help-17

[paper depot]

stop.

stop.

[lethal scarab]

which is ur base

[paper depot]

stop.

stop.

[lethal scarab]

can you please stop and let me explain my work please

[paper depot]

i want to see your entire process written out on paper.

[lethal scarab]

i want you to help me

[paper depot]

and i also want to help you.

but you have to write your process out on paper.

[lethal scarab]

so please

i dont even have to do this on my finals

is the answer even correct

[paper depot]

ok sure you don't have to do it on your finals.

but you do have to do it here.

[lethal scarab]

what

[paper depot]

because my goal is for you to walk out with an understanding,

[lethal scarab]

but i have an understanding

[paper depot]

and not merely as an answer-generating machine.

or to put this another way: the right answer matters much less than how you get to it. and you will benefit greatly from writing your work out in a clean manner.

even for simple equations like this one.

[lethal scarab]

that is not what im trying to use you as

here

here i have showed you

what i narrowed it down to

[paper depot]

ok

this is almost ok

[lethal scarab]

after this i typed in calculator

[paper depot]

but somewhat messy

[lethal scarab]

and solved

so can was the asnwer right

[paper depot]

this is how i would have preferred for you to write it out:

[lethal scarab]

please i wrote it out

its almost been an hour on this one equation

[paper depot]

this is bigger than this one question.

$-10 \cdot 8^{v+4.8} = -96 \ 8^{v+4.8} = 9.6 \ v + 4.8 = \log_8(9.6) \ v = \log_8(9.6) - 4.8$

[twin meteor]

Ann

[paper depot]

this is the kind of working that i wanted to see from you.

[lethal scarab]

did i still get the answer?

or was it wrong

[paper depot]

,w log_8(9.6) - 4.8

[twin meteor]

Results provided by WolframAlpha

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

Requested by ann_dec

[paper depot]

yes, answer A is correct.

[lethal scarab]

oh look nice it was right

imma try 16

[paper depot]

ok

write your work out neater than last time, please.

[lethal scarab]

can i please just do it my way

i showed u ik how to do it

[paper depot]

blech.

fine.

do it your way i guess, but i am not enthusiastic about it.

[lethal scarab]

is it B?

[paper depot]

yes

[lethal scarab]

Ok thanks very much

can i show u more equations

or no

its different concept now

[paper depot]

well, sure, we can go over those.

i will warn you though that being lackadaisical about your algebra can and will bite you in the ass and it will hurt.

[lethal scarab]

idk what that means

19-23

are any of them like the ones we did

am i missing something here

[paper depot]

they're kind of related but it's more like the other side of the coin now

these are logarithmic rather than exponential equations

[lethal scarab]

Ok

so how exactly

oh wait

[paper depot]

the ``basic form'' for these equations is $\log_a(x) = b$. and the key step is to raise $a$ to the power of both sides, getting $a^{\log_a(x)} = b$ whence $x = a^b$.

[twin meteor]

Ann

[lethal scarab]

i got it think

tell me if wrong

im gonna send

look

is it D?

[paper depot]

ok yes correct

[lethal scarab]

alr nice

im gonna try the others

wait

oon 20

on 20

what do io do with the 10-9

would i mkae it 1

[paper depot]

no

you would not make 10 - 9z into 1z

[lethal scarab]

huh

what would i do

[paper depot]

well imagine that the equation said $10 - 9z = -17$ and you needed to solve for $z$ (i.e. forget about any logarithms or exponentials; this is just a linear equation). what would you do?

[twin meteor]

Ann

[lethal scarab]

move the 10 over

then divide by 9

[paper depot]

you subtract 10, then divide by 9 (by -9 would be better but this way is fine too)

so do the same shit in your equation.

[lethal scarab]

yea so i got 3

what now

i got

log_12 a =3

is that right

wherw would i go from here

oh shoot

12 to the power of 3

is 1728

[paper depot]

12^3 = 1728 yes

[lethal scarab]

ok i get it

[paper depot]

use ^ (most likely Shift+6)

[lethal scarab]

D

Ok

ok what abt for 21

and 22

is it same

it look swierd on 21

[paper depot]

for 21 the strat is not wildly different

yes it does look weird bc of the negative sign under the log

but just do it normally anyway

probably best to get done with 21 before i tell you what is up with 22

[lethal scarab]

Ok

for 21

would i didved the -3

dang im stuck

idk

i dived the -3 and then what\

[paper depot]

you divided both sides by -3. what does your equation look like now?

[lethal scarab]

log_3 - 7v=2

[paper depot]

log_3(-7v) = 2

[lethal scarab]

how is times

is it

[paper depot]

it is not multiplication

log_3 is a function and not a numbeer

those brackets are important so it's not like you're taking log_3 of thin air and then subtracting 7v from it

[lethal scarab]

so log 2 divied by log 7?

[paper depot]

no

-7v = 3^2

[lethal scarab]

wait im triopping

wrong type

huh

why that

oh the

base u move over

idk when to do that tho

?

how do u know when to move the base over from the log

[paper depot]

you do not "move anything over".

[lethal scarab]

the 3 becomes the base for the 2 tho no?

[paper depot]

i really do strongly recommend never using that terminology again, because it is extremely prone to errors, misinterpretation, and misrememberance.

now:

i will also deliberately cover up the -7v by replacing it with a single letter, x.

