help-13

[heady mountain]

ah nvm this is wrong

um

<@&286206848099549185> (sorry for ping)

[cedar kiln]

@heady mountain Has your question been resolved?

[heady mountain]

.close

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[jagged stag]

where did the propability of 95% come from'

?

[muted bear]

Empirical rule

[jagged stag]

also why is 106.2 rounded up and but not 133.8?

[astral bay]

well it's "a number of students between 106.2 and 133.8"

106 and 139 are both not between 106.2 and 133.8

[jagged stag]

ohhh thank you

[jagged stag]

ah i got it now thanks

.close

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[cedar kiln]

Please don't occupy multiple help channels.

[cedar kiln]

@median plover Has your question been resolved?

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[vital valley]

It feels like I don't have enough given stuff to plan out masses

[cedar kiln]

@vital valley Has your question been resolved?

[vital valley]

<@&286206848099549185> I gained negative mass while solving this problem, now I can't find my center of gravity

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@vital valley Has your question been resolved?

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[lethal sequoia]

How to derive
a²= b²+c²-2bc cos(A)
b²= a²+c²-2ac cos(B)
c²= a²+b²-2ab cos(C)
Can it derived from single triangle? I checked so many video and they only focused on deriving one side anyone any idea or I'm just dumb

[gritty galleon]

yeah

[muted bear]

You do the same construction but focused on a different side

[gritty galleon]

it looks a lil wrong tho..

[inland ocean]

Yeah

It's a lot wrong

[lethal sequoia]

Like?

[lethal sequoia]

Wait I will fix jt

[hard ember]

For cosA, draw perpendicular from either B or C. For cosB draw perpendicular from either A or C. And similar for CosC

[lethal sequoia]

Like this?

[hard ember]

Yeah

[lethal sequoia]

What's next step?

Next step like this something?

,rotate

[cedar kiln]

@lethal sequoia Has your question been resolved?

[lethal sequoia]

,rotate

[lethal sequoia]

Is this correct?

[analog kelp]

YES

[lethal sequoia]

Oh

[cedar kiln]

@lethal sequoia Has your question been resolved?

[cedar kiln]

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[heavy mortar]

.reopen

[cedar kiln]

✅

[dire geode]

why are you reopening this

[cedar kiln]

@lethal sequoia Has your question been resolved?

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[storm junco]

hey just curious what a vector function with multiple variables is called,
im trying to google some stuff about them but cant find the name

[tranquil maple]

dont quote me on it, quite new here but my best guess would be a "vector transform function", but im sure someone will correct me if im wrong

[short blade]

usually defines a surface

[cunning loom]

Can you be more precise? You might mean a map of sets, or you might want the vector space structure, in which case you mean a linear map.

Like which function do you care about

[storm junco]

Honestly I'm not sure, I'm reviewing parameterized vector functions and thought what if there were more variables, not sure what it would mean geometrically or what it's goal would be

[short blade]

it is still a vector valued function

the domain has just changed from a 1 dimensional space to 2 dimensional

geometrically, it usually defines a surface as opposed to a curve

[storm junco]

Ah ok thats interesting I didn't know that, is there not a name for these specifically though? They're just called vector valued functions?
All I'm trying to do is get search results and learn off of that

[crystal raptor]

are you trying to generalise something like a map from R to R^2 where V(a) = <x(a),y(a)>?

[short blade]

a parametrized surface

it’s a specific kind of vector valued function

[storm junco]

Like I said I have no idea, just thought what if there were more variables ^

[storm junco]

Oh yep I'm finding stuff for that

That works out, thanks!

.close

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[agile linden]

Hi, I tried to solve this problem but I don't understand how to solve it

[agile linden]

these are the tries I did

[crimson sedge]

you integrated it instead of differentiating

[crimson sedge]

partial derivatives are incorrect

[crimson sedge]

again incorrect partial differentiation

[agile linden]

Can you explain it a little more about intergrating it?

[crimson sedge]

what do you mean

you did integration instead of differentiation

it's asking for $\pdv{P}{y}$

[wraith dagger]

kenfps

[crimson sedge]

and $\pdv{Q}{x}$

[wraith dagger]

kenfps

[agile linden]

oh, I read that wrong, I thought you meant to intergrate than differetiation

[agile linden]

I did this for the equation 2x-5y, which would be x^2-5xy+g(y)

but it was wrong

[crimson sedge]

Because you are NOT supposed to integrate

[agile linden]

is differetiation rate of change?

[crimson sedge]

A vector field $F(x, y)$ is conservative iff $\pdv{P}{y} = \pdv{Q}{x}$

[wraith dagger]

kenfps

[crimson sedge]

Do you agree?

[agile linden]

yes

[crimson sedge]

Then why are you integrating the P component of the vector function?

[agile linden]

I was following the notes I took in class

that's why I was a bit confused

[agile linden]

Is it possible you can give me an example?

[cedar kiln]

@agile linden Has your question been resolved?

[agile linden]

<@&286206848099549185>

[potent wave]

Do you know what function is P?

For step 2

[agile linden]

I do not

I thought it was P because of the equation

thats why I chose it

[short blade]

F(x,y) = Pi + Qj

[short blade]

read step 1 here

[potent wave]

Yeay

[short blade]

now in step 2, begin by identifying P and Q

and let us know what you think those are

[agile linden]

wouldn't P be (-5x+12y-4) and Q is (2x-5y)?

