help-23

[fair dagger]

you need to get it into the standard shape first.

otherwise obviously the result would disagree

can't be both right

[karmic monolith]

Would this be the standard shape?

[fair dagger]

yeah

now we just use standard quadratic formula

a = -tan(y)², b = -1, c = tan(y)²

[karmic monolith]

Hey I really appreciate the help, this is just getting more confusing so I'm going to ask my instructor the next time we have class, tyvm for your time!

[fair dagger]

Eh, we were essentially done 😮

https://en.wikipedia.org/wiki/Quadratic_formula

[karmic monolith]

I'll try, just seems like the result would be very different from the answer that's correct

[fair dagger]

hm let's see, I will try too

[karmic monolith]

(answer)

[fair dagger]

let me see

okay no

let's do this together

[karmic monolith]

Ok

[fair dagger]

(b +/- sqrt(1 - 4ac)) / (2a)

[karmic monolith]

plug this in? ^

[fair dagger]

so I get (1 +/- sqrt(1 - 4tan(y)^4)) / (2tan(y)²)

[karmic monolith]

Why would it be ^4?

[fair dagger]

,wolf s = sqrt(-s² + 1) t

[flat frigate]

Results provided by WolframAlpha

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[fair dagger]

Wolfram alpha finds the correct t

something went wrong in our calculcation

let's retrace our steps

[fair dagger]

Because both a and c are tan²

[karmic monolith]

Oh ok I see

[fair dagger]

I found the mistake

Here we were missing a square at the left hand side.

I wasn't paying enough attention either.

Then it's sin(y)² = tan(y)²(1 - sin(y)²)

thus sin(y)² = tan(y)² - tan(y)² sin(y)²

thus sin(y)² + tan(y)² sin(y)² = tan(y)²

thus sin(y)² (1 + tan(y)²) = tan(y)²

thus sin(y)² = tan(y)² / (1 + tan(y)²)

[karmic monolith]

K, I understand that

So we're going back and not doing quadratic formula right?

[fair dagger]

yeah it is unnecessary this time

there is no linear term

you could use it still but it would just be wasted effort

Does this make sense?

I think this should be right result

[karmic monolith]

I mean the part you just did does, overall no 😂

[fair dagger]

I guess it's a longer process

the essential step to get started is to use this extra variable y, so that y= tan^-1(xy) and there is no inverse trigonometry

after that it is equation solving

try remove square roots and fractions

[karmic monolith]

So the general steps for these types of problems would be to "expand" the inverse trig functions, then simply the resulting equation?

[fair dagger]

yes

[karmic monolith]

Alright.

[fair dagger]

One can play these kind of games for all trigonometry

[karmic monolith]

I'm gonna put this away for tonight since I've been working on the homework for 4 hours and I'm tired, thanks very much for your assistance

[fair dagger]

though, maybe that makes you happier, going from tan to sin or cos is maybe the most difficult

[karmic monolith]

Nothing in trig makes me happy 😂

Anyways, seeya, thanks again :)

[fair dagger]

You are welcome 🙂

bye

[karmic monolith]

.close

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[wide steppe]

determine the length of R-L

[wide steppe]

||ping me when ur here @ Helpers||

[fair dagger]

sqrt((12cm)² + (8cm)² + (5cm)²)

[wide steppe]

144+64+25?

[fair dagger]

and square root

,wolf sqrt((12cm)² + (8cm)² + (5cm)²)

[wide steppe]

Hm

Oh

[fair dagger]

there is not really much happening, just using a formula

[wide steppe]

Mk

Tysm

[flat frigate]

Results provided by WolframAlpha

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Requested by M8732#6402

[wide steppe]

Eh

oh

[fair dagger]

square was in the wrong place in the query

[wide steppe]

RL = 15,26 cm?

[fair dagger]

was doing 8 square cm instead of (8cm)²

ye

[wide steppe]

Oh

Ok

[lean otter]

to determine the length of RL, you need the length of NL
to get the length of NL->
NL^2 = 12^2+8^2
then its just RN^2+NL^2=RL^2

[wide steppe]

Ya thanks

(:

.close

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[royal skiff]

Confused about the final step to solve part f

[drowsy moss]

$5(5^k) = 3(5^k) + 2(5^k)$

[flat frigate]

Zybikron

[drowsy moss]

and as a point, you don't know that $5^{k+1} + 3^{k+1} = 2q$. That's what you're trying to prove.

[royal skiff]

Oh thanks

[flat frigate]

Zybikron

[royal skiff]

I think I got it

.close

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[timber atlas]

I’m having trouble drawing the diagram

[timber atlas]

I drew this much but I’m having trouble with builder ben’s part.

[safe radish]

@timber atlas Has your question been resolved?

[timber atlas]

<@&286206848099549185>

[drowsy moss]

Ben is looking over the opposite side of the building. So the wheelbarrow is 10.8m from the other side of the vertical

[timber atlas]

That?

<@&286206848099549185>

[safe radish]

@timber atlas Has your question been resolved?

[timber atlas]

.close

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[desert hinge]

I dont understand how to do any of these but 2.

[safe radish]

@desert hinge Has your question been resolved?

[safe radish]

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[real wyvern]

I am struggling to integrate to specifically get the integrated function to look at how it is asked

[real wyvern]

I don't know how to integrate the function as a whole so I integrate 20/6 and 20/h

and I get: 10/3h -20ln|h|

But how do I make into the 'show that' formula

[stoic dune]

See "separable differential equation"

[safe radish]

@real wyvern Has your question been resolved?

