help-10

[obtuse pebble]

@spice chasm Has your question been resolved?

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.close

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[tough vector]

i have a question for this series

[civic zealot]

Go for it

[tough vector]

okay so I want to find if its convergent or divergent so im going to use the intergal test

but obviously just by looking at it when n is 1 itll just be 0

so i guess my two questions are why cant i just take the intergal of this series from 1 to infinity anyways

[civic zealot]

You can

[tough vector]

or why cant i just change the 1 to a 2 instead and do the intergal test

[civic zealot]

You can

[wispy wadi]

you can think of something easier than that tbh, does the question ask you for the integral test ?

[tough vector]

okay because in this example he takes the derivative of the function first to find out if you can use 2?

yeah it asks for intergal test

[tough vector]

what would you use though just for reference

[wispy wadi]

serie of positive terms and ln(n)/n^3 <= n/n^3 = 1/n²

there are other easy ideas but this one seems... not tiring ?

[civic zealot]

Oh, it has to be positive and decreasing to use the integral test.
Yeah. So ignore the first non-positive term, then use the derivative to show decreasing over the remaining terms. then integrate

[tough vector]

we cant tell just by looking at it that its not positive and decreasing

?

so I have to show it everytime

because like obviously its continous and its positive but i have to show it decreasing is what your saying\

[wispy wadi]

it's quite obvious here but it's better to prove + sometimes you could get wrong if you don't look at it so you might as well take the habit

[tough vector]

okay that makes sense thank you

[slim lake]

such a nice pfp

[tough vector]

also did you have a test that would be better

[wispy wadi]

direct comparison test was nice here

[tough vector]

because I couldnt think of one

to a p series?

[wispy wadi]

ln(n)/n^3 <= n/n^3 = 1/n² and the serie sum 1/n² converges

[tough vector]

thats actually smart thank you

.close

[obtuse pebble]

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[narrow flax]

Need help with Q17

[obtuse pebble]

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[tardy vigil]

I’m not even sure where to start. I’ve tried some things like using cofunction identities and double angle identities and cosine and sine identities but am just not getting anywhere

I’m guessing I need to draw right triangles and do something with ratios

Idk

[daring rock]

Just an idea

All four cosines can be related to each other by sum/double angle identities

Seems like you may be able to use those identities to create a system of equations maybe

Plus cos(60) is already known as 1/2

Haven't really thought it through very much though

[obtuse pebble]

@tardy vigil Has your question been resolved?

[tardy vigil]

Yeah

[tardy vigil]

<@&286206848099549185>

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[obtuse pebble]

@timid silo Has your question been resolved?

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[zinc lake]

2. According to the data released by the internal affairs ministry,
   Japan's population totaled 125,927,902 as of January 1, 2022, down
   726,342 or 0.57% from the previous year. Japan's population has
   been decreasing 0.14% since 2013.
   a) Write an exponential function that models the population P(n),
   of Japan after 2013.
   ➤ b) Assuming that Japan's depopulation continues, estimate
   Japan's population in 2027.

[misty jetty]

you need to find the initial value which is the population in 2013

[zinc lake]

and then what's next?

[misty jetty]

ab^t

you have b and t

u just need a

which is the inital value i was talking about

b in this case is the rate of decay

or 1-0.14

0.84

t is the years after 2013

which is the 'x variable'

[obtuse pebble]

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[warm lynx]

Somebody help?

nvm got it thanks

[obtuse pebble]

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[rough bloom]

there is a 93.75% chance of the value 45 being returned, and a 6.25% chance of a 1 being returned. How do I calculate the average value of the function

[rough bloom]

(and how do i extrapolate this to a set of values and their chances)

[devout prairie]

.9375*45

And .0625* 1

[rough bloom]

the sum of?

alrighty good enough :D

!close

[devout prairie]

Yeah

.close btw

[rough bloom]

.close

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

How would I attempt this

[timid silo]

Are all shaded regions = in area

[tardy epoch]

,rotate

[timid silo]

no way riemann is here to help me with riemann sums

i don't understand what is supposed to happen her

[daring rock]

I'd venture to say the word "estimate" in the prompt is giving you permission to assume that the three shaded regions all have the same area

[timid silo]

ok

[daring rock]

And, you're told what that area is

[timid silo]

right

so

are negative areas a thing

is the area from a to b -8?

[daring rock]

Yes*

usually it's called the "signed area" to indicate that it can be interpreted as negative

but yes you're right, the integral from a to b is -8

[timid silo]

ok

so

area from c to a is 0?

