precalculus

[willow bear]

i mean your friend was right

[rich light]

Where’s the error

[serene heath]

u had a 1 that magically disappeared

[rich light]

dope

okay so I think this is right, but earlier I performed a magic trick on the 1 because I brought over the x^2 and the -1 from the right simultaneously

I understand the answer was ultimately wrong, but theoretically wouldn't I still be able to bring over both values simultaneously?

Nvm I see what you're talking about haha. Thanks

[sharp marsh]

What quadrant is (−13π/9) in?

[hexed ermine]

-9pi/9 is at 180 degrees

so.....

270 is at -13.5pi/9

[sharp marsh]

oh

so 2nd

[hexed ermine]

no

in between 180 and 270

so 3

[sharp marsh]

It's negative though

[hexed ermine]

oh whoops

[sharp marsh]

How did you know what -270 would be

?

[hexed ermine]

well 3pi/2 is 270

it is 2nd tho

just add 2pi to -13pi/9

so -13pi/9 + 18pi/9 = 5pi/9

we know at 4.5pi/9 its at pi/2 or 90 deg

and at 9pi/9 its at 180

so 5pi/9 is in between 90 and 180

[sharp marsh]

So for this Sketch each angle in standard position.

8pi/3

Would it be on the 3rd quadrant?

[hexed ermine]

lets subtract it by 2pi

so 8pi/3 - 6pi/3

giving us 2pi/3

this might be easier for you to work with

[sharp marsh]

So how do you decide between subtracting it

and adding it?

by 2pi

[hexed ermine]

if its higher than what youd like

or lower

get it into the range of [0, 2pi]

[sharp marsh]

uh

[hexed ermine]

so like 5pi

ok its higher than what id like so imma subtract 2pi

giving us 3pi

still higher

so -2pi

pi

this is a good value to work with

[sharp marsh]

o h

[daring yarrow]

@sharp marsh what i said wasnt wrong

(−13π/9) + a full circle (18π/9) = 5π/9

which is 100°

or -260°

which is in the second quadrant

[sharp marsh]

where is -13/9 coming from

[hexed ermine]

your problem lol

[rich light]

https://media.discordapp.net/attachments/696076755412451369/696111014772605098/image0.png?width=902&height=677 Errors?

[hexed ermine]

you dont add the terms

you multiply them

[rich light]

on the top?

[hexed ermine]

yeah

[rich light]

How’s this

[hexed ermine]

ok but why (x+3) uptop

[rich light]

🤦‍♂️

[steel venture]

can you show us the picture

[honest spear]

nevermind I was just thinking too hard on those notations and confused myself

: P

[sharp marsh]

is the coterminal angles of − 9π/4

Not -π/4 and -17π/4

[stuck lark]

they ARE

[sharp marsh]

oh

Why is it saying i'm wrong then

[stuck lark]

show the q?

[sharp marsh]

?

Determine two coterminal angles (one positive and one negative) for each angle. Give your answers in radians. (Enter your answers as a comma-separated list.)

(b)
− 9π/4

[stuck lark]

one positive and one negative

[sharp marsh]

oh

ok

[rich light]

Is it possible to solve for x from here on out?

Ignore the 3 in the upper right corner

[lilac pier]

You can square both sides of the inequality in order to remove that sqrt

[rich light]

Sounds good, thanks!

[fleet yew]

I was gonna say be careful squaring cuz it might be negative

But it says its greater than 0 so ur good

[vernal spindle]

when solving problems such as sin^-1(1) how do you handle the -1 power, since sin(1) itself is pi/2

[lapis mango]

Woah

[lilac pier]

sin(1) is not pi/2, sin(pi/2) is 1. sin^-1 (1) means at which angle(s) is sin = 1, and that's at pi/2

[fleet yew]

@vernal spindle the notation for this is really confusing but bear with me

$\sin^{-1}(x)$ is NOT the same as $(\sin(x))^{-1}$ or $\frac{1}{\sin(x)}$

[obsidian monolith]

AMD:

[fleet yew]

Instead

If $\sin(a)=b$ , then $a=\sin^{-1}(b)$, which is also sometimes notated as $a=\arcsin(b)$

[obsidian monolith]

AMD:

[vernal spindle]

I kind of see it

so sin^-1(1)=x is sin(x)=1

[stuck lark]

provided $a\in[-\tfrac\pi2,\tfrac\pi2]$

[obsidian monolith]

RokettoJanpu:

[vernal spindle]

about for tan^-1(-1), how do i know which point since multiple have it where cos/sin gives -1

[uncut mulch]

firstly tan is the ratio of sin/cos not cos/sin
and consider looking up ranges of inverse trig functions

[vernal spindle]

(-pi2,pi2)?

