help-26

[stiff egret]

Yea

[golden wigeon]

And luckily, we are also differentiating with respect to u rather than x

$\dv{u} e^u = ?$

[thorny flame]

King Leo

[stiff egret]

e^u still

[golden wigeon]

So heres what you have:
$$\dv{y}{x} = \dv{y}{u} \cdot \dv{u}{x}$$
$$\dv{y}{u} = e^u$$
$$\dv{u}{x} = \ln(2)$$

[thorny flame]

King Leo

[stiff egret]

So the answer is e^(x ln(2)) • ln(2)

?

[golden wigeon]

Correct

[stiff egret]

So if i had a^x would all the answers be e^(x ln(a)) • ln(a)

[wind whale]

close

dy/dx of a^x would be (a^x)(ln(a))

[stiff egret]

So this is wrong

[wispy gulch]

no its correct

its just not simplified

[stiff egret]

Ohh

Nvm yea

[stiff egret]

So this is also right just not simplified

[wispy gulch]

yea

[stiff egret]

K

[wispy gulch]

do uk how to somplify?

[stiff egret]

Umm not really but i understand why this one worked

Because its just inverseing one the steps we did earlier

[wispy gulch]

u use rule 3 and then rule 7

[unkempt basin]

that's an interesting way

i was taught to write y = a^x then lny = xlna but the underlying is the same thing I guess

[wispy gulch]

e^(x ln(a)) • ln(a) = e^( ln(a^x)) • ln(a) = a^x ln(a)

[wispy gulch]

even i gave the same suggestion

[stiff egret]

Hmm

[unkempt basin]

and then i use implicit diff

[wind whale]

might also help to review the chain rule. I know that I learned the chain rule before learning implicit differentiation

[stiff egret]

What's the chain rule

[wind whale]

its this part that King Leo was showing you. Its called the chain rule

basically it helps you find the derivative of composite functions

sometimes also referred to as f(g(x))

[stiff egret]

Ic

[unkempt basin]

do you have a teacher or are you just learning for fun

[stiff egret]

Im trying to test outa calc 1 and now i think i signed up for alot XD

[unkempt basin]

because i find it strange that you're learning d/dx 2^x before knowing what chain rule is

[stiff egret]

Learning for fun

[unkempt basin]

i see

that makes sense then

with a teacher, you'd probably be taught chain rule first

[stiff egret]

Makes sense yea i was just curious on how to deravative exponentials cuz i know those are related

[unkempt basin]

well if you know how to differentiate e^x, e is just a constant

so it's basically the same thing as 2^x but for e^x, lne is just 1

so you don't write that bit

[stiff egret]

Ic yea actually makes a lot of sense

[unkempt basin]

d/dx (a^x) = (a^x)lna

but yea knowing how to do it is a lot better than just memorising it

[stiff egret]

Fiturized are u in highschool

[unkempt basin]

i guess so

im in final year of college but that's the equivalent of high school in most countries

[stiff egret]

Oh ok

[unkempt basin]

in my country, we have two years of college which is the final years of high school for you

[stiff egret]

So like calc 1-2?

[unkempt basin]

yea

i probably know more than that

[stiff egret]

Ic

[unkempt basin]

i think I know content from calc 4

cuz of my education system

in England, we do minimum 3 subjects so we learn more in depth about them

compared to somewhere like the US

[stiff egret]

Yea my school is sorta dumb they teach what could be tought in like a month in a year

[unkempt basin]

😂

[stiff egret]

We have a whole 3 month unit on how to write exponential functions from graphs...

[unkempt basin]

like I forgot the derivative of 2^x, the other day but i could derive it again myself

so id definitely recommend knowing how to derive it

[unkempt basin]

huh..... why?

how many lessons per week?

[stiff egret]

Yea would u have a estimate on how much i would need to learn to test out of a calc 1 class. Because i know all the stuff we should learn like derivative integrals and finding areas under curve and stuff but is that all there is in the year

[stiff egret]

Like 5 hours a week

And idk people are dumb

[unkempt basin]

i have no idea what your situation looks like (im not really understanding the first sentence). also, i have no idea what calc 1 teaches.

im not from the US, assuming you're from there

[stiff egret]

Yeaa

I just want a goal or something to reach on what i should be able to do i understand you aren't from here but do you know a good level to get to at least starting off ig

[unkempt basin]

its probably better to ask someone more familiar with your education system

[stiff egret]

Makes sense

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[leaden totem]

can someone explain how this is wrong

[mint crescent]

What happened to y=

[leaden totem]

ohhhh

lmfao

yeah that makes sense sorry chat thanks

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[cedar wagon]

<@&268886789983436800>

Once again bro

[livid briar]

wait whats wrong with this 😭

[rigid ivy]

He was warned already not to ask troll questions.

