help-13

[pastel vault]

So angle Y can also be 20 degrees

[bronze sparrow]

so what do I plug in for the answers

[pastel vault]

This is how it would look like

[pastel vault]

20 degrees, blank, and 1 possible triangle

[bronze sparrow]

alright ill try it out thanks

[pastel vault]

no worries

[cedar kiln]

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[wraith dagger]

Ava

[worldly chasm]

Care to explain where each part comes from?

So, first let's consider how many ways you can choose the colors. There are 5 ways to choose the first and 4 to choose the second, and order does matter here, so that's 5P2 = 5*4.

You have two different colors 4 from 1 and 3 from the other

I might just be misremembering the formula for nPk

Yup

I remembered right

You're selecting 2 of 5 colors

Not dumb, just learning

Next, you choose 4 balls from the first and 3 from the second, but you can choose these in any order as well. 7C4 and 7C3.

But you have to permute these as well, and remove identical permutations, so that's 7C4 * 7C3 * 7!/(4!3!)

Yes

Hold up

You're missing a piece

You can select the 7 balls in any order

Oh nevermind

I don't think you care about order here.

Yes

Then that's the last thing

5P2 vs 5P1

So the last part was just me misinterpreting the prompt

Please disregard

0.36% I assume

[pastel vault]

Smart

[worldly chasm]

0.0036 = 0.36%

You're very welcome

[cedar kiln]

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[cedar kiln]

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[green crow]

I need help with this problem. I managed to solve the first part using |W|=K+U

[peak minnow]

All the kinetic energy would be converted to potential energy so
1/2mv² = mgh

[green crow]

And why is that

[peak minnow]

Energy cannot be destroyed or created

But can be converted to satisfy law of conversation of energy

[green crow]

Isn't law of conservation of energy

K_1+U_1 =K_2 +U_2 ? Unless I'm making a mistake because this equality is still relatively new to me

[peak minnow]

The law is initial energy is equal to final energy

In that example we have 4 energy

We can have more or less

[green crow]

In that equation does something go to zero based off the expression your wrote

[peak minnow]

Idk why the angle is given in the question, maybe I'm wrong

[peak minnow]

No?

[green crow]

Because my goal is to solve for a distance

And I also remember U=1/2 k(delta s) ^2 for a spring

[peak minnow]

Oh, then just search up projectile motion

It deals with this

I thought we needed to find how high up the kid went

[green crow]

I already solved the final velocity

[peak minnow]

Yeah there is a formula to find distance traveled by motion of projectile

Let me find it

[green crow]

I don't think that will work here

Because it's using a spring

[peak minnow]

Spring to give speed and energy, no?

[green crow]

Is Work a type of energy

[peak minnow]

Yes

[green crow]

I said when I made this post Work=kinetic energy +potential is that somewhat valid?

[peak minnow]

Yes

Should be

[green crow]

I was going to say, can I say work=kinetic-potential now?

[peak minnow]

I'm pretty sure there's a theorm of that kind

I think it is work energy theorm

[peak minnow]

Why

[green crow]

To me, it seems like it could work because the unit would still be joules

This what I was thinking. I calculated W from the previous part

[peak minnow]

Isn't that what u used to calculate vf?

[green crow]

Yes but I used W=K+U

[peak minnow]

Also isn't x distance of spring compressed or expanded

Ithought we were finding distance of boy

[green crow]

Hm, maybe I'm confusing something

[peak minnow]

I just think we can find horizontal distance via projectile motion

[green crow]

I would agree with that but the angle is not located near the spring it's somewhere else

[peak minnow]

Idk how u are getting spring in this

Actually I'm confused

The question is how far up the incline student went

[green crow]

Because usually in projectile motion its a person throwing a ball and sometimes its arched at an angle

[peak minnow]

That means we need to find at what point the student lost all the energy and stopped

[peak minnow]

Think from right side, from the ground u threw the boy

[green crow]

Well if the student stops then can I say K_final =0

[peak minnow]

Yes

So projectile motion doesn't work here btw

We can probably do something like this

W - drag force x distance = 0

[green crow]

I haven't encountered a problem where I had to find where the energy is lost at least not yet

[peak minnow]

And we find the distance

[green crow]

What's W here

[peak minnow]

Total energy

[green crow]

Isn't that the definition of work

force times distance

[peak minnow]

Yes

[green crow]

Is the distance here the goal distance

[peak minnow]

Yes

At that distance the boy loses speed and slows down

And stops

But maybe that's also wrong

[green crow]

I'm not sure how drag comes into play here I would need to know the density, drag coefficient, cross sectional area

[peak minnow]

We aren't given anything about drag

Yeah but if u read the hint it says surface is frictionless till incline

[green crow]

I'm tempted to Say F=-kx from hooks law

[peak minnow]

That means there is some friction in the inclined part

Hook is for spring not boy

[green crow]

Are you saying this turns into a inclined plane problem with sum of forces

[peak minnow]

Yeah something like that

Isn't there any hint for part b?

[green crow]

Unfortunately no

[peak minnow]

Can u screen shot the entire question

[green crow]

That is the entire question

[peak minnow]

It can't be

The hint gives more info then the actual question

Talking about part a

[peak minnow]

I think what i said about drag force should be valid in here

But we need stuff to find that

[green crow]

I think you are onto something it says coefficient of kinetic friction is 0.20

[peak minnow]

Ik

But we don't have the density or area

Density of air is kinda common but what about area

[green crow]

Maybe something like this

[peak minnow]

I don't like components of force

They suck

[green crow]

True

My guess would be to solve for acceleration

In x direction

[peak minnow]

Btw at what point did u draw those components?