[willow bear]

"pi2"

is your / key broken?

[vernal spindle]

thats what the internet told me

[stuck lark]

ouch

[uncut mulch]

^

[fleet yew]

It's \frac{\pi}{2} lol

[stuck lark]

to clarify the range of arctan is $(-\tfrac\pi2,\tfrac\pi2)$

[obsidian monolith]

RokettoJanpu:

[fleet yew]

How u do that

\t ?

Oh you dont actually need the {}

Thats cool

[uncut mulch]

fkn google didn't display the table properly. and/or varsity itself had some weird coding

[stuck lark]

so when computing arctan(-1), think in reverse. what angle x is between -pi/2 & pi/2 such that tan(x)=-1?

[vernal spindle]

the part confusing me is where is -pi/2

[stuck lark]

between quadrants 3 & 4

[vernal spindle]

thats not (3pi)/2 ?

[stuck lark]

3pi/2 & -pi/2 are coterminal

[vernal spindle]

ahh

so its a full line down the middle?

[stuck lark]

i have an idea of what you're saying but it's not well worded

[vernal spindle]

wait actually it's pi/2 to -pi/2 meaning Q2 and Q3 right

[stuck lark]

no, -pi/2 to pi/2. that sweeps out q4 & q1 (& +x axis)

[vernal spindle]

ahh i see

so that means Q1 and Q4 and all numbers are + and - since its including negatives? if that makes sense

[stuck lark]

your q doesn't make sense

[vernal spindle]

Ok, so I get its Q4 to Q1

Wait would it be (7pi)/4

[stuck lark]

the range of arctan is $(-\tfrac\pi2,\tfrac\pi2)$

[obsidian monolith]

RokettoJanpu:

[stuck lark]

$\tfrac{7\pi}4\notin(-\tfrac\pi2,\tfrac\pi2)$

[obsidian monolith]

RokettoJanpu:

[vernal spindle]

now im really lost

[stuck lark]

what angle x is between -pi/2 & pi/2 such that tan(x) = -1?

[vernal spindle]

im looking at this

[willow bear]

the parentheses in (7π)/4 are redundant jsyk

[vernal spindle]

the only points that can give any kind of one are either pi/4 or 7pi/4

[stuck lark]

tan(7pi/4)=-1 BUT 7pi/4 doesn't lie between -pi/2 & pi/2

btw tan(pi/4) != -1

[vernal spindle]

do i do a coterminal?

[stuck lark]

find an angle between -pi/2 & pi/2 that's coterminal w/ 7pi/4

[vernal spindle]

subtract 2pi from 7pi/4 ?

[stuck lark]

sure

[vernal spindle]

-pi/4

[stuck lark]

check tan(-pi/4)=?

[vernal spindle]

how would the cos,sin points be affected since its -pi/4 ?

is it still sqrt(2)/2,sqrt(2)/2

[stuck lark]

check em yourself

[vernal spindle]

how do I do that?

[stuck lark]

also were you ever taught how cos & sin relate to the x & y coords along the unit circle?

[vernal spindle]

isn't it just (cos,sin)

[stuck lark]

what's that tell you about the coords on the unit circle corresponding to coterminal angles?

[vernal spindle]

ill try; do you -360 from them?

[stuck lark]

that's not what i'm asking

if two angles are coterminal, what does that say about their corresponding (x,y) coords on the unit circle?

[vernal spindle]

they are the same?

[stuck lark]

so do the cos & sin values change when going from 7pi/4 to -pi/4?

[vernal spindle]

No?

[stuck lark]

question mark or period?

[vernal spindle]

No.

[stuck lark]

[vernal spindle]

Ohh my problem is I keep looking at pi/4 thinking I subtracted 360 and got -pi/4 there

I see it now

[stuck lark]

great. now focusing on what arctan does, arctan(x) gives an angle y between -pi/2 & pi/2 such that tan(y)=x

[vernal spindle]

mhm

I think that's the last part I'm struggling to understand for this question, I would have put 7pi/4, how do I know it's not within the range

[stuck lark]

-pi/2=-2pi/4, pi/2=2pi/4

[willow bear]

well how do you know 7 ∉ (-2, 2)

[earnest barn]

Don't tell me how to do it but

How can this be shown?

The numbers seem so arbitrary

I don't see what they would've done to get to them

I suppose I could show that 3b - 2a - 6c < 0 but I don't think that's the way to go about it

[willow bear]

maybe consider the fact that f(x) > 0 for all x

perhaps 2a - 3b + 6c could be expressed as a sum of values of f at suitably chosen points

[fleet yew]

@earnest barn

What is the quadratic formula?