$\log_3(x) = 2$.

[twin meteor]

Ann

[lethal scarab]

ohhhh

ok

[paper depot]

the right step here is raise 3^ both sides.

thus, $3^{\log_3(x)} = 3^2$

and after simplification, $x = 3^2$.

[twin meteor]

Ann

[lethal scarab]

so the answr is A?

[paper depot]

yes

[lethal scarab]

ok

so whats the gist with 22 u were telling me

[paper depot]

for 22 there's two routes:

one if you're confident with exponent laws, the other if you're not.

"confident" here means "i could get woken up by a SWAT team at 3 in the morning and placed at gunpoint to solve a problem with exponent laws and i could still do it flawlessly"

tell me which route to give you.

[lethal scarab]

it dont matter

whichever

what u thinks better for me

[paper depot]

ok

in that case, you need to either recall or (re)learn log laws.

properties of logarithms that help you simplify expressions involving them.

does that ring a bell or would you like a refresher

[lethal scarab]

no keep going

[paper depot]

these

that ring a bell?

[lethal scarab]

yes i think i get how to do it tho

[paper depot]

ok then show your work

[lethal scarab]

Ok

this good?

or no

23 looks even worse

[paper depot]

ok let's see

log_8(3/x) = 1 is correct

then 3/x = 8

again kind of sloppy the way you wrote it

but you seem to have gotten x = 3/8 correctly

[lethal scarab]

how i do 23

[paper depot]

same fashion

just be careful with how you handle that fraction

[lethal scarab]

Ok

is it A/

?

also i need help with these

[paper depot]

show work

[paper depot]

ok still somewhat messy but yes x + 5 = 64x is correct

yes, x = 5/63

[lethal scarab]

Ok

with the graphs

is the answer a on the first one

for 28

[paper depot]

yes

[lethal scarab]

how do i do 29

none of these options look good

[paper depot]

can you tell me how you did 28

like what was your thought process for choosing the right graph

[lethal scarab]

i mean simple

4 is your y point ur starting

then 2x is slope

so 2 rise

1 run

rise over run

but i feel like 30 doesnt have right answer choice

?

[paper depot]

-2 not 2

but yes ok

so identify the y-int first

in y = -3/4 x - 2

what's the y-intercept

careful with signs

[lethal scarab]

wym -2

so is 28 not A?

yea i also feel like theres not a correct answer choice for 29

[paper depot]

ah shit.

yeah, sorry, i got confused with trying to juggle the details for two questions at once.

let's do these one at a time

[lethal scarab]

but 28 is right no?

[paper depot]

no, for question 28 option A would have been y = -2x + 4. it is a line with negative slope.

its slope is -2

going 1 cell to the right makes you drop by 2 cells, not rise by 2 cells.

[lethal scarab]

huhh

[paper depot]

if you move right along the line you go down

[lethal scarab]

wait im so confused

oh n vm

im dumb

[paper depot]

im explaining what option A is like

and why it's wrong

[lethal scarab]

so its C

[paper depot]

the line with the correct slope is actually option C

[lethal scarab]

ye

[paper depot]

it still has the right y-int and it has the right positive slope 2

ok now this

what's the y-int here

[lethal scarab]

so what abt 29

-2

[paper depot]

right

[lethal scarab]

im assuming its B

[paper depot]

name all answer options which have y-intercept -2

ignore slopes for now

[lethal scarab]

b and d

[paper depot]

ok

[lethal scarab]

and according to the slope

the line moves correctly in b

so b

[paper depot]

the slope is negative so the line goes down and the correct graph is B, yes.

[lethal scarab]

but i feel like the graph is plotted wrong

alr

[paper depot]

why so

what strikes you as wrong here

[lethal scarab]

nothing mb

here what abt these

[paper depot]

ok, do you know what exponential function graphs look like generally

[lethal scarab]

eh

[paper depot]

in particular do you know how to tell if an exponential function y = a * b^x will be increasing (going up) or decreasing (going down)?

[lethal scarab]

acc yes

i forgot now but i think i remember

idk if i recall

[paper depot]

i think i might have to go now cause im a little tired and/or have some other stuff coming up

like irl

[lethal scarab]

dang is there any other ppl

can u just do this one rq please

im acc cooked

<@&286206848099549185>

[tardy cove]

first look at the y intercept of each function

so for 38) its +2 right?

then u need to figure out if the exponential function is decreasing or increasing

whenever u raise a fraction by an exponent it get smaller so for (1/2)^x its a decreasing

so find the decreasing line that has a y intercept of 2

wait i am wrong

for 38) theres only one decreasing so choose that

plug in 0 for x so

y = 5 x (1/2)^0 + 2
y= 5x1 + 2
y=7

Therefore graph should have y intercept of 7

@lethal scarab

[vocal sleet]

@lethal scarab Has your question been resolved?

[vocal sleet]

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[edgy timber]

Can anyone do this

[steep crater]

I would chain rule it

[bronze osprey]

logarithmic differentiation

which relies on the chain rule also

[steep crater]

@edgy timber give it a try

[scenic ravine]

or paramatrize it and then differentiate that

[exotic surge]

Edit: nvm I thought its integration 🤦

[vocal sleet]

@edgy timber Has your question been resolved?

