help-26

[rigid cloak]

See

We have fixed 1 member

[sonic rune]

yea

[rigid cloak]

Now did what you did in previous question

We have the sample space where 1 member is fixed

[sonic rune]

how would ido that with this

[rigid cloak]

See

U have to choose two membees

Members

So total should have been 12C2 right?

[sonic rune]

12c2?

[rigid cloak]

12 choose 2

[sonic rune]

oh yea

[rigid cloak]

To form a committee of 2 members

[sonic rune]

yea

[rigid cloak]

But now we have been given there is 1 girl

So our choices decreases

Now we need to choose

11C1

Only one member out of 11

So this is our total cases or u can say sample space

[sonic rune]

7 over 11

[rigid cloak]

Now we want the prob. That 2nd member is also a girl

So we choose 7C1

So these are the favourable cases

So now we divide favourable cases by sample space

So what we get?

[sonic rune]

7/11?

[rigid cloak]

Let's check

,w 7C1/11C1

[thorny flame]

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[rigid cloak]

No

Not the radiation constants

Aah

,w 7choose1/11choose1

[thorny flame]

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[rigid cloak]

What

What is 7 C 1

@sonic rune

Yea it would be 7/11

[sonic rune]

ye

[rigid cloak]

Enjoy

[sonic rune]

but thats nopt the answer

this is the answer

idk how to get to it

[rigid cloak]

7/15

Ok i see

We have 12C2

[sonic rune]

yea

[rigid cloak]

12x11/2

66 total cases initially

Now

They confirmed that one is girl

Oh

Got it

See they confirmed there is one girl

But they did not confirm which girl

So it will be

8C1x11C1/12C2

8x11/67

8x11/66*

@sonic rune try to think

I am going wrong i feel

Our earlier approach was correct

[sonic rune]

yea idk

[rigid cloak]

Check on the net

Idk how are they getting 15

When total people are 12

Idk

[sonic rune]

wait one sec

ok so ones a girl

and one might be a girl

idk

try working with this ig

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[magic elm]

Hey guys, I'm taking Additional mathematics igcse. in one of the topics i dont understand straight line graphs (like the whole thing) it includes dinstance between two points, midpoint, gradient, gradients of parallel and perpendicular lines, can someone give me a example of those type of questions and how to solve it step by step, I'd appreciate it so much

[magic elm]

The codes are IGCSE(0606) and 0 level (4037)

[restive inlet]

do they provide resources to you?

gradient or slope is the
(change in y)/(change in x)
between two points

[neon iron]

rise/run

[pearl fog]

yes

[restive inlet]

do the give notes/explanations of those above subtopics

[pearl fog]

meaning you move 4 up everytime you move 1 right

wdym

[restive inlet]

can you show your work for that?

you're not squaring properly

[pearl fog]

$(a+b)^2\neq a^2+b^2$

[thorny flame]

Skill_Issue

[restive inlet]

you're not allowed to square terms within () individually like that

wdym

what do you mean

can you show what you'd get in your next step

[serene talon]

wait

do you know the formula for (a+b)^2

[restive inlet]

that's exactly what we said you can't do

[serene talon]

what is it

[restive inlet]

no

$(a+b)^2\neq a^2+b^2$

[serene talon]

tell me what you think the formula is

[thorny flame]

ℝαμΩℕωⅤ

[restive inlet]

you can't square a,b individually like that and/or mess around with the sign in the middle

[serene talon]

have you heard of the distributive law

[restive inlet]

no

[serene talon]

no

try it out

[restive inlet]

forget about the square root

for now

[serene talon]

(7 + 3)^2 = 100, 49 + 9 = 58

no no

[restive inlet]

for now focus only on this component:
$$(4 - (-3))^2$$

[thorny flame]

ℝαμΩℕωⅤ

[serene talon]

a * (b + c) = a * c + b * c

that's the rule

[restive inlet]

and i'm going to remind you that

you can't square a,b individually like that and/or mess around with the sign in the middle

[serene talon]

now use that to solve (a + b) * (a + b)

• is times

negative sign?

[restive inlet]

if you want to expand stuff out, you'd need to apply stuff like the distributive property
but here, the simpler way would be to simplify the (4 - (-3)) first

[serene talon]

remember that a - (- b) = a + b

[serene talon]

you should probably also learn this

so using that formula, what do you get for (a + b) * (a + b)?

[restive inlet]

don't overthink

[serene talon]

no

ok try to simplify (a + b) * (a + b)

[restive inlet]

didn't realise you've mistaken written your indexes

no

[serene talon]

jus try this first

using the distributive law

exponentials

a ^ b

[restive inlet]

stuff like subscripts and superscripts

[serene talon]

b is the index

[restive inlet]

no

[serene talon]

NO

[restive inlet]

no

[serene talon]

simplify it using the distributive law

WHICH IS

(a + b) * c = (a * c) + (b * c)

try it

[restive inlet]

you've written $x^2,x^1,y^2,y^1$, but those should've been $x_2,x_1,y_2,y_1$, to indicate the first and second coordinates of each point

$$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$

[serene talon]

ok you have to understand

(a + b) * 2 = 2 * a + 2 * b

[thorny flame]

ℝαμΩℕωⅤ

ℝαμΩℕωⅤ

[serene talon]

but (a + b) ^ 2 IS NOT EQUAL TO a ^ 2 + b ^ 2

simplify (a + b) ^ 2 using the distributive law

to get the correct formula

try it

yes

[restive inlet]

and (a-b)^2 isn't equal to a^2 + b^2, or a^2 - b^2

[serene talon]

go on

mhmm?