[daring rock]

yes

but

stop using the word area

The integral from a to c is 0

[timid silo]

ok

is that like a displacement idea

[daring rock]

basically. When they use the word "area" in the last part, they're literally talking about area

as in, all positive

[timid silo]

ohhhh

so 24

[daring rock]

yeah

[timid silo]

and then c to 0 is 8

[daring rock]

yes

[daring rock]

it's actually 0 to c, but yes

[timid silo]

so the answers are

0
8
24

[daring rock]

yeah

[timid silo]

that was ez enough

thanks

[daring rock]

sure 👍

[timid silo]

i love it when u actually get help on this server

some times it's a wasteland

[daring rock]

your profile says you've been here since yesterday lol

but yeah, sometimes it might take a while

[timid silo]

i leave and join whever i need help

[daring rock]

ah fair enough

[timid silo]

@daring rock hi sorry r u still there

[daring rock]

,rotate

[timid silo]

Did I do this right

,rotate

[timid silo]

That’s 4x^2 btw

Not ^-2

[tardy epoch]

can't read the limits

[timid silo]

What limits

[daring rock]

the bounds of the integral

I think it says -17 and 4x^2 ?

[timid silo]

Oh

Yeah

It does

[tardy epoch]

looks right

[timid silo]

Ok

Thnx

[obtuse pebble]

@shrewd raft Has your question been resolved?

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[high lily]

have you tried anything at all

[cloud pond]

yep

I differentiated the function

[high lily]

what else

[cloud pond]

so i got y' = 1/2(x-2) using the chain rule

[opal marlin]

Can you try finding the slope of Q?

[cloud pond]

and did tried 4 = 1/2(6-2)

[high lily]

y' is not the same as y

[cloud pond]

yeah

[opal marlin]

That is the value of Q at x = 6

Not the slope

[cloud pond]

y' should be gradient

[high lily]

plugging in x=6 into your derivative function, gives the slope/gradient at that location

[cloud pond]

oh ok right

[pseudo ledge]

This is an interesting question @cloud pond
Could you lmk if you manage to solve it?

[cloud pond]

yep

[high lily]

can you continue from there

[timid silo]

is the answer 1

x=1

?

[opal marlin]

Please don't spoonfeed answers

[cloud pond]

so 1/2 x-1

[pseudo ledge]

oh

[cloud pond]

1/2 6 -1 = 2

[timid silo]

lol i was just verifying my answer ...

[pseudo ledge]

Differentiate the curve
Sub in x val of Q to find Gradient of the line

[opal marlin]

how would he know the answer when hes trying to solve it

[pseudo ledge]

@cloud pond have you learnt how to differentiate/ what a gradient function is?

[cloud pond]

yes

[high lily]

crap notation but ok

[pseudo ledge]

Aight so
When you differentiate the curve and put in the x value of Q, you’d get the gradient of the Q line right?

[high lily]

you've plugged in x=6 into your derivative
and you now know the gradient at Q is 2

[cloud pond]

yes

[high lily]

try continuing

[cloud pond]

y=2x+b

[high lily]

unecessary

[pseudo ledge]

Can I offer a hint 😭

[cloud pond]

4=2(6) +b

so b = -8

y=2x-8

[high lily]

you are not required to find the equations of the tangent lines or their intercepts

[pseudo ledge]

||look at the relationship between the two lines||

[opal marlin]

Do you know what two perpendicular lines have in common?

[cloud pond]

the intersect point?

[pseudo ledge]

No there’s summin else

About gradients

[high lily]

it was a poor question

what should have been asked was
do you know the relation between slopes of perpendicular lines

[cloud pond]

oh yea

negative reciprocals of eachother

[pseudo ledge]

Yes ‼️‼️

[high lily]

hence what will be the slope of the line passing through P

[cloud pond]

oh so -1/2

and then -1/2 = 1/2x -1

[high lily]

yes

[cloud pond]

x=1

y=1/4x(1-2)^2

=1/4x1

=1/4

p (1, 1/4)

[high lily]

don't use x for multiplication
and they only really wanted the x coordinate

[cloud pond]

oh yea true

thanks for the help guys

🙂

[obtuse pebble]

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[brazen thunder]

this is a question from my linear algebra assignment

This is my solution there are a few more problems like this and I don’t have means of checking my soul

sol*\

so I was curious if someone could let me know if this is the right way to go about it

ignore part c

should be no

<@&286206848099549185>

[obtuse pebble]

@brazen thunder Has your question been resolved?

[obtuse pebble]

@brazen thunder Has your question been resolved?

[obtuse pebble]

@brazen thunder Has your question been resolved?

[obtuse pebble]

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[sage dagger]

how would i do this quadratic equation

[sage dagger]

I need to find the range of values for k

x^2 + (3k - 4)x + 2(k + 4) = 0

it has real roots

do i expand brackets or no?

