help-27

[charred tusk]

huh pythag with 2 unknown sides?

[thin inlet]

i think there is a way to do this with similar triangles

as well

[charred tusk]

uhm i dont think im that advanced yet

[thin inlet]

yes

similar triangles is easier

[charred tusk]

all the info we got rn is that afb is a right triangle and that af is 5.12 fc is 2.88 and ac is 8

[charred tusk]

thats what im asking all along

so yea how does that work

[thin inlet]

angle ABF=angle FCB

do you see why

[charred tusk]

not rly

[thin inlet]

well, do you agree that angle BAF and angle ACB are complementary?

angle A and angle C are complementary

[charred tusk]

why exactly? the picture is not drawn to scale btw, but if its not the visuals then idk how your suppose to get this

im so confused rn

[thin inlet]

the sum of angles of a triangle are 180

[charred tusk]

right ofc

[thin inlet]

angle ABC is 90 degrees

[charred tusk]

right

[thin inlet]

meaning the sum of angles A and angle C must be 90 degrees (they are complementary)

to form a triangle ABC

NOW

look at triangle ABF

it CONTAINS angle A

it contains a right angle

meaning it ALSO must contain an angle with the same measure as angle C

[charred tusk]

i gtg do some work rq, could u do the problem while im away, thanks

[thin inlet]

no

[modern gulch]

Nah.

[modern gulch]

flip one of the triangles so both face the same direction

[charred tusk]

.close

[devout snow]

Channel closed

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Use .reopen if this was a mistake.

@charred tusk Has your question been resolved?

[winter torrent]

this is a very unkind way to behave in this server and may result in reduced engagement from active helpers

[charred tusk]

how? I just wanted to ask if they can the problem while im not online, is that a problem? Doesn't that maximize production? Also because i simply dont understand his approach so while im away he can do the explanation. So when I can back instead of going back and fourth I can see if his approach makes any sense to me, and also see if his appraoch would work by checking the answer, if it does good, if it doesnt thats another story.

the helper is meant to help ofc, and as long as I do get the help, i really dont think that the "engagement" would be all that important

[devout snow]

@charred tusk Has your question been resolved?

[tender cobalt]

"if it doesnt, thats another story"

if it doesnt, what would you do

@charred tusk are you willing to work with the problem step by step?

[charred tusk]

yea sure i just wanna know how i can find bf

thats my only question

[tender cobalt]

you do use similar triangles here

[charred tusk]

right but i dont quiet know how

[tender cobalt]

now first, identify which triangles youre seeing

[charred tusk]

ok

right and?

[tender cobalt]

to show proof that you can see the triangles, you should say which triangles youre seeing

[charred tusk]

well theres 3 triangles in total: abf, bfc, abc

but abc isnt important i dont think

so yea just bfc and abf

[tender cobalt]

yea youll only need 2 of the triangles to solve this

that works

now because of similar triangles, a factor k will multiply from FC to FB

and similarly the same factor will multiply from FB to FA

right

[charred tusk]

could u elaborate a little?

[tender cobalt]

FC * k = FB

and similar triangles guarantees that FB * k = FA

[charred tusk]

hm interesting

[tender cobalt]

this is because both triangles have the same shape and so should have the same ratios of sides

[charred tusk]

right ok

whats next

[tender cobalt]

now we gather what we know and try using algebra to solve this

what's FC and FA?

[charred tusk]

af is 5.12 fc is 2.88

8 when added up exactly

[tender cobalt]

FB, we dont know, and k we also dont know

we're looking for FB

[charred tusk]

right YEA thats my question

[tender cobalt]

k would be nice to know, but you wont need to

consider the following

FC * k = FB
FC / FB = k

FB * k = FA
FB / FA = k

FC / FB = k = FB / FA
FC / FB = FB / FA

if you havent seen this before, its a bit weird to combine at first

this is the few times though where each move has a geometric explanation

so if you need some more context to those steps, I can provide

[charred tusk]

yea please do

[tender cobalt]

first, we consider the ratio FC / FB

this is how many times bigger FC is than FB

[charred tusk]

sure

[tender cobalt]

now because of similar triangles, FB / FA must be the same ratio

the triangles are similar = they have the same shape = FB must be equivalently that many times bigger than FA

(you can consider a shape as being the same if the sides share the same relative size)

does this check out?

[charred tusk]

sure yea like dilation

[tender cobalt]

thats an idea there

therefore FC / FB = FB / FA

[tender cobalt]

the k here represents the ratio, the algebra is using k to match the equations together

by witing that they both equal k, the algebra can use this to show that both ratios are equal

are you with me so far?

[charred tusk]

my teacher did this question but didnt have a clear explanation but its like exactly the opposite of what u got

instead

its

af/bf = bf/fc

[tender cobalt]

oh whoops

I accidentally swapped the ratios

it still works just the same

[charred tusk]

it does?

