help-33

[real shore]

so this?

srry im a bit slow

english isnt my first language

but i know math

[vast jolt]

it's ok, maths is not mine

[real shore]

alr

basically

in this situation

the symbols do not change

because they cross-multiply

when you have a fraction that is equal to another fraction you cross multiply

it stated above that you can write R as R/1

so that you'll have 2 fractions

so R = ρ1/A --> R/1 = ρ1/Α --> (R x A) / 1 = ρ1/1 (cross mutliplying here) --> A/1 = ρ1/R (dividing with the assembler to get rid of it, in our case it's R because we want to keep A in the left side as we're making a formula that solves for A)

i hope you understand

lmk if you dont

[vast jolt]

i'm have to think about it and hope it sinks in at some point. thank for the help

[real shore]

Alr, basically

one last tip

when you use the triangle

is to hide the variable you are solving for

so if you are solving for c

hide c with your finger

and what's left in the triangle is what you should do

[vast jolt]

this part i understand.

[real shore]

and what don't you understand

or causes confusion

[vast jolt]

it's ok, i need to go now but thanks. i'll read it again tomorrow

[real shore]

👍

[marsh citrus]

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[marsh citrus]

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[silent marten]

Hiii guys

[silent marten]

can someone check my awnsers please

If E is a complex vector space, we denote by dimR E its dimension as a real vector space, obtained by restricting scalars, and by dim E its dimension with the initial complex structure.

Remember that Cn[x] denotes the space of polynomials with complex coefficients and degree less than or equal to n, and M(C, m×n) denotes the space of matrices with complex entries of size m×n

[tight furnace]

I think your answers for the polynomial ones are wrong

[plucky mantle]

1. ( \text{Dim}_\mathbb{R} \mathbb{C} = 2 )
   • Correct. As a real vector space, (\mathbb{C}) has a dimension of 2.
2. ( \text{Dim} \mathbb{C} = 1 )
   • Incorrect. The complex dimension of (\mathbb{C}) is 1, but its real dimension is 2.
3. ( \text{Dim}_\mathbb{R} \mathbb{C}^2 = 4 )
   • Incorrect. The real dimension of (\mathbb{C}^2) is 4, but the complex dimension is 2.
4. ( \text{Dim} \mathbb{C}^2 = 2 )
   • Correct. The complex dimension of (\mathbb{C}^2) is 2.
5. ( \text{Dim}_\mathbb{R} \mathbb{C}_2[x] = 4 )
   • Incorrect. The real dimension of the space of polynomials with complex coefficients and degree less than or equal to 2 ((\mathbb{C}_2[x])) is 6, not 4.
6. ( \text{Dim} \mathbb{C}_2[x] = 2 )
   • Correct. The complex dimension of (\mathbb{C}_2[x]) is 2.

[elfin berry]

☆⃞゙۪ ᪤hiִgh࣪nᧉ͟s̲s: 🫶🏻 𝖨𝘇𝗓𝕪 ܲ 𖧪

[plucky mantle]

7. ( \text{Dim}_\mathbb{R} M(\mathbb{C}, 4 \times 2) = 16 )
   • Incorrect. The real dimension of the space of (4 \times 2) matrices with complex entries ((M(\mathbb{C}, 4 \times 2))) is 8.
8. ( \text{Dim} M(\mathbb{C}, 4 \times 2) = 8 )
   • Correct. The complex dimension of (M(\mathbb{C}, 4 \times 2)) is 8.

[elfin berry]

☆⃞゙۪ ᪤hiִgh࣪nᧉ͟s̲s: 🫶🏻 𝖨𝘇𝗓𝕪 ܲ 𖧪

[copper raven]

did you put that in chatgpt ? @plucky mantle

there's easy mistakes on your thing

[plucky mantle]

No (?)

[plucky mantle]

Let me see

[copper raven]

with no. 2 at least

and no. 3

and no. 6

[plucky mantle]

one sec

Oh yeah

Let me rewrite on a paper

[copper raven]

and 7 also

[copper raven]

ok ok, we're just very wary of huge walls of text with trivial mistakes around here

that's the style of the gpt posting some ppl do

[marsh citrus]

@silent marten Has your question been resolved?

[marsh citrus]

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[fossil lotus]

So I got this differential equation $\frac{d^2}{dt^2}x(t)-x(t)=0$ and I've managed to get the general solution $x(t)=Ae^-^t + Be^t$, I've been given the boundary conditions of $x(0)=A$ and $\frac{dx}{dt}(0)=0$. From the first I get that $B=0$ but the second one implies that A is also 0 which makes no sense. What is going on?

[elfin berry]

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

[fossil lotus]

For the first condition it was $x(0)=A+B$ (since $e^-^0$ is 1), so B has to be 0

[elfin berry]

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

[fossil lotus]

The second condition just doesn't make anysense since for $\frac{dx}{dt}(0)=0$, $-Ae^-^t$ has to be 0 so A has to be 0

[elfin berry]

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

[marsh citrus]

@fossil lotus Has your question been resolved?

