help-17

[carmine leaf]

its t/T

[deft sedge]

ye

and they plugged in 5 for T

when t is usually used

to represent time

e^-kt

[carmine leaf]

can you just read what i said again for 2)

[deft sedge]

for exmaple

[plain minnow]

For t = 5 we have that t/5 = 1 so 2¹ (doubled once). For t = 10 we have that t/5 = 2 so 2² (doubled twice). See the pattern forming for every increment of 5 for the value of t this way?

[deft sedge]

T looks like its the constant 5. which is yrs. which is time

[carmine leaf]

it represents doubling time. which sure is time but its a constant

[deft sedge]

N is doubling time tho

or is that formula in gernal

[carmine leaf]

no N is initial value

[deft sedge]

the doubling time

N = 2

? 😄

[carmine leaf]

oh sorry

n is the 'doubling' factor

[deft sedge]

wth

[carmine leaf]

if you instead wanted something to triple

n would be 3

because doubling something is multiplying with 2 repeatedly and tripling is multiplying with 3 repeatedly

and T is the time it takes for that to happen once.

[deft sedge]

1+1

[carmine leaf]

then why are you confused 💀

[deft sedge]

bro

theres a fraction

in the exponent this is why im confused

and then they plug in 5 yrs (time) into the captal T

which is what they normally dont do

t=Time

here T=time?

t/5 is implied that T is 5

[carmine leaf]

t is time in the sense that its the time we observe that is moving

T is just a constant that is somehow related to the word time

namely the doubling time

but it doesnt change

so using 500 * 2^(t/5)

if you put t=5 which is to say, lets look 5 years into the future.

then it would have doubled

if you put t=10 which is to say, lets look 10 years into the future

then it would have doubled twice

[deft sedge]

thats a nothing burger. we didnt measure speed of stuff, rather exponential growth previously. but technically we did find stuff with speed using t aswell.
here they use T for time, which is frustrating. feels like all im doin in memorizing stuff and not really getting it

[carmine leaf]

ok. if you cant understand the difference between a constant being related to time and moving time

then

forget time or any unit really

t, moving

T, constant

thats it

who cares if its time, length, speed or w.e

one thing is moving

[deft sedge]

e^-kt. t is not moving

[carmine leaf]

and one thing is consant

yes t is moving

thats the whoel fucking point

of a variable

[deft sedge]

oh shit

u right

[carmine leaf]

theres a difference between someone asking you

[deft sedge]

but in this example

[carmine leaf]

to find something at a specific point t=5

[deft sedge]

T can also be moving

since it can be any number?>

[carmine leaf]

no not within the context of the problem

[deft sedge]

i understand

5x

for example

is constant

[carmine leaf]

5 is the constant

[deft sedge]

well

no its not

mb

y=5

here its t/5

which idk how to take that

worse an exponent

[carmine leaf]

ok so you know how you have

y=mx+b

m and b are consatnts right?

[deft sedge]

b is constant

and m if its serparte

[carmine leaf]

m is aswell

no

[deft sedge]

seperated

[carmine leaf]

m is also a constant

[deft sedge]

5x isnt constant tho

but 5 is

thus seperated

[carmine leaf]

yes but i didnt say

mx did i

i said m

[deft sedge]

ok

both constants

[carmine leaf]

yes. so just extend the same logic to T

it can techincally be any number

but its only 1 number within the context of the same problem

namely a constant

[deft sedge]

ok

fine

its a constant

should i graph this

1k^t/5

[carmine leaf]

500 * 2^(t/5)?

[deft sedge]

ye

[carmine leaf]

graph it

why not

[deft sedge]

looks like exponential growth function

but shifted

to the left

[carmine leaf]

it literally is an exponential function

[deft sedge]

whats the dif

between

e^x

and this sht

?

just a shift?

y use one over the other

[carmine leaf]

not even a shift its the exact same thing

[deft sedge]

ok how would u write 500 * 2^(t/5)?

in terms of e

[carmine leaf]

1000=500e^(k*5)

to find exponential rate

and then just do 500e^(kt) with that k

i guess a more direct way is $500e^{ln(2^{t/5})}$

[twin meteor]

Køter

[deft sedge]

i regret asking lol

[carmine leaf]

$500e^{\frac{ln(2)}{5}t}$

[twin meteor]

Køter

[carmine leaf]

with some log rules

well just 1 log rule

so k=ln(2)/5

[carmine leaf]

to answer this. it can be more intuitive to use 2 as a base if you are asked to find something that is doubling

but i dont think anyone really uses any other base than e

maybe in some applied math im not sure 💀

[deft sedge]

so its better to use

this formula

rather than

other one?

like its the simplest

[carmine leaf]

yes because its pretty clear how it works in terms for doubling

like t=5 is double once

but maybe tahts not clear with 500e^(0.1386294361t)

[deft sedge]

how would u solve for t tho 💀

[carmine leaf]

wdym solve for t

[deft sedge]

is t just like x

and thats the final formula?

[carmine leaf]

yes

exactly

[deft sedge]

lol

[carmine leaf]

t variable

you could be asked like at what t would the end value be 1200 for example

and you would need to solve for t

but you never fully replace t

[deft sedge]

ok but whats t

[carmine leaf]

its always just t

[deft sedge]

if its moving

[deft sedge]

solve for T u mean?