YES

FINALLY

[restive inlet]

its exactly that

if you want to expand stuff out, you'd need to apply stuff like the distributive property
like above
but here, the simpler way would be to simplify the (4 - (-3)) first

can you simplify
4 - (-3)

no

how are you getting 12

no

4 isn't being multiplied to -(-3)

you shouldn't say that
1 - 1 is -1 would you?

yes

same thing applies here

you shouldn't be multiplying
4 with
- (-3)

[serene talon]

no

its 4 - (-3)

[restive inlet]

remember that a - (- b) = a + b

[serene talon]

which is 4 + 3

no

[restive inlet]

the latter

[serene talon]

like said before

(a + b) ^ 2 is NOT a ^ 2 + b ^ 2

[restive inlet]

the first one was again exactly what we are saying repeatedly NOT to do

[serene talon]

also in this situation its easier to just add first

[restive inlet]

no

[serene talon]

remember?

[restive inlet]

again
4 - (-3) isn't 12

[serene talon]

yes

good

now simplify

YES

[restive inlet]

no

why -25

yes, but why are you doing -5^2

[serene talon]

-5^2 = -25

BUT

(-5)^2 is 25

cos negative * negative is positive

only if its inside

cos the ^ comes first

pedmas

powers before multiplication

and the - counts as multiplication

but in this scenario

you are taking (-5) ^2

and negative times negative is positive

YES

[restive inlet]

long way to go about it but yes

or the simpler way would be to simplify
6-5 first

[serene talon]

yes

and 1 ^ 2 is...

yes

yes

have you learnt surds?

like

simplifying square roots

if not, just leave it as sqrt(50)

[topaz sinew]

@magic elm Has your question been resolved?

[restive inlet]

wtf, why did all the messages get deleted

[pearl fog]

yeah wtf

[rapid wyvern]

got wiped for

nword in status

[pearl fog]

what the fuck 😭

but only some of it got wiped

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@magic elm Has your question been resolved?

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[rough steeple]

In this slide, how is it that you can just divide 196 to 980 to get 5?

In order to divide the bottom, wouldn't you also have to divide 0.2 by 196? Or can you just apply the division to that one term?

[rough steeple]

This slide was on a khan academy video I am self-studying for

[restive inlet]

0.2 * 14^2
is a single term

dividng that by 196 leaves 0.2

[rough steeple]

Why wouldn't it be a separate term if you still have 0.2 remaining after dividing by 196 on top and bottom

[restive inlet]

terms are expressions separated by + or -

considering something simpler
2 * 2
divided by 2
is 2, not 1 * 1

[rough steeple]

Ohhh, that makes sense, you would only have to divide the 2 at the end to get the same result either way

that makes sense, thanks

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[rough steeple]

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[topaz sinew]

✅

[rough steeple]

@restive inlet Sorry, one last thing. I get that it is the case. But why would you say terms are seperated by + and - but not / and *

[restive inlet]

by definition

well depends on context

[rough steeple]

how so?

[restive inlet]

here you could describe this fraction as a single term divided by another

[rough steeple]

but in another case this wouldn't be the case that it would be a single term divided by another

?

[restive inlet]

can you rephrase that

[rough steeple]

I am just asking in what situation would this (the thing I am replying to) not be the case. Because you said it was situational if I understand you correctly

[restive inlet]

depends onhow you describe stuff

[rough steeple]

what would be an example of a differing case

[restive inlet]

980 by itself is a term
0.2 * 196 by itself is a term
980/(0.2 * 196) is the quotient of two things that are individually considered terms, forms a term

[rough steeple]

Oh I see, so like as you said it would differ depending on how you see what is a term by how you describe it?

[restive inlet]

give this a read
https://simple.wikipedia.org/wiki/Term_(mathematics)

[rough steeple]

thanks 🙂

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[hazy bolt]

I'm confused about the definition of simple curves

[hazy bolt]

From what I know a simple curve is a curve whose not intersect itself

[celest bough]

If i am not mistaken a simple curve is exacly that

[hazy bolt]

but when I have to understand if a curve is simple or not I'm not sure

I have an example

I should explain the proprieties of this curve in polar coordinates (if it is regular, closed, simple)

My issue begin since I can say that it is closed since with a multiple of 2pi it returns same values

If i use a graphic processor I notice it doesn't intersect itself, but from definition if shouldn't be rho(theta1)=rho(theta2)

that is not what happens

(so it should be consider as simple curve)

[celest bough]

I don't want to confuse you cause i am not sure of my awnser here

Sorry, i talked about this topics some time ago ( also my English is not the best )

I hope someone can help you if not i can give you my perspective

[hazy bolt]

ok np, thank you anyway

while this is the definition I got

so in my case when t1 = 2pi and t2 = 8pi I got exactly that gamma(t1) = gamma(t2)...

<@&286206848099549185>

[elfin plinth]

Yes. What can I help you?

