precalculus

[novel dirge]

and now I am also worried that if I use P instead of whatever I am supposed to use

she will start tallking how I copied my homework from the internet

like she did once before, but luckily, I did it on my own

but she did catch like half my class

and was all smug

sorry for the rant

can you please explain how my notation doesn't work and how the one you wrote is different?

[stable pasture]

Is limits a part of precalculus or calculus?

[steel venture]

@novel dirge reviewing it it looks like mine doesn't actually work

the best way to approach it is to say

$P(A'\cap B'\cap C')'$

[obsidian monolith]

Buncho Maximos:

[steel venture]

precalc

[stable pasture]

sorry

[steel venture]

np

@novel dirge to explain why your notation fails we can look at two variations of it

$(P(A)\cup P(B))\cup P(C)$

[obsidian monolith]

Buncho Maximos:

[steel venture]

and $P(A)\cup (P(B)\cup P(C))$

[obsidian monolith]

Buncho Maximos:

[steel venture]

for the first

$P(A)\cup P(B) = P(A) + P(B) - P(A\cap B)$

[obsidian monolith]

Buncho Maximos:

[novel dirge]

the problem is that Idk what + and - do

[steel venture]

its just adding and subtracting

whoops

$P(A)$ will be a number between 0 and 1
@steel venture

[obsidian monolith]

Buncho Maximos:

[novel dirge]

not sure what that means either...

[steel venture]

just that $P(A)$

[obsidian monolith]

Buncho Maximos:

[novel dirge]

I feel like an idiot now for not understanding this

[steel venture]

will be a number between 0 and 1

don't, it can be confusing

but P(A) is just the probability something happens

[novel dirge]

so P(A) can be 0,6?

[steel venture]

yep

any number between 0 and 1

inclusive

[novel dirge]

ok

I understand that now

but not how it is connected to the other parts

or how I use that

[steel venture]

so now if we look at

$P(A)\cup P(B) = P(A) + P(B) - P(A\cap B)$

[obsidian monolith]

Buncho Maximos:

[steel venture]

• and - just mean add

and subtract

[novel dirge]

...

I never did any operations with groups

what can I google to see how this works?

the + and -

[steel venture]

its the same thing as saying 1 + 1

or .5 + .7

or x + y

and 6 - z

its just adding

and subtracting

[novel dirge]

ok, but how do I do the adding and subtracting?

How do I get this $P(A)\cup P(B) = P(A) + P(B) - P(A\cap B)$
?

[obsidian monolith]

Mr.Pancake:

[steel venture]

before we go forwards

i made a mistake

its not $P(A)\cup P(B) = P(A) + P(B) - P(A\cap B)$

[obsidian monolith]

Buncho Maximos:

[steel venture]

its $P(A\cup B) = P(A) + P(B) - P(A\cap B)$

[obsidian monolith]

Buncho Maximos:

[novel dirge]

ok

[steel venture]

but

it really means what it says

say we have two probabilities

A and B

the probability of A or B is = the probability of A + the probability of B - the probability of both

[novel dirge]

I kinda understand it when you write it and explain it like that

but I couldn't do it myself

that is the biggest problem

[steel venture]

well just remember

P(A) means the probability of A happening

P(AnB) is the probability of both A and B happening

P(AuB) is the probability of A or B happening, not both

[novel dirge]

so if I have A and B and just one of them turns out to be true I can write AuB?

[steel venture]

um no

usually when you work with these variables

A and B will be events

and you won't really get "results"

it mostly about the probabilities of the events

[novel dirge]

Yeah, but I have to somehow write it like that

in that format

[steel venture]

can you show me the question

what language is it

[novel dirge]

Serbian

[steel venture]

hm

can you translate it

[novel dirge]

yep

a. Ostvarila su se sva tri događaja A,B i C
b. Ostvario se bar jedan od događaja A,B,C
c. Ostvarila su se bar dva od događaja A,B,C
d. Od događaja A,B,C, ostvarili su se samo A i B
e. Ostvario se samo jedan od događaja A i B
f. Ostvarila su se dva od događaja A,B,C
g. Nije se ostvario nijedan od događaja A,B,C
h. Ostvarilo se najviše dva od događaja A,B,C.```

this is the problem

[steel venture]

which one specifically

[novel dirge]

Write it like the group expressions( I think that is the closest translation I can think of)

[steel venture]

i put it on google translate no worries

[novel dirge]

and then a to h are the different casses

[steel venture]

but whhich one are you talking about

[novel dirge]

well, all of them

[steel venture]

so if I have A and B and just one of them turns out to be true I can write AuB?
@novel dirge

[novel dirge]

but I would like to be able to figure them out myself

[steel venture]

when you said this

[novel dirge]

d

no

not d

e

e) Only one of the events A and B turned out to be true

[steel venture]

gotcha

so we have A B and C

and either A is true

or B is true

[novel dirge]

yes

[steel venture]

ah

i see

yes you were correct

I thought we were talking about probabilities but i think AuB is what your teacher is looking for

[novel dirge]

yes

I mean, it isn't her that made this, she just found this on the internet

but yes

that is the format I am supposed to write this in

so for e I can write AUB?

[steel venture]

yep

[novel dirge]

Can you help me with the rest?

I will try doing them first though

and you correct me if I am wrong

is that ok with you?

[steel venture]

sure

[novel dirge]

great, thanks

[carmine elbow]

When you're done helping him/her, can you help me with a problem?

[novel dirge]

For b, I am not sure if it can be AUBUC?

if one of 2 is AuB then shouldn't AuBuC be if it is 1 of 3?

[steel venture]

but it doesn't include a and b and c

or a and b

you want at least 1

[novel dirge]

and what does mine say?

[steel venture]

just 1

[novel dirge]

just one of the 3?