Also remember the b term is negative

[willow bear]

NO

"negative" does not mean "has a minus sign attached"

[earnest barn]

Ahh that's a good idea Ann

Yeah, I know negative doesn't mean a minus sign

I already figured out that

The graph of f(x) will not cross the axes

But I don't know how to apply that to the inequality

[night spade]

Can anyone please explain why the answer for Q16, part a) is 50?

[sharp marsh]

Find the length of the arc on a circle of radius r intercepted by a central angle θ. (Round your answer to two decimal places.)

Radius = 9 Central Angle = 150

I did 1/2(9)^2 * 5pi/6 and keep getting 106.03

but it's wrong

[uncut mulch]

what formula are you trying to apply?

[sharp marsh]

1/2r^2 * (Central angle)

[uncut mulch]

where are you getting that expression and what's supposed to give?

[sharp marsh]

Hold on

[uncut mulch]

sounds like you're mixing up area with arc length

[sharp marsh]

lemme screenshot

o

I see

So in this problem would I just do 9 * 5pi/6?

[uncut mulch]

yes

[sharp marsh]

oh

o

k

Thanks

[unkempt stirrup]

@uncut mulch the one I asked you about was the problem before this one.(sorry for pinging)

[uncut mulch]

not really inclined to assist with a problem several hours old

and the person may not be there

[unkempt stirrup]

Like may I know about it?

What I did was differentiated the equation and set it equal to zero and found the critical point and plugged in.

[uncut mulch]

are you doing part b)?

[unkempt stirrup]

no part a

[uncut mulch]

solve N(s) = 0

[unkempt stirrup]

you mean N'(s)=0?

[uncut mulch]

no

it is most likely that zero students will get a perfect score (or 0)

[unkempt stirrup]

they are asking for the maximum possible score right?

[uncut mulch]

1 sec. let me see how to do it properly

[unkempt stirrup]

sure

[sharp marsh]

Find the point (x, y) on the unit circle that corresponds to the real number t.
t = 7π/4

What 2 do

[uncut mulch]

trig is fine here

[unkempt stirrup]

precalc?

[uncut mulch]

depends on the level i guess

[unkempt stirrup]

sure

did you figure any way for that prob ramonov?

[uncut mulch]

probs not the proper way.

[unkempt stirrup]

wdym?

[uncut mulch]

i think i'm missing something

[unkempt stirrup]

mhm

mhm just let me know if you figure something out

[sharp marsh]

Oh what happened?

I got pinged

[uncut mulch]

determine standard deviation which will be 5.
~99.994% of scores would be within 4 standard deviations. i.e. below 28 + 20 = 48.
there's a small percentage remaining. max score is usually a nice round number so its likely to be 50
there's probs a better explanation.

[unkempt stirrup]

https://discordapp.com/channels/268882317391429632/363224154469826562/696295907431415829

[rich light]

Not sure where to go from here. I was told to put it on a number line, but I have no idea how to do that while it's a fraction

https://media.discordapp.net/attachments/664685113069535274/696432488431747202/image0.png?width=508&height=677

[daring yarrow]

isnt that it?

could multiply both sides by (x+1) i guses

but they want you to solve in fraction form

also @rich light make a number line for top and bottom part of the fraction alone

if you put them under one another you can get one for the fraction as a whole

or something idk

[daring yarrow]

,rotate

[daring yarrow]

@rich light

bruh does it make sense or what

[rich light]

not to me, naw haha

but nothing in math makes sense to me 🙄

Am I supposed to use that table to complete the rest of the question? or just to plot the fraction on a number line?

[daring yarrow]

i suck at explaining

[rich light]

Nah man I’m just super illiterate in this shit lol

[patent beacon]

@rich light
(x + 1/2)/(x + 1) is negative when exactly one of (x + 1/2) or (x + 1) is negative. We can split this into cases:

x + 1/2 is negative and x + 1 is positive:
Means x < -1/2 and x > -1

And there's three other cases to go the same way

Well, we don't care about two of those cases. We only care about when exactly one of those two terms are negative

[rich light]

Alright, I understand that part

But why would I look for the negatives right off the bat? I get that it gives you the answers lmao, but is there a theoretical reason or something behind it

[patent beacon]

Positive/Positive = positive
Negative/Positive = negative
Positive/negative = negative
Negative/negative = positive

So if we care about when a/b is negative, we look for:

• when a is negative
• Or when b is negative
  But not both

[viscid thistle]

a is negative XOR b is negative

Yeah

I'm a programmer

[ivory lake]

i have a precalc test that i need to take tonight or tomorrow morning

its gonna be a timed online test since we're all quarantined but i was wondering if anyone would help me out on the test

please dm me i need to review asap

[viscid thistle]

I won't

[daring yarrow]

I won't

@patent beacon my way of solving works too right?