[short blade]

F = Pi + Qj

what is i being multiplied by

[agile linden]

by P?

[short blade]

and what about in the problem they gave you

what is in front of the i

[agile linden]

(2x+5y)

[short blade]

so what would be the logical conclusion

what is P

[agile linden]

2x+5y

[short blade]

yea

now do you know how to compute a partial derivative?

[agile linden]

I believe so

might need a refresher

[short blade]

well what is dP/dy

(as in the partial)

[agile linden]

dx/dy

[short blade]

let me ask it again

what is $\pdv{P}{y}$

[wraith dagger]

maximofs

[short blade]

given P = 2x + 5y

[agile linden]

2?

[short blade]

no

take the derivative with respect to y

treating x as a constant

[agile linden]

5

well 5y

[short blade]

yes

no

[agile linden]

just 5?

[short blade]

just 5

what is the derivative of 5x with respect to x

[agile linden]

would that become 5?

[short blade]

yes

probably want to get a refresher on derivatives yeah

[agile linden]

yeah it's been awhile

[short blade]

but yeah anyway, we now know the partial of P with respect to y is 5

now take the partial of Q with respect to x

[agile linden]

would it become 1?

[short blade]

what is Q

[agile linden]

Oh, are we on a different equation?

[short blade]

no

remember F = Pi + Qj

we figured out P = 2x + 5y

what is Q

[agile linden]

(-5x+12y-4)

[short blade]

yes

so take the derivative of that with respect to x

[agile linden]

that why I was wondering if it's a different equation

it would be 5

[short blade]

no

[agile linden]

wait no

it would be 12

[short blade]

no

[agile linden]

wouldn't we treat y as a constant?

[short blade]

we’re taking the derivative of -5x + 12y - 4

with respect to x

[agile linden]

-1/6x^6?

[short blade]

ok let’s take a step back

and go to P again

because we missed something very slight

P is not 2x + 5y

P is 2x - 5y

so now

what is $\pdv{P}{y}$

[wraith dagger]

maximofs

[agile linden]

it's 5

[short blade]

no

what is the derivative of -5x with respect to x

[agile linden]

-5

[short blade]

yes

so what would dP/dy be

[agile linden]

-5

[short blade]

yes

now back to Q

Q is -5x + 12y - 4

what is dQ/dx

[agile linden]

-5

[short blade]

yes

that’s it

[agile linden]

Ooh okay, I need to practice my derivatives more

thank you!

.close

[cedar kiln]

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[echo sinew]

@slate lintel

[echo sinew]

lemme add the qns here one second hahah

i was thinking for time shld i be using displacement?

[obsidian coral]

Please don't ping specific people

[echo sinew]

er she was helping me earlier? and i got their permission to cont and ping em btw!

[slate lintel]

it's fine in this case

we were talking in DMs

[echo sinew]

cus they had to leave due to personal reasons ^^

mhm!

[slate lintel]

I think you have the tools you need for this! you had the velocity vector and you can use that to figure out when the ball lands, and then use that to figure out how far it goes

(yeah ik it lands 18m in this case but like in general for an angle theta and velocity v)

[echo sinew]

mhm

okie lemme try

[echo sinew]

okay so the velocity vector was <14cosθ,14sinθ>

hmmm

[slate lintel]

yeah but i think it wants you to prove it for an arbitrary velocity v first

so just vcos / vsin

[echo sinew]

hmm i dont understand?

[slate lintel]

it wants you to prove this first, right?

[echo sinew]

yep

[slate lintel]

that works for any velocity, not just 14 m/s

[echo sinew]

mhm

ohh so ur saying i shld use v instead of 14 so that it cld be applicable for any t along the travel?

[slate lintel]

yeeee

exactly

[echo sinew]

ahh i see

so it wld be <vcosθ,vsinθ>

from this do i er get r by integrating what i have here?

[echo sinew]

like the position?

[slate lintel]

figure out how long it's in the air

like what time does it hit the ground again

[echo sinew]

time hm

[slate lintel]

$p_y = p_0 + v_0t - gt^2$

[wraith dagger]

Hayley

[slate lintel]

because the only vertical acceleration is gravity

[echo sinew]

ouh

[echo sinew]

may i know what p stands for here?

[slate lintel]

position

i think some books use s for position for some reason

so we want to figure out when the y-position is 0 (when it hits the ground)

[echo sinew]

ooh

[echo sinew]

so this is like py = 0 when its on ground right

[slate lintel]

yeah

[echo sinew]

how exactly issit written out tho? like 0 = 0 + 14t - 9.8t^2?

[slate lintel]

yeah we'll get to that but for now just leave it as v and g

[echo sinew]

its not wrong to write it dis way right?

ouh

[slate lintel]

well it'd be v sin(theta) because we're talking vertical

[echo sinew]

Oh ya cus when its landing v might be diff

but g is always 9.8 right?

[slate lintel]

g is always 9.8 but you can just leave it as g since the target uses g, see it?

[echo sinew]

ooh okie got it!

[slate lintel]

but yes you can put in 0 for the p0 since it starts on the ground

[echo sinew]

oki

and py can be 0 too right?