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[hasty cairn]

How do I solve this?

also meisdeisme bot pinned me

so argue with it sorry

[frigid geyser]

hint: put things into sin and cos

[hasty cairn]

My teacher didn’t specify this outside of help videos…

He just taught me what the pythagorean identity is

why do teachers do this?

Just give you problems you weren’t told….how to solve in the lesson…

[frigid geyser]

partly to let you experiment

[lyric ingot]

Oh wait my bad

[hasty cairn]

Experiment……Uhhh I fail when I enter experiments

Like fail a lecture

so…..

[frigid geyser]

well, i suggest you get around with experimenting with your question, yes?

[hasty cairn]

wdym?

[frigid geyser]

time to solve this

[hasty cairn]

I experiment think of an answer and it’s wrong…

[lyric ingot]

.

[lyric ingot]

I will as well

[hasty cairn]

And I blame the lesson because it didn’t teach me how to solve…the problem….

[lyric ingot]

What did the lesson teach you then?

[ancient escarp]

expressing it in terms of sin and cos is one of the best ways to simplify trig when you're stuck

[hasty cairn]

Maybe I’m just bad at math…?

[ancient escarp]

stick to what camilleone said

[hasty cairn]

Pythagorean identity and tangent identity

[hasty cairn]

They didn’t tell me that in the lecuture

[frigid geyser]

now you know

lectures can't cover everything

sometimes you have to find out yourself

and right now you're finding out putting things into sin and cos is a great way to simplify trig

[lyric ingot]

Tangent identity like sinx/cosx??

[hasty cairn]

But they give me the problems….of the classwork not homework…that needs to me covered now!

[hasty cairn]

yea

[lyric ingot]

Thats mostly all you need then?

[frigid geyser]

yeah, if you know sec, tan, and cot, and tan = sin/cos, you pretty much have everything you need - just put things into sin and cos and simplify from there

[lyric ingot]

Im a bit stuck lol

[hasty cairn]

I’m really bad at it then….

[frigid geyser]

whut

[ancient escarp]

what have you tried so far

[lyric ingot]

||cosx/sinx + sinx/cosx idk how to simplify||

[ancient escarp]

your denominator is wrong

[hasty cairn]

It’s not

How could it be?

[ancient escarp]

2(sin(x)+cos(x))?

[hasty cairn]

Sinx + cosx

[frigid geyser]

specifically, the sin + cos part is wrong

[ancient escarp]

^

[hasty cairn]

Can’t be simplified

[ancient escarp]

that's not how common denominators work

you're multiplying each fraction by the other's denominator / denominator

so how did you end up adding them

[ancient escarp]

that's the same image...?

[ancient escarp]

where'd the two come from

[hasty cairn]

My ipad

[ancient escarp]

you also crossed out the cosines to cancel and rewrote it?

[hasty cairn]

no I moved down a page

[frigid geyser]

maybe it'd help if you wrote everything step by step

[hasty cairn]

look at the page numbers

[frigid geyser]

or maybe let's go back to basics

[ancient escarp]

that's cool but you wrote
$$\frac{1}{\cos(x)}\cdot 2\sin(x)\cos(x) \implies 2\sin(x)\cos(x)$$

[flat frigate]

a disappointing son

[ancient escarp]

which is what i was confused about...

[frigid geyser]

what is $\frac{a}{b} + \frac{c}{d}$?

[flat frigate]

Camilleone

[ancient escarp]

my two questions still remain

1.) how are you getting a 2
2.) why do you cancel the cos in the denominator but not the numerator

[hasty cairn]

$ \frac{a}{b} + \frac{b}{d} \text{ equals } \frac{a+b}{2(bd)} $

texit…

$$ \frac{a}{b} + \frac{b}{d} \text{ equals } \frac{a+b}{2(bd)} $$

[ancient escarp]

that ain't right

[flat frigate]

Nerdy_Coder

[hasty cairn]

texit

there you are

[frigid geyser]

that ain't right

[ancient escarp]

lol

[hasty cairn]

yes it is

[ancient escarp]

alright

i guess it's right

if you say so

[hasty cairn]

….why is it wrong?

[ancient escarp]

because that's not how adding fractions work

[hasty cairn]

you must multiply to a common demoniator

[ancient escarp]

yep

[hasty cairn]

bd is the smallest common demoniator

in variables

[ancient escarp]

why are you getting 2(bd)

and also a+b?

[lyric ingot]

Bro how does that work

[ancient escarp]

by that logic, $\frac12+\frac14=\frac{2}{16}$

[flat frigate]

a disappointing son

[hasty cairn]

bd + bd = 2(bd)

[ancient escarp]

you're not adding the denominators

you're adding the fractions over a common denominator

[hasty cairn]

but IDK what the demoniator is

[ancient escarp]

https://www.youtube.com/watch?v=N-Y0Kvcnw8g

[hasty cairn]

no the demoniator is a variable

[ancient escarp]

same rules apply

[hasty cairn]

I DID ECD!

what is the demoninator

$$bd^2?$$

[flat frigate]

Nerdy_Coder

[hasty cairn]

$$ \frac{a}{b} + \frac{b}{d} \text{ equals } \frac{a+b}{bd^2} $$

?

[flat frigate]

Nerdy_Coder

[lyric ingot]

What

When what where how?