[marsh citrus]

@fossil lotus Has your question been resolved?

[still temple]

I think the general solution is actually a complex function, innit?

I mean, $C^{-ix}$ is also a solution, so...

[elfin berry]

Palahoo

[marsh citrus]

@fossil lotus Has your question been resolved?

[still temple]

Did the problem give the general solution ?

Or you found it?

Could you paste the original problem, please?

@fossil lotus

[fossil lotus]

that was the problem

just solve the second order differential equation i gave with the boundary conditions

[fossil lotus]

yeah, but i made it into the real form which is what we were told to do

go from complex to real

[still temple]

Ok

But the A is the same A of your general solution?

Because, as I undestood it, you decided to call these constants of A and B

We could also write $x(t)=Se^{-t}+De^t$, for example

[elfin berry]

Palahoo

[still temple]

S and D are just constants

So it would be actually $A=S+D$ and $-S+D=0$. Therefore, $S=D$ and so $2S =A$, and we conclude that $S=D=\frac{A}{2}$

[elfin berry]

Palahoo

[still temple]

@fossil lotus

Sorry for my wrongly english

[fossil lotus]

yeah, it works like that

the complex solution can be transformed into the general solution where the constant are the same

[still temple]

Isn't that?

[still temple]

@fossil lotus

[marsh citrus]

@fossil lotus Has your question been resolved?

[fossil lotus]

Yeah, I see, thanks

[marsh citrus]

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[sturdy lagoon]

problem got mne stuck

[desert dirge]

where are you stuck

[sturdy lagoon]

since I was doing 5^2

or i mean

3^2 getting like 9 and then 9-3 = 6 but does weren't working

[desert dirge]

why were you doing 3^2?

[sturdy lagoon]

i mean 3^5

[desert dirge]

im not sure where that came from either

[sturdy lagoon]

oh

[desert dirge]

lets pull it back

[desert dirge]

whats the first step you think you should do, very first thing

[sturdy lagoon]

put 5 in the x spot

[desert dirge]

why?

[sturdy lagoon]

not sure I thought it was x

[desert dirge]

nono, youre solving for x

[sturdy lagoon]

i suppose if not

ah ok

-3 to both sides

[desert dirge]

good

then what

[sturdy lagoon]

mmm

not sure

[desert dirge]

$\sqrt[4]{x+3}=2$

[elfin berry]

AℤØ

[desert dirge]

throw an idea at me

[sturdy lagoon]

-3 both sides

3^4?

i know they get rid of the square root

[desert dirge]

cant -3 yet

you can get rid of the fourth root by raising both sides to the fourth power

[sturdy lagoon]

k so x+3=2

[desert dirge]

nope

you didnt do anything to the 2

[sturdy lagoon]

oh

x+3=16

then -3 both sides

[desert dirge]

good

[sturdy lagoon]

I see, thank you

.close

[marsh citrus]

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[marsh citrus]

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[tacit vale]

help

[tacit vale]

this is what i have so far idk if its correct

<@&286206848099549185> pls this is due soon 🙏

i will reward you with goodnight kisses

[late hare]

daymn this problem seems cool

@tacit vale

what's your reasoning so far

and you might consider writing big numbers on their simplified form

[tacit vale]

my bad g

[tacit vale]

so i need dc/dt
i have c in terms of a and b(constants) and theta (variable)
i derive my c eq and get dc/dt in terms of a,b,c (constants) and d0/dt (0 = theta)
i find d0/dt by finding how much the angle between earth and venus changes per day by finding ther angular velocity

so their periods are 365 and 255 respectively, so their angle changes by 360/365 and 360/255 per day, so the change in theta is
(360/365 - 360/255) deg/day = d0/dt

then just subtitute everything

giving me this

wait

mistake

[tacit vale]

this is

-360/365 + 360/255

negative

[late hare]

you can get c from a and b

[tacit vale]

and theta, yes, i did that. c is in the denominator

OH

[late hare]

what's missing now

[tacit vale]

nevermind

i need to square root the denominator

[tacit vale]

whats missing is idk if i did this right

can someone validate my answer

this is graded and w a group and theyre all lost

so i dont wanna be the reason they fail yk

[late hare]

ah well i won't be doing the calculations to that 😂

[tacit vale]

fair enough

[tacit vale]

issue was the value was too smal

w he sqrt it looks alr

[tacit vale]

😭

i need to know if its right

[late hare]

you need to realize C doesn't have to be big anyway

[tacit vale]

ik but it is

and thats irrelevant

the distance between earth and venus changing at 2.7 km per day is 100% wrong

[tacit vale]

this is wrong too smh

idk

its too big

i think

maybe idk

26 million miles per day is a bit much

[late hare]

bro if it wasn't 3 am and me laying in bed i would try and help

when is this due

[tacit vale]

nw bro

tonigt

2.5 hours

[late hare]

sheet

but i was thinking imagine a scenario where earth is just in front of venus right

venus will surpass earth after a while

but for a while C will get smaller and grow again but very low rates

this is C we valuating

[tacit vale]

c may get smaller sometimes but in the given position c would grow

but even so, c would not grow at a rate as high as i calculated

[late hare]

hmm

[whole socket]

my brain is having an aneurysm looking at that- time to whip out the ti-89

[tacit vale]

it actually might be possible for it to be as high as that idk

bro i fixed a mistake and got a bigger answer

37.7 million miles/day

[hasty ruin]

aren’t you mixing degrees and radians

360 and π in the same expression is sus

[tacit vale]

yes i realized

thanks

looks better now

[marsh citrus]

@tacit vale Has your question been resolved?

