help-10

[halcyon hull]

so should i calculate the area of 2s to 3s under the graph?

?

but how do i find that

[shy wasp]

if you calculate the area between 0 to 2s, what do you think that'd give us?

[halcyon hull]

the area of the top triangle

[shy wasp]

Yes and what physical quantity does that represent?

[halcyon hull]

um

displacement?

[shy wasp]

yes exactly

for the first two seconds the area under the graph is 1/2 * 2 * 5

i.e. 5

[halcyon hull]

but since its under the graph

[shy wasp]

so that means we moved 5 m ahead

[halcyon hull]

wouldnt it be -5

[shy wasp]

the v is always positive in the first 2 seconds if you notice

the area we are calculating is over the x-axis

[halcyon hull]

oh

since you said under the graph, i thought you meant in the -y axis

[shy wasp]

so now you know what to do next

calculate the area

from 2s to some t second such that the area cancels out our current displacement

[halcyon hull]

by under the graph, do we mean under the slope/line?

like, the region under it?

or does it mean under the x axis

[shy wasp]

yes the region bound by the x-axis and the line/graph

sorry if my language wasn't clear earlier

[halcyon hull]

so would it mean something like this?

heres a better picture

[shy wasp]

yes exactly, A1 is the area under the graph from 0 to 2s

[halcyon hull]

ohhh

i see

[shy wasp]

and the area is positive

hence the positive displacement

[halcyon hull]

cna it be any t value? after 2?

can*

[shy wasp]

no, you have to calculate it

do you need another hint?

[halcyon hull]

before that,

to find the position or displacement that makes t=0, should the area after t=2 be the t value that cancels out the area between t=0 and t=2?

[halcyon hull]

in other words, should it be the value that cancels out the area of a1 in this image?

in other words*

[shy wasp]

YESSS

[halcyon hull]

ooo i see

[shy wasp]

after t = 2 the displacement should be -5

since after t = 2s, the v is always negative, that means that the penguin is returning

[halcyon hull]

why -5?

[shy wasp]

so after some time it will cross the original starting position

[halcyon hull]

i see

[shy wasp]

Because we calculated A1 = 5, that's why

[halcyon hull]

ohh and its -5 because it cacels out the max displacement of 5?

[shy wasp]

YESSS

[halcyon hull]

ok

between t=2 and 3, the area is -2.5

so should i add another 2.5s to balance it out?

[shy wasp]

so there goes our -2.5

[halcyon hull]

since a1 is 5

[shy wasp]

this means that between 2s to 3s, then penguin travelled 2.5 m towards the starting position

what would adding 2.5s mean now?

[halcyon hull]

he goes back to the starting pos?

[shy wasp]

after 3s our velocity is constant and it is equal to -5

how much time will the penguin take to travel -2.5m?

[halcyon hull]

1 second

[shy wasp]

it is travelling at -5 meters per second

[halcyon hull]

oh

so half a second

[shy wasp]

yes

[halcyon hull]

woahhh wait a minute

i think i finally figured it out

im gonna try to explain it

[shy wasp]

yes please

[halcyon hull]

so basically, first we find the area under the graph

so from t=0 to t=2

that area was 5m/s

but then it goes constant at t=3

between t=2 and 3, the area is -2.5m/s

after that, the velocity is 5m/s constantly

[shy wasp]

that area was 5 meters, the units

[halcyon hull]

and since is 5 m in 1 second, then id have to do have a second for it to be -5m/s

and -5m/s cancels out 5m/s

so the answer is 3.5s

[shy wasp]

yes perfect explanation

[halcyon hull]

thank you so much

this was a lifesaver

have a nice day/night

[shy wasp]

so next time you can just calculate the area

and if the area is under the x-axis, that area would be negative, ie negative displacement, otherwise positive

then proceed according to the question

[shy wasp]

you too

also don't forget to .close if your query has been solved

[halcyon hull]

.close

[obtuse pebble]

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[celest zodiac]

is the symbol in front of ((x,y),(a,b)) a d?

[celest zodiac]

also what does s represent and d?

[timid silo]

d is a distance function

delta (not s) is a bound on the distance

[royal shard]

could also be a $\partial$

[warm shale]

martin3125

[royal shard]

since the thing on the right of ((x,y),(a,b)) is a delta

this "partial" (which is also kinda a delta i think) can mean partial differentiation, but also the "bound" of something as Toby said

for example $\partial M$ would be the "edge" or "bound" of the set M

[warm shale]

martin3125

[royal shard]

but in this case i think Toby is right

tldr: ignore what i wrote

[sacred lichen]

definitely since this is the epsilon delta definition, so we require a metric d

[celest zodiac]

ahh okay

also what does the upside down A and backwards 3 represent

does the backwards 3 mean all real numbers

[royal shard]

the A means "For all"

the 3 means "There exists"

there is also an option for the 3 with a "!" behind it
that means "There exists exactly one"

oh wait

i meant $\exists$

[celest zodiac]

in this case since its A3>0, its for all what?