[hazy bolt]

I'm not getting clearly the definition of a simple curve

I wrote also an example I'm not getting

[thorny flame]

x_Shadow_x

[opal vault]

I think you're missing the definition of the curve

I'm guessing your curve is in 2D

[hazy bolt]

yeah

[opal vault]

but rho is not a curve here

it's a function

so if you wanna define a curve it's gonna be $(\nu, \rho(\nu))$

[thorny flame]

rafilou2003

[opal vault]

that's you're curve in 2D

and so for the curve to "intersect"

[hazy bolt]

yeah as a parametrization of the graph

[opal vault]

you need their first coordinates to equal

so $\nu_1 = \nu_2$ if it ever intersects

[thorny flame]

rafilou2003

[opal vault]

so you never have $(\nu_1, \rho(\nu_1)) =(\nu_2, \rho(\nu_2))$ with $\nu_1 \neq \nu_2$

[thorny flame]

rafilou2003

[hazy bolt]

why?

for example if I put 2pi or 8pi....

[grand ether]

what's happening here

[hazy bolt]

So that's exactly this

[thorny flame]

x_Shadow_x

[opal vault]

If you input 2pi and 8pi, then is $(2\pi,\rho(2\pi)) = (8\pi,\rho(8\pi))$?

[thorny flame]

rafilou2003

[hazy bolt]

Oh ok I got what you mean

Since - in ( - ,rho(2pi)) and ( - ,rho(8pi)) are not the same

thank you so much

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[olive lichen]

I have $\frac{d^2x}{dt^2} + 2\gamma \frac{dx}{dt} -w_0^2x = f(t)$ where $\gamma, w_0, t \in \mathbb{R}$. f is real valued as well.\

I managed to solve it by assuming x and f both have fourier transforms but my problem is proving the differentiation property. How do I go about proving $xe^{iwt} \eval_{-\infty}^{\infty} = 0$?

[thorny flame]

KySquared

[olive lichen]

I used integration by parts to arrive at $xe^{iwt}\eval {-\infty}^{\infty} -iw\int{-\infty}^{\infty} xe^{iwt} dt$

[thorny flame]

KySquared

[topaz sinew]

@olive lichen Has your question been resolved?

[topaz sinew]

@olive lichen Has your question been resolved?

[topaz sinew]

@olive lichen Has your question been resolved?

[olive lichen]

<@&286206848099549185>

[opal vault]

as t -> infinity?

[topaz sinew]

@olive lichen Has your question been resolved?

[olive lichen]

Yeah it would be as t tends to infinity
Sorry I just now saw this

[opal vault]

sorry to say

but just like (-1)^n

it doesn't have any limit

[olive lichen]

But the whole point is that xe^(iwt) must tend to zero or else the differentiation would fail

So it must go to 0

[opal vault]

"it must"

but it doesn't

[olive lichen]

There's no way
The differentiation operator becomes a multiplicative operator when transitioned with a Fourier Transform

[opal vault]

can you show your entire work?

[olive lichen]

That operation would be divergent otherwise

[olive lichen]

Okay
One moment

Using integration by parts\
$\begin{aligned}
\mathcal{F}{ \frac{dx}{dt}} &= \int_{-\infty}^{\infty} \frac{dx}{dt}e^{i\omega t} dt\
&=xe^{i\omega t} \eval_{-\infty}^{\infty} - i\omega\int_{-\infty}^{\infty} xe^{i\omega t} dt\
&=xe^{i\omega t} \eval_{-\infty}^{\infty} - i\omega X(\omega)
\end{aligned}$\
For this to make sense, $xe^{i\omega t} \eval_{-\infty}^{\infty}$ must tend to zero\

Because the Fourier transform is possible, we know that $\int_{-\infty}^{\infty} \abs{x(t)} dt < \infty$
Therefore, $\abs{x(t)} \eval_{-\infty}^{\infty} = 0$. Thus, $\lim_{L \to \infty} \abs{x(t)} \eval_{-L}^{L} = 0$

[thorny flame]

KySquared

[opal vault]

wait x is a function of t

that wasn't the most obvious

but uh

going from this

to this

loses too much information

[olive lichen]

?
I just took its derivative

[opal vault]

just say that |x(t)| -> 0 at infinity and |x(t)| -> 0 at -infinity individually

and wait a moment

[olive lichen]

So $\lim_{t \to \infty}\abs{x(t)} = 0$ and $\lim_{t \to -\infty }\abs{x(t)} = 0$?

[thorny flame]

KySquared

[opal vault]

yeah but I'm getting sus of something

[olive lichen]

?

[opal vault]

what are your conditions on x

[olive lichen]

At this point x is an unknown function
At this point all we assumed is that it is absolutely integrable and thus took a Fourier transform of the differential equation

[opal vault]

ok but

there are absolutely integrable functions that don't converge to 0 in extremes

[olive lichen]

But I thought absolute integrability meant a FT exists

That's what it says in the book anyways

[opal vault]

well yeah it does

but FT existing doesn't mean that x(t) -> 0 at +/- infinity

[olive lichen]

Hmm
Is there some other way I can go about this then?

Other than just making the second assumption "assume x(t) tends to zero as t tends to infinity"

[olive lichen]

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[grim jacinth]

Mind sending your other steps? I can see why the complex stuff might turn up in the bottom, but I’d like to see how you get there

Never had a problem where that was the case

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[steep cloak]

d question correct?