[hasty cairn]

foto

[frigid geyser]

no

[hasty cairn]

forgot

$$ \frac{a}{b} + \frac{b}{d} \text{ equals } \frac{ba+db}{2(bd)} $$

[flat frigate]

Nerdy_Coder

[hasty cairn]

$$ \frac{a}{b} + \frac{b}{d} \text{ equals } \frac{ba+db}{bd^2} $$

[flat frigate]

Nerdy_Coder

[frigid geyser]

both are wrong

[hasty cairn]

$$ \frac{a}{b} + \frac{c}d} \text{ equals } \frac{da+bc}{cd^2} $$

[flat frigate]

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

[hasty cairn]

hello

[frigid geyser]

you're just guessing at this stage; i suggest watching the video first

[hasty cairn]

Who?

I did just a minute ago see the vid

[onyx pilot]

what

$$ \frac{a}{b} + \frac{b}{d} \text{ equals } \frac{ad+b^2}{bd} $$

[flat frigate]

ShadoeStorm

[onyx pilot]

$$ \frac{a}{b} + \frac{c}d} \text{ equals } \frac{da+bc}{bd} $$

[flat frigate]

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

[safe radish]

@hasty cairn Has your question been resolved?

[lyric ingot]

You can use cot(x) + tan(x) = csc(x) * sec(x)... but I don't think they taught you that

[hasty cairn]

oh

[hasty cairn]

They didn’t

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[lean otter]

Can anyone help me to solve this problem?

[lean otter]

<@&286206848099549185>

[worthy hemlock]

You have to wait at least 15 minutes before pinging helpers

[stone mesa]

I need help guys been studying for my exams and we were given a practice sheet(not graded) and I just cant find the answer
The length of a rectangle is 12 m more than thrice its width. The perimeter of the rectangle is 200 m. Find the length of the rectangle.
its been almost an hour

<@&286206848099549185>

[ancient escarp]

not.

your.

channel.

for the third time

#❓how-to-get-help

[oblique nova]

derivative of a vector-valued function, in this case r(t), is found by taking the derivative of each component

$$ \dv{\textbf{r}}{t} = \begin{bmatrix} \dv{}{t} \left( \sin(t)\cos(t) \right) \ \ \dv{}{t} \left( \frac{1}{t^2+4} \arctan(2t+1) \right) \end{bmatrix} $$

what happened to my sint cost

[flat frigate]

anamono

[oblique nova]

there we go

anyways that would be the first step in solving (a)

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

Is it solved yet? I think I'm very dumb at this point.

[oblique nova]

nah im giving you the steps to do it

[oblique nova]

this is the first step you would do for solving part a

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[distant steppe]

what do I predict

[distant steppe]

?

[trim venture]

sign of quotient

like positive divided by positive is a positive quotient

and negative divided by divided by positive is negative

like wise we have to find the sign of quotient

[distant steppe]

so for a) it'd be a negative quotient right

[trim venture]

yes

also

[trim venture]

we also have to calculatr the quotient in later part

[distant steppe]

what does positive times negative give you

negative right

[trim venture]

yes

[distant steppe]

sorry what do you mean by this

[trim venture]

we have to also wind the value of quotient

(1/5) / (-2 / 5) =?

[distant steppe]

Ohh so I predict the quotient (answer) first then I do the actual question to get the quotient (answer)

[distant steppe]

-1/2 or -0.5 ?

[trim venture]

yes

[trim venture]

any of them

-1/2 would be better ig

[distant steppe]

thanks

.close

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[ancient escarp]

division

[zinc crown]

use .close if you have no more questoins

[safe radish]

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[hot herald]

man this is the stupidest goddamn question and i'm getting mad about it but math isn't my strongest suit so can someone explain to me how

x^2-2xy+y^2 is (x-y)^2 simplified

i just can't seem to process it and i'm getting annoyed lol

[fervent dome]

$x^{2}-2xy+y^{2} = (x-y)^{2}$

[flat frigate]

謝墨離

[slow sequoia]

anyone here

can u help me

Elizabeth had {1}{4} meters of green ribbon. She used {1}{2} of it for her project in Science. How many meters of ribbon did she use?

[worthy hemlock]

Because (x - y)^2 = (x - y)(x - y)

[worthy hemlock]

#❓how-to-get-help

[slow sequoia]

idk

[hot herald]

man that doesn't really help

[slow sequoia]

i am new

[hot herald]

okay go to #help-17 @slow sequoia

i like can't seem to understand like hsnkshxkejdj bc jxvjjcj jvjvjckvjcjc

[worthy hemlock]

Do you know the process of factoring?

[hot herald]

yeah

[worthy hemlock]

Because that's all that is

[hot herald]

like i get that

^2 is two of them

i just

don't get it

in this one

like i just don't understand and i don't get it

[hot herald]

i just don't get where the two (x-y) came from like in the equation is it the 2xy or is it the x^2 + y^2 like odnbfkduckduckgo

djjdkduucskcicj

[worthy hemlock]

It's a special case that's why

Like these

[hot herald]

oh okay just lemme see

ya we never learned about the special case so that's probably why i'm so goddamn confused

[worthy hemlock]

https://youtube.com/watch?v=6QQJoDshUt8

[hot herald]

okay that makes a little more sense

so if it were (s-t)^2 it's just s^2 - 2st + t^2

[worthy hemlock]

Yes

[hot herald]

okay thank you

you help

sorry i was just losing my shit over it

thank you

.close

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[proper sage]

foil dat hoe

[hot herald]

how do i foil it

it's in a weird way to be foiled

[proper sage]

what if i write it like

you know how a^2 = a * a

[hot herald]

yes

[slow sequoia]

Elizabeth had 4LaTeX: \frac{1}{4} meters of green ribbon. She used LaTeX: \frac{1}{2} of it for her project in Science. How many meters of ribbon did she use?