[green crow]

Acceleration in y should be 0

[green crow]

When you gave me the idea of it being like a inclined plane problem

[peak minnow]

No i mean on graph

[peak minnow]

Here?

[green crow]

That was always there

[peak minnow]

Cuz if so then the components are wrong

[green crow]

The angle

The fbd is for the blue circle

[peak minnow]

Yeah which theta do u have

Oh

OK

[green crow]

Blue circle is boy

[peak minnow]

Still if we use this

The hint is useless

Cuz we are given friction constant

[peak minnow]

What's uk

[green crow]

Coefficient of friction

[peak minnow]

And fk is friction?

[green crow]

I'm thinking solving acceleration and use a kinematic equation

[green crow]

Yes

[peak minnow]

Which kinematic equation u are going to use?

[green crow]

I don't know yet but I only know v_f

[peak minnow]

Wait which axis component did u use?

Y?

[green crow]

I got a=3.92 m/s^2

[peak minnow]

Im thinking we could do this

[green crow]

Makes sense because the boy isn't deceleration

[peak minnow]

but instead of drag force we use friction force

[green crow]

In this problem can I say something about the initial velocity?

[peak minnow]

But the floor has friction that means the boy should slow down not speed up

initial velocity is 0 when he was at rest

Like way in the beginning

Wait I'm thinking maybe

Force of friction = constant x force

Find friction

And then
Ke + Pe - friction x distance = 0

[green crow]

Am I on to something?

[peak minnow]

Maybe

I don't get it why is acceleration positive when the boy stops in the track

[green crow]

I'm not good at rounding but is 32.653 meters two significant figures

[peak minnow]

33.0

[green crow]

If I put 33 and it turns out its not right then I'll have one attempt left so not sure if I want to risk it

[peak minnow]

Maybe not

Close this one and reopen so it gets prioritize and ask about how to round off

Someone else might help

[green crow]

Lets go. You were right

[peak minnow]

Whose better

[green crow]

So it turns out all that vector stuff was actually right

[peak minnow]

Oh ok, tbh can u pls explain what u did to solve

Im clueless

[green crow]

Did you understand how I got the forces in the free body diagram

[peak minnow]

Yes

The fx and fy

[green crow]

So I got the idea to solve for a_x because mainly from previous problems I've done. And a_y=0 because the boy isn't accelerating upward

[peak minnow]

But ax is horizontal should there be some theta stuff in here too?

The boy is moving on an inclined surface

Nvm the question is solved thats the good thing

!done

[cedar kiln]

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

[green crow]

I think this the hardest problem I've done ever

But thanks to you I was able to get an idea on how to solve it

.close

[cedar kiln]

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[patent verge]

is someone able to help me solve this? i need to minimize it with simplex method

[patent verge]

I've tried it like 3-4 times and keep getting the same answer which is wrong so I just want to look at someone elses work or something and see where I'm going wrong

[coral jewel]

what have you tried

[patent verge]

my work is messy but i can send it

That's trying the same question 2 times

and this is the actual answer so i dont know what im doing wrong

<@&286206848099549185> if anyone that knows how to do it can do it and show their work thatd be really cool cuz i just dont know where im going wrong

[cedar kiln]

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[cedar kiln]

@patent verge Has your question been resolved?

[cedar kiln]

@patent verge Has your question been resolved?

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[dull shard]

I'm doing mechanics / maths at A level and in an answer it says initial velocity of A = final velocity of B (they are 2 particles connected by a light inextensible string over a smooth pulley)

Is it also true that final velocity of A = initial velocity of B because the string is inextensible?

[flint plinth]

do you have a picture?

offhand i would think their velocities are always equal (if you set up the coordinates appropriately)

[dull shard]

yeah one sec

like this

[flint plinth]

what do the two a ->> arrows indicate?

[dull shard]

acceleration in that direction

[flint plinth]

is that defining the coordinates for each block?

[dull shard]

no there is no coordinates it's like suvat and I got the answer I was just wondering because Vb = Ua is Ub = Va

[flint plinth]

ok well if you define the velocities with the convention that block A's positive velocity points to the right and block B's positive velocity points down, then aren't they always equal? assuming the string doesn't stretch or compress

[dull shard]

okay yeah that was my question ty

I didn't know if it was a rule

[flint plinth]

that's usually the assumption in this kind of problem

[dull shard]

ty

.close

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[vocal robin]

i dont understand the point slope form. with y = mx+b, I can visualize understand each component. I have found no explanation of the corollary y-y1 = m(x-x1) form, other than that it is derivable through some algebraic manipulation and can describe arbitrary points against a given x1,y1 and slope. Still, the equation is meaningless to me when I look at it, and it seems most of the resources out there do not care about making it meaningful for students.

[modest lichen]

I mean it is just one step away from the original equation.