[cinder sequoia]

!original

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Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[sudden temple]

it is not

[cinder sequoia]

yeah, firstly you should understand that [a,b] is a <= x <= b, and (a, b) is a < x < b

so if you have [-5, infinity), what should that be as an inequality?

[steep cloak]

do i need to show where 0 is

[cinder sequoia]

where did the 0 come from?

oh on the graph

it could be a good idea

[cinder sequoia]

i just realized i misread your answer, your inequality is correct

i am once again stupid please disregard that

your inequality is almost correct, but notice that the -5 is next to a square bracket, meaning that it's included in the inequality

[topaz sinew]

@steep cloak Has your question been resolved?

[steep cloak]

OK I FIXED IT

correct?

@cinder sequoia

also i have this question

solve x

im not so good in doing logarithm

[cinder sequoia]

maybe it helps to combine the RHS using log rules

log(a) + log(b) = log(ab)

[steep cloak]

logs are all in base 5 can i eliminate the log

how do i do that?

and like make them kind of like exponent

@rigid cloak

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[clever citrus]

Can someone explain whats going on here

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@clever citrus Has your question been resolved?

[whole geode]

@clever citrus the sign changes because the direction of dr is opposite between the two curves

Essentially, dr in curve 1 is negative dr of curve 2.

[clever citrus]

so like one is going counter clockwise while the other is going clockwise (neg)

is that what it means?

[whole geode]

That is one specific example of what it is getting at

[clever citrus]

because I only remember putting this in my notes

[clever citrus]

the second case right?

where the sign changes

im not sure what the first case means

where there is no sign change

[whole geode]

If you have a regular scalar function it doesn't care about direction

[clever citrus]

[whole geode]

It's only when you have a vector function dotted with the differential that the direction matters

[clever citrus]

I seee

ty

[whole geode]

Ye!

Yw I mean

[clever citrus]

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[noble fable]

Can anyone take photo and send solution for integration of x sin inverse x?

[noble fable]

If u send the solution I can understand it faster rather than explanation it would time to understand

[restive inlet]

try searching online

[noble fable]

?

It takes time

[restive inlet]

we aren't here to do the work for you

[noble fable]

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[icy sky]

it takes time to do the solution, unfortunately

[prisma mesa]

Took me 4 secs to find an integral calculator with steps. And it only took that much because i have quite a bad internet

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[fossil igloo]

How do I solve question d?

[fallow igloo]

Show the entire question

[fossil igloo]

I

I only get to solve 0.580pi and 1.420pi

[vernal matrix]

Be careful of dividing by things that could turn out to be zero

You need to also consider sin(x) = 0 as well and add the solutions to that to whatever you find

[fossil igloo]

I see

the answer:

[vernal matrix]

0, pi and 2pi from here, the remaining two you have

[fossil igloo]

If that how do I do it?

[vernal matrix]

Well, either don't divide by sin(x), keep it there, and do the rearrangements basically in the same way you have, but at each step (and notably the bottom line as per here) you'll have a factor of sin(x), so you consider cos(x) = 2, cos(x) = -1/4 and sin(x) = 0 (the latter getting you the missing solutions)

Alternatively, at the point you divide by sin(x) (or any other variable for that matter), check where it could be zero (that is, in this case, solve sin(x) = 0), and list those, then find all other solutions and add those to the list too

[vernal matrix]

So at this point you keep the 0, pi and 2pi noted, and then...

[vernal matrix]

...when you find the other two, that's all the solutions for you

[fossil igloo]

Let me try it

Very tricky eh

Thank you for your help!👍

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[peak kelp]

Is it possible to ask for an explanation to what is integration

[peak kelp]

Ping when here

[clever citrus]

They already gave you pretty much everything.

first plug everything in the first principle formula

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@peak kelp Has your question been resolved?

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[somber plover]

F

[somber plover]

F

Anyone???

[wet harbor]

tried L'hopital ?

or is it not allowed

[somber plover]

Not allowed

Plus haven't read it's on the syllabus of class 12 I'm at 11class

[wet harbor]

i don't think that would be an issue xD lhopital is so easy to use

but it's not allowed so we can't do that

[somber plover]

Yeah

derivative isn't taught to us yet

[wet harbor]

maybe we can try using the properties of ln

i didn't solve it, i'm giving you ideas, since that seems like the only thing obvious

[somber plover]

Pls help broo

I don't know how to start

Nvm I got it

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[chilly walrus]

these two are the same

[opal girder]

Yes

[chilly walrus]

and these two are right angled

[opal girder]

correct

[chilly walrus]

so?

[opal girder]

i still dont get the mathematics behind it

[chilly walrus]

this angle is 180 degrees + dtheta

[opal girder]

ohhhhhhhhh

[chilly walrus]

here i give two 90 degree angles = 180 degree, what's left is dtheta

[chilly walrus]

yeah it's a bit non obvious but it works out

[opal girder]

okok i got it

yeah my geometry is terrible man

thank you so much

[chilly walrus]

no problem!

[chilly walrus]

if you are done, type ".close"

[opal girder]

.close

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[ripe ermine]

how do i break it into partial fractions?