[slate lintel]

yep! because we wanna know when it hits the ground again

[echo sinew]

i got t = v/g

i didnt get the v^2 tho its normal right o.0

cus i will prob multiple it later with smth else ig

[slate lintel]

did you remember the sin theta?

sorry i forgot to put that in there we want to use $v_y$ not just $v$

[wraith dagger]

Hayley

[echo sinew]

mhm sin2θ

oop icic

[slate lintel]

like this thing should say vsinθ*t in the middle

[echo sinew]

hm btw why are we doing dis im a bit confused

[slate lintel]

we're trying to figure out how long the ball is in the air

so we can figure out how far it goes

[echo sinew]

okok

[slate lintel]

bc like there's nothing slowing itt down horizontally

[echo sinew]

mhm

[slate lintel]

so it'll keep going until it hits the ground

[echo sinew]

okie now i got v/g(sinθ) = t

Displacement = velocity . time

[slate lintel]

hmm i get $t = \frac{v\sin\theta}{g}$

[wraith dagger]

Hayley

[slate lintel]

or is that what you mean

[echo sinew]

so im guessing i have to us <vcosθ,vsinθ> here smwhere

[echo sinew]

ooop yep haha

i placed the constants tgt xD

[slate lintel]

yeah so now we figure out how far it goes in the x direction in that time!

[echo sinew]

so its easier to visualise

[slate lintel]

and yeah displacement = velocity * time

[echo sinew]

issit like what we have here times vcosθ?

[slate lintel]

yeah

[echo sinew]

OOOH I understand

i got the thing proveddd

tysmm!!!!

[slate lintel]

[echo sinew]

ur really patient lol appreciate ur time & help ^w^

[slate lintel]

okok then the next part is pretty easy just put in the numbers and crank the wheel

[echo sinew]

hhaha i did that first alr lol xD

[slate lintel]

yeah that makes sense haha

[echo sinew]

so i need to rewrite cus the current one is messy

AHHA

okie imma close the channel so others can use it

tysm have a grt day ^w^

.close

[cedar kiln]

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[safe violet]

what is the integration of e^-2x

[safe violet]

I have to do this one

[dreamy sleet]

You can use integration by parts

[safe violet]

for that one also?

[dreamy sleet]

You can use integration by parts here

[dreamy sleet]

For this one, you can use the integral of e^x

[safe violet]

Yeah for that I need to know what is the integral of e^-2x

@dreamy sleet ??

[cedar kiln]

@safe violet Has your question been resolved?

[safe violet]

<@&286206848099549185>

[crimson sedge]

It is 1/(-2)e^(-2x)

[safe violet]

#help-13 message
isn't this?

did you say something?

[crimson sedge]

If you diff, it becomes -2e^(-2x) and idk where the c is from..

But if you integrate it is -1/2e^(-2x)+C

[safe violet]

ok I confused myself on it

so can you check my complete answer?

#help-13 message

[crimson sedge]

Ok, here's the issue, with integration by parts, what you have is $\int udv=uv-\int vdu$.\
So it should be:
$\int x^2e^{-2x} dx=x^2(-\frac{1}{2}e^{-2x})-\int-\frac{1}{2}e^{-2x}2x dx$\
Where we've chosen $u=x^2$, $dv=e^{-2x}$ so $du=2x$ and $v=-\frac{1}{2}e^{-2x}$

[wraith dagger]

qingfengmingyue

[safe violet]

is this for the first part?

damn

[crimson sedge]

Yes, for the first time you apply it

You need to apply it twice in total

[safe violet]

oh

so I chose v instead of dv

hah?

dv = e^-2x

but I let v = e^-2x

[crimson sedge]

Yes

[safe violet]

let me try one more time

[crimson sedge]

Yes, that looks right to me

[safe violet]

thank you sir

[safe violet]

.close

[cedar kiln]

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[cedar kiln]

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[valid shuttle]

Hi, i am reading a textbook right now and i dont understand this simplification. Can someone explain it please

[crimson delta]

the first step is an arithmetic progression

[vague rapids]

This turns into a diophantine equation, solutions of which are written below

[valid shuttle]

The bottom one is the first equation

[vague rapids]

They did the factorization because (not stated but as it seems) a and k both are integers. So you can solve for them easily by factoring

[frosty ocean]

Write all the a's together and the numbers next

[vague rapids]

Since 2000 is to be expressed as product of two terms, we would get a limited cases (here 20)

[frosty ocean]

a+(a+1)+(a+2)+...+(a+k) = (a+a+...+a) +(1+2+3+..k)

There are k+1 a's

and there is sum up to k natural numbers

You can simplify this

[vague rapids]

Also you could think of an A.P.
The given series is an A.P. with first term=a and last term=(a+k) and common difference=1 and there are (k+1) terms
So apply the formula of sum which is (n/2)(a+l)
You would get ((k+1)/2)(a+a+k)

[frosty ocean]

Where are the dots?

[frosty ocean]

SImplify further

[valid shuttle]

Sorry but i dont get it

[frosty ocean]

Repeated addition is multiplication

For example, 2+2+2+2 = 4 times 2 = 8

[vague rapids]

from 1 to k there are 'k' numbers and there is one 'a' for every number. Also there is one extra 'a'.
So in total there are k+1 number of 'a's

[frosty ocean]

So what is (a+a+...+a) when there are (k+1) number of terms?

