help-27

[wicked turtle]

that's the key fact anyway

the rest is algebra and simplifying

[quaint hearth]

thank you so much

[wicked turtle]

sure

cheers

[quaint hearth]

im going to class tomorrow and proving it by derivatives

[wicked turtle]

haha

[quaint hearth]

just to annoy him

[wicked turtle]

👍 \

[quaint hearth]

have a great day/night

[wicked turtle]

you too, later

[quaint hearth]

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[restive river]

I calculated the initial height to be 0.3125m, the radius is 0.125m, and the weight of the ball is 0.0055kg, will the ball be able to fully go around the loop and go to the next output ?

[pastel pasture]

Conserve mechanical energy between the point at which you drop the ball and the bottom of the loop

And I'm pretty sure if the velocity at that point is $\sqrt{5gR}$ then it would complete the loop

[woven radish]

NEONPerseus

[restive river]

I calculated the velocity to be approx 2.5m/s using conservation of energy, so I think it can make the loop. And mechanical energy stays the same I believe

Thanks so much

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[cinder idol]

how do i solve 17(c)?

[weak cove]

What have you tried so far?

[cinder idol]

rewriting in cosx and sinx

[weak cove]

okay where did that get you to

[cinder idol]

a wrong answer

[weak cove]

lol

well it shouldn't bring you to a wrong answer

so some algebra must've went wrong along the way'

[cinder idol]

heres the working out

[devout snow]

@cinder idol Has your question been resolved?

[inner sand]

At your second line @cinder idol

cosAcosB + sinAsinB using our knowledge on trigonometric formulas what is it equal to

It should be faster imo that way than using the product to sum formula, but works finely this way too

You also have to write somewhere the angles when cos and sin are equal to zero before you proceed to this point / exclude them in the end

Also the mistakes comes from the assumption that -cos(x) = cos(x) from what I see

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[brazen widget]

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[proud nimbus]

<@&286206848099549185>

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[small raptor]

https://gyazo.com/52bfdb077647d6fda52683b7ac4cceac

[small raptor]

no clue how to do error propagation

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[small raptor]

https://gyazo.com/acd265bf67684cd6e42e8228c4059777

[small raptor]

https://gyazo.com/01a722a6ef4e0fc7a72a31f1211594c5

using these error propagation

i dont get any of the asnwers that are there

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[mystic musk]

I am wondering which chapter is this cuz I have never seen this before

[small raptor]

Idk if this is even in maths coz this is physics, but if it were in i'd guess it'd be under probability and statistics

[mystic musk]

oh u are done with calculus already?

[small raptor]

But i dont think in high school you would have to worry about that

[small raptor]

No im still studying it, way pass that part tho

[mystic musk]

which part are u now?\

[small raptor]

Im doing physics as a degree

but physics is heavily maths based

so i only need to learn the calculus i need for physics

rn im doing Multi var and vector calc

[mystic musk]

oh

I am doing vector calc soon too

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[manic sedge]

There was a question where we had to find x+1/x when x³+1/x³=110 well I found the solution by using this

(x+1/x)³=x³+1/x³+3x.1/x(x+1/x)
=>(x+1/x)³=110+3(x+1/x)
Let x+1/x be a so
=>a³-3a-110=0
I had to find one factor so I did it by factor therorem and got that one solution is 5 so 5 should be the ans but my exact question is If a³-3a-110=0 then =>a(a²-3)=110 So a should be equal to 110 and (a²-3)=110?

[manic sedge]

I think it's not but still

My teacher did it that way

[winter patrol]

=>a(a²-3)=110
So a should be equal to 110 and (a²-3)=110?
very wrong

[manic sedge]

Exactly

[winter patrol]

My teacher did it that way
very questionable,
do you have a pic of exactly what they did

[manic sedge]

And so how can I factor a³-3a-110=0
Without using hit and trial method

[gray swallow]

a^3

[manic sedge]

I don't have a pic but he did
a=110 and a²-3=110 so a is also equal to √107

[manic sedge]

Typing mistake

[gray swallow]

ye

[manic sedge]

I was like what the f

[gray swallow]

understood

[gray swallow]

yeah this makes no sesnse

because if both were equal to 110

then a*a^2-3=110^2

[manic sedge]

Yeah we can just put the values and see that it's not true

[gray swallow]

yeah

weird teacher

[winter patrol]

any product property isn't a thing

[manic sedge]

Very weird and also he can't even solve this question I showed him how to solve that and at the end he couldn't even factor 🥲

[winter patrol]

time to get a new teacher

[manic sedge]

Yeah so if
a(a²-3)-110=0 how could I solve this without any hit and trial method is there any equation like Quadratic formula but easy?

Not that general cubic equation

Rather than this cause this dosen't have and bx² term

[gray swallow]

i mean

its okay

you can just continue solving it even withou ta bx^2 term

assume that b=0

[manic sedge]

Can't do anything he is a school teacher I can't 🥲

[manic sedge]

Hm

[gray swallow]

just like when solving 2x^2 +6

we cab assume that b=0

[manic sedge]

Ok

And still have to use general cubic equation

[winter patrol]

i suppose you could consider what happens if you assume that a will be a nice integer value

[gray swallow]

yeah

i dont like it very much either so hopefully someone can give a better solution

[winter patrol]

in which case you'd consider factor pairs of 110

[manic sedge]

Yeah hit and trial method I did it that way

[winter patrol]

the solution for these types of problems is usually nice

[manic sedge]

Hmm

Great thanks guys

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[little onyx]

How to find the trigonometric ratios of obtuse angles in terms of triangles ( not graph)

[warped relic]

Hmm

In terms of triangles, it needs sine rule and cosine rule

[little onyx]

Whats that

[warped relic]

$\frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c} = \frac{\sin(\text{angle})}{\text{side opposite to angle}}$

[woven radish]

VulcanOne

[warped relic]

This is sine rule

[little onyx]

Does it mean the length of the Side opposite to the angle?

