help-10

[tired sierra]

thanks juke

.close

[obtuse pebble]

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[thorny dust]

need help on making sure this is right

[sweet edge]

,w .3/(2003-1999)

[warm shale]

Results provided by WolframAlpha

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

Requested by Jukelyn#9848

[sweet edge]

yup

[thorny dust]

thank you

.close

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[tall arrow]

The wedge price for blackcurrants is 75, while the kilo price for sugar is 15. We are going to mix blackcurrants and sugar to make blackcurrant jam with a kilo price of 51. How much sugar is in the jam?

[tall arrow]

<@&286206848099549185>

[silver stump]

question pls

@tall arrow

[tall arrow]

The question is asking for the amount of suger in the jam

[silver stump]

I mean

where is the question

[tall arrow]

isn't it here

[silver stump]

sorry

[tall arrow]

is it not visible

oh, ok

[silver stump]

yes it wasn't
I'm switching between multiple tabs
currently in my uni class

getting bored
this teacher is teaching us about NoSQL and SQL databases
oooof

what does wedge price mean @tall arrow

?

so anyway
whatever that means
we would take the blackcurrants and sugar in a particular ratio

such that the price ends up being 51 for the kilo

so part of that kilo would be sugar

and the rest would be blackcurrants

you get that much
but now how exactly would we find that what ratio they are in?

[tall arrow]

yeah take your time
I apologise for the delay, something came in the way

the wedge price would essentially just mean price. The kilo price for sugar is 15

[tall arrow]

i'm unsure
can't we argue that there's more than oen answer?

to the question

considering that you can put different combinations

nevertheless the book says that it's only one answer; 40% sugar

with that in mind, the ratio that we might divide them in is 10%'s

[silver stump]

yes
that's possible
and the combos would be in a range

[tall arrow]

okay, but assuming that we want to get to the book's answer of 40% total sugar, what equations do we need to set up in order to solve this

[silver stump]

read the last line

now you can't take negative

nor can you make it such that there is no sugar

and the weight is also to be kept 1 kg

think a bit more on it

you're close

[tall arrow]

s = sugar
b = berries

1.5s + 7.5b = 51
s + b = 1

that should be it

[obtuse pebble]

@tall arrow Has your question been resolved?

[obtuse pebble]

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[mighty tulip]

for this

would u use the index law?

[timid silo]

Index law ?

[mighty tulip]

index law

[timid silo]

What is that ?

[mighty tulip]

index law is uh

how do i explain it

Index laws are the rules for simplifying expressions involving powers of the same base number

[timid silo]

By the way it is written mathematically ?

Ohh

$10^{9-5}$

[mighty tulip]

yes ik

but what do u do after

[timid silo]

$\frac{3.45×10^9}{5.3×10^5} = \frac{3.45}{5.3}×10^{9-5} = \frac{3.45}{5.3} × 10^4$

[mighty tulip]

oh dam

[timid silo]

The fraction approx evaluates 0.65

,w 3.45/5.3

[mighty tulip]

so then would u multiply the answer to the scientific notation?

[timid silo]

Or u can also use 10³

[mighty tulip]

ye it would be 103

[timid silo]

Oh ok

[mighty tulip]

tyty

[timid silo]

Then make it 10³ for scientific notation and u would get kilo

10³ = kilounit

[mighty tulip]

what is dat

[timid silo]

1km = 10³m
1kg = 10³g
If u wanna use scientific notation for 10³ u can just write kilo unit

the gram unit turns into kilogram

[mighty tulip]

but im trying to convert it into metres

[timid silo]

The meter unit turns into kilometres

[timid silo]

Similar methods

[mighty tulip]

kk

.close

[obtuse pebble]

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[thorny dust]

need help finding range and domain

[thick gyro]

hmm,

i assume those arrows

[thorny dust]

yes

[thick gyro]

are also apart of the graph

anyway, remember that the domain is the lowest and biggest x value the graph could take

and the range is the lowest and biggest y value the graph could give you

@thorny dust whats the biggest y value you see

[thorny dust]

3

[thick gyro]

whats the lowest

[thorny dust]

-2?

[thick gyro]

no

the arrow on the left

you can see that its pointing down

[thorny dust]

-8? or is it -10

[thick gyro]

the arrow is indicating that the left part of the graph will go on forever

if thats the case and its pointing down

what do you think the lowest y value is

[thorny dust]

-inf

?

[thick gyro]

yh

so whats your range

[thorny dust]

-inf and inf

[thick gyro]

no

not inf

remember

[thorny dust]

ohhh

[thick gyro]

.