[livid briar]

ohh

[half edge]

@neon iron your making me MAD now

[craggy haven]

.close

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[rigid ivy]

love this

[cedar wagon]

Kirby lock in

[livid briar]

tyty

locking in for my stats test 🙏

[cedar wagon]

Just eat the test and you'll get the knowledge of it, that works in ssb

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[vast night]

i need help with g and n

[rigid ivy]

for g, just an intuitive idea is fine

[vast night]

The answer is supposedly 0 but I'm not sure why

[rigid ivy]

there's no good reason

[hoary burrow]

Just follow the graph as it goes to the right

[rigid ivy]

the reason they want you to use is "the height looks like it's approaching zero as x gets bigger"

[cedar wagon]

And n ask you for which x, the graph is under the x axis

[rigid ivy]

It's a "you gotta eyeball it" problem, but there's no definitive way to answer it

[vast night]

but if it were zero, wouldn't the middle portion of the graph not be intercepting the x-axis since it's the horizontal asymptote

[rigid ivy]

the limit toward infinity is zero. Which means "as x keeps growing, y keeps getting closer to zero"

[vast night]

so it doesn't matter if the function passes 0

?

cuz there's an x-intercept of (-1,0)

[hoary burrow]

We're only interested about how the function behaves as it approaches infinity

[rigid ivy]

No. You only care about end-behavior

[vast night]

Yea if the end-behavior to both negative infinity and positive is 0, than there's a horizontal asymptote of 0 right?

[rigid ivy]

correct

But not just zero. Any number. If either goes to, say 1, 2, pi, e, whatever. Then you will have a horizontal asymptote at that same value

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[solemn steeple]

hi i think i did something wrong in this process… i dont think i cant divde 180 by -4 lol

[broken niche]

What are you trying to do, convert to degrees?

[solemn steeple]

yes

[raven field]

you should multiply by 180/pi not equate

[broken niche]

Specifically 1 radian equals 180/pi degrees so you should substitute radians with (180/pi) degrees

[solemn steeple]

okwait im solving it out

would it equal -900 degrees??

[raven field]

can you show how you got that

[raven field]

at least simplify 180 with the 3 would be easier to multiply

[solemn steeple]

is it right

??

[raven field]

no

[solemn steeple]

then how do i go about it

[raven field]

your process is right it's just you messed up multiplying

[solemn steeple]

what

where did i mess up

you cancel out the pis

[raven field]

yea but 180*(4/3) doesnt give 900 tho

[solemn steeple]

no i did

180*-4

then divded by 3

im so confused im so sorry

[raven field]

nah it's ok

[raven field]

gives -720
divide by 3 you get -240

[solemn steeple]

what

then i think my calculators wrong

😭

[raven field]

damn

[solemn steeple]

thank you

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[placid anvil]

PreCalculus Polar Graphs, Polar Equations, Polar Coordinates, Rectangular Equations stuff

[placid anvil]

I got a test tomorrow and quite a lot to review on this unit

So far all I know is x = r cos theta, y = r sin theta, r^2 = x^2 + y^2, and tan theta = y/x. I also know that Polar Coordinates include (r, theta). Polar Equations are such that have r and theta as variables. Rectangular equations are ones that have x y and integers. To convert rectangular equation to a polar equation, you replace x with r cos theta and y with r sin theta. When converting rectangular to polar, you substite x and y for its equals and then solve for r. I'm not sure how to do it in vice versa though. This is just the first confusion and I'll go over the rest once I receive the initial assistance

Example question is Conver the equation from polar to rectangular form. 1. r = 7 (I think you solve this by squaring both sides to get r^2 = x^2 + y^2). 2. theta = 2pi/3 (I think you solve this by substite theta in tan theta = y/x to find y and x and then use y and x to find r^2 to make the equation r^2 = y^2 + x^2. Ill send more later

[placid anvil]

Please correct if any of my understand is incorrect or unclear in anyway.

[acoustic pecan]

your ideas sound good to me, except the last phrase, theres no need to do that bit after the tan

[raven field]

by "not sure I can do it in vice versa" you mean when converting from r,theta to x,y or x,y to r,theta ?

[placid anvil]

when converting polar to rectangular form

<@&286206848099549185>

😭

[raven field]

it's better to have an r^2 term like let's say for example
r=3sin(t) [using t instead of theta]
multiply by r for both sides
r^2=3rsin(t)
remember that y=rsin(t) and r^2=x^2+y^2
x^2+y^2=3y

[placid anvil]

I see

Can you explain this question

r = 7

Polar to Rectangular

[acoustic pecan]

your idea for how to do that one was fine

same with the second one except that last unnecessary bit

[raven field]

there is different ways to think of it
like a circle with raidus 7
or square both sides and sub for r^2

[placid anvil]

Oh so I was right

Yea so for the second one though

We're not given an r value

We just given t = 2pi/3

[raven field]

yea just sub in tan(t)=y/x

[placid anvil]

so tan(2pi/3) = y/x

• sqrt 3

-sqrt3*

[raven field]

yep

[placid anvil]

so y = sqrt 3 , x = -1?