[carmine leaf]

no

the difference just is t doesnt have only one value

but you can be asked sometimes to take out one specific value of t

but after you do that its not like you can replace it in the formula

it still just goes back to t

you just found one specific point

[deft sedge]

like finding t

in expo growth

[carmine leaf]

ye like find at what t 500 grows to 1250

[deft sedge]

or y rather

oh

[carmine leaf]

so 1250=500*2^(t/5)

you find your t

but the function is still 500 * 2^(t/5)

you just found out specifically how long it takes for 500 to grow to 1250

[deft sedge]

ok

makes sense

so how would u find t there

1250/500

then what

[carmine leaf]

you can do 2 things. take ln both sides or log_2 both sides

[deft sedge]

which is fastest or easiest

or both

idk

[carmine leaf]

log_2 prob makes more sense

[deft sedge]

multiplying by log 2?

log_2

to both sides?

[carmine leaf]

youre not multiplying

but yes. do log_2 to both sides

[deft sedge]

what r u doin then

🔸 🔸 🔸 🔸

[carmine leaf]

idk what its called specifically tbh. but multiplying is for numbers

log_2 is not a number is a function

i just say 'take log_2 of both sides'

[deft sedge]

mkay

[carmine leaf]

so how would it look?

[deft sedge]

log_2 (1250/500) = log_2 (2)^ (Log_2(t/5))

that right?

[carmine leaf]

what is the 500 doing on RHS

[deft sedge]

no clue

maybe it wanted to stay for some reason

[carmine leaf]

and fix some parenthesis on RHS aswell.

[deft sedge]

felt at home

it cloned itself

got rid of it anyway

[carmine leaf]

RHS, sorry

[deft sedge]

idk man tbh

how to fix the parenthesis

[carmine leaf]

if you have a=b and you take log_2 of both sides

you end up with

log_2(a)=log_2(b)

what is b here?

in your example

youve been doing it with ln a lot why is log_2 any different 💀

[deft sedge]

its not

but like

e is easier

and so is ln

but ok

one sec

[deft sedge]

looks kinda sus tbh

esp on RHS

[carmine leaf]

yes because you are only taking log_2(2)

and then taking that to a power of t/5

you want to take log_2 of EVERYTHING on the RHS

not just some things

bro, i keep fucking up my left and rights 💀

[deft sedge]

how about now

[deft sedge]

lol all good

[carmine leaf]

no. its $log_2(2^{t/5})$

[twin meteor]

Køter

[carmine leaf]

you just put everything inside log_2

not just 2 not 2 and then the exponent

everything as it is

put it in

so now what?

[deft sedge]

log_2 (1250/500) = log_2 (2^(t/5)) ?

whats next?

sleep ninja

LOL

jk

pls hurry tho

HAHAHAA

[carmine leaf]

yes. maybe also put (t/5) so its obvious that everything is in the exponent

[deft sedge]

i have classes

need to sleep

[carmine leaf]

well log_2(2^x)=x

like how log_e(e^x)=x

should be obvious then

[deft sedge]

fk man

log properties

at this time

not fun

its 7am

almost

8

[carmine leaf]

log_a(a^x)=a

you just cancel the log and the base

because log_a and a^x are inversese

youv edone this a million times

[deft sedge]

u mean x

[carmine leaf]

with ln

[carmine leaf]

yes sorry

ln(e^x)=x

[deft sedge]

ok

[carmine leaf]

youve done this so many times

any base

works

not just e

log_3(3^x)=x

so just cnacel them

[deft sedge]

log_2 (1250/500) = t/5

[carmine leaf]

yes

[deft sedge]

now what

i swear

im not lying

[carmine leaf]

you cant be serious 💀

[deft sedge]

on not knowing

LOL!!23eo;2qj3r

[carmine leaf]

just multiply both sides by 5

[deft sedge]

aahhk

5(log_2 (1250/500)) = t

thats how long

[carmine leaf]

yes

im glad we didnt do ln on both sides 💀

[deft sedge]

well

[carmine leaf]

would need to do change of base formula and everything

[deft sedge]

looooool

what did we find

since t is moving

[carmine leaf]

we found that one specific value of t

[deft sedge]

how long from

500-1250

?

[carmine leaf]

exactly

[deft sedge]

cooooooooooooooool

thanks koter

[carmine leaf]

basically

500 * 2^(t/5)

,w 500 * 2^(log_2(1250/500))

[twin meteor]

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Requested by Køter#5504

[deft sedge]

whats that

what we found

?

1250 days?

yrs?

[carmine leaf]

no i just put in the t we found into 500 * 2^(t/5)

to make sure its correct

,w 5*log_2(1250/500)

[twin meteor]

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Requested by Køter#5504

[deft sedge]

sick

but

what was t

like

what did we find

so 6 yrs

[carmine leaf]

6.6 units of time

[deft sedge]

coools

[carmine leaf]

sure

[deft sedge]

koter my boy

my eyes r burning

l0l

thanks man

keep being awesome bro

always appreciate ur time helping me

i try my best to follow u

and not be impatient

or rude

u know

like with ahem ann

😄 😄

[carmine leaf]

nws

you can tell me if im overexplainin

so save time is ok

[deft sedge]

of course bro

i always do

if it happens

but ye

i got this set now

ty my g

[carmine leaf]

np

[deft sedge]

pick this up another time

gn g

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[vast shale]

hi frens

[vast shale]

how come at the bottom right, the first one is only w instead of w^2

if i use the quotient rule, i always think / g^2

i found something exactly the same, just with the w^2

anyone able to help me out?