[thorny dust]

-inf and 3

[thick gyro]

yh

now for the domain, whats the lowest x value you think the graph can take

remember that the left and the right part of the graph go on forever

cuz thats what the arrow is for

[thorny dust]

so now would that be inf

[thick gyro]

well lowest

[thorny dust]

.close

[obtuse pebble]

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[shadow cave]

if someone can help with this function would be appreciated just ping me ill be there

[high lily]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[shadow cave]

ive got an answer but its wrong

[high lily]

show work

[shadow cave]

this is the answer but i got no idea how they got to that

[high lily]

you didn't expand (x-1)^2 properly

[shadow cave]

isnt it just x^2 +1

[royal basin]

no, it is not just x^2 + 1.

(a+b)^2 ≠ a^2 + b^2.

[shadow cave]

sorry i dont understand

what formula is that

[high lily]

its freshmans dream stating what not to do

are you implying that
(1+1)^2 is 1 + 1 = 2?

[shadow cave]

no but

[high lily]

and/or just as bad
are you implying that
(1-1)^2 = 1 + 1 = 2

[shadow cave]

i mean when you have (x-1)^2 i thought you just square all the terms inside of the brackets

[high lily]

because that's effectively what you did

that's a hard no, like with those numerical examples where that clearly doesn't work

to expand, recall the definition of squaring and distribute properly
or apply binomial theorem

[obtuse pebble]

@shadow cave Has your question been resolved?

[obtuse pebble]

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[hexed agate]

e^(-x) cos(x) = 0

How should I solve this? Newton's method? Taylor series?

[royal basin]

much easier

zero-product property

[timid silo]

you can disregard e^(-x) as that is never zero

[hexed agate]

ok so cos x = 0
x = pi/2 +- npi

this is the solution?

[timid silo]

yea

[hexed agate]

The graph looks very odd

[thick gyro]

wdym

that graph looks normal

have you never done second order differential equations?

[hexed agate]

not muxh

sorry I was very confused by desmos

.close

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[gleaming ridge]

The answer should be 1/4 right? I just have to calculate F(4) - F(2)

[gleaming ridge]

In fact, the probability can never be greater than 1 right? I am a bit confused

[obtuse pebble]

@gleaming ridge Has your question been resolved?

[warped fulcrum]

What are you exactly confused about?

[gleaming ridge]

The answer for this question is given 4

ie P(2 <= X < 4) = 4?

That seems wrong as we know that P(-infinity < X < infinity) = 1 right?

[gleaming ridge]

F(4) - F(2) = 1 - (1/2 + 2/8) = 1/4 should be the answer?

[warped fulcrum]

Oh yes that's weird. The probability is always in the interval [0, 1] 🤔

[gleaming ridge]

right

I am guessing they misprinted 4 instead of 1/4

but not sure

[warped fulcrum]

Yes I'm guessing so too

Since I also get 1/4

[gleaming ridge]

ah, cool. It confirms my suspicions

Thanks for the help

[warped fulcrum]

No worries

[gleaming ridge]

.close

[obtuse pebble]

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[fathom nebula]

Can someone help me with the first limiti

Limit*

Better photo

[old halo]

simplify the fraction on the right and replace x+3 with t

you should have a definition of an euler's number

[fathom nebula]

Like this?

[old halo]

almost

9-x^2=(3-x)(x+3)

so (3-x)/(3-x)(x+3) = 1/(x+3)

[fathom nebula]

Mmhhh so at this point x+3= e?

[old halo]

no

lim(x+3)=lim(x) as x->inf

so this limit becomes lim(1+1/t)^t as t -> inf

[fathom nebula]

Mmmhh ok

[old halo]

lim(1+1/t)^(t-3)=lim(1+1/t)^t to clarify

[fathom nebula]

Ah ok

And at this point the results will be e^2

[old halo]

hmm

i don't think so

[old halo]

this is e by definition

e^x = lim(n->inf) (1+x/n)^n

you have (1+1/n)^n

[fathom nebula]

Ok arrived at this point my ask is, I have to semplify something?

[old halo]

what equation do you have right now?

[fathom nebula]

1+1/t^t

[old halo]

(1+1/t)^t?

[fathom nebula]

Ops sry

[old halo]

no worries

this is e by definition

[fathom nebula]

I arrived at this point

[fathom nebula]

And you told me to modify to this

[old halo]

nonono, return back to x

[old halo]

yes, the denominator

oops

[old halo]

you had different numerator

x-3 instead of x

[fathom nebula]

And the nominator will become also 3-x

?

[old halo]

yes, divide both halves of a fraction by x-3

[fathom nebula]

At this point I can semplify the x-3 with x+3?

[old halo]

x-3 with 3-x

and carry out the minus sign

[fathom nebula]

Wut

[old halo]

the answer will be e^-1 sorry

[fathom nebula]

So we arrive at this point

[grizzled shore]

,rotate

[fathom nebula]

Right?