[steel venture]

yes

[novel dirge]

but it is supposed to be at least 1

oh, there is an at least

[steel venture]

yes

[novel dirge]

damn

this changes it

for g, where nothing happened, would it be A'nB'nC'?

[steel venture]

yeah

[novel dirge]

I still can't figure out b

can you help me?

[steel venture]

yeah sure

so think of it this way

if we want at least 1

then there can't be a scenario where none happen right?

[novel dirge]

yes

but since it says at least 1 even more can happen

like 2 or even 3

[steel venture]

yep

[novel dirge]

right?

[steel venture]

just not 0

how do we write 0 happening

[novel dirge]

A'nB'nC'?

[steel venture]

good

[novel dirge]

so (A'nB'nC')' ?

[steel venture]

yep

[novel dirge]

😆

for d, where out of A,B and C, only A and B happened, it would be (AnBnC')?

since it doesn't include C

[steel venture]

👍

[novel dirge]

it is?

[steel venture]

yep

[novel dirge]

great

but I have no idea what to do if at least 2 of 3 are true

[steel venture]

which letter is that

[novel dirge]

c

c. Ostvarila su se bar dva od događaja A,B,C

at least 2 events happened

[steel venture]

you could say something along the lines of
(AnBnC')u(AnB'nC)u(A'nBnC)u(AnBnC)

[novel dirge]

so that is either any 2 are true or all are true?

[steel venture]

yep

[novel dirge]

and if only 2 are correct, I just drop the last part?

(AnBnC')u(AnB'nC)u(A'nBnC)

[steel venture]

yep

[novel dirge]

and for the last

were max 2 of 3 are true

that means that 1 is always false?

so can that be A'nB'nC' ?

[steel venture]

that is just the probability that all 3 don't happen

but you have the right idea

max of 2

means that

you can never have how many

[novel dirge]

3

so 1 and 2 are possible

[steel venture]

but 3 isn;t

so how do you write that all three together is not true

[novel dirge]

(AnBnC)' ?

[steel venture]

👍

[novel dirge]

sorry, but is it possible for you to just quickly check this?

just to make sure I didn't mess up somewhere

and thank you for the help and patience

[steel venture]

looks good

[novel dirge]

great

thanks

[hushed sorrel]

im not US

but is precalc including complex numbers?

bc i have a complex number q idk where to post

[uncut mulch]

semi-advanced algebra. use one of the free general questions channels denoted by greek letters

[hushed sorrel]

ok thx xx

[silk sail]

Can someone help me with writing this function?

Suppose the value of Tech Advanced stock (in dollars) is given by g(t)where t represents the number of days after January 1, 2014. Express the values of the following stocks in terms of the function
g

The value of VirtualTech's stock, p(t), is always the same as the value of TechAdvanced's stock two weeks later.

[sour hemlock]

I think your question is incomplete

[silk sail]

Here is the entire question. I just cant figure out C.

[sour hemlock]

p(t) = g(t+14)

[silk sail]

Ok thank you that makes sense, idk why that one was seeming so confusing to me

[novel dirge]

@steel venture sorry for bothering you again

but I just checked again and am a bit confused with the second answer

where at least 1 needs to be positive

(A'nB'nC')'

isn't that the same as just (AnBnC)?

[willow bear]

no

it's $A \cup B \cup C$ actually

[obsidian monolith]

Ann:

[novel dirge]

(A'nB'nC')'
is equal to that or that is the correct solution to my problem?

because at least one has to be true

out of 3

[willow bear]

i'm saying $(A' \cap B' \cap C')' = A \cup B \cup C$

[obsidian monolith]

Ann:

[novel dirge]

oh

but is that correct for my problem?

that at least one of the 3 is true?

[willow bear]

idk

i cbf to scroll all the way up

[novel dirge]

you don't have to scroll up

I have to write a formula using set operations

to express that out of 3 events -A,B and C, at least 1 happened

and the formula is what I got with the help of Buncho

but now when I tried to check it again, by myself I got confused

same for the one where no events are true, where we wrote - A'nB'nC'

can you please also check them, if it isn't a problem?

[novel dirge]

anyone?

[viscid thistle]

I don't understand why don't you use OR ?

[novel dirge]

how should I use it?

[viscid thistle]

You have 3 events right?

[novel dirge]

yes

[viscid thistle]

And you want that one or two or three events happened?

[novel dirge]

in one problem, at least 1 of the 3 happened

but can be more

and in the second problem, 0 happened

[viscid thistle]

And what do you have to write?

[novel dirge]

I need to write a formula that does this, but with set operations

[viscid thistle]

The U is a sort of "OR" for the sets
In French we call it "union" I am searching the English word

[novel dirge]

I think it is also Union

[viscid thistle]

What do you call a set operations, U isn't one of them?

[novel dirge]

in my language we call it something similar

it is

https://www.cs.sfu.ca/~ggbaker/zju/math/set-oper.html

in one problem, either A or B is true

so the solution is to write AUB

Idk if I am overthinking about the 2 problems I asked here about, but the more I think about them, the less I understand them

[viscid thistle]

So why don't use it in your problem?

AUBUC

[novel dirge]

and would that be the correct solution?

so that at least one is true?

[viscid thistle]

Yes why wouldn't it be?

Yes

[novel dirge]

and what about the second one?

where none are true

is that one correct?

[viscid thistle]

Repost it I didn't see it

[novel dirge]

A'nB'nC'

none of the 3 are true

[viscid thistle]

Yes

Or it can be '(A U B U C)

[novel dirge]

and if at least 2 are true (AnBnC)' ?