considering QII

[raven field]

you are solving for the equation so the result should be an equation tho

[raven field]

multiply both sides by x

[placid anvil]

but we dont know r

[placid anvil]

why?

tan 2 pi/3 = -sqrt3

[acoustic pecan]

you dont need to know r, has nothing to do with this

[placid anvil]

how can I write a rectangular equation without r

[acoustic pecan]

youre writing an equation in terms of x and y

why would you need r?

thats for a polar equation

[placid anvil]

isnt that what x and y =

[acoustic pecan]

sure, but you dont need to find r if youre only looking for x and y, which you have right here

[placid anvil]

To my understanding a polar equation would be smt like r = 2 sin theta and rectangular would be like x + y = (?)

y = -sqrt3*x then

[acoustic pecan]

a rectangular equation is just in terms of x and y
a polar equation is just in terms of r and theta

[acoustic pecan]

yeah, thats it

[placid anvil]

so thats the final answer?

[acoustic pecan]

yup

[placid anvil]

mb

my teacher has a notes sheet

and it has like an explanation

but all the questions are empty

ok so thats settled

ill get my follow up questoins

r = 2 sin theta

how do i solve that

[acoustic pecan]

multiply both sides by r i would say

[placid anvil]

r^2 = 2 sin theta *r

[acoustic pecan]

then you have an r^2 and an rsint

[placid anvil]

2y = r^2

woah

so x = 0

[acoustic pecan]

back up

where did that come from

[placid anvil]

oh nvm

[acoustic pecan]

r^2=x^2+y^2

[placid anvil]

well

yea so if r^2 = 2y + 0

y = sqrt(2y)

[acoustic pecan]

no

nothing here says that x=0

follow the maths

[placid anvil]

r^2 = 2y

[acoustic pecan]

r^2=x^2+y^2

so its just x^2+y^2=2y

[placid anvil]

oh

so for my rectangular form answer

would I write r^2 = 2y

or x^2 + y^2 = 2y

[acoustic pecan]

this

[placid anvil]

ok

[acoustic pecan]

^

[placid anvil]

r = -6 cos theta

r^2 = -6 cos theta r

r^2 = -6x

y^2 + x^2 = -6x

?

[acoustic pecan]

indeed so

[placid anvil]

wow

maths

r = -2 csc theta

r = -2/sin theta

r sin theta = -2

y = -2

uh

[acoustic pecan]

yup

[placid anvil]

is that it?

[acoustic pecan]

thats it

[placid anvil]

so thats my final "rectangular form" answer

[acoustic pecan]

it is

[placid anvil]

what is rectangular form even mean

[acoustic pecan]

i call it cartesian

but its just in terms of x and y

thats all it is

[placid anvil]

so

from polar equation

do you get polar coordinates from that

(r, theta)

[acoustic pecan]

yeah
like with some y=f(x) i have points (x,y)
for a r=g(theta) i can have some (r, theta)

or etc

[placid anvil]

so like r = -2 sin theta

it would be (-2 sin theta, theta)

[acoustic pecan]

exactly

[placid anvil]

not that i would ever have to find that

alr

so polar equations only graph circles?

[acoustic pecan]

no, they graph anything you can graph in the x-y plane really

though theyre more convenient for some things

[placid anvil]

in the form of r = (?) sin/cos theta

[acoustic pecan]

that would be a circle i believe

[placid anvil]

alr

so say r = 4 cos theta

r^2 = 4 r cos theta

x^2 + y ^2 = 4x

to get a circle equation in (x-h) ^2 + (y-k)^2 = radius^2 form

[placid anvil]

On the notes sheet from here it goes to (x-2)^2 + y^2 = 4

So from our original equation we derive a circle radius, and a vertex

[acoustic pecan]

yeah thats completing the square, though i dont think its really necessary

it would be a center rather than a vertex

[placid anvil]

yea

well

idk

it says vertex on my sheet but i reallly have no clue

so

is there like an equation

like how did they go from x^2 +y^2 = 4x to (x-2)^2 + (y^2) = 4

like how'd they know to add 4 by both sides to make a perfect square

which would be rectangular form

[acoustic pecan]

x^2-4x = [x^2-4x+4] -4 (because we want (x-2)^2
=(x-2)^2-4

then move the 4 over

[placid anvil]

no likee

they had x^2 -4x + y^2 = 0

so disregarding the y^2

we have a^2 + bx = 0

how'd they know to add c to both sides

to make a perfect square

[acoustic pecan]

i just did this right here

[placid anvil]

i dont understand

nvm i got it

[acoustic pecan]

in a perfect square we have (x+a)^2=x^2+2ax+a^2

so here in x^2-4x 2a=-4 so a=-2

then a^2=4

[placid anvil]

i was looking for (b/2)^2 = c

So in the equations r = a cos theta or r = a sin theta, the shapes are only circles right

[acoustic pecan]

for that specific form yeah

if i do this for example:

