{\sqrt{1 - \sin q}}{\sqrt{1 + \sin q}}$

[woven radish]

knief

[river pebble]

i think you simplified $\sqrt{1-sin^2(q)}$ wrong

[woven radish]

hye

[misty crest]

sqrt{1-sin^2}

[tawny thorn]

use this

[blissful igloo]

ok first of all write angle pls

[solar goblet]

what am i looking at?

[misty crest]

$\sqrt{a^2-b^2} \neq a - b$

[dim lantern]

It's a-b right?

[misty crest]

🤔

[river pebble]

nope!

you can simplify using a trig identity tho!

[woven radish]

knief

[dim lantern]

Ok

[river pebble]

but just on this - this would mean $\sqrt{5^2-4^2} = 5-4=1$

[woven radish]

hye

[river pebble]

not correct logic

[dim lantern]

Ohh

Then how do i solve

[misty crest]

sin^2 + cos^2 = 1

[dim lantern]

Inside the root?

[river pebble]

first, did you successfully get to
$\frac{\sqrt{1 + \sin q}}{\sqrt{1 - \sin q}} + \frac{\sqrt{1 - \sin q}}{\sqrt{1 + \sin q}} = \frac{2}{\sqrt{1-sin^2(q)}}$?

[woven radish]

hye

[dim lantern]

Ya

[river pebble]

so do you understand how to use the trig identity above to simplify the denominator?

[dim lantern]

Then it'll be cos^2

Got it

[river pebble]

yep

so do you see how to get to the final answer now?

[dim lantern]

2sec?

[tawny thorn]

yes

[river pebble]

yeah but in this case they don't simplify to sec

[dim lantern]

2/cos nvm

[tawny thorn]

sec(x) = 1/cos(x)

[dim lantern]

And i cant solve this one as well

[tawny thorn]

you have : tan²(A)+1 = sec²(A) , from this you get : cos(A)=1/(sqrt(1+tan²(x)))

[dim lantern]

And then i can use the other identity

[tawny thorn]

sin(A) = tan(A)/(sqrt(1+tan²(A)))

[dim lantern]

Cos² sin² one

[tawny thorn]

then you put tan(A)= sqrt(2)-1

[dim lantern]

Ya this one easy as well

Woah

Thanks y'all

[tawny thorn]

No problem

[dim lantern]

.close

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[lost laurel]

We can start by paramatrizing the ellipse

to do we we have $x= \sqrt{5} \cos(t); y =\sqrt{5} \sin(t); z = 3 - \sqrt{5} \sin(t)$

[woven radish]

What a wonderful world !

[lost laurel]

so I have $r(t) = ( \sqrt{5} \cos(t), \sqrt{5} \sin(t), 3 -\sqrt{5} \sin(t))$

[woven radish]

What a wonderful world !

[lost laurel]

So $r'(t)= ( - \sqrt{5} \sin(t) ,\sqrt{5} \cos(t), -\sqrt{5} \sin(t))$

[woven radish]

What a wonderful world !

[sly blaze]

and now substitute (1,2,1) in yea

you've pretty much done it

was there something you were struggling with on it

[lost laurel]

so I then have $(1,2,1) + \lambda r'(1,2,1)$

[woven radish]

What a wonderful world !

[lost laurel]

not really

just wanetd my answers verified

[sly blaze]

r is only in terms of t so you don't have r'(1,2,1)

[lost laurel]

at, right

[sly blaze]

so use the original parameterisations

for the x=1,y=2,z=1

[lost laurel]

so, uh

hmm

I havr to figure out what t is now

[sly blaze]

yes you're finding the parameter t corresponding to (1,2,1)

or rather, if you substitute your (1,2,1) into the parameterisations you used for x,y,z you can then find the direction vector this way

e.g x=sqrt(5) * cos(t)

so cos(t) = 1/sqrt(5)

and you'll end up with the value r'(t) which corresponds with (1,2,1)

if you understand

[lost laurel]

the issue is

I also have $2 = \sqrt{5} \sin(t)$

[woven radish]

What a wonderful world !

[sly blaze]

That's fine

[lost laurel]

how

[sly blaze]

if you look you have cos(t) = 1/sqrt(5) and sin(t) = 2/sqrt(5)

so cos^2 + sin^2 = 1

[lost laurel]

ooh

[sly blaze]

it's consistent

[lost laurel]

right

my bad

yea

[sly blaze]

good to check though when you're doing it in exams in stuff

tbf i thought i did it wrong as well lmao cause it does not look like those would be the same t

[lost laurel]

$t = arcsin(\sqrt{1}{5})$

[woven radish]

What a wonderful world !

[sly blaze]

yes although in this question you can get away with not evaluating t directly since your r(t) only uses sin(t) and cos(t) which you already know the values of

either way gets you the same answer

[lost laurel]

cool

tahnks

.close

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[raw karma]

I don’t understand how 2a = 8 but then 1 a = 3

[raw karma]

Could someone please explain this to me

[winter patrol]

what illegible monstrosity is this

its $2^a$, not $2a$

[woven radish]

ℝαμOmeganato5

[winter patrol]

whos work is that?