[valid shuttle]

(K+1)*a

[frosty ocean]

Right

What is sum up to first k natural numbers?

[valid shuttle]

I dont know

[dire geode]

Which part don't you know

[valid shuttle]

.

[dire geode]

1+2+3+...+k=?

Have you learned arithmetic progression sums

[valid shuttle]

Not yet

Is this good?

[vague rapids]

yes

[frosty ocean]

This is the AP sum

Why did you say "not yet" to riemann's question?

[valid shuttle]

Because we havent learnt it in school yet

[frosty ocean]

To continue from our work

[frosty ocean]

= (k(k+1))/2

[valid shuttle]

Is it this formula?

[vague rapids]

yes

[vague rapids]

.

[valid shuttle]

Okay now i get it thank you guys

Just one more thing

[valid shuttle]

Where does this formula come from?

[dire geode]

You prove it with induction

https://brilliant.org/wiki/induction/

[cedar kiln]

@valid shuttle Has your question been resolved?

[valid shuttle]

Thanks for everything guys

[cedar kiln]

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[wanton summit]

Show that nc0 + nc1 + nc2 + nc3...ncn = 2^n

[wanton summit]

i dont really understand what they did here

[tropic oxide]

they find |P(S)| in two different ways

[wanton summit]

how do they count teh subsets of S

in a different way

can you explain that to me

[tropic oxide]

there are two ways

1. count nCk for the number of k-element subsets of S, sum for k from 0 to n
2. two options for each element: either it belongs to the subset or it doesn't => 2 * 2 * ... * 2 [n copies]

which one do you want explained

[cedar kiln]

@wanton summit Has your question been resolved?

[wanton summit]

both

wait no not the sum one

[tropic oxide]

which one first

[wanton summit]

i dont think they did that

2

[tropic oxide]

bruh what

see me saying they did this
"i dont think they did that"

are you trying to piss me off on purpose here

[wanton summit]

sorry but

doesnt that involve sigma notation

[tropic oxide]

sigma notation is just that: notation

they wrote it the longhand way

it doesnt change the fact that theyre still adding some shit up

[wanton summit]

right

could you please explain both

[tropic oxide]

which one first

[wanton summit]

start with 1

[tropic oxide]

k

[wanton summit]

and forgive me but

[tropic oxide]

for any k between 0 and n inclusive, the number of k-element subsets of S is equal to nCk

[wanton summit]

im not familiar with sigma

[tropic oxide]

you dont NEED to be

[wanton summit]

okay im just

letting you know

[tropic oxide]

do you agree or disagree

[wanton summit]

yes that makes sense

its how we define nck isnt it

from n elements choose k

[tropic oxide]

yes thats one of the definitions

the possible sizes for a subset of S, which has n elements, are all integers from 0 to n inclusive

agree or disagree

[wanton summit]

yes

[tropic oxide]

the number of 0-element subsets is: nC0
the number of 1-element subsets is: nC1
the number of 1-element subsets is: nC2
.........
the number of n-element subsets is: nCn

adding all this up, we get that the count of ALL subsets of S is
nC0 + nC1 + ... + nCn

agree or disagree

[wanton summit]

yes

by the way can i also think about this as how many ways you can arrange n digits into a number with up to n digits

and showing that that is 2^n

is that word problem equivalent to this series

[tropic oxide]

no and yours is poorly stated enough that i dont want to touch it

dont ask me to either

does this explain method #1? yes or no

[wanton summit]

yes

but how do they get to 2^n

[tropic oxide]

you misspelled "ok, now on to method 2?"

anyway

there are n elements in S

when constructing a subset of S, for each element you make a binary choice: either it belongs to the subset or it does not

agree or disagree

[wanton summit]

yes

agree

[tropic oxide]

ok

since these choices are independent of each other, we multiply their counts together

to get the product of n copies of 2

agree or disagree

[wanton summit]

yes thats the multlipication principle

[tropic oxide]

the product of n copies of 2 is 2^n

agree or disagree

[wanton summit]

agree

[tropic oxide]

does this explain method #2, yes or no

[wanton summit]

yes

you find the number of subsets from 0 to n, add them up

deduce that something is either part of it or it isnt

and use multlipication principle n times

[cedar kiln]

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[crimson sedge]

How do I know which is a min and which is a max?

[vague rapids]

Mininmum would be smaller than Maximum

[vestal bear]

true

[cedar kiln]

@crimson sedge Has your question been resolved?

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[boreal trail]

is it jus me or textbooks wrong/wording of q is scuffed

[boreal trail]

if you win the 10% of $20

u gain $18 dont u ?

and if u win nothing $-2

cuz $2 spend

[tropic oxide]

it's the worked solution that is scuffed

[boreal trail]

but when calculating the expected, they only included the -2 when losing, but not 20-2

[boreal trail]

would it be 18 * 0.1 then

[tropic oxide]

should be yes

[boreal trail]

k ty

.close

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[crimson sedge]

How do I know which equation is suitable for the question?

I still don’t understand

[cedar kiln]

@crimson sedge Has your question been resolved?