[viscid thistle]

Ok ok wait

[novel dirge]

ok, sorry

[viscid thistle]

To make the think easier

Try to think like if U was "or", n was "and"

' was "not"

Now try to retrieve the solution you find with this method

[novel dirge]

The thing is, I don't have much problems solving if the entire formula is true or not

I have problems writing it

which doesn't really make sense

[viscid thistle]

The problem is "at least two event are true"

Right '

?

[novel dirge]

yes

[viscid thistle]

I have a formula but it is quite long

Try to find a formula with the method I told you

[novel dirge]

is the one I wrote here wrong?

[viscid thistle]

Repost it

[novel dirge]

no

wait

I did that one

I asked about the one with max 2

so out of 3 events max is 2 correct

[viscid thistle]

What?

There can't be 3 events correct?

[novel dirge]

no

2 is the highest number of correct events

I wrote it incorrectly the last time

[viscid thistle]

Okay

So

With and, or, and not

[novel dirge]

and the formula for that is (AnBnC)'?

[viscid thistle]

A' U B' U C'

Yes

It works

[novel dirge]

the one I sent?

[viscid thistle]

They are the same

[novel dirge]

great

thank you

[viscid thistle]

Np

[slim finch]

@toxic nova can you help me with question again pls 🙂

https://cdn.discordapp.com/attachments/363224154469826562/695844882635423744/Screenshot_67.jpg

[patent beacon]

What's the question?

[rich light]

Are you able to help w/ this? @patent beacon

I can’t factor any more from there correct? So I head right to sign analysis + interval notation?

[patent beacon]

@rich light
Okay, so you want to know when
(2x + 1)(x - 3) is positive

That happens when they're both positive, or when they're both negative

[rich light]

why do I want to know why it's positive in that case vs negative in the last example we talked about?

like when you helped me with my other question

[twilit shadow]

Can someone please tell me what I did wrong?

[patent beacon]

If ab > 0
Then a and b have the same sign.

If ab < 0
Then a and b have opposite signs

@twilit shadow
Check your multiplication on (x + 4y)(x + 4y)

[twilit shadow]

Thanks

I see what I did

[tardy ridge]

use pascals triangle

[rich light]

This look right?

@patent beacon

[patent beacon]

If 2x + 1 ≤ 0 and x - 3 ≤ 0,
Then x ≤ -1/2 and x ≤ 3
Which is satisfied by x ≤ -1/2

But the logic is correct, x ≥ 3 and x ≤ -1/2 are the solutions to this problem

[obsidian monolith]

Display results online and refine query

Requested by Kaynex#7520

[patent beacon]

You can see the solution where the parabola goes above the x-axis

[rich light]

okay yeah, I see what I did wrong

thanks

in terms of interval notation, would it be (3, ♾️ ) (-1/2, ♾️ ) @patent beacon ?

[patent beacon]

(-inf, -1/2] U [3, inf)

The U means "union", which is just a way to combine the sets into one.

[rich light]

okay, why did you get -inf for the first?

[toxic nova]

@slim finch yes

[vapid torrent]

A vase has 10 black balls in it and 10 red balls, a second vase has 10 red and 20 black. If a coin is tossed, and its heads, 2 balls are taken out of the first vase, if its tails, 2 balls are taken out of the second vase

How can I create a probability space for this?

[karmic wave]

@rich light the negative infinity comes from the less than part

[rich light]

gotcha

[harsh cipher]

Hello

[uneven dawn]

Could anyone hop in voice chat for a few minutes and walk me through a radical function problem?

Like this one ^

[harsh cipher]

lose root and solve

[uneven dawn]

just square both sides?

would the right then just be 5x + 10?

Its also asking for the sum of solutions so how could it just be one answer?

I think i got it actually, I need to do sqrt(5x-3) * sqrt(5x-3) * -1^2

?

I struggle to even find a video solving a problem like this one

[daring yarrow]

(a+b)² = a²+2ab+b²

@uneven dawn

also

https://www.youtube.com/watch?v=d24-xDTuVX4 @uneven dawn

very similar problem

[viscid thistle]

https://cdn.discordapp.com/attachments/446817824338608139/697265386252402718/unknown.png How are you supposed to do ii of this question? Just had this on a test I completed around an hour ago but ii was the only question I couldn't answer 😦

[daring yarrow]

is M the middlepoint of the circle

[viscid thistle]

It's a parabola and yes, M is the midpoint

Meaning M should be where I've plotted the red dot

[uncut mulch]

consider: (p+q)^2

[viscid thistle]

Is it something to do with the p^2 + q^2 / 2 bit

[uncut mulch]

(p+q)^2 expands to p^2 + q^2 + 2pq which is related to that and the given relatationship p + q = pq

[viscid thistle]

p^2 + q^2 + 2(p+q) ?

[uncut mulch]

yes

[viscid thistle]

Is that supposed to be on the y coordinate tho

Oh wait never mind

I think I got it

y = x^2 / 2 ?

[uncut mulch]

you're missing something

[viscid thistle]

Crap

Was i close

What do I have to do next after that expansion then

[uncut mulch]

(p+q)^2 = p^2 + q^2 + 2(p+q)
x^2 = p^2 + q^2 + 2x
what is p^2 + q^2 in terms of y?

you ignored the part involving 2(p+q)

[viscid thistle]

In terms of y, wouldn't 2y = p^2 + q^2

[uncut mulch]

yes

[viscid thistle]

or the other way around

Hmm alright I'll try this out now, and I'll see if I can get the answer

I think I got it

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

[uncut mulch]

are you sure its +?

[viscid thistle]

Oh my bad minus

But is that right tho?

[uncut mulch]

if its a minus, and not a plus. yeh

[viscid thistle]

Fuck how did I not get that thanks for the help

[uncut mulch]

if your comfortable with the relation between (p+q)2, p^2 + q^2 and pq, its easier to work from y.