[edgy timber]

I think all steps are correct,

[bronze osprey]

,w (d/dx sqrt((x^2 + 1)/(x^2 - 1))) - (-2x)/((x^2-1)sqrt(x^4-1))

[bronze osprey]

I haven't read your steps but the final result is correct

yeah steps look all good

[vocal sleet]

@edgy timber Has your question been resolved?

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

So I played this game in the past and just got back into it, which involves conveyor belts, splitters which devide items evenly into 2 outputs and importantly, conveyors can join back up to combine their flow. (the game is called Assembly Line (2) for anyone wondering)
But when I played it about a year ago it got me thinking about if it were possible to split a stream of items on the conveyor belt into any ratio of outputs using only 50/50 splitters and feedback loops. I think I proved it was, but I'm not sure if I got there all the way, I lost my notes on it and I never verified it with someone else. It also wasn't very formally described, so I'd love to see it written down proprely.
Statement/question: Is it possible to split a constant stream on a single conveyor line (I) into two streams A and B, where the flow of A and B can form any ratio of natural numbers, while using only normal (50/50) splitters?
**Example 1: **
Just using 1 default splitter, you can split the input I into 2 equal streams of 0.5. So A=B=0.5. And their ratio A/B=0.5/0.5=1.
**Example 2: **
Using 2 default splitters, you can split the input in half at splitter (1), where 1 side leads to the first output A and the other half leads into another splitter (2). One of the sides of splitter (2) goes into the other output B, but the other half goes back to the start and gets added to our original input as feedback.
Now this is where the fun math starts. Because of the white line (see picture), what goes into splitter (1) is not just the input of 1, but also whatever came out of the line from splitter (2).
It's easy to see that the line coming from splitter (2) is only a quarter of what went into (1), and whatever went into (1) is 1 + whatever came out of (2).
To make our lives easier, let's say what goes into (1) is T (for total of what into 1), which gives us:
T= 1+1/4T
We can solve this.
T-1/4T = 1
3/4T = 1
T = 4/3
So we know because the input and the white line came together, 1.33... is entering (1).
Half of T, or 0.66... exits from output A.
1/4th of T, or 0.33.../s exits from output B.
And the other 1/4th of T goes back to the start to fill I (=1) back up to 1.33...
Which makes the ratio B/A=1/2.
Because you now know you can split the stream into 2/3rds of the input and 1/3rd of the input, you could use this configuration from example 2 as a splitter instead of (1) or (2) and get other ratios for A and B.
The question is if it is possible to make A and B any possible ratio.

[green totem]

The game does include advanced splitters where you can assign them to output any ratio, but I'm asking this out of mathematical interest, not to gain an advantage.

[vocal sleet]

@green totem Has your question been resolved?

[green totem]

Is this the right place for this question or should I go somewhere else for an elaborate proof?

[vocal sleet]

@green totem Has your question been resolved?

[willow pike]

hmm

[willow pike]

can you explain why you're saying T = 1 + 1/4T

like i get that T is the rate of flow into the splitter

but why 1/4T

[small yarrow]

ig no cuz the fraction needs to be in format of sum of powers of 1/2

[willow pike]

i think maybe

[small yarrow]

ig

[willow pike]

it can be used like binary

probably

[green totem]

a flow of "1" is coming from the input I
And whatever joins together with I to make T is whatever comes out of (2)
The amount which comes out of (2) is a quarter of what went into (1) as T in the first place

[willow pike]

after I get it

[green totem]

that's what chatgpt tried to give me, but it keeps struggeling with the infinite recursion part, which is incorrect

[willow pike]

then isn't it 1 + T/4 = T

[green totem]

same thing, yes

[willow pike]

oh you meant it like that

[green totem]

oh right, I forgot there are different ways to write fractions over whole numbers

[small yarrow]

look, i am not sure but lets take 1/37 as an example and try to obtain that with sum of powers of 1/2

OH WAIT

dude

[willow pike]

yes

exactly

[green totem]

you're not limited by sums of powers of 1/2

[small yarrow]

why tho?

it can split in more than 2 ways?

like 1 to 3?

[willow pike]

ok so if i simply take it

we can split the number 1 into 1/2 and 1/2

[small yarrow]

aren't we suposed to be limited to sums of powers of half?

[willow pike]

and a half into 1/4 and 1/4

and so on

[small yarrow]

cuz like it can only split 2 ways

[willow pike]

and here we can pick any two sets of sums of two powers from some of all possible numbers

[green totem]

I'm trying to find the configuration again of how I did it in the past, but that's how I'm stuck.
I'm pretty sure I found a way that if you have a fraction 1/n I found a way to make the fraction 1/(n+2) first.
And then with recombination there was some way to get a fraction 1/(n-1) as well, which basically meant you could make any fraction, but I forgot how

[willow pike]

and connect the unnecessary ones back to the start

[small yarrow]

and any fraction then can be obtained, provided for a non termination decimal, u need infinite conveyers

[small yarrow]

dang, i might not know that

we need pro guys here

[green totem]

yeah, I'm assuming infinite conveyors, splitters and an infinite 2d working area

[small yarrow]

yeah then yes

see if fraction is terminationg, we can easily get as sum of finite conveyers

[green totem]

but like, each fraction should be able to be made with a finite amount of splitters

[small yarrow]

oh

[green totem]

not infinite sums, just recursion

[small yarrow]

look, according to me only fraction with terminating decimal can be obtained, am not too sure about the same for non terminating

and i aint big brain enough to try to derive the recursion u obtained

oof

u sure tho that the recursion exists?