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[real night]

d

[quick moth]

Did you draw that picture?

That just doesn’t look right at all

[real night]

yea i was just like doing a refernce

i know it isnt right

[quick moth]

Ph ok

This a vector problem

Try to redraw the problem to correctly identity the resultant vector

[real night]

p

prob more like this

[quick moth]

No direction is different than angle

It’s like bearing

K nvm picture doesn’t matter

[real night]

i struggle with these problems

sorry

[quick moth]

But a 90 degree direction

Is diffirent than 90 degrees

[real night]

oh

[quick moth]

Here’s a reference

A direction of 0 degrees is straight forward and if you think about it makes she’s e

Sense*

Wait what unit are you learning right now

Are you doing vectors?

[real night]

well im on law of cos

and sines

[quick moth]

Do you know vectors

That’s what I use but whatever

[real night]

I lowkey forgot it

[quick moth]

If you are doing law of cosines

You should try to redraw a picture

[real night]

how so

[quick moth]

This is how u did it

I did*

Not you

You know that angle is 70

Because if you reference the photo above at whar a direction of 200 is

You know the smaller sliver is 20 degrees

So the actual angle must be 70

[real night]

im like

getting and not getting it

where did u get the 70 degrees from

[quick moth]

Here

So I drew the direction of 200 degrees in blue right

And it goes two full quadrants

Which is 180 degrees

[real night]

OH

[quick moth]

So 200-180=20

[real night]

makes sense

[quick moth]

90=20+y

[real night]

so

[quick moth]

Y= 70 degrees

[real night]

does it really like

[quick moth]

Now you just do law of cosines

[real night]

matter which way u do this?

for instance

couldnt u just do normal unit circle rotation?

[quick moth]

no

…

[real night]

well i mean

[quick moth]

You could transfer the direction angle into a normal angle if you like

At the start

But it would just be 3 steps instead of 2

[quick moth]

No

Because then your angle would be 20 degrees

[real night]

oh that makes sense

[quick moth]

So now do law of cosines

[real night]

ye

1 sec

let me just re draw my shit

and then il do it

[quick moth]

What did u get

[quick moth]

Woah that’s not the right picture

But

Right answer

[real night]

i re did it

so

i made the triangle right

and stuff

[quick moth]

Alright 🙏

Have a good one

[real night]

thank u

u helped a lot

[quick moth]

Anytime

[marsh citrus]

@real night Has your question been resolved?

[marsh citrus]

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[quaint arrow]

So i began by finding the gradient of the function < 1, 2y, -2z-1 >.

The gradient of the line is <-1, 2, 0 >. I know that the Normal line of the plane must be parallel to the gradient of the line. Thus <-1, 2, 0 > = lambda <1, 2y, -2z-1 >. Am i on the right path?

Then i think it's clear that lambda is -1 (by looking at the x coordinates). So <-1, 2, 0> = <-1, -2y, 2z+1>

I understand everything except for how x = -5/4 was calculated

.close

[marsh citrus]

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[ornate crown]

How to get a girls heart

[ornate crown]

Frfr

[fervent yarrow]

u def don't need math for that

[placid oak]

[ornate crown]

No but i need you

[placid oak]

These channels are for math questions, please move to #discussion or #chill

[ornate crown]

1 + 1 = i love you

True or not

[placid oak]

No, 1+1 = <@&268886789983436800>

[ornate crown]

Bro

Does math even exist

Are we in a simulation

Im just tryna have fun

Pls dont ban me

Im 13

[rancid prism]

@ornate crown do not troll in help channels.

.close

[marsh citrus]

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[still temple]

question regarding a)

would the equations [
\m\cos\theta\p{T_1+T_2} = 0 \
T_1 \m\sin\theta = T_2\m\sin\theta
]
be valid?

[still temple]

theta being like

[upbeat pasture]

It rotates

m*omega^2 r

not 0

[still temple]

ah right radially

[upbeat pasture]

Yes

[still temple]

would the equations [
\m\cos\theta\p{T_1+T_2} = ma_c \
T_1 \m\sin\theta = T_2\m\sin\theta +w ]
be valid?