[viscid thistle]

If it's a test you are supposed to do it by yourself @ivory lake

[ivory lake]

oh my fault

didnt know that

NVM then

[astral mountain]

[queen root]

hi

i need help on this word problem: Two trains leave a city at the same time and travel along straight highways that differ in direction by 72 degrees. One train averages 40 mph and the other averages 55 mph. How far apart will the cards be after 45 minutes? Round to the nearest tenth of a mile.

please help me solve this asap

[fleet yew]

@queen root first find the distance traveled by each train

[viscid thistle]

Is my answer cprreect

[viscid thistle]

ahh. Next after I finish precalc, calculus! Yay finally... been about 6 months since algebra, bout time.

[blazing raven]

lol

nice

[karmic wave]

@viscid thistle is your answer to what cprreect?

[bronze juniper]

I have 4 problems in pre-calc (Pythagorean identity) and would be very appreciative if someone could help me out. thanks

[jagged glade]

@bronze juniper do you know how to draw the right angled triangle with sinθ1 = 11/61

[bronze juniper]

no

[jagged glade]

It's okay, drawing helps to visualize the problem

Do you recall the definition of sinθ?

[bronze juniper]

no. its been awhile

[jagged glade]

,w sine

[obsidian monolith]

Display results online and refine query

Requested by Biscuit#5208

[jagged glade]

Opps

https://images.app.goo.gl/j2h6QDVQEASwWicM6

Please take a look on the picture I found

[bronze juniper]

yeah i understand that

@jagged glade i just watched a video and i understand it now. thanks for your help though!

[jagged glade]

You're welcome!

[viscid thistle]

what is deg mode used for?

on a calculator

[craggy nest]

Dog mode? I think degrees....

[viscid thistle]

could you explain

specifically for what

@craggy nest

sorry, im new to trig

[craggy nest]

I'm not sure but I am guessing that deg mode is used to switch your calculator from radians to degrees

Or some other angle unit to degree

[novel dirge]

How do I find which member of the binomic formula doesn't contain x?
The formula is (x^2 + 1/x)^n and the sum of the first 3 coefficients is 46

and can someone explain me how to know when to use combinations, when permutation and when variations?

[vernal spindle]

I'm trying to find the exact values for sin^2(x)-cos^2(x)-sin(x)=0 now but I'm confused where to start for this one

[uncut mulch]

try expressing everything in terms of sin(x)

[vernal spindle]

I can't find any of my formulas that I can apply to do that?

[uncut mulch]

what about the most common trig identity

[vernal spindle]

sin^2(x)+cos^2(x)=1?

[uncut mulch]

yes that

[vernal spindle]

would it be; sin^2(x)-1

[uncut mulch]

would what be that?

[vernal spindle]

what i replace it with

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

[novel dirge]

Can someone explain to me how this is done?

especially how the underlined one appeared?

[unkempt stirrup]

@novel dirge

[novel dirge]

here

yes

[unkempt stirrup]

i cannot texit

[novel dirge]

when I did it on my own, I put it as 4^(-12+4k)

[unkempt stirrup]

ah leave this

so you have (1/4x)^12-k

right?

@novel dirge

[novel dirge]

yes

[unkempt stirrup]

so the power can be split individually

we dont care about 1, so it can be written as $\frac{1}{4^{12-k}\cdot x^{12-k}}$

[obsidian monolith]

FinalBoss:

[unkempt stirrup]

there you go

cause 1 raised to any power is just 1

[novel dirge]

yes

but can't we just write it as (4x)^-1?

and then ^12-k

[unkempt stirrup]

wdym?

[novel dirge]

afaik 1/x = x^-1

[unkempt stirrup]

we basically like to combine the x terms

so we make sure they stay together

[novel dirge]

ok

[unkempt stirrup]

so now when you multiply by $x^{k-12}$ in the numerator and denominator we get $x^{k-12}$ on the top right?

[obsidian monolith]

FinalBoss:

[unkempt stirrup]

cause the denominator would cancel

[novel dirge]

where is x^k-12 in the numerator?

[unkempt stirrup]

i said you should multiply

[novel dirge]

and how did k and 12 reverse?

wasn't it 12-k, but now it is k-12

[unkempt stirrup]

yeah i just said that you need to multiply by $x^{k-12}$ in the numerator and denominator we get $x^{k-12}$ on the top

[obsidian monolith]

FinalBoss:

[unkempt stirrup]

the denominator x power term would cancel as we multiplied with its conjugate power

$x^{12-k} \cdot x^{k-12} = 1$

[obsidian monolith]

FinalBoss:

[novel dirge]

is this what you meant?