[raw karma]

Mine…

Wait a minute

It’s 2 to the power of a?

I did not realise that

[winter patrol]

the way you're writing your 6s makes it resemble 8s

[raw karma]

Not nice ):

[raw karma]

But anyway, it slipped my brain that it was 2 to the power of a so I get it now

[winter patrol]

the way you structured your work is also not ideal

a bit repetitive and all over the place

$2^a \times 3 \times 5^2 = 600 \
2^a \times 75 = 600 \
2^a = 8 \
a = 3$

[woven radish]

ℝαμOmeganato5

[raw karma]

I was trying to do it more methodically by figuring out what 3 x 5 squared was and then working back from there

.close

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[spice jetty]

Can someone help me find a function that can be integrated using the Lebesgue-Integral but not with the Riemann-Integral

[wicked rover]

have u tried anything?

[stone stump]

its very weird that such an example didnt come up in your class

[wicked rover]

such examples usually motivate measure theory in the first place...

[supple knot]

this is actuallly an exercise in googling

or reading your real analysis or functional analysis textbook

[devout snow]

@spice jetty Has your question been resolved?

[raven leaf]

who's Dirichlet?

[stone stump]

why is dirichlet?

[supple knot]

how is dirichlet?

[raven leaf]

where is Dirichlet?

(probably six foot + rubble from world war ii under)

[devout snow]

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[orchid wasp]

Find center of 10x^2+2x+y^2=1

[orchid wasp]

I did 2x(5x+1)+y^2=1

But idk what now

[winter patrol]

undo

you want to complete the square for the
10x^2+2x

factorising the way you did isn't helpful here

[orchid wasp]

So how do I get rid of the 10

10(x^2+2x/10)?

[winter patrol]

yeh

[orchid wasp]

How do I find the major diameter of it

I got (x+1/10)^2+y^2=0.11

[winter patrol]

how'd you get that

[orchid wasp]

From the 10(x^2+2x/10)+y^2=1

[winter patrol]

show exactly what you did

[orchid wasp]

I complete square of x and then added right side 10 times (1/10)^2 and alsodivided the left side

[winter patrol]

can you take a pic of your work
or type the equations you went through

[winter patrol]

you didn't divide the left side properly

(you didn't divide the y^2 term)

also, dividing by 10 isn't ideal and isn't really needed if you just want the centre

[orchid wasp]

Oh

So what I do to find the major diameter with it

[winter patrol]

you want to get the equation in standard form,
and have 1 on the right side of the equation

[orchid wasp]

Ok so it’s (x+1/10)^2/0.11 +y^2/1.1=1

?

[devout snow]

@orchid wasp Has your question been resolved?

[devout snow]

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[jagged lily]

Erm

[jagged lily]

Not a clue where to start at all

[lusty sapphire]

Feels like telescoping

That's my guess anyway

[devout snow]

@jagged lily Has your question been resolved?

[devout snow]

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[junior lagoon]

hey can someone help breakdown this question for me, I should have a question similar to this on the actual exam and I want to know how to approach a question like this, Thanks

[south fiber]

which part are u struggling with

[junior lagoon]

I want just a general walk through in how to approach and solve a question like this

[south fiber]

ok

well for the first one it's asking you to find an expression for h which is labeled in the picture

we have a sidelength of the trapezoid which can be considered a hypotenuse to find H

we're also given the top side length which is b cm long as well as the bottom which is 10 cm short

to find the basw of the triangle you would have to subtract 10 from B then divide it by 2

do you follow so far

[junior lagoon]

yes

[south fiber]

so we know that pythagorean theorem is a^2 + b^2 = c^2

c^2 being the hypotenuse

in this case c=20

[junior lagoon]

and would b be 10-b/2 ?

[south fiber]

so the full equation for this specific triangle would be ((b-10)/2)^2+h^2=20^2

[south fiber]

yes exactly

so the only unknown is H

you can use algebra to isolate it

this also plays in to the next question to find the top length of the trapezoid

[junior lagoon]

so we treat b as like the function of h (idk what we call it)

[south fiber]

yeah the equation

since you don't have to find the actual number you can just use an expression

[junior lagoon]

so (b-10/2)^2 - 400 = h^2?

[south fiber]

yep exactly

h^2

square root the whole thing

[junior lagoon]

(b-10/2)-20=h?

[south fiber]

not exactly because you have to square root the entire expression on the other side of x

u didn't independently

[junior lagoon]

ah ofc

[south fiber]

so it would be ((b-10)/2)^2-400)^1/2=h

[junior lagoon]

okay cool

[south fiber]

^1/2 js means square root

[junior lagoon]

yup

[south fiber]

can u do the rest?