[pale perch]

if the numerator is h and the denominator is g

then the first one is gh'/g^2=h'/g, (the w cancel), i think you just forgot to write it by accident

[vast shale]

no im stuck for like 1 hour on this with headache

[pale perch]

i mean on your last line, the first part of the numerator should be pbw(bw+c)^{p-1}

you didnt write the w, you just wrote h' rather than h'g

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[flat anvil]

Can someone help me please, I don't know how to do this.
ABC is a right-angle triangle inscribed within (O) (B=90). D is on the chord that doesn't have B. M is on DC chord. N is on BC chord. The intersection of DC and AM is E, the intersection of BC and AN is F, the intersection of DN and BM is K. Prove E, F, K is linear.

[paper depot]

,rccw

[flat anvil]

I learn about some quadrilateral inscribed within a circle in this unit, I learnt thales, pythagoras, congruent triangle, three special lines in a triangle, and other stuff. The question usually doesn't require other theorem but sometimes you can use other theorem to solve it faster.

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@flat anvil Has your question been resolved?

[flat anvil]

<@&286206848099549185>

[flat anvil]

btw the note above the picture means nothing, it's from another problem

[proper glacier]

i can’t help you with this question but, what is the actually question that u are asking or information that u need?

[flat anvil]

the actual question is prove K is in the middle of EF if DM and BN chord equal

In the actual question, I manage to prove if DN intersect BM at K'. OK' meet EF at the middle point, K. I can't prove K' and K is the same point.

[flat anvil]

This is the picture for the actual question

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[vast shale]

sir

where does the w come from then?

[pale perch]

Youre doing the quotient rule, the numerator is h'g-hg', and g=w

Also, you should open your own channel for this

[vast shale]

derivative of bw +c
with b and c being constants and w the x.

isnt that just like derivative of ax+c
so 1*a x ^(1-0) + 0

for h' that is

[pale perch]

open a new channel

this is someone elses

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[eager walrus]

is this a quiz

[grand pine]

isn't it just A?

or am i tripping

because that's the only think i see that's been factorised out by something (10)

tell me if it is

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[midnight knoll]

Is this correct for part 1 b?

[lament lichen]

Yes that's fine

[midnight knoll]

Thanks. This makes it not continuous and not differentiable correct?

<@&286206848099549185>

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[wanton mesa]

q: a>0 find the smaller of the two solutions to

[wanton mesa]

I tried b^2 -4ac formula but couldnt solve it anyways any ideas?

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[ebon current]

can we determine this?

[ebon current]

.close

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[wispy heart]

any idea why it says "F > F_0.025"? coz 1.208 is not greater than 4.03.

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[unreal sparrow]

Why is the answer to part C equal to 0 < theta < pi/2 and not -pi/2 < theta < 0?

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[unreal sparrow]

<@&286206848099549185>

[lusty fox]

because it never gets there

for example halfway through at -pi/4, 2theta = -pi/2 but then sin2theta = -1

and you cant do that in a square root

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[unreal sparrow]

Ok so then what explains the limit at +pi/2? Why can’t it be above pi/2?

[lusty fox]

cause it already finishes

it can it will just cycle again

[unreal sparrow]

oh ok thanks

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[modest musk]

I found out the ratio of the bases, what do I do next?

I mean of those smaller triangles in there.

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[prisma imp]

Pls anyone could

Help?

[river nebula]

Don’t close your channel so fast haha give us time to answer! Haha

[prisma imp]

K sry

[river nebula]

Since they are similar, means their corresponding lengths are in prior Tokyo each other.

You know about similarity?

[prisma imp]

a bit

Yh

[river nebula]

Yeah so what can we say about the ratio of the lengths of the 2 parallelograms ?

What’s the ratio?

[prisma imp]

15:24:60

Idk

. close

.close

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[stuck girder]

Hi

[stuck girder]

Why is the integral of x sin x = x cos x + sin x + C

[fluid obsidian]

either find the derivative of right or integrate x sin x

[stuck girder]

Ok

Thank you

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[grim swallow]

Is this rational or irrational

,rrcw

,rccw

[grim swallow]

Pls tell

Fast

[keen umbra]

rational

[grim swallow]

How

The number is repeating

[keen umbra]

cuz we can let x=43.123456789

[grim swallow]

Ok

[keen umbra]

100000000x=43123456789.123456789

[grim swallow]

Ok

[keen umbra]

minus the both of them

[dim quail]

just use definition

[keen umbra]

but then you can see that the repeating decimals get minused out

[dim quail]

irrational numbers do not repeat nor terminate

[grim swallow]

But they are still continuing

After 9 there will be another 123456789 and another

[dim quail]

yes

which means it is non terminating but repeating

thus it is rational

[grim swallow]

Ok I thought only the numbers that can be represented in the form of p/q

Thanks

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[dim quail]

.reopen

[vocal sleet]

✅

[dim quail]

yes that is also the definition

[grim swallow]

Ok

[dim quail]

but we cant find the exact value of p and q here

so we opt to the second definition

[dim quail]

which is this

[grim swallow]

What is 2nd

Tell

[dim quail]

this

i told

[grim swallow]

Ok

Thanks

👍🏻👍🏻

[dim quail]

yw

[grim swallow]

Yw?

Oh you welcome

Ok

.close

[vocal sleet]

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[stray bay]

What is this asking for? I've done the answers but I'm assuming they're wrong. Specifically D - H.