$y = \frac{p^2+q^2}{2} = \frac{ (p+q)^2 - 2pq}{2}$

[obsidian monolith]

ramonov:

[viscid thistle]

I recognised the fact that there was a p + q and a p^2 + q^2 and I did it halfway through but then I stopped because I thought i was overthinking it

Grrr

There goes 3 marks

[uncut mulch]

if you did some sort of squaring of (p+q) you'd likely still get 1

[viscid thistle]

Unfortunately not

Bad habit of overthinking things

[vernal spindle]

How can I find sin(195) without just typing it in a calculator

[past meadow]

sin is taken based on the angle from the x axis, and its pi/6 from the x axis

but its negative in that quadrant

so -sin(pi/6)

pi/6 being 15 degrees obviously

wait im wrong

pi/12

not pi/6

but same concept

[vernal spindle]

erm

what do you mean from the x axis

[past meadow]

like in terms of the unit circle centreed on the origin

[vernal spindle]

isn't sin the y axis since its (cos,sin) ?

[past meadow]

the angle is measured from the x-axis

is what i meant

the actual value is the y coordinate

idk im sorry if that was poorly explained

[lilac pier]

@vernal spindle What sneaky means is that sin is negative in the third and fourth quadrants. We have the angle 195 degrees atm so it's in the third quarant and it's negative. Now we always go for the shortest possible angle from the x-axis.

[past meadow]

Thanks for clarifying, it was probably a bad explanation

[vernal spindle]

Ok I get that now, so we know sin is negative

[fleet yew]

P(no spades) = 1 - P(Spades)

[vernal spindle]

1-Sin(-195)?

[lilac pier]

No no

[fleet yew]

What

[vernal spindle]

I don't know

[fleet yew]

Seems simple

Just trying to find sin(-195°)

Right?

Or is there something im missing

[uncut mulch]

no minus

[lilac pier]

you can try writing it as sin(180 + 15)

[vernal spindle]

can i then break it to sin(180)+sin(15)

[lilac pier]

Uh no

looks like you hvaen't been taught the addition formula

[vernal spindle]

ohh

sin(x)cos(y)+cos(x)sin(y)

[lilac pier]

this will get you -sin15

[vernal spindle]

how'd you get that

[lilac pier]

use the formula

[vernal spindle]

This is a basic question but whats the easiest way to calculate sin15

[fleet yew]

Exactly?

[vernal spindle]

yeah

[fleet yew]

Use half angle identities

[vernal spindle]

Why a half angle identity?

[fleet yew]

Sin(15) = sin(30/2)

What

[jagged inlet]

wrong chat

[vernal spindle]

So far I have

-1(sqrt((1-cosx)/2))

[fleet yew]

Yep

[willow bear]

uh

what's that -1 doing there

[vernal spindle]

cos(180)

[willow bear]

okay my bad

you were going for sin(195°) originally

i didn't scroll up

okay alright but

you don't have x, you have 30

presumably

since sin(15°) is what you want

so yknow

$\sin(15\dg) = \sqrt{\frac{1 - \cos(30\dg)}{2}}$

[obsidian monolith]

Ann:

[willow bear]

this is what you would have at this point

if you chose to use the half angle identity for sine that is

which is not the only way to go for 15° in particular mind you

there's one that may be a little bit easier in the sense of not requiring as much fucking around with roots within roots

and to reiterate kaynex's suggestion, it'd be rewriting sin(15°) as sin(45° - 30°)

and then using the angle difference identity

[vernal spindle]

i think thats easier i lost myself within all the roots

I got sqrt(6)-sqrt(2) over 4

[willow bear]

(sqrt(6) - sqrt(2))/4

yes

[vernal spindle]

negative^^

[willow bear]

that is correct

and now to get back to the original problem of sin(195°) you'll multiply that by -1

[vernal spindle]

wow i actually understand this that all helped a lot

[vernal spindle]

I'm trying to solve this problem now

What exactly am I suppose to try to find

[fleet yew]

@vernal spindle what do you think you're supposed to find

[vernal spindle]

x?

So I need to get x alone

[fleet yew]

Yeah

Here is a secret hack

You can write cos^2(x) as 1-sin^2(x)

[vernal spindle]

3(1-sinx=2(1-sin^2(x))

[fleet yew]

Use the distributive property

[uncut mulch]

theres something more efficient

[vernal spindle]

Divide?

[uncut mulch]

no

consider rearranging the equation to the form:
stuff = 0
and factorising (1-sin^2(x))

[vernal spindle]

oh subtracting?

[uncut mulch]

is part of it

[vernal spindle]

3(1-sinx)-2(1-sin^2(x))=0

[tight ravine]

q11

i have no idea how to do

[uncut mulch]

in use

[vernal spindle]

How could I factor 1-sin^2(x)?

[uncut mulch]

1-sin^2(x) = 1**^2** - sin**^2**(x)

[vernal spindle]

(1-sinx)^2 ?

[uncut mulch]

no

a^2 - b^2 = ...?

[vernal spindle]

(1+sinx)(1-sinx)

[uncut mulch]

yes

$3(1-\sin(x)) - 2(1-\sin(x))(1+\sin(x)) = 0$

[obsidian monolith]

ramonov:

[uncut mulch]

can you factorise that further?

[vernal spindle]

would the 1-sin(x)'s be combined to (1-sin(x))^2

[uncut mulch]

wdym?

[vernal spindle]

wait i think thats wrong

erm

i dont see anything else

[uncut mulch]

do both terms have a common factor?

[vernal spindle]

sinx?

[uncut mulch]

no

[vernal spindle]

I don't think they do then

[uncut mulch]

(1- sin(x)) perhaps?

[vernal spindle]

I'd take that all the way out?