[unkempt stirrup]

you multiply by x^{k-12}

it would lead the denominator to 1

(the x term)

$\frac{1}{4^{12-k} \cdot x^{12-k}} \cdot \frac{x^{k-12}}{x^{k-12}}$

[obsidian monolith]

FinalBoss:

[unkempt stirrup]

ya there you go

is it clear?

[novel dirge]

this part yes

but what now?

and what about the second part of the problem?

[unkempt stirrup]

could you specify which part you have doubt?

[novel dirge]

for this one I am not sure what to do after this step

and the second par is (-2x^2)^k

[unkempt stirrup]

yes so this can be written as $ (-2)^{k} \cdot x^{2k}$

[obsidian monolith]

FinalBoss:

[unkempt stirrup]

right?

law of indices

[novel dirge]

the problem asks me to find out the coefficient that stands next to x^3 in this binomic formula

[vapid torrent]

what is the exception for this rule: P(AB) = P(A)P(B) (probability)

[unkempt stirrup]

@vapid torrent this channel is occupied.

[novel dirge]

never did law of indices

we barely did anything in school and my teacher made me dislike math

that is why I now struggle with this

[unkempt stirrup]

like $ (a \cdot b)^{n} = a^{n} \cdot b^{n}$

[obsidian monolith]

FinalBoss:

[novel dirge]

oh

I know that

[unkempt stirrup]

yeah so your "a" is -2 and "b" is x^2

[novel dirge]

didn't recognize that was it

[unkempt stirrup]

now again using law of indices you can combine the x terms

do you recognize that?

[novel dirge]

what exactly?

oh, I do

or not

[unkempt stirrup]

$ x^a \cdot x^b = x^{a+b}$

[novel dirge]

I do in the solution, but not in what I wrote

[obsidian monolith]

FinalBoss:

[unkempt stirrup]

you should have a multiplication sign between them so that this works out

[novel dirge]

I only have this

[unkempt stirrup]

yeah we are now talking about the next term

[novel dirge]

oh, ok

thought you ment to use something from the first part

[unkempt stirrup]

nope

so you get this?

[novel dirge]

I think so

(-2)^k * x^2k

[unkempt stirrup]

what next do you need?

$ (-2)^{k} \cdot x^{2k}$

[novel dirge]

what to do next

[obsidian monolith]

FinalBoss:

[unkempt stirrup]

yeah so the previous $x^{k-12}$ can be merged with the present new $x^{2k}$ as they have a multiplication sign

[obsidian monolith]

FinalBoss:

[unkempt stirrup]

law of indices?

[novel dirge]

wait

[unkempt stirrup]

yeah

[novel dirge]

what x^(k-12)

ohh

[unkempt stirrup]

the previous term

[novel dirge]

I can rewrite it

to get that alone

[unkempt stirrup]

combine them

[novel dirge]

wait a sec, I will send you the pic in a min

[unkempt stirrup]

yeah $ x^a \cdot x^b = x^{a+b}$

[novel dirge]

want to try activating my brain

[unkempt stirrup]

sure

[obsidian monolith]

FinalBoss:

[novel dirge]

is this good?

,rccw

[unkempt stirrup]

yeah great

[novel dirge]

great

what next?

[unkempt stirrup]

umm what do you need?

[novel dirge]

the problem asks me to find out the coefficient that stands next to x^3 in this expanded binomic formula

[unkempt stirrup]

lemme look

[novel dirge]

👍

[viral zealot]

@novel dirge Substitute k =5

[unkempt stirrup]

so now you have $x^{3k-12}$

[novel dirge]

why 5?

[obsidian monolith]

FinalBoss:

[viral zealot]

OK you explain finalboss

[novel dirge]

I also have the other 2 parts, without the x

[unkempt stirrup]

you have to figure them itself!

your question says to find the coefficient of x^3

[viral zealot]

@novel dirge question is x^3 coefficient right?

[novel dirge]

I am just asking if I can just leave them alone or if I need to solve them, too

[unkempt stirrup]

@viral zealot lemme explain

[novel dirge]

and yes, coefficient that is next to x^3

[viral zealot]

@unkempt stirrupOk

[unkempt stirrup]

so yeah coefficient basically means the stuff other than the variable you asked for (here x^3)

like the coeff of x^2 in 42* x^2 is 42

@novel dirge

[novel dirge]

now I am confused

[unkempt stirrup]

ah what part?