,w graph r=10sin(15theta)

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[acoustic pecan]

it gets a bit different

[placid anvil]

limacon

or

no its just a rose

a limacon would be r = a+-b cos theta / sin theta

since n is odd in ur equation, it'll always have exact n # of petals

in r = a cos( n theta)

or sin

[acoustic pecan]

yeah, if its even youll get double

[placid anvil]

alr so in equations like r^2 = 4 sin 2theta

its 4 petals

[acoustic pecan]

thats different

[placid anvil]

oh

we square root?

or

nvm

well r= +- sqrt 4 sin 2 theta

so idk what'd u do from there

to find ur shape

[acoustic pecan]

in that case i think you might get 2

[placid anvil]

lemniscates

r u a math major

[acoustic pecan]

maths and physics

but i deal with more abstract stuff these days

[placid anvil]

does this server pay u

to like tutor

[acoustic pecan]

nope

[placid anvil]

u get like volunteer hours or smt?

[acoustic pecan]

more like i just pop on whenever i feel like it

same with everyone else

[placid anvil]

for the lemniscates it says petal length = sqare root of coeff

what does that mean

sqrt of a in r^2 = a^2 cos ntheta

[acoustic pecan]

the coefficent here would be the 4

[agile harness]

nope just a pastime lol

[placid anvil]

so in this equation

well

yea you would sqrt a^2 in the form of r^2 = a^2 sin/cos ntheta

to find a

or

yea might things wrong

it says r^2 = a^2 cos ntheta

but I think a lemniscate is r^2 = a cosntheta

where sqrt a = the length

[acoustic pecan]

theyre the same thing

[placid anvil]

yeah

[acoustic pecan]

you just changed what 'a' was

[placid anvil]

i realize that now

it just writes the equation weird

alr so r = 4 + 3 cos theta

would be a convex limacon?

or a dimpled limacon

,w r = 4 +3(costheta)

[acoustic pecan]

likely a dimple

,w graph r = 4 +3cos(theta)

[placid anvil]

so r^2 = 16 cos 2 theta is a lemniscate

r = 4 sin 3theta is a rose

r = sin theta is a limacon

or cardiod

,w graph r^2=16cos(2theta)

,w graph r = sin(theta)

,w graph r = sin(theta)

[thorny flame]

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[placid anvil]

so

r = a sin theta is a top or bottom quadrantal circle

r = a cos theta is a left or right quadrantal circle

r = a + b cos theta where a/b = 1 is a cardiod

a/b <1 is a Limacon inner loop

2 > a/b > 1 is dimpled

and a/b >= 2 is a convex

then r = a cos/sin ntheta is a rose with n odd petals or 2n even petals

then r^2 = a sin/cos (ntheta) is a lemniscate where petal length is sqrt of coef

@acoustic pecan

and for graphing r^2, we do have to put +- cooridinates for r ?

[acoustic pecan]

this one im not certain, but likely yes
i never really worked with a negative r back when i did this but ive seen its allowed so probably

but overall sounds good

[placid anvil]

alr last unit

the equation z = 3+4i would be a

ik (a,b) are points in z = a+bi

but what would the equation be

[acoustic pecan]

you mean in the complex plane?

it would just be a point, thats a number

[placid anvil]

idk i have 6.5 Complex Nimbers in Polar Forms ; DeMovire's Theorem as the name of the unit

and it say example 1

plot each complex number in the complex plane

and #1 is a = 3+4i

where (3,4) are points

[acoustic pecan]

yup, thats all it is

[placid anvil]

alr ill prob just review the rest tmr on my own cuz its just demoivres theorem which is just memorization

thanks man for the help

[acoustic pecan]

no worries

[placid anvil]

i sent u a friend request so if ur fine with it i can send u a dm for future questions

thanks anyways tho

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[solid summit]

how do i find the lcm

[sweet shard]

Factor each one of them

[topaz sinew]

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[cerulean sun]

not really a math equation, but i do need help with something physics related. we're tasked to make a leyden jar using a jar, saline water, nail, aluminum foil and electrical tape. however it wouldn't work so i tried to following

• remade the jar
• switched my cap to plastic
• shrunk the area that the foil is occupying
• sanded the pvc pipe
• i even made an electroscope to see if my pvc pipe rubbing is working, and it is actually
  but it still wouldn't work. im honestly stumped. can anyone help? ill try anything as long as it's within the instructions of our teachers T-T

[cerulean sun]

<@&286206848099549185>

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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.

[sweet shard]

#old-network for physics server

#old-network message

[cerulean sun]

Ah thank youu

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[restive plank]

I got D, wanted to verify

[pseudo jetty]

lgtm 👍

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[versed locust]

as = a^2 + a + 1/ a + 1

[versed locust]

What happens when i try move the a on the left to the right

like what does it do?

[jovial goblet]

Is that an s?

[versed locust]

ye?

