help-23

[solid shell]

Hopefully that will get you started

[jagged carbon]

oh I see, appreciate it

[safe radish]

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[coarse anvil]

I have found the negation of the statement and was going to try contradiction to prove it, but not sure entirely how to disprove it

[calm storm]

I wouldn't suggest contradiction personally

[coarse anvil]

How else could i do it?

[quasi sentinel]

you can find the minimum of y(y+2)

[coarse anvil]

What wuold i do from there

[quasi sentinel]

that proves the statement

i think

[coarse anvil]

Ohh yeah i get it, thanks

[safe radish]

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[supple elbow]

Could anyone help me calculate Fs?

Were not given l, only the mass of the ball thats on a string (drawing)

[vital bison]

If you could translate from Dutch what its saying

[supple elbow]

And theta= 28degrees

[vital bison]

Actually its possible to figure out with context

[supple elbow]

“A metal ball with mass is attached to a string, mass of the ball is m=627 g and g = 9.81 ms^-2. Theta is 28degrees with the ceiling. Calculate Fs with the given formula”

[vital bison]

Do you know how dot products work?

[supple elbow]

fg * l * cos theta? ithink?

[vital bison]

Do you not have access to the length of the string?

[supple elbow]

no i don

t

[vital bison]

Oh well we don't need it so I think we're good

[vital bison]

Yes, and what would theta be?

[supple elbow]

28 + 90?..

[vital bison]

perfect

[supple elbow]

how could we get rid of the l?

[vital bison]

You're diving by l anyways

It will take care of itself

[supple elbow]

is it like this?

[vital bison]

yes it is

[supple elbow]

whats the difference between |x| and ||x||?

|| x ||

uuuhm ||.x.||

double | hahahaha

[vital bison]

|x| represents the absolute value of a real number while ||x|| represents the magnitude of a vector

[vital bison]

Write \||x|| instead

Basically whatever you have inside |x|, make it positive and get rid of the pipes

[supple elbow]

Thanku! last question (b), theyre asking me what the range [a,b] is the solution of the dot product can be in

but i dont rlly understand that one either

[vital bison]

You need to find the maximum and minimum value of the dot product basically, which you can do by varying the angle between the two vectors

[supple elbow]

i know that cos can be min -1 and max 1

but idk how i can continue frm that

[vital bison]

and your dot product was $||\vb{F_g}||\cdot||\vb{l}||\cos \theta$

[flat frigate]

jewels!

[vital bison]

so what's the range of it, if cos(θ) may be varied from -1 to 1

[safe radish]

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[pure grail]

anyone help? i need to proof this using mathematical induction. idk what to do next with the summation of combination(n+1, k)

[safe radish]

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[pure grail]

.close

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[delicate pollen]

how do i solve if this serie convergence or diverges

[left willow]

see what lim n->infty n^3(6sin(1/n)-sin(6/n)) is

[safe radish]

@delicate pollen Has your question been resolved?

[delicate pollen]

ill try

is it 0 @left willow

i did lim of n^3(6sin(1/n)) - lim of n^3(sin(6/n)) and its inf-inf ?

[left willow]

you can't subtract infinities/undefined terms

[left willow]

how about you try a change the variable t=1/n

[delicate pollen]

ill try

[safe radish]

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[delicate pollen]

ive found 0

.reopen

[safe radish]

✅

[left willow]

sus

[delicate pollen]

is that wrong

[left willow]

oh wait is it

[delicate pollen]

so does that make this serie converges

[left willow]

lim t->0 (6sin(t)-sin(6t))/t^3=6cos(t)-6cos(6t)/3t^2=(36sin(6t)-6sin(t))/6t=36-1

nth term converges to non zero term, thus series diverges

[delicate pollen]

oh i did as t goes to inf

why is it t goes to 0

[left willow]

we changed the variables

t=1/n
lim n->infty is equivalent to lim t->0+

[delicate pollen]

ahh got it

does anything change if n^3 becomes n^2

[left willow]

nth term converges to 0 for n^2, so we need to do some extra work to see if it converges or diverges

[delicate pollen]

im confused by this, if the limit is 35 then doesnt that mean it converges?

[left willow]

the nth term converges, but if its not 0 the series does not converge

[delicate pollen]

so if nth term converges but not = 0 then serie diverges?

[left willow]

if the series converges then nth term converges to 0

thats all

I'm sure you must have seen this somewhere at the start of the chapter

[delicate pollen]

yea

then how did we find this serie divergence? cant it be non

[left willow]

what?

[delicate pollen]

if the nth term converges but not 0, then serie diverges?
and if nth term converges and its to 0, then serie converges?

[left willow]

first statement is tru second is false

[delicate pollen]

alright

so when n^3 is n^2 we get nth term converges and its to 0

then how do we follow up after this fact

[safe radish]

@delicate pollen Has your question been resolved?

[left willow]

idk, seems like lot of work to do

[ripe shore]

Do you realise the series n^3 diverges. If yes then the bracket term is 6.sin(x)-sin(x). Sin is between 1 and -1 so bracket tern is between 7 and -7. So the given series is between -7n^3 and 7n^3 both of which diverges Hence this also diverges

[safe radish]

@delicate pollen Has your question been resolved?