[proper sage]

(s - t)^2 = (s - t) (s - t)

[hot herald]

RUEL GET UR OWN CHANNEL

[proper sage]

awwww shit now you can foil it

[hot herald]

stop SENDING TJEN IB mINE

ok wait

wait yeah

ok let me process that

[proper sage]

foiling isn't the hard part

its the part where you take apart the square

[hot herald]

so basically

(x-y)^2 is

yeah it's just x-y x+y no?

[proper sage]

not quite

[safe radish]

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[lilac hull]

I have an optimisation problem to find when pigs should be harvested for maximum mass of meat. I have the data for population to time (it decreases) and length to weight, and length to time. How will I find time to weight?

[lilac hull]

@ me if u got an idea

[safe radish]

@lilac hull Has your question been resolved?

[safe radish]

@lilac hull Has your question been resolved?

[safe radish]

@lilac hull Has your question been resolved?

[inland ivy]

You could linearly interpolate the data ig

Aka fill in the gaps between datapoints with a straight line from one point to the next

Mathematically

@lilac hull

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@lilac hull Has your question been resolved?

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[heady basin]

Closed previous channel but am restock

Anyone help please ?

Any formulas i create are self suggestive

[sacred quiver]

middlepoint of AC = M: isnt AC = 2, because MB to B = 1 and we have a 90 degree angle which includes BM and MC, so MC is 1 too?

MCB is an isosceles triangle

[heady basin]

Is it ?

[sacred quiver]

i guess

ABC = 1 then

[sacred quiver]

oh wait

im talking about the picture above bruh

[sacred quiver]

are you talking about this?

[heady basin]

Yeh its drawn in a tricky way

[sacred quiver]

ahh

[heady basin]

Bac Disney necessarily equal 45 in the top oicture

Mob doesn’t have to be iscoeles

[heady basin]

This is another way of drawing it

<@&286206848099549185>

??

<@&286206848099549185>

[broken yew]

I didn't think it should be as hard as this

Do note the answer should be the same no matter what angle x is (from how the question is given)

So you could cheat and set it to 45

I think in your diagram you have too many letters.

[heady basin]

That would give area of 2

[broken yew]

You probably only needed 2

[heady basin]

Shouldn’t it be +1

[broken yew]

indeed.

So I only need 2 letters to label these sides right?

But then I haven't used pythagoras on the outside triangle

This lets me form an equation

$$(x^2+1)+(y^2+1) = (x+y)^2$$

[flat frigate]

Shuri2060

[broken yew]

$$x + y + 1 = xy$$

[flat frigate]

Shuri2060

[broken yew]

hmmm then I could think about a few things

I could try out similar triangles, for example

[plucky elk]

what's the argument against setting x=y

[broken yew]

none.

[plucky elk]

from symmetry/tangent

[broken yew]

just if you want to prove it.

You don't do that.

[broken yew]

it's not true in general.

The given diagram is just bad.

[heady basin]

Ive tried everything using similarities it doesn’t get u anywhere

[plucky elk]

angle ABC is a right angle and the dotted line is a perpendicular bisector

[broken yew]

https://i.imgur.com/J3yYq7L.png

I'm going to borrow this

c / 1 = 1 / d

[plucky elk]

so triangle ABC is an isosceles triangle, making it a 45-45-90 triangle

[broken yew]

Can you see why? @heady basin

From similarity

[broken yew]

What??? How do you know this

[broken yew]

https://i.imgur.com/oHfYohJ.png

^fits the requirements, isnt isos.

So in my version we have xy = 1

ok?

wait no it isnt

[heady basin]

Yes i saw that

I got cd as 1

And the area as ABCD/2

[broken yew]

xy = 1, in which case ive done an error above, 1 sec

[heady basin]

And since cd is 1 ba/2 is the area

[broken yew]

$$(x^2+1)+(y^2+1) = (x+y)^2$$

[flat frigate]

Shuri2060

[broken yew]

$$2 = 2xy$$

[flat frigate]

Shuri2060

[broken yew]

Oh, I already had that fact from this equation 🤦

$$xy = 1$$

[flat frigate]

Shuri2060

[heady basin]

Yes that’s what i mean

Whatever similarity u use

[heady basin]

You get facts already established

[broken yew]

That's fine, we're not done yet.

[plucky elk]

yup i'm wrong

[broken yew]

There are several ways we can express the area of this triangle

$$A = \frac{x + y}{2} = \frac{\sqrt{(x^2+1)(y^2+1)}}{2}$$

[flat frigate]

Shuri2060

[broken yew]

ok?