[novel dirge]

like the coeff of x^2 in 42* x^2 is 42
@unkempt stirrup this

[viral zealot]

Use text @unkempt stirrup

[unkempt stirrup]

coefficient of $x^2$ in $42\cdot x^2 $ is 42

[obsidian monolith]

FinalBoss:

[novel dirge]

I get that now

[unkempt stirrup]

yeah good

[novel dirge]

but how do i apply that in the ptoblem?

[unkempt stirrup]

wait

so you have been asked the coefficent of x^3 and what we have is $x^{3k-12}$

[obsidian monolith]

FinalBoss:
Compile Error! Click the reaction for details. (You may edit your message)

[unkempt stirrup]

mhm

[novel dirge]

I thought I have to find this

[unkempt stirrup]

yeah that would be at the end

lemme explain dude

[novel dirge]

ok

[unkempt stirrup]

we need coefficient of $x^3$ and we have with ourself $x^{3k-12}$

why is this happening

[obsidian monolith]

FinalBoss:

[unkempt stirrup]

ok so there we go

do you get this?

[novel dirge]

yes

[unkempt stirrup]

so what value of "k" makes $x^{3k-12}$ as $x^3$

[obsidian monolith]

FinalBoss:

[novel dirge]

5

[unkempt stirrup]

right

so plug that k= 5 everywhere you have in your equation

[novel dirge]

I understand that now

but am still confused about the other part of the problem

[unkempt stirrup]

$T_{6} = 12 \choose 5 \cdot 4^{-7} \cdot (-2)^5 \cdot x^3$

[novel dirge]

the 1/4 and -2

[unkempt stirrup]

omg

$T_{6} = {12 \choose 5} \cdot 4^{-7} \cdot (-2)^5 \cdot x^3$

[obsidian monolith]

FinalBoss:

[unkempt stirrup]

here you go

@novel dirge

[novel dirge]

👍

[unkempt stirrup]

right?

[novel dirge]

I believe so, can't really prove othervise

[unkempt stirrup]

ah you cannot do this?

just put k as 5

cause you just said that yourself.

[novel dirge]

wtf

I just saw my teachers result

[unkempt stirrup]

yes go on

[novel dirge]

it is 99/64

[unkempt stirrup]

we need to simplify further

[novel dirge]

I have no clue wtf can be 99/64 here

[unkempt stirrup]

we need to simplify further

yeah now you have to work with calc

[novel dirge]

what exaclty?

[unkempt stirrup]

whats 12 choose 5

its 792

[novel dirge]

so I just have to plug 5 everywhere instead of k and just solve that?

[unkempt stirrup]

right?

yeah cause you just figured out that letting "k" as 5 would give you the x^3 term

[novel dirge]

oh

in that case I know how to do that

[unkempt stirrup]

yeah simplify

do it and check the answer

[novel dirge]

will do

Thank you very much

you better than my teacher

which isn't that hard, but still

you really saved me

[unkempt stirrup]

ah nah man

wait do you want to see a magic?

@novel dirge

[novel dirge]

always

[unkempt stirrup]

,w (12 choose 5) * (4^-7) * ((-2)^5)

[obsidian monolith]

Display results online and refine query

Requested by FinalBoss#1305

[unkempt stirrup]

here you go

[novel dirge]

nice

didn't know the bot can do that

[unkempt stirrup]

it can do anything

is everything clear to you now?

[novel dirge]

yes

[unkempt stirrup]

try it once again and you will be done!

[novel dirge]

feel kinda dumb for not being able to figure this out on my own

def need to practice a ton for college entrance exams

starting to panic just thinking about it

[unkempt stirrup]

Nah man everything is just new to everyone, the thing that matters is the amount of efforts you put in.

[novel dirge]

any tips on how to study and practice math efectively?

I need to cover 3 years of hs in 3 months

maybe 4

[unkempt stirrup]

I too am in HS

and tbh its hard to cover up all the part of math

[novel dirge]

yep

that is why I am already in panic mode

idk how march passed without me doing anything

now I feel like I am out of time

[unkempt stirrup]

you going to uni?

[novel dirge]

finishing hs in a few months

then need to do the entrance exams

[unkempt stirrup]

Just one thing I could say, "practice makes a man perfect"

and this server is there to help too!

[novel dirge]

I see

My emergency plan now is to practice math for at least 2-3h daily

and hope that I can cover everything

[unkempt stirrup]

you can if you are determined

btw cya.