[warm shale]

martin3125

[royal shard]

but i think you mean the $\varepsilon$

[warm shale]

martin3125

[royal shard]

that is just a variable

we often use epsilon and delta in these things

[celest zodiac]

ohhhh okay

thank you sm this was so helpful <3

[royal shard]

so here we could also write: $\forall\varepsilon >0\exists\delta >0: (0<d((x,y),(a,b))<\delta)=>|f(x,y)-L|<\varepsilon$

[warm shale]

martin3125

[royal shard]

which would mean:
"For every epsilon greater 0 we can find a delta greater 0 such that |f(x,y)-L|<epsilon is true for every x,y that satisfy 0<d((x,y),(a,b))<delta"

[celest zodiac]

omg thank you so much for breaking it down

its so much clearer <3

[obtuse pebble]

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[timid silo]

f(x) = 2/x

is it correct to say that
f c R² (c means "is a subset of")

If we look at the definition of a function, then it's not. It would mean that every element in R is related to exactly one element in set R.
But zero is not related to any element, so this is not true.
This is true:
f c R\{0}xR

but R\{0}xR is a subset of RxR, and f is a subset of R\{0}xR, then f is a subset of RxR. This is transitive property of "c" (is subset) relation.

But because of the definition of a function we can't say that f is a subset of RxR.
Or is it that, f is a subset of RxR but f is no longer a function because the definition is not satisfied, so it is just a relation.

So, for f(x) = 2/x , is this true or not:
f c R²

TLDR: Anyway... the thing that bothers me is this
If this is a function f(x) = 2/x
is it true that f is a subset of RxR?
If not, then it seeems to me that "transitive property" of "subset relation" is broken.

some set f is a subset of R\{0} xR, and R\{0} xR is a subset of RxR, then f is a subset of RxR.

[warm canopy]

what definition of function are you looking at that is making you think f cant be a subset of RxR

particularly, what is it about the definition that is making you think it cant be a subset of RxR

[timid silo]

it can't be a subset of RxR, because in definition we say that every element from the first set must be related to exactly one element of the other set.
If we say that f is a subset of RxR, then it must (by definition) have all real numbers related to exactly one element. But zero is not related, so f is not a subset of RxR.
... or f is a subset of RxR but it's not a function, just a relation.

[warm canopy]

so the definition is telling you it must be a subset of R/{0} x R, which is a subset of RxR, i dont see the issue here

f being a subset of RxR doesn't mean you now must have (0,something) in its definition

[obtuse pebble]

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[obtuse pebble]

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[sage dagger]

I know this is a simple question but I get confused with what goes where, I would like some help with a couple of questions in 6 and 7

[timid silo]

Okay what's troubling you

[sage dagger]

I’m fairly new to this log stuff and I often forget what stuff goes where in log form

like for example part a

Does that mean log base y = 3^x or is it the other way around I don’t quite understand

@timid silo

[timid silo]

Okay so @sage dagger

[
y = a^x \implies \map {\log_a} y = x
]

[sage dagger]

ah

okay

So 6a) log base 3 of y = x?

and for y= x^2 would be log base x of y = 2?

[restive acorn]

Yes

Log gives you the exponent, so to speak

[sage dagger]

What about Q = e^x

since that’s ln

[restive acorn]

Ln is log base e

Just a different notation for that one

So

lnQ = x

In that case

[sage dagger]

Ah okay that makes sense

And for question 7 it’s just reverse engineering and doing the opposite

[restive acorn]

Yes the reverse of taking a logarithm is "exponentiation by the same base"

Thats the way i describe it

So problem 7a

5 = 3^x

3^[logbase3(x)] = x

In the nitty gritty of the algebra

[sage dagger]

Where does the 5 go?

[restive acorn]

This is another example

Not same problem

[sage dagger]

Ah

Right

[restive acorn]

This is the answer

[sage dagger]

Oh yeah

So the exponent is sort of like what the log equals to?

[restive acorn]

So you get

3^ [logbase3(5)] = 5

In that step

[sage dagger]

I have a question, why does the 3 get raised to the log? I’ve seen that form before but I don’t understand why we do it

[restive acorn]

Because to cancel out the log

You have to raise it as a power

You exponentiate it by the same base

[sage dagger]

Alright that makes sense

[restive acorn]

The reason you show that is understand what you did mechanically to get from one line to the next

You raised both sides up as exponents

[sage dagger]

Okay I think I get it now

Thank you for the help

.close

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[sage dagger]

I’m not sure how to start

Do I have to get this in index form first?

[late stump]

which question in particular are you stuck on?

[sage dagger]

Just in general

I haven’t done this type of question before

Where I solve for x in a log form equation

[late stump]

okay this would be a good start

especially for the more concise problems like a and b

[sage dagger]

So for part a

x = 10^6?

[late stump]

yup, correct!

and the same idea applies for b

[sage dagger]

Okay.

What about for c

I mean

d

Where x is now the base

[late stump]

well, how did you get the index form for question a?

you used the base of the log as the base of the index, right?

[sage dagger]

Yeah

[late stump]

so the same reasoning applies here

[sage dagger]

Okay

@late stump 64 = x^6?

[late stump]

yes! now, you can just take the 6th root of both sides

[obtuse pebble]

@sage dagger Has your question been resolved?

[obtuse pebble]

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[icy viper]

Hello, I'm a Chilean student. I apologize in advance if my English is bad. I'm seeking help to solve the following problem (it's translated into English, I hope I did it well... :c)

[icy viper]

While doing some research to solve the problem, I came across graphs (not sure if they are called the same in English), Eulerian and Hamiltonian paths, but I still can't solve the problem.

[polar fossil]

yes they're called graphs in English as well
and a hamiltonian path is what you want

[icy viper]

It really is like a combination of a Hamiltonian and an Eulerian path. I need to go through all the points and lines.