[delicate pollen]

im guessing the thumbs downs mean no

[ripe shore]

See bracket term(6sin(1/n)-sin(6/n)) is irrelevant it is a like a constant whose value is always between 7 and -7. If n^3 diverged both 7n^3 and -7n^3 divergea

[left willow]

What you did here was very non rigorous and it was already answered, the follow up question was does \sum n^2(6sin(1/n)-sin(6/n)) converge?

[ripe shore]

No same proof will do

n^2 diverges

It is not non rigorous

I am bounding the series between two known divergent series

[left willow]

No wut, bounding between -infty and +infty doesnt mean it diverges

[ripe shore]

OK then use the property that for any positive number n I can find x such that x^3=n . And all these functions are increasing on natural number

[left willow]

by your reasoning would you say [(\sum^\infty 6\sin(1/n)-\sin(6/n) ] diverges? because [\sum^\infty 1] diverges?

[ripe shore]

This sequence also diverges yeah although proof needs to be changed

[left willow]

no this series converges

[ripe shore]

No it doesn't. Tell me the where it converges and why

[ripe shore]

Interesting. OK I was wrong.

[left willow]

replacing n^3 by n, that series also converges

tho seems like it doesn't converge when replaced by n^2

[ripe shore]

See for n>0 sin(1/n) is positive right So (6.sin(1/n)-sin(6/n))>1 for all n. Hence the summation should diverge. You are adding positive number greater than 1 at each step.

[left willow]

????

does anything that you wrote make sense

[ripe shore]

Sin(1/n) is positive for n>0 ?

I see the fault

[left willow]

plus integral calculator already confirms that it converges, so not sure what your trying to achieve

[left willow]

b4 I leave, just saying n^2 case is pretty non trivial(but checking integral calculator confirms that it diverges) , so you don't need to think to much about it, the exercises you might face will ask you something pretty direct( don't quote me if I missed some simple trick)

[dusky pivot]

Thank you for the clarification. Indeed, handling the product of two polynomials with high degrees can become quite complex, especially when it comes to calculating the coefficients. In most practical exercises or applications, you will typically deal with polynomials of reasonable degree, and the process I described should work effectively. For extreme cases or complex mathematical situations, specialized software or symbolic calculators can be valuable tools.

If you have any more questions or need assistance with any specific mathematical problems in the future, please feel free to ask. Good luck with your studies and any exercises you encounter!

[safe radish]

@delicate pollen Has your question been resolved?

[safe radish]

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[soft sentinel]

\frac{2z}{z+3}=\frac{3}{z-10}+2

[soft sentinel]

i need help fir this problem

solve this equation

[sleek plank]

$\frac{2z}{z+3}=\frac{3}{z-10}+2$

[flat frigate]

890s

[soft sentinel]

yes it this one

[sleek plank]

cross multiply to get rid of the denominators

[soft sentinel]

10 x 2z and 3x 3 ?

i think you can use this method

[sleek plank]

2z • (z-10) = 3(z+3) + 2(z+3)(z-10)

[soft sentinel]

this is what you meant?

[sleek plank]

yea

[potent rain]

nevermind

[sleek plank]

because we multiply both sides by (z+3) and (z-10)

since the 2 doesn’t have a denominator

[potent rain]

cant you cross multiply as the are equal?

[sleek plank]

because the 2 is there, we can’t

we also need to multiply the 2

[potent rain]

ok

so we multiply both denominator and numirator

so therefor itll be 2z(z-10) / (z+3)(z-10) = 3(z+3)/(z-10)(z+3) + 2(z+3)

[soft sentinel]

alr i got 2z^2 - 20z = 3z + 9 + 2z + 6

[potent rain]

wait so am i wrong?

[sleek plank]

don’t forget to also multiply the 2 by (z-10)

[sleek plank]

yea

[soft sentinel]

and now since i have this result i have to change places in 1 or 2 number

[potent rain]

bro how am i wrong?

because we multiply both numirator and denominator

and also the 2 with z+3

[sleek plank]

if you want to ask a question, please get your own help channel

refer to #❓how-to-get-help

[sleek plank]

the 2 also needs to be multiplied by (z-10)

[potent rain]

i have a maths exam tomorrow morning which consists of Algebra of this kind. I am participating as a way to practice

[sleek plank]

yes but you asking questions is impeding me helping @soft sentinel

[potent rain]

sure buddy my bad

[soft sentinel]

i did?

[sleek plank]

since you multiply the left side by (z-10) to get rid of the denominator there

but since 2 doesn’t have a denominator of z-10

it doesn’t cancel out

[potent rain]

hey bro is it possible if we add the 2 to 3/z-10 by finding common denominator and then cross multiplying and solving for z?

[sleek plank]

you could

but cross multiplying would be trickier

especially with rational functions

[potent rain]

if we add two will we need to find common denominator between 2/1 and 3/z-10

[soft sentinel]

2z^2 -20z = 2z^2 -20z + 3z + 9 + 2z + 6

????,

i did 2 twice

[sleek plank]

not the 2z, the 2 on the left side of the original equation

the one that was added to 3/z-10

[soft sentinel]

my head exploded

im lost

[sleek plank]

so we start with this
$\frac{2z}{z+3}=\frac{3}{z-10}+2$

[flat frigate]

890s

[sleek plank]

and then we cross multiply

[soft sentinel]

okay

[sleek plank]

giving us

$\frac{(2z)(z+3)(z-10)}{(z+3)} = \frac{(3)(z+3)(z-10)}{(z-10)} + 2(z-10)(z+3)$

[flat frigate]

890s

[sleek plank]

you get it so far?