[novel dirge]

https://tenor.com/view/the-simpsons-ralph-hi-wave-gif-5247865

[lapis mango]

Question: How would you apply DeMoivre’s Theorem to (1+I)^5

I ended up splitting it in powers and solving because I’m confused about how to apply it

[patent beacon]

Remember DeMoivre's says:
[cos(t) + isin(t)]^n = cos(nt) + isin(nt)

[lapis mango]

So since r isn’t there we don’t use r

[patent beacon]

That's just what DeMoivre's says. In this case, you'll want:
[r(cos(t) + isin(t))]^n = rⁿcos(nt) + rⁿisin(nt)

Note that's just
(ab)ⁿ = aⁿbⁿ

[lapis mango]

.... I feel dumb

Thank you

[patent beacon]

Why feel dumb? I think it just clicked

[lapis mango]

That too

I didn’t really consider any other approach then what I had in front of me

Actually I think I’m a bit confused again

[stuck lark]

show whatcha got so far?

[lapis mango]

I’m confused how to progress from here

Since I don’t know what r is

[stuck lark]

computing (1+i)^5. try converting 1+i to polar first

[viscid thistle]

hello

quick question

how do you calculate tangent ratio by head

in your head?

[lapis mango]

as in opposite over adjacent or in reference to angles @viscid thistle

Also @stuck lark thanks for that

using this: https://www.intmath.com/complex-numbers/4-polar-form.php helped out

Honestly when it came to the problem it was just that Webassign was refusing to accept my answer

[vernal spindle]

I'm trying to solve this

but how do i find cos(x)=3/5 since 3/5 isn't in the trig circle

[lilac pier]

@vernal spindle You can let cos^-1 (3/5) = theta

This way you get cos(theta) = 3/5 by applying cos on both sides

Now you can make a right angled triangle using the given sides, you should know cos = adjacent / hypotenuse

[vernal spindle]

A question i've had about the triangles is how do you know which is adjacent and which is suppose to by the hypotenuse

isnt the hypothenuse always the longest side

[lilac pier]

The hyp is always fixed, it's the larger of all 3 sides

adjacent and opposite depend on the angle

[vernal spindle]

so in this case how would i know which

[lilac pier]

you can let it be any angle other than the 90 degree angle

doesn't matter

[lilac pier]

In this case the angle in question is the top one

b/w 3 and 5

[vernal spindle]

sin90/5 = sinX/3

?

[lilac pier]

?

[vernal spindle]

I think I used the wrong formula

[lilac pier]

actually I don't think letting cos^-1 (3/5) = theta does any good

rip

[vernal spindle]

what else can i do

[uncut mulch]

pythagoras

[vernal spindle]

a^2+b^2=c^2?

[past meadow]

yes

[vernal spindle]

I got 4

is cos^-1(3/5) = 4?

[uncut mulch]

no

4 is the length of the 3rd side

arccos(3/5) returns the angle at the top
what is the sin of that angle?

[vernal spindle]

Law of sines?

[uncut mulch]

something simpler

why not just apply basic trig ratios directly since you have a right triangle

[vernal spindle]

like what?

[uncut mulch]

arccos(3/5) returns the angle at the top
what is the sin of that angle?

the sin of a non-right angle of a right triangle gives what: ?

[vernal spindle]

opp/hyp?

[uncut mulch]

yes

[vernal spindle]

so the sin of the angle is 3/5

[uncut mulch]

no

the angle arccos(3/5) is indicated at the top right?

[vernal spindle]

yeah

[uncut mulch]

which side is opposite that?

[vernal spindle]

4

[uncut mulch]

what's the hypotenuse ?

[vernal spindle]

5

[uncut mulch]

so what would be the sin of that angle?

[vernal spindle]

sin(4/5)?

[uncut mulch]

no

[vernal spindle]

oh that is the actual sign of it so just 4/5?

[uncut mulch]

sine or sin not sign

[vernal spindle]

oops

[uncut mulch]

but yes 4/5

[vernal spindle]

I have another problem to practice because I don't fully get this yet

so hyp = 5 and opp = 4

and adj=3

3/5 ?

[uncut mulch]

yes

[vernal spindle]

I made myself a study guide 🙂

[rich light]

well organized ngl

[vernal spindle]

The part that bothers me is the very bottom right corner, I don't know of a way of showing the expressions are equal without making it confusing

sin(a)=b = sin-1(b)=a

I don't feel thats easy on the eyes to see whats equal to what

[uncut mulch]

that's a bit dodgy by itself

[vernal spindle]

but that expression is true right?

[uncut mulch]

if your intention was to use $\iff$ then no. you would need to restrict the intervals

[obsidian monolith]

ramonov:

[vernal spindle]

but sin(a)=b is equal to sin-1(b)=a ?

[uncut mulch]

consider what happens if
a = pi

[vernal spindle]

sin(pi)=b?