[green totem]

it sounds harder than it is once you get the general idea, in the first step I always used the same idea of 2 splitters in the same feedback loop as in my example 2. What changes is that I can just use previous ratios obtained as a splitter in the diagram and the math stays the same as well.
After a while there was a pattern that occurred, which I'm trying to recreate atm

[small yarrow]

ahh

[small yarrow]

bruhhhh

i thought only 1 unit is available 💀

[green totem]

Writing on paper is a bit easier, the top half is the same as example 2.
In the bottom half, I used the module of example 2 as splitters in the same configuration.
The ratios of A and B are now 2/3 or 3/2, which already disproves the powers of 2 thing.
Also we could make other configurations with 50/50 splitters of (1/3)/(2/3) splitters.
As well as we now unlocked splitters that divide into (2/5) and (3/5)

[green totem]

the input into I always stays 1

[small yarrow]

no i thought like u cant add 1 and 1/2

i was wrong

mb

[green totem]

ooh

[small yarrow]

but how will u prove that we can obtain any fraction?

[green totem]

I noticed not having a name for the recursion/recombining part was annoying, so imma call it R.

[green totem]

There was a pattern I noticed in the past, but I'm trying to find it again.

Also if someone more theoretically gifted in maths sees this and can write this down better, I'd love to see it.

[green totem]

This is the proof of concept of how splitters A and B would look if you used the (1/3)/(2/3) splitters for A and B.
But again, we assume an infinite grid to build on, so you don't have to construct all of these and I don't care how many splitters it would take. It's just to show how A and B can be made from the previous splitter layouts

Correction: just realised A and B in the image should be named (1) and (2), my bad. The actual output A goes north and the output B goes east.

This is the point were things can be generalised I think

[real bridge]

By assuming the splitters can be any number, wouldn't that imply the answer to be true since you can split the input into any number

And just redirect the splits into A or B, or a feedback loop (another splitter)

[green totem]

The thing is, I'm not assuming a splitter can be any number. I'm starting from only a splitter that splits 50/50, and I'm trying to figure out if we can make any number with that.

I'm working on the generalisation

I've worked it down to only having to find prime ratios

[real bridge]

Was talking about the splitters count being any length which i think you kinda did there

[green totem]

I'm basically there with general method 1 also

[real bridge]

With like the feedback loops, that creates a new splitter i think

[green totem]

probably, the annoying part is the testing if something makes a new splitter or not.
The testing at the start takes a long time until you can generalise it into a new method with fixed math so the testing process speeds up.

Like, this is how long testing takes now, and I think this proves you can make prime numbers as long as you have the even number before it.
Which is always a multiple of smaller primes.
So I think this finally proves you can make any splitter n

This now makes it so I can use any splitter with ratio (1/n)/((n-1)/n)

But like, what if I don't want 1/37, but 19/37 or smth

Oh right, I can split it into A=1/37 and B=36/37
If I then split B into BA= 1/36 and BB=35/36, the actual flow of BA = (1/36)*(36/37) or also 1/37.
I can keep splitting off into 1 less every output (so BB into BBA=1/35 and BBB=34/35)
After doing this 19 times, I have 19 streams of 1/37.
If I combine all of those, I have a stream of 19/37.

Alright, I think that proves it again

.close

[vocal sleet]

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[upper tendon]

!status

[vocal sleet]

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

[upper tendon]

I solved quadratic equation

the first one gives

$sec{\theta} {+-} tan{\theta}$

[twin meteor]

bagelguy3

[upper tendon]

the second one gives

$-tan{\theta} +- sec{\theta}$

[twin meteor]

bagelguy3

[upper tendon]

but I still cant seem to be able to determine theta

oh wait

sec(theta) is always greater than tan theta in magnitude

that means

hmm

<@&286206848099549185>

.close

[vocal sleet]

Channel closed

Closed by @upper tendon

Use .reopen if this was a mistake.

[tough aurora]

Did you solve it?

[upper tendon]

no

nobody helped so i gave up

[tough aurora]

You want to reopen?

I can help

[upper tendon]

.reopen

[vocal sleet]

✅

[tough aurora]

so, you have to keep in mind where θ is to understand which to choose for the two ±

[upper tendon]

ok look

for the first one

theta is in the first quadrant and second quadrant

secant is positive in both

so therefore for the first quadratic equation

I meant second quadratic equations

u know what i'll just call it 'x' instead of theta

-tanx +- secx

-tanx + secx > -tanx - secx

therefore -tanx + secx is alpha

and -tanx - secx is beta_2

[tough aurora]

It says θ is between -π/6 and -π/12, so it's always the 4th quadrant

[upper tendon]

as for the first equation

WAIT WHAT

I read pi/12

:(

[tough aurora]

that sucks

[upper tendon]

pi/6 is 30

and pi/12 is 15

so theta is basically between -15 and -30

OH yes

4th quadrant

mb

yes

so that means

secx is positive and tanx is negative

that means

[tough aurora]

yh

[upper tendon]

first one

secx - tanx is greater

so its

secx + tanx = alpha_1

[tough aurora]

yes

[upper tendon]