[opal vault]

hint : +2x^2 - 2x^2

(the denominator)

[ripe ermine]

what do i do with the x^5 tho

[opal vault]

long division

x^5 = ...(1+x^4) + ...

[ripe ermine]

im completely clueless😭

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[bold meadow]

Can someone explain to me the limit rule in this question

[bold meadow]

I get the answer but don’t understand the limit rule

Nvm

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[pine flame]

There are two lines

[pine flame]

(2a + b)x + (a + 3b)y + (b - 3a) = 0

And Mx +2y + 6 = 0

They are concurrent

Need to find the value of M

How do I solve this without the family of line formula?

<@&286206848099549185>

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[surreal mural]

any attempts thus far?

[grand blade]

so for the first 2 inequalities im assuming that im just using the definitions of max norm, euclidean norm and manhattan norm, but im not so sure

ig u can use cauchy-schwarz to do euclidean norm and manhattan norm inequality

<@&286206848099549185>

[surreal mural]

right, for the first one I would just apply the definitions and square both sides, assume |x_i| to be the max from the max norm and the inequality should follow pretty easily. For the second one I would once again apply the definitions and square both sides, then use induction for d to show the inequality (maybe there is any smarter idea?). Cauchy schwarz works well for the third one and the forth one is similar to the first one

[topaz sinew]

@grand blade Has your question been resolved?

[grand blade]

wait for the second inequality did u mean the third since thats the one that introduces d

?

[surreal mural]

$\sum_{i=1}^d |x_i|^2 \leq \left(\sum_{i=1}^d |x_i| \right)^2$

[thorny flame]

rbit

[grand blade]

oh lol mb

thanks for the tips im not very experienced at L.I proofs yet

[surreal mural]

i bet theres a smarter way but it works i guess

[grand blade]

yeh ok i might have a search around havent had much thought into this yet and still got days b4 its due

but cheers

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[sharp swift]

Hi, i just joined the server, and i could use some help progressing in math in general. I have almost completed a course on khan academy( i dont know if you know it) called algebra 2, with basic trig and some other stuff

[sharp swift]

i was thinking calc or sum but i really dont know

[frail fiber]

u can try basic calc now

[ionic oar]

precalc comes before calc

it has important concepts that you need to know

[restive inlet]

done exponents and logs yet?

[ionic oar]

but aren't there more courses between algebra and calculus?

[sharp swift]

yes the basics of it

[restive inlet]

algebra 2 and precalc have a lot of overlap

[cinder sequoia]

mainly trig might be something new

[ionic oar]

i'm not american

so you help

[sharp swift]

ok ill look at it

thanks guys

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[cinder sequoia]

do you have a question?

!redir

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This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[restive inlet]

<@&268886789983436800> they just keep coming

[cinder sequoia]

i tried giving him a chance :(

[restive inlet]

if it's their first post, no tolerance imo

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[fallow torrent]

How do I verify the following inequality in this interval $\left[-\pi, \pi \right]$?

$\sin(2x) - \sin(x) < 0$

if I am not mistaken $\sin(2x) = 2\sin x \cos x$

How do I proceed?

[thorny flame]

Shadow91518

[opal vault]

So you can factor

And think about the signs of A and B when A*B < 0

[fallow torrent]

Can you elaborate?

[snow nimbus]

How about x = 0.5?

[topaz sinew]

@fallow torrent Has your question been resolved?

[vernal matrix]

(If you have a product being negative, what can you say about the factors?)

[autumn willow]

one of em is negative

if there r 2

so sinx is negative

or 2cosx-1 is negative

or u can say sinx is -ve or cosx<1/2

[vernal matrix]

(true, but that question was meant for OP to answer )

[autumn willow]

o

mb

[fallow torrent]

I am trying to understand, but I think I have some gaps with trigonometry.

Anyway
How come $2\cos x - 1$?

Ps: don't write abbreviated, I'm Italian

[thorny flame]

Shadow91518

[autumn willow]

@fallow torrent

[autumn willow]

,rotate

[autumn willow]

now for that to be zero

either sinx must be 0

or 2cosx-1 must be 0

do u get it?

[fallow torrent]

yes, I got it.

thank you

[autumn willow]

so from -pi to 0

sinx is negativwe

and cosx needs to be less than 1/2 in the range or 0 to pi

so i think its wrong

maybe

@vernal matrix

cuz its 1 at cosx=0

and also at x=0

the eqn becomes 0

cuz of sinx

so maybe it does not hold true

[topaz sinew]

@fallow torrent Has your question been resolved?

[fallow torrent]

so I have $\sin x > 0$ and $\cos x > \frac{1}{2}$

[thorny flame]

Shadow91518

[fallow torrent]

$0 < x < \frac{\pi}{2}$ and $- \frac{\pi}{3} < x < \frac{\pi}{3}$ but I want only the negative part because at start is < 0 right?

[thorny flame]

Shadow91518

[autumn willow]

its lesser than

<@&286206848099549185>

[silent siren]

Be very careful when you cancel.