[soft sentinel]

wait hold on

im watching to understand

ohh i got it

we give to everyone the same things

[sleek plank]

yep

on the left, the z+3 cancels out and on the right, the z-10 cancels out

[soft sentinel]

the thing down are getting canceled?

?

[sleek plank]

$(2z)(z-10) = 3(z+3) + 2(z-10)(z+3)$

[flat frigate]

890s

[soft sentinel]

just understood why it got canceled

because they where arleady abov e

[sleek plank]

yep

[soft sentinel]

so you crossed on it

[sleek plank]

yes

now we can simplify

[soft sentinel]

and now it sound easy

i will try

[sleek plank]

alright

[soft sentinel]

2z^2 -20z = 3z + 9 + 2z - 20+ 2z+6

i either have to keep simplify or change the method of calculus

i will try to simplify

[sleek plank]

it should be $(2z^{2}-20z) = (3z+9) + (2z^{2}-14z-60)$

[flat frigate]

890s

[soft sentinel]

the end is different

i will try to understand why

[sleek plank]

you have to use foil for the 2(z-10)(z+3)

[soft sentinel]

use foil?

[sleek plank]

First
Outer
Inner
Last

it's a way to remember how to multiply a term in the form (a+b)(c+d)

we multiply the first of each: a*c

then the outer: a*d

then the inner: b*c

and then the last: b*d

and add them all

[soft sentinel]

like z x z then z x 3 and you do 10 x z and 10 x 3?

something like that?

[sleek plank]

could you use latex?

[soft sentinel]

i don't have it

i will do it on a scrap paper

[sleek plank]

ok

[sleek plank]

we have a bot that does latex for you if you want

[soft sentinel]

i may check later

[sleek plank]

just surround ur latex in dollar signs

[soft sentinel]

anyways i found z^2 + 3z - 10z - 30

by doing ab cd

[sleek plank]

we can simplify the 3z-10z in the middle

giving us -7z

[soft sentinel]

and then i multiply all of that by 2?

[sleek plank]

yep

[soft sentinel]

ohhhhh 2z 2 - 14z + 60 i got

damn

i mlean -60

[sleek plank]

yep

now you can add it all together

[soft sentinel]

and now we have the first part of the exercice

hmm

[sleek plank]

here's the full equation

$2z^{2}-20z= 3z+9 + 2z^{2}-14z-60$

[flat frigate]

890s

[soft sentinel]

i replaced some numbers

2z^2- 2^2 = 3z -14z-20z-9-60

[sleek plank]

ok, now simplify

[soft sentinel]

z^2=-9z-51

finally

that was long as heck

[sleek plank]

the z^2 should have canceled out to 0

[soft sentinel]

oh

[sleek plank]

cuz 2z^2-2z^2

[soft sentinel]

so -9z=-51

[sleek plank]

yep

that's it

[soft sentinel]

epic

[sleek plank]

epic

[safe radish]

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[tidal imp]

!nosols

[safe radish]

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

[icy lance]

not really helpful if they dont know how you get to it

[hushed yoke]

Agreed

[brazen grove]

Help!

[tidal imp]

Also it’s not even correct because you need it in decimals

so lmao

[hushed yoke]

lol

[tidal imp]

Anyway

[daring swan]

ca

https://tenor.com/view/cat-reach-reaction-bamtoro-gif-26399871

yep

1/2 is

lmo

lamo

[brazen grove]

(2/3) to the power of 3 please 🙏🏾!!

[hushed yoke]

alright can i get help rather than having people flame one another in my current help channel?

[daring swan]

https://tenor.com/view/lego-batman-batman-sad-sad-batman-sad-walk-lego-batman-sad-gif-23138196

[hushed yoke]

114/73= 1.5616

[daring swan]

lmao wtf when is that gonna be useful

[tidal imp]

Ok hold on

Is d the noise rating?

[hushed yoke]

I believe so

[daring swan]

gonna go to a rock concert, "hmm, yes i hear that this rock concert is about 20,000 times as intense of a mouse squeaking at night"

[tidal imp]

Ok, we’re solving for P here, not d

[daring swan]

lmao bro df

[tidal imp]

Since we’re comparing the intensities of the sounds

[tidal imp]

Try solving for P

[hushed yoke]

Ah got it

$114=10\log\left(\frac{P}{10^{-11.4}}\right)$

[flat frigate]

xNiden

[hushed yoke]

$P$ = $10^{-11.4}*10^{11.4}$ = $1$

[flat frigate]

xNiden

[hushed yoke]

$73=10\log\left(\frac{P}{10^{-11.4}}\right)$

[flat frigate]

xNiden

[hushed yoke]

$P$ = $10^{-11.4}*10^{7.3}$ = $10^-4.1$

[flat frigate]

xNiden

[hushed yoke]

$\frac{1}{10^-4.1}$ = $12589.25$

[flat frigate]

xNiden

[hushed yoke]

.close

[safe radish]

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[lean otter]

what does : Equivalently, the field of complex numbers is algebraically closed.

mean

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

.close

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[lost nexus]

I'm stuck on this problem

[icy lance]

have you written out the vector w-cv?