-tanx - secx = beta_2

so alpha_1 + beta_2 = 0

[tough aurora]

no hold on, α₁ is the bigger root

[upper tendon]

yes

[tough aurora]

so sec-tan

[upper tendon]

oh yeah

wait

secx + tan=

ahh mb

I wrote it wrong

secx - tanx = alpha_1

=

-2tanx

😭

jee questions

how tf am i supposed to do this in 2 mins

I couldn't do it in 20

the only one i was even able to solve was this

[tough aurora]

well, I don't know if there's any quick trick
but if you mind the interval where θ is it shouldn't take much time
be careful when reading the exercise text

[upper tendon]

yeh

thx

.close

[vocal sleet]

Channel closed

Closed by @upper tendon

Use .reopen if this was a mistake.

[tough aurora]

np

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[dim crow]

Hola, estoy intentando comprender exactamente de qué trata la técnica de análisis de datos llamada ASCA o ANOVA-SCA. Tengo varias preguntas al respecto: es una técnica de 3 vias? O es solo de una tabla? Por qué se llama ANOVA-SCA si en vez de usar el SCA usa el PCA?

[vocal sleet]

@dim crow Has your question been resolved?

[wanton bloom]

Translation:

Hi, I'm trying to understand exactly what the data analysis technique called ANOVA-SCA or ANOVA-SCA is all about. I have several questions about it: Is it a three-way technique? Or is it just a table-based technique? Why is it called ANOVA-SCA if, instead of using SCA, it uses PCA?

[vocal sleet]

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[fiery mortar]

do i just find the dirivative of the top function then find the letters

[tawny nacelle]

yeah

[flat python]

Yes

[fiery mortar]

i must eb doing something wrong then

[flat python]

how is D wrong

interesting

[tawny nacelle]

you seem to have forgotten the second term in the product rule

you've found $f(x)g'(x)$, but the derivative of $f(x)g(x)$ is $f'(x)g(x) + g'(x)f(x)$

[twin meteor]

higher!

[tawny nacelle]

also... I think you forgot to apply the chain rule as well

[flat python]

just use the quotient rule?

try to write everyhing out

[tawny nacelle]

when you differentiate (x + 5/x)^-1, you gotta differentiate the inside stuff too

[flat python]

step by step

do you know the quotient rule?

or only the product rule

[fiery mortar]

yes

[flat python]

well then just write everything out step by step

quotient rule works better here

[fiery mortar]

whats the chain rule

[flat python]

$d/dx f(g(x)) = f'(g(x)) \cdot g'(x)$

man latex sucks

[paper depot]

\cdot

[twin meteor]

Hylke54

[paper depot]

\dot is for putting a dot above a letter

$\dv{x} f(g(x)) = f'(g(x))\cdot g(x)$

[twin meteor]

Ann

[flat python]

thanks ann

[slender ravine]

$f(x) = \frac{x^2}{x^2+5} \
f(x)` = \frac{(x^2 + 5)(2x) - (x^2)(2x)}{(x^2+5)^2} \
simplify$

[twin meteor]

quadius

[slender ravine]

cant you just do that

[flat python]

almost! it's actually devided by x^2 + 5/x !

[slender ravine]

i just did lcm in the denominator so its correct

[slender ravine]

you mean x + 5/x

[flat python]

damn

[fiery mortar]

how do i apply the chain rulew

is this f'(g(x))

im very confused

.close

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[fiery mortar]

am i on the right track

[silk osprey]

just split it

$\frac{x + 3}{x + 2} = 1 + \frac{1}{x + 2}$

[twin meteor]

knief

[silk osprey]

so your derivative should be -1/(x + 2)^2

you messed up your quotient rule, the derivative of x + 3 is 1 and the derivative of x + 2 is 1 not x

numerator should be x + 2 - (x + 3)

= -1

[fiery mortar]

oh

[silk osprey]

then do you know what you’re solving for

if it’s perpendicular to y = x then what should the slope be

[fiery mortar]

-1

so i can solve it using the quotient rule

[silk osprey]

wdym

[fiery mortar]

i would use the quotient rule in this problem tho

?

[silk osprey]

you don’t have to

but obviously it works

[fiery mortar]

so whatever i plug into x the final answer should be -1

so -3,-1

[silk osprey]

yuh

[fiery mortar]

i might be cooked

i have 2 quizes tommorrow about this stuff each worth 10% of my total grade

gonna suck

but thanks for your help

.close

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[grave flame]

I need help with this

I dont know where i went wrong

.close

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[compact raven]

This is Grade 10 Geometry. I don't understand picture 1,3,4 (both questions in pic 4).
I have done the working out for picture 2 however I'm unsure if it's correct. I am also missing two reasons (There are two big "Helps" written next to them).

[vocal sleet]

@compact raven Has your question been resolved?

[compact raven]

I mean I understand this but I still have no clue on how to solve for x

So x=20?

Now I do not know how to prove F1= 270

Do I use congruency?

[scenic crow]

the angle between xa aand BA seems to be 90°

[compact raven]

Yeah but can I state the reason as "parallel sides" ?