[warped relic]

Yeah

[little onyx]

Okay i get it thanks..
but still i wanna know how u get the value out of the sine of that obtuse angle....
If i was to say... sine 234°..how would i figure it out?

[gray swallow]

you put it into your caulculator

[little onyx]

Ofcourse thats one way

But people back then didnt have any canculator did they?

How did they find out

[warped relic]

You can find it using angle addition identities

[little onyx]

So wuts that

[warped relic]

$\sin(A \pm B) = \sin(A)\cos(B) \pm \sin(B)\cos(A)$

[woven radish]

VulcanOne

[warped relic]

You can take 234° and break it down to known angles

Like 60°

45°

90°

30°

Those angles

You know their sines and cosines

[devout snow]

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[hardy tartan]

I need help with this

*** A circle has a radius of 3.8 cm and an arc of 4.5 cm. Calculate the central angle.***

How to you calculate the central angle?

[polar chasm]

What's the formula for arc length given radius and central angle?

[restive river]

Do you know the relation between all three of those

[hardy tartan]

B • r : 2

Right?

[polar chasm]

What's B?

[hardy tartan]

4,5 cm

And radius 3,8 cm

[polar chasm]

Well then your formula is incorrect

[hardy tartan]

Why

[polar chasm]

where central angle is in radians

[hardy tartan]

I don’t understand it

[restive river]

dont type words in math mode, either don't or write \text{...}

[hardy tartan]

Can y’all help me

What shall I do?

[polar chasm]

Do you know formula to calculate arc length given central angle and radius?

[hardy tartan]

No

[polar chasm]

Btw do you work in radians or degrees?

[hardy tartan]

B • R : 2

[polar chasm]

$\frac{B\cdot r}{2}$ equals what?

[woven radish]

MathIsAlwaysRight

[polar chasm]

I mean I don't get your formula

$s = r \theta$ this is formula for arclength. $\theta$ is angle and $r$ is radius. $s$ is arc length.

[woven radish]

MathIsAlwaysRight

[polar chasm]

Have you ever seen this?

$s = r \theta$

[woven radish]

MathIsAlwaysRight

[polar chasm]

Have you ever seen this?

[hardy tartan]

No

[polar chasm]

okay, do you work with degrees or radians?

[hardy tartan]

Idk

[hardy tartan]

Just help me

[polar chasm]

degrees is e.g. 360° and radians is 2pi

[polar chasm]

https://www.mathsisfun.com/definitions/arc-length.html

[devout snow]

@hardy tartan Has your question been resolved?

[supple knot]

.close

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[hardy tartan]

.reopen

[devout snow]

✅

[supple knot]

stick to one channel

.close

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[high chasm]

im slightly confused on (ii) when it comes to deciding whether a matrix is in row-echelon form

why does the above matrix satisfy (ii)?

firstly i have labelled in green the pivots, r they correctly labelled as pivots?

[covert geyser]

yes

the leading ones

[high chasm]

i see

would these be the pivots for this matrix?

[covert geyser]

no

that matrix wont work

that line on the right means that that systems of equations for the last line is able to be solved by just 0

ie ax^2 + bx + c = 1

but 0 0 0 means its 0x^2 + 0x + 0 = 1

but 0 cant be equal to 1

[high chasm]

r these just the pivots then?

[high chasm]

mhm

[covert geyser]

yup those are the pivots

but the matrix dont work xD

did u guys cover R1 and R2 and stuff?

[high chasm]

its still in row - echelon form tho

[upper schooner]

Well yea - but more that you’re showing the system is inconsistent

[high chasm]

ye

im not really bothered about if they work rn, im just concerned regarding the row echelon form property

[covert geyser]

yeah that leading one at the very end will still be a pitvot

[high chasm]

wdym

[covert geyser]

cus its still a leading value

[high chasm]

which leading one?

[covert geyser]

this one

[high chasm]

bruh i said that

and u said no

[covert geyser]

💀

mbmbmb

[high chasm]

dw dw

so these r the pivots?

[covert geyser]

studying for 3 midterms gets to u

yes

[high chasm]

klkl

[covert geyser]

they are pivots xD

[high chasm]

calm thx

[winter brook]

??? wait, correct if I'm wrong but why would anything in the augmented column be a pivot?

[covert geyser]

a pivot doesnt have to do with an augmented column

the definition of the pivot is that its just the first non zero entry of a row

[winter brook]

Okay, when I think "pivot" it refers to the values on the diagonals that u use to eliminate the stuff below it in the column so mb, I see

I REFER TO**

[covert geyser]

yeah thats rref

[high chasm]

so when we consider pivots, we can consider the augmented column if needs be?