[stray bay]

I've graphed it and used multiple different calculators but it's giving me the same answer of 0. What is this process called?

[lime gorge]

Can u show ur work?

I’m seeing a couple ones that aren’t right

[lime gorge]

I don’t think u could get this thru graphing

[gray python]

That gives an idea of the answer at best

[lime gorge]

Maybe it’s called something like “evaluating functions at non numerical inputs”?

[stray bay]

Showing my work is pointless because I know for a fact that I've done the wrong process here.

[lime gorge]

Ok Lemme give u an example

[stray bay]

I just need to know what the question is asking for here because I don't know the terminology for this question.

[strange crater]

it's asking you to "plug in" those values

so for "f(0)" you replace every "x" with "0"

for "f(9)", you replace every "x" with "9". and so on

[stray bay]

Yes, I understand those parts.

It's just the '-x', and 'h-x', and so on.

[strange crater]

replace "x" with "-x"

[stray bay]

Doing that in a calculator gives me a 0.

So I assume I'm doing that wrong.

[strange crater]

if $f(x) = \frac{x}{x^2 + 1}$ then $f(-x) = \frac{(-x)}{(-x)^2 + 1}$. do the arithmetic with that

[lime gorge]

Don’t do it with a calculator

[twin meteor]

cwatson

[stray bay]

Ah, I see.

One moment, let me try to replace all the variables.

[strange crater]

don't use a calculator here, because you're not really "calculating" anything here, in the same way as parts a, b

[stray bay]

I'm not?

[strange crater]

not really, because there will still be some x's in there

[stray bay]

Let me try this rq.

I got 0 again

[strange crater]

please show your work then

[stray bay]

1: Factored for each term -fx = -x/x^2 + 1

[strange crater]

what do you mean factored?

[dim quail]

tf is -fx lol

functions dont work like that

[stray bay]

Oh

[dim quail]

$f(-x) \neq -fx$

[twin meteor]

bettim

[sudden compass]

in this case thats true, no?

[stray bay]

I removed unecessary parenthases

[strange crater]

it's irrelevant, and not how OP should approach the problem IMO

[stray bay]

factored, applied the product rule to -x

[strange crater]

look here again

[stray bay]

raised -1 to the power of 2

multiplied x^2 by 1

zzz...
is there a khan academy video on this?

[carmine leaf]

f(x) does not mean f multiplied with x

[stray bay]

I think that would be better because I'm clearly doing something wrogn.

[strange crater]

you should have gotten $f(-x) = \frac{-x}{(-x)^2 + 1} = -\frac{x}{x^2 + 1}$. which you seem close to getting I think

[twin meteor]

cwatson

[stray bay]

hm...

[strange crater]

do you see why? Just replace "x" with whatever is in the parentheses of "f(x)". For part d, that is just "-x"

it's the same thing you did for "f(0)", "f(9)", etc. except you're not replacing with a specific number

[stray bay]

is it asking for the value of 'x' here?

[strange crater]

no, it's not asking for any value there. you can't get a value for x

[dim quail]

value of f(x)

[stray bay]

ok, and is it possible to find this answer by looking at a graph with the equation shown?

[strange crater]

it might be, but that's a harder way IMO

what I mean is, in this case you might be able to tell but not in general

[stray bay]

I don't understand. I can't wrap my head around not using graphing for this. Isn't f(-x) a transformation?

[carmine leaf]

yes reflection across y-axis

they are asking what that function will look like

[stray bay]

well, it would barely change

[carmine leaf]

sure but you have f in terms of x

so now you just need to write out f in terms of -x

and that would be the expression for f reflected across the y axis

[stray bay]

y=\frac{-x}{\left(-x\right)^2+1}

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

[carmine leaf]

on both sides

[stray bay]

how do I use the bot thingy

[carmine leaf]

$ on both sides

[stray bay]

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

[twin meteor]

Piessons

[stray bay]

this is how I graph it

[carmine leaf]

sure

[stray bay]

and this is what it looks like

[carmine leaf]

ok

[stray bay]

would it just be -1

[carmine leaf]

no it would be the expression of the graph you just drew

[stray bay]

but this is the expression

[carmine leaf]

yes so thats the answer

[stray bay]

which is the part that confuses me

really?

[stray bay]

for this?

was it really that simple

[carmine leaf]

finding what f(-x) is equal is equivalent to finding the expression for f(x) reflected across yaxis

yes

[stray bay]

oh.

[carmine leaf]

you can put 1*x^2 instead also

[stray bay]

I guess that solves my question then, thank you

[carmine leaf]

sure do you understand -f(x) aswell

and the others?

if you are familiar with graph transformations then just think of them like that

[stray bay]

hmm

$f\ne :0$

[twin meteor]

Piessons

[stray bay]

is what I got for -f(x)

[carmine leaf]

do you know what transformation -f(x) is?

[stray bay]

$-f\left(x^2+1\right)=x$

[twin meteor]

Piessons

[stray bay]

not quite in a graphing sense

[carmine leaf]

you know f(x) does not mean f multiplied with x

it means the function f evaluated at x

you cant treat it as multiplication and just 'factor' things out and in

you are still trying to find -f(x) which is relfection across x-axis btw

if f(x)=x/(x^2+1)

can you see a way to get an expression for -f(x)=?

[vocal sleet]

@stray bay Has your question been resolved?