[upbeat pasture]

mg my friend mg

in 2nd equation

[upbeat pasture]

Indeed

[still temple]

okay so [
T_2 = \f{T_1\m\sin\theta - w}{\m\sin\theta}
]

[upbeat pasture]

Yes

[still temple]

this gets me 30.9 N

[runic temple]

Isnt the tension in the lower cord 0

[still temple]

does that sound logical

[runic temple]

Oh nvm

[upbeat pasture]

Well did you take g=9.8 or smt?

coz g=10 its 30N

[still temple]

yeah 9.81

[upbeat pasture]

Yeah then it's correct

[still temple]

okay so for b)

we want to find $\ds v = \f{2\pi r}T$, or rather find the missing, $T$ i presume?

[runic temple]

a_c = v^2/r

[runic temple]

This is the formula that got me through circular motion

[still temple]

isnt 2pir/T revolution/s though?

[runic temple]

Yeah

[still temple]

yeah

so jsut

okay wait so

[
a_c = \f{\m\cos\theta\p{T_1 + T_2}}m
]

[still temple]

we know all of that

we also know [
a_c = \f{4\pi^2 R}{T^2}
]

[still temple]

so solve for T, plug into this equation, and thats our answer?

is my thought process right

[runic temple]

Sounds right

[still temple]

okok lemme do it and see if i get an insane result

oh wait

i just realised that tension and period would have the same letter

i guess i should change period to t

ok

i got

3.52 rev/s

lets see if this is correct

oh no my textbook got 0.7 rev/s

hmm

what did i mess up

,rccw

[still temple]

this is what i have

wait hold up

what i derived is just the velocity m/s

do i do $\ds \f{v}{2\pi r}$ next

[runic temple]

Yeah

[still temple]

wait wait w

why does that formula make sense

[runic temple]

f=1/t = v/2pir

[still temple]

[runic temple]

f = rev/s

[still temple]

i mean i guess so

yeah got it

what does go slack mean

[runic temple]

No tension

[still temple]

so like

what do i do here

[runic temple]

Use these equations

[runic temple]

Or jump to this point

[still temple]

so [
T_1 \m\sin\theta = w
]

[runic temple]

Yup

[still temple]

so now

[
a_c = \f{\ff{w}{\m\sin\theta}}m
]

[still temple]

right

[
v = \s{\f{gR}{\m\sin\theta}}
]

[still temple]

[
f = \f1{2\pi r}\s{\f{gR}{\m\sin\theta}}
]

is this it

[still temple]

ah yuck

lemme reopen since this channel is a bit convoluted by this point

.close

[marsh citrus]

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[serene cloud]

Hello

[tall pewter]

Hello

[serene cloud]

Can you help me with this one?

I don't remember how to solve it..

[still temple]

i'd agree in most situations to resort to using the limit of e

[serene cloud]

why?

[still temple]

but this is just like, annoying. I'd recommend performing a taylor series expansion or l'hopital to ease the process

[serene cloud]

I didn't learn lhospital

[still temple]

okay i see

so we have to use the limit of e

[serene cloud]

yeah

so

What do I do next?

[midnight halo]

dont you have to solve the limit of the exponent of e?

[still temple]

wait

hold on

this is all in fact unnecessary

[serene cloud]

i have to solve this one

[still temple]

evaluate the limit for the base and the exponent separately

i.e.

[serene cloud]

what I showed you is my unfinished ...uhm

answer

[still temple]

figure out [
\lim_{n\to\infty} \f{2n^2-n+1}{3n^2-2}\
\lim_{n\to\infty} \f{n^2 -1}n
]

[still temple]

can you tell me what both of those are

[serene cloud]

yeah

the one on top = 2/3

[still temple]

yes

[serene cloud]

and 1

[still temple]

1?

[serene cloud]

i think

[still temple]

are u sure?

the power in the numerator is bigger

[serene cloud]

infinite

[still temple]

yes

indeed!

so you have (2/3)^infinity

do you know what the graph of $\ds \p{\f23}^x$ looks like as $x$ approaches infinite?

[serene cloud]

nope

wait

i know

[still temple]

mhmm

what is it

[serene cloud]

is it somehow decreasing?

[still temple]

mhmmm

its an exponential function

[serene cloud]

yes

[still temple]

when an exponential function has a base 0 < b < 1 its decreasing

b > 1 its increasing

so what can u say about the function as x goes to infinity?

[serene cloud]

Oh, God, we learned with this 2 formulas 🥹🥹

what you are saying its new a bit

[still temple]

what im saying is like

precalculus knowledge haha

so its not something new

or shouldnt be

[serene cloud]

😭😭

its not like that

[still temple]

anyways like what is ur limit \

do you know

[serene cloud]

only, that I was taught through a method

what limit?

[still temple]

.

whats the answer

[serene cloud]

dont know

[still temple]

,w graph (2/3)^x from -10 to 10

[elfin berry]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by lexqa

[still temple]

what about now

[serene cloud]

aa

and what do I do with it?

[still temple]

just what is x as it approaches infinity

what is [
\lim_{x\to\infty} \p{\f23}^x
]

[still temple]

thats what im asking of you

[serene cloud]

0?