[tender cobalt]

geometrically, youre still considering a ratio and matching them

FC / FB = FB / FA isn't wrong geometrically, right?

simply flipped

but still correct

[charred tusk]

i believe so, when u do cross multiplying i think its still the same

[tender cobalt]

algebraically, you can get to either statement by raising both sides to the -1th power

doing this flips fractions around

(a/b)^-1 is b/a

[charred tusk]

yea so after that im guessing u get the bf squared and u just roo the other side

[tender cobalt]

wait wait

[charred tusk]

with the maf*mfc

[tender cobalt]

FC / FB = FB / FA
(FC / FB)^-1 = (FB / FA)^-1
FB / FC = FA / FB

[charred tusk]

wut

[tender cobalt]

I said -1st power

you didnt predict what would happen

[tender cobalt]

you can see I just ^-1 both sides and I get your version

so its just as correct

[charred tusk]

ah alr but yea regarding the similarity

its just remembering

[tender cobalt]

its not going to be easy remembering three fractions

instead its easy to remember that corresponding sides should match up

and their ratios should be equal

[charred tusk]

ah right we just consider a ration of x or k

and then we just make the equation make them equal to each other

ah ok comparison

alr so just remember the ratio

[tender cobalt]

you can do ratio between same sides of a triangle yea

these ratios are called "trigonometric ratios"

AF / FB and BF / FC are both "trigonometric ratios"

[charred tusk]

so aslong as we know 1 of the sides is the same we can do this

[tender cobalt]

this particular method

but its not like the sides being that different would break a lot

[tender cobalt]

anyways the sides matching now now allows you to solve for BF

do you know how?

[charred tusk]

oh yea i did it alr just needed some clarification on the similarity part

[tender cobalt]

thats good

you did mention cross-multiplying so I shouldve known

[charred tusk]

ty for actually unlike the other 2

[tender cobalt]

np

oh btw

[charred tusk]

i found that to be much easier

[charred tusk]

yea?

[tender cobalt]

do you want to know why the function f(x) has to have two negative signs in it?

why that wouldnt cancel?

[charred tusk]

elaborate

[tender cobalt]

first, we need an easier way to think of this "greatest integer function"

usually this is phrased as "the greatest integer that is less than x" which can be simplified

if you have a decimal x, you round down towards -∞

3.5 would round down to 3
-3.5 would round down to -4

[charred tusk]

right

[tender cobalt]

that is what the "greatest integer function" does

[charred tusk]

right

[tender cobalt]

for this raeson its more commonly known as the "floor function"

and using the better symbols of:

,,\lfloor x\rfloor

[woven radish]

mtt07734

[tender cobalt]

(this is to be compared with the "least integer function" or "ceiling function" which forcibly rounds towards +∞)

(which is $\lceil x\rceil$)

[woven radish]

mtt07734

[tender cobalt]

you can tell these apart, right?

[charred tusk]

right

[tender cobalt]

now for that double-negative

if you place a positive number x in there,

the - sign moves it into the negatives

because of this, the floor function rounding it towards -∞ will move it away from 0

the outside - sign moves it back into the positives

-floor(-3.5) = 4

compare this to if you just used floor normally

floor(3.5) = 3 rounds towards 0 instead

so the reason for that double-negative is to get the floor function to round up instead of down

[charred tusk]

oh yea i thought u wrote 4

it just makes the opposite of whats inside

[tender cobalt]

yea my mistake, thats a typo

interesting that trick

its just like the ceiling function isnt it

[charred tusk]

yea yea i got a little confused by what u mean but yes

[tender cobalt]

this is a way you can define the ceiling function

[charred tusk]

haha i mean yea i thought u meant like power to negative 1 or smth

but yes

[tender cobalt]

,,-\lfloor-x\rfloor=\lceil x\rceil

[charred tusk]

this is pretty easy to understand

[woven radish]

mtt07734

[tender cobalt]

so its a trick because you didnt learn the ceiling function yet

but they wanted you to develop that intuition on your own

bit of a tall order but its not that bad

[charred tusk]

haha yea i thought u meant power to negative one or smth, but yes this is a pretty easy concept

[tender cobalt]

slick

nah its just regular - signs

[charred tusk]

anyways uh yea regarding gif

[tender cobalt]

?

[charred tusk]

our class sped through it in like 2 days did a test and just went to trig after it

[tender cobalt]

oh thats your abbreviation

[charred tusk]

so we didnt

rly

go in depth into ceilling

we had a few word problems and stuff and just eneded there

anyways its better to know then not

[tender cobalt]

yea

the way it sounds,

this is one of those things youll just know

so they give you a few problems to easily slip that into you

[charred tusk]

right and uhhh smth regarding trig

i dont know why im not getting it this seems so easy

ive did this on khan so many times but like this seems so weird rn

[tender cobalt]

go post it

[charred tusk]

very easy

2 sides given

idk why

i got opposite and hypo

[tender cobalt]

man I never know why they dont teach you this

youre my only lead rn

consider the following

then you try to answer whats happening after I show this

youve figured out that $\sin(?)=\frac{13}{31}$

[woven radish]

mtt07734

[tender cobalt]

now you need to somehow figure out the angle inside

[charred tusk]

OHHH

wait its like that??

oh sh

[tender cobalt]

multiplication and division isnt going to help you here

[tender cobalt]

not always

[charred tusk]

1 sec i might remember this

[tender cobalt]

the sine of this angle is 13 / 31

if only there was a way to

invert this sine

go from 13 / 31 to the angle

[charred tusk]

we havnt started the unit yet

but yes let me try smth rq

[tender cobalt]

alr

(I hope youre not just testing sin(1°), sin(2°), sin(3°)... in the calculator)

[charred tusk]

ofc not

[tender cobalt]

lol

[charred tusk]

there was smth regarding like recipricols or smth

im not sure if its this

[tender cobalt]

its not cosecant

[charred tusk]

or was it negative sin

[tender cobalt]

its not negative sine

thats just the sine but negative

[charred tusk]

oh neither?

yea

isnt that how u solve it

hm

[tender cobalt]

this has to be a separate function in its own right

subtraction is the inverse to addition

division is the inverse to multiplication

and a special function is the inverse to sine

functional inverse

[charred tusk]

ok lol i forgot everything i learnt online tell me the solution

[tender cobalt]

its called arcsin(x)

and written as $\sin^{-1}(x)$ or $\arcsin(x)$

[woven radish]

mtt07734

[tender cobalt]

the -1 is the function inverse kind of -1

people usually are not happy seeing a number like that and its not an exponent

it can lead to confusion

like how $f^{-1}(x)$ isnt $\frac1{f(x)}$

[woven radish]

mtt07734

[charred tusk]

it was this ha

[tender cobalt]

???