[quasi current]

use the discriminant rules

D = b^2 - 4ac

it will have real roots when D > 0

a = 1, b = 3k-4 and c = 2k + 8

[sage dagger]

yeah ik im supposed to use the discrininant thing but do i expand brackets

[quasi current]

(3k-4)^2 - 4(1 * 2k+8) > 0

that should be your equation

[sage dagger]

what about the x term outside the brackets

[quasi current]

thats disregarded when doing the discriminant

[sage dagger]

ah okay

[quasi current]

but expand and solve it as a quadratic inequality

you should get two sets of solutions for the real roots

[sage dagger]

9k^2 - 32k - 16

[quasi current]

complete the square or use the quadratic formula

[sage dagger]

alright

so -4

and

4/9

so

[sage dagger]

x > -4? and x > 4/9?

[quasi current]

The parabola will have real roots when k < -4/9, k > 4

[quasi current]

yep

[sage dagger]

but im wrong then

i did the signs wrong

[quasi current]

I believe so yeah

give me a sec

[sage dagger]

why is it -4/9

[quasi current]

ill recheck what i did

[sage dagger]

mk

[quasi current]

if you factorised

you should get (9x+4)(x-4)

thus x = -4/9 and x = 4

[sage dagger]

ohhh

[quasi current]

=> k > 4 and -4/9 > k

[sage dagger]

so real roots is

<0

or

0>

[quasi current]

its both value

look on desmos

[sage dagger]

is discriminant for real roots greater than 0

[quasi current]

yes

[sage dagger]

or equal to

but it doesnt say if roots are unequal

just says real

so i think it means real and equal

[quasi current]

D > 0 2 real roots
D = 0 2 repeated roots (one roots)
D < 0 2 complex roots (no roots)

[quasi current]

it means two real roots

thus when D > 0

[sage dagger]

oh alright

and i got one more question if thats okay

if something like

[quasi current]

thats alright

[sage dagger]

(x^2 + 3x + 18) + k(x + 2) = 0

same thing but with no real roots

similar process?

[quasi current]

expand the kbracket and simplify to the form

ax^2 + bx + c

[sage dagger]

alright

x^2 + 3x + kx + 2k + 18

[quasi current]

okay list out your a, b and c terms

[sage dagger]

so

a = x^2

[quasi current]

and do the same thing but this time its D < 0

[quasi current]

no a = 1

[sage dagger]

oh ye

mb

[quasi current]

we just want the coeficcient

[sage dagger]

wait is b 3x + kx?

is that one term

im confused

[quasi current]

B is 3 + k

[sage dagger]

oh

and then c is 2 + 18?

[quasi current]

2k + 18

we ignore the x's

[sage dagger]

right

and then we use b^2 - 4ac again?

[quasi current]

yup but this time

we're doing b^2 - 4ac < 0

[sage dagger]

okay

k^2 - 2k - 63

or do we add the < 0 at the end

so i got

2k + sqrt(4k^2 - 252k) / 2k @quasi current

[quasi current]

no thats seems incorrect

give me a sec

[sage dagger]

i thought so as well

[quasi current]

(3+k)^2 - 4(1*2k+18)

k^2 + 6k + 9 - 8k - 72

simplify that and that should help

[sage dagger]

i got

k^2 - 2k - 63

[quasi current]

k^2 - 2k - 63

now solve the inequality

k^2-2k-63< 0

[sage dagger]

alright

[quasi current]

which should give you the range betweem -7<k<9

[sage dagger]

ohhh

right

so when do we need to use the quadratic formula?

tysm for the help

btw

i appreciate it

[obtuse pebble]

@sage dagger Has your question been resolved?

[obtuse pebble]

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[onyx gyro]

I'm struggling to wrap my mind around how to prove double inequalities like this

[onyx gyro]

without just inputting a bunch of random numbers, that is

[spice citrus]

you could multiply the entire equation by x

to get a quadratic

you're allowed to do this because x is positive so it won't affect the inequality sign

[onyx gyro]

ok...

aaaaah

k i'll try that, thanks!

[obtuse pebble]

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[obtuse pebble]

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[humble wind]

need help with b

[obtuse pebble]

@humble wind Has your question been resolved?

[obtuse pebble]

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[unreal nexus]

Hey guys

[unreal nexus]

Help me pls with test

one minute

I'll send

[humble wind]

no

[earnest elk]

No we can’t help you cheat on tests

[humble wind]

you'll be banned for that

[unreal nexus]

Guys, test from my tutor

I dont get how to solve

[humble wind]

nope

[unreal nexus]

Um

[humble wind]

skip for now then

come back to it later

[unreal nexus]

It was helpfully

[unreal nexus]

Yeah, I need to solve it today

[humble wind]

not helping you cheat bro good luck

[unreal nexus]

Thank you, very helpfully server

[unreal nexus]

You too

[humble wind]

wc

do .close

[earnest elk]

The pleasure is all ours

[humble wind]

can anyone help with b please

not a test

can send you proof

[unreal nexus]

do .close

How to close it?