[polar fossil]

you don't need to go through all the lines if i'm reading that correctly

[icy viper]

between each point if there are stations, in the image only the terminals and the combinations are considered, but in between there are more stations.

[polar fossil]

oh right

technically you don't need to travel from Puente Cal y Canto to Santa Ana

or a few others

[obtuse pebble]

@icy viper Has your question been resolved?

[icy viper]

I talked to my teacher and he told me that the answer is 6 paths, but I still don't know why. He also told me that connections can be made from any station (not just combinations or terminals)

<@&286206848099549185>

[jovial lynx]

de que universidad es

[icy viper]

uc

[jovial lynx]

suerte si alguien te responde wn, sigo con lo mio

[icy viper]

😭

[obtuse pebble]

@icy viper Has your question been resolved?

[jovial lynx]

asi amix?

esto es con el algoritmo blossom

[obtuse pebble]

@icy viper Has your question been resolved?

[jovial lynx]

@icy viper

[icy viper]

Perdón

[icy viper]

Si así es el mapa

[jovial lynx]

https://brilliant.org/wiki/blossom-algorithm/

[obtuse pebble]

@icy viper Has your question been resolved?

[obtuse pebble]

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[rough nexus]

Is there an algebraic way to form a quadratic equation such that:

1. There are two positive integers as solutions when the equation is equal to 0;
2. There are two positive integers as solutions when the equation is equal to -1

[rough nexus]

I've tried to form two equations with 4 different roots and tried to solve for anyone of them, but in the end I made no progress and it's just a mess and a bunch of identities.

[static furnace]

i'm not sure about the second one off the top of my head, but the first one should be straight forward right?

[polar fossil]

i think the idea is to make it satisfy both conditions at once

[static furnace]

ohhhhhhhhhh

yes that would complicate things

[polar fossil]

you need $\sqrt{b^2 - 4ac} \in \Z$ and $\sqrt{b^2 - 4a(c+1)} \in \Z$ if I understand this right

[warm shale]

Hayley

[polar fossil]

so you need uh

two squares that are some distance away from each other

[zenith raft]

you need more than that i think

[elfin burrow]

what about $-\frac{1}{3}x^2+\frac{1}{3}$?

[warm shale]

tushar

[static furnace]

both roots there are not positive. you're left with x^2=1 which has -1 and +1 as roots

[elfin burrow]

oh, i thought they just wanted integer roots

you can just translate it to the right though

[static furnace]

both positive unfortunately

[elfin burrow]

simply $-\frac{1}{3}(x-3)^2+\frac{1}{3}$

[warm shale]

tushar

[static furnace]

wait this works though

[obtuse pebble]

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[fallen needle]

how would I solve this question?

[grizzled shore]

what have you tried

[fallen needle]

I'm confused with where to go

I know that there needs to be more than 12 days in this sample of rainy days

[grizzled shore]

how did you get 12

[fallen needle]

0.4 times 30

[grizzled shore]

right

[grizzled shore]

if i write it as P(rainy days > 12) are you comfortable with this notation?

[fallen needle]

👍 '

[grizzled shore]

are you also comfortable with saying that rainy days are bernoulli trials with parameter (probability) = 0.3?

[fallen needle]

not rlly

[grizzled shore]

have you heard of bernoulli trials?

[fallen needle]

now i do haha

[grizzled shore]

ok um

bernoulli trials just means they are binary

so either it rains or it doesn't

[fallen needle]

understandable

[grizzled shore]

if we have a collection of n independent bernoulli trials we call it a binomial distribution

which is what you have here

we would say find P(X > 12), given X ~ B(30, 0.3)

[fallen needle]

ahh ok ok

[grizzled shore]

do you see where they all come from

[fallen needle]

umm number of trials = 30, probability = 0.3?

[grizzled shore]

yeah

if we have n bernoulli trials with probability k

then the number of "successes" can be described by a binomial distribution written as B(n, k)

[fallen needle]

ahhhhh i got it now

thank you

.close

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[granite notch]

I have a function, f(x), which must contain the point (0, pi/2) and have a horizontal asymptote at y = 1. However, I want to change it so that f(x) approaches y = 1 faster. Any ideas how I can do this? (ideally by adding a constant variable I can tweak)

[night moth]

make it something like f(nx) probably? where n is greater than 1

[rigid pine]

I don't know if this is cheating but:
f(x) = 1, x ≠ 0
= π/2, x = 0.

[rich silo]

Yes

You see those two 2's, the greater the value the faster the decay/damping

[night moth]

mhm

[granite notch]

ah, this works much better

fyi, I'm trying to fit a curve to arctan

[night moth]

the factor n essentially changes the rate of change of the function by a factor of n (when you differentiate f(nx), it is n * f' (nx) )

[granite notch]

this is where I'm at now

black is atan

this function is bad

[night moth]

maybe you could use taylor series to approximate tan and then switch out the x and y?

[granite notch]

I was actually just trying to do an approximation using just the function 3x^2 - 2x^3, but it involves me solving for y in this equation

the results are quite good if I do manage to get it working tho

[night moth]

wow how did you come up with that?

[granite notch]

starting from the smoothstep function (modified to be from -1 to 1)

this can be chained together and stretched some more to create false sin and cos

then I divided fsin by fcos to get ftan

[granite notch]

after expanding and simplifying, I got this

[obtuse pebble]

@granite notch Has your question been resolved?