[stray bay]

I've ran out of time. I'll have to guess.

[vocal sleet]

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[flint lantern]

Given $x+y+z=3$, prove the inequality

$$\frac{x}{16+y^3}+\frac{y}{16+z^3}+\frac{z}{16+x^3}\geq\frac{1}{6}$$

[twin meteor]

random-internet-guy

[flint lantern]

normally, for these kinds of inequalities, I'd just use cauchy-schwarz to add up all the complicated denominators, but the equality case for this inequality happens at x=0, y=1, z=2 (and similar cases), so that wouldn't work

[vocal sleet]

@flint lantern Has your question been resolved?

[vocal sleet]

@flint lantern Has your question been resolved?

[vocal sleet]

@flint lantern Has your question been resolved?

[vocal sleet]

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[neat phoenix]

I am trying to attempt this problem

[neat phoenix]

This is my working

But for some reason my friend gets a different answer to me

<@&286206848099549185>

[vocal sleet]

@neat phoenix Has your question been resolved?

[vocal sleet]

@neat phoenix Has your question been resolved?

[neat phoenix]

.close

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[placid osprey]

$$M=c\left( \frac{1kg\ m^{2}}{s^{3}} \right)^{\alpha } \left( m\right)^{\beta } \left( \frac{m}{s^{2}} \right)^{\gamma }  $$

[twin meteor]

dgh

[placid osprey]

How do we go from that to $$\alpha =1,\ \beta =-\frac{1}{2} ,\ \gamma =-\frac{3}{2} $$

[twin meteor]

dgh

[placid osprey]

In a well-defined physical problem, the given quantities are power $P$, distance $L$, and gravitational acceleration $g$. The task is to determine a mass $M$. The answer will have the form $M=c P^\alpha L^\beta g^\gamma$, where $c, \alpha, \beta$ and $\gamma$ are constants. Determine $\alpha, \beta$, and $\gamma$.

[twin meteor]

dgh

[maiden iron]

M is in kg

P is in kg m²/s³

L is in m

g is in m/s²

$[M] = c[M]^{\alpha} [L]^{2\alpha} [T]^{-3\alpha} \times [L]^{\beta} \times [L]^{\gamma} [T]^{-2\gamma}$

[twin meteor]

Ｈｅｒｅｌｓ

[maiden iron]

you regroup them

$[M] = c[M]^{\alpha} \times [L]^{2\alpha + \beta + \gamma} \times [T]^{-3\alpha -2\gamma}$

[twin meteor]

Ｈｅｒｅｌｓ

[placid osprey]

Not sure what you did there

[maiden iron]

Dimensional analysis

[placid osprey]

okay so why did you replace P and g with M and T

why alpha iin exponent

[maiden iron]

Power isn't in the international system of units

so I split it into mass and times

[maiden iron]

from this, you should obviously see a system of equations to solve for alpha, beta and gamma

[placid osprey]

Power is defined as work per time, where work is force times distance. Therefore, we can express power in terms of mass (force has units of mass times acceleration), length, and time

$P=\frac{W}{t} =\frac{Fd}{t} =\frac{ma\times d}{t} =\frac{m}{t} \times \frac{d}{t^{2}} =\frac{m}{t^{3}} \times d^{2}$

[twin meteor]

dgh

[maiden iron]

i forgot length too

But I actually did it

P = f.v
v : m/s
f = ma : kg.m²/s
so P : kg.m³/s²

hmm maybe I did a mistake somewhere

ah my bad

$[M] = c[M]^{\alpha} [L]^{3\alpha} [T]^{-2\alpha} \times [L]^{\beta} \times [L]^{\gamma} [T]^{-2\gamma}$

[twin meteor]

Ｈｅｒｅｌｓ

[placid osprey]

Starting from $M=c P^\alpha L^\beta g^\gamma$, we can use dimensional analysis to determine the units of each term. We know that the units of power $P$ are given by $[P]=[M][L]^2[T]^{-3}$, where $[M]$ represents the unit of mass and $[L]$ represents the unit of length. Thus, we can substitute this expression for $P$ in the original equation to get:

$$M=c(M[L]^2[T]^{-3})^\alpha L^\beta g^\gamma$$

Expanding this expression and collecting terms, we get:

$$[M] = c[M]^{\alpha} [L]^{2\alpha + \beta} [T]^{-3\alpha} [g]^\gamma$$

Note that we have used the fact that $[g]=[L][T]^{-2}$.

We can further simplify this expression by collecting the terms involving $[L]$ and $[T]$ to get:

$$[M] = c[M]^{\alpha} [L]^{2\alpha + \beta + \gamma} [T]^{-3\alpha -2\gamma}$$

[twin meteor]

dgh

[placid osprey]

What do you think

[maiden iron]

a mistake on the dimension of P

[placid osprey]

The correct units of power in the International System of Units (SI) are $[P]=[M][L]^2[T]^{-3}$, where $[M]$ represents the unit of mass, $[L]$ represents the unit of length, and $[T]$ represents the unit of time.