[charred tusk]

i remember now

[tender cobalt]

no its not

thats not "negative sine"

thats "inverse sine"

[charred tusk]

bad wording but yes i meant the same thing

right its the inverse

but yes yes

[tender cobalt]

• negative sine: -sin(x) would be additive inverse, it with sin(x) adds to 0 (not what youre looking for)
• reciprocal sine: 1/sin(x) or csc(x) would be multplicative inverse, it with sin(x) multiplies to 1 (not what youre looking for)
• inverse sine: sin^-1(x) or arcsin(x) would be functional inverse, it with sin(x) composes to x

[charred tusk]

haha yea

[tender cobalt]

I prefer to use arcsin so I dont type ^-1

[charred tusk]

wrong wording

[tender cobalt]

now for the problem

[charred tusk]

i got it

its easy

[tender cobalt]

very nice

similarly theres arccos, arctan, arccsc, arcsec, and arccot

a functional inverse for the six trig functions

[charred tusk]

alr let me speed throgh ill see if i got any difficult stuff before i head to sleep

[tender cobalt]

slick

good job so far btw

[charred tusk]

ive already learnt this online last year but just forgot most of them

alr its enough for today ima head to bed

alr thanks mate

[tender cobalt]

np

[charred tusk]

.close

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[magic pine]

what is your question

[upper schooner]

(you also don't need to find y', the bottom half is sufficient - then put the c1 you found back into the original)

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@nimble pier Has your question been resolved?

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[kindred plume]

does anyone know how to do proofs?

[solar goblet]

angle jkm is not congruent to angle mlk

[kindred plume]

u sure? my teacher was able to solve in class today with my classmate while i was working on other things

[urban jungle]

This can be proved by symmetry

Hint: find the rhombi

[devout snow]

@kindred plume Has your question been resolved?

[kindred plume]

uhhh

I know JK, KL, and LM are congruent,

[kindred plume]

i don’t know what any of this means 😭

[urban jungle]

Focus on the perpendicular bisectors and what they tell you about the surrounding triangles

A proof by symmetry is where you show that the figure is symmetrical and use that to show that certain things are congruent

[kindred plume]

Would symmetry mean reflective?

I’m sorry we haven’t had real math class in like 3 weeks 😭

[urban jungle]

We could have multiple kinds of symmetry but it would be reflective in this particular case

[kindred plume]

would right here be verticals?

where the markings are)

[kindred plume]

or what even would be reflective? i have no clue at all, these are the only things i have written down

[devout snow]

@kindred plume Has your question been resolved?

[devout snow]

@kindred plume Has your question been resolved?

[devout snow]

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[devout snow]

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[visual maple]

$-2x^2-6x+2=mx+6 \ -2x^2-6x-mx-4 = 0 \ -2x^2-x(6+m)-4=0 \ \Delta b^2-4ac = 0 \ (m+6)^2-4\cdot -2 \cdot -4 = 0 \ m^2+12m+36-32 = 0 \ m^2+12m+4 = 0 \ \ $then use quadration formula: $\ \frac{-12\pm\sqrt{128}}{2} \ -6\pm 4\sqrt{2}$

[woven radish]

pixel

[visual maple]

what now

is that the value of m

just between whatever i got?

[urban harbor]

yes, that's what you solve for when you do the quadratic formula for a quadratic in m isn't it?

[visual maple]

i got it

woohoo

just had to plug it back in

[urban harbor]

I am proud, seems like you've got the hang of this now and no negative errors anymore 😄

[visual maple]

all thanks to you

so yeah

im up to week 5 content

woohoo!

just 2 more weeks

and im all caught up with like math methods

then i'll have math specialist

💀

i do two different math subjects

fun

[urban harbor]

great

lucky dog

[visual maple]

thing is

i was on vacation

and i came 4 weeks late into school

i had an exam first week back

and assignments due

so its been really hard

so yeah

im still catching up

its week 6 now

so my brain rn is overloading with like all the info

from all different subjects

but like theres nothing else that I can do apart from try my best

so like seriously thank you for being patient with me

😭

[urban harbor]

keep working on it, seems like you are doing well, you have like actual algebra skills and ability to retain information for more than 10 minutes which is more than i can say for a lot of math students i come across 😅

[visual maple]

oh dw once i get to math specialist which is the higher level of math and i start doing vectors

im gonna make you hate me

🗣️ 🔥

vectors WAS easy last year (year 10)

but this year, the stuff they're doing now is a lot harder

we have like 121237 different formulas, its crazy

[urban harbor]

there's like like 0 memorization when it comes to vectors, everything is super intuitive

i cant think of any formulas you need to remember

[visual maple]

yeah it should be fine either way

just gotta like yk

[urban harbor]