(1-sinx)(3-2(1+sinx))

[uncut mulch]

= 0

[fleet yew]

Got your signs mixed up

(1+sinx)(3-2(1-sinx))

[uncut mulch]

no the signs were fine

[lethal warren]

@tight ravine the points where the tangents intersect the graph are just the points where the gradient of the graph is -2/25

you just need to find dy/dx and make it equal to -2/25

[fleet yew]

Oh nvm

Didnt see those parantheses

[uncut mulch]

$3(1-\sin(x)) - 2(1-\sin(x))(1+\sin(x)) = 0 \
(1-\sin(x))( 3 - 2(1+\sin(x)) = 0$

[obsidian monolith]

ramonov:

[uncut mulch]

can you continue from there?

[vernal spindle]

i dont see where i can go next

[uncut mulch]

zero product property

something i'm pretty sure you've applied countless times already

[vernal spindle]

oh

(1-sinx)=0 (3-2(1+sinx))=0

pi/2 and pi/6 ?

[uncut mulch]

what interval are you supposed to solve for?

[vernal spindle]

sin?

how would i know?

[uncut mulch]

they'd tell you

if they don't you'd need to provide the general solution

[vernal spindle]

would pi/2 and pi/6 be the general solution?

It just says solve the equation

[uncut mulch]

those would only be 2 of the solutions

[vernal spindle]

I'd have to keep adding?

[uncut mulch]

there is also more than 1 solution to
sin(x) = 1/2 in [0,2pi)

look up general solutions for trig and how to write them

[vernal spindle]

I I think she only wants the solutions between 0 - 2pi because we solved 2sin^2(x)-3sinx-1=0 and she said the solutions were pi/2 and 3pi/2

pi/2, pi/6, 5pi/6

[uncut mulch]

2sin^2(x)-3sinx-1=0 are you sure that was the correct equation

[vernal spindle]

+1

[uncut mulch]

x = 3pi/2 isn't a solution to 2sin^2(x) - 3 sin(x) + 1 = 0

[vernal spindle]

I don't know then, she had us rewrite it as 2t^2-3t+1=0 to make it look easier too

[uncut mulch]

depends on your preference

if you have no issue viewing it as a quadratic in sin(x) then sub isn't needed

[vernal spindle]

Let's see if i understand this, for 1+sinx=2cos^2(x) I got 3pi/2, 7pi/6, and 11pi/6

[uncut mulch]

,w solve 1+sinx=2cos^2(x) for 0<=x<=2pi

[obsidian monolith]

Display results online and refine query

Requested by ramonov#0921

[vernal spindle]

😦

[uncut mulch]

feels like you weren't careful with signs

is that supposed to be
1 + or 1 - sin(x)?

[vernal spindle]

I put 1+sin(x) instead 1-sinx when i factored it out

[uncut mulch]

either way, there's some issue with signs,
show your work

[vernal spindle]

I got it now pi/6 and 5pi/6

I had (1+sinx)(1-2(1+sinx)) after i factored it out instead of (1+sinx)(1-2(1-sinx)

[uncut mulch]

missing a ) at the very end but ok

[vernal spindle]

I'm working on my formula sheet but I can't figure out how you get the vertex for intercept form

[uncut mulch]

axis of symmetry/ x-coord of the vertex will be the average of the zeros

[harsh cipher]

good afternoon

ok....

range y=f(x) = 0<=y<=4

range of y= rootf(x) + 2

[willow bear]

P

A

R

E

N

T

H

E

S

E

[harsh cipher]

hahah

Hello Ann

[willow bear]

S

did you mean $\sqrt{f(x)+2}$ or $\sqrt{f(x)} + 2$

[harsh cipher]

where did I do it wrong?

[obsidian monolith]

Ann:

[harsh cipher]

second

[willow bear]

these are sqrt(f(x) + 2) and sqrt(f(x)) + 2 respectively

[harsh cipher]

I will take a note, write the closing sign after x

then +2

Thank you!

y=sqrt(f(x))+ 2

anyways

the reason why y>=2 is we sub in 0 for f(x) ?

so 2<=y<=4

[willow bear]

0 ≤ f(x) ≤ 4 implies 0 ≤ sqrt(f(x)) ≤ 2 implies 2 ≤ sqrt(f(x))+2 ≤ 4

[harsh cipher]

I did not get that...

could you explain it easier please?

[willow bear]

this is as elementary as i could possibly go

so give it another dozen rereads

[harsh cipher]

sure

will do

[rare zephyr]

No. 3

Ive expanded the expression and substitute tan with sin/cos then substitute them back into my equation but uhhhhhh now Im stuck

I look back at the question just to have myself more confused on what to do

[uncut mulch]

draw some triangles

[rare zephyr]

Right angle triangles?

[uncut mulch]

apply the given information

[rare zephyr]

Owhhhh

Got it

I know what to do know thx

[uncut mulch]

note: make sure to watch the signs and quadrants

[brittle fable]

when taking the composition ( f  ∘ g ) ( x ) and we call the new function h(x)

is h(x) effectively the same as f(x) if we change the domain of f to be the same as the domain of g?

[willow bear]

no

h(x) is f(g(x))

[obsidian monolith]

khaled014z:

[viscid thistle]

what

1. your question doesn’t make sense

2. this isn’t the right channel

3. don’t double post

[cedar cosmos]

Becuz the dg cancels out

[vernal spindle]

I'm solving cos(2t)=sin(t) I turned it to 1-2sin^2(t)=sin(t) then subtracted sin(t) getting 1-2sin^2(t)-sin(t)=0 but now im stuck

[tawny nacelle]

its a quadratic in sin(t)

[pale bison]

personally that's too much work

maybe use\$\sin(\theta)=\cos(\theta-\frac\theta2)$

[obsidian monolith]

Publius:

[willow bear]

pub

big wrong

[vernal spindle]

I was seeing if I could get a, b, c but i dont think i can because the 1- infront

[willow bear]

huh?