Ive only really used it for faster algebraic manipulation (sometimes) and some numerical methods when coding

$m = (y-y1) / (x - x1) \rightarrow y-y1 = m(x-x1)$

[wraith dagger]

UsingApp

[modest lichen]

idrk waht u mean tho

[pastel vault]

So $y = m(x - x_1) + y_1$ and so the slope of this line is just $m$

[wraith dagger]

south

[pastel vault]

Now if you substitute (x, y) = (x1, y1) into the equation, you get 0 = 0

So (x1, y1) lies on the line

And also if you have a slope and one point on the line

That determines the unique line: only one line is possible

[vocal robin]

why does y = m(x-x1) + y1

[pastel vault]

(If you get something like 0 = 2 then this is false, so (x1, y1) wouldn't lie on the line)

[pastel vault]

Add y1 to both sides

[vocal robin]

that does not mean anything to me

i want visual intuition not algebraic manipulation

[pastel vault]

We're just finding the slope of the line

[pastel vault]

Visual intuition and the algebra go hand in hand

Notice I didn't expand everything out

You just have to look at the coefficient of x

You see m(x + ....) so there is an mx term in there

And the rest are constants

[vocal robin]

y = m(x-x1) + y1 does not make sense to me

[pastel vault]

The reason for this relies on the visual intuition

[vocal robin]

because y - y1 = m(x-x1) does not make sense for me

[pastel vault]

What about this doesn't make sense

[vocal robin]

it means nothing to me. it is just letters

[modest lichen]

well m is just a ratio between change in y and change in x

[pastel vault]

Alright how about we expand it then

$y = mx + (-mx_1 + y_1)$

[wraith dagger]

south

[pastel vault]

So the slope is m

And the y-intercept is -mx1 + y1

[vocal robin]

why (-mx

[pastel vault]

$m(x - x_1) + y_1 = mx + m(-x_1) + y_1$

[wraith dagger]

south

[vocal robin]

that did not help

"the slope times the difference of the first coordinates of two points + the end coordinate of one point equals th slope times the first coordinate plus the slope times the negation of the other first coordinate..."

like, do you see how that is super hard to follow and meaningless visually

where I am at currently.

just letters, being re-arranged to be more letters. that doesn't do anything for me in terms of comprehension

[pastel vault]

The slope is m

You have to understand that the slope is just the number in front of the x

[vocal robin]

i know what all of the letters mean

[pastel vault]

The y-intercept isn't very useful I know

[vocal robin]

how they are arranged in the equation is not intelligible to me

"the slope times the difference of the first coordinates of two points + the end coordinate of one point equals th slope times the first coordinate plus the slope times the negation of the other first coordinate plus the other second coordinate"

i can't comprehend this sentence

first, the fact that x1 and y1 are on two different sides of the equal sign is very bizarre for me

like you two equations being mixed together

[pastel vault]

Okay so

We can let $(x_2, y_2)$ be anywhere

[wraith dagger]

south

[vocal robin]

yeah

[pastel vault]

So we can just let $(x_2, y_2)$ be the point (x, y)$

[wraith dagger]

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

[pastel vault]

So by definition, the slope $m = \frac{y - y_1}{x - x_1}$

[wraith dagger]

south

[pastel vault]

We just subbed in (x2, y2) = (x, y)

[vocal robin]

yes i understand that

[pastel vault]

Then just rearrange to get $m(x - x_1) = y - y_1$

[wraith dagger]

south

[pastel vault]

$y - y_1 = m(x - x_1)$

[wraith dagger]

south

[pastel vault]

So what this tells you is that the point-slope formula is just another way of expressing the slope between two points

Except one of the points is (x, y) where x and y are variables

[vocal robin]

rearranging it gets it into a form that does not make sense to me

[pastel vault]

You can always rearrange it back to this

[vocal robin]

the problem is that I want to understand equations not memorize them

[pastel vault]

And I'm telling you this

[vocal robin]

but it is an expression that does not seem comprehensible on it's own

y = mx + b is comprehensible on its own

zero algebraic manipulation required

[vocal robin]

right now my understanding is "these two forms are equivlanent, but I have no idea how this second form makes coherent sense and definitely could never derive it myself from scratch"

[warm crescent]

you multiply both sides by (x-x1)?

[vocal robin]

thats just the algebraic trick to switch between the two forms right?

deriving second form by manipulating the first form also does not feel "from scatch" either

[warm crescent]

why not

-mx1 + y1 is a constant so you can rename it

[vocal robin]

we don't learn y = mx + b by first deriving it from the second form

we are able to learn it from first principles

is it not possible to do it the other way around?

derive point slope form before deriving point intercept

[warm crescent]

yes u can

y-b=m(x-0)

m=(y-b)/(x-0)

[vocal robin]

y-b=m(x-0)
this I also do not fully get

anyway, just going to memorize the point slope equation to pass the exam and then forget it forever. thats what american schooling wants me to do. hate algebra so much

otherwise I will spend 3 days searching for intuition-focused resources that do not exist. every single time I encounter an equation in preparation for my exam

[warm crescent]

you know (0,b) is a point

just plug x=0 to see y=b

and plug x=x1 to see y=y1 is also a point

LHS solve for m

RHS solve for b

then use both

ok step by step

y=mx+b

y-b=mx

[vocal robin]

even going from y = mx + b to y -b=m(x-0), makes me lose all semantic context

I just turn off my brain and go into dumb algebra mode "subtract by both sides..."

and am no longer thinking about the problem

but just stupid symbols on paper

[warm crescent]

know what ur right it maybe its not even necessary

oh

[cedar kiln]

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[distant wyvern]

how do we get this factorization

[distant wyvern]

i get expanding rhs proofs it, but given just the lhs as a function how would we factorize it

[raven shard]

(not relevant)

[pastel vault]

Write this as $\frac{n^6 - 1}{n - 1} - n(n^2 + n + 1)$

[wraith dagger]

south

[raven shard]

there are innumerably many ways you can do this but without knowing about the factorization beforehand you wouldn't think it's possible

[pastel vault]

Then you can see (n^2 + n + 1) will factor

Yeah but it's all based on tricks

Like the question becomes 'how did you know to write it like that'

[wraith dagger]

themadchessplayer

themadchessplayer

[distant wyvern]

yesh i get this

[distant wyvern]

:<

.close

[cedar kiln]

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[raven shard]