(must mean ive forgotten them)

[visual maple]

get this math subject out of the way, THEN ill actually start my next one

just got 2 more weeks worth of content to do

oh week 5 is easy

[urban harbor]

https://tenor.com/view/get-back-to-work-gif-1126008344771276194

[visual maple]

🔥

[urban harbor]

what are you waiting for, get to more studying

[visual maple]

😭

bathroom break and ill get back to work

also I really like how you explained stuff, is it fine if I add you and ask you stuff in dms?

or do u have like a server that I could join and ask you there

[urban harbor]

appreciate it, but i prefer not to get DMs, feel free to ping me when you open a help channel here and if i'm around ill check it out, but not excessively

[visual maple]

alright thats fine

thanks again for the help

see you around

.close

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[visual maple]

(bathroom break) 😄

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[restive river]

"Show that the points A, B, and C with position vectors, a= 3i -4j -4k, b= 2i -j +k, c = i -3j -5k respectively form the vertices of a right-angled triangle."
in this question how would i know the direction of right and triangle sides SO i can know whats the initial point and whats the terminal point of vector AB For example, if I want to find the vector AB, then AB = (x2-x1)i + (y2-y1)j + (z2-z1)k. Here, x2, y2, z2 are the terminal points and x1, y1, z1 are the initial points. My question was: how would I know what's the initial point and what's the terminal point? A can also be x2 or B can also be x2 it depends on the vector AB direction

[devout snow]

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[steady lake]

just want a more experienced set of eyes on this to make sure i'm not making any crucial mistakes

[steady lake]

the assignment is to convert a point to polar coordinates

i'm iffy on the value i got for r

[eager nova]

Sqrt(b^2+c^2) is not equal to sqrt(b^2 * c^2)

[steady lake]

the other answer i got for this was sqrt(0)

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[spark ruin]

ot sure how lagrange theory could be used here

eres my attempt

!occupied

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[spark ruin]

!help

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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[thin inlet]

do you know how to phrase this in terms of a constrained optimization problem?

i.e. maximize () subject to the constraint ()

[spark ruin]

im not sure what ur asking, generally when i do these types of questions involving lagrange, i put T(x,y,z) and the constraint equation G(x,y,z) together as L(x,y,z,lambda) = T(x,y,z) - lambda*G(x,y,z), then partially differentiate them to x,y,z and lambda.

so far all questions ive done leaves the partial derivitives of x,y, and z only in terms of x/y/z and lambda, where i can substitute into lambda's partial derivative and then resubstitute into equation to see which values give a critical value, clearly not the case here and i cant think of any method other than this

[spark ruin]

i think ive done what u described here

[spark ruin]

top left of photo

[devout snow]

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[devout snow]

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[flat raft]

In this problem i was able to obtain the color the spoke if we wanted to use atleast 2 colours. This is x^p - x

But i cannot obtain the number of paintjobs if two paint job are the same if we can obtain one from the other by rotation

[flat raft]

<@&286206848099549185>

[halcyon idol]

Can you please resend the image attached, as it is not of high quality or precisely, the text is not readable

[little totem]

can you describe all of the possible ways in which two paintjobs are considered the same?

suppose i gave you
RRBYG and YGRRB

are these the same paintjob? why or why not?

[halcyon idol]

Thank you good sir

[flat raft]

They are the same because if you think of paintjob in terms of the diagram of a wheel given the text, you can obtain one from the other simply by rotating

[little totem]

ok so how many paintjobs are identical to RRBYG then?

[flat raft]

idk

[halcyon idol]

The theorem of number theory that this problem proves is known as Burnside's lemma or the Orbit-Stabilizer theorem, which provides a method for counting orbits of a group action.

[little totem]

i dont think thats the theorem it proves

[halcyon idol]

It is as far as I remember

[little totem]

use the same visualization you just described

[flat raft]

It is fermat theorm

[little totem]

i think burnside lemma is far more abstract than this problem instance

[flat raft]

There are p-1 similar paintjobs

[halcyon idol]

Fermat's little theoram?

[little totem]

perfect!

now can you imagine how we would group these paintjobs?

suppose you made boxes

and in each box, you put all of the same paintjobs together

[halcyon idol]

Using at least two colors: k^p ways.

[little totem]

what do you notice about these boxes?

[flat raft]

yes

[little totem]

you forgot -p on the end for substracting the single color case

[flat raft]

There a 6 in each box

[little totem]

a 6?

[halcyon idol]

My bad

[little totem]

you mean p-1?

[flat raft]

I meant p

[little totem]

oops yeah p not p-1 haha

so the number of distinct paintjobs is the number of boxes

you have the total number of paintjobs counting the rotations

you know how to group them into boxes

how many boxes are there?

[flat raft]

$\frac{x^p - p}{p}$

[woven radish]

schrödinger

[halcyon idol]

This represents the number of boxes or equivalence classes. So, the number of boxes is given by the sum:

[
\frac{1}{p} \sum_{r=1}^{p} \gcd(r,p)
]

[woven radish]

KavyaSahai

[little totem]

bingo

[little totem]

this expression must be an integer

and you can then turn this into a modular arithmetic statement

that proves fermat's little theorem

[flat raft]

hav u read the book?

how did you knw the solution

[little totem]

i have not read the book, but this does look like a book i want, from the pages you sent

[halcyon idol]

No, I haven't but surely would read, if I would improve my English

[little totem]

i am a combi enthusiast, the combi problem at base is very straightforward and relatively simple

the relation to fermat's little theorem mostly comes from two observations:

1. familiarity with fermat's little theorem, recognizing its algebraic form

2. im pretty used to the idea of applying integer arguments in combi. often times you need to sanity check your answers, and if something you need to count gives an answer that cant even be an integer, you know you messed up

if you apply the integer argument, little theorem follows pretty directly

do you mind sharing the book source?