[obtuse pebble]

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[uneven night]

What should I do her someone please help me

[fierce lagoon]

Factor out 1/4 and then use u-sub

[uneven night]

Do u mean like that

[fierce lagoon]

Well yeah that'll work

Let $u = \frac{x+2}{2}$

[warm shale]

Umbraleviathan

[uneven night]

So 2du=dx?

[fierce lagoon]

Yeah

And then x = 2u - 2

So it becomes $2\int \frac{2u - 3}{u^2 + 1}du$. Split the fraction

And then you're gonna have to do another sub (w-sub)

[drifting badger]

Lmao, sorry, I thought your 2's were 3's

[uneven night]

I get this

[drifting badger]

Alphabet sub

[uneven night]

-2-1 is equal to -3 right?

[fierce lagoon]

Yeah I made a typo

[warm shale]

Umbraleviathan

[uneven night]

And now?

Another sub?

[fierce lagoon]

Did you split the fraction

[uneven night]

Now I will do another sub on the left

Right?

[fierce lagoon]

Yes

Perfect

[uneven night]

Finally

So

Is that the same answer?

[fierce lagoon]

Similar

Yeah

[uneven night]

No it’s not

Something is wrong

[fierce lagoon]

What was your sub

My brain is trying to think

Your sub should've been w = u^2 + 1

[uneven night]

Yea I saw that

[fierce lagoon]

And then ln|w| = ln|denominator|

[uneven night]

And I u should factor out the 2

From the beginning

I should do it like that

Right?

Then I should use u sub

[fierce lagoon]

Well no

[uneven night]

Isn’t that the same thing?

[fierce lagoon]

The w-sub should always result as ln|w| (after integrating)

[fierce lagoon]

What you have up here is correct

[uneven night]

What should I alter to get right answer then

[fierce lagoon]

Well, for the left integral, set w = u^2 + 1 so that dw = 2u du

[uneven night]

I did that

[fierce lagoon]

Just let dt = 2udu

Then it becomes $2\int \frac{1}{t}dt$

[warm shale]

Umbraleviathan

[uneven night]

Didn’t we do that?

[fierce lagoon]

Hm

[uneven night]

I think from the beginning it was wrong

I should write it like that instead

[fierce lagoon]

ln|t| = ln|u^2 + 1|

Hm

[uneven night]

Then after i would use u sub

[fierce lagoon]

Yeah and then it should work, idk why it doesn't

Probably because we had factored out 4/4 but that shouldn't matter

[uneven night]

But can I set that equal to each other

I mean is that true

[fierce lagoon]

Hm I guess you have to multiply the inside by 4 again

To make up for that original factoring

[uneven night]

But it should work when we did it first time

So this is right

This first thing is right well?

[fierce lagoon]

Let me check

Because they only differ by a constant??

[uneven night]

Yes

[fierce lagoon]

You're fine

Yeah they differ by a constant

[uneven night]

So after we get this

[fierce lagoon]

You're fine

You just differ by a constant of 2ln(4)

[uneven night]

Whattt

So I did right???

[fierce lagoon]

So what you have is correct

Yes

[uneven night]

How?

I mean they say the answer should be ln x^2+4x+8

[fierce lagoon]

Up here

The graph of f'(x) and the integrand are the same

You see what they have is your answer but the argument is multiplied by 4

But that means that, by log rules, can be separated as + 2ln(4)

[uneven night]

Ohhh

Yeaaag

[fierce lagoon]

So your answer differs by a constant

[uneven night]

So my answer is correct

Thank u so muchhhh

Sorry for taking ur time

[fierce lagoon]

Lol no

Problem*

No problem

[obtuse pebble]

@uneven night Has your question been resolved?

[obtuse pebble]

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[restive fractal]

a soccer team has 3 possible outcomes when playing a game, win, loss or draw, chances of winning is 80%, a loss 15% and a draw 5% calculate the chance of the team winning the 2nd match

[obtuse pebble]

@restive fractal Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[jagged gull]

how does line hitting some axis called?

[devout prairie]

Intercept

[jagged gull]

.close

[obtuse pebble]

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[jagged gull]

if I am solving a problem at precision at 0.000

can I say it is at thousandth precision

or one in thousandth

[obtuse pebble]

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[dreamy steeple]

m not able to prove induction

[spice bay]

What did you get for a_n+1 ?

[dreamy steeple]

i found general formula for a_n

then trying to find induction as in k

[spice bay]

Do you remember what a_n+1 is?

[dreamy steeple]

yes its given

[spice bay]

And to calculate a_n+1 you need a_n, which is your hypothesis

[dreamy steeple]

yes

should i try to replace a_k + 1 as the answer of a_n+1

[spice bay]

Not sure what you are trying to do

You have the proposition that a_k = x/(kx+1)

By induction you need to prove that a_k+1 = x/((k+1)x+1)

[dreamy steeple]

yes

[spice bay]

But a_k+1 is given as you said

so just plug the values

[dreamy steeple]

ok i got it

thanks

[spice bay]

perfect!