Substituting this expression for $P$ in the equation $M=c P^\alpha L^\beta g^\gamma$, we get:

$$M=c(M[L]^2[T]^{-3})^\alpha L^\beta g^\gamma$$

Expanding this expression and collecting terms, we get:

$$[M] = c[M]^{\alpha} [L]^{2\alpha + \beta} [T]^{-3\alpha} [g]^\gamma$$

Using the fact that $[g]=[L][T]^{-2}$, we can further simplify this expression to:

$$[M] = c[M]^{\alpha} [L]^{2\alpha + \beta + \gamma} [T]^{-3\alpha -2\gamma}$$

[twin meteor]

dgh

[maiden iron]

you didnt fix it

[placid osprey]

Can you show me lol

[maiden iron]

P is energy over time

oh wait a sec im dumb ._.

but let see it again

energy is kg x m/s² x m = kg.m²/s²

so energy over time is kg.m²/s³

yea you were right

[placid osprey]

okay one sec

To determine the values of $\alpha$, $\beta$, and $\gamma$, we need to use dimensional analysis to match the units on both sides of the equation:

$$M=c P^\alpha L^\beta g^\gamma$$

Using the expressions for the units of the given quantities, we can rewrite the above equation as:

$$[M] = c([M][L]^2[T]^{-3})^\alpha [L]^\beta ([L][T]^{-2})^\gamma$$

Expanding and simplifying the right-hand side of the equation, we get:

$$[M] = c[M]^{\alpha}[L]^{2\alpha+\beta+\gamma}[T]^{-3\alpha-2\gamma}$$

Equating the exponents of $[M]$, $[L]$, and $[T]$ on both sides of the equation, we get a system of three equations:

$$\begin{cases} 1=\alpha \ 0=2\alpha+\beta+\gamma \ 0=-3\alpha-2\gamma \end{cases}$$

Solving this system of equations, we get:

$$\alpha=1, \quad \beta=-\frac{1}{2}, \quad \gamma=-\frac{3}{2}$$

Therefore, the expression for the mass $M$ in terms of the given quantities is:

$$M=cP L^{-\frac{1}{2}} g^{-\frac{3}{2}}$$

[maiden iron]

yes

[twin meteor]

dgh

[placid osprey]

is that right

[maiden iron]

it is

[tropic canopy]

bro just typed random letters and he answered "yes" in 2sec

[placid osprey]

not sure if the 1, 0, 0 is right

[maiden iron]

since all that is of a dimension of a mass

so we have alpha = 1

we dont have a length and a time

so their exponents are equal to 0

[placid osprey]

In the expression $M = c P^\alpha L^\beta g^\gamma$, we can see that $M$ has dimensions of mass ($[M]$), $P$ has dimensions of power ($[P]$), $L$ has dimensions of length ($[L]$), and $g$ has dimensions of acceleration ($[T]^{-2}$).

Since the dimension of mass is only present in $M$, we can say that the exponent of $M$ in the expression must be 1, i.e., $\alpha = 1$.

On the other hand, we don't have any other dimension present in the expression apart from mass, so the dimensions of length and time must cancel out in the final expression for $M$. This means that the exponents of $L$ and $T$ must be zero.

Hence, we get $\beta + \gamma = 0$ and $-2\alpha - 2\gamma = 0$, which leads to $\alpha = 1$, $\beta = -1/2$, and $\gamma = -3/2$.

[twin meteor]

dgh

[maiden iron]

should be good, but I prefer this

its more clear

[placid osprey]

\begin{itemize}
\item To determine the values of $\alpha$, $\beta$, and $\gamma$, we need to use dimensional analysis to match the units on both sides of the equation:
$$M=c P^\alpha L^\beta g^\gamma$$

\item Using the expressions for the units of the given quantities, we can rewrite the above equation as:

$$[M] = c([M][L]^2[T]^{-3})^\alpha [L]^\beta ([L][T]^{-2})^\gamma$$

\item Expanding and simplifying the right-hand side of the equation, we get:

$$[M] = c[M]^{\alpha}[L]^{2\alpha+\beta+\gamma}[T]^{-3\alpha-2\gamma}$$

\item Equating the exponents of $[M]$, $[L]$, and $[T]$ on both sides of the equation, we get a system of three equations:

$$\begin{cases} 1=\alpha \ 0=2\alpha+\beta+\gamma \ 0=-3\alpha-2\gamma \end{cases}$$

\item Solving this system of equations, we get:

$$\alpha=1, \quad \beta=-\frac{1}{2}, \quad \gamma=-\frac{3}{2}$$

\item Therefore, the expression for the mass $M$ in terms of the given quantities is:

$$M=cP L^{-\frac{1}{2}} g^{-\frac{3}{2}}$$

\item In the expression $M = c P^\alpha L^\beta g^\gamma$, we can see that $M$ has dimensions of mass ($[M]$), $P$ has dimensions of power ($[P]$), $L$ has dimensions of length ($[L]$), and $g$ has dimensions of acceleration ($[T]^{-2}$).

\item Since the dimension of mass is only present in $M$, we can say that the exponent of $M$ in the expression must be 1, i.e., $\alpha = 1$.

\item On the other hand, we don't have any other dimension present in the expression apart from mass, so the dimensions of length and time must cancel out in the final expression for $M$. This means that the exponents of $L$ and $T$ must be zero.

\item Hence, we get $\beta + \gamma = 0$ and $-2\alpha - 2\gamma = 0$, which leads to $\alpha = 1$, $\beta = -1/2$, and $\gamma = -3/2$.
\end{itemize}

[twin meteor]

dgh

[placid osprey]

@maiden iron Is this right?

Just to recap!

Probably only change the g dimension? ([L][T]^-2)

And I think you mentioned ssomoething about SI-units earliier, right?

[vocal sleet]

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[ruby rivet]

Hii can someone verify my answer for this problem?