[junior lagoon]

I was just wondering

[south fiber]

im gonna be right

bakc

[junior lagoon]

thats all g

[south fiber]

ok back

[junior lagoon]

could i use:
(h/20)=sin(x)

[south fiber]

wait hold on i gotta go again

[junior lagoon]

and then make it
h=20sin(x)

this could follow with the next question to finding b

like if we have another variable say a as the adjacent side of the right angle triangle

we could do a/20=cos(x)

a=20cos(x)

b=10+2a for both right angles?

[south fiber]

yep

[junior lagoon]

ok

so on section c

c. Show that the cross-sectional area of the box guttering, cm2, is given by

A=200sin(x)(2cos(x)+1)

how would this be approached

would it be by basically inputting all this into the area of a trapezium?

[south fiber]

yep

once you do that js use uv'+vu' to find the derivative for question D.

are you allowed to use a graphing calculator?

[junior lagoon]

id assume a question like this would be tech active

so prolly

[south fiber]

it's possible without one but it just saves a lot of time

for question e

[junior lagoon]

hm

just back to question c real quick

i have no clue how to get to that

[south fiber]

sorry i cant help rn

i have to go

im so sorry 💔

[junior lagoon]

thats all good thank you so much for the help btw

helps a ton bro

[devout snow]

@junior lagoon Has your question been resolved?

[devout snow]

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[devout snow]

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[rugged crater]

how can i find the critical points of this derivative?

i have a calc exam tomorrow and i’m a little lost

[lethal pollen]

Multiply both sides by x^2

So then we get $e^{6x}(6x-1)=0$

[woven radish]

denzio321

[lethal pollen]

Obviously no real values of x fufill e^(6x)=0

So we solve for 6x-1=0

[lethal pollen]

Since we multiplied by x^2 in the first part here it is good to note that if any of our solutions is 0 we must reject that solution

But since none of the solutions are 0 we're fine

@rugged crater

[rugged crater]

okayy thank you so much

.close

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[hushed void]

How many of the positive divisors of 7200 have square roots that are integers?

[hushed void]

Do i factor it?

[near jolt]

sure, try it out

[hushed void]

2^5.3^2.5^2

Soo 6.3.3

[near jolt]

following this, can you write out the factorization a divisor can have?

[hushed void]

Positive

[near jolt]

but we want them to have interger square root

[hushed void]

Hmmm how can we do that🤔

[near jolt]

first lets try and think about this

[hushed void]

You mean full square?

😓

[near jolt]

what do you mean?

[hushed void]

My english is bad bro sry

[near jolt]

for example a divisor can have the prime factorization 2^3.3^1.5^1

i am asking you to generalize this

[hushed void]

Ah

[near jolt]

what must a divisor's prime factorization be?

[hushed void]

This?

[near jolt]

not really.... Maybe lets go with something simpler

How did you prove the "number of divisor" formula?

[near jolt]

that is, how do you know there are 6.3.3 possible divisors?

[hushed void]

Yes because he asked for positive ones

[hushed void]

İdk ıts rote

[near jolt]

OK I think for you to solve this question you would need to know the idea behind the proof. It's important

The divisor of 7200 must be of the form 2^a.3^b.5^c, where, 0<=a<=5, 0<=b<=2, 0<=c<=2, can you see why?

[hushed void]

What does a b and c mean

[near jolt]

variables

[hushed void]

But do we need them? 2 3 5 isnt ennough?

[near jolt]

a is any integer between 0 and 5

b is any integer between 0 and 2

c is any integer between 0 and 2

[near jolt]

because we need a way to find the divisors

[hushed void]

Hmmmmmm

[near jolt]

for example, 2^4.3.5 is a divisor

[hushed void]

İ think you just give me the way to solve it 1 min

This?

[near jolt]

yeah

[hushed void]

Bro can you tell me a bit about the prof

[near jolt]

because you have a nice way to write the divisors

if a number "n" is a divisor of 7200, what must the prime factorization of "n" be?

[hushed void]

Ummm

[near jolt]

2 divides 7200 only 5 times, how many times can 2 divide "n"?

[hushed void]

😓idkkk

N mean what

Sry man im a bit stupid

Lets skip thisbro

[near jolt]

"n" is sort of a random integer we don't know much information about (we simply know n is a divisor of 7200, that is not enough information to determine what n is)

for example "n" can be 2, 4, 8,....

[hushed void]

2^n ?

+-1

[near jolt]

try and avoid using the same variable for different stuffs.
"n" cannot be 64 because 64 is not a divisor of 7200

[hushed void]

yeah that makes sense

[hushed void]

that makes sensetoo

@near jolt bro i think you explained but some words doesnt realy translate to turkish i think thats why we cant understand each other..

Wanna skip this?

[near jolt]

yeah I was just trying to guide you to the same solution.

the solution i have is the same as theirs

[hushed void]

Wanna see my circle question?

i want to see otherpeople solutions

İf u want

[near jolt]

maybe you haven't learned variables but idk, id think all schools at some point teaches them

[near jolt]

sure

[hushed void]

What is the length of the radius of the great circle?