[covert geyser]

ideally we use reduced row echelon form where each row has a leading 1 as its pivot to help us find what all the values for x are

[winter brook]

^

[covert geyser]

if u want

[winter brook]

that's what I was thinking

[covert geyser]

a pivot is just the first non zero entry

[high chasm]

kl

[covert geyser]

heres a good image

[winter brook]

yeah so the last pivot is a pivot but also useless

[covert geyser]

yes

[high chasm]

ye kl

[covert geyser]

makes the system impossible xD

[winter brook]

oh wait, I'm so dumb, this makes sense

[high chasm]

would this be NOT row echelon since in the 2nd row the 1 is not to the left of the 1 in the third row?

acc wait no

it is to the left

i meant the 1 in the 2nd row is not to the left of the 1 in the first row

?

[winter brook]

nope

all the entries below a pivot has to be 0

[upper schooner]

right

[high chasm]

so its not row echelon since the pivot in each row is not further to the right than all the pivots in all the rows above?

kl

thx

[upper schooner]

Basically this here effectively

[high chasm]

isnt that also reduced row echelon?

or is that below and above

the pivot

has to be 0

ie the whole column par the pivot

[winter brook]

nope reduced row echelon is a bit different

reduced is when u have the row echelon form and then go in reverse order to elimnate the entries ABOVE in each column to have only diagonal entries of 1s in the matrix

[high chasm]

ye so its what i said above

the entire column par the pivot has to be 0

[winter brook]

but isn't not "also"

[high chasm]

and also has to be in row echelon form

[winter brook]

they're different

[high chasm]

ok

kl

[winter brook]

reduced is further

[high chasm]

okok

cheers

[winter brook]

yuh

[high chasm]

has anyone done simplex algorithm here?

cuz im getting flashbacks

[winter brook]

ur getting simplex algo flashbacks from row echelon form?

[high chasm]

ye

[upper schooner]

,w simplex algorithm

[woven radish]

Results provided by WolframAlpha

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Requested by chartbit#3171

[winter brook]

ahah, gg I guess?? I don't see the relation

[high chasm]

like transforming a matrix into reduced row echelon

[upper schooner]

useless

[high chasm]

it seems similar to simplex algorithm

well

very similar in fact

[upper schooner]

Don't remember doing it, but I see what you mean

[high chasm]

like the whole pivot and row operations

to make everything 0s and 1s

u do the exact same thing

in simplex

[upper schooner]

Aha, Gaussian elimination as well too

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[runic reef]

Find exact values for sin(2theta). cos(2theta), and tan(2theta) using the information given.

[runic reef]

cos(theta)=13/85, theta in QIV

...

[random urchin]

Use formulas for double angles, for example cos2x = 2(cosx)^2-1

[runic reef]

2*(4/85)*(13/85)

2x(84/85)x(13/85)

,calc 2 times 84/85 times 13/85

[woven radish]

The following error occured while calculating:
Error: Undefined symbol times

[runic reef]

,calc 2 * (84/85) * (13/85)

[woven radish]

Result:

0.30228373702422

[runic reef]

,fraction 0.30228373702422

(13/85)^2

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[heady orbit]

Help plsssss

[heady orbit]

I dont understand at all lmao

[slate narwhal]

Find the angle

Or side length and use it to solve for x

I assume based on what you get for an answer is the direction you go

[devout snow]

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[cloud gust]

Hello!
I have a graduate-level statistics question.
Suppose you wanted to create 2x2 tables (filled with random values) such that the proportion (let's call this "Q") of these tables that produce significant chi-square test results (p-values of less 0.05) follows a specific distribution.
How would you approach this?

[cloud gust]

In particular, I'm hoping to create a symmetric distribution for Q.

What I have so far:

1. generate T number of tables (I've been using T=6)
2. assign random binomial values to each cell
   a. probability = 0.5
   b. the number of trials is chosen from a random uniform distribution
   i. the max and min for the uniform distribution here is just arbitrary for computational purposes. I've been using U[10, 1000]
3. Run a chi-square test on all T (6) of the tables and record which results have a p-value < 0.05 in a binary vector. (e.g. <0,1,1,0,0,1>)
4. Calculate the proportion ("Q") of those T tables which had significant chi-square results (so for <0,1,1,0,0,1> we'd get Q = 0.5)
5. Repeat steps 1-4 for J repetitions (I've been using J = 1000) and plot the distribution of the proportions (the E's)

I was expecting either a uniform or binomial distribution for Q. Instead however, the proportion of the tables that return a significant p-value is very skewed.

I've run it over and over again an the median is always way above 0.5
I wonder if this is being induced by step 2.b. (using a uniform distribution to select the number of trials for the random binomial sample). I don't have a good argument for why I'm doing that in the first place. I'm simply required to specify the number of trials and wanted to pick something random so I went with uniform.
My other suspicion is that somehow my p-value threshold of 0.05 is causing this, but I don't think that's the case. I'm just grasping at straws.

Here's the code I've written so far (in R) if that helps:
https://pastebin.com/C6di7KBT

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[tiny linden]

I need a sanity check

What is the right answer I need to double check

<@&286206848099549185>

It takes 10-20 seconds

[livid fiber]

Hello!

[tiny linden]

Hi!

[livid fiber]

Needing the right answer

Whats your answer?

[tiny linden]

It’s just 1 minus the 2 numbers right?

[livid fiber]

Correct.

Any thing else you need help with?

[tiny linden]

Alright so it’s 0.1

[livid fiber]

Yep.

[tiny linden]

Nothing else thank you

[livid fiber]

Yw!

[devout snow]

@tiny linden Has your question been resolved?