[obtuse pebble]

@granite notch Has your question been resolved?

[obtuse pebble]

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[worthy dagger]

I’ve been given homework involving sequences and series, and financial mathematics, and this questions, specifically the last part has stumped me

[worthy dagger]

This is what I’ve got thus far

I’ve got the answer to the last questions within the questions, but I need help on how to actually figure that out

[obtuse pebble]

@worthy dagger Has your question been resolved?

[obtuse pebble]

@worthy dagger Has your question been resolved?

[obtuse pebble]

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[slow phoenix]

btw these are the workings

i feel like i went wrong somewhere in the first picture & ill appreciate your patience and guidance helping me ^^

[slow phoenix]

if required ill send a new pic of the working

if the above one isnt clear

[buoyant ingot]

Sup

[slow phoenix]

hello TvT

ive been stuck on this question for damn long

[buoyant ingot]

How long

??

[slow phoenix]

3 hours TvT

[buoyant ingot]

(O_o)??

How

[slow phoenix]

issit easy o.0

IDEK

cause i keep doing stoopid mistakess

[buoyant ingot]

Get a calculator

[slow phoenix]

help me TvT i wanna sleep i wanna finish this qn and set myself freeeee

[slow phoenix]

really i alr have one

is there any actual helpers here ?????

i saw 3 ppl typing and someone here said get a calculator TvT

[buoyant ingot]

TvT

[slow phoenix]

uk how to solve?

[buoyant ingot]

TvT

[slow phoenix]

i desperately need help TvT

<@&286206848099549185>

:">

[obtuse pebble]

@slow phoenix Has your question been resolved?

[vapid hamlet]

TvT

[obtuse pebble]

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[timid silo]

Hello, I have this question here that I could use some help with. It's just a substitution problem:

[timid silo]

Oh.

I realized that I need to try again before asking. Found an error in my work.

.close

[obtuse pebble]

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[echo gazelle]

I'm rlly confused about lower and upper RHS and LHS estimates

[echo gazelle]

So RHS is always upper/overestimate and LHS is always lower/underestimate?

[late stump]

is this for riemann sums?

[echo gazelle]

yes

[late stump]

then not necessarily

look at this example

https://www.storyofmathematics.com/wp-content/uploads/2021/05/left-and-right-riemann-sum-graph.png

the left and right only really denote which corner is touching the graph

[echo gazelle]

ah

[quaint lily]

right would be overestimating for increasing functions

and vice versa

[echo gazelle]

I'm getting confused whether over or underestimate starts at the beginning of the chart when t=0 For ex this chart

t | 0 2 4 6 8 10
v | 0 1 3 4 4 6

if it was n=5, △t = 2 in my notes i have RHS = upper estimate LHS = lower
LHS = △t (v(0)+v(2)+v(4)+v(6)+v(8)) and
RHS = △t (v(2)+v(4)+v(6)+v(8)+v(10))

but for all the practice problems i've done i've followed that generally and keep having to swap my answers for LHS and RHS for the correct answer. How do you know which formula to use? Are those formulas correct for RHS and LHS or are they swapped?

What im trying to get at is how do you know which formula to use for upper or lower estimate? Is lower estimate always v(0) + ...

[late stump]

if the function is decreasing over the interval, left would be overestimating

if it’s increasing, then right is overestimating

[echo gazelle]

okay

[echo gazelle]

which formula is for overestimating and which is for under?

[late stump]

again, depends on the graph you’re dealing with

[late stump]

look at what tan and i said

[echo gazelle]

Okay

ty for helping me understand it better i watched a video and i get what you're saying better now

.close

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[undone sentinel]

how do i solve this

[obtuse pebble]

@undone sentinel Has your question been resolved?

[glossy basalt]

I have read this a few times and i still don't get it, is that the whole question or am i missing something?

[undone sentinel]

this is the entire question

these are the choices

[glossy basalt]

sorry, i really don't know, wish that i can help😔

[undone sentinel]

its an sat question if that does anything

[undone sentinel]

this is their explanation if you can understand it

[glossy basalt]

back, now reading

oh......... i think i get it now

the wordings are so... ambiguous

[glossy basalt]

i think the point is
"8 kilometers in one hour"

argh, it's so weird

okay, i give up, this is so weird

[obtuse pebble]

@undone sentinel Has your question been resolved?

[obtuse pebble]

@undone sentinel Has your question been resolved?

[timid silo]

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[tight patio]

If the 2nd term of an arithmetic sequence is 10 and the 7th term is 35, find the 11th term.

[tight patio]

I got 260

is that correct

[errant lark]

Nope

!show

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

[errant lark]

So that we can point you where you went wrong.

[tight patio]

a11 = 10 + (11 - 1) * 25
a11 = 10 + 10 * 25
a11 = 10 + 250
a11 = 260

[tight patio]

i did this

[errant lark]

What is the formula for $a_n$ ?

[warm shale]

.enemagneto

[tight patio]

a11 = a1 + (n - 1) * d

?

a11 is the 11th term of the sequence,

• a1 is the 1st term of the sequence,
• n is the position of the term,
• d is the common difference between consecutive terms.

I have notes

[north smelt]

yes must be Like thay

[errant lark]

a11 ? or an ?

[tight patio]

the formula is an

sorry

[brave bramble]

So worth saying that a2 and a7 are not consecutive.

[errant lark]

Okay.

[tight patio]

wdym?