[vernal spindle]

I think im trying to apply the wrong method

So far I have 1-2sin^2(t)-sin(t)=0

[willow bear]

oh that. sorry, didn't see your message earlier

what if i were to tell you that this is the same as (-2)sin^2(t) + (-1)sin(t) + 1 = 0

and that this is indeed a quadratic in sin(t)

[vernal spindle]

Would it be

-2sin^2(t)-2sin(t)+sin(t)+1

[willow bear]

where'd the =0 go

anyway if you wanna solve this by factorization go ahead ig

[vernal spindle]

-2sin^2(t)-2sin(t)+sin(t)+1=0

I think i made it worst, i have -2(sin^2(t)+sin(t))+(sin(t)+1)

[novel dirge]

yesterday I got help here with writing set operation expressions for probability

but when I sent it to my teacher, she said that 2 are wrong

and that I should use + instead of U and that I shouldn't use n at all

e. Of the events A and B, only one happened
f. Of the events A,B and C, 2 happened

for e I wrote (AUB)

and for f

(AnBnC')U(AnB'nC)U(A'nBnC)

but it looks like they were somehow wrong

she didn't want to tell me what was wrong but told a friend of mine who copied everything from me

[vernal spindle]

I solved the -2sin^2(t)-2sin(t)+sin(t)+1=0 using the quadratic equation and got -1 and 1/2

[willow bear]

you got sin(t) = -1 and sin(t) = 1/2.

[vernal spindle]

How do I know it's sin(t)= thats where i got lost

[willow bear]

what else could it be? what were you solving for?

you can make a substitution, u := sin(t)

then your equation will become -2u^2 - 2u + u + 1 = 0

and you'll get two sols: u = -1, u = 1/2

does that make sense to you

[vernal spindle]

ohh i see it now

so 3pi/2, pi/6, and 5pi/6

[willow bear]

👏

[vernal spindle]

What can i do next in cos(2x)cos(x)+sin(2x)sin(x)-1=0

[heady jewel]

whats cos(A+B)

try relating it to your q

[vernal spindle]

cosxcosy-sinxsiny

[heady jewel]

whats cos(2x+(-x)) now

[vernal spindle]

Why not cos(x-y)?

[heady jewel]

lmao cos(x-y)=cos(x+(-y))

but you can be stubborn and directly use cos(x-y)

go on

[vernal spindle]

cos(2x-x)-1=0

[heady jewel]

👍

[vernal spindle]

What do I do with that

[heady jewel]

can you solve cos(x)=1?

[vernal spindle]

How did you get that

[heady jewel]

whats x+x

[vernal spindle]

2x?

[heady jewel]

whats x+x-x

[vernal spindle]

x

[heady jewel]

=2x-x

[vernal spindle]

Ohh I see it now

cos(x)=1

so 0?

[heady jewel]

do you know that cosine of any multiple of 2pi=1

better get a general form

so like x=2pi*k where k \in the integers

would be the general form

[cosmic cargo]

Hello! This is my first time in this server is anybody able to help me with a trigonometry problem 😌

[heady jewel]

[cosmic cargo]

How do I find the intersection points of polar equations r = 2 and r = 4sin2theta?

[vernal spindle]

I can’t find my mistake

[lilac pier]

4th line

He did

[tardy ridge]

I thought he was using a trig identity

that's just the question

whoops

yeah fourth line you have to subtract

you can't multiply

[lilac pier]

@vernal spindle The mistake is that after the 3rd line, you should've added sinx - 1 on both sides, then on the RHS you have 0.

after that you can factorize

[vernal spindle]

ohh

I need to get rid of the subtraction and get multiplication so i can find the zeros right?

[lilac pier]

If you mean factorize, then yeah factorizing is better

After adding sinx - 1 on both sides, you get

$[1 + \sin(x)][1 - \sin(x)] + \sin(x) - 1 = 0$

[obsidian monolith]

Sup?:

[vernal spindle]

then, 1sin(x)((1+sin(x))-1)

[lilac pier]

What can you do now

[vernal spindle]

?

if i take out sin(x)

[lilac pier]

From where?

No need to factor out sin x

Just use this

sinx - 1 = -(1 - sinx)

Now you have -(1-sinx) in both the terms, so factor this out

$[1 + \sin(x)][1 - \sin(x)] - [1 - \sin(x)] = 0$

[obsidian monolith]

Sup?:

[vernal spindle]

factor out (1-sin(x))?

[lilac pier]

Yes

[vernal spindle]

Would it be (1-sinx)((1+sinx)-1)

[lilac pier]

Yes = 0

Now either (1-sinx) = 0, (1+sinx) - 1 =0

or both

so check it out

[vernal spindle]

for 1-sinx=0 i got pi/2

[lilac pier]

Okay.

For the other one?

[vernal spindle]

0, pi, 2pi

[lilac pier]

Or you can write npi, where n is any integer.

For the first one, you can write pi/2 + 2npi, where n is any integer.

[vernal spindle]

I keep understanding the problem but when i see the next problem i don't get it again :/

like now for cos(x)sin(x)=cos(x), i took -cos(x) and got cos(x)sin(x)-cos(x)=0 and im stuck

[patent spruce]

whats a pi?

[hollow oar]

id oint knw o

i need help with 7th grade math

[patent spruce]

can yo please hepl explen

[hollow oar]

is this for only colege kids i need help my teacher called my mom

[patent spruce]

i got a lunch detenction at hom

[hollow oar]

that stink

[patent spruce]

and now i hove bruses on my behynd

[hollow oar]

what category is 7tyh grade math in

[lilac pier]

@vernal spindle Factor out cosx

@hollow oar Try the sections pre-university

[vernal spindle]

cosx(sinx-1)=0?