[old halo]

not quite

[fathom nebula]

Because I semplified the x-3 with 3-x

But I don't get it a lot the carry out of the -

[fathom nebula]

Ah ok

1-1 semplified? So will remain only t^t-3?

[fathom nebula]

Oh Right there was also this possibility

[fathom nebula]

1/e?

[old halo]

yes

[fathom nebula]

Ah ok

[old halo]

I made a sign mistake the first time

[fathom nebula]

Final results I guess

Thk you very much 😊

.close

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[low pagoda]

is this correct?

[dark stirrup]

Do not have multiple channels open

Close this one or your first one

[obtuse pebble]

@low pagoda Has your question been resolved?

[low pagoda]

.close

[obtuse pebble]

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[obtuse pebble]

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[distant schooner]

A = (hw+k)/w

[distant schooner]

how do i make w the subject

i can make the w^2 into subject

but not w

[daring rock]

what do you have for w^2 ?

[distant schooner]

A/h+k

[gleaming ridge]

Multiply both sides by w and simplify

[distant schooner]

yes that's what i did in my head

so

Aw=hw+k

then from this point on i was stuck

tried hw+k/h+k

[gleaming ridge]

How are stuck, move everything with a w to one side then take common

[distant schooner]

so

nah i can't make it so that there's one w

idk how

[gleaming ridge]

What happens if you bring hw to left hand side

[distant schooner]

yah that's what i did wait hang on

so

Aw/hw = k

then i'd have to get rid of A and h

so w/w = Ahk

which doesn't work

[gleaming ridge]

No, no, no

hw is getting added to k

So, going left side it would be substracted

Aw - hw = k

Let that sink in

[distant schooner]

oh wait yeah

ok

and then

move the a and h to the other

[gleaming ridge]

Then can you take w out as common

[distant schooner]

OH SHIT

damn my dumbass

thanks man

[gleaming ridge]

Suree

[obtuse pebble]

@distant schooner Has your question been resolved?

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[timid silo]

h

[timid silo]

When 4 m3 of water leave the cylindrical tank, the level of water in the tank goes down by 1.5 mm. What is the radius of the tank?

<@&286206848099549185>

Recently it got timed out by 17 mins

[high lily]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[timid silo]

1

@high lily

This is the question

@high lily

[gloomy vector]

stop pinging them repeatedly lmao

say the volume is x, if x-4m^3 the water level goes down by 1.5mm

and you can try to use the formula for the cylinder

[timid yacht]

make a sketch and visualize

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[gloomy vector]

bruh

[timid silo]

Ok if x-4m3 the level is x-1*10to the power -9)

<@&286206848099549185>

<@&286206848099549185>

[gloomy vector]

dude stop spam pinging helpers

[timid silo]

ok sory

can answer question

[gloomy vector]

im too lazy to calculate

[timid silo]

pls

me : (

[gloomy vector]

well

[grand ore]

Ok i can

What r u doing

[gloomy vector]

check the pin it has the question

[grand ore]

I need to know the level of ur maths question to see if i can help

[grand ore]

Sorry, not for my level

[timid yacht]

did u do it?

[timid silo]

yES

[warm shale]

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

[timid silo]

What is t

[gloomy vector]

wait

[timid yacht]

show me

[gloomy vector]

basically its
x=$\pi$r^2h,
x-4m^3=$\pi$r^2(h-1.5mm)

and convert the numbers

not sure why latex doesent work

[warm shale]

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

[timid silo]

πr²t = 64

t = 4m - 1.5mm

t = 3.9985

Convert mm to m

1.5mm equals 1.5x10-9

equals 3.999999999999

Yes can take that too

nearly 4

But u have to reduce it

Cant take 4

4 was already the height

ok so 4 is the height?

[timid yacht]

no

[timid silo]

3.9985 might cancel with π

u mean 3.999999999

[timid silo]

No

[timid yacht]

1: what is the shape of the missing water?

[timid silo]

4 - 0.0015 = t

Hey wait how u got the r

[timid silo]

Dosent matter, but here it's cylindrical

[timid silo]

U gotta calculate r

[timid yacht]

it does, dont help if you dont understand either

[timid yacht]

what is this?

why do u subtract this?

[timid silo]

it says level go down

hmmmmmmmm

πr²t=4×4×4
r = √(64÷(π×t))

Take π as 3.14

[timid silo]

Yes

[timid silo]

It says

so what is volume?

[timid silo]

@timid yacht

[timid yacht]

yes

[timid silo]

Volume remains the same as b4

[timid yacht]

but we dont know the height of the tank at all

[timid yacht]

yes, we have a big pool in our garden, and want to lower the level of it by 1,5 mm

[timid silo]

Its level decreases because when u put cube in cylinder , cylinder takes a bit more space

[timid yacht]

and by doing it we let out 4 m^3 water

[timid silo]

64 is the water vol

[timid yacht]

Q: what is the area of our pool?