(though not to be completely disappointing the complex number approach as mentioned by south does shed some light on this matter

for instance, you can deduce that x^8 + x^7 + 1 is factorable over the integers for the exact same reason)

[pastel vault]

Wow yeah I see it, that's amazing

[raven shard]

numbers 🥰

[distant wyvern]

math is brutal and fun at the same time

[pastel vault]

So yeah that generalises to x^(3k + 2) + x^(3k + 1) + 1

[raven shard]

and you can replace one of the ks with a j 😅

[pastel vault]

Really? wow

Oh right complex number approach

[distant wyvern]

.reopen

[cedar kiln]

✅

[pastel vault]

Yeah I wasn't thinking of complex numbers at first

[distant wyvern]

could i get some elaboration on this

[pastel vault]

By definition $\omega^3 = 1$

[wraith dagger]

south

[pastel vault]

So $\omega^5 = \omega^3 \omega^2 = \omega^2$

[wraith dagger]

south

[pastel vault]

So we have $\omega^2 + \omega + 1$ in total

[wraith dagger]

south

[pastel vault]

$= \frac{\omega^3 - 1}{\omega - 1} = \cdots$

[wraith dagger]

south

[distant wyvern]

$= {\omega^2 + \omega +1}$..?

[wraith dagger]

SirGareth

[pastel vault]

Yes, omega^4 = omega

[distant wyvern]

ahhh

i get it now

thanks

.close

[cedar kiln]

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[pastel vault]

No worries

[cedar kiln]

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@crimson sedge Has your question been resolved?

[crimson sedge]

.close

[cedar kiln]

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[gloomy surge]

"A certain component is critical to the operation of an electrical system and must be replaced immediately upon failure. If the mean life of this type of component is 110 hours and its standard deviation is 30.5 hours, find the minimum number of these components to stock so that the probability that the system is in continual operation for the next 2000 hours is at least 0.95."

I don't know what equation(s) to use.

[upper abyss]

Time until failure is usually modeled with an exponential distribution

That being said, you're looking for the distribution of parts used within 2000 hours

Then again they are giving you the SD here, so they must not want you to use an exponential distribution

What class is this for?

[gloomy surge]

Statistics

central limit theorem unit

[upper abyss]

Ah okay. We're assuming the time until failure is normally distributed

Then let's say we bought 20 of them. We can create a new random variable called the "sample mean" which represents the average time they all fail.

By the Central Limit Theorem, the sample mean's distribution is:

• Normally distributed
• Mean is the same as the original mean
• SD = σ/√n

Where σ is the original SD and n is the number of parts bought

[cedar kiln]

@gloomy surge Has your question been resolved?

[gloomy surge]

.close

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[kindred hornet]

how would i find this

[dreamy void]

hmm

[wraith dagger]

𝔸dωn𝓲²s

[dreamy void]

a = -7

[wraith dagger]

𝔸dωn𝓲²s

[dreamy void]

So the question is which R makes sense

So that in either comes -4 out

-7 - R = -4
or
R - 7 = -4

yea I think the latter equation does more sense, I gave also enough hints lol

[cedar kiln]

@kindred hornet Has your question been resolved?

[dreamy void]

lmao

[kindred hornet]

thanks

i will try this

so from (-3,3)?

[dreamy void]

no

-4 is not even in that interval

Either the left or the right side must be -4

[kindred hornet]

oh

but what about the other point

[dreamy void]

based on R you figure that out

[kindred hornet]

R is 3

[dreamy void]

Ok

So that would mean

(-7-3,3-7) = (-10,-4)

[kindred hornet]

ohh

[dreamy void]

the other equation gives us -3 which gives us the interval (-4,-10)

So the first one (-10,-4) does more sense

and since at x=-4 converges

it should be (-10,-4]

[kindred hornet]

oh yeah

it says its wrong though

[dreamy void]

then -7-x=-13 should make sense

x = 6

(-7-6, 6+(-7)) = (-13, -1)

[kindred hornet]

it says thats also wrong

[dreamy void]

then idk lmao

[kindred hornet]

yeah its pretty weird

[cedar kiln]

@kindred hornet Has your question been resolved?

[raven shard]

a radius of convergence is a positive number...

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[unique crane]

why is this wrong

[unique crane]

shouldnt it equal 0 somewhere there

[cyan turtle]

piecewise function ?

(dont quote me tho, not 100% on your context)

[unique crane]

no piecwise function, thats just for me to visually see the range

on desmos

[unique crane]

but what im not quite sure is how between [-2, 2], this is not true

[cyan turtle]

no no i mean f(x) could be a piecewise function *unless the problem/what youre doing says it cant be

[unique crane]

oh yes

it could be

i see what you mean

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[hidden badge]

helloo how might i find the sum of this infinite series?
$$\sum_{n=1}^\infty{(\frac{3}{n(n + 1)} + \frac{1}{2^n}})$$
it is not a geometric series, and trying to find the limit of its partial sum formula with l'hopitals is much too hard to be the solution

[wraith dagger]

declspecl

[buoyant latch]

Perhaps partial fraction?