[little totem]

thanks!

[flat raft]

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[full snow]

1.Why is there a "+ " plus & not a "-" minus sign.
2. How do we decide which one is alpha and beta?

[restive river]

1- they factored the - as well
2- the letters are arbitrary. Choose whatever you want

[full snow]

1. wdym, please expand
2. alright, so they're just variables & i can switch alpha & beta if i like

[restive river]

,, -a -b = -(a+b)

[full snow]

wish i was a consonant & not a mere variable

[full snow]

oh

is this an alegbraic identiity?

[restive river]

its calling factoring

[full snow]

i dont get it

[restive river]

the opposite of distribution

[full snow]

what is factoring

[restive river]

whats ur native language btw

if it is not english

could be easier to communicate thru that

[full snow]

just do it in english

[restive river]

ok

[full snow]

i know factoring

but i'd like u to elaborate

[full snow]

as in just the term, forgot the meaning

[full snow]

ohhh

[restive river]

it goes both ways

basically

[full snow]

so like breakdown vs making it in its whole form

gotcha

[full snow]

though how the + come here?

[restive river]

yeah sure u can think of it that way

[restive river]

its because

- is shorthand for writing multiplication by -1

like uh

[full snow]

oh nvm

im stupid

it multiplies and ends up being like LHS either ways

[restive river]

,align
-a-b = -1\2a + -1\2b &= -1\8{a+b} \ &=-\8{a+b}

[full snow]

mhm

[full snow]

oh wait

[full snow]

if it was minus instead the value would change

[restive river]

minus where exactly

woops wrong reply

[restive river]

meant this

[full snow]

the thing we're talking about

[restive river]

u gotta be more specific

[full snow]

lol

[restive river]

confusion

[full snow]

no.1

[restive river]

yeah what about it

[full snow]

what we've been discussing all this time

[restive river]

what part of it do u want to be minus'd

[full snow]

#help-27 message

[restive river]

ok so u r asking if -4(x-2)?

[full snow]

-4(x - 2) would be 8 and not -8

is that why

[full snow]

yeah

[restive river]

yes

[full snow]

omg

[restive river]

-(a-b) = -a+b

[full snow]

maff aint for me

[full snow]

yeah so like synonyms

[restive river]

lmao

[full snow]

thanks oylo

bye

[restive river]

bye

[full snow]

.close

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[full snow]

i dont think im thankful

[devout snow]

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[calm maple]

So

[calm maple]

I need to think of a sum of an that converges, sum of bn diverges and lim an/bn = 1

I need some help

[devout snow]

@calm maple Has your question been resolved?

[cosmic wing]

https://www.youtube.com/watch?v=TQjEhCGG0yQ&t=11s

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[leaden nimbus]

Can someone help me with this problem my teacher demonstrated? Specifically the highlighted part?

[unreal thunder]

The chat is occupied

[leaden nimbus]

Literally everything else makes sense

[cunning mica]

Oh oop

Sorry

[leaden nimbus]

Np dude

[sinful gust]

cos(x) ranges from -1 to 1. can you think of a value of cos(x) that when squared is greater than 0?

[unreal thunder]

@leaden nimbus are you wondering why -1<=cosx<=1?

[sinful gust]

hint: ||for all real x, x^2 >= 0||

[leaden nimbus]

Im wondering why when my teacher squared everything-1 became 0

[dense lynx]

a square of a real number cannot be negative, so the lower bound of -1 is replaced with 0

[leaden nimbus]

Ohhhhhhhhhh

That's a bruh moment srry guys I can't use my brain

[dense lynx]

u good

[leaden nimbus]

Tysm everyone

. Close

oh wait nvm dubs pls continue

[unreal thunder]

all g

[leaden nimbus]

.close

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[bright juniper]

4 of the same generators in 3 hours spend 15 dollars worth of fuel. How many dollars worth of fuel would 3 generators spend in 6 hours?

[gusty shale]

Perhaps compute the amount spent per generator per hour?

[twin thunder]

take the amount of doller spent by one generator in 1 hour be x

then compute

[desert compass]

1 generator in 3 hrs uses 15/4 dollars
1 generator in 1 hr uses 5/4 dollars
3 generators in 1 hr uses 15/4 dollars
3 generators in 6 hrs uses 45/2 dollars

[devout snow]

@bright juniper Has your question been resolved?

[bright juniper]

Wrong

[desert compass]

ah i see i had made a calculation error

@bright juniper is it ok now

[devout snow]

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[bright juniper]

.reopen

[devout snow]

✅

[bright juniper]

Not still need gelp

Help

@desert compass

[desert compass]

uh is it wrong?

[bright juniper]

What is wrong

Your answer yes

@desert compass

[desert compass]

what's the correct answer

[bright juniper]

There are 4 answers nor sure wich one is correct

[desert compass]

can u send em?

[bright juniper]

Ohh 45/2 is correct

[devout snow]

@bright juniper Has your question been resolved?