[ruby rivet]

I got 5/pi inches, but many other people have gotten different answers.

Here's my solution:

25 = pi*d_2

d_2 = 25/pi

20 = pi*d_1

d_1 = 20/pi

Increase = d_2 - d_1

Increase = 25/pi - 20/pi

Increase = 5/pi inches

Help!?!?!

[dark kiln]

5/pi sure sounds right

also it's a famous thing about a string wrapped around earth

if you have a string wrapped around earth and you want to increase its length so that it's 1 meter in the air

you would just need 2m * pi more string

[ruby rivet]

Exactly but this fucking math educator says that it is 25/pi 😭

that's her working

literally makes no sense

[dark kiln]

cool

[vocal sleet]

@ruby rivet Has your question been resolved?

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[cinder temple]

ello

[buoyant dagger]

Hi

[cinder temple]

ok so, is there a way to reprensent the plans

[buoyant dagger]

Yes by using a variable

[cinder temple]

how about two variables

[buoyant dagger]

Okay

X and Y

[cinder temple]

right

so write an equation for the memberships like y=mx+b

where m is slope and b is interceopt

intercept*

[buoyant dagger]

y = 9 + 1x
and
y = 27x

[cinder temple]

ok

so solve it

y=y

[buoyant dagger]

And then you equal both?

[cinder temple]

so 27x=9+x

yea

[buoyant dagger]

x = 9/26

[cinder temple]

uh

that's right

but now i'm second guessing myself since it's a fraction

wait

y=27x is wrong i think

[buoyant dagger]

x + 27y?

[cinder temple]

no

it's unlimited discs so x doen't matter

y=27

[buoyant dagger]

So no variable?

But It says 27 dollars per month

[cinder temple]

yes but

it's only in the timespan of one month anyways

[buoyant dagger]

Okay

9 + 1x = 27

[cinder temple]

yes

[buoyant dagger]

X = 18

[cinder temple]

ya

and what's y?

[buoyant dagger]

Each option costs?

[cinder temple]

um

well you only need x and y

[buoyant dagger]

Oh wait

27?

[cinder temple]

ye

so the first box is...?

[buoyant dagger]

27?

[cinder temple]

yes!!

and what's the second box??

[buoyant dagger]

18

Okay thank you

[cinder temple]

no problem!

[buoyant dagger]

.close

[cinder temple]

adios

[vocal sleet]

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[molten notch]

how to solve?

and also wtf is a -45˚

how tf does it have a negative

wtf

[vast shale]

its probably a hyphen, not a negative sign

[molten notch]

what's a hypen

[vast shale]

so the angles of the triangle are 45, 45 and 90

[molten notch]

wuz a hypen

[vast shale]

,w hyphen

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Toby#2275

[molten notch]

i see

soooo

wait let me draw

@vast shale

right?

[vast shale]

yup

[molten notch]

WWW

so how do i solve for the hypo

[vast shale]

pythagoras

[molten notch]

can you teach me that

i learn quick

[twin meteor]

Toby

[molten notch]

wow

so is C always the hypo?

[vast shale]

yes

[molten notch]

ok so B and A are the legs

which of B and A is the adjacent

not shown right?

so they're both just legs?

[vast shale]

it doesn't matter since a^2+b^2=b^2+a^2

[vast shale]

yeah

[molten notch]

i see so to solve for the hypo

[vast shale]

[molten notch]

i just do 5^2 + 5^2?

wait then what if one of the legs are missing?

[vast shale]

then squareroot

[molten notch]

squaroot?

how

[vast shale]

if you have b and c, then you can rearange for a

[vast shale]

it's c^2 on the right

so you need to squareroot to get c

[molten notch]

wait so lets say i have b and c

i do a^2 = c^2 - b^2?

[vast shale]

yup

[molten notch]

ohh

nice thanks

.close

[vocal sleet]

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[ember tendon]

prove that for every non-directed connected graph there exists a node that when removed along with its adjacent edges the graph stays connected\
hint: pick a node $x$, consider the node with the maximal distance from an arbitrary node $x$ and use it in the proof

[twin meteor]

metnal

[ember tendon]

im confused as to how the hint even relates to the question..

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[fossil ermine]

I try my best to explain my problem in a good manner to understand cuz i only know the german math vocabulary not the english i can just guess em, my problem is to find the extreme points. in my Paint maths u can see that my extreme points were (3/2 / 135/64) and (-3/2 / 81/64). but in the solution of my book its (0/0),(-4/4)

[fossil ermine]

where is my problem? what did i do wrong?

9. a)

[gleaming sluice]

You only need the 1st derivative

Set it equal to 0 ans you have a quadratic, solve for x, you'll get 2 points, and then plug them to get both x, y pairs

[fossil ermine]

i did on the left

i got 0 and -4

omg

im so stupid

bruhhh

@gleaming sluice thyx

.close

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[leaden lance]

Stuck on grouping

On the 3rd line. How does my teacher take out (a) as a gcf?

<@&286206848099549185>

[sour ermine]

That should be

3a²-9a not 3a²+6b²

3a²-9a
3a(a-3)

[mighty walrus]

^ You have to factor it with same terms

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[carmine lion]

How can I prove that det(M1M2) = det(M1)det(M2)

[carmine lion]

I just want the most simplest answer without numerically proving it.