Looks like they are both positive so you can just split the series apart assuming it converges

[main needle]

the second best trick besides geometric series formula is telescoping sums, where you rewrite a fraction and hope almost everything cancels

here you can write 3/n/(n+1) as 3/n-3/(n+1), and when you write out a few terms of that notice that a bunch of things cancel out

[hidden badge]

ahh genius i totally forgot those existed... and did you get that rewrite through partial fractions? because i did get that through partial fractions but maybe you did something different
$$\frac{3}{n(n+1)}$$
$$n(A + B) + A = n(0) + 3$$
$$A = 3$$
$$B = -3$$
$$\sum_{n=1}^\infty{\frac{3}{n(n+1)}} = \sum_{n=1}^\infty{\frac{3}{n}} - \sum_{n=1}^\infty{\frac{3}{n+1}}$$

[wraith dagger]

declspecl

[main needle]

yea it's like always with partial fractions

[hidden badge]

oh gotcha LMAO time to reread the telescoping series chapter 😅

thanks 🙏

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[nocturne ether]

how do you solve 5?

[cedar kiln]

@nocturne ether Has your question been resolved?

[main needle]

you'd have to look at how the y column changes between values, like if it always went up by a number or doubled every time

neither of those are true though so there's another way to test if there's any polynomial fit, which is taking successive differences over and over

[nocturne ether]

so keep going past the second difference?

or am i misunderstanding

[main needle]

for second differences I got like 5.1 5.3 5.2 if that's what you got too

since those are about the same and on the second layer you'd say it's a quadratic fit

[nocturne ether]

hmmm

okay i don’t think i understand

this is actually my brothers problem his book is giving a very weird and specific way of solving it

but it’s not working

[main needle]

like if you keep taking the differences
0 3 11.3 24.7 43.3
3 8.3 13.4 18.6
5.3 5.1 5.2

which means 5.2/2*x^2+bx+c is a decent fit

[nocturne ether]

ohhhh

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[north smelt]

What am I doing wrong? I have done this problem 4 times my paper is about to rip

[zenith plinth]

which question

11 or 12

[zenith plinth]

is x = -1 in questionn no 11

[nimble mulch]

You multiply 2x by (x+3) in the start

which is 2x² + 6x

Instead of 5x

And secondly the -(x²-9) bracket

after you subtracted the x²

it became -9 leftover

but it'd be +9

since you have - times -

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[twilit linden]

Why cant a polynomial ring be a field? The classic explanation I see is that f(x)=x has no inverse, but in a prime field wouldn't the inverse just be x^{p-2} given by Fermat's little theorem?

[stiff totem]

your coefficients are from the prime field

x is an indeterminate so any multiplicative inverse is going to end up being x^{-1} which isn't a polynomial

[cedar kiln]

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[prisma gull]

say Z_5[x], do you consider the p(x)= x^3, q(x)= 1 same?

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[light inlet]

Hi

[light inlet]

Correct?

[wanton mountain]

Yes

y = F(x) right?

If the answer is yes, then everything is in order

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@light inlet Has your question been resolved?

[light inlet]

Yee

[light inlet]

Alr tyy

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[lavish prawn]

so, i need to build this natural succession such that a_i = 1/(4i +2), a_(i+1) = 1/(4i-2) , a_(i+2) = 1/(4i-6), a_(1+3) = (4i-10) ... and so on but my issue is, since the 4i is always constant i'm having a lot of issues to come out with the succession, is it even possible to build such succession?

[lavish prawn]

the reason i want to find such succession is that in for this exercise, using the topmost property which i already proved should be sufficient to prove part iii-)

since part iii-) 1/((2i-1)(2i+1)) = (1/(4i -2)) - (1/(4i +2))

if i manage to find a succession a_n such that a_i = (1/(4i + 2)) and a_(i+1) = 1/(4i - 2) then i'm done

for part II-) it was easy, just let a_i = 1/i but part 3 is trickier

[opal ridge]

a_n = -1/(4n+2)

[lavish prawn]

then a_i = 1/(4i + 2) which i want, but a_(i+1) = 1/(4i + 6) which i don't

[opal ridge]

but a_(i-1) = 1/(4i-2)

if you prefer a_n = -1/(4n-2)

[lavish prawn]

ugh, let me think about it for a minute

i don't understand how a_n = -1/(4n+2) would help me solve the exercise, sure i can get a_i = -1/(4i+2) and put it in the sum but when i put a_(i+1) i don't get the expected result to put inside the sum to follow the topmost "theorem"

[opal ridge]

.

it's just shifting indices by one

[lavish prawn]

why for the second exercise i could just make a_n = 1/n to get both a_i = 1/i and a_(i+1) = 1/i+1 but for this one gotta use some a_n = something but both a_i and a_(i+1) doesn't get me what i need for what's inside the sum?

[opal ridge]

I don't understand your question

here, you can check that a_(i+1)-a_i gives you exactly what's in the sum

[lavish prawn]

oh, i see it now, i wasn't thinking about it right

i had to find a_n such that a_(i+1) - a_i = (1/(4i -2)) - (1/(4i +2))

how did you find a_n by the way?

[opal ridge]

your formula basically looked like a_n=1/(4n+2), it was just inverted (hence multiplying by -1) and indices shifted by one (hence a_n=-1/(4(n-1)+2) = -1/(4n-2))

[lavish prawn]

alright, thanks for the help man, i was stuck with this for almost an hour, i've been learning about induction for 2 days and it involves a lot of successions and the like

.solved

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[crimson sedge]

P

[crimson sedge]

There are 2 groups, namely group A and group B. Group A consists of 6 men and 3 women with an average height of 168cm. Group B consisted of 3 men and 6 women with an average height of 160cm. If male students are taller than female students, which of the following changes would cause the difference in the average heights of the two groups to be the smallest?

(A) Move 2 boys from group A into group B and 2 girls from group B into group A.

(B) Add 2 boys to group A whose height is the same as the 2 boys in group A.

(C) Move the 2 shortest women from group B into group A

(D) Move all males from group A into group B and move all females from group B into group A

(E) Add 2 women to group B whose height is the same as the average of women in group A

Anyone have an idea?