[devout snow]

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[rustic hare]

Hi can someone take a look at my work and see what I did wrong in the trig sub

[rustic hare]

I got 2 different answers when I tried to integrate this intergal

I first tried with trig sub and then tried with u sub

[ocean silo]

why did you cancel out the $2cos(\theta)$'s after substituting everything

[woven radish]

Triaxyz

[rustic hare]

Oh I see what I did wrong

I should of multiplied them

[ocean silo]

yes

[rustic hare]

Thanks

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[rose pumice]

can some one help please

[rose pumice]

i have the answers but still am confused

its just part b i am confued on

[supple knot]

• Ask your math question in a clear, concise manner.
#help-27 message

[soft umbra]

What’s your confusion?

[wicked turtle]

on line 5, how did your +4 become -6?

[soft umbra]

About the reliability of your answer?

[rose pumice]

sorry i just cant figur out whats happening

my eyes are accting up a bit sorry

[supple knot]

you can read discord but not the solution?

you need an optometrist, not internet strangers and certainly not doing math

[rose pumice]

OH wait i get it now

where there is a y you put what y equals

sorry i be dumb some times

[rose pumice]

i think the answers where photo copied and the paper its on its looking good and my head hearts when looking at it

!close

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[shadow quest]

How is this equivalent

[acoustic leaf]

you can "rationalize the denominator" on the first one by multiplying by $\frac{\sqrt{3}}{\sqrt{3}} = 1$

[woven radish]

cloud

[shadow quest]

?

[acoustic leaf]

multiplying any number by 1 gives the same number back

[shadow quest]

Yea

[acoustic leaf]

and any number divided by itself is 1

[shadow quest]

Yea

[acoustic leaf]

so you can multiply numbers by 1 (in a specific way) to make them look different (even though they are the same number)

[shadow quest]

Okay

…

[acoustic leaf]

in this case, multiplying either of them by $\frac{\sqrt{3}}{\sqrt{3}}$ is the same as multiplying by 1, and will make one of them look like the other (after simplifying)

[woven radish]

cloud

[shadow quest]

Oh ok

I see

Thanks

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[abstract prism]

.

claim

[abstract prism]

identify the surface who's vector equation is

please ping me when you respond, I have discord open in the background

<@&286206848099549185>

<@&286206848099549185>

very helpful

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[idle coral]

Hello, does anyone know how to solve this equation? I don't know how to continue

[scarlet sequoia]

it should be 3u there

(and in the next few lines as well)

[idle coral]

a

thanks

[scarlet sequoia]

also

$$\Big(\frac25\Big)^{u^2} \neq \Big(\Big(\frac25\Big)^2\Big)^u$$

[woven radish]

Modus

[scarlet sequoia]

RHS is just (2/5)^(2u)

not u^2

[idle coral]

hhmm, why not?

[scarlet sequoia]

is 2 * u the same thing as u^2

this is clearly what I'm trying to say

[idle coral]

What could I do now?

[scarlet sequoia]

last line is still incorrect

I recommend doing

25/4 = (2/5)^(-2)

after u-sub

and then just compare the powers

find u, come back to x that's it

$$\Big(\frac25\Big)^{u^2+1}=\Big(\frac{25}4\Big)^{2-3u} \iff \Big(\frac25\Big)^{u^2+1}=\Big(\frac{2}5\Big)^{2(3u-2)}$$

[woven radish]

Modus

[idle coral]

smart idea

[scarlet sequoia]

good

[idle coral]

thanks

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[bright juniper]

What value dose n need to have so the equation dosent have an answer

[magic halo]

wdym

i dont understand

[dense jay]

10^x-20=n-10^x+1

10^x[11]-20=n
10^x=(n+20)/11

[bright juniper]

i need to find the value of n so the equation dosent have any answers

[rustic jetty]

shouldnt it be 10^x = (n+20)/25?

Either way n is the same

[rustic jetty]

What values can 10^x never take?

[bright juniper]

it can never be minus

@rustic jetty

[rustic jetty]

It is strictly greater than 0

(it can never take 0 as well)

so when (n+20)/25 <= 0, there will be no solution

[bright juniper]

ok do i solve that now?

[rustic jetty]

yea

[bright juniper]

got n<=-20

@rustic jetty

[rustic jetty]

yea

gj

[bright juniper]

so thats the answer?

[rustic jetty]

yup

[bright juniper]

so n cant be more or equal to 20?

[rustic jetty]

when n <= -20, the equation will have no solution

[devout snow]

@bright juniper Has your question been resolved?

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[stoic crystal]

First missing value in table should be 4, and second should be -5

is my graph correct?

also is the curve smooth enough?

<@&286206848099549185>

😭

help

[gleaming thorn]

It looks correct 👍

[stoic crystal]

is it smooth enough though?

So for c)i) I am getting x=0, x=2.55

is that correct

anyone pls help? 😭

[whole iron]

I would use desmos graphing/graphing calculator to validate your answers (appropriate if your allowed to use a graphing calculator) OR plug the values you got back into the function to see if it equals 0 (appropriate if allows your teacher does not allow you to use a graphing calculator)

It would seem that your answers are c(i) are correct.

Also, if you look on the graph you drew, you see that the function is equal to 0 at x = 0 and x ~ 2.5

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@stoic crystal Has your question been resolved?

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[bright juniper]

In the diagram i got a triangle with 3 circles the circles radius is 5cm i need to find the triangles perimeter

[hot steeple]

What have you tried?