[jovial goblet]

So u want to divide both sides by a

[versed locust]

but how would i divide the right side

like what does diving by a on right side do? does it litterly make it another fraction or

[jovial goblet]

Divide each individual term by a

What are you solving for though? Like s in terms of a?

[versed locust]

yes

[jovial goblet]

Yeah so just divide each individual term by a and then put it all over a common denominator

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[brisk ice]

how do i show this is 1/sqrt(e)

[sinful flint]

notice that

$\left(1+\frac1{x}\right)^{x^2}=\left(\left(1+\frac1{x}\right)^x\right)^x$

[brisk ice]

yea

should it be to the power of x again tho?

[sinful flint]

yes, mb

[thorny flame]

Bonk

[sinful flint]

do you know the definition of e?

[brisk ice]

no but i do know the limit for that

yea i get e^x

but then i just get 1

,w e^-x*(((1+1/x)^x^2)) limit when x approaches infinity

[thorny flame]

Results provided by WolframAlpha

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Requested by oakdt

[brisk ice]

idk what i’m ding wrong

doing

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[brisk ice]

<@&286206848099549185>

[broken niche]

You cannot combine them because when you take the limit of an expression you have to consider the entire expression at once rather than taking the limit of small parts separately

What you have shown though is the right side of the product goes to infinity and clearly the left side goes to 0 so you have 0*inf aka an indeterminate form

[brisk ice]

yea but how do i go about solving it

[vernal shard]

Try e^(lim ln(…))

That is,
[ e^{ \lim_{x\to\infty} \ln \left( e^{-x} \left( 1+\frac1x \right)^{x^2} \right) } ]

[thorny flame]

shsgd

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[neon iron]

To those who joined the IMO, at what age/grade did you first start practicing for the IMO? And how did you practice?

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[neon iron]

<@&286206848099549185>

[neon iron]

<@&286206848099549185>

[sweet shard]

This isn't a math question

Don't use help channels for surveys

[fallow heart]

And also, is that so urgent that you've pinged helpers three times? 😅

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[feral bluff]

!nospam

[feral bluff]

.close

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[lofty gate]

Help me pls, guys. I don't understand why it turned out to be 60

[restive inlet]

60° is acute and
cos(60°) = 1/2

[timber crystal]

That diagram is an insanely bad representation of what the question is describing

[lofty gate]

so the angle is 60?

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[pearl swift]

hey

[pearl swift]

need help with bearing

how does bearing work

.rename

[sweet shard]

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

[pearl swift]

ok thx

.close

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[white ridge]

hello

.reopen

uh

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[subtle flame]

hmm

[subtle flame]

AD=BC=AS=SB=DC
need to calculate the degree of angle ASB
8th grade olympiad.

[wispy gulch]

your attempt?

[subtle flame]

i tried, i did nothing, i am bad at geometry

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@subtle flame Has your question been resolved?

[pallid zenith]

draw the line AB and note similar triangles

in addition, we can say that ABCD is isosceles since the diagonals are equal

[placid merlin]

label the angle theta, and try to solve for other angles using isosceles triangles and the fact that the angles of a triangle sum to 180

[pallid zenith]

wait...

all of the lines are equal?

it’s isosceles bash then

[timber crystal]

A lot of primary-secondary Olympiads are just brute forcing algebra

[subtle flame]

of red, yes

can you write the full solution please?

[placid merlin]

how about we go through it

lets start with SAD

actually, lets go with ADS

we assume the beginning angle is theta

what is the angle ADS

[placid merlin]

by the beginning angle, i mean ASB

traingle ADS is an isosceles triangle, so we can find what angle ADS is

[subtle flame]

and what angle is?

[timber crystal]

We labeled ASB as theta. If the triangle is isosceles, then what would be angle ADS?

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[rigid ivy]

you should find the range of f for each piecewise part

that will help you decide

[narrow karma]

oh, got it, thanks!

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[sick dome]

Hi this is a question on induction

[sick dome]

I'm not really confused with the induction part, but rather how the formula on the right hand side is used

So when I see this question, I know base case = 1

On the left hand side I get $\frac{x^1-y^1}{x-y} = 1$

[thorny flame]

Minnie

[sick dome]

Since the question tells me to prove that a) is true its pretty obvious the right hand side should be 1 as well

But I don't get how to uhh really use that series looking thing on the right hand side

My first reaction is $x^{1-1} + x^{1-2} \cdot y + \ldots + x\cdot y^{1-2} + y^{1-1}$

[thorny flame]

Minnie

[sick dome]

Which is obviously not 1

Then my second reaction is $x^{1-1} + y^{1-1} = 2 \neq 1$

[thorny flame]

Minnie

[sick dome]

So is the base case just $x^{1-1} = 1$??

[thorny flame]

Minnie

[keen raptor]

how many terms should there be in the sum when n=1

[sick dome]

I don't know

thats what im confused with

[keen raptor]

How many should there be for a general n?