I believe, the right version will be most helpful, so I will try that

$$A = \frac{\sqrt{x^2y^2+x^2+y^2+1}}{2}$$

[flat frigate]

Shuri2060

[broken yew]

hmmmmmmmmmm

[heady basin]

It’s so strange this question

[broken yew]

xy = 1

[heady basin]

Nothing like any other bmo2 or Olympiad question

[broken yew]

$$A = \frac{\sqrt{x^2+y^2+2}}{2}$$

[flat frigate]

Shuri2060

[broken yew]

Then I can equate it to the other area equation

$$(x+y)^2=x^2+y^2+2$$

[flat frigate]

Shuri2060

[broken yew]

Which just leads me back to xy = 1

hmmmm

[heady basin]

Yup

See what i mean

[broken yew]

yah sure, but dont wanna give up yet lol

[heady basin]

I tried Heron’s formula to try and get something else

It doesn’t give u anything new either

Then i tried 1/2 ab sin c

Same result

[broken yew]

https://images-ext-1.discordapp.net/external/thUDxX5KeUOPscRpwZHgp8F_Lu6kQMaq3TfqG9J2fok/https/i.imgur.com/J3yYq7L.png

if the angle is 30 degrees do you still get 2

i hope.

[heady basin]

You do

Wait

Maybe not

[broken yew]

https://i.imgur.com/rK7FYV6.png

good point, obviously the area should depend on x

And I think we exhausted all the information given

Therefore, question broken

Likely missing the fact its a perpendicular bisector

oh right and the area goes to infinity

As you let the angle go to 0

🤦

[heady basin]

Yes seems like it

[plucky elk]

give your instructor the solution making the area $\pi$

[flat frigate]

riemann

[plucky elk]

and say, "great problem! i had no idea you could be this creative"

[heady basin]

?

[plucky elk]

Set pi = area = (x^2+1) / (2x) and solve for x

$x = \pi \pm \sqrt{\pi^2 - 1}$ should do it

[flat frigate]

riemann

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[quasi bison]

[4; 0; 0; 0] = 4 * [1; 0; 0; 0]

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@lean otter Has your question been resolved?

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[dry lake]

find t such that (t,5) lies on the line through (0,3) and (-8,0)

[dry lake]

i dont know where to start

[ancient escarp]

a line has a constant slope

you're given two points, find the slope of the line

[dry lake]

is the slope 3/8

[ancient escarp]

yep

[dry lake]

what next

[ancient escarp]

find the equation of the line

[dry lake]

so y=3/8x + 3

[ancient escarp]

looks good

[dry lake]

should i graph in now

[ancient escarp]

you could if you want to

or you could plug in what you know

you know when x=t, y=5

then solve for t

[dry lake]

could you tell me how to slove for t with that

[ancient escarp]

you found the equation of the line correctly

$y=\frac38x+3$

[flat frigate]

a disappointing son

[ancient escarp]

plug in x=t and y=5

since you're solving for when t when the line is at the point (t,5)

[dry lake]

oh

so t=16/3

[ancient escarp]

looks good 👍

[dry lake]

ok tysm

.close

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[dense garnet]

Here Is my process to getting this answer

wait 1 minute while I type it up

Step 1: find y’ y’= 2/(x+1)^2

Step 2: Re write line they gave: y=x/2 -2 slope=1/2

Step 3: y’ = slope so: Plug in y’ = 1/2 —> 1/2=2/(x+1)^2 x=1

Step 4: Plug x=1 into original y function y=(1)-1 / (1)+1 —> y=2/0 = 0

[lean otter]

It wants the parallel line

[dense garnet]

So that means I use slope = 1/2 correct?

so I set y' = 1/2 to find x

Wait, It is asking for equations notice the plural. Am I missing another equation? and How do I find it?

Oh I think I get it. Here is the graph

I found the bottom one at -1/2

Now I need to find the top one.

What is a good process to finding the top one?

Ok I got the answer

But that was from guessing 3.5.

What is a "repeatable" method I can take to solve the top one?

.close

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[lean otter]

Hello

[lean otter]

How to solve 2.6 c?

<@&286206848099549185>

<@&286206848099549185>

[stoic dune]

You should have a formula for projecting vectors

[lean otter]

Yes, but what direction i means here?

Formula is here:

[safe radish]

@lean otter Has your question been resolved?

[vale linden]

occupied?

[lean otter]

Nope. Please continue

.close

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[blazing bloom]

I’m stuck on what I have to do next, I have to simplify

[plucky elk]

use double angle formula

[blazing bloom]

alright one sec

[blazing bloom]

thanks:)

was looking for them

i did that, still no clue on how to continue

oh wait

nvm

no nvm i don’t know

[plucky elk]

you subbed in all 4 terms?

[blazing bloom]

yeah

[plucky elk]

what do you get

[blazing bloom]

lemme write it more clearly

with subbed you mean that i used those formulas to change my previous stuff right?

don’t know english terms

[long copper]

i believe so

like replacing sin(2x) with 2sin(x)cos(x)

[blazing bloom]

yeah

sorry for the bad handwriting

[long copper]

What you've done is almost correct but you're aiming to break them down into 1 trig function. So instead of using cos(2a) = cos^2(a) - sin^2(a) i would use either cos(2a) = 1 - 2sin^2(a) or cos(2a) = 2cos^2(a) - 1 (my advice would be to use the cos(2a) = 1 - 2sin^2(a)). Also remember that you're breaking down -cos2y and -cos2x so the signs will switch

the stuff for the sin(2x) and sin(2y) is fine

[blazing bloom]

alright i’ll try it again

thank you:)

[long copper]

np

if you get to the correct simplification you should notice that a lot of things cancel out

[blazing bloom]

so with the 1-, the cos(2y) would become 1 - 1 + 2sin^2(y)? or what did you mean?

[long copper]

that's what i meant

the 1- cos2x would simplify to 1 - (1 - sin^2(x)) which is 1 - 1 + sin^2(x) as you have stated

= sin^2(x)

[blazing bloom]

yeah i just realised

one min

is that correct?