[devout snow]

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[torpid horizon]

Can somebody help me solve the net ionic equation of solid aluminum will react with a solution of iron (ii) chlorate

[torpid horizon]

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[heady canyon]

hi

[heady canyon]

8 men repair a road for 40 days. how long will it take 24 men to finish the job if they work at the same rate

so i multiply 8 and 40

its 320

then if i divide 320 by 24

its a decimal

13.3

what do i do

[placid rover]

leave it as a fraction

A quicker calculation would have been to notice 24 is 3 times as many men

so it should take 3 times less

40/3

[heady canyon]

ohh

i see thank uu

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[viral lynx]

https://i.gyazo.com/1c7acbf33b74f46d56ccefc975eb7ce8.png

[viral lynx]

The derivative is (8x^7)(e^(-4x))+(x^8)(-4e^(-4x)), but im not really sure what to do next

I know normally I would get the roots to get the mins and maxes but this equation is kinda fucked up so im not sure how to do that

[neat night]

set it equal to 0 and factor

[viral lynx]

I got (x^7)(e^(-4x))(4)(2-x)=0

So the critical points are x = 0 and 2 ?

Ok ye I got the answer

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[viral lynx]

https://i.gyazo.com/ff0e23e3e898d4c29225738cd2b37108.png

[viral lynx]

Im pretty far with this one but idk how to figure out if its x > a or x < a

[sick fulcrum]

let's see the work?

[viral lynx]

I know that the derivative is (e^x)(1-10e^x) so the critical points are ln(1/10) and 0

The thing is the way I normally figure out if its x > a or x < a is I plug in numbers into the original equation and see if theyre greater than or less than the critical points

But with this specific equation its really hard to tell

Like idk what ln(1/10) is but I know its negative

[sick fulcrum]

,calc log(1/10)

[woven radish]

Result:

-2.302585092994

[sick fulcrum]

great first step

you can take a second derivative btw

[viral lynx]

what does that do

[sick fulcrum]

so this can tell you about your stationary points

your stationary points might be increasing or decreasing

[viral lynx]

sry what are stationary points

I probably know but they might be called something different from where I learned them

[sick fulcrum]

basically stationary points are where your derivative equals to 0

f'(x) = 0

[viral lynx]

so the points on the x axis?

[sick fulcrum]

so if you have some x value that's a stationary point

f'(x) = 0 for there

and you might not be able to tell if it's increasing or decreasing

so you need to look at f''(x)

[viral lynx]

(e^x)(-10e^x) like that?

[sick fulcrum]

i'll assume you differentiated it correctly

so yeah

[viral lynx]

Hope so 😅

[sick fulcrum]

if there are any zeros of f'(x) you'll plug them into

this equation here

and then see if it's increasing

although

notice something interesting

this is just -10 e^2x

which is literally always negative

so the stationary points (if they exist) are all decreasing

[viral lynx]

What even is this btw

its a quadratic, right?

[sick fulcrum]

sorry I gtg someone will be around soon

[viral lynx]

np thx for your help so far

[sick fulcrum]

not exactly a quadratic

[viral lynx]

okok

still not really sure what to do next...

[weak cove]

Hi odie

what do you need help with

[viral lynx]

I worked out the derivative is (e^x)(1-10e^x) so the critical points are ln(1/10) and 0

I was wondering how to figure out if its x > a or x < a

Derpz said I should get f''(x) so I got (e^x)(-10e^x) but im not really sure what to do with this

[weak cove]

hold on, how exactly is the critical point 0?

[viral lynx]

From e^x = 0

[upper schooner]

e^x is never zero

[weak cove]

e^x never equals 0

e^0 is 1

Odie, a function f(x) is increasing over all the intervals where f'(x) is positive, or >0

so find f'(x) - which you have already done

the critical points help you determine where the f'(x)=0

this tells you when the function flips from increasing to decreasing

so once you have f'(x)

and you know where it flips (the critical points)

you should be able to construct the intervals for where f'(x) > 0

[viral lynx]

so my only critical point is ln(1/10)?

[weak cove]

well tell me again what f'(x) is

[viral lynx]

(e^x)(1-10e^x)

[weak cove]

and that you are certain of?

[viral lynx]

uuuuh I think so?

idk now haha

[weak cove]

then yes there will only be 1 critical point and it is ln(1/10)

well I just am not double-checking it for you lol

I bet its right

[viral lynx]

ye I think so

[viral lynx]

just the way you said that made me think I was wrong lol

[weak cove]

yah I can see that lol

[viral lynx]

What do you mean construct the intervals

[weak cove]

like figure out what they are

using your logical thinking ability

[viral lynx]

Well I know that ln(1/10) is negative but I dont know what specifically

(I know that I could use a calculator but im not allowed one in my exam so I wont use it)

[weak cove]

why does it matter if a point is negative?

you are trying to decide if the function is increasing

not if it is positive

here's an idea

where f'(x) = 0 the slope is 0, and the function is neither increasing nor decreasing

you found that f'(x)=0 only at one point

the critical point

ln(1/10)

that means the function changes from increasing to decreasing only one time

and the change happens at that point

so

you figure out now

before that point

was it increasing or decreasing

because then afterwards it will be the opposite

and you will have your intervals

[weak cove]

is that wrong? or just sus

https://tenor.com/view/sussy-amog-amogus-sus-amogsus-gif-21543526

[upper schooner]

At most one time

x^3 vs x^2

[weak cove]

true

that's true

because the function can have a horizontal slide

[upper schooner]

Anyways the second derivative can help you determine if it's a max/min

[weak cove]

probably easier to just plug in a point right before hand

idk though

that's just what I would do

[upper schooner]

'tis fair, that should work, at least in this case(!)