[obtuse pebble]

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[uneven night]

Hey can I get help with this integral

I don’t know what to do next

[obtuse pebble]

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[uneven night]

Hey can I get help with this integral

[uneven night]

I don’t know how to move forward

.helpers

/help

/helpers

[teal turret]

!15m

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[uneven night]

Sorry

[teal turret]

Hold on a sec, Lemme try it out

[timid silo]

yeah i kind of solved it

there are some ways to do it

[uneven night]

Hey lexqa u are a such good teacher!

I want the easiest way please

[timid silo]

i think trig sub is an idea, partial fractions maybe with some substitution

the way i did it was straight up double u subbing

[timid silo]

💀 debatable

[uneven night]

Okay but can I ask u a quick question before this

[timid silo]

sure

[uneven night]

I did this integral

But the answer says I should get this

[timid silo]

oh gosh thats a lot of things

yeah what about ti

[uneven night]

As u see my ln is not the same

So where did I do wrong

[timid silo]

okay i wont lie @uneven night me reading through that and figuring out what u did wrong is going to take longer than me just writing out how to do it and you finding out from there

because the library closes soon 💀

so i will do that rq

[uneven night]

Okay I’m so sorry

I don’t wanna take ur time

I will try on the first problem instead. Thank u anyways.

But if u have time later, I would appreciate if u just can let me know why I didn’t get the same as the real solution😊

[timid silo]

@uneven night

,rotate

I don't even know if I got it right first try let me see

[timid silo]

Oh seems like I did, nice

[timid silo]

Sorry this is a bit messy, because had to rush it a bit

But ask if u don't get something

[obtuse pebble]

@uneven night Has your question been resolved?

[uneven night]

Omg u are so kind u literally did it, I just don’t understand the fourth line

I understand the first three lines

Where did the 2 come from?

[timid silo]

yeah i knew u would get confused over that

so basically

thats a thing you usually do with

linear-to-quadratic integrals

as in, integrals with a linear numerator and a quadratic denominator

the idea is that, in the form it is in, it is hard to integrate it

so we split it into two fractions we can deal with easier

to "split it" you need to make it so that you have one integral with a 1 numerator, and the remaining integral something you can handle better

[uneven night]

So did U add 2 or multiply?

[timid silo]

no actually, i just said x - 1 = 1/2(2x+4) -3

@distant moth this is your time to shine!

https://www.youtube.com/watch?v=VUaujlcAosQ

@uneven night

this is a good video to understand the meaning of what i said

[uneven night]

I’m so confused so u got in the third line integral of x-1/x^2+4x+8 how did u move forward to 2( x+4) / 2 (x^2+4x+8 ) and then -3

[timid silo]

yeah haha it is veryy weird for someone who has no knowledge of it

[uneven night]

Im so sorry but I don’t get it

[timid silo]

also it is x+2 if u r factoring out that 2

[uneven night]

I will watch the video also

[timid silo]

try and watch this video

and then come back to me

[uneven night]

Then my solution

Why didn’t it worked

I didn’t do exactly like u did but I used u sub

[timid silo]

uhh let me try and see

[uneven night]

I just used factored out the 4 and used u sub two times

[timid silo]

the part where u factored out the 4 from the fraction

seems

sketchy

also do u not get the same result if u expanded the brackets in ur ln?

[uneven night]

I get x^2+4x+8/4

If I expand it

[obtuse pebble]

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[timid silo]

Oh lmao

Reopen

[uneven night]

So u didn’t see anything wrong right?

Someone actually told me I did right

[obtuse pebble]

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[uneven night]

@timid silo I don’t what happend I just got thrown out hahaha

[uneven night]

So u said I can’t cancel the fourths like that?

@timid silo can u please explain then where did the 3 come from? I feel so stupid

[timid silo]

sorry a bit busy right now, but it is just another representation of what x-1 is

[uneven night]

Not a problem take ur time

So I watched the video and he took basically 2ax+b and multiplied with (px+q) but I don’t understand why set p/2a

Sry @timid silo I can’t agree with ur methoood I don’t understand it. I have looked into it 10 times I don’t get it. I watched also the video It doesn’t feel like my brain is understanding it. Anyway thank u so much.

[timid silo]

Well, it isn't anything crazy at all

It's like what you would do for partial fractions, Substitution, etc.

It takes a bit to rationalise it

If you want to there are always ways to do them other than that

You just have to look harder for it

[uneven night]

Then why didn’t my method work

It’s so beautiful solution but I don’t get why I get wrong in Ln

These 2 where did they vanish

So I don’t really understand how u got from red to green to blue

[timid silo]

I factorised the numerator?

It eliminated both 2's?

2(x+2)?

[uneven night]

Wait u don’t have a 2 in x

How did u factor out 2 when u only have x+4

In numerator

[timid silo]

From red I replaced x-1 with 1/2(2x+4) -3 like I have already said thrice. To blue is just factorisation

[timid silo]

I have 2x+4...