[errant lark]

But here, you used a2 in place of a1. That's wrong.

[tight patio]

we dont have the 1st term???

what?

[errant lark]

You aren't given the first term.

You can calculate it though.

[north smelt]

d is not 25?

[tight patio]

diff is 25

[errant lark]

No.

[tight patio]

35-10

?

HUH how

[north smelt]

d = 5

a1 = 5

[tight patio]

why?

ohhh

what

i get it

I understand now

[north smelt]

a11 = 5+50 = 55

[tight patio]

yeah

i was gonna show that

tysm

@north smelt

how did u get a1 tho?

[north smelt]

you are welcome

[tight patio]

.close

[obtuse pebble]

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[north smelt]

a1+d = 10
a1+6d = 35
5d = 25
d = 5
and a1 = 5 too

if a2 = 10, then a2 = a1+d = 10

if a7 = 35, then a7 = a1+6d = 35

[obtuse pebble]

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[obtuse pebble]

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[sage dagger]

I’m trying to solve for x, is this right so far

[timid silo]

yes

[sage dagger]

And all I have to do now is square root both sides right

[timid silo]

yes

no

they're both already square roots

[shadow tangle]

x^(1/2)=sqrt(x)

[sage dagger]

But don’t I have to get x on it’s own

[timid silo]

yes

[sage dagger]

Oh

Wait

I can square both sides

Right

[shadow tangle]

yep :)

[sage dagger]

so then it’s just x = 3

Alright cool

Ty for the help

[shadow tangle]

np

[sage dagger]

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[stoic elm]

how do i do this one?

[fathom flicker]

what is the missing angle

[stoic elm]

62

[fathom flicker]

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[fathom flicker]

does this give you any information about the triangle?

[stoic elm]

<C and <b are the same

[fathom flicker]

Sure

and does that give you any info about the type of triangle?

or about the possible side lengths?

[stoic elm]

its isosceles?

[fathom flicker]

yes

meaning what

[stoic elm]

2 equal sides and 2 equal angles?

[fathom flicker]

yes

and what are the equal sides then?

[stoic elm]

ac and ab?

[fathom flicker]

yes

so what is the answer then?

[stoic elm]

AB is congruent to AC

.close

[obtuse pebble]

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[minor parcel]

I need help with a

[obtuse pebble]

@minor parcel Has your question been resolved?

[obtuse pebble]

@minor parcel Has your question been resolved?

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[slim garnet]

Can someone help me with 10a? I have done a little bit so far but It’s a little messy and I don’t know where to continue

[obtuse pebble]

@slim garnet Has your question been resolved?

[slim garnet]

<@&286206848099549185>

[main elm]

you want the area?

@slim garnet

[slim garnet]

yes

@main elm

[slim garnet]

@main elm

<@&286206848099549185>

[untold trail]

which triangle do u want to find the area of ?

@slim garnet

[slim garnet]

A

The first one

[untold trail]

but theres 2 triangles

are you trying to find the area of teh whole figure?

[slim garnet]

yup

Sorry, I’m in class @untold trail

[untold trail]

no worries

lemme take ae look

[obtuse pebble]

@slim garnet Has your question been resolved?

[untold trail]

is this what was given?

[slim garnet]

Yup

The angle on the top is 115

Because they are supplementary

I got my teacher to do it

Thanks for your time

.close

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[vivid cedar]

am lost on how to do this

i have 0 clue

well i know that radius of circle=sqrt(13) and center is 3,-4

but beyond that idk

OH WAIT

no wait

its tangent

so you have to take a pointon the circle

but

hmm

yea im lost

[violet sentinel]

I'm assuming that you haven't worked with derivatives yet?

there's an easier geometric way anyways 🙂

ok so how do we find the slope between the two points?

[vivid cedar]

slope formula

y2-y1 over x2-x1

[violet sentinel]

bingo

so we have two points right?

what's the center?

[vivid cedar]

(3,-4)

[violet sentinel]

great!

so that's the slope between the points?

[vivid cedar]

but its tangent right so we cant use the cneter

[violet sentinel]

no you'll see what we do

[vivid cedar]

ok

the slope is

[violet sentinel]

just the process 🙂

[vivid cedar]

2

wait

yea

slope is 2

[violet sentinel]

ok great

so we know the slope is 2 🙂

no wait

[vivid cedar]

its ok take ur time

[violet sentinel]

yeah it's not two

$m = \frac{-2 - (-4)}{6 - 3} = \frac{2}{3}$

[warm shale]

mellowdramallama

[violet sentinel]

so the slope is 2/3

[vivid cedar]

but $m = \frac{-4-2}{3-6} = \frac{6}{3}$

[warm shale]

a_cheesefox

[vivid cedar]

oh wait i see waht i did wrong

yea mb

im dumb

[violet sentinel]

not dumb, just learning!

[vivid cedar]

https://cdn.discordapp.com/emojis/828699971738009627.gif

yay

[tardy epoch]

not dumb, just exploding

[violet sentinel]

ok so, for now don't worry about how we found this secant line, but this is the line that connects the two points, right?

pop quiz, what does a tangent point to a line mean?

[vivid cedar]

its the point where the line touches the circle

[violet sentinel]

exactly

[vivid cedar]

yay

[violet sentinel]

it only touches that point specifically

something like this, right?

now what do you notice about the lines? 🙂

[vivid cedar]

they are perpindicular

[violet sentinel]

😮

and how do we find the slope of a perpindicular line?