[hollow oar]

im 7th grade

[lilac pier]

Yes

[vernal spindle]

pi/2 and 3pi/2

[lilac pier]

both should be pi/2

Now for sinx = 1, you can write x = pi/2 + 2npi, where n is any integer.

Try to think of what you can write for cosx=0

[vernal spindle]

I think i just need to find the values between 0-2pi

[lilac pier]

Oh okay

[vernal spindle]

For cosx-5sin(2x)=0, first i applied cosx-5(2sinxcosx)=0

[lilac pier]

Good.

You got the hang of it

[vernal spindle]

I'm kind of stuck here

[lilac pier]

Factor out cosx

[vernal spindle]

cosx(1-(5(2sinx))) loke that?

no

erm

[lilac pier]

$\cos(x)[1 - 10\sin(x)] = 0$

[obsidian monolith]

Sup?:

[vernal spindle]

I got cosx=0 and sinx=1/10

[lilac pier]

yeah

[vernal spindle]

I know cosx=0 is pi/2 and 3pi/2

but sinx=1/10 i dont know

[lilac pier]

@vernal spindle Sorry was gone

[acoustic harbor]

the arcsin of (1/10)

[lilac pier]

Well it's just arcsin(1/10) like nelson said and if you look at the graph of the sin function, that's the only solution in the range 0-2pi

[acoustic harbor]

doesn't -arcsin(1/10) work

[vernal spindle]

Is that the same as sin^-1(1/10

[acoustic harbor]

yes

[vernal spindle]

So the answer is pi/2, 3pi/2, and sin^-1(1/10)

[vernal spindle]

For sin^2(x)+sin(x)-2=0 I got sinx=-2 and sinx=1

What do I do with sinx=-2

[dusk elm]

How can you isolate x?

[vernal spindle]

take sin^-1

[dusk elm]

applying the inverse operation seems like a good idea

[vernal spindle]

so x=sin^-1(-2) is it?

I know sinx=1 is pi/2

[dusk elm]

I think you may have done the problem wrong because sin will always take a value between -1 and 1

[tardy ridge]

What

What is the problem

[vernal spindle]

To solve this

[tardy ridge]

well then it means it's not a solution

[dusk elm]

Took me a second but solving you get sin(x)=-2 so there is not solution for that one

so you only solution is pi/2 for [0,2pi]

[vernal spindle]

ah makes sense

[vernal spindle]

I'm struggling with this problem now

[viscid thistle]

if it has that shape

that means we can achieve our goal through simple translations

so the vertex is currently at (0,0) for 2x^2

[vernal spindle]

yea

[viscid thistle]

and it also contains the point (1,2)

no, 2x^2 contains (1,2)

[past meadow]

yeah i realised what you meant lmao

[viscid thistle]

we want it to contain (1,1)

(1,2) is 1 unit above (1,1)

so what do we need to do?

oh wait

scratch that

the final equation will come in the form 2(x-h)^2 + k

where (h,k) is the vertex

if (h,k) is the vertex, what is h?

@vernal spindle

[vernal spindle]

one of them is 0 right

since its on the y axis

[viscid thistle]

h is 0

so the final equation will be in the form 2x^2 + k

the point (1,1) is on this equation

this means that 1 = 2(1)^2 + k

what is k?

[vernal spindle]

Do I plug in the other values and solve for k

[viscid thistle]

yes

i just did that

[vernal spindle]

how did you get the 1 for y

[viscid thistle]

it says that (1,1) is on the equation

[vernal spindle]

ohh

k=-1

[viscid thistle]

yes

[vernal spindle]

y=2(x-0)^2-1

and the equation for this graph would be y=1/2(x+2)^2-3 right

[viscid thistle]

yeah

[full garden]

hey guys very silly question but can someone please explain how i can get that 165 point? like I was able to get the rest of them but for some reason couldn't figure out the 165

[patent beacon]

It's half way between 120 and 210

So their average

[full garden]

omg i actually love u wth i was watching this tutorial and the guy was like we have to add 90 to the period and stuff omg thank u so much that helps alot

[patent beacon]

120 → 165 is a quarter period. So +45

[full garden]

omg yes i think that's what he was mentioning ❤️ thank you so much love u bro thank you so much really

[vernal spindle]

I'm trying to solve this problem again for practice but i cant figure out how to get sin and cos for the first one, so far i turned sin(2x) to 2sin(x)cos(x)

[lilac pier]

you have tan theta, you can make a right angled triangle with the given sides, find the unknown side

find sin and cos theta

[vernal spindle]

3^2+4^2=x^2 ?

[lilac pier]

Yeah

[vernal spindle]

ah i got it

but what about the negative

[lilac pier]

in the second quarant sin is positive, so keep it as it is

only cos will be negative

[analog relic]

$\tan{\theta}=-\frac{3}{4} \implies \frac{opp}{adj}=\frac{3}{4} \implies $ By Pythagoras' Theorem, $hyp=\sqrt{opp^{2}+adj^{2}}=\sqrt{3^2+4^2}=5 \implies \frac{opp}{hyp}=\frac{3}{5}$ and $\frac{adj}{hyp}=-\frac{4}{5}$ (because it is in the second quadrant) $\sin{2\theta}=2\sin{\theta}\cos{\theta}=2(\frac{3}{5})(-\frac{4}{5})=-\frac{24}{25}$

[obsidian monolith]

Lσνιηg✧Sσνєяєιgη:

[viscid thistle]

Why tan is negative but opp and adj are both positives

[steel venture]

its unit circle

[uncut mulch]

dodgy explanation

[willow bear]

i was about to say just that

i could repeat my suggestion of doing it purely algebraically

using tan^2(x) + 1 = sec^2(x) to get sec^2(x), then from that get cos^2(x), then cos(x), then sin(x), then everything else

[fleet yew]

@viscid thistle he's just talking about the sides of the triangle

lengths are positive numbers

[viscid thistle]

Ah yes I'm stupid

[viscid thistle]

But if tan is the quotient of two lengths how can it be negative

[fleet yew]

by convention. we assign negative values to tangent in the second and fourth quadrant

[willow bear]

uh no

tan(x) is just sin(x)/cos(x)

don't call it a convention when it isn't

[fleet yew]

i mean i was thinking more along the lines of that up and right being positive is a convention in the cartesian coordinate system

[willow bear]

nothing to do with tan

[viscid thistle]

When we use tan with radians I'm ok that it can be negative

But when we work with triangles

If tan is a quotient between two lengths how can it be negative? Or how can cos or sin be negatives?