This from my testbook

İ started like this

[near jolt]

how do you know O, O2,E is a straight line?

[hushed void]

I guess that part is not visible because the circles are tangent to each other from the inside.

[near jolt]

oh ok

right

[hushed void]

soo what can we do next?

[wheat pawn]

well, for a start, i'd get AB

[hushed void]

@wheat pawn

[wheat pawn]

where do you get x, x+1 and x+2 from?

[near jolt]

radius is fixed they set it to x+5

OG=OE=OF=x+5

[hushed void]

Yep

[wheat pawn]

okay, are you allowed trigonometry?

[hushed void]

Yay

Whatever you want

[wheat pawn]

Okay. Then i'd get the angle at B

[hushed void]

Bro are u sure cuz trigo make this gonna long and hard

[wheat pawn]

and then you solve the quadrilateral O, O1, B, O3

[hushed void]

What about this

[wheat pawn]

you got O1 to B being 5

you got O3 to B being 3

you can compute the angle at B

[hushed void]

Hmm🤔

[wheat pawn]

with that you can use pythagoras twice on the bottom right triangle, and the left right triangle

and that should give you x (on a pretty annoying equation tbh)

[near jolt]

I don't think they have extracted enough enformation to solve for x

[hushed void]

Ye i dont think soo

[near jolt]

there should be many solutions possible for x

[wheat pawn]

should give him 2 solutions, at most

[hushed void]

Lets focus on the bottom triangle?

[wheat pawn]

lemme add names for extra points for clarity

you know the angle of B comes from the 6-8-10 right triangle

so you know the three trig functions on b

cosB=6/10, sinB=8/10

(you can work with this instead of the angle itself)

since you have that from B, you can compute O3 to P, and PB

since you have PB, you know that O1 to B is the O1 to P plus PB

that's also OQ

[hushed void]

🤔

[wheat pawn]

since QP is x, you also know O3 to Q

so now you have the triangle O, O3, Q, and thus can solve x

with pythagoras

[hushed void]

What u find for x

[wheat pawn]

i mean, i'd have to actually do those steps to get the value

[hushed void]

Bro inthink we should focus 3-4-5 not 6-8-10

[wheat pawn]

congruent triangle, same angle, same sin/cos/tan

[hushed void]

🤔🤔🤔hmmmmmmm

Sin^2alpha+cos^2alpha=1 ?

[wheat pawn]

cosB = 6/10 as i said before.
and cosB = BP/BO3 = BP/3

that gives you BP = 1.8

same with PO3=2.4

can you continue from there?

[hushed void]

ill try

[near jolt]

where did you get OO1 perpendicular to O1B?

[hushed void]

Yeah we try solve it

[near jolt]

or is that just a angle you drew(that isn't a square)

[hushed void]

@wheat pawn idk bro it seems not a good way

İ cant countinue it

İm stuck at where you stop

Help

[wheat pawn]

Since you have PB = 1.8, PO1 = 5-1.8=3.2
Since you have PO3=2.4, and PQ=x, QO3=2.4-x
Since you have those two, you now have the triangle
O, O3, Q, which is right triangle, and you have both legs QO3, OQ=PO1 and the hypothenuse x+2.
Use pythagoras on the triangle. Get x.

you'll have (2.4-x)^2+3.2^2=(x+2)^2

[near jolt]

this is based on the false assumption that OO3 is perpendicular to O1B

I roughly drew a graph on geogebra they are almost perpendiculat but not quite

about 83 degree ish

[wheat pawn]

no it isnt, and in the drawing you very clearly see how i drew them explicitly not perpendicular

[near jolt]

wait then how is QO3=2.4-x

[near jolt]

mb i mean OO1 perpendicular to O1B

mb i mistyped

[wheat pawn]

i'm assuming that OO1 is perpendicular to BO1 (which is), PO3 is perpendicular to BO1 (by how i defined P), and OR to PO3 (again, by how i defined OR)

[near jolt]

why is OO1 perpendicular to BO1

the graph I drew shows an angle to 83 ish

[wheat pawn]

O1 is the midpoint of AB

[near jolt]

but how does that say OO1 perp BO1?

[wheat pawn]

Circle centered on O is tangent to circle centered in O1 with radius O1F=5

[hushed void]

Ye

[wheat pawn]

thus the center of both O and O1 lie on the midline of AB

[near jolt]

by the same logic you can say OO3 is perp to O3 B but it isnt

[hushed void]

Btw guys if im not wrong this is the solution

[near jolt]

tangency of circle does not imply that OO1 is per to AB

[wheat pawn]

tangency of circles means that the tangency point and the centers are in the same line

[hushed void]

Hmm🤔

[near jolt]

yeah that is true

but that still doesn't tell us anything about AB being tangent to OO1

[wheat pawn]

im using this

[near jolt]

yeah im aware

[near jolt]

but by the same logic, why isn't OO3 perpendicular to BO3?