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[thin vale]

this seems like it has a rather complicated proof

Here's what I was able to find online

https://math.stackexchange.com/questions/60284/how-to-show-that-detab-deta-detb

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[noble hollow]

I'm not sure how to apply the general derivative rule to this function

[boreal remnant]

Can you find the derivative of sqrt(u)

[chilly scroll]

Use the chain rule

[noble hollow]

The rule that we're using for this problem is this one

[chilly scroll]

That’s the chain rule

[noble hollow]

Okay

[chilly scroll]

Square root is equal to the power of 1/2

So you could rewrite it as: (x^2-12x+46)^1/2

[noble hollow]

f(u) is x^2 - 12x +46

?

[chilly scroll]

No f(x) is that thing to the power of 1/2

[noble hollow]

Alright then

[chilly scroll]

So bring down the 1/2

Then subtract the exponent by 1

And multiply it by the derivitive of the interior

So final answer should be: (1/2)(2x-12) (x^2-12x+46)^-1/2

I put the derivative of interior in front cause it’s easier to type it but it could be both

But remember you could still like simplify this further

By using algebra

[noble hollow]

(1/2) ((x^2 - 12x +46)^ -1/2)

[chilly scroll]

Close. You have to multiply it by the derivative of the interior function

Hence the name chain rule

So multiply this answer by the derivative of (x^2-12x+46) and you should be set I think

But then you have to use algebra to simplify it but theoretically it is correct mathematically. Prolly won’t register as right tho cause it prolly will only accept simplified versions

[chilly scroll]

@noble hollow your missing one part to it

[noble hollow]

(1/2) ((x^2 - 12x +46)^ -1/2)(2x-12) ?

[chilly scroll]

Yeah

[noble hollow]

How do I distribute the 1/2 and the -1/2?

Do I multply everything in the parenthesis by 1/2 first then distribute the -1/2?

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[noble hollow]

.close

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[quiet grove]

Hahahaha to find the margin of error you need to calculate the standard deviation, but for which one? The sample proportion or population proportion? The sample proportion, because that’s where we get the value for it from in order to find a bunch of other stuff. In order to find the standard deviation for sample proportion we need to find the sample size if we don’t have it. We know how to find it with the confidence interval. Lower Bound + Upper Bound / 2 = the Sample Proportion. So 0.186 + 0.254 = 0.440. 0.440 / 2 = 0,22. The confidence level is 95% or 1.96 as a normal value (I think maybe as Z-Score, IDFK). But we don’t have the sample size. One we to find margin of error which we are asked to find is by multiplying the critical value by the Sample Size. In order to find the Sample Size we need to known the standard deviation! In order to find the standard deviation you need to know three other formulas! It never ends and I just want to know the answer!!!

[rapid swift]

margin of error of what? the sample mean?

[quiet grove]

95% confidence level. The confidence interval is from 0.186 to 0.254. What is the Sample Proportion and the Margin of Error? Well I can averaging Lower and Upper Bounds of the Conference Interval to find the Sample Proportion, shown as 0.186 + 0.254 = 0.440. There are two numbers here, so 0.440 / 2 = 0.220 or ≈ 0.22. Sample Proportion, denoted as ^p^ or p-hat, = 0.22. Now how do I find the Margin of Error, as well?

[rapid swift]

95% confidence level of what?

the sample mean?

[quiet grove]

Yes!

95% = 1.96

Idk why it just does

That’s what the chart told me

[rapid swift]

are you asking where the 1.96 comes from?

[quiet grove]

NO

Okay look.

We now have ^p^ = 0.22. Confidence Level of 95% is shown as 1.96. The Confidence Interval is from 0.186 to 0.254. With the given information and I don’t know what other formula(s), how do I find the Margin of Error?

[quiet grove]

^^^

I’ve been looking for answers the past three hours!

[quiet grove]

^^^^^

[rapid swift]

isn't the margin of error just the distance from the point estimate to one end of the confidence interval?

[quiet grove]

I’m the wrong person to ask, so I’m really hoping that question is rhetorical.

Is it?

[rapid swift]

I think people use the term "margin of error" a little loosely, so I never got really clear on that

most statistics I did never used that term; we talked about confidence intervals instead

[quiet grove]

Well I’m gonna try this and see what happens.

[rapid swift]

but in the few questions I've seen on this server, that's what they usually mean

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[vast shale]

How do you change unnatural numbers to natural numbers?

[karmic imp]

Just remove un

[shrewd dagger]

exorcise them

[rugged vortex]

Wdym unnatural

Just divide the unnatural number by itself

[vale quartz]

So basically you just

[vast shale]

Hm?

[vast shale]

Ooh okie

Huh

I actaully don’t know guys

[rugged vortex]

What is an unnatural number

Like

[vast shale]

I failed Maths since 3rd grade

[rugged vortex]

Did you mean irrational

[vast shale]

Like decimals, fractions

Not 1, 2, 3, 4

Idk

[rugged vortex]

Oh

I mean they themselves cannot become integers

So you can't

You can input them into an absolute value floor function

[vast shale]

So I just have to leave them at this?

[vast shale]

What’s that. I think I need to do it

[rugged vortex]

Wwair what's the actual issue here

If you can't simplify it further, just leave it as is

[vast shale]

Cause the question said

“To natural numbers”

And my teacher was also like “Hey that’s only the working”

but anyways

I found k but now I need to find t

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[vast shale]

can anyone help?

anyonee?

[trim walrus]

Let them all be equal to k

[vast shale]

ok next