You could do the following...
Let $F(x) = \sin\left(2x\right)-\sin\left(x\right)$
\begin{enumerate}
\item Solve $$ F(x) = 0 $$
\item Solve $$ \frac{dF}{dx}(x) \eval_{F(x)=0} > 0 $$
\item Now, given two solutions, $x_1$ and $x_2$, you know $F(x) > 0$ for all $x \in (x_1,x_2)$ if $\frac{dF}{dx}(x_1) > 0$ and $\frac{dF}{dx}(x_2) < 0$
\end{enumerate}
That is, given two consecutive points where $F(x_1) = 0$ and $F(x_2) = 0$, then $F(x) > 0$ in the region $x \in (x_1,x_2)$ if $\frac{dF}{dx}(x_1) > 0$ and $\frac{dF}{dx}(x) < 0$.

[thorny flame]

Shuba

[silent siren]

Make sense, or want me to explain it further?

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[fading bay]

I selected ADE but the answer seems to be wrong ?

[wary tulip]

those all sound ok

[golden blade]

Consider F maybe

[fading bay]

Integral test? but cos function is increasing sometimes

[golden blade]

doesn't look fun to integrate that too either

[golden blade]

my guess is E is wrong

If we took the converging series 1/n³ for example then n³ would cancel out but 3-cos(n³) doesn't exist

but that's just one specific example

my point is that the cosine term may be bothering for limit comparison test

[fading bay]

You were right

[golden blade]

thanks

[fading bay]

may i have help with another question ?

[golden blade]

sure

[fading bay]

im having trouble with the last part

[golden blade]

Hmm okay

Well my guess is they want you to calculate $$\int_0^{0.76} \frac{7}{3}x^3-\frac{49}{6}x^7 : dx$$

[thorny flame]

𝔸dωn𝓲²s

[golden blade]

Maybe you did a smol mistake somewhere?

Among the calculations, possibly?

[fading bay]

0.080990392 is my answer

[cursive patrol]

no

you have F(x), so you do not want to integrate F(x) again

it suffices to plug in your value of x

[golden blade]

right

[thorny flame]

𝔸dωn𝓲²s

[golden blade]

you heard the man

[fading bay]

so i plug in x to my macluarin series

[golden blade]

yes

I oversaw that this was you integral actually

[fading bay]

7/3*(0.76)^3 - 49/6*(0.76)^7 = -0.17174689

[golden blade]

pretty much

but I noticed how off this approximation is

,w Integrate[sin(7x^2),{x,0,0.76}]

[thorny flame]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by adoniszgks

[fading bay]

my calculator only gives me up to 8 decimals

[golden blade]

,w 7/3 (0.76)^3- 49/6(0.76)^7

[thorny flame]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by adoniszgks

[fading bay]

add an zero

[golden blade]

😂

yea

[fading bay]

damn that works 😂

thank you @cursive patrol and @golden blade

I have one more attempt to answer until I can't may i verify my coefficients?

[golden blade]

Have you noticed, you can use the geometric series formula?

[fading bay]

yes

[thorny flame]

𝔸dωn𝓲²s

[golden blade]

Can you tell me your first value you had previously?

[fading bay]

0

[golden blade]

how did you get that?

[fading bay]

This was not what i had when i used the geometric series

i had something else then i inputted my x for my coefficients, 0 for c0 was the only one correct

[golden blade]

x is not your coeffcient

[thorny flame]

𝔸dωn𝓲²s

[golden blade]

your c_ns are the factor before x

they are like separate

n = 0 not x = 0

[thorny flame]

𝔸dωn𝓲²s

[golden blade]

You are basically considering the first 5 terms from 0 to 4 of your sequence c_n

[fading bay]

okay

my function is (-1/2)^n * (1/2)x^n+1

[golden blade]

yea it's the same

[fading bay]

ive gotten 0, 1/2, -1/4, 1/8, -1/16

[golden blade]

i just pulled out the -1

[fading bay]

oh ok

[golden blade]

wait

how 0?

You are evaluating $c_0 = \frac{1}{2} \cdot (-1)^0 \cdot \left ( \frac{1}{2} \right )^0$

[thorny flame]

𝔸dωn𝓲²s

[fading bay]

Honestly Idk I put 0 and the website gave me that correct

[golden blade]

but you also had 0 previously and it was marked red???

😂

[fading bay]

that screen was before

I swear i had 1/2 for the first one and it was red

[golden blade]

ah ok

[fading bay]

I don't understand why

c_0 = 0

[golden blade]

neither me

👁️👄👁️

[fading bay]

anyway thank you for the help

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[hidden belfry]

I dont understand how that square root of 2x^2 becomes 4x^2

[vernal matrix]

Common denominator the numerator terms

[hidden belfry]

Yes but how do you compute this. I cant seem to find it

Nvm

I just saw the loght

Light*

Thanks

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[bold kernel]

I’m confused by this a bit. Since when are the rows of matrices representing planes?