[timid silo]

64m³

[timid silo]

L×B

[timid silo]

It is L x H

Oh area

Sry

I told the vol

[timid yacht]

64??

?????

[timid silo]

Spreeter

please simplify it

my brain got damaged

Wait...

Oh damn

I read the question slightly wrong

[timid silo]

Vol = 4m³

Sry guys

[timid yacht]

we know the volume of the missing water(here:air)

and that its a cylinder

what is the volume formula?

for this kind of shape

[timid silo]

V = πr^2h

[timid yacht]

yeah

h is ..?

[timid silo]

x - 1.5mm

[timid yacht]

no

look at the picture

[timid silo]

wdym im right

[timid yacht]

its 1.5 mm

[timid silo]

huh

its the air

not h

[timid yacht]

we know the volume of the air(missing water)

and the height that goes to that is 1.5 mm

[timid silo]

OOO

ok

ok

t = height of water in cylinder = (cube_root(4)-0.0015)

[timid yacht]

still wrong, please let me finish

[timid silo]

0.0015 is the height and 4 is the volume

how u can combine them

[timid yacht]

4m^3= 0.0015m * pi * r^2

solve for r

[timid silo]

There are basically two ways to solve this , for r , either u take air volume or water , I'm taking water so

[timid silo]

This is correct for water vol method

[timid yacht]

we dont know the remaining water at all

so no

[timid silo]

Isn't the question like , the water didn't lowered cuz it was taken out , it was lowered by level cuz it changed chape

Shape*

So we need to do it with water vol method

It was never given in question that water was taken out

[timid yacht]

nope

[timid silo]

It only said that shape of container changed

[timid yacht]

bro, are you trolling here?

[timid silo]

?

Ok sry I'll let u finish

[timid yacht]

read the question carefully

[timid yacht]

it says so in the first sentence

[timid yacht]

@timid silolook here

[timid silo]

ok

omg

thank u

[timid yacht]

it wasnt incorrect, just TOO correct 😄

[timid silo]

lol

[timid yacht]

@timid silosee where you did go wrong now?

[timid silo]

I have an another question

[timid yacht]

how would u start

[timid silo]

Finding the area

[timid yacht]

this is the same type as the previous

just an other info is missing

which is the volume now

[timid yacht]

yes

which is a what kind of shape?

[timid silo]

Rectangle

[timid yacht]

so the area is?

[timid silo]

Area = W x H

1.2 x 0.3 = 0.36m^2

[timid yacht]

yes

and the length is?

[timid silo]

length is 1.2m

[timid yacht]

no, i mean we had area* height before

just now its horizontal

so area*length

use the flowrate

[timid silo]

3metres

[timid yacht]

per?

[timid silo]

1 minute

[timid yacht]

and we have how many minutes?

[timid silo]

Um in hr?

60 minutes

[timid yacht]

so the length becomes...

[timid silo]

180m/hr

[timid yacht]

yes, 180

so multiply them

[timid silo]

So V = Area x length?

[timid yacht]

length height width, whatever, we just need a 3rd dimension

[timid silo]

180*0.36 = 64.8m^3/hr

MAN u are an amaizing tutor

[timid silo]

Yea i see now , I was connecting this to a similar question that i solved in past , i didn't read carefully i gotta say that , thanks for clearing the problem

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

Channel closed

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[obtuse pebble]

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[finite mango]

i dont understand this qn.. can anyone help?

[obtuse pebble]

@finite mango Has your question been resolved?

[twilit pelican]

do you know inverse laplace transformations?

[finite mango]

kinda

[twilit pelican]

hmm okay

so there's a property known as the first shifting theorem

[finite mango]

i just use the tables

[twilit pelican]

thats fine lol

the first shifting property states that

$\mathcal{L}^{-1}{F(s-a)} = e^{at}f(t)$ where $f(t)$ is the inverse of $F(s)$

okay never mind

[finite mango]

okay

[warm shale]

blanket

[twilit pelican]

there we go

holy

we were already given the inverse of F(s) which is sin(2t)

[finite mango]

is it C?

[twilit pelican]

uhh

yeah

sorry was blind for a sec

[finite mango]

damn

can i confirm with you afew more laplace questions?

[twilit pelican]

ye

[finite mango]

isnt a b and d the same?

[twilit pelican]

not quite

well

[finite mango]

can i not use it as a product?

[twilit pelican]

laplace is a linear transformation

[finite mango]

oh

[twilit pelican]

but we cant split it up multiplicatively

[finite mango]

but we can split it up when the operators are + and -?