[weak cove]

What are you studying chartbit?

[upper schooner]

Nothing

Well I am covering some stuff in my spare time but me no student no more haha

[weak cove]

oh, what did you study?

[upper schooner]

Maths covered mostly analysis and algebra towards the end

[weak cove]

I would've been surprised if it wasn't math considering the context lol. What was your favorite class

[upper schooner]

I don't actually know probably Banach spaces I liked the most [tl;dr part of analysis]

[weak cove]

thanks for that last bit lol, idk what I was expecting but of course I've never heard of it

Did you enjoy studying math? and if you don't mind me asking what are you doing/plan to be doing as a career afterwards?

[upper schooner]

Quite did [and the fact I lurk here probably shows I still quite enjoy it to some extent lol]
At the moment I do development work really(!) Thinking of maybe doing more study but depends on life

[viral lynx]

I think I got it, x > ln(1/10) because I subbed in x=0 into e^x - 5e^2x and got -5
-5 < ln(1/10) < 0 so the graph is increasing
is this correct ^^ ?

[weak cove]

well, again do you want to know if the function is increasing or if it is positive

substituting into f(x) will tell you if it is positive, but substituting into f'(x) will tell you if it is increasing

that's interesting. I am going to start studying math next year, I am quite looking forward to it

Hopefully I like it as much as I think I am going to

do you have any advice you would give me? math specific or just general advice for college?

[upper schooner]

Ermm probably nothing too deep that I can think of at the moment, just that university math can often be a shift of mindset from school math, and if you have the opportunity to cover some university-style course if you can (can make the transition much easier, some people struggle with it at times!)
That said, you'll probably be fine I guess, if you're looking forward to it(!)

[viral lynx]

https://i.gyazo.com/c7f53daf8ce6e87de02777729c0bd061.png

I got the correct answer thanks guys!!

❤️

[weak cove]

no problem

nice job

Yeah, I've taken about two years while I was in high school, just through calc4 and diff eq linear algebra

so nothing too serious

I am trying to get used to analysis from MIT OCW because I hear all the time that it is super difficult

and I have more than enough time during my break year right now to do some studying

Just very bored, one of the reasons I am in here a lot too

[upper schooner]

Relatable haha
But otherwise yea think you'll be fine - a lot of the time first analysis courses can be the toughest ones [not sure what they put in that one] but once you've done that, if you survive it's usually nice after then, at least from my experience

[weak cove]

that's good to hear

its been like proof stuff so far, induction and like introduction to cardinality of sets

like that kind of things

but I am just beginning

thanks for the chat

[viral lynx]

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[icy depot]

I m having trouble finding domain of k(x)

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[sharp parcel]

Can someone help me with d

[devout snow]

@sharp parcel Has your question been resolved?

[sharp parcel]

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[magic pine]

petition to ban vague questions from schools

[magic pine]

what have you tried

[devout snow]

@fast wyvern Has your question been resolved?

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[small raptor]

https://gyazo.com/1f9c60dca7f0a2033074d16bc4b9e157

[small raptor]

Im using error propogation formula but im not getting any of the answers

[devout snow]

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[untold vector]

Are both terms equal?

[scarlet sequoia]

yes

[warped relic]

Yes

[flat wigeon]

Yes

Can someone please help me btw in 23

[untold vector]

when we take ab to numerator then why it needs to be ab and not -ab?

[scarlet sequoia]

wdym by taking ab to the numerator

[untold vector]

@scarlet sequoia so here when I took ab to RHS shouldn't it be -abx?

[scarlet sequoia]

what you did is right (obv y and ab aren't 0)

if you want to have minus on the RHS you can do that also

then it would be

(b^2 + ac)y = -abx

it's equivalent form

in fact it depends on what you multiple both sides by

-ab or ab

[untold vector]

So the minus sign would be with numerator or with denominator at a time.

[scarlet sequoia]

generally

[woven radish]

Modus

[scarlet sequoia]

you can put minus wherever you want

[lofty glacier]

what is the question?

[untold vector]

ah okay, good to know this. I didn't know that.

[untold vector]

Modus already answered, thanks.

Many thanks @scarlet sequoia 👍

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[spiral hornet]

In 2019 and 2020, we invested 400 euros in the bank at the beginning of each month. We will withdraw the saved funds in the following way: with equal withdrawals of b euros at the beginning of each month in 2022, 2023 and 2024 and with another withdrawal of c = 3,000 euros at the end of 2024. Calculate b if interest is paid in a conformal manner with monthly capitalization and if the annual depreciating interest rate is 3% all the time.

[devout snow]

Please don't occupy multiple help channels.

[spiral hornet]

Can you please help me

Nobody helped me

Please

?

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@spiral hornet Has your question been resolved?

[devout snow]

@spiral hornet Has your question been resolved?

[spiral hornet]

.close

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[midnight mantle]

Hi! I was wondering if someone could help me (I'm also new here so I'm sorry if I do smth wrong). I need to construct a rectangular triangle in which one corner is 45 degrees, and the hypotenuse is 6,5cm long.
NOVO

[buoyant rock]

Rectangular as in one side is 90°?