[dreamy sleet]

is y = -sqrt(x + 1) + 2 suitable?

you can check this by putting y = 3 and seeing if x is real

[cedar kiln]

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[crimson sedge]

.reopen

[cedar kiln]

✅

[crimson sedge]

X=0 then

[dreamy sleet]

Can you show your working

[crimson sedge]

@dreamy sleet

[dreamy sleet]

the fifth step is wrong

[crimson sedge]

How?

@dreamy sleet

[dreamy sleet]

-1 ≠ sqrt(1)

[crimson sedge]

Huh

[dreamy sleet]

because the square root function only gives the positive output

[crimson sedge]

So then how would I solve it properly?

[dreamy sleet]

Can you rewrite your working?

[crimson sedge]

Why, you’re right

[dreamy sleet]

Can you solve this for y = 3?

[crimson sedge]

No?

[dreamy sleet]

Why not

[crimson sedge]

Because of this

Idk Im confused

[dreamy sleet]

this is right

by the way I think it’s meant to say ”y >= 2” instead of “x >= 2” for part ii

[dreamy sleet]

so with that in mind, can -sqrt(x + 1) + 2 ever output 3?

[crimson sedge]

No

[dreamy sleet]

so what is the correct inverse?

[crimson sedge]

Sqrt(x+1)+2

[crimson sedge]

So this is for y<=2?

@dreamy sleet

[dreamy sleet]

yes

[crimson sedge]

So if I want to find the suitable equation I just see if each equation is true for a value in the given domain or range?

@dreamy sleet

[dreamy sleet]

you keep pinging me after I don’t reply for a few seconds

can you not do that

[crimson sedge]

Oh sorry

[dreamy sleet]

not sure what that means

[crimson sedge]

All good

Thank you

[dreamy sleet]

has your question been resolved?

[crimson sedge]

.close

[cedar kiln]

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[vapid knoll]

i got this log question i dont even know where to begin cant see anything

[tropic oxide]

this looks mistyped

is that cube root of 2 meant to be the base of the log?

[vapid knoll]

i wrote it in this machine from my book

i dont think so

i'll send the original

[proven thistle]

Let us assume

[proven thistle]

Naww

That's not base lol

[vapid knoll]

🤔

[tropic oxide]

the question is either ugly or just unsolvable

either way you should talk to your teacher and clarify

[proven thistle]

Ajax try solving it twice

One with ³√2 as base and one without

Do you need help in any?

[vapid knoll]

what do you mean without the base?

[vapid knoll]

here

[proven thistle]

Put ³√2 as a coefficient of a

In 1 case

[vapid knoll]

all right

so i got $log_2^{1/3} {(a)} = b$
and $ log_4 (a^3) = log_{2^2} (a^3)$

oop

[proven thistle]

2 power 3/2

[wraith dagger]

ajax4074

[vapid knoll]

you mean
$log_2^{3/2}$ ?

[wraith dagger]

ajax4074

[vapid knoll]

by this:

[proven thistle]

Yes

Yes

[thick hill]

i got 3/2 (log⁡ a-log⁡2 )

[vapid knoll]

-log2 ?

[vapid knoll]

wait not that

[vapid knoll]

... (log a...)?

oh i see

[proven thistle]

They're doing the Second case

But im 100% sure you just need to do case 1

[vapid knoll]

but not this?

what was that for

[proven thistle]

?

You could do it any way lol

[vapid knoll]

im soo lost rn...

[cedar kiln]

@vapid knoll Has your question been resolved?

[fallen moat]

the answer would be pretty ugly
still stuck?

[vapid knoll]

well how ugly are we talking..?

i got options

[fallen moat]

oh interesting

[vapid knoll]

ugliest is 3b/2

4b
2b
b
b/2
3b/2

[fallen moat]

is that log is of base 10 or e?

anyways

[vapid knoll]

i suppose its 10

cant we figure out what they meant to write depending on the options?

[fallen moat]

with those options, I'm pretty sure it's $\log_{\sqrt[3]2}a=b$

[wraith dagger]

biscuityxd

[fallen moat]

instead of that

[vapid knoll]

how do you see that?

[fallen moat]

because with the original one, the answer is way uglier than these answers

[vapid knoll]

i cant see a relation between $\sqrt[3]2}$ and base 4

[wraith dagger]

ajax4074

i cant see a relation between  $\sqrt[3]2}$ and base 4
```Compilation error:```! Extra }, or forgotten $.
l.57 i cant see a relation between  $\sqrt[3]2}
                                               $ and base 4
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2020/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]```

[fallen moat]

and if it's this, one of the answers is correct

try to use $\log_ba=\frac{\log a}{\log b}$

[wraith dagger]

biscuityxd

[vapid knoll]

intriguing...

[fallen moat]

do tell if you are still stuck @vapid knoll

[vapid knoll]

so $\log_{\sqrt[3]2}a=\frac{\log_4a}{\log_4\sqrt[3]2}a}$

[wraith dagger]

ajax4074

so $\log_{\sqrt[3]2}a\=\frac{\log_4a}{\log_4\sqrt[3]2}a}$
```Compilation error:```! You can't use a prefix with `\endgroup'.
<to be read again> 
                   \endgroup 
l.57 so $\log_{\sqrt[3]2}a\=\frac
                                 {\log_4a}{\log_4\sqrt[3]2}a}$
I'll pretend you didn't say \long or \outer or \global or \protected.```

[vapid knoll]

what

im not even gonna try to edit that shit wait a second

[fallen moat]

lol

[vapid knoll]

consider that 3 up there

[fallen moat]

ok

[vapid knoll]

now i could go on and say this:

but is it really necessary

nvm if i do say that:

[fallen moat]

nah nah

[vapid knoll]

where

[fallen moat]

the second step is incorrect

[vapid knoll]

how

[fallen moat]

try mutliplying the numerator and denominator by 3

[vapid knoll]

u mean take its cube or?