[vivid cedar]

make it so that they multiply to -1

so the slope is -3/2

[violet sentinel]

😮 😮 😮

wow

holy crap

we have the slope

rest is easy peasy

so now we just use the point slope formula with our given point on the circle?

[vivid cedar]

so

$(y+2)=-3/2(x-6)$

[warm shale]

a_cheesefox

[violet sentinel]

yep!

now put that into y = mx + b form

[vivid cedar]

but it says point slope form

so

which one is right

[vivid cedar]

$y=-3/2x-8$

[warm shale]

a_cheesefox

[violet sentinel]

close but your b is wrong

[vivid cedar]

oh wait

$y=-3/2x-7$

[warm shale]

a_cheesefox

[violet sentinel]

so close!

wrong sign

[vivid cedar]

oh yea

[violet sentinel]

remember to distribut the negative

[vivid cedar]

$y=-3/2x+7$

[warm shale]

a_cheesefox

[violet sentinel]

bingo!

[vivid cedar]

but hsould i put it in point slope form

cause the thigngie asks for point slope

[violet sentinel]

oh okay yeah

so then don't simplify lol

[vivid cedar]

ok

thanks

.close

[obtuse pebble]

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[violet sentinel]

np! Just to verify

[vivid cedar]

yay

[obtuse pebble]

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[lyric iron]

Linear Algebra 2:

[lyric iron]

I'm completely lost lol. Brain is fuzzy from doing this assignment all day xD

I'll take any help I can get

[fathom flicker]

start with the properties of a linear transformation

what are they?

can you verify them for both pieces of your piece-wise function?

that'd sort out 1.

[lyric iron]

i believe they are additivity and

homogeneity?

[fathom flicker]

Well I'd need you to explain what those are

how could you verify something is additive or homegeneous

[lyric iron]

additivity: f(u+v) = f(u) + f(v)

[fathom flicker]

mhm

[lyric iron]

homogeneity: f(ku) = kf(u)

[fathom flicker]

f?

[lyric iron]

sorry, accidentally hit enter

[fathom flicker]

aha

yes that looks good to me

so now verify that your transformation here satisfies both those properties

[lyric iron]

I've never had to do this with a piecewise before, so this is kind of a scary question to me

[shell wave]

which bit do u need help with

[fathom flicker]

start with evaluating case 1

u, v, are in M_2(R)

then evaluate case 2

u, v are M

then case 3 is...

?

u in M_2(R), v is M (note that this is without loss of generality the same as a potential case 4 of v in M_2(R) and u is M)

[lyric iron]

wait, what is case 2?

[fathom flicker]

u=v=M

[lyric iron]

where are you getting those cases from, I'm confused, sorry

[fathom flicker]

well they are the cases from the piecewise function are they not?

you have to different input options

either your input is M_2(R)

or M

now if you want to verify these for two different inputs, u, and v

what are all the possibilities of where u, and v can come from?

well they can both come from M_2(R)

or they can both come from M

or one can come from M_2(R) and the other from M

[brave bluff]

I don't think that is a piecewise function. I think it the function that sends the matrix M \in the set of 2x2 matrices to AM (in the set of 2x2 matrices)

[lyric iron]

i am now even more confused

xD

[fathom flicker]

ohhhhh

[gilded needle]

i was about to comment the same thing

[fathom flicker]

Okay well you can remember the part about evaluating if the function is additive/homgeneous, but forget the part about the cases

because our function isn't piece-wise

so ignore most of what I said

[gilded needle]

it's a bit weird to use the brace like that, hence the confusion

[lyric iron]

So wait, why isn't it piecewise

that looks like a piecewise

[fathom flicker]

that's what I thought

[gilded needle]

no look at the two arrows

one is $\to$, the other is $\mapsto$

[warm shale]

Bungo

[fathom flicker]

the function is from M_2(R) to M_2(R)

and it maps M to AM

[gilded needle]

it's saying, f maps from domain M2(R) to codomain M2(R), and is defined by f(M) = AM

yep

[lyric iron]

ohhh okay i see

[fathom flicker]

new notation unlocked

cool

well anyways, now that you know what the function is actually saying, just verify those two properties

i.e take two arbitrary matrices from the domain

etc...

[lyric iron]

So, would i do something like f(M1+M2) = f(M1) + f(M2) where M1 and M2 are arbitrary matrices

[gilded needle]

side note and unpleasant memory, i had an analysis prof who used the same $\to$ arrow to mean both things, it was confusing as hell sometimes, ok end side note and thanks for reading haha

[warm shale]

Bungo

[gilded needle]

yep

and then check scalar multiplication as well

[fathom flicker]

I'd cry

[lyric iron]

sigh... my profs notes are so poorly written, i have no idea what im doing

[fathom flicker]

just so we are clear here, writing it isn't enough. you have to show that this equality is true

[gilded needle]

btw, what this problem is hinting at is that we often don't even bother making a distinction between a matrix M and the linear map that it induces

[lyric iron]

yeahh i know, but I just don't know how to prove this in detail

[fathom flicker]

pick $$A=\begin{bmatrix} a & b \ c & d \end{bmatrix}$$ and $$B=\begin{bmatrix} e & f \ g & h \end{bmatrix}$$

[warm shale]

austinu

[fathom flicker]

now what is $$A+B$$?