[midnight mantle]

yes

one corner is 90 degree, but I must make the other corner 45 and the hypotenuse 6,5cm

[buoyant rock]

Then do the same with the other side you don’t know

Then knowing all side and angles should make it pretty easy to construct

That’s not a 0 btw it’s o for opposite

So just plugging into a calculator really

[honest aurora]

please dont give the full answer. you can do this without any trigonometry

[honest aurora]

since you know that one corner is 90 degrees, another is 45 degrees, can you find out what the angle of the last corner must be?

[midnight mantle]

I hope I understood the question right (im sry english isnt my first language) but it isn't set. But I think since the inside corners of the triangle must always equal 180, it should be 90,45,45

[honest aurora]

thats correct

now there is this property we have when we know that 2 angles are the same in the triangle. do you know what this property is?

[midnight mantle]

I've been thinking but I can't really thnink of an answer

[honest aurora]

thats ok. maybe try googling "isosceles triangle"

if you look at some pictures, you may be able to notice it

[midnight mantle]

so

Should the cathetus (I dont really know the word for it in english, but basically the two shorter sides of the rectangulard triangle) be the same lenght?

[honest aurora]

exactly yes, have not heard the term "cathetus" but for all i know it may be correct in english. but now we know that the two shorter sides will have the same length

so now is the last important part that you will need to know to do this problem. since we have a 90 degree angle in the triangle, there is an important formula that we can use to relate the lengths of the triangle

[midnight mantle]

pythagora?

[honest aurora]

yep

so we have $a^2+b^2=c^2$ right?

[woven radish]

Duh Hello

[honest aurora]

and using what we know about this particular triangle, what can we say about $a$ and $b$?

[woven radish]

Duh Hello

[midnight mantle]

that they should be half of c? ( im really sorry if this is wrong, I suck at maths)

[honest aurora]

no problem, but not that isnt exactly right. what we found earlier is that the two smaller lengths are the same right?

[midnight mantle]

yes

[honest aurora]

so mathematically we can say that $a=b$

[woven radish]

Duh Hello

[honest aurora]

does this make sense to you?

[midnight mantle]

basically a equals b

[honest aurora]

yep, do you know what we should do next then?

[midnight mantle]

honestly

no

[honest aurora]

well we know that $a$ and $b$ are equal, and we have the pythagorean theorem saying $$a^2+b^2=c^2$$ could we perhaps replace $b$ with $a$ here?

[woven radish]

Duh Hello

[midnight mantle]

I think we could?

[honest aurora]

yep, do you know what that equation would become when we do that?

[midnight mantle]

Im starting to feel dumb here, but still no

[honest aurora]

well since $b=a$ then $b^2=a^2$, so we can just replace the $b^2$ in the equation

[woven radish]

Duh Hello

[midnight mantle]

okay

I think I got that part

[honest aurora]

so what does the equation become after that?

[midnight mantle]

inverted? Im sry theres a proper word for it here but I cant really find it in english

[honest aurora]

i doubt that inverted is the correct word

it is supposed to look very similar, just that b^2 is supposed to be a^2 now

[midnight mantle]

a square plus b square is same as b square plus a square?

[honest aurora]

that is true yes, but not exactly what i am looking for

what we want to do is we want to remove b from the equation entirely

and we can do this because we know that a and b are equal

[midnight mantle]

so

wait

a² × 2 = c²

[honest aurora]

perfect

now we are close to the answer

[midnight mantle]

yayy

[honest aurora]

so we know what c is already right? so we just need to find out what a is

[midnight mantle]

i just wanna say, thank you for having the patience to do this

[honest aurora]

no worries 👍

so now how would you go about figuring out what a is?

[midnight mantle]

basicaly

so

a²×2=6.5

that should mean that

[honest aurora]

dont forget its c^2

[midnight mantle]

a²×2=6.5²?

[honest aurora]

yep

[midnight mantle]

okay

just gimme a moment

I got 4.5

4.6*

[honest aurora]

yep thats correct

and since you know a=b then you know that b is also 4.6

so now you can attempt to draw the triangle

[midnight mantle]

yay

it worked out

thank you soo much

I have a test and this was the only thing I couldnt figure out

[honest aurora]

no problem 👍

good luck with the test

[midnight mantle]

thank youu

[honest aurora]

also since you are new here. once you dont need this channel anymore you can type .close to free the channel up for someone else

[midnight mantle]

I was just wondering how to do that

thank you again

.close

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[short iron]

Question 84, I've got this far but not sure how to factor whats left

[short iron]

[\frac{d}{dx}(\frac{1}{3}x^3+2x^2+2x)=1]
[\frac{1}{3}(3)x^{3-1} +(2)x^{2-1}+2(1)x^{1-1} = 1]
[x^2+4x +2= 1]
[x^2+4x+1= 0]

[woven radish]

dopediscorduser

[sonic smelt]

Completing the square is what I would recommend

[short iron]

Alright thank you

.close

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[loud spindle]

So for this I implicit differentiated

[loud spindle]

Then put in the value of 1 for x

To get 1/2

For gradient

But I am not really verifying? Or am I

[quaint citrus]

So there’s 2 parts to this question

1. Verify that (1,1) lies on the curve

2. find the gradient at (1,1)

Did u do 1. ?

[loud spindle]

No

I don't know how

I skipped to 2. Tbh

[quaint citrus]

Just plug in (1,1) for x and y into the equation

Then see if the left side equals the right side

[loud spindle]

I get 1 when 1

Yea

[quaint citrus]

Ok yea so as long as u show that, I’m sure that should be enough for the first part

[loud spindle]

Ohh dam

Does a gradient of 1/2

Seem right?