[wheat pawn]

hm

[near jolt]

im saying that sure, OO1 F are on the same line

but this tells us nothing about its relation with AB

[arctic field]

OO1 and O1B are not perpendicular

[wheat pawn]

yeah that is indeed a faulty assumption

[near jolt]

im amazed you drew that on desmos

[arctic field]

i know the equation of the circle

[wheat pawn]

why not? it's one of the fastest ways to check constructions

[hushed void]

Soo guys the problem is wrong?

[arctic field]

no

[wheat pawn]

no

[arctic field]

you're just solving it wrong

[hushed void]

Why my answer is wrong 🤔

[arctic field]

which one

[arctic field]

this one?

[hushed void]

.

[arctic field]

it's right though

[hushed void]

Yes

[hushed void]

Do you have a idea for shorter

i feel like i solved itwrong

[arctic field]

i didn't do it that way and i can't be bothered working out what you've done

[wheat pawn]

they were referring to the assumption i was making from misinterpreting one of your early lines

[hushed void]

Nvm bro this bullahit skip?

This problem is just

Bullshit

Lets skip

.end

.close

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[near jolt]

yeah it looks right

[devout snow]

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[icy zinc]

Basic statistics question

I'm confused about what the interquartile range means. It's measuring the spread of the middle 50% of the dataset right?

Suppose the dataset is
1 2 3 4 5 6 7 8 9 10

The Q1 to Q3 goes from 3 to 8. That includes 6 numbers which is 60 % of 10 not 50%

Suppose the data set is

1 2 3 4 5 6 7

Q1 to Q3 is from 2 to 6

Thats more than 50 %

Maybe spread of the middle 50% is supposed to be an inexact definition?

[devout snow]

@icy zinc Has your question been resolved?

[mild comet]

If the size of the data set is small, then it may not be very close to 50%

according to wikipedia, this is the exact definition

[devout snow]

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[willow cypress]

This system of equations thing is killing me 💔

[iron kindle]

,rccw

[willow cypress]

so I know I gotta make like BA=RN

Uhm, so I wanted to get rid of the fractions right

but I realized, idk what to multiply by for each equation

Should I treat each equation as it's own thing

So multiply the top one by 5 and bottom one by 4

Or should I do both so multiply everything by 20

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[worthy heath]

for a function $f(x)=x^2sin(\frac{1}{x})$ find f'(x) at x=0

[woven radish]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[wind mason]

[worthy heath]

💔

[wind mason]

We must prove the limit at x->0 first, no?

Or is this just a calc question?

[worthy heath]

i messed up my basics about continuity and differentiability

[worthy heath]

ok i know function tends to 0 at x=0

so it is continuous

[wind mason]

Yep.

[worthy heath]

so its derivative must exist at x=0

right???

[twilit comet]

squeeze theorem

[worthy heath]

what other 2 functions would you assume?

[wind mason]

I mean, not necessarily.

But the point is to say that (no continuity)->(no derivative). The converse is not 100% true.

[worthy heath]

ohhh

so what is happening here?

we know its continuous

how can i prove whether it has a derivative at x=0 or not

[wind mason]

(You calculate it.)

[worthy heath]

wait

so can i do this

$\lim_{h\to 0} \frac{h^2sin(\frac{1}{h})}{h}$

[woven radish]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[wind mason]

Yes.

[worthy heath]

so this is 0

so derivative at x=0 is 0

[wind mason]

Wait.

[wind mason]

How exactly did you get this?

[worthy heath]

f(0+h)-f(0)/0+h-0

[wind mason]

Ah.

Yeah, okay.

[worthy heath]

butt wait

if i differentiate it

and then take lim x to 0

then LDE

how

[wind mason]

Well, because of the lone $\sin(\frac1x)$ term.

[woven radish]

;(

[worthy heath]

cos*

but

[wind mason]

Ah. Well, same thing.

[worthy heath]

derivative exists at x=0 right?

we just found its value too

[wind mason]

Yes.

[worthy heath]

then why does differentiating and doing it messes it up

[wind mason]

Let me double check, I might be mixing up my values.

[worthy heath]

okay

wait does a function being differentiable at x=0 and having a derivative at x=0 mean two different things?

[wind mason]

Eh.

[wind mason]

Yeah, I had my values mixed up.

It’s undefined, notice what happens in the limit.

[worthy heath]

so f'(0) does not exist?

this is confusing

[supple knot]

,w d/dx x^2 * sin(1/x) at x=0

[woven radish]

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[supple knot]

indeed

oh

[worthy heath]

then whats wrong with this?

f(0)=0

which is known

[supple knot]

wolfram says undefined because the function is undefined at x=0

[supple knot]

you didn't say this in the beginning

[tawny thorn]

so the limit is defined ?

[worthy heath]

the derivative is 0 at x=0 but the derivative is discontinuous at x=0 ????

[supple knot]

yea this is correct now that we know f(0) is defined to be 0

[worthy heath]

oh my bad

[supple knot]

yea if it helps here's the visualization

[worthy heath]

.