[sick dome]

n terms

[keen raptor]

Right

So what about n=1

[sick dome]

does that mean i only take the first term??

which would be

[sick dome]

this

as base case

[keen raptor]

Yes

and that checks out since (x-y)/(x-y) = 1

[sick dome]

thats what i assumed because i knew it had to result in 1

but can i conclude that generally, for n= x terms, I calculate using the first x terms in the formula?

[keen raptor]

$x^{1-1} = y^{1-1}=1$

[thorny flame]

Dreyuk

[keen raptor]

calculate using the first n terms yes

[sick dome]

okay thank u

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[wooden osprey]

note that moon gravity = 1/6 earth gravity

it is to me, though the problem should have mentioned it

[loud oasis]

you have to either know the local value of g on the moon, or the mass and radius of the moon

[agile harness]

look in the back of the book

it will have the value of g on the moon

radius of the moon

mass of the moon etc

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[dapper frigate]

how do i find the area?

I have done 4-1
and then 3 as the width
1/2(3)(30000)

[sinful flint]

!occupied

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[sinful flint]

!ss

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[rigid prawn]

uh

oops

can someone please help me with 23

so what i did was i turned a=logb of a^b

then i plugged in to the thing

i get logb of logb of a^b

is the answer just a^b or did i do something wrong

[half edge]

$b^a = a^b$

[half edge]

why dont u just take log_b of both sides

[rigid prawn]

i did

so like a=logb of a^b

[rigid prawn]

^

[half edge]

not seeing how u got that

[rigid prawn]

wait what

[half edge]

oh

well

[rigid prawn]

oh i think i wrote it weirdly

like logsubscriptb

then a^b

[half edge]

$a = \log_b {a^b}$

[half edge]

this is what u got?

[rigid prawn]

yeah

[half edge]

ok hmm

now what should you do

[rigid prawn]

i like plugged it in to the logb a thing

but i dont know if thats what im supposed to do

[half edge]

why dont u just use the log exponent rule

[rigid prawn]

like b*logb a^b?

[half edge]

yes

now what

[rigid prawn]

plug it in?

[half edge]

$a = b\log_b a$

[half edge]

what are we trying to solve for @rigid prawn ?

[rigid prawn]

logb a

[half edge]

do u see that anywhere here

[rigid prawn]

oh yeah

[half edge]

so what should you do

[rigid prawn]

uhh

a/b?

so like

plug in x

and then divide?

[half edge]

we're on 23

right?

[rigid prawn]

oops yeah

but like

i would substitute the logb of a

for a variable

like x

[half edge]

whats wrong with this

[rigid prawn]

sorry im not really sure

[half edge]

wdym lol? im pretty sure thats the answer

[rigid prawn]

oh ok

[half edge]

why are u confused? did u expect it to be something else

[rigid prawn]

no i just thought like your response implied that it was incorrect

my fault

thanks a ton though you helped me a lot

[half edge]

np

[rigid prawn]

cya

.close

[half edge]

cya

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[sick dome]

does e) imply base case is n= 2?

even tho the top part says for all positive integers n? {1,2, ... }

[rigid ivy]

I suppose so

They couldv'e written the problem better

[sick dome]

yeah...

[sick dome]

thank u for confirming!

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[rigid ivy]

Oh, hey @sick dome, did you solve your even-odd problem?

[sick dome]

I'm not sure if I did it right but I tried both the induction way and just the long division algorithm way

I'm seeing my prof tmr so I'll ask him about it

[rigid ivy]

Yeah, I played with the problem a bit last night. I understand it better now, at least

[rigid ivy]

[sick dome]

oooh

true math euthusiast

[rigid ivy]

I used trichotomy of numbers. If $a\ne b$, then $a<b$ or $a>b$.

[thorny flame]

SWR

[rigid ivy]

I made a set of all $k$ such that $p>2k$, then asserted it had a maximum element.

[thorny flame]

SWR

[rigid ivy]

The proof was simple after that

[sick dome]

I believe the knowledge here is out of my reach because I have no clue what is happening aha

[rigid ivy]

It's hard to prove, because you have to know what you are allowed to know

[sick dome]

is there some form of contradiction you used?

[sick dome]

you are right

Feel free to ping me after you're done typing - I'll be finalizing a few more proofs before bed time

but I will definitely check back

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[rigid ivy]

I basically did this: If $p$ is not odd, then $p\ne 2k$ for every $k\in\bZ$. If $p\ne 2k$, then either $p<2k$ or $p>2k$ must be true. Consider the set of all $k\in\bZ$ such that $p>2k$. This set must have a maximum element since $2k$ is bound above by $p$ (this part is a bit tricky to prove). Let $m$ be this largest element. Then $p\le2(m+1)$. But it must be $p\ne2(m+1)$ or else $p$ would be even. So, $2m<p<2m+2$. Meaning $p$ has to be equal to $2m+1$

[thorny flame]

SWR

[sick dome]

WHOAAA

[rigid ivy]

@sick dome I think that's the strat you want, but there are a certain number of facts I used there where I am not sure if you are allowed to use them in your proof.