[still temple]

yes!

thats ur solution

[serene cloud]

uhm

a sec

through the method learned at school, he gave me this

it*

and u=its 0...

🤣🤣🤣

wow

@still temple Thank youuu

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[marsh citrus]

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[still temple]

<@&286206848099549185>

i need help lol

[rose sorrel]

Ayoo you so hot baby girl😆

[still temple]

??

[rose sorrel]

Lol jokin

[rain ravine]

??

[rose sorrel]

You pinged me lol.

[still temple]

ok so help

<@&286206848099549185>

[rose sorrel]

Hey I am back

[still temple]

and you arent doing shit

[rose sorrel]

Gimme time baby

[still temple]

STOP

.close

[marsh citrus]

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[rose sorrel]

I was just joking around

[marsh citrus]

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[heady atlas]

textbook says this diverges - im confused because it is a small finite series

[sharp mauve]

its a finite number right?

so it diverges i guess

[untold wyvern]

what

[sharp mauve]

idk lol im f bad at math xD

[unborn condor]

you got the sum of the terms [4^(n-1)]/3^n

[heady atlas]

found a and the ratio

[unborn condor]

so 1/4 * (4/3)^n

[devout mauve]

they forgot to put the +...

[heady atlas]

to be a = 1/3 * 4/3

i mean ratio = 4/3

[unborn condor]

since (4/3)^n does not tend to zero, the sum is divergent

[heady atlas]

the i did a(1-x^n) / (1/x)

[sharp mauve]

oh wait this isnt improper intergrals

mb

[heady atlas]

wouldnt that be if it was an infinite series

but here n = 4?

i dont see dots

[brave marsh]

It's probably implied that the series goes on forever.

[heady atlas]

the other questions have +...

[brave marsh]

Otherwise it doesn't really make sense to ask for convergence.

[heady atlas]

yeah

thats wat i thought

like the question doesnt apply

[unborn condor]

what is the actual full instructions

[heady atlas]

heres the next problem with the dots for reference - if they were implied why are they in these

[unborn condor]

i'm... not sure. It could be missing the dots as oversight

or it could be missing the dots because it's not supposed to be a series

[heady atlas]

i thought that for a moment too

but then why would they say it diverges

it makes sense to me they forgot to add dots given the series to inifinity does diverge

anyway thanks for the help, I thought I was going crazy for a moment

.close

[marsh citrus]

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[hearty hedge]

I want to know how to solve it :|

[hearty hedge]

like, steps

[midnight estuary]

Find the domain of this function

log_a(x) is defined for all x > 0

[hearty hedge]

:/

[hearty hedge]

i know that
but how 4x-x^2 can equal 0 so I can make interval

[midnight estuary]

It's not equal to zero, the domain of this function is a open interval

Just solve for domain and you'll get your answer

[hearty hedge]

i mean like
(x,inf)

[midnight estuary]

(f+g)(x) is defined for all x in dom(f) intersection dom(g)

[hearty hedge]

oh
thats half the answer
but as i told
im stuck with how to solve..
log(4x-x^2)
log cant give number equal to or smaller than 0
so it must be more, therefore the domain is (the number that makes it zero, inf)

if 4x-x^2
equals to 0, it will be undefined

[midnight estuary]

yes

So just do greater than zero and solve for x

[hearty hedge]

wait, do i have only to guess some numbers and trying?

:/

second log cant be 2

[midnight estuary]

No just solve the inequality (4x-x^2) > 0 for x

[hearty hedge]

oooh

its (0,4)

[midnight estuary]

Yes

[hearty hedge]

and cant be 2 cuz of ln

[midnight estuary]

Not just 2, you need to take intersection with the domain of ln(x-2)

[hearty hedge]

(2,inf) ig

so the answer is (2,4)

alr thx

[midnight estuary]

Yes!

[hearty hedge]

tysm

[midnight estuary]

Welcome 😄

[hearty hedge]

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[marsh citrus]

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[sullen ice]

how do i show this?

[sullen ice]

obviously i know its not actually divisible here

so i need to re-word a whole bunch of stuff

but how do i show that i want to show that 2k^2-1 is or is not divisible by 4

[untold wyvern]

your trying to show that the top implies the bottom?

[sullen ice]

if that makes sense

yes

[untold wyvern]

So im just going to rewrite the question using the definition of divisibility. If $2(2k^2-1)=4c$, is it true that $(2k^2-1)=2c’$ for some integers $c,c’$

Does this hint you anywhere?

[sullen ice]

shouldnt it be 2c

[untold wyvern]

oh oops

[elfin berry]

Iusgnol

[untold wyvern]

This would just be the question, of course its quite obvious here that $c=c’$

[elfin berry]

Iusgnol

[sullen ice]

is there a way to also do this with transitivity property?

which i know is just doing the same thing here

[distant peak]

no

[sullen ice]

oh

[distant peak]

I mean, transitivity would imply that :
4 divides 2(2k² -1) and 2(2k²-1) divides something, therefore 2 divides 2k²-1

it makes no sense I think

[sullen ice]

it just implies that 2 divides 2(2k^2-1) which is true

[distant peak]

anyway,
2(2k² - 1) = 4c => 2k² - 1 = 2c

and its done

[sullen ice]

ahh i like that way

its the same thing obviously

[marsh citrus]

@sullen ice Has your question been resolved?