[uneven night]

Oh okay sry now I understand the blue and the red but theee greeen is so hard

Soorry it’s nothing against u it’s just me

I’m so new to this

[timid silo]

It's fine, it is just that if you expand 1/2(2x+4) -3 you get x-1

So you didn't change anything

[uneven night]

Yea I can see that

[timid silo]

It is a common method to solve linear to quadratic integrals like I told u

[uneven night]

So u took 1 over that because it’s x-1

I mean if it was x-2 would u take 2/2(2x+4)

[timid silo]

Wdym take "1 over"

[uneven night]

Divid*

So we had x-1 in the beginning right in the numerator

So u wrote that like 1/4x+8

[timid silo]

Jo because in that case

U sub is

Enough

For the denominator

Oh wait I just realised what u mesnt

Ah yeah we take 1/2(2x+4) -4

The whole point of this is because 2x+4 can be gotten rid of with the derivative of the denominator

The reason why I did all of that is just to do that when u set u=x^2+4x+8

[uneven night]

Yea

Though is there any rule here to be applied

How do u see that is equal to each other

[timid silo]

I mean

Just common linear transformations ig?

U can represent x-1 with anything

[uneven night]

U know what I will leave that problem I will stay with my solution even if I get little wrong

Thank u anyway

I wish I saw the problem like u do

Are u a professor or something ?

U are so good at math

I still can’t see anything wrong with my solution

U told me also it looked good so I will go on with this

What do u think?

Do u understand my solution?

Or do u see anything that seems unusual?

[timid silo]

No lmao

[timid silo]

Heavily debatable

💀

[uneven night]

https://youtu.be/ngnH0ZSyPfk

This problem is maybe similar

Or Am I wrong

Maybe I can see what he is doing and try do the same

[obtuse pebble]

@uneven night Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[keen tiger]

this is the mentioned example so P(met 2) = 1/2 x 1/2 = 1/4 and P(two came) = 2/3 x 2/3 = 4/9 But I can't figure out how to get the intersection of these 2 events (needed for the conditional probability in the question)

[obtuse pebble]

@keen tiger Has your question been resolved?

[keen tiger]

<@&286206848099549185>

[obtuse pebble]

@keen tiger Has your question been resolved?

[obtuse pebble]

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[strong plaza]

what is -3-0+-10

[rigid fog]

So what’s -3-0

[obtuse pebble]

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[lofty thunder]

Hi, I need help understanding the steps to solve this radical equation

[lofty thunder]

Do I combine liked terms? Add the 21 to 4?

[daring rock]

Are they both under the radical sign? Or is the +4 on the outside?

[lofty thunder]

both under it

[daring rock]

Then sure, you can combine them

[lofty thunder]

this kinda looks like factoring?

leading co eff , but there's a 2

Am I allowed to do that under a radical?

[sudden wadi]

this may be dumb but cant you sqaure both sides to solve for p

[lofty thunder]

I don't have a clue

[sudden wadi]

cause then youd have the left and side without the sqaure root and p2 on the other side

move that p^2 to the other side and then you have a quadratic that = 0

[lofty thunder]

uhh

does the factoring thing I mentioned works too?

[sudden wadi]

hold on ill draw it out

you can porbably do that but idk where it would get you

[lofty thunder]

what happens to that 2p?

[sudden wadi]

you subtract p^2 from both sides

[lofty thunder]

ah

and I can't subtract by 25 because it's not a liked term

[sudden wadi]

its just to create a quadratic cause then you can use quadratic formula or factorise to solve for p

[lofty thunder]

hmm

then I do the leading co eff factoring

[sudden wadi]

whats that?

[lofty thunder]

find 2 numbers that multiplies to 25 but sums into -14?

[sudden wadi]

yeah you can do it that way or just use quadratic formula

[lofty thunder]

though it looks like it won't work

guess I have to use the quadratic formula

[sudden wadi]

yeah i think you have to use quadratic formula

[lofty thunder]

good so far? lol

not sure at this point

or I think I can simplify by 2?

[sudden wadi]

you can turn -(-14) to just 14

and yes you can divide it all by 2

so youd have 7 +- 2√ 6

[lofty thunder]

ah so the bottom becomes 1 removing the fraction

there we go'

Thanks

.close

[obtuse pebble]

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[misty valley]

Anyone familiar with Hypothesis Testing

[misty valley]

College level stats?

[obtuse pebble]

@misty valley Has your question been resolved?