[twilit pelican]

yeah

[finite mango]

ah

[twilit pelican]

had it been the sum of those functions and (a) was the sum, then that wouldve been the right answer

[finite mango]

i see

so its d now right?

[twilit pelican]

yeah

[finite mango]

okayy

[twilit pelican]

as a side note, we can usually rule out (a) and (b) or similar answers that look like it because of this fact

i can't recall any examples where it would result like that, but usually u can say "nah"

[finite mango]

also is this qn wrong?

[fierce lagoon]

I'm sorry side tangent here. I am also a victim of auto correct

Loving the name

[finite mango]

tell me your story

[fierce lagoon]

"Homomorphism -> homophobic"

[twilit pelican]

OH MY GOD

NO WAY

😭

[finite mango]

i cant find any inverse laplace transforms with this 2nd shift thing examples online, only normal laplaces

[twilit pelican]

id literally cry if that happened holy

[finite mango]

math is cancelled

[twilit pelican]

right

okay anyways sorry lol back to the q

[finite mango]

its fine, but shouldnt inverse laplace have the s function, function of s or s variable idk how to phrase it

[twilit pelican]

i was going to say

i think that's a typo

LOL

[finite mango]

HAHAHA

im so confused

[twilit pelican]

if it is, then id probably guess d to be the answer there

hopefully you've done laplace of the unit step func

[finite mango]

well this laplace is not in the syllabus im just studying online

[twilit pelican]

ah

[finite mango]

unless if you miss the exam then it is

[twilit pelican]

i see

well

[finite mango]

ive seen unit step function videos

[twilit pelican]

actually wait none of these would be right then

[finite mango]

it should be (2t^2 - t) right?

[twilit pelican]

unironically none of these are right imo

[finite mango]

or t(2t-1)

[twilit pelican]

well

no this is literally just a typo

i cant

what

LOL

[finite mango]

😳

[twilit pelican]

uhh

[finite mango]

is this more than just a typo

[twilit pelican]

yeah

no i

my brain is malfunctioning

[finite mango]

work of a drunk man setting this question

[twilit pelican]

soo

hold on

$\mathcal L{(t - 1)(2t - 1)u(t - 1)} = e^{-s}\mathcal L {((t + 1) - 1)(2(t + 1) - 1)}$ \
$=e^{-s} \mathcal L {t(2t + 1)}$

MAN DIE

[warm shale]

blanket

[twilit pelican]

oh wait

B

B

BBBB

AGL;JFHNL;DFBG

[finite mango]

🤔

[timid silo]

.

[twilit pelican]

well

still incorrect that should be laplace, not inverse

but wtv

[finite mango]

ans B?

shouldnt it be {2t^2 - t} tho

[twilit pelican]

yes it is but does this breakdown make sense to you?

you substitute the shift in for all the ts you see in the f(t) function

[finite mango]

OH

[twilit pelican]

so the 2(t+1) - 1 = 2t + 2 - 1 = 2t + 1

then we just have a regular t

so 2t^2 + t

[finite mango]

okay made sense

i thought its just only the one that is the same

[twilit pelican]

no yeah so you shift all your ts

this is a good example

their g(t) is the non unit step function

[finite mango]

then shouldnt this qn be {2t^2 + t}

[twilit pelican]

that would be c

[finite mango]

i dont even know whats option d

[twilit pelican]

option d is madness

dont worry about that one

[finite mango]

[twilit pelican]

but do you see how we would then rewrite

$(t - 4)(2t - 3)$ as $((t + 4) - 4)(2(t + 4) - 3)$

[warm shale]

blanket

[twilit pelican]

because of the step of 4

[finite mango]

oops i didnt multiply in the 4 by 2

man my brain aint braining

[twilit pelican]

lol0ol

[finite mango]

wow!

simple algebra

[twilit pelican]

sometimes

algebra aint that simple anymore

it just gets ruf

[finite mango]

dedge

okay last, but ive done it

[twilit pelican]

no ur chillin

[finite mango]

this would be d

[twilit pelican]

..

is that an e^s or e^t

?

[finite mango]

e^t id guess

on the question right?

[twilit pelican]

ye

which one did you do?