[fallen moat]

$\log_ba=\frac{\log a}{\log b}≠\log a-\log b$

[wraith dagger]

biscuityxd

[vapid knoll]

whaat

[fallen moat]

you ask how

[vapid knoll]

oh shit its not on the same log ok

nah nah ok i see that

[vapid knoll]

but i dont see how multiplying by 3 would help

[fallen moat]

use $n\log a=\log a^n$

[wraith dagger]

biscuityxd

[vapid knoll]

oh... right...

im so blind ?

[fallen moat]

nah, you're brain stuck only 😛

[vapid knoll]

oop a got lost there

[fallen moat]

use this

[vapid knoll]

ok 1 sec

[slate lintel]

girl r u writing on your floor 😭

[fallen moat]

lol

[vapid knoll]

no this is paper guys

wait

[fallen moat]

looks good

[vapid knoll]

but then it gets fucked again excuse my french

[fallen moat]

lol

[vapid knoll]

is that equal to that

i bet not

[fallen moat]

log_4 (2) means 2=4^(?)

[vapid knoll]

oh yeah

1/2

as ^

so sqrt

i was going by heart

it's not taught that way here exactly

ok so that's $2 log_4a^3$

[wraith dagger]

ajax4074

[vapid knoll]

if this is b than $log_4 a^3 is b/2$

[fallen moat]

yes

[wraith dagger]

ajax4074

[vapid knoll]

holy shit this took so god damn long

[slate lintel]

use \log btw

[vapid knoll]

thanks for suffering w me

[vapid knoll]

idk that what?

trial /log4

trial $/log2$

[wraith dagger]

ajax4074

[vapid knoll]

?

[slate lintel]

$\log_3(x^5)$ vs $log_3(x^5)$

[wraith dagger]

Hayley

[vapid knoll]

what's the difference?

[slate lintel]

it's not slanted and doesn't look like three variables being multiplied together it's a small thing

[vapid knoll]

ooh

[vapid knoll]

looks like i actually am blind now...

nvm thanks @fallen moat

also @slate lintel for the contribution...

biscuity did you just-

lmao

ok that's it for today...

.close

[cedar kiln]

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[hollow wharf]

hello

[cedar kiln]

@hollow wharf Has your question been resolved?

[slate lintel]

draw a quick sketch

30° is a special angle that you should know the side length ratios

[cedar kiln]

@hollow wharf Has your question been resolved?

[olive gust]

@hollow wharf

to determine the length of each tile needed for the gabled roof we can use some trigonometry heres how we can calculate it
1consider one side of the gabled roof we have a right-angled triangle formed by the slab (base) one side of the roof (height) and the slope of 30°
2Since we know that both sides of the roof have an angle of inclination of 30° this means that both sides are congruent.
3Let's label the length of each tile as x meters
4 using trigonometry we can find out that x = height/sin(30°)
to find the height which is half the width or slab length since it meets in exactly in middle divide 24 m by 2: height = 24/2 = 12 m
now plug these values into our equation: x = (12 m) / sin(30°)
alculate sin(30°) ≈ 0.5 and substitute it back into our equation x ≈ (12 m) / 0.5 x ≈ 24 m
therefore to order tiles for this gabled roof before its structure is finished you will need tiles with a length approximately equal to 24 meters each

[cedar kiln]

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[vapid knoll]

so...

[vapid knoll]

oops

tried the change of base formula

[slate lintel]

what did you get when you tried the change of base formula?

[vapid knoll]

$/log_2 (7) = /{log_5(7)}\frac{log_5(2)}$

[wraith dagger]

ajax4074
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

toby____

$\log_2(7)=\frac{\log_5(7)}{\log_5(2)}$

[vapid knoll]

great thanks

yeah then we add the bottom:

$\frac{\log_2(7)}{\log_5(7)}=\frac{\log_5(7)}{\log_5(2)\cdot \log_5(7)}$

[wraith dagger]

ajax4074

[vapid knoll]

if they simplify (i dont remember if they can)(they being $log_5(7)$ 's)

[wraith dagger]

ajax4074

[vapid knoll]

then we would get $\frac{1}{\log_52}$

[wraith dagger]

ajax4074

[fallen moat]

yea, and this is the simplified form of a

[slate lintel]

log5(7) is just a number so yes you can divide top and bottom by it

[vapid knoll]

oh we CAN simplify??

[vapid knoll]

ok great

[slate lintel]

,tex .log rules

[wraith dagger]

Hayley

[slate lintel]

hmm this one doesn't show but there's a way to simplify it a bit further

[vapid knoll]

yeah the fives go we get 7s and they simplify each other i guess

[vapid knoll]

but this didnt help

[slate lintel]

1 = log5(5)

[vapid knoll]

yeah right right

come on hayley im not that bad...

so i need the argument to be 10 that's why im so confused

[fallen moat]

hint2:2*5=10

[vapid knoll]

i was so confused

[vapid knoll]

but

idk which one you meant but...