[quaint citrus]

Show me ur work for the implicit differentiation

[loud spindle]

Let me write it down

Sec

[quaint citrus]

Ok

[sacred rock]

Check your dms, I got the same result when I did it yesterday 🙂

[loud spindle]

Ohh okay tyy!!!

I appreciate itt

[loud spindle]

Ohh cool tyy!!

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[limber torrent]

solve the Diophantine equation and explain it to make like I'm a fifth grader

[grizzled yew]

Post it

[limber torrent]

.close

[devout snow]

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[restive river]

Hello

[restive river]

Need help with the following exercise

I dont understand why we obtain sin(1)

I mean the Red part

Does anyone Need more details?

[devout snow]

@restive river Has your question been resolved?

[restive river]

Hello man, no It has not been solved

[devout snow]

@restive river Has your question been resolved?

[restive river]

Man no

[main python]

1-x goes to 1 so sin(1-x) goes to sin(1)

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[restive river]

i cant seem to finish this work, i dont know how to do it

it is algebra 2

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[supple knot]

pick one problem you're stuck on.

and explain where you're stuck

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[near barn]

Hi!

[near barn]

https://research-repository.griffith.edu.au/bitstream/handle/10072/58722/88799_1.pdf?sequence=1

are the 3.3 and 3.4 connected here?

[devout snow]

@near barn Has your question been resolved?

[devout snow]

@near barn Has your question been resolved?

[devout snow]

@near barn Has your question been resolved?

[prisma canopy]

Yes. 3.3 describes a function for r in terms of theta and h. They then integrated w.r.t theta

[near barn]

would you be able to explain how to determine # of toilet papers using the archimedes spiral? im quite confused what happens here:

[prisma canopy]

Well, if you have the lenght of the spiral, then you only need to divide that by the length of the toilet papers to get number of toilet papers right?

[near barn]

how do i determine length of spiral using this method tho

[prisma canopy]

Like, imagine you're at the centre of the toilet paper roll right? And let the radius of the paper roll be = R. Now we imagine that we're starting at the part where the toilet paper starts.

So from here, we introduce a couple of new variables. Let's say the angle you make as you go along the toilet paper roll is theta. Then as you keep going outwards to the edge of the toilet paper, your distance from the edge reduces by a roughly the thickness of the toilet paper right?

@near barn

[near barn]

"Then as you keep going outwards to the edge of the toilet paper, your distance from the edge reduces by a roughly the thickness of the toilet paper right?" why so?

[prisma canopy]

Because the toilet paper has thickness? Like, take a piece of string or a ribbon or whatever and wind it across some cylindrical object. How is it that the wound ribbon/string is gaining thickness? It's because it's overlapping itself right? So each time you wind it around the entire cylinder, the thickness increases by h, where h is the thickness of the ribbon/string/whatever else you're using.

Also, checking the paper again, go to close to the bottom and see figure 3. It might help you visualise it

[near barn]

Ah right makes sense now. every one roll, the distance decreases by the thickness right

[prisma canopy]

Yup

[near barn]

how can that concept be used to determine number of toilet paper then?

whats your chess rating btw haha

[prisma canopy]

So the formula r(theta) = R + theta * h/2pi is clear to you?

[prisma canopy]

Meh I haven't played chess in a while, and even at my best I was only ~1200 on chess.com

[near barn]

no not really. why does radius increase by theta h / 2pi?

[near barn]

thats pretty solid

[prisma canopy]

First, note that theta is in radians here. Next, as you said, the toilet paper increases by 1 paper thickness over a full rotation aka 360 degrees or 2pi radians. Thus, the increase in h over unit angle theta is 1/2pi

[near barn]

yup makes sense now. and is it common knowledge that spiral equation is r(theta) ?

[prisma canopy]

Nah, but it is expected to be able to derive such an equation quickly

[near barn]

can i just start by saying "equation of a spiral = r(theta)

or would i have to prove it

[prisma canopy]

Depends on context

In most schools/colleges/universities or what have you, it'll probably require a proof

[near barn]

how would I prove that?

[prisma canopy]

We just did?

The entire "wind it around the cylinder and it increases by h*theta/2pi" etc. is a valid proof

[near barn]

oh wow yeah youre right

lol

yeah i get it now.

what happens in the 3.4 section then

[prisma canopy]

That's where we take the integral, basically winding it around the cylinder 2n times and noting how much length it took to do so

[near barn]

what does Si look like in the diagram? quite unclear

[prisma canopy]

That would be Si

[near barn]

is Si = h/

[prisma canopy]

theta * h/2pi? Yeah

[near barn]

is Si the area or the length?

[prisma canopy]

Length

[near barn]

i have this

so every time length li is reached, a new segment Si is created? @prisma canopy

and why does li = ri delta theta

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[topaz blade]

so

i have to find the point where the resultant forces line of action intersects the line AB, in feet from A

i have Fr, and the angle of Fr off the x-axis

obviously

but how do i find where the line of action intersects?

is Fr not a free vector?

[devout snow]

@topaz blade Has your question been resolved?

[topaz blade]

<@&286206848099549185>

i think i figured out that the resultant force acts on the point where the 2 forces intersect

but thats as far as i got

.close

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[noble roost]

helo

[noble roost]

need help with this

let me upload work

one sec

slow network

there

, rotate

[noble roost]

, rotate

[inner sand]

Hmm over here

[noble roost]

yuh

[inner sand]

Isn’t the formula ax^2 +bx + c = a(x-x1)(x-x2)?