[finite mango]

the image provided is so

[twilit pelican]

fr

LMFAO

[finite mango]

my ans is d

[twilit pelican]

well

be happy because ur right

[finite mango]

😏

[twilit pelican]

absolutely cracked at the laplace

[finite mango]

laplace genius right here

[twilit pelican]

my god

so good

[finite mango]

thanks to you tho

[twilit pelican]

i think i need ur autograph

[finite mango]

🔥

[twilit pelican]

[twilit pelican]

naa all you

just guided u a bit

[finite mango]

you helped alot

guess how long i did for this thing

[twilit pelican]

i guide others to a treasure i already got to but forgot about

imma guess

3 hours

[twilit pelican]

.

never mind

sorry

holy shit 😭

[finite mango]

😎

[twilit pelican]

is that 12 days? 💀

[finite mango]

im cool like that

[twilit pelican]

i mean

you just went over a lil bit

not a lot

just a bit

[finite mango]

10plus

[twilit pelican]

its okay

justa little bit

nw

[finite mango]

gotta utilise all the time given

[twilit pelican]

absolutely

[finite mango]

thanks!

[twilit pelican]

the time's there, we gon use it

np! glad i could assist

[finite mango]

btw i added you but i cant seem to get it across

❤️

[finite mango]

.close

[obtuse pebble]

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[wise hedge]

My question is:

[wise hedge]

Does every angle on the unit circle have a 90 degree angle?

[high lily]

that question doesn't make much sense

[undone sparrow]

What kind of angles

[high lily]

wdym by angles having a 90° angle

[wise hedge]

Angles that end up in quadrant II?

So anything in quadrant II would theoretically be > 90 degrees.

On a unit circle, that is.

[undone sparrow]

What do circles have to do with this

[wise hedge]

I am supposed to "superimpose" a right-triangle in each of the quadrants inside my brain?

Bruh, you're asking questions that make you appear to be unaware of the unit circles purpose in trigonometry.

[undone sparrow]

Probably

[high lily]

your question is very poorly phrased to understand what you want

[wise hedge]

I was first taught about sine, cosine and tangent on right-triangles only. But now they are using the same functions on any angle triangle. That's where I have been lost.

[high lily]

reference triangles can be drawn on the unit circle
and those reference (right) triangles will have a 90° because they're right triangles if that's what you mean...

[wise hedge]

Yes, that's what I'm asking, if I am expected to superimpose a right-triangle on the unit circle in the 4 quadrants.

As needed.

Because that's the only way my brain understands sine, cosine and tangent is with the existence of a right-angle.

[high lily]

on the unit circle
cos(t) and sin(t) give the x and y coordinates respectively

[wise hedge]

Yes I understand that but those are ratios between 2 legs of a right-triangle.

[high lily]

and tan(t) = sin(t)/cos(t)

[wise hedge]

Yes, I know that equation also.

[high lily]

right triangle trig definitions are a much simplified version of the unit circle definition

[wise hedge]

That's my problem is understanding sine, cosine and tangent as ratios between 2 legs as we move away from the right-triangle context.

on a 180 degree angle there is no longer an "opposite" leg.

[high lily]

you said you knew everything i just typed about the unit circle

[wise hedge]

I do but you just reprinted equations, that doesn't equate to my understanding of the concepts.

I have the equations.

[high lily]

on the unit circle
cos(t) and sin(t) give the x and y coordinates respectively

[wise hedge]

Right

[high lily]

when it comes to the unit circle, you shouldn't really concern yourself that much with opp/adjacent anymore

[wise hedge]

Okay, you're telling me something that is helping but that's what my problem is, getting away from how I was taught from the beginning.

[high lily]

if you really wanted to, you can use reference angles/triangles

(which uses stuff with opp/adjacent in combination with other things)

[undone sparrow]

I just treat it as a point, 180 would lie on the negative part of the X axis on the unit circle so it's cosine would be -1

[wise hedge]

That's my point once again, I was asking before, "Do I need to visualize a right-triangle rotated to a specific point so that the idea of sine, cosine and tangent once again makes sense in my brain?

[high lily]

know the rules and visualise in a way that you're comfortable with

knowing the theory, you can approach it in multiple different ways and ultimately stick to the one you're most comfortable with

can you give a specific problem and i can guide you through multiple potential approaches

[wise hedge]

here's my point. when I first started learning about sine, cosine, and tangent it was on a right-angle so the relationships were easy to see and understand. But when you start talking about going past 90 degrees those definitions no longer make sense so I'm trying to understand them when going past 90 degrees on a unit circle.

Let's say the angle is 270 degrees of a unit circle so we're talking about 3/4ths rotation around the circle and they ask for cosine/sine at that point.

[wise hedge]

$\pi3/2$ radians

[warm shale]

j4w4

[high lily]

on the unit circle
cos(t) and sin(t) give the x and y coordinates respectively
cos(270°) gives the x-coord on the circle at that angle of rotation which is 0
sin(270°) gives the y-coord which is -1

[wise hedge]

Okay, the 0 degree makes sense because the line is completely vertical. And I guess since we're talking about a point on the line, yes -1 makes sense also. Okay.

So the x,y on a vertical line seems easy enough to figure out just intuitively but I'll bet it's much harder to do without an easy start.