[long copper]

yeah i think so

yeah it's correct

you can start cancelling some things now

[blazing bloom]

alright is there anything else i can do? to simplify further

alright

[long copper]

gimme a sec just need to check that what i've done is right

[blazing bloom]

alright

thanks btw

[long copper]

just realised in your denominator you've got a y where it should be 2 cos^2(x)

[blazing bloom]

oh did i write it wrong

my bad

pretty tired😅

[long copper]

hang on just got a bit more to do

[blazing bloom]

alrighty

[long copper]

ok so you can cancel some things out, starting with all the 2s since you have 2 2s on the top and 2 2s on the bottom

[blazing bloom]

alright one sec

from here on out i can just scrape the ^2, sin y and cos x right

i’ll be left with sin x / sin y

[long copper]

now since you know that sin^2(y) = (sin(y))^2 you can simplify sin^2(y) on the top to sin(y) and cancel the sin(y) on the bottom. You can do the same for the sin^2(x) and the sin(x)

[blazing bloom]

wait what am i saying

[long copper]

I see what you mean, i just realised you've accidentally written cos^2(x) instead of sin^2(x) on the denominator

[blazing bloom]

oh

going to the toilet, gimme a minute

[long copper]

ok

[blazing bloom]

i’m rewriting everything atm won’t be long

[long copper]

ok

[blazing bloom]

i’m cancelling out rn

[long copper]

You should end up with cos(x)sin(y)/cos(y)sin(x)

[blazing bloom]

yup

[long copper]

now you can simplify further

using the identity that tan(x) = sin(x)/cos(x)

[blazing bloom]

got it

[long copper]

since it's cos(x)/sin(x) it should become cot(x)

because you're basically breaking that fraction down into:
cos(x)/sin(x) * sin(y)/cos(y)

your final answer would then be either:
cot(x) * tan(y) or
tan(y)/tan(x)
both are the same thing so i don't think it really matters but you never know

[blazing bloom]

so cot(x)*tan(y) right?

alright

thank you man:)

[long copper]

np

[blazing bloom]

.close

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[last onyx]

Hi

[last onyx]

How I should know for example that Z = 3i is a cubic root or a square root?

Please can you help me? @lean otter

anyone

[broken yew]

huh?

what kinda question is this

if I asked you 'is 2 a cubic root or a square root?' could you answer???

question don't make sense.

[last onyx]

Look at this example

And this

Do I should know that Z is a square root or it's a cubic root?

or it's from question?

@broken yew

[broken yew]

Read the question.......

Find the cubic roots OF 3i

Find all the numbers that cube to make 3i...

[last onyx]

Ok it's from question not from my own

and here why we should write for k = 0, k = 1, k = 2

Why?@broken yew

[broken yew]

I can't see.

look up how to do it...

[last onyx]

and this you should see

._.

Please

[broken yew]

if you don't understand why we're plugging in those values of k

I don't think you understand what exactly it is you're doing

[last onyx]

You can say like this

[broken yew]

A complex number has argument between 0 to 2pi

What else can I really say.....

Adding any multiple of 2pi to the argument will not change the number

[last onyx]

Ok so without writing for k = ...
for k = ...

True?

[broken yew]

no idea what you mean

honestly, if you're confused try looking up a tutorial

[last onyx]

Why I should write for k = 0, for k = 1

ok what I should know the number of "k" that I should use it?

Ok

Thx

.close

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[clever dust]

helpp

[safe radish]

@clever dust Has your question been resolved?

[safe radish]

@clever dust Has your question been resolved?

[valid grove]

1. is just draw two parallel lines AB and EF and transversal line CD

2. Is draw two intersecting lines and place a point C that doesn’t lie on any of those lines

3. Can be something like this (Sorry for horrible drawing, hope you can do better)

[clever dust]

ohhhhh okayy

[valid grove]

Did you get 1&2?

[clever dust]

no, im still working on it :((

i don't know how to do it

[valid grove]

Do you know term transversal?

[safe radish]

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[muted pond]

im stuck on this question I have to setup a system of equations but im just confused

[valid grove]

I guess ur question is incomplete

[muted pond]

how do i setup a system of equations with the given variables

<@&286206848099549185>

[safe radish]

@muted pond Has your question been resolved?

[safe radish]

@muted pond Has your question been resolved?

[quasi bison]

@muted pond are you still here

[safe radish]

@muted pond Has your question been resolved?

[muted pond]

yeah

[quasi bison]

do you want to ask the question that you actually wrote, i.e. "how do i set up a system of equations with the given variables?"
or do you actually want to just ask "how do i solve this problem?"

[muted pond]

how to set it up I already know how to solve them

[quasi bison]

...

i mean ok like

you haven't introduced any variables, that's the problem.

once you say what you want your variables to represent, then we can talk about setting up equations.

||like, perhaps you would like to have a variable representing the amount of 25%-butterfat cream, and a variable representing the amount of 3.5%-butterfat cream, but you haven't said it yet||

[muted pond]

yeah

but how would i write it down would it be like .25x .035y?

[quasi bison]

no

you want to use the names x and y for them, yes?