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[clear plume]

What's the difference between Fcdf and Tcdf on a TI-83 calculator?

[amber birch]

F is for the F distribution

t is for the t distribution

they are both CDFs or for cumulative probabilities

[marsh citrus]

@clear plume Has your question been resolved?

[clear plume]

ty

[marsh citrus]

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[copper ice]

does anyone what happens if i add x1 to left side and x2 to right side? x1<x2 and are in this domain (-inf, -2] ?

[copper ice]

will the sign of the inequality change?

[still temple]

okay so

do you think it'll change

@copper ice

[copper ice]

no it shouldnt

[pale crag]

Take 2 numbers from ]-inf, -2[ and give it try

[still temple]

yea, give it a try

[pale crag]

I wad always do that when i have problems like this

Always try

[still temple]

same, even tho I know the answer, I'm too scared I'll misremember something

[pale crag]

Can u do that @copper ice

[copper ice]

yea

[pale crag]

Lol, || Most thing make me scare if i miss (-) ||

[pale crag]

Alr , try go write it here , feel free to ask if u need help

[still temple]

same

[copper ice]

ok, so i substituted x1 and x2 with number from given range and i can see that the sign doesnt change, but help me to understand in general x1 and x2 form without take number and substituting

[still temple]

doesn't?

-4<-3

[pale crag]

-4 < -3
Right @copper ice

Delete it

[fickle forum]

GIRL

EXPLAIN MINE

[pale crag]

Go step by step 🥺

[pale crag]

Sorry i'm Not girl

[fickle forum]

boy

[still temple]

calm down

[pale crag]

Pls open channel 👀

[fickle forum]

I DIDIIDIDI

[pale crag]

Then go to ur channel

[fickle forum]

im sorray 😪

[fickle forum]

i did

[pale crag]

@copper ice pls back to here

[copper ice]

here

[still temple]

so does it change or no

[pale crag]

It was < and become >

[copper ice]

now i want to add x1 =-4 to left side and -3 to right side

[copper ice]

yeah, so it doesnt change

[pale crag]

So
-8 < -7

[copper ice]

but how do i understand without take number and substituting

[pale crag]

You should think where is the domain that they give it to u

[still temple]

it literally does change from one direction to another, what do you mean

[pale crag]

If the domain in negative like they give u ]-inf , -2[
As we try we can see that it's Change

[copper ice]

i mean this doent change

[still temple]

oh my bad

[still temple]

memorize or trial and error

[pale crag]

[copper ice]

.close

[marsh citrus]

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[iron cliff]

all i know is that the exterior angles are 450

[still temple]

do you know what regular means in regular pentagon

[iron cliff]

yes

[still temple]

what does it mean

[iron cliff]

all sides and angles are even

[still temple]

correct

can you take a wild guess at the degrees of the interior angles

all of them

does anyone know the answer for this

[iron cliff]

180

[still temple]

not here.

[still temple]

where

[still temple]

nope

[still temple]

category above this, named available

[still temple]

huh?

[iron cliff]

in help-forum

[still temple]

#help-35

[cobalt sentinel]

@still temple #❓how-to-get-help

[still temple]

click on that and send a message there

[iron cliff]

so back tot the problem

how to i get interior angles from exterior ones

[still temple]

you don't care about the exterior ones in this case

[tepid fulcrum]

a

[still temple]

yes

but let him figure it out

[tepid fulcrum]

oke

[iron cliff]

oh ok

[still temple]

can you tell me the sum of all the interior angles

it's not 180

[iron cliff]

360

if not idk

no it cant be that

[still temple]

yes

it can be that

[iron cliff]

oh

right

[still temple]

so

[iron cliff]

ok then

[still temple]

can you go from here

[iron cliff]

5 divided by 450?

[tepid fulcrum]

where do you get the 90+?

[iron cliff]

nvm thats the exterior angles sry

[still temple]

try again

[iron cliff]

360 divided by 5

[still temple]

correct

[iron cliff]

oh wow

[still temple]

it's 72

[iron cliff]

thanks again man

ill propably be back soon

[still temple]

you're good

just try to understand why

[iron cliff]

thanks

.close

[marsh citrus]

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[thick lintel]

Hi! I have this question to solve and I wanted to check if my solution was valid

[thick lintel]

i tried to bound the sequence above and I think (1/(n+2))^n <= 1/n^2 for all n >= 1

hence by this theorm

the series absolutely converges, therefore converges

[bold ice]

Show that by induction and would be correct

[sour glade]

wait how can you use induction in this, this is just series

[thick lintel]

i said that (1 + (1+n))^n >= 1 + n(1+n) by bernoulli's inequality

i think they're asking me to show my upper bound is true for all n >=1

[sour glade]

looks correct to me

[thick lintel]

Oki dokey

[bold ice]

Weren't you supposed to use the sqrt test?