[misty valley]

<@&286206848099549185>

[nimble blaze]

According to 7, the significance level is 0.10, so we have a confidence interval of 90%. This means we are confident for numbers between -1.64 to 1.64. Imagine our significance level was 0.05: that means our confidence interval would be from some number less than 1.64 to some number greater than 1.64, thus being outside of the confidence interval for the significance level for 0.10. Similarly, if our significance level was 0.15 (greater than 0.10), then there would be a confidence interval inside the confidence interval for the 0.10 significance level. This means that a p-value of anything less than our significance level will lead to values outside our confidence interval, and a p-value greater than our significance level will lead to values inside our confidence interval. We call this our "confidence interval" because it is between the bounds of our confidence interval that we are confident we don't have to reject the null hypothesis. Thus, if a something is likely to fall outside our confidence interval, then our null hypothesis is likely incorrect and we reject the null. A p-value that is less than the significance value would mean that values are likely to fall outside the confidence interval. If this is the case, we reject the null hypothesis. **Likewise, a p-value greater than the significance level means the values are likely to be within the confidence interval, which would mean we don't have to reject the null (in formal terms, we "fail to reject the null" ** -- this doesn't mean we accept the null, it only means we can't reject it; there's a slight difference )

[obtuse pebble]

@misty valley Has your question been resolved?

[obtuse pebble]

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[hollow kestrel]

I tried using the distance formula, but having an exponential makes it difficult

[crystal solstice]

This isn't a distance formula question

[crystal solstice]

Call the point on the function $(a,e^{10a})$ and write the area of the rectangle as a function of a. Then maximize that function

[warm shale]

FLORIDA MAN

[hollow kestrel]

ok how would I do that? Like a limit as x approaches a?

[crystal solstice]

No, a is a constamt

Can you write the area in terms of a?

What's the length in terms of a? The height?

[hollow kestrel]

yea they dont give me anything except the corner is (9,0)

[crystal solstice]

We're calling the point where the rectangle touches the function (a,e^10a)

The y value of that point is the height of the rectangle

[hollow kestrel]

oh ok

then the length would be 9-e^10a right?

[crystal solstice]

Not quite

[hollow kestrel]

woops 9-a

[crystal solstice]

Yeah

So the area is (9-a)e^(10a) where 0<a<9

Now maximize that function

[obtuse pebble]

@hollow kestrel Has your question been resolved?

[hollow kestrel]

Thanks for the help @crystal solstice appreciate it.

.close

[obtuse pebble]

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[broken jay]

could someone explain how i would do this?

[timber fox]

any multiple of 12 would just be the same time wouldnt it

my brain is trying to process

[broken jay]

it has something to do with mod and stuff i think

1 would be 5:00 still no?

[timber fox]

thats what amm sayinnn

cus its a multiple of 12

2 would be

5 + (17-12)

3 would be

also remember our number system stops at 12 and starts at 1

[broken jay]

wait why 5+?

[timber fox]

because your passing 17 hours

and if the clock runs from 1 - > 12 then wouldnt it go 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10

[broken jay]

ohh

[timber fox]

so a multiple of 12 makes it remain the in same spot

using that we can get 103 = 8*12+7

so it would be 4 + 7

or 11 :)))

I think XD

[broken jay]

it kinda makes sense

[timber fox]

yeah

[broken jay]

still a little confused :')

[timber fox]

im just thinking if its 12 or 24...

cus its saying hours and the clock runs from 1 -> 12

idk

lots of brain stuff

[broken jay]

no clue

[nocturne minnow]

It's an analogy clock, only 12 hours on that

[timber fox]

is the method correct then?

[broken jay]

:-:

[timber fox]

not too bad

get your trees going

ez clap

[broken jay]

trees

[timber fox]

was the previous one correct?

[broken jay]

im not sure

gotta submit it all

:")

[timber fox]

ohh I see

well have fun with the trees

[broken jay]

ill try thank you

.close

[obtuse pebble]

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[wise kettle]

How do I show things that seem obvious like this?

like its clear that 2pim is just more rotations, and k=l mod 2 means same reflection, but idk how to put it as a proof

[random depot]

ok hol on

[obtuse pebble]

@wise kettle Has your question been resolved?

[wispy wadi]

use inverses
p_(-omega+theta) = r^(l-k)
so now you get an equality between a rotation and a reflection

this leaves you with very few possibilites

[obtuse pebble]

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[whole dock]

How do i solve this

[wispy wadi]

wdym by solve, there is no unknown ?

[whole dock]

I have to find a vector perpendicular to a and b vector say d vector and that d vector has to have a dot product of 15 with c vector

[wispy wadi]

oh you want to find that vector

well, it's just a x b here

calc a x b

[whole dock]

But the dot product wont give 15 with c vector

[wispy wadi]

then why did you write (a x b).c = 15 ?

[whole dock]

12th

Question

[wispy wadi]

ok, so it's lambda * (a x b) for some scalar lambda

you need to determine it using that c.d = 15

[whole dock]

I dont understand

[wispy wadi]

calculating a x b is a pain but I guess I'll do it

[whole dock]

Actually

Have calculated it

Its 32î - j + 4k

[wispy wadi]

and (a x b).c, do you have it ?