[muted pond]

yeah

[quasi bison]

then you would first write:

Let x = amount of 25% butterfat cream, in gallons
Let y = amount of 3.5% butterfat milk, in gallons

exactly like this

[muted pond]

ok i did

[quasi bison]

and now we can set up and write down the two equations

we have two pieces of data:

• the mixture has to have a volume of 50 gallons
• the mixture has to contain 12.5% butterfat

each of these will get translated into one equation

[muted pond]

it says that the 25% is from cream though and 3.5% from milk

[quasi bison]

okay so you can correct the definitions of x and y if you wish

here

[muted pond]

how do i plugin the 12.5

[quasi bison]

you don't.

[muted pond]

maybe the end result has to be 12.5?

total?

[quasi bison]

i would suggest starting by translating the first piece of info into an equation.

[muted pond]

x + y = 50?

[quasi bison]

yes, good.

[muted pond]

im so stupid honestly

[quasi bison]

...do you want to have me continue to help you, or do you want to beat yourself up?

[muted pond]

help

[quasi bison]

okay

so we have taken care of the total volume of the mixture

now we need to take care of the butterfat

calculate how much butterfat there will be in the mixture, in terms of x and y

[muted pond]

28.5

[quasi bison]

twenty-eight and a half gallons?

and why doesn't that depend on x or y?

[muted pond]

would the other equation be x + y = 12.5%?

[quasi bison]

no

...okay, are you willing to set the problem aside for a moment? i want to check if you're missing something by asking you a question that doesn't have to do with this directly

[muted pond]

sure

[quasi bison]

let's say we have those same types of dairy products (25% cream and 3.5% milk)

and we mix 9 gallons of the cream with 1 gallon of the milk

how much butterfat is in the mixture by volume?

[muted pond]

10 gallons?

[quasi bison]

no, that's how much cream there is in total.

i'm asking you how much butterfat the cream contains.

of course, i don't want just the number, but also how you calculated it.

[muted pond]

oh you mean whats 25 percent of 9 gallons

[quasi bison]

no, "25 percent of 9 gallons" will not answer my question either.

[muted pond]

2.25 gallons will be buttermilk?

or cream

[quasi bison]

2.25 gallons is the fat content in 9 gallons of cream, at best.

and again, this does not answer my question.

we don't have 9 gallons of cream, we have 9 gallons of cream mixed with 1 gallon of 3.5% milk

and i am trying to get you to calculate how much fat is in that mixture

[muted pond]

9x + 1y

[quasi bison]

what did i say about setting the problem aside

we don't have any x and y anymore.

[muted pond]

then would it be a total of 28.5% of fat

[quasi bison]

no

no!!!

okay, look, at least you were able to say that 9 gallons of the 25% cream contains 2.25 gallons of fat
how much fat is in 1 gallon of the 3.5% milk?

[muted pond]

.035

gallons

[quasi bison]

0.035 gallons, yes.

you should not omit the leading zero from decimals between 0 and 1 like you just did.

so

i ask you again

we have 9 gallons of cream mixed with 1 gallon of 3.5% milk

how much fat is in this mixture

[muted pond]

2.60

[quasi bison]

no

2.25 + 0.035 is not 2.60

0.035 is not the same as 0.35

[muted pond]

2.285

[quasi bison]

there we go

so like

you understand that to find the amount of fat in a mixture, we would calculate how much fat is contributed by each component of the mix, and then add all that up

yes?

[muted pond]

yeah I understand now

[quasi bison]

this excruciating calculation i took you through had better not been for naught

so

we have a mixture of
x gallons of cream (25% fat)
y gallons of milk (3.5% fat)

how much fat is in this mixture?

in gallons, obviously.

[muted pond]

50

[quasi bison]

?????

read what im asking you again

i am NOT asking you for the total volume of the mixture

i am asking you:

[muted pond]

28.5%

[quasi bison]

NO!!!!!

what the fuck!

you claim to "understand" and then keep stepping on the same fucking rake again

to find the amount of fat in a mixture, we would calculate how much fat is contributed by each component of the mix, and then add all that up

you read this

and it came in one ear and out the other by the looks of it

[muted pond]

no I just dont get the question u just asked

[quasi bison]

what part of "how much fat is in the mixture?" do you not get

[muted pond]

2.285

[quasi bison]

bruh

ok

fine

fine

you know what

lets do this excruciating calculation all over again

we have a mixture of
x gallons of cream (25% fat)
y gallons of milk (3.5% fat)

how much fat is in x gallons of 25% cream?

[muted pond]

well idk how to do it since we dont have the number of gallons

[quasi bison]

we do

it's just not a "number" anymore

it's a variable

but that doesn't stop you - shouldn't stop you - from doing the same things to it that you would do to a number

[muted pond]

.25x?

[quasi bison]

0.25x, yes.

do not omit the leading zero from decimals unless you want to set yourself up for arithmetic fuckups.

now,

how much fat is in y gallons of 3.5% milk?

[muted pond]

.035y

0

[quasi bison]

0.035y, yes.

so how much fat is in the mixture in total?

[muted pond]

how do I figure that out

[quasi bison]

we would calculate how much fat is contributed by each component of the mix, and then add all that up

and then add all that up

[muted pond]

but its x and y

[quasi bison]

is there something stopping you from writing down the two terms and putting a plus sign between them

are you allergic to plus signs perhaps

[muted pond]

0.035y + 0.25x

0.285xy?

[quasi bison]

this, yes.

[quasi bison]

but not this.

[muted pond]

ok

[quasi bison]

consider: just because you see two terms that look vaguely similar and have a plus sign between them,
does not mean this is a signal for you to "combine" them in some way you don't understand and will likely screw up.