[lost nexus]

yes

[icy lance]

do you know any ways to show vectors are perpendicular?

[lost nexus]

a • b = 0?

[icy lance]

jsut write c as c in the ()

[icy lance]

exactly

[lost nexus]

Like this?

[lost nexus]

,rotate

[icy lance]

i believe so

[lost nexus]

how do i proceed from here

since i have a c in some of the terms

[icy lance]

collect them, you have [(2-2c)i+(1-5c)j].[2i+5j]=0

then just do the dot product

[safe radish]

@lost nexus Has your question been resolved?

[safe radish]

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[lost nexus]

@icy lance sorry i had to go eat dinner with the fam

[lost nexus]

Are these right?

The bottom there

[icy lance]

nononono

i think youve greatly misunderstood what the dot product is

its not a multiplication like (x+2)*(x+5) or something

theres no such thing as i^2, j^2 and k^2

[lost nexus]

oof

[lost nexus]

umm

[icy lance]

thats why i said to gather the i and j terms

[lost nexus]

Would it be like this instead

[icy lance]

no

did we have [ (2-2c)i + (1-5c)j ] . (2i+5j)?

[lost nexus]

yes

[icy lance]

okay, then its this but get rid of the i and j

[icy lance]

this has no unit vectors in it, in fact it has no vectors in the result

youre only multiplying the corresponding components coefficients

[lost nexus]

so it just becomes 2(2-2c)+5(1-5c)=0

right

[icy lance]

yeah

[lost nexus]

then algebra to find c

[icy lance]

indeed so

[lost nexus]

gotcha

got 9/29

seems like it was right

thanks for the patience and the help!

[safe radish]

@lost nexus Has your question been resolved?

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[icy lance]

no worries

[safe radish]

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[outer flame]

hi i need help with naive bayes

[outer flame]

this is the question

this is the frequency table i did

and these are the calculations

i was wondering did i do the calculations correct or what am i missing ?

appreciate any help

[safe radish]

@outer flame Has your question been resolved?

[safe radish]

@outer flame Has your question been resolved?

[safe radish]

@outer flame Has your question been resolved?

[safe radish]

@outer flame Has your question been resolved?

[safe radish]

@outer flame Has your question been resolved?

[cunning jolt]

.reopen

Hi Helpers I want to complete circles lesson this week this is day 1 should i start sums or should i do something that is important

.close

[safe radish]

@outer flame Has your question been resolved?

[safe radish]

@outer flame Has your question been resolved?

[safe radish]

@outer flame Has your question been resolved?

[jovial shale]

Let me take a look

Assuming all are independent, wouldn't the predictions be (i) True (ii) True (iii) True

[spark island]

hi

[steel lion]

how do i solve this?: there are 2 cars, they need to drive 100km. one of them is 54km/h faster and needs 25mins less to drive the 100km. i need to solve this in a quadratic equation, i need the speed and the time (for the 100km) for both cars

[safe radish]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[silk token]

@steel lion I suppose that it speed is constant so I would set both equations:

First of all the data we have, the fastest vehicle is the second one
v2=v+54
xi1=xi2=0
xf=100km
tf1-25=tf2

The speed is constant so I can calculate it with the equation of the average speed (I guess that how you say it in English) because it doesn't change

∆x/∆t=v

And then you just put in that equation the values we already have

For the first one (the slower)
100/t=v

For the second one
100/(t-25)=v+54

I hope it helps you (@steel lion )

[steel lion]

@silk token thanks for the answer but it is wrong and i already got ot right

[silk token]

Oh my bad, it says minutes and not hours 😭

[edgy cloud]

Looks like the question this was originally opened for didn't find any takers, it's time to recycle the channel.

.close

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[lilac sequoia]

Is this correct? I'm not fully sure if I got the maximum error right

Should my maximum error be 3/2, or is 0.45 right?

[safe radish]

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[lilac sequoia]

<@&286206848099549185>

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[pale mortar]

Is there a way to prove this?

[pale mortar]

I tried plugging 0°, 15°, 30°, 45°, 60°, 75°, and 90° into theta and it all seems to hold true.

[hard crest]

this is cool

find the coordinates of C and B

and then P and Q

[flat frigate]

oogaooga

[pale mortar]

Is this like an actual rule for trigonometry or something?

[hard crest]

it's definitely close to one

,tex .double angle

[flat frigate]

Haylsune Miku

[hard crest]

oh yeah

[pale mortar]

Aha I see

[hard crest]

good job

[pale mortar]

Thanks!

How do I close this

[hard crest]

type .close

[pale mortar]

.close

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[reef quartz]

can someone help me with b(i)?

[pulsar pecan]

consider part a, but with x = pi/7

[reef quartz]

right

oh fuck

on the right side it's sin(pi)

which is 0

and then you divide -sinx by 2sinx

-1/2

thanks for the help

.close

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[dapper grail]

Struggling wiht getting the limits of my function... please help!