Can I assume f = 90 even though I still have to prove f1 = 270

[scenic crow]

no

look at the picture k sent

[compact raven]

ok

[scenic crow]

so as you can see

x would be d and why would be f correct

[compact raven]

I don't follow

[scenic crow]

this picture right

[compact raven]

Yeha

[scenic crow]

you could look at the picture you sent

and 2x could be D oand 2y could be f right?

[compact raven]

D is 180 no?

So x cant = D

And I don't see how f could = 2y

[scenic crow]

D on the picture k sent i apologise for the confusion

[compact raven]

oh my bad

Alright yeah

[scenic crow]

You got it?

[compact raven]

Nah sorry

[scenic crow]

so

a line crossing 2 perpendicular lines right

[compact raven]

On the pic K sent
2x = D
therefore D = H But what is H in this scenario

[scenic crow]

I mean a line between 2 parallel lines

[compact raven]

yeah

[scenic crow]

D is not only H

[compact raven]

it is also A

and E?

[scenic crow]

yup

and if 2x is E and 2y is F?

[compact raven]

2y = f
therefore 2y = G
2y = C

oh

[compact raven]

180

[scenic crow]

yup exactly

[compact raven]

oh so that means x is 45 and so is y

tyty

[scenic crow]

Not necesseraly

[compact raven]

:(

[scenic crow]

you just know that both of them together are 90°

[compact raven]

therefore

[scenic crow]

after all why couldnt x be 42 and y 48

[compact raven]

i see

[scenic crow]

Hm does it matter for this particular problem where x+y is all you're looking for?

[compact raven]

nah I need to prove that F1 = 270

[scenic crow]

exactly

[compact raven]

repost so we dont have to scroll

[scenic crow]

yup

so

if 2x + 2y = 180°

2hich we have proved already

[compact raven]

yes

[scenic crow]

what do we need to calculate F

x and y or only their sum?

[compact raven]

only their sum no?

[scenic crow]

yup

[compact raven]

2x+2y = 180
x+y = 90
therefore F= 180 - 90
F= 90
F1 has been proven to be 270

Ill rewrite everything and everything that has been proven in to my book later

[scenic crow]

Well done

[compact raven]

I still have the other two to do

but first

[scenic crow]

4 and 5?

[compact raven]

can u check my work on this one I did

[compact raven]

yeah

I am missing two reasons so im sceptical that this even works

the two missing reasons are indicated by "help"

[scenic crow]

what can you tell me abvout the triangles DEA and CDB

[compact raven]

There sides are parallel indicated by the one arrow, two arrows, and three arrows

[scenic crow]

thus the angles inside should be...

[compact raven]

The same

[scenic crow]

thats exactly right

[compact raven]

But dont I need a reason?

Can i really just write 'parallel sides'

[scenic crow]

well if their sides are parallel and their angles are the same they are...

[compact raven]

I see

thanks now on to 4

[scenic crow]

i was aiming at congruent

[compact raven]

Im slow

So i say congruent?

[scenic crow]

no worries you are doing good

[compact raven]

ty

[scenic crow]

perhaps im being to vague

[compact raven]

Nah i shouldve have thought of htat

since they are congruent

[scenic crow]

its something you get better at with practice

and then ABD would also be congruent right?

[compact raven]

yes

[compact raven]

For these two I don't even know where to start

[scenic crow]

It usually helps to start with the definitipon pf the parallelogram

so that we know what we ned to prove that something is a parallelogram

[compact raven]

Alright heres what I have

Ok i dont hava definition but I have something else basically a definition

The diagonals of parallelogram bisect each other. The oppposite angles of a parallelogram are equal. The interior angles add up to 360. The opposite sides of a parallelogram are parallel and equal.

So wait

Its given A1= A2 and the same with C. So we have A1 = C1 A2 = C2

[scenic crow]

yup

[compact raven]

But we cant say that because we havent proven that it is a parallelogram right?

Brb

[scenic crow]

well the first sentence is that ABCE is a parallelogram

[compact raven]

back

[compact raven]

Oh i see

So from here what needs to be done?

How do I go about it

[scenic crow]

Well can we assume that the lines AD anf FC are parallel?

[compact raven]

Yes since there angles are the same

[scenic crow]

so what is d1 equal to

[compact raven]

F1

[scenic crow]

and thus d2 is equal to

[compact raven]

F2

[scenic crow]

so

A1=C1 and D2=F2

and A1 is opposite of C1 and D2 is opposite of F2...

[compact raven]

Therefore AFCD is a parallelogram

[scenic crow]

yup

well done

[compact raven]

angles of a parallelogram is the reason right?

[scenic crow]

yes

The oppposite angles of a parallelogram are equal

as you said yourself

[compact raven]

yeah

Now onto 5

[scenic crow]

good

now

where would you start

[compact raven]

D = C2 angles opp. equal sides
A2 = B angles opp. equal sides

no wait

let me edit that

oh

AB ll DC

so

ABCD is a parallelogram because opposite sides of a parallelogram are parallel and equal.

[scenic crow]

how do we know that AD and BC are parallel

[compact raven]

Is that not given?

[scenic crow]

is it?