[worthy heath]

understood

[supple knot]

,w diff x^2 * sin(1/x)

[woven radish]

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[tawny thorn]

that why its undefined

because , cos(1/0) is undefined

cos(infinity)

[worthy heath]

hm

i cant wrap my head around it

[worthy heath]

yes i know what ur saying, but i wanna know if i got the derivative as 0 then why is it discontinuous there?

[supple knot]

the function is differentiable, but not continuously differentiable

meaning the function f(x) can have a derivative at a point, but the derivative f'(x) need not be continuous at that point

[worthy heath]

ohhh it makes sense now

yeah yeah i got what youre saying

thank you so much

have a nice day

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[devout snow]

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[wispy geyser]

am i right with B here?

[wispy geyser]

<@&286206848099549185>

[supple knot]

!15m

[devout snow]

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.

[wispy geyser]

my bad.

just in a rush 😭

[supple knot]

yea that doesn't matter

[rustic jetty]

,w 3 + cos(x) = 2cos^2(x)

[woven radish]

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[rustic jetty]

yeah correct

[wispy geyser]

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[violet vault]

I'm trying to follow my notes from class and is this right?

[violet vault]

Cause I don't understand why I have 4A+Ax^2 ?I thought it would be like Ax^2+2Ax+2A+Bx+C=x^2-2x+1

[dense jay]

Ax^2+Bx+C original
2Ax+B first
2A second
0 third and so forth

[fourth]+2[second]+[original]
[0]+2[2A]+[Ax^2+Bx+C]

[violet vault]

What abou the first? Whta kind of a rule is this?

I don't understand the pattern

[dense jay]

your equation only has the fourth derivative, the second derivative and the original y in it

so you only need y''''=0, y''=2A, y=Ax^2+Bx+C

[violet vault]

Ooohh right right

[violet vault]

Thank you!

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[wind mason]

Is it mathematically correct to say that an axiom can overpower another?

[jagged harbor]

what does it mean for one axiom to overpower another axiom

[wind mason]

Like, for example, $\int_{-a}^a\frac1xdx=0$ for $a\in\mathbb{R}$. However, an integral over an area with a vertical asymptote technically does not exist, right? So why is this the case?

[woven radish]

;(

[supple knot]

where are the axioms in that statement

[wind mason]

$\int_{-a}^af(x)dx=0$ for $f$ and odd function, $a\in\mathbb{R}$.\
An integral over a vertical asymptote does not exist.

[woven radish]

;(

[wind mason]

I feel like I am getting the concept itself wrong.

[jagged harbor]

that's not an axiom

[jagged harbor]

also, the premise is incorrect

[supple knot]

,w int -1 to 1 of abs(1/sqrt(x)) dx

[wind mason]

But 1/(x+1) is not odd?

[woven radish]

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[supple knot]

oh

odd

[faint gorge]

abs value so the square root doesnt yield negative

[tall knoll]

it's even actually

[empty flame]

Yes mb that's why I deleted

[supple knot]

,w int -1 to 1 of 1/sqrt(x) dx

[woven radish]

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[supple knot]

checkmate real analysts

[acoustic leaf]

you can integrate an odd function with a vertical asymptote and have it converge (in which case the integral must be 0), e.g. x^(-1/3), but this integral is just divergent

[wind mason]

💀

Lol

[faint gorge]

lebsgue integrable

[faint gorge]

riemann

[wind mason]

Hmm.

,w plot 1/cbrt(x)

[woven radish]

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[wind mason]

How does it diverge?

Maybe my calculator misled me.

,w integrate from -1 to 1 1/x

[woven radish]

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[wind mason]

Odd.

Guess desmos broke.

[acoustic leaf]

as with any integral involving a vertical asymptote, you have to split up the integral up so that you integrate, e.g. from -1 to 0 and 0 to 1. if both integrals converge themsevles then the integral converges overall

[wind mason]

But I mean, my question still holds.

[wind mason]

Ok.

[jagged harbor]

your question has seemingly nothing to do with your example

[wind mason]

Yes, desmos broke.

[jagged harbor]

poor desmos

[wind mason]

How does that work for 1/cbrt(x)?

[acoustic leaf]

,, \int_{-1}^1 x^{-1/3} \odif x = \int_{-1}^0 x^{-1/3} \odif x + \int_0^1 x^{-1/3} \odif x = \lim_{b\to 0^-} \int_{-1}^b x^{-1/3} \odif x + \lim_{a \to 0^+} \int_a^1 x^{-1/3} \odif x

[woven radish]

cloud

[wind mason]

Oh. This.

[acoustic leaf]

this is a fairly standard way of treating improper integrals

[wind mason]

I forgot it.

💀

Okay, guess I was just being stupid.