[dapper grail]

UPDATE: Remembered I had to inverse the order of integration since integral of sin(y^2) is not possible by hand

Are these limits correct?

<@&286206848099549185>

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[earnest python]

need help understanding the epsilon delta proof

[earnest python]

for functions of 2 variables

i understand the general premise and procedure

heres the example from my textbook:

what would happen if i incorrectly say L=1? for ex

[stoic dune]

Then you wouldn't be able to get |f - L| arbitrarily small

The big deal here is that |f(x,y) - 0| can be made as small as any choice of ε

[earnest python]

mmm ok

so i would still get a relationship betwwen epsilon and delta but its up to me to make sure it makes sense in the context

?

let me try to work through the math

i think i understand now that I cant

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[dusk basin]

How do I prove that for any n > 2, the symmetric group S_n is not cyclic

[pulsar pecan]

A cyclic group must be abelian

Prove that it is not abelian

Therefore it must not be cyclic

[dusk basin]

Thank you so much

[stoic dune]

You could show that S3 is not abelian, and that S3 is a subgroup of any permutation group

[dusk basin]

I thought that by why would that logic not apply with S2 < S3 < S{n>3}

[stoic dune]

An abelian group has abelian subgroups.

Taking the contrapositive of that:
"If a subgroup is not abelian, the original group wasn't either"

I think you might be trying to suggest that:
"If a subgroup is abelian, then the original group was too"
But that's false

[dusk basin]

I understand thank you

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[balmy wasp]

How do you factorise? Could someone help me work through a question and explain?

[sudden forum]

consider a b and c the respective coefficients.
then split b into sum of two numbers such that the product of these two numbers is equal to the product of a and c

[flat frigate]

n11

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[coarse trail]

how would i solve this? im really bad at basic algebra

[hasty wagon]

do you know partial fractions?

[coarse trail]

i do not recall

[hasty wagon]

oh.... lemme grab a site for you

https://www.mathsisfun.com/algebra/partial-fractions.html

[coarse trail]

oh yes i do know that

i got to this but im unsure what to do after. am i able to cancel the x+4s and x-3s out or no because its being added on the top?

[hasty wagon]

nah

the other side

[coarse trail]

so then what do i do

[hasty wagon]

the other way round

[coarse trail]

is A1 and A2 the zeros of the expression its being multiplied to

[hasty wagon]

$\frac7{(x+4)(x-3)}$ into something like $\frac{A_1}{(x+4)}+\frac{A_2}{(x-3)}$

[coarse trail]

so is A1 -1

[flat frigate]

Biscuity

[hasty wagon]

where A1, A2 are constants

[coarse trail]

are they just 7

for this one

7/(x+4) + 7/(x-3)

[hasty wagon]

nah

it's not that easy

did you read through the site?

I always recommend mathisfun for it's clear instructions

[coarse trail]

so i followed it

[hasty wagon]

yes?

[coarse trail]

now im at

[hasty wagon]

ohhh, you mixed up the negatives

[coarse trail]

oh

is one of the 1/x+4 supposed to be negative

because that would mkae sense

[hasty wagon]

yes
and 1/(x-3) is positive

careful!

[coarse trail]

okay i found what i did wrong

i did the A1 and A2 right but accidentally switched the (x-3) and (x+4) a the last part

[hasty wagon]

oh...

[coarse trail]

i understand how to do it now thank you for helping

[hasty wagon]

Cheers!

[coarse trail]

how do i close this again

.close

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[wind fog]

whats the equation of a sets of point which are always same distance apart from (b,0) and the y axis

[stoic dune]

Such points would be (x,y) such that:
√[(x - b)² + y²] = |x|

Squaring both sides:
(x - b)² + y² = x²

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[blazing pilot]

Help

Induction question

[quasi bison]

you can (re)prove $\sum_{k=1}^m k = \frac{m(m+1)}{2}$ and then recognize that $$\sum_{k=n}^{n^2} k = \sum_{k=1}^{n^2} k - \sum_{k=1}^{n-1} k$$

[flat frigate]

AnnGhost

[rain bear]

seems like u gotta use $\sum_{n=1}^k(n) = n(n+1)/2$

[quasi bison]

your n and k are the wrong way around above/below the sigma

[quasi bison]

this would be the least painful way i think btw

[flat frigate]

Rootsyl

[grim plover]

cant he collect all the n terms together, and the natural numbers together, and add n^2 at the last?

[icy lance]

n is the upper limit, k is the index

[quasi bison]

still wrong notation

[quasi bison]

longer than my approach imo

[grim plover]

hmm true ig

[blazing pilot]

Ohhh I see

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[normal dew]

help with question 5 please

<@&286206848099549185>

[somber jewel]

draw a diagram

,rccw

[lean otter]

!15min

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[normal dew]

sorry \

[somber jewel]

@normal dew

[normal dew]

doin it now

im confused with it saying it is due west of town z

and with the size

question

[somber jewel]

due west means town Y is west of Z

[normal dew]

ok thanks

but the size question what is it asking for?