[compact raven]

Well i assumed it was because there was a line showing congruency

Showing that they are congruent

[scenic crow]

that line means that they are equal in length

not that they are at the same angle

lets start from the beginning

with a bit of logic

all of the problems until now were solved using angles so this one will be too

and to prove that this is a parallelogram we have to prove that A = C and B = D

lets start with A=C

[compact raven]

i know D = B but we first have to prove that ABCD is a parallelogram so I dont what reason to give

[compact raven]

ok

A1 = C1 alt angles?

[scenic crow]

that is correct

[compact raven]

the same with A2 =C2 alt angles

[scenic crow]

correct

now what is B equal to

[compact raven]

B = D but we need a reason :(

[scenic crow]

look at the triangle ACB

[compact raven]

B = A2

angles opp equal sides

[scenic crow]

the sum of the angles is 180° right?

[compact raven]

Yes

[scenic crow]

so

so

A2 + C1 + B = 180

[compact raven]

yes

[scenic crow]

so

B=180° - A2 - C1

no what is D

using the same logic

[compact raven]

A1 + C2 + D = 180
D= 180 - A1 -C2

[scenic crow]

and as we said before

A1=C1 and A2=C2 right

[compact raven]

yea

oh

[scenic crow]

so if we replace all A1 with C1 and all A2 with C2

[compact raven]

it will be the same

therefore

B = D

[scenic crow]

exactly

well done

[compact raven]

ABCD is a parallelogram because the opposite sides of the quadrilateral are equal and parallel.

[sharp pilot]

@compact raven do you know that if both pair of opposite angles of a quadrilateral are equal then it's a parallelogram? Yes or No

[compact raven]

yeah

[sharp pilot]

Ok then it's done

[compact raven]

yeah but we had to prove that ABCD is a parallelogram

you cant just say that

you need proof/evidence

which we now have

[sharp pilot]

You just got B=D and you knew before that A=C so by that we proved its a parallelogram

[compact raven]

Yeah

Ok well ima write all this down now

@scenic crow Thanks alot man!

[scenic crow]

You're welcome have a good one

[compact raven]

You too!

🫡

[scenic crow]

use .close when you feel like you wont need this channel anymore to free it up for others to use

[compact raven]

Can I use it but still read message history?

[scenic crow]

yes it saves

[compact raven]

Alright will do thanks

.close

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[compact raven]

<@&286206848099549185>

This channel wont go to the top category

.close

.reopen

[vocal sleet]

✅

[compact raven]

.close

[vocal sleet]

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Use .reopen if this was a mistake.

[compact raven]

man

[scenic crow]

it goes away after some time dont worry about it

[compact raven]

oh alr ty

[compact raven]

.

.reopen

[vocal sleet]

✅

[compact raven]

hey sorry to disturb again

How do I write this all out

Must i draw what k sent as proof of corresponding angles

how do I write out that x+y = 180?

Everything before that

too

[scenic crow]

they are parallel lines

so 180° -2x = 2y

[compact raven]

so x + y = 180 reason: parallel sides?

[scenic crow]

=> 180° = 2x+2y

[scenic crow]

well its a line crossing two parallel lines

[compact raven]

So I don't have to give a reason ?

given?

[scenic crow]

if you have to write something id say equality of corresponding angles

[compact raven]

in this case then what is x corrosponding to?

y?

[scenic crow]

2x is corresponding to 180° - 2y

you have to look at the ones on the opposide sides of the line

[compact raven]

i see thanks man

[scenic crow]

you're welcome

[vocal sleet]

@compact raven Has your question been resolved?

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[high dawn]

why is this question incorrect? I'm using 2-sampTint on the TI-84
x1=87
x2=96
sx1=12
sx2=8
n1=37
n2=30

[glass oak]

well we need to see what you got for the t-values

[high dawn]

the confidence intervals? or something else

[glass oak]

I would start by writing out the formulas you're using

[glass oak]

well how did you do this problem

[high dawn]

my professor just wants us to use the calculator

uh

through calculator

2-samptint

i'll send what i put in

[glass oak]

oh. do you h ave instructions for it?

[high dawn]

yee

[glass oak]

http://tibasicdev.wikidot.com/2-samptint

[high dawn]

he told us in class to use the calculator functions for the work so we don't have to do by formula

[glass oak]

got it

[high dawn]

this is what i'm getting

[glass oak]

reading up on the calculator command

[high dawn]

ty 🥹

[high dawn]

?

[glass oak]

just copy and pasting from the ti84 manual

[high dawn]

ohh ok

website might just be wrong as well because it's a free website thats had a couple of wrong questions throughout the sem

but i just wanna make sure

[glass oak]

thinking

I don't see any mistakes

in your work

[high dawn]

it might be the website then

i'll email professor thank u 🥹

[glass oak]

[high dawn]

how do i close help channel?

[tall rain]

.close

[high dawn]

.close

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[limpid mauve]

Hey, I had an exam a couple of days ago, and I need y'alls opinion on this

[limpid mauve]

They said the energy of a photon is given by this

Which confused me, because in my experiance, this is not a you can write formulas

Or relationships

[pure marten]

I think it just comes from E = hf

[limpid mauve]

Yeah

[pure marten]

What do you mean by this then

[limpid mauve]

The way they use the brackets

[pure marten]

¯_(ツ)_/¯

The point of any notation is its convenience

[plush condor]

I think those are just the units

[pure marten]

You can drop the parentheses too if you want

[plush condor]

Energy in Mega electron-Volts and wavelength in picometer

[limpid mauve]

That is what I think now

[plush condor]

and the r could be for radiation?