.close

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[jagged harbor]

so where's the axiom

[acoustic leaf]

so for the first theorem, it needs to be amended to say "if it converges", and the second one is just wrong

and i think you mean "theorem" rather than "axiom"

[near jolt]

you could say something like the axiom "f is differentiable" overpower "f is continuous" (this is pretty informal)

[jagged harbor]

and my point of asking what it means for one thing to overpower another is, it's unclear what you want it to mean, so no, it could never be used rigorously

it'd still sound silly even if you did define it

there's nothing wrong with sounding silly, it's endearing

but your question was about correctness, so I'd say it's not correct

[near jolt]

im assuming they meant that is is possible for an axiom to imply another but not the other way

[jagged harbor]

it's unclear.

[near jolt]

its probably one of the less "silly" way of interpreting it

[jagged harbor]

the use of the word "overpower" is itself silly

attempts to decipher silly language are not intrinsically silly

it is silly to use silly words like "overpower" in a rigorous context

like, you don't say one thing overpowers another while assuming you sound rigorous

[near jolt]

nah id figure we would see a lot more of that after this generation becomes mathematicians

we already got hairy ball etc

[supple knot]

d's nuts theorem

[jagged harbor]

no, there's a blatant failure in using imprecise language in a situation where precision is mandatory

[near jolt]

oh i just meant more like defining things using silly words

[jagged harbor]

you're missing the point

it's not your fault

[near jolt]

now i get what you mean by using imprecise language

i just misinterpreted to using silly words

[jagged harbor]

all is well

if the original question asker had defined overpowering to be your best guess, I'd say it's fine but wouldn't catch on

but defining a whole new object, yeah go nuts

it's a "dominating set", sounds good, thank you mx. graph theorist

[devout snow]

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[acoustic garnet]

so here im trying to solve a line integral for F(x,y) along the curve C which is given by the equation y=x^2 with 0 less or equal to x less or equal to sqrt(pi) in the direction of increasing x

but i end up on an integral which i think is way to complex for what i think its supposed to be for solving by hand, so im assuming i did something wrong

[sullen island]

have you checked if G is conservative or not ?

[acoustic garnet]

G is indeed conservative

[sullen island]

well you have F = G + something

line integral of G is easy to compute

so you're really left with the line integral of that something to compute "manually"

[acoustic garnet]

which has something to do with it being conservative?

[sullen island]

yes

the line integral only depends on the endpoints yada yada

[acoustic garnet]

alright, thank you alot. Ill try it out and see if i got it

[devout snow]

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[hybrid grotto]

Is this a valid solution to this trig substitution problem? The actual answer looks much different than mine and i assume its because instead of solving for theta via arcsec, they used a triangle to go back to the x world, though i find this way much easier if its correct thank you

[misty crest]

yea no.

check your numerator after substituting

[hybrid grotto]

Hmm where did i go wrong?

[misty crest]

major mistake

[hybrid grotto]

Oh ok let me look

[hybrid grotto]

Im confused if this form (the top one) is used wouldnt it replace the numerator?

Sorry if i sound like a noob trig sub is my kryptonite

[misty crest]

yea this is why you shouldn’t try to memorize whatever you write down without understanding what you’re writing

what you’ve written is the three different scenarios in which youd perform a trig sub and the appropriate substitution for each scenario

what you’ve written is not what the sqrt term simplifies to, it’s the substitution youd make if you saw a sqrt of that form

[hybrid grotto]

Oh my i see where ive gone wrong its still under the sqrt and i must replace the x in it not the whole sqrt

Correct?

[misty crest]

yes

[hybrid grotto]

Ahh thank you that clears a lot of things up, and side note how would you intuitively approach this problem? Rather than memorizing my professor just wrote these forms down without any intuition which kind of ticked me off

[misty crest]

well the motivation behind all of these substitutions stem from the pythagorean identities

in particular
1 - sin^2 x = cos^2 x
sec^2 x - 1 = tan^2 x
1 + tan^2 x = sec^2 x

so when we do the substitutions sec tan and sin

we get a perfect square

which cancels the sqrt making things nicer to work with

[misty crest]

and from these we can attach the appropriate constants

[hybrid grotto]

So would this correlate to the second identity? Since we are subtracting x^2 - a?

[misty crest]

yes

[hybrid grotto]

Ok ok makes sense

[misty crest]

the reason why we attach the factor of a is so that upon squaring we have a factor of a^2 inside the sqrt which we can then factor to get one of these identities

[hybrid grotto]

I see, thats much more clear thank you again

[misty crest]

you’re welcome

[hybrid grotto]

How do I close?

[misty crest]

!done

[devout snow]

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

[hybrid grotto]

!done

[devout snow]

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

[hybrid grotto]

.close

[devout snow]

Channel closed

Closed by @hybrid grotto

Use .reopen if this was a mistake.

[hybrid grotto]

Thanks again mate

[misty crest]

you’re welcome sir

[devout snow]

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[haughty dirge]

hi i just wanted to check if my homework was correct and if not could you explain what i did wrong

[misty crest]

integralcalculator.com

[haughty dirge]

thanks

[restive river]

np always glad to help

[misty crest]

thanks

[obtuse violet]

Guys I have an integral question it's been two hours and idk what I am doing wrong it's in french but here's what I can translate cuz it's too long

I need to use IBP

I got to this integral when I tried to begin from F(x)-F(1)

Any ideas please???