[somber jewel]

angle

[normal dew]

ohhhh thank you

.close

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[normal dew]

im confused on what to do

[quasi bison]

do you know how to solve linear equations in general?

[normal dew]

i do but i need a brush up on it

[quasi bison]

ok in that case time to jog your memory

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

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[quasi bison]

do you remember such algebraic moves as "add ___ to both sides" or "multiply both sides by ___"? @normal dew

[normal dew]

yes

[quasi bison]

ok

right

would you like us to go through some simpler examples or do you want to try the one you came with

[normal dew]

simpler would help please

[quasi bison]

ok

i'll want you to write out your work in full and verbalize all actions you take

here's the first equation for you to solve:
x + 4 = 11

[normal dew]

oh wait do i minus 4 on each side

[quasi bison]

you subtract 4 from both sides, yes.

write that out in full.

[quasi bison]

ok that'll do

[normal dew]

thanks for the help

[quasi bison]

what?

we're not done here are we

[normal dew]

oh

[quasi bison]

i was going to give you another one to solve now

[normal dew]

rightio

[quasi bison]

8x = 96

[quasi bison]

ok yeah you divide both sides by 8

great

let's now up the ante a little bit

2x + 1 = 91

[quasi bison]

ok

it looks like you've recalled enough

can you now try the one you came with, i.e. 3x - 5 = -6?

[quasi bison]

,rccw

[quasi bison]

ok yeah seems good

[normal dew]

thank you

👍

.close

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[dreamy dagger]

i cant eliminate the i, how to solve this?

[merry sleet]

first the expression of r can be greatly simplified no?

[dreamy dagger]

ys

[quasi bison]

i is meant as the imaginary unit here right?

also can you show your work thus far

[dreamy dagger]

ys

[dreamy dagger]

wait

[dreamy dagger]

i can only simplify to this step

but i cant eliminate the i in the denominator

so does it no answer?

[merry sleet]

is tamago your second account?

[dreamy dagger]

ys

[merry sleet]

i^6 =?

[dreamy dagger]

-1

ohhh

my problem

oh wait

.close

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[lean otter]

hey can someone help me with integers

[icy lance]

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[unkempt salmon]

Can someone help me with 3 and 4?

[unkempt salmon]

In 3 I tried factoring, adding them by multiplying by denominator

I tried all I could but cant find answer

[plucky elk]

Did you try putting it over a common denominator first

[unkempt salmon]

Sqrt(x) = (sqrt(2)+1) i also tried that

[plucky elk]

(x^(3/2) * x^(3/2) - 1) / x^(3/2) = ?

[unkempt salmon]

Uhh wait what

[unkempt salmon]

How did you get that

What does it mean tho

[plucky elk]

I did this

[plucky elk]

Just a common step to get towards simplification

[unkempt salmon]

I just cross multiplied to get ((x sqrt(x))^2 - 1)/(x*sqrt(x))

[plucky elk]

That's the same thing I wrote yes

,tex .exp rules

[flat frigate]

riemann

[plucky elk]

And √x = x^(1/2)

[unkempt salmon]

So why did you do 3/2?

Instead of 1/2?

[plucky elk]

x = x^1

[plucky elk]

Apply first row

[unkempt salmon]

But I am still not close to final answer tho

[unkempt salmon]

Bro I am even farther from the answer than when I started 💀

[plucky elk]

Simplify it more

[plucky elk]

Using the first row again

[unkempt salmon]

Yes now thats (x^3 - 1)/(x^(3/2))

[unkempt salmon]

Sqrt x = (sqrt(2)+1)

Now what do I do 💀

[plucky elk]

Calculate other parts of this expression

Whenever you get stuck, go back and see what else you haven't calculated yet

x^3, x^(3/2)

[unkempt salmon]

Other parts?? Thats the question itself

Uhmm

[plucky elk]

You broke down the problem into smaller parts

[plucky elk]

Using this

[unkempt salmon]

So its ((sqrt(2)+1)^6)/(sqrt(2)+1)^3)

[plucky elk]

So calculate this for your x

[plucky elk]

How did you get this

If it's true, it's easier to just evaluate the beginning expression rather than putting over a common denominator

[plucky elk]

x^(3/2) - 1/(x^(-3/2))

[unkempt salmon]

I still dont get what you meaj

[unkempt salmon]

How do I evaluate the first expression

Thats the question I am asking here

[safe radish]

@unkempt salmon Has your question been resolved?

[unkempt salmon]

Nope

[safe radish]

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[grim plover]

Can I continue on like this? Is this allowed?

[grim plover]

(I omitted the limit and the 1/x^2 part)

[final halo]

how do you know that for example N/2 is an integer

[peak estuary]

you will have to do some stuff with floors but roughly speaking the idea would work. but I dont see this being useful

[grim plover]

I am trying something, perhaps it might be useful

[grim plover]

Wdym

[grim plover]

Stuff with floors?

[final halo]

$\sum_1^M$ does not make sense if M isnt in {1,2,3,...}

[flat frigate]

ΣΑCu

[peak estuary]

$\floor{\frac n 2}$

[flat frigate]

Denascite

[grim plover]

Well for some big a and constant b, N/b^a must approach an integer

[grim plover]

Ohh wait I get it. This fixes what @final halo says

While what i write as the upperbound may not always been an integer, it's floor will be

.close

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[lean otter]

This one fast lol
https://i.imgur.com/7MNPP03.png
Is the answer mean value theorem?