[sick dome]

wait thats so smart

So you bounded it between two even numbers

which has to be an odd number

[rigid ivy]

I actually was curious about this problem before you had asked about it. So seeing it get brough up made me even more curious

[sick dome]

Ah

I feel like I would have to use induction within your proof

[rigid ivy]

yeah. I had to realize that them not being equal means one must be less than the other. Then I thought about the maximum such element. But proving that a maximum exists, I think, is the hardest part

[sick dome]

First show there is some finite set

then induction to show it has a maximal

[rigid ivy]

If you can use induction, then you can prove a maximum k exists. If you cannot use induction, then you must assume as true that a maximum k exists.

Induction, and bounded subsets of integers having min/max elements are equivalent statements. One can prove the other, but you cannot prove either independently.

(Well, you can if you fall into set theory and use the axiom of induction, but then you still need to take some inductive principle as axiomatically true)

[sick dome]

Ah

I feel like for the first week of proofs class the professor definitely does not expect this much

[rigid ivy]

In that regard, you should def ask your professor. It would be helpful to know which, in your class, is the axiom and which follows as proof: induction, or bounded integer subset

[sick dome]

I will ask him tomorrow

[rigid ivy]

probably not

[sick dome]

ahaha

[rigid ivy]

It's a hard proof imo.

Not hard .. but hard

[sick dome]

teeddious

and looong

I think that is how I would think of it

[rigid ivy]

it's on the short side compared to what some proofs can look like

[sick dome]

definitely

[rigid ivy]

long for a first-timer though, absolutely

[sick dome]

Anyways I must return to my remaining two questions

[rigid ivy]

Definitely in the high-tier for new students

sure sure

take care

.close

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[sick dome]

you too

[topaz sinew]

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[elfin parcel]

Find x (ping when respond please)

[elfin parcel]

I tried to cube both sides but I got stuck on that

[thorny elbow]

yea you wouldnt cube in this case cause its a sum

[elfin parcel]

What am I supposed to do then?

[thorny elbow]

no clue brother

i just know cubing aint it

maybe thats wrong too lol

lemme think about it

how far did u learn

did you learn calculus yet?

[elfin parcel]

Nope

[thorny elbow]

how far are you then

itll help when i try to see what method to use

[elfin parcel]

9th grade

[thorny elbow]

ye whats the topic you learned for these questions

[thorny elbow]

^^

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[olive charm]

I need help understanding how to solve this:

[coarse herald]

!status

[topaz sinew]

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

[olive charm]

step 1

[coarse herald]

can you calculate the length of the segment DE

[olive charm]

7cm?

[coarse herald]

yes

and now what kind of triangles are DEC and DEA

[olive charm]

right triangles

[coarse herald]

yes, so lets consider one of the triangles eg: DEC, DE = 7cm, DC = 9cm so how can you find out EC?

[olive charm]

5.7?

[coarse herald]

yeah well idk the value but you used pythagoras i assume?

[olive charm]

yes

[coarse herald]

now can you do the same for DEA and find out AE?

[olive charm]

3.9

[coarse herald]

great and AC is just?

[olive charm]

9.6

[coarse herald]

are you sure

[olive charm]

uh

yes

[coarse herald]

oh you chaned 5.6

yeah now it seems correct

[olive charm]

oh ok

do you mind helping me with this one too?

[coarse herald]

sure but you need a bit of visualisation to see this

[olive charm]

ok

[coarse herald]

you mind marking the vertices?

[olive charm]

i forgot what vertices are

[coarse herald]

nice

the corners basically

[olive charm]

oh ok

8 vertices

[coarse herald]

no like can you mark them as abcdefgh

[olive charm]

oop my bad

uh then im not sure

[coarse herald]

wait ill do it

[olive charm]

ok...

[coarse herald]

here

now what is the value of AC

[olive charm]

25?

[coarse herald]

???

[olive charm]

uh...could you maybe explain what you mean...

im a little stupid

[coarse herald]

you understand the labelling of the vertices right

[olive charm]

yes

[coarse herald]

ok

now forget efgh

and focus just on abcd

like look at the figure from the top if you understand better

now what is AC

[olive charm]

sorry i still dont get it

[coarse herald]

okay wait

ill draw it better

do you agree that this is the top view?

[olive charm]

yes

[coarse herald]

now what is ac

[olive charm]

ohhh 12.5?

[coarse herald]

??

[olive charm]

like half as in half the square?

[olive charm]

or just the line?

[coarse herald]

where did the half come from

[olive charm]

i dont know, im confused

[coarse herald]

no the length of the line segment AC

[olive charm]

ohh ok

[coarse herald]

like you did in the previous question for the right triangles

[olive charm]

im not sure how to figure that out...

[coarse herald]

pythagoras?

[olive charm]

right

[coarse herald]

💀

[olive charm]

sorry!

7.1

???