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

[cedar kiln]

@crimson sedge Has your question been resolved?

[frail bolt]

Who helps me solve this problem with

[young heron]

Average height of women - 152 and men -176(hope u can do the rest

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[stuck grotto]

confused

[hollow yacht]

q ^ 0 = 1 if q≠0, so you get p*1 = 10

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[latent fox]

Someone explain pls

[peak minnow]

do u know long division?

[latent fox]

Yh

[peak minnow]

ok so divide x^2 by 16x^2 + 1

[latent fox]

Oh because the numerator has a polynomial power equal to the denominator?

[peak minnow]

yes

[latent fox]

Ok I will try that

[peak minnow]

for partial fractions the polynomial power of numerator needs to be lower then denominator

[latent fox]

But I dont understand something

I tried to integrate this equation with a substitution

And I got a different answer

[peak minnow]

which substitution u used?

[latent fox]

Why is my answer different

[peak minnow]

u sub doesnt work here btw

trignometric sub works here

[latent fox]

@peak minnow

Whats wrong with this ^

[peak minnow]

everything?

u cant just square something without doing the inverse

[latent fox]

Why, all I did was square the top and bottom

[peak minnow]

like that i can just do this

$2 = 2 $
$\newline 2^2/1^2 = 2$
$\newline 4/1 = 2$

[wraith dagger]

JustToPro

[peak minnow]

4 = 2

if u are squaring something , u need to take a square root

$$2 = 2$$
$$(\sqrt2)^2 = 2$$
$$2 = 2$$

[wraith dagger]

JustToPro

[latent fox]

Bruh i dont get it

[peak minnow]

do u know why we can divide and multiply a number ?

[latent fox]

Why

[peak minnow]

well cuz we are also doing the inverse

division is inverse of multiplication

same way if we are doing square root we need to inverse it

and inverse of square root is square and vice versa

well whatever , u sub doesnt work here

trignometric sub works here

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[true jetty]

what happened in the numerator

[livid hound]

factored out 2^n

[true jetty]

would this be correct

[livid hound]

no

you've written $2n$ instead of $2^n$, those aren't the same thing

[wraith dagger]

ℝαμΩℕωⅤ

[true jetty]

oh sorry the 2n should be 2^n

okay okay ty for pointing that out, you may have saved me from a couple minuses there

tyvm!

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[hazy kite]

Can someone help me with validation to an answer?

[hazy kite]

well i've been studying AP and stumbled into the fact of transforming an AP into a series of sums, making every term of the AP equal to the terms on the sum.

[hazy kite]

like how the upper limit of the sum equals a_n and the number shown in both terms are equal

<@&286206848099549185>

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@hazy kite Has your question been resolved?

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[ocean mural]

n choose 6 = 28989675

any clever way to do this by hand other than bash

[ocean mural]

tried using the algebraic formula for n choose k but like you just get that
n(n-1)(n-2)...(n-5) = 720 * 28989675

[fair geyser]

there's probably nothing

it's like n! = 6402373705728000

no clever way

[simple egret]

i meannnnnn

u can do prime factorization

[ocean mural]

idk I had it in a contest td and had like 5 mins

[fair geyser]

with a calculator?

that wouldn't help much i guess

,calc(720^(1/6))(28989675 ^(1/6))

[wraith dagger]

Result:

52.47220410016

[ocean mural]

no calc lol

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[manic mountain]

does this equal square root of x

[nocturne reef]

[frigid sinew]

This is x^2/4 so yes basically sqrt of x

[nocturne reef]

$\sqrt[m]{x^n} = x^{\frac{n}{m}$

[wraith dagger]

TayBee
Compile Error! Click the reaction for more information.
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[manic mountain]

is this function the same as y=sqrtx

[frigid sinew]

Yes

[manic mountain]

how come the domain of the first function is all real numbers while the second function is x >= 0, x ∈ R

[nocturne reef]

Because if you square a negative number it becomes positive

Which expands the domain to include negative x

But sqrt(x) doesn't have that luxury

Remember the domain of the square root is all non-negative real numbers

[manic mountain]

yes thank you

so they're not equivalent functions

but algebraically they are

[nocturne reef]

Yeah

Identical for non-negative x, only defined for (x^2)^1/4 for negative x

Your root would be equivalent to $\sqrt{|x|}$

[wraith dagger]

TayBee

[nocturne reef]

In terms of domain

[manic mountain]

thank you

.close

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[crystal halo]

so finish the problem but its wrong anyone can tell me why

[crystal halo]

it is possible book is wrong and has been before

wait i think i did integral wrong

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[kindred sapphire]

I need help

[patent belfry]

Alright ill call 911

Where is your emergency

[drowsy harbor]

xD

bro why is this fucking role still here

[patent belfry]

I think he died

[drowsy harbor]

lol

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[sleek galleon]

Hey, how would i switch the bounds for this? I assume you have to since cos(x^2) is an impossible integral?

[sleek galleon]

Sorry to whoever was about to help me an hour ago with this, my computer totally stopped working

[random shale]

The z is independent so you can do it first

Then it becomes a double integral

Plot the region defined by bounds of the double integral

And use that to swap the bounds

[sleek galleon]

How can I just do the z first without changing the bounds of the inner integral?