The right side when you simplify the quadratic

In the nominator

Over here

[noble roost]

yea

$-2(x^2+3x-9)$

[woven radish]

Mathematician

[inner sand]

Yeah, shouldn’t the last equation have -2 in front of the brackets?

Ja

[noble roost]

oo

but -2x^2+6x+18 can also be written in ax^2+bx+c

shouldnt i look at this equation and make it equal to 0?

[inner sand]

Yee

But then doing x-x1 x-x2 shouldn’t it have a infront

like a(…)(…) when doing that

[noble roost]

when i equal this equation to 0 i get something else for x

so i cant do it because a>1?

in ax^2+bx+c?

[inner sand]

Here also a two went missing

I think

[noble roost]

$6x-2(x^2-9$

[woven radish]

Mathematician

[noble roost]

$6x-2x^2+18$

[woven radish]

Mathematician

[inner sand]

No no look upper equation 6x + 2

in the first fraction

[noble roost]

ohhh

yeahh

so $6x-2-2(x^2-9)$

[woven radish]

Mathematician

[noble roost]

$6x-2-2x^2+18$

[woven radish]

Mathematician

[noble roost]

$-2x^2+6x+16$

[woven radish]

Mathematician

[inner sand]

Ye

[noble roost]

that gives me something sqrt and not nice to see

[inner sand]

+2

[noble roost]

so yeah its -2x^2+6x+20

[inner sand]

6x + 2

Yep

That has nice roots

And it simplifies to -2(x-5)(x+2)

[noble roost]

this is x=-2 or x=5

[inner sand]

where -2 is the a infront of x^2

[noble roost]

does it matter?

[inner sand]

Yes

[noble roost]

why?

[inner sand]

since when we divide by it

the < changes

[noble roost]

but i dont divide it

i look at

-2x^2

[inner sand]

ok then your sign when doing interval method

are different

since then you start with -

rather than +

[noble roost]

i dont understand

shouldnt i equal the equation
-2x^2+6x+20=0?

[inner sand]

Yes and find roots

And simplify it

But it isn’t simplified as (x-5)(x+2)

Since when we foil it it’s different equation

Our is simplified as -2(x-5)(x+2)

Now why it’s important

When you do interval method you look for the sign of your highest x

[inner sand]

which here is a +, but in reality is a -

So you start with a - and switch from right to left

[noble roost]

let me understand the first step

[noble roost]

you said i need to simplify it?

[inner sand]

Isn’t this what are you doing in the left side first?

Same like there

[noble roost]

i have now -2(x+2)(x-5)

[inner sand]

Correct yeah

[noble roost]

over (x-3)(x+3)

and that is smaller and equal to 0

[inner sand]

Yes

[noble roost]

now what do i do

?

[inner sand]

What did you do in your left side when you were at this point?

Interval method right?

[noble roost]

im not sure

[inner sand]

What's confusing you

[noble roost]

since i have -2(x+2)(x-5)

dont i need to make it equal to 0?

[inner sand]

Yeah and we find the roots 5 and -2

And put them on the line with 3 and -3

[noble roost]

right

so whats the problem

[inner sand]

With what sign do we start

[noble roost]

smaller and equal

[inner sand]

I mean in the interval line

We write plus/minus

[noble roost]

-3

[inner sand]

What is our rightmost sign

plus or a minus

[noble roost]

a plus

[inner sand]

Why you think so?

[noble roost]

because when i put -3 in the graph i put a number in that domain which for example lets say is -4 as x and then solved the equation with it and got a positive numbero

oh wait this is for the left side

for the right side its 5

and i set nubmer 6 as and solved

and still got positive

[inner sand]

You sure?

-2(x+2)(x-5)/ (x-3)(x+3)

[noble roost]

i didnt understand you

[inner sand]

Don't we have this -2(x+2)(x-5)/ (x-3)(x+3) <= 0

Like do we agree we have this

[noble roost]

yeah

[inner sand]

Substitute 6 in

[noble roost]

wait dont we need to divide by -2 and then its still 0 in rhs and sign switchees?

[inner sand]

Yes

Exactly

[noble roost]

so i dont need to make it equal to 0?

[inner sand]

What I meant here

No no no, you are missing the point

[noble roost]

i think i am

[inner sand]

Imma try not make it confusing

In this -2(x+2)(x-5)/ (x-3)(x+3) <= 0 we have this

if we divide by -2 and the sign flips, we have

(x+2)(x-5)/ (x-3)(x+3) >= 0 and this

In both cases the solution's the same, but in the first we take the ones with -, in the second the ones with a plus

What you did in your work is do the interval, forget about the coefficient -2 and solve your equation that is <= 0 and take the ones with +

If it had -2 everything changes

[noble roost]

so i dont need to make it equal to zero but simplify it first

[inner sand]

You can equal it to zero straight up if it doesn't confuse you to make a mistake

Gotta be extra careful on these

Like forgetting that -2 when finding the roots of the nominator and rewriting it really did change the outcome

Equaling it to zero only gives you the roots, they don't change whether you simplify it or not, just makes it easier to read, what's important is the signs

Since we want an easy way to see what's the sign before x^n to know if we start with a plus or a minus from the right

[noble roost]

yeah

[inner sand]

What?

I'm not clicking right now, where did I say this?

There is no difference if we equal -2(x+2)(x-5) to 0