[willow bear]

if you're working with triangles then your angles are constrained to (0, pi/2) and your trig functions are all positive

[viscid thistle]

Oh ok

[proper pulsar]

This is hard lol

[onyx pagoda]

I have a question if someone can dn me that would be great

[lilac pier]

[willow bear]

i dm'd them about it.

[south forum]

how do i solve #1?

[lost mesa]

do you know what a composition of functions is

[south forum]

no i just started doing this

im so confused

[lost mesa]

ok so

[willow bear]

http://google.com/search?q=composition+of+functions

[lost mesa]

$$(f \circ g)(x) = f(g(x))$$

[obsidian monolith]

andrwxlin:

[south forum]

o ok

so.

do i just

plug in the x

with 4

and multiply

f x g

[lost mesa]

no

theres a difference between the closed circle and opened circle

the closed circle means multiply (i can't remember the latex code for it)

the open circle means composition

[willow bear]

no, you don't multiply anything

[lost mesa]

you take the result of g(x) as your input for the function f(x)

[willow bear]

to calculate g(f(4)), the first step is to calculate f(4).

[south forum]

omg thats not a multiplication sign im so dumb

[lost mesa]

but in your problem it's g(f(4))

^

[south forum]

ohh ok

[tropic hare]

Mhm.

[south forum]

omg im crying

ok so like

what do i do to start it then

[willow bear]

did i not say exactly what to do?

[lost mesa]

$$(g \circ f)(x) = g(f(x))$$

[obsidian monolith]

andrwxlin:

[lost mesa]

you're given f(x) and g(x) and x=4

[south forum]

yes

[lost mesa]

to calculate g(f(4)), the first step is to calculate f(4).

[willow bear]

on the other hand, some people choose not to listen to me. or they choose not to tell me when something i've said confuses them, instead opting to remain silent.

bit of a shame, really.

[lost mesa]

you take the result of f(4)

and plug that into function g

[south forum]

ohh ok that makes more sense thx, and srry @willow bear didnt see but
thanks both of you for the help

[tardy ridge]

here is a nice challenge question related to your question @south forum

[south forum]

thx @tardy ridge

[carmine elbow]

I need help with a half-life problem

[vague minnow]

Hey, I have a question on how to solve this equation. I think I've made some progress, I got a simpler looking equation, but can't figure out what to do next

[tardy ridge]

@vague minnow I think i got it, do you have the answer so I can verify

[vague minnow]

no I don't unfortunately

oh wait

actually I do

yeah I do

what did you get

[tardy ridge]

well I haven't checked it over

one of them is 1 and -3

then I have a quadratic formula thing

which I haven't simplified yet

[vague minnow]

hm

the teacher didn't give the answer

but wolframalpha says the answer is 1/2

[tardy ridge]

hmmm ok I will try it again

weird

I have -3 and 1

[vague minnow]

hm, -3 and 1 shouldn't fit because the log_x thing on the left has x as base, so x should be positive and not equal to 1 🤔

[tardy ridge]

you are right

but I still have 1

do you want to see what I did

[vague minnow]

sure

[vague minnow]

hmmm

it all makes sense

[tardy ridge]

well no x=1 means log 1 = 0

so it's undefined

but it looks like I can't get the 1/2 solution

[vague minnow]

oh I think I got it

[tardy ridge]

what did you do

[vague minnow]

I think you made a small mistake

it should be log(3-2x) = -log(x)

[tardy ridge]

OHHHHHHHH

ah well that's good

[vague minnow]

thanks for your help dude, very much appreciated 🙂

[tardy ridge]

Glad it worked

[carmine elbow]

Can you help me with my problem?@tardy ridge

[tardy ridge]

what is your problem

[carmine elbow]

I’ll show you

[tardy ridge]

I dunno if I can solve it

ok

[carmine elbow]

I’m doing number 14 and I’m trying to find the decay constant k

[tardy ridge]

I have not learned this

[carmine elbow]

Oh

Do you know anyone on this server that knows half life?

[vague minnow]

oh half-life

I'm not familiar with English terms but.

is it nuclear physics?

it seems like it

[tardy ridge]

I've only seen half life 3 memes

[vague minnow]

ha

yeah

isn't that where they got the name from

[carmine elbow]

🤷🏾‍♀️

[vague minnow]

hm

when I plug in the half-life

I need to convert it into seconds, right?

cause that's like the standard thing

cause if that's the case

then it would be this

[pale pond]

can someone help me with my trig identity stuff

i'm not sure wehre to0 start

[hardy abyss]

try factoring out sec^2(x)tan^2(x)

[pale pond]

thank you

[pale pond]

can someone help me with 7

[lost mesa]

try multiplying $\frac{\tan x}{1 + \sec x}by\frac{\sec x-1}{\sec x-1}$

[obsidian monolith]

andrwxlin:

[lost mesa]

actually i got a faster way for you

express tan and sec in terms of sin and cos

@pale pond