[lucid crypt]

matrixes are just another way of writing a system of equations

for (a), the matrix says

x-2y+3z=0 for the first row

and so on for the second

that is how you get planes

[bold kernel]

Hmmm

I guess I just thought the rows and columns of matrices were always vectors

But now it seems like the rows are planes

And a plane takes three vectors to build so it’s confusing me

It’s like one row contains three vectors if that makes sense

[brave fox]

they aren't planes, they are planes through the origin

[bold kernel]

Planes through the origin yes

Still though those are planes yes

Not vectors

I’m just surprised by that

At first glance looking at a I felt like I was seeing three linearly dependant vectors

[brave fox]

in 3d planes through the origin and vectors are close to interchangeable

[bold kernel]

But it turns out it’s three planes through the origin or rather the same plane three times

I was just struggling with the interpretation of colinearity

[bold kernel]

Ahhh

Why is the through the origin part so important

For it to be similar to a vector

[brave fox]

but actually i realize that in this case they are talking about solution space

so the set of all vectors that satisfy the equality

and it happens to be a plane

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[cursive patrol]

u cannot simplify that much

why? that is the best way of expressing a value like this

[flint stump]

!original would help explain stuff better

[topaz sinew]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[cursive patrol]

well u can just say a = 0.463647609...

[solemn matrix]

thats in radians

the answer Mqnic gave was in radians is what i meant

[cursive patrol]

so what's the issue with just using a = arcsin(1/sqrt5)

[solemn matrix]

unless you want integrals or infinite sums u wont get much better

[cursive patrol]

cos(arcsin(x)) is sqrt(1 - x^2)

[flint stump]

arcsin is just the inverse function of the sine function

so sin(arcsin(x)) is just x

so you can just further apply the sin^2 + cos^2 = 1 identity for some angle A whose sin(A) = x

[flint stump]

and that converts it to this

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[cunning sedge]

I tried a bunch of things but the answer was never an option

[fallow igloo]

Show your work

Did you obtain it from ChatGPT?

[vernal matrix]

Really? I wonder what the <@&268886789983436800> think about that?

[spice owl]

Could anyone help me with this

[fallow igloo]

!occupied

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[somber plover]

4

[vernal matrix]

,rccw

[vernal matrix]

Were you able to do everything before that point?

[rigid schooner]

Since they are equal, the discriminant must be equal to 0

Therefor you are left with an equation

[somber plover]

I reached here

[rigid schooner]

Wait a second

$D = b^2 - 4ac$

[thorny flame]

Randel_

[rigid schooner]

Use this

since the roots are equal, D = 0

[somber plover]

Yeah I did it

Then I reached to
4c²+4m²a²=3m²c²

[rigid schooner]

Check your calculations

You did not put the 4

[somber plover]

🤦🏻‍♂️

Ohh I'm sorry

[rigid schooner]

It's alright

[somber plover]

Thankyouuu

[rigid schooner]

no problem 👍

[somber plover]

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[rigid schooner]

have a nice one

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[whole spade]

i wasn't able to solve it

[whole spade]

what i did was multiply by the denominator conjugate

but i didn't get far

[wet harbor]

why not just divide in both numerator and denomenator by x

see where this leads

[whole spade]

like factor out x and divide?

i don't think i can divide x by sqrt(x)

[wet harbor]

like multiply both by 1/x

the numerator becomes 1

and what happens to the denomenator ?

[whole spade]

denominator divided by x?

[wet harbor]

yes

simplify it and write it down

[whole spade]

this?

wouldn't that just be the same thing

[wet harbor]

yes now put the x inside the roots

with proper adjustments

[whole spade]

so i'll have 1/x^3 for the first and 1/x^2 for the second

so basically i factorize whats inside the roots

by the largest power

[wet harbor]

yes

then try calculating the limit again and see if it's easier

[whole spade]

my issue is with the 7-x though

if x > 7 then i would get a - value in the root

which doesn't work

but the limit does work with - inf

yeah it worked

theres no limit for +inf but there is for -inf

[wet harbor]

does it really matter whether or not it's positive or negative?

it will be in the form of 1/x

so it goes to zero either way

[autumn willow]

r yall kinda gettin something like x/0

??

[whole spade]

yeah but 1/x^2 goes to zero quicker than 1/x so i would have a negative in the square root

[whole spade]

no?

[autumn willow]

ok nvm then

do u know what the answer is?

[wet harbor]

honestly getting the x inside the root when it's negative is very suspicious as well

i recommend that you make a substitution intead of x being -y or something then do it again to ensure that it's positive

[whole spade]

what do you mean

[wet harbor]

you have two limits

one going to infinity

the other is going to -inf

so you solve it twice

once for positive infinity

and another where you make a substitution to change the limit again into positive infinity

there you will simply replace each x with -x

that would be the second limit

[whole spade]

huh

so after i solve for + inf

i have to change all the x with -x to solve for - infinity?

[wet harbor]

i would recommend it

here is what i'm worried about

if you leave it like that and solve for -inf

you put the x inside the root

if there is any value at which what is under the root was negative and x was negative too then this would be invalid

as $\sqrt{ab} = \sqrt{a}\sqrt{b}$ is wrong if both a and b are negative

[thorny flame]

Mohamed Mohsen

[whole spade]

yeah i see what you mean

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[wet wind]

Help pls

[wet wind]

Need help with these

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[hushed bough]

Can anyone help me find the value of JN

[stray harbor]

any angles given ?

[hushed bough]

No

[stray harbor]

Set up the ratio of corresponding sides

JK/KL = JN=NM

so 18/12 = X/Y

[hushed bough]

so JN = (3xNM)/2 ?

[stray harbor]

yeah but should first set that JM= JN + NM = X + Y

[stray harbor]

and than you know that x = 3y/2

[hushed bough]

alright

thanks

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[neon iron]

Can a line be parallel to itself? I want to prove a relation of two lines equivalence. But if it is, then it don't follow the definition of parallel lines right?