What would be the way to figure out the cosine and sine for 265 degrees?

[high lily]

265° isn't a special angle so you'd pretty much need the aid of a calc

to efficiently get an accurate approximation

[undone sparrow]

sin(265) = sin(270-5) = -cos(5) so you can visualize it as a right triangle with 5 degree angle

[wise hedge]

Yes, there we go, always that key advice to visual the right triangle in that specific quadrant, is that right?

So, not throwing everything away that was learned in the beginning.

[high lily]

i mentioned at the very start that you could do that if you wanted

[wise hedge]

Yes, if I want, but isn't that a good idea?

[high lily]

if you're comfortable with that, sure why not

[wise hedge]

Is that not suggested by many a teacher so that the students can make the transition to the unit circle easier?

[undone sparrow]

Sure you can also do it in the original angle's quadrant if you want, but it could be trickier

[wise hedge]

I don't need tricky yet, I need easy. 🙂

[high lily]

as you get more experienced you won't need the full unit circle

[wise hedge]

I figure to just rotate a right-angle into the quadrant where the terminal line is and visualize it being right there.

[high lily]

just a reference triangle and consideration of quadrants for the appropriate signs

[undone sparrow]

Be a pro and use a calculator

[wise hedge]

Thank Your for the help, I'm sure it will all fall into place as I continue to mosh through the training.

.close

[obtuse pebble]

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[timid silo]

.open

Hey can anyone help me with this question:

[timid silo]

Why do some equations have their degree as not defined , ex Q 3

[tardy epoch]

sine is a function that can take in any real value, in this case e^x is that value

similarly, the exponential is a function that can also take in any value x

so their composition, sin(e^x) is a function

you can convert to degrees, but it's more useful to think of the argument as radians

[timid silo]

Thanks brother

.close

[obtuse pebble]

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[full oar]

I don’t get it help :’)

like for question O

63 + 63 = 126

and then 180 - 126 = 54

but idk how to find x and y

[high lily]

you've yet to apply properties related to parallel lines

[obtuse pebble]

@full oar Has your question been resolved?

[timid silo]

X is 63

Y is also 63

[obtuse pebble]

Channel closed

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[full oar]

.reopen

[obtuse pebble]

✅

[full oar]

wait how?

[timid silo]

Learn about properties of parallel lines

Check out YouTube

[obtuse pebble]

@full oar Has your question been resolved?

[obtuse pebble]

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[plain grove]

Hi guys how can I determine the equation for mass

[ruby path]

I hope we have some context following

[timber wind]

What do u mean by that?

[sharp pecan]

surely

[ruby path]

$$\frac{E}{c^2}$$
$$\frac{F}{a}$$
$$\rho V$$

[warm shale]

NEONPerseus

[ruby path]

[sharp pecan]

{\displaystyle {\begin{aligned}E_{\rm {rel}}={\sqrt {(m_{0}c^{2})^{2}+(pc)^{2}}},!\end{aligned}}}

[warm shale]

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

[sharp pecan]

solve for m and you get another funny mass equation I guess

[plain grove]

Here I can show you exactly what I Lena

Mean*

L=length of fish

The attachment are loading…..

[obtuse pebble]

@plain grove Has your question been resolved?

[plain grove]

Finally!! Here

[obtuse pebble]

@plain grove Has your question been resolved?

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[shadow lava]

would you consider this 0? if f(x) is 0, and g(x) is 3

0*3, right?

[pine sail]

Yes.

[shadow lava]

alright thanks

.close

[obtuse pebble]

Channel closed

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Use .reopen if this was a mistake.

[shadow lava]

.reopen

[obtuse pebble]

✅

[shadow lava]

sorry just to be clear, f(g(1))

like this

is it still talking about limits?

or just the value of f(g(1))

[pine sail]

Just that. If there's no limit mentioned.

Since g(1) is well defined.

[shadow lava]

but f(1) is not well defined

g(1) = -1

f(1) = 0?

[pine sail]

f(1) is zero isn't it. But f(g(1)) doesn't require you to use f(1)

f(g(1)) = f(-1)

[shadow lava]

ooo

right right

[pine sail]

Now your graph doesn't specify if f(-1) is defined or not. I'm assuming it's zero

[shadow lava]

the answer is 0 because it lands here specifically

[pine sail]

Yes. Then zero.

[shadow lava]

but normally there is a filled or hollow dot, no?

when drawing on graphs

this line just ends without a dot

that is considered normal?

[pine sail]

Well I thought you drew it.

[shadow lava]

nope lol

[pine sail]

So well you know.

[pine sail]

Haha then sure.

It's zero.

[shadow lava]

maybe it's bad practice to end lines like that

should be crystal clear, included or not included, if the line ends

[pine sail]

No that just seems like it's hand drawn.