[gritty viper]

well you use the z bounds for the integral

but notice the z bounds don't depend on x,y and the x,y bounds don't depend on z

[sleek galleon]

hmm, ok so you can just move the z bounds to the front but since x depends on y, that's why you'd plot the xy bounds

[gritty viper]

Yeah

Integral bounds define your region in a very natural way: here it's 0≤z≤4, 0≤y≤1, 2y≤x≤2

you can swap these bounds around any way you want and they still define the same region

[sleek galleon]

And if they don't depend on each other, then you can evaluate them in any order

ok that makes sense

[gritty viper]

yeah

[sleek galleon]

thank you!

[gritty viper]

np

[sleek galleon]

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[steep steeple]

trying to understand this question and i dont understand what to do man 😭

[carmine bronze]

May I assume correctly that you know how to do derivatives?

[cedar kiln]

@steep steeple Has your question been resolved?

[steep steeple]

for the most part yes

sorry for the delay in response

stomach decided it was time to kill itself

[carmine bronze]

Do you understand what the implications are for a derivative to be positive, negative, or zero?

[steep steeple]

I do not directly recall

[carmine bronze]

The derivative can be used to calculate the instantaneous rate of change of a function (IROC) at any given point of the function.

[steep steeple]

oh that

yeah I believe I do understand that then

[cedar kiln]

@steep steeple Has your question been resolved?

[steep steeple]

actually

wait

i figured it out

.close

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[crimson sedge]

can someone please explain why we don't draw the negative part of the graph ?

and also do we always integrate from the axis to the x intercept? I am a little confused on how to figure out the bounds of integration

[void glen]

4-x^2 looks like this, we dont really need the negative bit because we're just examining our enclosed area

[void glen]

and this depends on what the enclosed area is

[cedar kiln]

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[gloomy crater]

how to do

[humble karma]

Can you think of any examples, or counter examples?

[gloomy crater]

wait for arc length u just take a particular cross section and find arc length of that right

[humble karma]

The arclength is just the length of the curve. Here the curve sits on the unit sphere.

What kind of simple curve has, say, length 2pi on the sphere?

[gloomy crater]

radius 1 circle

[humble karma]

Yeah, so any great circle / meridian right?

Now, knowing such a curve has length 2pi, are there curves that are longer?

[gloomy crater]

so if the curve covers the entire sphere then the arc length is just the volume

which is 4/3pi

wait nvm

wait yea

[gloomy crater]

i meant arc length is just the surface area

which is 4pi

[humble karma]

It's a nice idea, but it presupposes that such a curve even exists, which isn't particularly trivial.

[gloomy crater]

ye true

[humble karma]

There is simpler. If we know that a meridian is a curve of length 2pi, can you imagine any curve longer than a meridian? (Hint ||a meridian wraps once around the sphere||)

[gloomy crater]

one that wraps multiple times

[humble karma]

Yeah

[gloomy crater]

is there a limit to how big the arc length could get?

[gloomy crater]

so its false

[humble karma]

Well just from this construction, you could have it wrap around as many times as you want, so it should be as big as you want. If instead we want a curve that doesn't intersect itself, then you could imagine a spiral that wraps around the sphere, that would also be > 2pi

[gloomy crater]

ty

.close

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[livid hound]

won't be anything that covers "ALL" / the entirety of currently known math

[plucky merlin]

I think they were saying that as in there isn't going to be a book that covers all levels of math

mainly since math diverges into different "categories" and even "sub-categories" of mathematics

[livid hound]

something like what

[plucky merlin]

https://archive.org/details/DemidovichEtAlProblemsInMathematicalAnalysisMir1970/page/n3/mode/2up

This was pinned in #calculus

[livid hound]

#book-recommendations

would be a better place to ask

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[regal nebula]

I am currently confused on how mw^3 got into the denominator. I understand that when condensing or expanding, subtraction is division. But there is subtraction between log m and 3log w, so is it not division again?

[livid hound]

it is division again

that's what you get when you divide again (after applying power law for the 3log(w)

[regal nebula]

Alright sounds good

Thx

Wait so is 3log w being divided by log m?

[livid hound]

its not

$\blue{5\log(p) - \log(m)} - \red{3\log(w)} + \log(p)$ \
$\red{3\log(w)}$ is being subtracted from the $\blue{5\log(p) - \log(m)}$

[wraith dagger]

ℝαμΩℕωⅤ

[regal nebula]

I get it now

Thx

.close

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[dense peak]

Can someone explain my mistake?

[peak minnow]

sin^2 - 1 isnt cos^2

u = sec , works here

[dense peak]

wdym?

[peak minnow]

there are 3 trignometric subs

[dense peak]

oh shoot i used the wrong one

what sbout this one ?

[peak minnow]

thats wrong too

[dense peak]

yeah, figured so

where did i mess up

[peak minnow]

shouldve been x = sintheta

[dense peak]

ah man i think i keep mixing them up

i'm going to try doing them again, can u stick around for 10 mins or so?

[peak minnow]

sure ping me

[dense peak]

why is this wrong ? @peak minnow

[peak minnow]

dont know

seems correct tho

just that why is there still a 9

when we already have 9/9 which is equal to 1

[dense peak]

oh right but when i submit it without it's still wrong

[peak minnow]

ok i see it

[dense peak]

what is it

[peak minnow]

sec theta is x/9

[dense peak]

yea

[dense peak]

oh wow yea

it's right now

okay ima work on the other one now

[peak minnow]

ok , gl

[dense peak]

can u still help me if i struggle w smt if i ping u

[peak minnow]

ok

[dense peak]

Do you know where i messed up ?

[peak minnow]

seems fine

[dense peak]

what do i do now?