[thick lintel]

I don't think i've learnt that one tbh

ive learned the theorm i posted above, and the d'alembert ratio test

[bold ice]

hm

[thick lintel]

well i will close it for now and reflect on this

thanks guys

.close

[marsh citrus]

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[near thunder]

Help with Gauss-Jordan elim

I feel like I’m going wrong somewhere

[bold ice]

seems correct

[near thunder]

How do I continue

[bold ice]

Go back to system form

[near thunder]

Try and eliminate the -5?

And so on

[bold ice]

How would u eliminate the -5

[near thunder]

Oh I guess the -2

[bold ice]

You can do that

[near thunder]

So R1 + R3(-2)?

Or +2

I meant +2

The first row is correct but the 2nd and 3rd are reverted

<@&286206848099549185>

[marsh citrus]

@near thunder Has your question been resolved?

[near thunder]

<@&286206848099549185>

.close

[marsh citrus]

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[tough kernel]

i need help on question 3 and 4

[tough kernel]

starting from 3b

[marsh citrus]

@tough kernel Has your question been resolved?

[marsh citrus]

@tough kernel Has your question been resolved?

[marsh citrus]

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[hasty flicker]

How is

log(x^2) = 2 log (x)

[hasty flicker]

I don’t get it

[main idol]

That's just a property you should memorize

,tex .log rules

[elfin berry]

riemann

[main idol]

Along with these

[hasty flicker]

Oh

I see, my bad.

Thanks man

[sour glade]

here is an intuitive explanation for why that is
logb(x) is how many times do you need to multiply b with itself to get x

[hasty flicker]

Ye

[sour glade]

Lets say you get x by multiplying b with itself a times or x = b^a then for x^2 = x*x = b^a *b^a = b^(2a) that means you need to multiply b 2a times with itself

[hasty flicker]

Ohhhh

[sour glade]

for all other properties that are listed there, you should take some examples like base =2 and take n = 8, 16 , 4 etc and understand that way you won't have to memorise

[hasty flicker]

Thanks dude

Appreciate the help

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[sturdy lagoon]

wasn't sure which fraction it was

[marsh citrus]

@sturdy lagoon Has your question been resolved?

[sullen ice]

Are you familiar with linear algebra?

[sturdy lagoon]

Not sure

I just wasn't sure if they wanted me to simplify it further

[sullen ice]

It says you're an undergraduate

[sturdy lagoon]

yea

[sullen ice]

So then I think they expect you to use matrices

[sturdy lagoon]

never heard of that actually

terms i haven't heard?

🐶 💔

[sullen ice]

https://youtu.be/6eXl7sNZC9c?si=UxlKq8TK8qk1S2SV

[sturdy lagoon]

alright

thank you

.close

[marsh citrus]

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[slate blaze]

what am i doing wrong this is confusing me

[quick moth]

Oh the first one

You can’t take out a 6

Start by diving everything by 2

[slate blaze]

wait u dont use it from the question?

u can do any number?

[quick moth]

Oh well i kinda see what you did

I would start by taking out a 2

So it becomes 2(3x^2+7x+2)

[slate blaze]

like this?

[quick moth]

Not like that

I mean just take out a 2 from the original equation

And then factor like normally

[slate blaze]

why do we take it out tho

[quick moth]

It just makes it easier

Smaller numbers

[slate blaze]

why was this incorrect

this is confusing me

[quick moth]

(6x+12)(6x+2) is wrong

[slate blaze]

14 = 12 + 2
24 = 12 × 2

[slate blaze]

whats the mistake

[quick moth]

I don’t quite know where you got those factors

You should be getting (2x+4)(3x+1)

[slate blaze]

how 😭

[quick moth]

I don’t how you got those because you didn’t show your work

The correct answer should be 2(x+2)(3x+1)

Oh I see what you did

When you that method

You rewrite 14x as 12x +2x

[slate blaze]

isnt that what i did

[quick moth]

Not at all…

It should be rewritten as

6x^2 +2x+12x +4

https://youtu.be/kIFGLRXes5o?si=NOHaYGPm5fUM6JOM

Watch this video

[slate blaze]

okk

[slate blaze]

.close

[marsh citrus]

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[marsh citrus]

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[grave chasm]

how do i solve this

[grave chasm]

i need to find f1,f2 and f3

[nova solstice]

What are you solving for?

[grave chasm]

magnitude of the forces f1,f2 and f3

@sharp wing how do i find the angle

[marsh citrus]

@grave chasm Has your question been resolved?