[lean otter]

Or an answer- I imagine theres a lot of ways to do this :3

[hard crest]

sure MVT will work

[lean otter]

ty!

.close

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[safe radish]

@opal bramble Has your question been resolved?

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[spring basin]

pls help me

[spring basin]

need help with algebra 1

idk whats about it

but one of the questions are like l = prt

and i make it r = l/pt

but how do i show work

[mystic leaf]

what do you mean?

[tiny wraith]

You have to say you divided by p and by t

That's probably enough

but it kinda depends on what you are taught

[normal moss]

"Solving for a variable", "Isolating a variable" ...

Same thing as with something like $2x = 2$

[flat frigate]

USS-Enterprise

[normal moss]

You divide by 2 to get x

Here you just divide by pt to get r

[spring basin]

ohh ok

thanks

[normal moss]

No problem 🙃

[spring basin]

$what is this command thing$

[normal moss]

Latex

[spring basin]

super cool

[normal moss]

Symbol for math

Makes writing math here possible

[spring basin]

nice

[normal moss]

😄

[shy lynx]

Nice for writing fractions: $r = \frac{l}{pt}$

[flat frigate]

Morrow

[normal moss]

Or anything and everything math related really

Even 2x = 2 looks non-mathy for me

[spring basin]

how do you learn to type this

[normal moss]

It's not that hard

Pretty logical

[spring basin]

how to write fractions

[normal moss]

$fraction = \frac{numerator}{denominator}$

[flat frigate]

USS-Enterprise

[spring basin]

$r = \frac{L}{pt}$

[flat frigate]

sarabcool

$r = \frac{L}{pt}$

[normal moss]

Exactly

You can get a bit more complex

$\forall(a, b) \in \mathbb{R}^2: (|a| < 1 \land |b| < 1) \implies |a + b| < |1 + ab|$

[flat frigate]

USS-Enterprise

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@spring basin Has your question been resolved?

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[fleet vessel]

I'm having trouble deriving the product to sum formula of sinxsiny.

My process is this:

[fleet vessel]

nevermind, I'm braindead.

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[stiff hemlock]

Hey, what's up? Can someone help me solve this extreme value problem? (A rectangle should be bent from a wire of length 50 cm, which encloses an area of maximum content. Calculate how to choose the length and width.
2) A rectangular plot should have an area of 400 square meters. How long should the sides of the rectangle be chosen so that the perimeter of the rectangle is minimized?)

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@stiff hemlock Has your question been resolved?

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@stiff hemlock Has your question been resolved?

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

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[ashen juniper]

this guy is doing the cross product of vectors

can anyone tell me why it is i determinent -j then +k?

what determines wether the outside of the determinent of the vector will be positve of negative

[final halo]

+-+
-+-
+-+

is the pattern, this is just how you calculate the determinant when expanding about rows/columns

[ashen juniper]

can you choose any of these?

[final halo]

if youre doing cross product and have i,j,k along the top row then you must expand along the top row

[ashen juniper]

so i must do +-+?\

[final halo]

yeah

[ashen juniper]

thanks\

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[final halo]

https://en.wikipedia.org/wiki/Laplace_expansion

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[quasi yoke]

could someone explain this for me?

[hearty egret]

wow

clear ?

[quasi yoke]

ahh okay thank you

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[fleet vessel]

I'm trying to derive the sum-to-product formula for sinusinv, but I'm running into trouble. Here's my work, please excuse the handwriting:

I see that, if I multiply both sides by -1, I'll end up with the intended formula: cosx - cosy = -2sin((x+y)/2)sin((x-y)/2), but I'm wondering if there's some way to set this up where I don't have to multiply by -1 at the end. Or, alternatively, does it matter if I do? Can I just use cosy-cosx = 2sin((x+y)/2)sin((x-y)/2) instead?

[deft pivot]

there is absolutely no problem in multiplying both sides of an equation by any number but 0

[main mural]

omg it's "sin(u)sin(v)", i read that as "sinus inv[erse]"

[fleet vessel]

Sorry about that, I'll keep that in mind for next time.

[fleet vessel]

Got it, thanks. Is there a reason that the given formula appears to have been multiplied by -1, instead of being left in the form that I've derived?

For context:

[deft pivot]

yes that's what i was going to say

first terms in all other formula starts with x and second term is y

i think that the reason

[main mural]

exactly

you got (cos(y)-cos(x)) = -(cos(x)-cos(y))

[fleet vessel]

Got it, so it's likely just to keep all of the first terms in the same order of x and y?

Thanks for the help! 🫡

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[wispy flint]

Use direct comparison test to show if it converges or diverges

[wispy flint]

the cos^2n is confusing me, how do i know what to compare this series to?

[final halo]

cos^2(n) <= ... ?

[wispy flint]

idk

[final halo]

name literally any number that cos^2(n) is always less than or equal to

[wispy flint]

i honestly have no idea

of like what value cos^2(n) holds

[final halo]

okay then think of the graph of cos(x) and the graph of cos^2(x)

[wispy flint]

ok i know the cos(x) graph but idk cos^2(x)

[final halo]

okay well what is cos(x) always less than or equal to

[wispy flint]

1

[final halo]

okay and do you think squaring all values of cos(x) could cause them to be greater than 1?