[misty crest]

!occupied

[devout snow]

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).

[obtuse violet]

Ohhh I see sorry and thanks

[misty crest]

you’re welcome

[haughty dirge]

and last question i hope i am very confused on how to solve this cause my teacher can’t explain correctly

.close

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[proper path]

hey i was wondering how is this wrong?

[dense rampart]

cuz the top isnt power 1

[proper path]

do i have to do long division then?

[dense rampart]

try taking away an x from the fraction, making it a coefficient of that thing

that thing can use ur decomp

[proper path]

k

[dense rampart]

how did it go

[proper path]

idk what im doing rn 😭

[tribal cairn]

do long division first

partial fractions only work properly if the numerator is at least one degree less than the denominator

[proper path]

good idea

okay so i got 1+ (72-x)/(x^2 + x -72)

oh wait i got it i was overthinking it the whole time :/

thanks everyone

.cloe

.close

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[winter patrol]

yes

it's not quite in standard form yet

you currently have 15x^2 and 5y^2 on the numerators

express those fractions with just x^2 and y^2 on the numerators

(your denominator will look unappealing)

no

$\frac{15x^2}{49} = \frac{x^2}{\what}$

[woven radish]

ℝαμOmeganato5

[winter patrol]

$\frac{5y^2}{49} = \frac{y^2}{\what}$

[woven radish]

ℝαμOmeganato5

[winter patrol]

multiply both numerator and denominator by those respective values

what did "you" mean?

the values you said

1/5 for the fraction with y

1/15 for the fraction with x

can you show mean what you mean / what you intend to do

[drifting mauve]

😔 do you know what the standard equation of an ellipse is? @orchid wasp

then just try to reduce it into that

😔

[winter patrol]

because if you just multiply by 1/5, you'd now have 1/5 of what you started with

[drifting mauve]

key idea here is that the standard form requires right hand side to equal 1

[winter patrol]

the manipulation should only change the way the expression is presented
and not the value of the expression itself

because what you're describing just outright removing the 5 and 15 from the fraction

[drifting mauve]

can't help much besides outright spoonfeeding but try to make the RHS equal 1

[winter patrol]

multiplying $\frac{5y^2}{49}$ by $\frac15$ gives $\frac{y^2}{49} \redneq \frac{5y^2}{49}$

[woven radish]

ℝαμOmeganato5

[drifting mauve]

there are elementary algebraic skills 💔 at least you learned now

[winter patrol]

multiplying numerator and denominator by the same value, is essentially multiplying by 1
which doesn't change the value of the expression
is one of the baic principle of fraction simplification

[devout snow]

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[devout snow]

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[lethal pollen]

u sub u=sin x

[old spade]

give me a second

yeah im not getting anywhere with that

im not even able to place du in the equation

[lunar eagle]

try taking cosx as u and use by parts

[old spade]

so my dv is ln(sinx)?

wait no

[lunar eagle]

if u take cosx as t u get int

0 to 1

ln (√1-t^2)dt

[old spade]

ahhhh

hold on

im getting integra [0 to 1] log(1-x^2) as one term

which will be undefined for x = 1

[devout snow]

@old spade Has your question been resolved?

[old spade]

<@&286206848099549185>

[novel dew]

@old spade

[old spade]

i dont understand

how has the limit been brought in?

[novel dew]

also ln(1-x^2)=ln((1-x)(1+x))

[old spade]

tried that way as well

[novel dew]

since we can’t evaluate the integral at 1 we just evaluate the integral as it approaches 1

[old spade]

came across the same issue where ln(1-x) became undefined after splitting

[novel dew]

use log rules

[old spade]

hmm im not familiar with this

[old spade]

is there any other approach directly from here?

[novel dew]

just do ln(1-x^2)

if the limit doesn’t exist or approaches infinity we say the integral diverges

if not then the integral is equal to the limit

[old spade]

alright, give me a minute let me see what i can do

nope not getting it

i can't solve lim x->1 ln(x-1)

is 0xlog0 somehow defined?

[faint zinc]

It's a 0 × inf indeterminate form

Can solve using L'Hopital or other techniques

[old spade]

i figured the question out just now

log 0 came in 2 terms and canceled out

im pretty sure i can do that

log0 - log 0

and the answer seems right

.close

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Channel closed

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[devout snow]

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[sand isle]

i have a q

[sand isle]

lets say we got an equation

{ x is an element of R : x^2 <= 36 }

so we'll get values of x ranging form around [-6,6]

[pale dove]

<=

not >=

[sand isle]

sorry

yes

and so

the correspondent result will be {0, 1, 4, 9 , 16 , 25 , 36}

but real number encompasses all numbers

[sand isle]

So is this the codomain

[pale dove]

yeah?