[marsh citrus]

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[hard flame]

i need help solving this

[hard flame]

suppose $50000 is divided into 2 bank accounts. one pays 6% simple interest per year and the other pays 8%. After 3 years, there is a total of $10,363.25 in interest in the 2 accounts. how much was invested in each accounts(round up to the hundredths).

this to be exact

<@&286206848099549185>

[long coyote]

Hey wsg

[hard flame]

hii

[long coyote]

That is kinda long haha

[hard flame]

i am kinda confused where to start here

[smoky parrot]

Assume that one of the bank accounts has x amount of money

Then the other has 50000-x

[long coyote]

Ill let this man cook

[smoky parrot]

Substitute them in as principals for respective accounts, and you'll get two values in x

Add the two equations, set it equal to 10363.25, and solve for x and 50000-x

[hard flame]

X x 0.06 x 3 + (50000-X)0.08 x 3=10,363.25

is this right?

0.18X + 1200 -0.24X = 10,363.25

-0.06X = 9163.25

huh??

i think i made something wrong

ig i need to learn this by my self

[marsh citrus]

@hard flame Has your question been resolved?

[marsh citrus]

@hard flame Has your question been resolved?

[marsh citrus]

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[storm fern]

Given an arithmetic sequence of Sn = 5n^2 - 6n as the sum of n first sequence, then U5 = ...?

[proven vault]

what is U5

[storm fern]

the 5th number of the sequence

nevermind i got it

S(n) - S(n-1) = Un

[marsh citrus]

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[marsh citrus]

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[finite niche]

Four fours is a mathematical puzzle, the goal of which is to find the simplest mathematical expression for every whole number from 0 to some maximum, using only common mathematical symbols and the digit four. No other digit is allowed. The puzzle require that each expression have exactly four fours.
Some people choose to allow the use of certain functions or even decimals in-before the 4s to get the value 0.4. But i consider that cheating.
I've currently solved every number up to 39 but here i am stuck. How can i get this to work?

39 = 4 4 4 4

[finite niche]

Solutions for 1-32 as example:

[quaint elm]

doesn't look like anyone's found a solution without decimal

oh wait

[finite niche]

i found one using Gamma function but i consider that not being really rightful

[quaint elm]

i found one using only the symbols you use

... but the meaning of the exclamation mark is slightly different

[finite niche]

you mean double Factorial?

[quaint elm]

yeah

[finite niche]

yeah i'm a bit on the edge for that one, it's the same thing as with using gamma function, doesn't really feel right to use it

[quaint elm]

it seems fine to me

i had a program to find these once

[finite niche]

i find the beautiful thing in this math puzzle is to make it as simple as possible.
I can use double factorial but it's such a niece thing to use that i've never came across it in any math courses at Uni.
I can also solve it with gamma function by doing (4!+sqrt(4))*(4/gamma(4)).
But that feels like cheating.

[quaint elm]

gamma feels worse

since it's basically saying what if i had 3 instead of 4

[finite niche]

i agree, but the same thing is with double factorial.
It's like taking factorial and saying "lets just ignore some numbers".
I was considering Cos & Sin if it would be more acceptable solving it with that.
Somehow getting 1.5 from it and multiplying it with 26

Or using more than one root, like a solution i found on 33. Where i took third square root of 4 and then ^4! + sqrt(4) then dividing everything with sqrt(4) to gain 33

[quaint elm]

third square root
cube root?

[finite niche]

no i used third square root

TLDR it basically gives me (4^3 + 2)/2

[quaint elm]

,w third square root

[elfin berry]

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[finite niche]

to make it more clear

but thinking if can solve it in a similar way but for 39 instead

sorry, i miss spoke.

i took 8th root of 4. i took square root of 4 three times

2^3

this to cancel out 4 & 2 from the 4! to get 4^3

<@&286206848099549185>

[marsh citrus]

@finite niche Has your question been resolved?

[marsh citrus]

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[main mica]

a photo by the size of 12cm X 18cm is in a frame the frames S is same as the photoes i need to find the width of the frame

[mighty bramble]

the frames "S"?

[main mica]

no

area

s is area

[mighty bramble]

what other info is given

[main mica]

frame and the photo have the same size area

@mighty bramble

<@&286206848099549185>

[marsh citrus]

@main mica Has your question been resolved?

[tiny crown]

Are they testing whether you know which one is "width"?

[main mica]

.close

[marsh citrus]

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[marsh citrus]

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[marble summit]

hi, let g a strictly decreasing and continous function such as g(a)=0 and A={x in R such as g(x)<0}

How to prove that for all r>0 the ball B(a,r) has elements of of A ?

[marble summit]

please ?

[marsh citrus]

@marble summit Has your question been resolved?

[sinful flint]

Fix $r>0$ and consider $B(a,r) = (a-r , a+r)$. We know that g is strictly decrerasing so $g(a+ \frac{r}{2})<0$. But $a+\frac{r}{2} \in B(a,r) \cap A$

[elfin berry]

milo_schwartz

[sinful flint]

Actually I don't really know if g continuous is necessary

[marble summit]

why g(a+r/2) exists. ?

@sinful flint

[echo bough]

wdym why it exists?

a+r/2 is a real number right?

and g is defined on R

[sinful flint]

I don't know the domain, I thought g was defined in R