[wispy flint]

no

[final halo]

great so you know that cos^2(x) <= 1 for all x

[wispy flint]

mhm

[final halo]

so now use that to compare

[wispy flint]

so u compare with 1/n^(3/2)

[final halo]

yeah

[wispy flint]

converges bc p series

[final halo]

yes

[wispy flint]

which is bigger, 1/(n^3/2)?

[final halo]

huh?

[wispy flint]

nvm

ty

.close

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[gray snow]

prove sqrt(3) is irrational

[gray snow]

let sqrt(3) = a/b

i get to point where
3b^2 = a^2

but i dont know how to proceed

like a lot of proofs just say
since a^2 is a multiple of 3, a is a multiple of 3, but i dont understand why

[steep lily]

think about prime factors

if you write 3 as a product of prime factors, when you square it the powers of those factors must be even

e.g. $2\cdot3^2 \cdot7$ has square $2^2\cdot3^4\cdot7^2$

[flat frigate]

Edward II

[steep lily]

so if 3 divides a^2, because it must have come from a, 3 must have a power of at least 2 in the factorisation of a^2

so it must have had power at least 1 in the original a

that is 3 divides a

[steep lily]

.

[gray snow]

ohh i see

thank you

.close

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[modest rain]

This graph's behavior near the asymptotes is shown in table below.
So my quesiton is why is x-> 2- is y -> negative infinity when the leftmost part of the graph is still going up as it approaches asymtope of 2?

[main mural]

y -> positive inf

i think you made a mistake

[modest rain]

Let me check

Ok I think you're right, but I'm just confused how this + and - notation works

[main mural]

2- means from the left side, 2+ means from the right side

[modest rain]

Left side/right side meaning the graph's leftmost / rightmost function? Or the direction where the graph is going?

So this would be the left side?

I think that's what you mean

[main mural]

yes

[modest rain]

Hmm ok

[main mural]

imagine printing out the graph and cutting it with scissors in half exactly at 2

the right half will be 2+

the left will be 2-

[modest rain]

Ok makes sense

So just to check my understanding, -2+ would be y -> infinity, since for it to approach -2 it have to keep going up to y infinity? (infinity is unreachable in this case but yk what I mean)

[main mural]

(-2)+ would be y -> -infinity

(-2)- would be y -> +infinity

[main mural]

what's this function so i can plot it too?

[modest rain]

1/(x+2)(x-2)

[main mural]

nvm got it

$\frac{1}{\left(x-2\right)\left(x+2\right)}$

[flat frigate]

artemetra

[main mural]

let's gooo

[modest rain]

I checked my table and it's correct?

[modest rain]

But why is this? If we were trying to approach it this way we wouldn't get closer to the asmytope of -2?

[main mural]

you can think of (-2)+ as being just greater than -2

and of (-2)- as being just smaller than -2

[modest rain]

So you would have to go backwards on the rightside to get to -2?

Hence -infinity?

[main mural]

if you have $(-2)+ = -2 + \epsilon$, where $\epsilon > 0$ is some tiny value, you'll see that as $\epsilon$ gets smaller, y tends to $-\infty$

[flat frigate]

artemetra

[modest rain]

So essentially this?

[main mural]

uh

let me think

[main mural]

YES

yes

this is what i was trying to explain lol

[modest rain]

Ok I think this clears everything thank you

.close

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[steep lily]

epsilon delta as always

for the first, you use the fact that if f(x) > M (which it is near a by assumption), then 1/f(x) < 1/M and use that to find a useful δ, similarly for g and then algebra of limits

the third I need to think about

(and the second because that one might not be that straightforward, but I'm not sure)

[shadow blade]

I think that question is weird infinity/infinity is Indeterminate form

its about 4

[steep lily]

yes, so that limit isn't necessarily 0 and hence isn't to be selected

[shadow blade]

infinity^infinity = infinity?

if yes, then option 3

[steep lily]

yes, but that's not a proof

I do think the third is pretty straightforward though, again by saying you get f(x) > any M, g(x) > any N by choosing appropriate δ and then you can choose those so that M^-N < ε

.

[steep lily]

yep f(x)=g(x)=1/x make the second limit e-1

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[steep lily]

.

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[toxic crow]

HELP

[toxic crow]

.reopen

The population of rabbits on an island is growing exponentially. In the year 1998, the population of rabbits was 9400, and by 2006 the population had grown to 32000. Predict the population of rabbits in the year 2011, to the nearest whole number.

HOW DO I DO THIS SHIT

[little mesa]

you've probably been taught some form for exponential growth like $N(t)=Ae^{kt}$

[flat frigate]

chlamydia

[little mesa]

if you sub t=0, you get A as the initial population, with some multiplier k that you find with another given population value

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[brittle gorge]

Hi helpers, I was having some trouble with this proof and any help would be much appreciated

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@brittle gorge Has your question been resolved?

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[trail fjord]

How do I solve this, do I start by removing the parenthesis or whay?