[whole dock]

Thats what idk how to solve

I got 7 solving it but i need the answer to 15

I guess i somehow have to relate the cross product and c vector

[wispy wadi]

take d = 15/7 (a x b)

[whole dock]

Why is that necessary

And where did that come from

Im confused sorry

[wispy wadi]

idk if it's necessary, but if you did the calc right, it's a vector that works
d = 15/7 (a x b)
so d.c = 15/7 (a x b).c = 15/7 * 7 = 15
and d is perpendicular to both a and b because, well, it's colinear to a x b so by def

[whole dock]

Ohhh

So it doesnt matter if i

Multiply a scalar to a vector perpendicular

It will still be perpendicular

[wispy wadi]

? what's multiply (a x b) is 15/7 here

there's no i ?

oh you mean I

[whole dock]

Yes I

As in me

[wispy wadi]

and so yeah, your geometry properties hold
if two things are colinear then they are orthogonal to the same things

and a vector d is naturally colinear to x*d for whatever scalar x

that's the def of colinear

[whole dock]

How can two things be colinear and orhogonal to the same thing

[wispy wadi]

no, colinear between themselves => orthogonal with the same set of things they're orthogonal with

like if you take two parallel line, a third one perpendicular to the first is also perpendicular to the second

[whole dock]

Alr so tell me this

If i have a vector and b vector

And the cross product is d vector

And if i multiply scalar lamda to d vector it will still be perpendicular to a and b?

[wispy wadi]

yeah

and that's natural you need to imagine the scene in your head

d and x*d for whatever scalar have the same direction

so if d is perpendicular to both a and b, then so is x*d

[whole dock]

I see

[whole dock]

So the both lines must be parallel

In order for them to be perpendicular to the same line

[wispy wadi]

yeah that's an equivalence

[whole dock]

But in this case the lines are not parallel yet they are perpendicular to the same line

[wispy wadi]

because the three are not in the same plane

they span the 3d space

but x*d is colinear to d

so x*d, d, and whatever vector a are necessarily in a same plane

and your usual 2d geometry holds

[whole dock]

Do you mean dot product when you say x*d

[wispy wadi]

no, x is a scalar

like 2, 2d

[whole dock]

Ohhh

I understand

So the multiplication with a scalar

Just elongates the line without changing the plane

[wispy wadi]

yeah

[whole dock]

Thanks i understand now

.close

[obtuse pebble]

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[distant hinge]

ABC is a triangle inscribed within a circle with center O, CM⊥AB M∈AB, BN⊥AC N∈AC. Prove AO⊥MN.

[obtuse pebble]

@distant hinge Has your question been resolved?

[obtuse pebble]

@distant hinge Has your question been resolved?

[obtuse pebble]

@distant hinge Has your question been resolved?

[obtuse pebble]

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[ebon prairie]

Is it normal for my unknown variables to not equal my steady state values?

[ebon prairie]

For example, according to the EXCEL calculations of my steady state: my values for a = 0.074, b = 0.063 and c = 0.091. But after finding my equations and doing it through that, my values show to be a = 0.1047381546, b = 0.089775611 and c = 0.0897755611

[tardy epoch]

Very confusing what you're asking

Can you show the actual problem

[ebon prairie]

Ah- this is the Markov Chain Model calculation stuff.

It's an assignment, so idk how to explain it all together-

This is my Transition Matrix.

This is my initial matrix.

And this is my Steady matrix.

The steady matrix was calculated through excel, which generated my probability values for me.

But, to verify this, I have to solve for all the values algebraically.

To do that, I found my equation that equals to 1:

And these are my other 3 equations.

[obtuse pebble]

@ebon prairie Has your question been resolved?

[obtuse pebble]

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[timid silo]

POQ and RT are two parallel straight line.

Find the equation of straight line RT,

Find the x-intercept of straight line RT,

[timid silo]

I dont have any idea how to solve this...

[royal basin]

parallel lines have the same slope

[timid silo]

Idk how to find the slope as well since I dont know the coordinate of PQ or RT

[royal basin]

you do not need to know the coordinates of P, Q, R or T

you know two points on line PQ

(0,0) and (2,6)

[timid silo]

Ah I see

Now I know where my problem is

Tqvm!

.close

[obtuse pebble]

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[north cove]

Can someone help me get started with this problem?

[north cove]

I can't just brute force my way through it

[zenith raft]

i think you can look at quadratic residues mod 4

and find when n > 1 that won't be a perfect square

because it will be 2 mod 4

[north cove]

Right

[zenith raft]

does that mean you got it or do you want more detail? lol

[obtuse pebble]

@north cove Has your question been resolved?

[obtuse pebble]

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[weary canopy]

Hello. I need some help with the following problem:

[weary canopy]

Have I done something wrong? If not, what is the answer to the last equation?