[fallen moat]

hint3: (log_5 (5))/(log_5 (5))=1
use all 3 hints on this.

[fallen moat]

this is wrong btw

[vapid knoll]

ok great 2*5 =10 it is

[vapid knoll]

yeah that's why i said that

[fallen moat]

i see

[fallen moat]

use all 3 hints on this

[vapid knoll]

and i am trying to

[slate lintel]

oh right you're the one that writes on their desk

[vapid knoll]

floor if you will

where does the 2 go if i multiply it by 5?

[slate lintel]

i'm trying to figure out why you need it to have an argument of 10

[vapid knoll]

im trying to figure out how i dont

[slate lintel]

anyway you have $a = \frac{\log_5(5)}{log_5(2)} = \log_2(5)$

[wraith dagger]

Hayley

[fallen moat]

opps

haha, my bad

it's 3am, i guess i need to sleep soon

thanks @slate lintel

[vapid knoll]

go sleep sir

[fallen moat]

final hint:

[vapid knoll]

yeah i see

[fallen moat]

$a=\frac1{\log_52} \implies \log_52=\frac1a$

[wraith dagger]

biscuityxd

[vapid knoll]

i get 1/5a which is wrong

oh to get 10 i multiplied the argument with 5

but that would go in front thus wouldn't give me $\log_510$ but $5\log_52$

[wraith dagger]

ajax4074

[vapid knoll]

Scheiße

so the question is now:
how do i get $\log_510$ out of $\log_52 $ which is $\frac1{a}$

[wraith dagger]

ajax4074

[vapid knoll]

i CANT SEE

wait

do i change the bases

again

[slate lintel]

$\log_5(10) \neq \log_5(2)$...

[wraith dagger]

Hayley

[vapid knoll]

ok hayley i need to write it that way thoo?

[slate lintel]

I'm very confused on what you're trying to do

[vapid knoll]

im trying to write $log_510$ as $a$

[wraith dagger]

ajax4074

[vapid knoll]

which we simplified to be $\frac1 \log_5(2)$

[wraith dagger]

ajax4074

[vapid knoll]

are we even capable of changing the argument of a logarithm

[vapid knoll]

could i explain or is it still confusing

[vapid knoll]

shit this is wrong

[slate lintel]

what do you mean write $\log{5}(10)$ ``as'' $a$?

[wraith dagger]

Hayley

[vapid knoll]

isn't that what the question says?

it has the same 'question sentence' as the one we did before

[vapid knoll]

oop i didnt write it there previously...

[vapid knoll]

well maybe the options of the question will help you understand

1/(a-1)

1/a

a-1/a

you get the idea...

the others are similar as well

i cant believe im stuck HERE.

[slate lintel]

ahhh

ok um i guess what I would do is now that we have a value for a, try putting it into each of those and do more log rules

to see if it results in something equal to $\log_5(10)$

[wraith dagger]

Hayley

[slate lintel]

by the way, in english we'd usually say ``write $\log_5(10)$ in terms of $a$''

[wraith dagger]

Hayley

[vapid knoll]

ooh

thanks

[vapid knoll]

but i tried that already?

well i couldn't get to 10 when it was the base so i changed its form to get it in argument

well not 10 but 5

let me show it

$\log_5(2) = \frac1{\log_2(5)}$

[wraith dagger]

ajax4074

[vapid knoll]

i went back to this

then i tried multiplying the denominator and numerator by 2

[slate lintel]

do it the other way, take your a and plug it into the choices

[vapid knoll]

but then again it doesn't give me 10 but 25

[vapid knoll]

ooh that's what you meant...

[vapid knoll]

well when i do that i have to do $\frac1{\frac1{log_5(2)}-1}$

[wraith dagger]

ajax4074

[slate lintel]

here's a useful thing: $\frac1{\log_ab} = \log_ba$

[wraith dagger]

Hayley

[vapid knoll]

now that i think about this

[vapid knoll]

what if i added $\log_5(5)$ to this

[wraith dagger]

ajax4074

[slate lintel]

👀

[vapid knoll]

so the arguments would be multiplied?

5*2 giving me 10

so a + 1 ?????????

that is not an option...

[slate lintel]

[vapid knoll]

is this wrong

[slate lintel]

did we say that $a = \log_25$ or $a = \log_52$?

[wraith dagger]

Hayley

[vapid knoll]

second one ... right?

biscuity wrote it up there somewhere

[vapid knoll]

duck

[vapid knoll]

ok we'll go your way ...

[slate lintel]

no you had the right idea

[vapid knoll]

just the wrong values

[vapid knoll]

this option is so close that im starting to develop trust issues with myself rn

,rotate

[slate lintel]

you saw earlier that $\log_5(2) + 1 = \log_5(10)$, right?

[wraith dagger]

Hayley

[vapid knoll]

yeah...

oh so $\frac1{a}+1?$

[slate lintel]

yeah!