[odd pagoda]

parallelism is often defined in a way such that a line is parallel to itself

its just much more useful that way

[neon iron]

Is it a convention?
I had learner that paralel lines are those which never intersect to each other
But a line intersects with itself on all points

[opal vault]

The more common definition is that two lines are parallel if there exists a line perpendicular to both

[odd pagoda]

its often either included as a special case or it follows naturally from some other def

[coarse tusk]

I guess that makes it an equivalence relation

[odd pagoda]

or you can define it as both lines being cosets of the same linear subspace. that won't mean anything to you but in that case also a line is parallel to itself

[odd pagoda]

sadly that only works in the plane

[opal vault]

true

so yeah cosets make more sense

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[lone kraken]

Is there a word for replacing a modulo? Like I have $a x+t \bmod p = 0$ and replace it with $a x + t + kp = 0, k \in \mathbb{Z}$. Feel like there's a word for that.

[thorny flame]

Trapture

[inner wren]

!15min

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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.

[lone kraken]

wha

[vernal matrix]

Did you even ping helpers to begin with

[lone kraken]

i did not

[vernal matrix]

So what's the relevance of the factoid then, @inner wren?

[torpid sparrow]

No

[coarse tusk]

Probably a troll

[vernal matrix]

(as for if there's a word, I don't know of one myself)

[coarse tusk]

You could say residue

ax+t has residue 0 mod p

Well I guess this still has the word mod

[lone kraken]

Ah, I don't need a word I just thought there was a standard term

[torpid sparrow]

The term is the one you want to replace

[coarse tusk]

Mod is the standard term

[lone kraken]

So far I just wrote "I replace this with that so we can do blah"

[coarse tusk]

By the way modulus and modular arithmetic are not connected

mod doesn't mean modulus

I don't think it does anyway someone can correct me on that

[lone kraken]

What else would it mean?

[coarse tusk]

Modulus usually means | |

[lone kraken]

It's modulo

[coarse tusk]

Yes it's modulo

Not modulus

[lone kraken]

Yeah, no 😄

[coarse tusk]

Ah wait you wrote modulo I thought you wrote modulus

My bad

[lone kraken]

All good

[brave fox]

quotient group/ring

[torpid sparrow]

[ruby tree]

In "mod n", n is called the modulus

[coarse tusk]

it is?

I've never heard that

[ruby tree]

By some people

[torpid sparrow]

They are more connected than me with reality

[ruby tree]

But it's also absolute value for complex

[vernal matrix]

(well, modular arithmetic is the one of the first examples of these )

[lone kraken]

I'm sticking with integers, so no danger of confusion there

[coarse tusk]

Also like every other measure I think uses the word modulus to mean || ||

[lone kraken]

sometimes a rational sneaks in but i know how to deal with those

Anyways, seems like there's no word for whatever I'm doing, thanks folks 😄

[ruby tree]

Modulo means "with respect to a modulus of", according to Wikipedia

So modulo n (or mod n) is "with respect to a modulus of n", where n is indeed the modulus

[coarse tusk]

damn

[ruby tree]

Apart from some variation on the word "equivalence", I don't think there's a specific word for what you want

[lone kraken]

.close

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[desert apex]

Is there a proof of
the factors of a number are unique till till the first factor is smaller than or equal to it's square root
If we start factoring in ascending order of the first factor
It's used in the sieve of erathosthenes

[odd pagoda]

I am not quite sure what you mean by that sentence

do you mean the fundamental theorem of arithmetic?

that the prime factorisation is unique?

if you just have factors then they arent unique, there are lots of ways to factor a number

[topaz sinew]

@desert apex Has your question been resolved?

[desert apex]

No what I meant was for every composite number their factors are repeated twice
For example factor pair of 36

1 x 36
2 x 18
3 x 12
4 x 9
6 x 6 <-- square root of 36
9 x 4 <-- after square root all factors are repeating
12 x 3
18 x 2
36 x 1

My English is bad,
But I hope you get my point

Is there a proof that for any composite number the factor pairs are repeating

[vernal ledge]

yeah i get you

i use that a lot in checking if a number is prime

hmm

[desert apex]

I tried looking for the proof
But couldn't find it

[thorn violet]

I mean I think it’s kind of too simple to have a famous or named proof.

You can prove it by showing that if q is a factor of n, then n divided by q is also a factor of n.

[desert apex]

No what I meant was the factors are unique till the square root of a number,
After that the factors are repeating

[odd pagoda]

either q or n/q is <=sqrt(n), otherwise their product would be bigger than n

well they are repeating because they appear in the different order

3x12 appears before sqrtn and 12x3 occurs afterwards

[desert apex]

Yea I know but I thought that there might be a proof

[odd pagoda]

well I mean this is a proof basically

[thorn violet]

We just proved it together 🙂

[desert apex]

But this is just observation

[neon iron]

would this not be a postulate?

i think thats the word

i could be wrong

[odd pagoda]

well and we explained why it happens

a proof is basically just a proper explanation of why something always happens

[thorn violet]

Observing facts and drawing conclusions are what we mean when we call something a proof.

[desert apex]

I meant like a rigorous proof

[odd pagoda]

this is rigorous