So I thought it was you without realising.

[shadow lava]

or maybe it only needs hollow dot, if not included

[pine sail]

But we'll guess not!

[shadow lava]

otherwise no need to draw solid dot

[pine sail]

Definitely.

[shadow lava]

alright

so one last question

what if this f(x) landed on a slope

not asking for limit, just f(0)

for example this one on the left

I would just mark it as 3?

actually the question would be f(0), not f(x)

I'm guessing 3 would be correct here.. just want to make sure for slopes

[pine sail]

f(0) is 3 yes.

What slopes?

[shadow lava]

the curve

max/min

[pine sail]

What about it

[shadow lava]

it's not a line anymore

[shadow lava]

like this

where it ends

it's continuous

i guess that's fine, no problem for showing output of f(0)

[pine sail]

Of course not. Because it's Just one particular value.

[shadow lava]

yeah

so g(x) = 3

after calculating for g(x) now i'm looking at f(3)... so am I ignoring 2 now?

i already looked at left side of 2 to obtain 3 for g(x)

I still need to incorporate left side of 2 somehow into this question for f(3) somehow?

f(3) is 0

but left side of 3 is rising from 0 towards 1

I think I'm just looking at left side of f(3) now.. which would be 0
x -> 2- is only there for g(x)... but just want to make sure.

@pine sail mission accomplished

thank you!

https://tenor.com/view/smile-okey-we-did-it-mission-complete-mission-accomplished-gif-22504886

[pine sail]

You're welcome! I didn't do much!

[shadow lava]

moral support

.close

[obtuse pebble]

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[timid silo]

need help

[obtuse pebble]

@timid silo Has your question been resolved?

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[obtuse pebble]

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[obtuse locust]

idk how to go about solving this inequality

[sweet edge]

try factoring the top first

and then cancel stuff out

just keep track of what x cannot be from the beginning

[obtuse locust]

would it simply just be (x)(7x-26x-1)

[sweet edge]

,w factor 7x^2-26x-1

[warm shale]

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[sweet edge]

damn

mmm

complete the square then factor...?

[obtuse locust]

yeah see thats what i got lol

hmmm

[sweet edge]

u can't factor an x from -1

,w complete the square 7x^2-26x-1

[warm shale]

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[obtuse locust]

damn

[sweet edge]

,w expand 7(x-13/7)^2-176/7

fucking fuck shit

[warm shale]

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[sweet edge]

mb, I can't seem to get what I wanted

I'll mess with the problem a bit and come back

[tranquil sonnet]

You have 2 cases: the numerator is non positive and the denominator is positive, or the numerator is positive and the denominator is negative

You’ll have a system of inequalities for each case

[obtuse locust]

so what would i do for that?

[tranquil sonnet]

Solve them

[obtuse locust]

okay yeah so im looking for my zeros and vertical asymptotes?

[obtuse locust]

im not sure what to do with my 7x^2

when i look it up it suggests i use the quadratic equation?

[obtuse pebble]

@obtuse locust Has your question been resolved?

[spice chasm]

make a number line

note down the zeros of the quadratic numerator

@obtuse locust

on the number line mark the zeroes, 3, and 7/9 in least to greatest form

test a number from each region to see its sign

then once you have tested every region, the answer will be the regions that are negative

[spice chasm]

the first root is approximately -0.038, the second root is approximately 3.752, so graph the first root at the left of the number line, then graph 7/9, then 3, then 3.752

[spice chasm]

it’s not relevant as you can’t factor the numerator either way

[obtuse pebble]

@obtuse locust Has your question been resolved?

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[distant schooner]

so

for 1st = l

2nd = l(2/3)

3rd = l(2/3)(3/4)

4th = l(2/3)(3/4)(3)

but then for a)

i dont know how to make it in terms of l

[obtuse pebble]

@distant schooner Has your question been resolved?

[distant schooner]

.close

[obtuse pebble]

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[quartz charm]

question d)
line 4
why is it (x^3 - 1^3) and not just (x^3 - 1)

[sweet edge]

1^3 =1

they just wrote it like that for the sake of the video and intuition I guess

[quartz charm]

what if it was (x^3 - 6)

would it become (x^3 -6^3)

[dark stirrup]

It would become $(x^3-{\sqrt[3]{6}}^3)$

[warm shale]

SWR

[sweet edge]

no, it's the- yeah

re-watch that part of the video (if it's from a video) and then listen closely to why they did that

[quartz charm]

its from class, my teacher said finish the rest at home 🤷‍♂️

what does the stuff to the left of it mean

[sweet edge]

the left is just the general rule for difference and sum of cubes

[quartz charm]

ok thank you

.close

[obtuse pebble]

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

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