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[digital egret]

help please

Ive been at this for like 2 days and I cannot get the answer the teacher provided

I tried this and I gave up

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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

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[fleet jay]

Can someone help

[vapid nest]

What are the function types given by ur teacher

[fleet jay]

2^x

Exponential

That’s what it says

E(x)=2^x exponential

[shrewd thistle]

Polynomial is a something raised to a (constant) power
Exponential is something raised to a (variable) power

That's exponential

[vapid nest]

Hmm

[shrewd thistle]

https://www.math.utah.edu/~wortman/1050-text-ef.pdf

[shrewd thistle]

reading

[fleet jay]

Alright thx

[shrewd thistle]

just looking at it, it's probably going to be logarithmic

[fleet jay]

U change the x’s to -x?

[shrewd thistle]

If you plug in -x, then the graph would shoot off to the left

this has increasing y values along with increasing x values

[fleet jay]

F(-x)=(2)^-x

For the reflection over y axis

[shrewd thistle]

that would shoot off to positive infinity on the -x axis

[fleet jay]

Oh

But don’t u want to be positive infinity

To match the plot diagram

[shrewd thistle]

sort of (temperature won't increase to infinity, but that's not the point of the exercise)

But you want it to do so as x increases

if you're looking at the -x axis, x (temperature) would be decreasing

[fleet jay]

Can u possibly change the 2 to 1/2

I have no clue how to do this

[shrewd thistle]

It helps to recognize what certain families of graphs look like

this is a logarithmic graph

and this is a graph of the sqrt(x)

[fleet jay]

So would u transform up?

[fleet jay]

So like f(x)=-2^x+y value

[shrewd thistle]

Where x = 0, you'd simply add where you want your y value to start

in this case, it's 35, right?

[fleet jay]

Yea

[shrewd thistle]

So now you've got (the function you need to find) + 35

That gets you your starting point

[fleet jay]

Wait so it’s -2^x

[shrewd thistle]

Just like in the graphs I shared above

You're guessing 😉

[fleet jay]

Yea I got no clue lmao

[shrewd thistle]

Think about -2^x

[fleet jay]

I’m confused about the transformation

[shrewd thistle]

"transformation" is a fancy way of saying, "What do you add/subtract/multiply/divide in order to shift the basic shape of the graph to this thing over here"

Take the graph of x for example

If you just add 10, what happens? The transformation is an increase of 10 on the y-intercept

[fleet jay]

So would it be f(-x)=-2^x+35

[shrewd thistle]

First off, it won't be negative

think about what happens to -2 when you raise it to a series of powers

What's -2 ^ 1?

Just -2, right?

[fleet jay]

Yea

[shrewd thistle]

What happens with -2^2?

[fleet jay]

-4

[shrewd thistle]

It's +4

[fleet jay]

Oh yea

[shrewd thistle]

a negative times a negative gives you a positive

[fleet jay]

Mb

So the x will be negative

[shrewd thistle]

so if you use a negative coefficient with a variable exponent, your answers will end up very different from one sequential exponent to the next

negative exponents give you negative results, positive exponents give you positive results

so the exponents 1, 2, 3, 4, 5, 6...

[fleet jay]

So 2^x+35

[shrewd thistle]

The graph goes sideways from what you'd expect

[fleet jay]

Or is x-1?

Hold up lemme use a graphing calc

[shrewd thistle]

please do

lol

you'll notice that if you just raise a constant to a variable exponent, the graph immediately shoots off to infinity

whereas the graph you're given in the problem has a very gradual increase into infinity

[fleet jay]

I have no clue how to do this

Can u gimme a hint

[shrewd thistle]

As I said, I believe it's going to be a logarithmic function

[fleet jay]

@shrewd thistle -E(x)=2^x+35

[fleet jay]

Yo?

.close

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[worthy oxide]

here is the question i am triying to do

i do not know where to start i am a little behind on a lesson so any ehlp is appreciated

[devout snow]

@worthy oxide Has your question been resolved?

[scarlet mauve]

@worthy oxide does this help

[worthy oxide]

What do you do about the angles in the cos

like the cos(0)

let me look at the problem tho

this helps a lot

thank you very much

[scarlet mauve]

lmk which part u are struggling on, its hard to do the entire problem on my laptop t-t

[worthy oxide]

ok

yea it all makes sense

yea ok

yea

but now i am still confused on how you get rid of the angle sign inside of the sin

i did trig sub and got u^4/4 plug in sin(angle)

but i dont know what to do with the angle

[scarlet mauve]

show me exactly

[worthy oxide]

also i dont know if the next part is incoorect

ok

i will send pick

sorry my work is mesy

but where i am at this part at the bottom right

idk if i messed up the trig sub

but where i am confused is getting rid of the angle inside the sin

[scarlet mauve]

recall that x=3sintheta

therefore x/3 = sin theta

[worthy oxide]

ok so would i just convert it to x/3 instead of sin ?

[scarlet mauve]

yes

[worthy oxide]

ohhhh ok

ok

yea that makes sense

[scarlet mauve]

since the original question was in term of x, you want the final answer to be in terms of x

[worthy oxide]

my final answer is 162(((x/3)^4)/4)+C

is this correct?