[warm shale]

austinu

[fathom flicker]

now transform A+B

[lyric iron]

A+B

a+e b+f
c+g d+h

[fathom flicker]

verify that it is the same as transforming A, added to transforming B

see how that works?

[gilded needle]

that is a superhuman effort, i'd just write "by the rules of matrix multiplication" haha

[fathom flicker]

it is a proofs class after all though I am guessing XD

[lyric iron]

no it isnt even a proofs class

[fathom flicker]

okay, so take f(A+B)

[gilded needle]

no i agree, it's safer and quite likely what is intended, to do what you said

[fathom flicker]

abstract lin alg?

right?

[lyric iron]

its supposed to be lin alg 2

[fathom flicker]

ya

proofs class

[lyric iron]

I got a B in lin alg 1 last sem

and this is making no sense to me hahaha

different profs though

[fathom flicker]

there's always time to turn it around

[lyric iron]

yeah, im hoping so

[fathom flicker]

Okay so I'll try to sum up what I meant earlier in a hopefully clearer way that you can try out on your one

[lyric iron]

so, f(A+B) do i apply M -> AM?

[fathom flicker]

yes

and then take f(A) and apply that

and take f(B) and apply that

sum them

[gilded needle]

in fact, part (a) of this problem notwithstanding, you've graduated from boring calculation shit to fun conceptual shit

[lyric iron]

is M the same thing as F(A+B)?

[fathom flicker]

We take our arbitrary matrices $A$ and $B$. $\newline$ We find $f(A)$ and $f(B)$. Then we can find $f(A)+f(B)$.$\newline$Now seperately we take our matrices $A$ and $B$ and we add them. Then we find $f(A+B)$. $\newline$Now take these two pieces you did seperately, and show they are equal. i.e showing that $f(A+B)=f(A)+f(B)$

[warm shale]

austinu

[lyric iron]

So, I think im struggling with what the stuff inside the curly brace means

[fathom flicker]

Still not understanding what the function is you mean?

[lyric iron]

yeah, like what does the mapping stuff mean

[fathom flicker]

Okay

so top line here

it is telling you

[lyric iron]

inside the f:{

[fathom flicker]

yes

top line is telling you that

that our function takes 2x2 matrix and transforms it to a 2x2 matrix

does that make sense to you?

[lyric iron]

yep!

[fathom flicker]

the second line tells you how it transforms it

[lyric iron]

that makes sense

[fathom flicker]

it says that it takes this 2x2 matrix, and left multiplies it by A

that is the transformation

[lyric iron]

okay, so it takes a 2x2 and multiplies it by A

[fathom flicker]

left multiplies

yes

[lyric iron]

okay, that makes sense

that help a lot

I just had no idea what the function was saying lol

[fathom flicker]

Yes, so now hopefully what you should be able to do here is (first just be able to see that this is clearly a linear transformation) but secondly verify it on your own with 2 arbitrary matrices.

[lyric iron]

Yep, ill give that a go

Ok, i got it. now 2

ty for the help ❤️

.close

[obtuse pebble]

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[pearl charm]

I have a matrix question

[gilded needle]

do you have any thoughts on how to proceed?

[pearl charm]

no

i have no idea

[gilded needle]

ok, do you know the concept of calculating a normal vector by using a cross product?

[pearl charm]

umm, not rlly

can you walk me through?

[gilded needle]

well given any two vectors in the plane, if you take their cross product, you get a vector that is normal to the plane

does this sound familiar?

[pearl charm]

yes, yes

how does that relate to this problem

or like

how do i solve the problem using that

<@&286206848099549185>

.close

[obtuse pebble]

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[royal basin]

you're supposed to post your question first and only THEN ping helpers

and that only after 15 minutes pass without a response

[grizzled shore]

Do you have any ideas how you might approach this?

Do you perhaps have some better idea than just brute force

Do you know the Chinese remainder theorem?

[rocky goblet]

how many solutions are there to 4A+8B = 2 where A and B are natural numbers between 1 and 100?

[grizzled shore]

@tidal heart stop disrupting other people.

[rocky goblet]

(yeah my idea is that he's going to get completely the wrong answer)

[grizzled shore]

Bee he’s just trolling

[rocky goblet]

because those are my pronouns

neither

hmm

[shadow tangle]

Bruh

Mod ping?

[grizzled shore]

<@&268886789983436800> @tidal heart

[rocky goblet]

i'm just going to repost this

https://cdn.discordapp.com/attachments/903430332324405288/1133291076229267456/image.png

i'm not a "bro"

[grizzled shore]

Anyway @sleek snow I don’t know the CRT so I’ll let bee help you

[rocky goblet]

alright so

(i... don't really see how this solution is using CRT?)

[grizzled shore]

Personally I’d equate them (not 2) then try find pairs that are close to Integers and check

But idk how good this method is

But I’m sure there’s a better method

[rocky goblet]

13A-18B = 2, so 13A=18B+2, which means 18B + 2 must be divisible by 13

...oh i see the relevance of CRT
we're looking for a number that's equal to 2 mod 5, and equal to 0 mod 13

[rocky goblet]

so 52

[clear tinsel]

Ya I know

[rocky goblet]

...wait hang on i'm now not sure if that's actually true

ah yes it is

so

13A = 18B + 2

if we subtract 13B from both sides
13(A-B) = 5B + 2

so whatever this number is, it's divisible by 13 because it's 13(A-B), and it's equal to 2 mod 5 because it's 5B+2

it's 5B+2

if you look at 5B+2 mod 5, the 5B disappears, but the 2 is still there

[obtuse pebble]

@sleek snow Has your question been resolved?