[limpid mauve]

But I was unsure if they meant momentum × meter

[limpid mauve]

It's gamma

[heavy yoke]

in my opinion it would be better to write it as [ E_\gamma = \frac{\qty{1.24}{\MeV\pm}}{\lambda} ] but it's still reasonably clear what's meant here

[plush condor]

Oh... fair enough

[limpid mauve]

I juat have bad hand handwritting 😅

[plush condor]

Nah, your handwriting is much better than mine

dw

[pure marten]

im sure you can multiply a quantity and a unit

[twin meteor]

cloud

[limpid mauve]

My problem is that they mixed the symbols and units

[limpid mauve]

(I might have spent like 2.5 min on it with a ruller so it would be clear)

[plush condor]

I feel you on that one brotha,

[limpid mauve]

Like that is not how it looks normaly xd

Though

I wrote this 3

This is the best 3 I've written in my life

(Forget the Ω)

[limpid mauve]

Yeah

But tbf

I was severely sleep deprived

So what they did to be lazy, got me really confused

[limpid mauve]

Yeah, that would have been an acceptable way

Well anyways, that's it

.close

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[soft lodge]

yo

[soft lodge]

so for parts of a circle the radius is 14cm

i have 2 x π x r

2 x 22 x 14 divided by 7

i understand what the 2 and the 14 is

but why 2 x 22

[edgy kayak]

22/7 is an approximation of π

it is a good one here because things nicely cancel out

[soft lodge]

ok so for all questions 22/7 works out

[edgy kayak]

generally approximations depend on the question

[soft lodge]

ok

[edgy kayak]

but depending on your level, 22/7 should be fine if the questions are non calculator

[soft lodge]

ok so i have here

the radius is 35cm

i did 2 x 22 x 35 divided by 7 and got 220 as the circumference

[edgy kayak]

yes

[soft lodge]

right

and i got 7700 as the area

is this right

[edgy kayak]

what formula did u use

[soft lodge]

i used this

π r to the power of 2

22 over 7 multiplied by 35 x 35

[edgy kayak]

,calc (22/7)*(35^2)

[twin meteor]

Result:

3850

[soft lodge]

shit mb

ty

i got one more

the radius is 91m

not cm

do i have to make changes or na

[edgy kayak]

what do u think

[soft lodge]

i mean

i think u either treat it the same or turn it into cm

[edgy kayak]

yes

[soft lodge]

which one

i did it normally and got 572 as circumference'

@edgy kayak

[edgy kayak]

yes

[soft lodge]

and for the area i got 26,026

is this correct

[edgy kayak]

u dont need to keep asking for verification if u know the formula

[soft lodge]

yeah but i could make mistakes

[vocal sleet]

@soft lodge Has your question been resolved?

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[bold oriole]

Is anyone able to confirm that I have done this correctly, I have to do a quick presentation in class tomorrow about this question and want to make sure I did it right

[bold oriole]

.close

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[vocal timber]

One of the options that is my answer is option B, how do I identify which other option is correct

[mint badge]

well if your solution already matches up with an answer

then u already solved it?

idk what ur asking tbh

[paper depot]

it's a select all that apply question

[mint badge]

i se

see

[silk rampart]

Plug a value

[paper depot]

• A is out because it has no sqrt(x) in denominator (or indeed any such factor anywhere)
• C is out because it's clearly not equal to B (same denom, different num)

[mint badge]

good strat tbh

[paper depot]

D you can write the cot as cos/sin and simplify

and see it actually is equal to B

[vocal timber]

Let me do

[silk rampart]

@vocal timber just plug x = pi^2/16

or pi^2/36

[vocal timber]

How did you think of this value?

[silk rampart]

can't plug 0

can't plug pi^2/4

[silk rampart]

it is just D though

you don't even have to plug anything

do what ann said

[vocal timber]

Yes

That's what one can do in mind without writing

There's one more question like this

if y = sin³x
A) 3sin²xcosx
B) 3sinxcosx
C) 1.5sinxsin2x
D) none

Option A matches on solving

But there's other options to

[silk rampart]

it's AC

[vocal timber]

C

[silk rampart]

use sin(2x) = 2sin(x)cos(x)

[vocal timber]

How

How do I apply

[silk rampart]

just plug it 😭

[vocal timber]

there is cosx common to both which I can plug?

[silk rampart]

no, u see sin(2x) in option C

you just replace it with 2sin(x)cos(x)

and you shall see that it matches option A

[vocal timber]

Oaky

.close

[vocal sleet]

Channel closed

Closed by @vocal timber

Use .reopen if this was a mistake.

[paper depot]

honestly meh

[silk rampart]

i know but sometimes it works

[vocal sleet]

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[fiery path]

Anybody knows why not opt b)

[vocal sleet]

@fiery path Has your question been resolved?

[flat whale]

Picking the two remaining balls are not independent of the first 8 balls

There are combinations of 10 where the 2 last balls could be from the first drawing of 8 and vice versa

[fiery path]

Umm , I don't get it 😭?

[flat whale]

Do a simpler example with just 1 box of 3r and 2b and selecting 3 balls

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[flat whale]

Should be 9 in the simpler example

But your method would over count