[scarlet mauve]

oh shoot my bad. i forgot add the manipulation on the 2nd step it should be 2(27sin^3theta) * 3cos(theta) * 3cos(theta) (dtheta)

@worthy oxide

[worthy oxide]

oh ok

yea let me try one more time

i think i better understand the first steps

or i understand it better

[scarlet mauve]

yeah. its nice how x is already given to guyz already

[worthy oxide]

alright this is my answer

it does not look right tbh

it is the one at the top right

.close

[devout snow]

Channel closed

Closed by @worthy oxide

Use .reopen if this was a mistake.

[devout snow]

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[ocean quest]

So I'm having a lot of trouble with this. The parameter variation technique has yielded (v1)' = -t^-6e^6t and (v2)' = t^-7e^6t but I don't know how to integrate these to get my v1 and v2 functions

[ocean quest]

My Diff EQ course has an awful textbook

[devout snow]

@ocean quest Has your question been resolved?

[ocean quest]

<@&286206848099549185>

[urban jungle]

I’d recommend solving this by substitution; i.e. taking y = Ae^Bt(Ct^-5 + Dt^-4 … Gt^-1 + H) it might be less messy than integrating those

[ocean quest]

I thought that wasn't possible for powers in the negatives

[urban jungle]

Sure it is, you can do it for anything

Oh and you’d only have to go up to a t^-5 term

It’s a bit guess-ish but integration would probably give you a similar sort of thing since you’d have to do repeated integration by parts

[ocean quest]

The god-awful textbook put this type of problem under the Variation of Parameter chapter and then didn't give any information on how to solve it using the method that leads you straight into a brick wall

[tender cobalt]

also you didnt use variation of parameters correctly

you forgot to multiply by g(t)

small typo but it entirely gets rid of your e^

@ocean quest try consulting this instead to learn variation of parameters https://tutorial.math.lamar.edu/classes/de/VariationofParameters.aspx

I didnt learn it this way but it should at least give you some justification for whats happening

what matters more is this at the end

[ocean quest]

thank you so much

[tender cobalt]

for y'' - 6y' + 9y = 0, the homogeneous version of your DE,
y1 and y2 are the two homogeneous solutions

W(y1, y2) is y1 y2' - y2 y1' in case you dont know that
g(t) is the t^-7 e^3t (the bit at the end)

[tender cobalt]

just in case you need to do a hard integral though, this suggestion works wonders and makes sure youre at least not doing an integral, only derivatives

Ive "cheated" a few problems by just doing that and choosing a good guess

remember that your guess needs to be general enough to work

[ocean quest]

Well, looking back, this stems not from me using the variation of parameters incorrectly, but me being a complete idiot and factoring out the wrongly signed solution from the auxiliary equation

Thank you both so much!

[tender cobalt]

np

[devout snow]

@ocean quest Has your question been resolved?

[devout snow]

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[lament schooner]

how do i solve this with

[lament schooner]

subs u

[desert compass]

let 1-x = u^2

u will get integral (14(1-u^2)^2du)

[devout snow]

@lament schooner Has your question been resolved?

[lament schooner]

im trying to do that

im abit confused

i ended up with 14x^2 * u

[desert compass]

u need to substitute x with u

1 - x = u^2
so x = 1 - u^2

[lament schooner]

yes

and since i have a x^2 left

uh

do i make it x^2 = 1-u^4

[desert compass]

no no

x^2 = (1 - u^2)^2

[lament schooner]

ah right crap

how do i get rid of the x^2 then

[desert compass]

just plug this in the place of x^2

[lament schooner]

ah

i got 14 integral of u- 2u^2 + u^3

[desert compass]

ah well u forgot the u in the denominator

that'll cancel a common factor of u

[devout snow]

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[long trout]

hello! how do you find an equation of a circle that is tangent to two lines and internally tangent to the equation of another circle?

[hasty saffron]

!original

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[long trout]

oh this is my problem

let me give more context

the equations of the two straight lines are y=0, y=(12root(6))x/19+ (24root(6))/19

and the equation of the other circle is (x-1/2)^2+(y-24/(4root(6)))^2=(35/(4root(6)))^2

if someone could help by using latex on the equations, that would be helpful as it would be much easier to understand

[long trout]

is this better

[hasty saffron]

Yes

[long trout]

ok, thank u

[devout snow]

@long trout Has your question been resolved?

[hasty saffron]

This might be a bit long, but assume the center as (h,k)
Now you have this point and two lines, so try finding the distance of this point from these two lines||(it will be the radius of the circle)||, you should get the relation between h and k
If two circles touch internally then the distance between their centers is the magnitude of the difference between their radii

[long trout]

yes, i found the relation between h and k

idk what to do about the two circles touching internally part

[hasty saffron]

I wrote it in the last lines

[long trout]

so lets say we have the equation (x-1/2)^2+(y-24/(4root(6)))^2=(35/(4root(6)))^2

the equation of the circle internally tangent to this is (x-h)^2+(y-k)^2=r^2

[hasty saffron]

Right

Also

The unknown circle touches the x axis(coz it's the tangent), can we say that the y coordinate of the center is the radius?

Note that this would be the distance of the point from x axis, which is the radius

[long trout]

yes i think u can say this

[hasty saffron]

So k=r then, which simplifies it a bit here

Now you have the centers of 2 circles and their radii
The positive difference between their radii will be the distance between the two centers

Does that make sense?

[long trout]

hmm, that makes sense

let me try that

[hasty saffron]

Sure

Don't forget, you have the relation between h and k too

[long trout]

yes

[long trout]

now i have found the equation