[peak minnow]

integrate the cos/sin and use ur value of theta

cos/sin = cot btw

wait found another mistake @dense peak

[dense peak]

wehre

[peak minnow]

missing sec^2

[dense peak]

do i use powers of trig to break up sec^3theta or just make it sin over ocsine

[peak minnow]

i think u sub works here

actually nvm

both work (i think)

[dense peak]

Idk what to do now

@peak minnow

[peak minnow]

srry had to go

i think u = cos works?

[dense peak]

like this ?

[peak minnow]

yes

also thats 1/u^2 now

[dense peak]

i think i messed up somewhere

[peak minnow]

doesnt seem like it

[dense peak]

it was wrong when i submitted it (these are just practice questions, so no penalty)

[peak minnow]

u can simplify that further tbh

u get sqrt (x^2 + 9) + c when simplified fully

[dense peak]

it was wrong

[peak minnow]

hm

[dense peak]

i think i messed up somewhere

[peak minnow]

can u send the question?

[peak minnow]

that isnt the question u sent tho

[peak minnow]

yeah here write $\sqrt{x^2+9} + c$

[wraith dagger]

JustToPro

[peak minnow]

and check fi that is right or wrong

[dense peak]

it was wrong

[peak minnow]

,w \int \frac{\sqrt{9+x^2}}{x} dx

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by justtopro

[peak minnow]

ok we are off by just the 3 tanh^-1 (.....)

[dense peak]

yea how tho

[peak minnow]

,w d/dx -3tanh^-1 (\frac{\sqrtx^2+9}}{3})

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by justtopro

[peak minnow]

,w d/dx tanh^-1 (\frac{\sqrtx^2+9}}{3})

[wraith dagger]

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Requested by justtopro

[peak minnow]

ok what ever

[dense peak]

to me it looks fine but idk why its wrong

[peak minnow]

idk how to solve that tbh

$$\int \frac{1}{\cos^2 \theta \sin \theta}$$

[wraith dagger]

JustToPro

[dense peak]

I tried this

[peak minnow]

what did u let u equal to?

[peak minnow]

firstly there are only 2 cos thetas

and not 3

[dense peak]

oh yeah idk why i j see that now

idk what to do

[peak minnow]

me neither

close this one and open a new one maybe , someone better might be able to help with that

[dense peak]

do yk this one

[cedar kiln]

@dense peak Has your question been resolved?

[cedar kiln]

@dense peak Has your question been resolved?

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[gray sorrel]

find

[gray sorrel]

.close

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[wanton mountain]

Why are these equality valid?

[cedar kiln]

@wanton mountain Has your question been resolved?

[wanton mountain]

@old ridge

i stuck so hard in this step and im gona cry

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@wanton mountain Has your question been resolved?

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[boreal ice]

i got the wrong answer. my working was:

r^2 θ = 57.6 => 1
r θ = 12 => 2
1 / 2 => r = 4.8
r = 4.8 -> 2
θ = 2.5

[grand forge]

r^2 theta is not 57.6

that would be 115.2

[boreal ice]

noticedit

.close

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[lethal jungle]

can someone help me with this question i'm not sure why you times the sectors degrees by six and why you need the arc for that specific bit

[lethal jungle]

<@&286206848099549185>

[cedar kiln]

@lethal jungle Has your question been resolved?

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[timid sparrow]

I have no idea how to start solving this

[coral jewel]

use angle addition formula

[timid sparrow]

what is that

im not native english so it might have another name for me

[ruby tree]

the 2nd is leg is just 1?

[timid sparrow]

ye

[olive warren]

You have to solve for x right?

[timid sparrow]

yep

[coral jewel]

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

[ruby tree]

you can theoretically substitute sin(x) with sqrt(1-cos^2x) and then solve but idk how easy that is

[wraith dagger]

فطر

[timid sparrow]

uhh

could you simplify that

[coral jewel]

wdym simplify this is the formula

[timid sparrow]

i will try that

[coral jewel]

no

[ruby tree]

what will that do here?

[coral jewel]

too complicated

[timid sparrow]

oh ok its this

[ruby tree]

maybe but you can

[coral jewel]

divide each hand by some number such that angle addition formula is applicable

[ruby tree]

not even too complicated. you get the cos sqrt3/3 from the other side and then ^2 both sides and you have a 2nd degree polynomial if you just set cosx=u

[timid sparrow]

what is alpha and beta in this context

[coral jewel]

angles

[timid sparrow]

i know

but what values do they take

[coral jewel]

theres actually a really fast way to solve these type of questions

namely $a\sin x+b\cos x=c$

[wraith dagger]

فطر

[coral jewel]

divide each hand by $\sqrt{a^2+b^2}$

[wraith dagger]

فطر

[coral jewel]

then it will match up with the formula perfectly

[olive warren]

Yea this is probably the right method

[coral jewel]

i was taught this method during my trig chapter and its one of the best thing i learned

[ruby tree]

so what to divide with?

[coral jewel]

but anyway, try this

[coral jewel]

reading comprehension 0

[ruby tree]

?

my method is way easier

[coral jewel]

doubt so, i firmly believe mine is more useful for a wider range of these type of questions

but im no socrates, i am here to help, not to argue

[ruby tree]

well if it is only this type of questions then subsituting sin with sqrt.... always solves it

[coral jewel]

@timid sparrow

[timid sparrow]

im trying skycrew method first

i didnt really understand yours

[coral jewel]

sure, tell me how it goes

[ruby tree]

in this particular example its not even a second degree polynomial as the 1 from each side will cancel out

[coral jewel]

i didnt say its a second degree polynomial what

[lethal jungle]

no my question wasn't solved

[ruby tree]

nah i said it, and i am just now saying that it wont be

[coral jewel]

!occupied

[cedar kiln]

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).

[timid sparrow]

i got to this