[coarse herald]

yess

now wait let me label it in the originial question

do you agree that the figure is like this now

[olive charm]

yes

[coarse herald]

and what is the value of angle ACG

[olive charm]

56.6???

[coarse herald]

what 😭

[olive charm]

im very confused...

[coarse herald]

can you visualise the triangle acg

[olive charm]

yeah

[coarse herald]

and ac is?

[olive charm]

7.1

[coarse herald]

yes

cg is?

[olive charm]

8?

[coarse herald]

its given in the figure look closely

yeah!

ok so now you have 2 sides of the triangle acg

and what is angle acg

[olive charm]

15.1?

hello?

[coarse herald]

is acg like this

you have to look from a different perspective

[olive charm]

oh ok

[coarse herald]

its not easy to like see

so now can you tell the angle acg

[olive charm]

yes

15.1

[coarse herald]

how

[olive charm]

well thats what i got from 8 and 7.1

u never answered after that

[coarse herald]

i asked the angle not the length

[olive charm]

oh my bad

[coarse herald]

you used pythagoras meaning its 90?

[olive charm]

yes

[coarse herald]

yeah well then its correct

but you cant just use pythagoras beceause you need to first prove that acg is a right angled triangle

[olive charm]

how are you supposed to prove that? do you have to do that 180 degress thing?

[coarse herald]

no thats what ive been trying to ask you

whats the measure of angle acg

[olive charm]

i dont get it still...

[coarse herald]

its 90

thats why you were able to use pythagoras

but its fine if you dont understand this is just the distance from the origin of a point in 3d

you basically use pythagoras twice first to find AC then to find AG which is the length required

[olive charm]

ok...

[coarse herald]

<@&286206848099549185> if someone can help her better please do

[minor harness]

So do u know that the length of the longest diagonal is √l^2+b^2+h^2

@olive charm

[olive charm]

huh??? i dont get that at all

[minor harness]

Ok so this is the formula for finding the length of the diagonal of a three dimensional shape

If u use this

And take l as 5 b as 5 and h as 8

So the answer would be like √114

Or approx 10.67

[olive charm]

what?

[olive charm]

.

[minor harness]

?

[olive charm]

im lost from here

[minor harness]

Why?

[olive charm]

.

[minor harness]

Send the question again

This one

[olive charm]

ok

[minor harness]

Radical is underroot u know that right?

[olive charm]

nope

now i do

[minor harness]

Ok then u know

Now

So

[coarse herald]

i tried deriving that for her but shes getting stuck while calculating AG

[olive charm]

why are there 2 otherpeople typing?

[minor harness]

See the 5 in side is length

The other 5 inch side is breadth

And the 8inch side is height

[olive charm]

ok

[minor harness]

So using the formula √l^2+b^2+h^2

Can u find out the answer and leave it underroot?

[olive charm]

ill try

[minor harness]

Underoot symbol is there in the beginning

[olive charm]

√144

[minor harness]

How

5^2

=?

[olive charm]

i just did what i understood

[minor harness]

5^2=?

[olive charm]

8x8, 5x5, 5x5

[olive charm]

25

[minor harness]

The other 5^2?

[olive charm]

25

[minor harness]

Then the 8^2

?

[olive charm]

64

[minor harness]

25+25?

[olive charm]

50

[minor harness]

50+64?

[olive charm]

114

[minor harness]

There u are

[olive charm]

oh

[minor harness]

But it's underroot

[olive charm]

i wrote the wrong answer

[minor harness]

Bye gtg

[olive charm]

in the beginning

[minor harness]

Hope u understand

[olive charm]

thank you

<@&286206848099549185> I need help with this problem...

[sharp shard]

First you shall find the length of the side of the triangle that has the other side as 5 and 4

[olive charm]

6.4

???

[sharp shard]

Yes

Now we have the length of the bottom side of the quadrilateral

[sharp shard]

Now find the right side

[olive charm]

is the answer overall 4.8?

[sharp shard]

If you're talking about value of x I'll have to check

[olive charm]

im talking about the value of x

[sharp shard]

Wait a min

[olive charm]

ok

[sharp shard]

Yes

You're correct

[olive charm]

oh ok

[sharp shard]

Good job

[olive charm]

yay

i have like 6 questions left

is that ok?

[sharp shard]

Bring them on

[olive charm]

ok

also what is radical form?

[sharp shard]

Like it's simplest form

Let's say you have √50 you write it as 5√2

[olive charm]

oh ok

[sharp shard]

Here

[sharp shard]

We need to take this triangle

[olive charm]

ok

so then we would subtract the 1

from the 5

[sharp shard]

Exactly good

And the bottom length will stay the same

[olive charm]

4^2+6^2?

[sharp shard]

Yes

[olive charm]

so the answer is 7.2

?

[sharp shard]

Don't go into decimals

[olive charm]

oh

[sharp shard]

It was √50

[olive charm]

i had √52

[sharp shard]

Oh yes sorry