[obtuse pebble]

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[timid silo]

where did i go wrong

when i plug in my values they're true to the equation

but the answer sheet is different

this ist he equation they gave

[obtuse pebble]

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[obtuse pebble]

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[obtuse pebble]

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[vital star]

How to do this question?

Use fraction to represent 0.202323232323...

[royal basin]

do you know in general how to convert from repeating decimals into fractions?

[vital star]

I know the answer but i need the step to solve this question

Does it has any step to solve it?

Or we could just writing the answer down?

Could i solve like this?

x=0.202323232323...
100x=20.2323232323...
99x=20.03
x=20.03/99
=2003/9900

[high lily]

,calc 2003/9900

[warm shale]

Result:

0.20232323232323

[obtuse pebble]

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[obtuse pebble]

@rose tinsel Has your question been resolved?

[grim nexus]

can you use calculator ?

[obtuse pebble]

@rose tinsel Has your question been resolved?

[obtuse pebble]

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[echo gazelle]

is F(3) just -9?

[wooden cipher]

Explain your thought process

[echo gazelle]

Bc the area between 2 and 3 is 9 and negative

[wooden cipher]

Not quite

[echo gazelle]

Would it be 0 then bc its at y 0?

[wooden cipher]

So whats your answer for (a)? (Trust me, it will make sense)

[echo gazelle]

-2

[wooden cipher]

How can you rewrite the expression in (a) using fundamental theorem of calculus

[echo gazelle]

3
| f(x)dx = -2
0

[wooden cipher]

Thats the integral

[echo gazelle]

oh

[echo gazelle]

oh uh F(B)-F(A)

[wooden cipher]

what are B and A?

[echo gazelle]

B is 3 A is 0

[wooden cipher]

right

so F(3)-F(0)=?

[echo gazelle]

-2

[wooden cipher]

You can find F(3) because you know F(0)

[echo gazelle]

5

[wooden cipher]

Yup!

[echo gazelle]

thank you

i have a quick question for part a

[wooden cipher]

Yeah whats up

[echo gazelle]

when writing out the steps to solve it
would i write it out as
0 2
| f(x)dx + | f(x)dx
2 3

then write 0|2 out as

0 1
| f(x)dx + | f(x)dx
1 2

3 + 4
=7

or just as 3+4

[rough stratus]

😳

[wooden cipher]

Your bounds are flipped but yeah

[echo gazelle]

wait what

[wooden cipher]

integral from 0 to 2 is positive 7, integral from 2 to 0 is negathve 7

[rough stratus]

You wrote the limits the wrong way round

[echo gazelle]

oh

wait im so lost

[rough stratus]

Why?

[echo gazelle]

i thought i wrote it as 0 to 2?

[rough stratus]

From 0 to 2 means the 0 should be in the bottom

And the 2 on top

[echo gazelle]

oh

okay ty

[rough stratus]

Np*

[obtuse pebble]

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[obtuse pebble]

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[flint hull]

Am I on the right track? This is a memory less property for Gamma Distribution problem

[royal basin]

what's the distribution of Y and how are you given it

[flint hull]

The distribution for Y?

Would it be from j+1 to infinity?

[polar fossil]

presumably Y is a random variable that follows an exponential distribution with parameter lambda?

[obtuse pebble]

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[obtuse pebble]

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[sacred violet]

Hey Guys, calc 2 question: why is this function differntiable in P(0,1,0) df/dx and df/dy has 1/2*(sqrt(abs(x*y))*something else) for me so I would get 1/0 singularity -> so it would be differntiable for me

[sacred violet]

or do I have to check differntiable criteria with this criterium

[obtuse pebble]

@sacred violet Has your question been resolved?

[haughty coyote]

That is the definition of differentiability, not a "criterium".
Proving the partial derivatives exist is only enough if they are also continuous functions (in which case f is C^1)

[sacred violet]

yes you are right

it's the defintion

but how can I show the function is differentiable in a point

[sacred violet]

.

[haughty coyote]

Choose a method. It's probably C1 so you can choose

[sacred violet]

because as stated

[haughty coyote]

Though then you need to do continuity checks

[sacred violet]

df isn't defined

for my point

but the solution says it's differentiable

have i taken the wrong partial diff

[haughty coyote]

Actually looks like you have
df/dx = 1/(2 sqrt(|x|) ... not sqrt(|xy|)

But yes that means df/dx is not computable that way

However df/dx = 0 nonetheless by the definition

[sacred violet]

it's sqrt(xy)

polynominal rule with 0.5

ah wait

[haughty coyote]

Can't differentiate sqrt(x) at 0 anyways

[sacred violet]

I got it because the y is in the nominator

[haughty coyote]

Since e^z - 1 is still 0

[sacred violet]

.close

[obtuse pebble]

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[mighty carbon]

I need help with:

[royal basin]

what do you know about tangents drawn to the same circle from the same point?

[mighty carbon]

just a sec

is it that segment DE and EG are perpendicular?

[royal basin]

that is correct by itself but doesn't answer my question

when you have a point and a circle, you can draw two different segments both starting at that point and tangent to the circle,
as DE and DF are in your picture.
what do you know about those two segments?

[mighty carbon]

are they congruent?

[royal basin]

i don't know, are they?