[wraith dagger]

ajax4074

[vapid knoll]

ohhhhh thank youu

@slate lintel

[slate lintel]

[vapid knoll]

.close

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
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[crimson sedge]

I have no clue how to actually solve this

I know the scale factors but I don’t get how the scale factors help with solving

[long arrow]

ratio of areas = ... (?)
ratio of volumes = ... (?)

[crimson sedge]

The ratio of areas is a weird decimal

[long arrow]

yes but I mean it's scale factor squared and if it comes to volumes is scale factor cubed, right?

[crimson sedge]

Wait how do I find the ratios of areas?

[long arrow]

just area of the bigger solid divided by area of the smaller solid

(or the other way around)

[crimson sedge]

3.3344

1,755/3.3344 =526

[long arrow]

3.3344 is the ratio of the areas

that's true

and now explain what you did

[crimson sedge]

Divided the areas

[long arrow]

yes and then? (2nd line equation)

why volume divided by this

[crimson sedge]

Idk

[long arrow]

okay lemme explain

so we know that ratio of the areas is equal to scale factor squared

let's say scale factor is "k"

hence, 3.3344 = k^2, you agree?

[crimson sedge]

Yes

[long arrow]

from here, what's the value of "k"

[crimson sedge]

1.826

[long arrow]

yep

what we also know is that ratio of the volumes is equal to scale factor cubed

[crimson sedge]

Oh so

Basically then cube 1.826

And divide 1,755 by that

Thanks

.close

[cedar kiln]

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[limber snow]

Is this good enough?

[mighty drift]

Unless your teacher asks for that much detail, you don't need that many steps for computations. Doing a few simple steps at once is fine

[limber snow]

ngl i just be worried because sometime my TA deducts marks because the "proof isnt clear"

but other then that is it correct?

[mighty drift]

Except

The conclusion sentence

[limber snow]

oh

whats wrong with it

isnt it for all n in N?

[mighty drift]

We have shown that (the sum up to n+1) = (the fraction in n+1), not (the fraction in n) + the cube

Because that's the hypothesis, not the conclusion

[limber snow]

ssorry i dont get it

wym not the cube

i did it

(k + 1)^3

[mighty drift]

Look at the way you should write it

You should state you indeed successfully arrived at the conclusion, hence proving P(n+1) follows from P(n)

Yet you wrote basically P(n), not P(n+1)

[limber snow]

in my conclusion right?

[mighty drift]

Yes

Oh

I read too fast

You did the same at the start

[limber snow]

so thats wrong too?

[mighty drift]

You mistated the goal of the inductive step

[limber snow]

shit i dont rlly get the mistake

should i have done

k^3 + (k+1)^3 =

not

1^3 ...... k^3 + (k+1)^3

[mighty drift]

Is this your first proof by induction?

[limber snow]

yes

we just started

[mighty drift]

That's a big computation to start with

[limber snow]

wait

the assumption

one second let me fix n send again

[mighty drift]

You should state what you want to arrive at

Which is the same as the hypothesis, but replacing k by k+1

[mighty drift]

So that you get this

[limber snow]

like this?

[mighty drift]

Yes

[limber snow]

thanks alot

.close

[cedar kiln]

Channel closed

Closed by @limber snow

Use .reopen if this was a mistake.

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[limber snow]

Can I get a hint for this one

[limber snow]

im stuck

[solid juniper]

fermat’s little theorem might be helpful

or maybe that will overcomplicate it

[limber snow]

never heard of it before i can look into it if it helps

[solid juniper]

if you’ve never heard of it then you probably can’t use it

[limber snow]

oh yeah true

[solid juniper]

but it would be nice

[limber snow]

so should I

[solid juniper]

snow why are you hmming me so much?

[sacred grail]

flt seems too complicated for this

[solid juniper]

it’s not that bad T_T

[sacred grail]

you should consider using the induction hypothesis

[short blade]

maybe consider the remainders modulo 5

[solid juniper]

you can reduce 64 and 9 mod 5 then just check n=1,2,3,4 or something by flt

[sacred grail]

isnt it just -1 + 1 though

[limber snow]

9k^+1 -1 ?

[sacred grail]

anyway we seem to be doing it induction way so you should insider the induction hypothesis into your calculation

[limber snow]

i did that before
(64(64^k)) + 9k) but got stuck with that aswell so idk

[solid juniper]

ok you got me there

[still barn]

The strategy I would use is to notice that 64-55 = 9 and 55 is a multiple of 5

[sacred grail]

64^k + 9^(k - 1) is divisible by 5, write it as a multiple of 5

[limber snow]

wait where did u get 55 from doe

[limber snow]

alr

so i can rewrite it as 5m where m is an integer

right

[sacred grail]

yes

then insert that into your calculations

[limber snow]

wait uh

which calculation

[sacred grail]

this one

[limber snow]

but its k+1

[sacred grail]

well your next step makes it 64^k and 9^(k-1)

[sacred grail]

64^k + 9^(k-1) = 5m

[limber snow]

wait what

how

i dont see it at all

[sacred grail]

i am talking about this

[limber snow]

yes

[sacred grail]

substituting 64^k + 9^(k-1) = 5m into that would do you a lot of good

[limber snow]

im sorry but im confused right now.

how would i even transform my equation into that

64^k + 9^k-1

[sacred grail]

what to do

[limber snow]

ok got it