[muted pond]

ok

so how do i figure how much fat

there is

[quasi bison]

0.035y + 0.25x is the amount of fat in the mixture in terms of what went into it

in terms of what we want to get out of it,
there should be 50 gallons of cream with 12.5% fat content

therefore the total amount of fat should be what

[muted pond]

it should be equal to 12.5 percent?

[quasi bison]

12.5 percent of what?

[muted pond]

fat

[quasi bison]

.....

[muted pond]

0.035y + 0.25x = 12.5%

[quasi bison]

no

12.5% of 50

12.5% of the volume of the blasted mixture

[muted pond]

6.25 gallons

[quasi bison]

there we fucking go

so now

after so much time wasted

we have

x + y = 50
0.25x + 0.035y = 6.25

[muted pond]

im just not reading what ur writing down correctly or something

[quasi bison]

i don't know how to combat that from my end.

[muted pond]

thanks for helping me

i can see it now

[quasi bison]

don't thank me for what i didn't do

clearly i did not help you in the fucking slightest.

[muted pond]

i just didnt understand questions that you were asking me

i dont think thats how u teach

[hushed mirage]

@quasi bison if youre that stressed maybe take a break from helping? for your own sake. theyre trying their best yknow

[quasi bison]

you are too polite to me

you could've and should've told me off in a much more direct and rude manner

[quasi bison]

.close

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[pulsar condor]

what do the exterior angles sum to?

[final loom]

How about the sum of interior angles of a regular n-gon??

[pulsar condor]

Welp, google or check your notes

Not much can be done if you don't know either

<@&268886789983436800>

[thin wraith]

Thanks

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[placid kelp]

so i was able to construct bijections for all three parts

my question is just in terms of showing that each is a bijection

is is sufficient to find an inverse function for each rather than go through 6 nearly identical proofs of injectivity and surjectivity?

or is there some other way to shorten the proofs and make them less tedious

[safe radish]

@placid kelp Has your question been resolved?

[placid kelp]

<@&286206848099549185>

[safe radish]

@placid kelp Has your question been resolved?

[plucky elk]

as long as the inverse function has the right domain (R) and range (a,b), yea

i forget how to do (c). seems less straightforward

[placid kelp]

so as long as i can show that any real input is defined, and show that it has the range (a,b) thats good?

[placid kelp]

i did the following

$\varphi_3:[0, 1)\to (0, 1)$

$\varphi_3(x)=
\begin{cases}
1-\frac{1}{n+1},&x=1-\frac{1}{n}\text{ for some }n\in\mathbb{N}\
x,&\text{otherwise}
\end{cases}$

[flat frigate]

nix

[placid kelp]

kind of a hilberts hotel sort of situation

[plucky elk]

yea it's tedious but necessary. for example $\arcsin(x)$ doesn't have range $\mathbb{R}$ even though that's the domain of $\sin(x)$

[flat frigate]

riemann

[placid kelp]

true but sinx is only a bijection from [-pi/2,pi/2]->[-1,1]

[plucky elk]

ah yea that's right

[plucky elk]

i guess then you really do need to do all 6 proofs haha

[placid kelp]

riiiiip

[plucky elk]

honestly doing this is almost the entire work for 1-1 and onto

[placid kelp]

fair enough lol

i may ask my prof if he wants a fully rigorous proof of them being bijections or if i can just say "this is obviously a bijection" haha

[plucky elk]

your prof: 🙄 haha

[placid kelp]

alright thanks for the help @plucky elk 🙂
it is much appreciated

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[lean otter]

why wouldn't (a u b) be 0.90?

[broken yew]

Some people have both visacards and mastercards

[lean otter]

so union is strictly 1 or the other?

[broken yew]

no....

draw a venn diagram

[lean otter]

oh so union doesn't include the shared ground

[broken yew]

wtf

?????

look up what the definition of union is

[lean otter]

literally this

which would be .90

all A and all B

[broken yew]

why are you counting the middle twice

[lean otter]

so I cut out the middle?

@lean otter you will count it once for one of them. Basically just take the union and subtract the common area. 0.90 - 0.25 will get you the value you need. Sorry for the other guy being rude about it

[lean otter]

ty

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[subtle pumice]

im trying to solve a question using this

[subtle pumice]

the answer is 48 and I was thinking it could be 12*4 but I dont see how a + would turn into a *

[elfin yew]

Break up the integral into a sum of integrals? @subtle pumice

$\int_{a}^{b} [f(x) + g(x)] dx = \int_{a}^{b} f(x) dx + \int_{a}^{b} g(x) dx$

[flat frigate]

sills

[subtle pumice]

theres no g(x) though

these are the laws I know

[elfin yew]

You need this rule

[elfin yew]

g(x) = 4

\😉

[subtle pumice]

huh could you explain in a bit more simpler terms what to do

[elfin yew]

$\int_{1}^{10} [f(x) + 4] dx = \int_{1}^{10} f(x) dx + \int_{1}^{10} 4 dx$

[subtle pumice]

ohhhh the integral of 4 here is 36

[flat frigate]

sills

[subtle pumice]

for some reason, we did not learn to do the integral of just a number

[elfin yew]

That's weird

I'm so bad at Latex sorry

I'm trying to learn while helping

$\frac{d}{dx}ax + C = a \newline \int a =ax+C$

[flat frigate]

sills

[subtle pumice]

oh my bad I did learn that 2nd equation, I just didnt know I could apply it here

thanks for your help

.close