[uncut mulch]

the word equal here is inappropriate. the word equivalent would be better suited but still not quite true.

i.e. sin(pi) = 0
however sin^-1(0) isn't pi

for sine the angle needs to be restricted to [-pi/2, pi/2]

[vernal spindle]

So writing the restrictions along with it is better?

[uncut mulch]

yes

[vernal spindle]

cos is [0,pi]?

[uncut mulch]

yes

[vernal spindle]

why does csc say y cant be 0?

[uncut mulch]

what's the definitions of csc?

and (arccsc)

[vernal spindle]

1/sinx

[uncut mulch]

and the 'inverse'?

[vernal spindle]

erm

inverse of 1/sinx?

[uncut mulch]

arccsc(x) = arcsin(1/x)

i.e.
y = arcsin(1/x)

can 1/x ever be 0?
hence can y be 0 and will it be part of the range?

[vernal spindle]

makes sense

is this better :>

[uncut mulch]

open interval for cot

[jagged rampart]

I need help

How to do 1(a) (ii)

Hellooo

[viscid thistle]

what have you tried

[jagged rampart]

I forgot how inverse of g works

The g-¹

[viscid thistle]

solve $y=\frac{x}{3}-4$ for $x$

[obsidian monolith]

mátt:

[viscid thistle]

that's literally it

[jagged rampart]

What the heck that's so cool

I'm new here btw

[vernal spindle]

I'm trying to solve question a

So far I expanded to 2sin(x)cos(x)

[willow bear]

theta, not x

but continue

or is this where you're stuck?

[vernal spindle]

That's where im stuck

I don't know if im suppose to expand more or what

[willow bear]

consider that $\cos^2(θ) \cdot (1+\tan^2(θ)) = 1$

[obsidian monolith]

Ann:

[willow bear]

this will let you find the value of cos(θ) (minding the sign of course), and from that the value of sin(θ), and from there it should be clear

does anything need further explanation

[vernal spindle]

erm

where did you get that from

[stuck lark]

don't make me do \trig

[willow bear]

this is 1 + tan^2(θ) = sec^2(θ) but in a slightly different form

or alternatively this is just the Pythagorean identity with cos^2(θ) factored out

does that answer your question?

[stuck lark]

"but how does this help me"

[vernal spindle]

I'm trying to process it

[willow bear]

what part of what i am saying is tripping you up if any

[vernal spindle]

Where you got that expression

[stuck lark]

*equation

[vernal spindle]

I expanded my sin(2x) to 2sin(x)cos(x) and then your equation is totally different and i cant find how

[willow bear]

ok you seem insistent on replacing theta with x

whatever

[vernal spindle]

I dont have a theta button

[willow bear]

you've expanded sin(2x) to 2sin(x)cos(x) cool

so now remind yourself of what you know and what you wanna know

you know the value of tan(x)

[vernal spindle]

tan(x)=-3/4 and its in QII

[willow bear]

and x is in quadrant II.

so what you want now is a way to extract the values of sin(x) and cos(x)

since once you have those, you will be able to calculate 2sin(x)cos(x)

that make sense?

[vernal spindle]

yeah

[willow bear]

yeah so

to that end i'm making use of the pythagorean identity

rewritten in a way that involves tan(x)

a known

[vernal spindle]

this?

[willow bear]

sure

if you noticed, i wrote this earlier:

this is 1 + tan^2(θ) = sec^2(θ) but in a slightly different form

i hope my implicit use of the commutative law of addition and my multiplication of both sides by cos^2(θ) is not enough to throw you off kilter

can you continue from here or are there other things you need explained

[vernal spindle]

how did you get that from cos though is the part im over looking

[willow bear]

what do you mean by "get that from cos"

[vernal spindle]

Is there one step being changed or two?

and why did you put 1 on the other side, we don't want it =0?

[willow bear]

how does one "change" a "step"

are you upset that i wrote $1 + \tan^2(\theta)$ and not $\tan^2(\theta) + 1$?

[obsidian monolith]

Ann:

[vernal spindle]

I don't think thats the part throwing me off

[willow bear]

we don't really care much here about either side being zero

okay let me put it this way

do you understand how i went from $$1 + \tan^2(\theta) = \sec^2(\theta)$$ to $$\cos^2(\theta)(1 + \tan^2(\theta)) = 1?$$

[obsidian monolith]

Ann:

[vernal spindle]

To be honest, I half see it but I also don't

I'm over looking something

[willow bear]

all i did was multiply both sides by cos^2(θ)...

[vernal spindle]

Ohh because sec is 1/cos so it would be multiplying 1/cos by cos/1 giving 1

[willow bear]

uh

yeah sure

anyway

do you understand how to find the value of $\cos^2(\theta)$ from the equations $\cos^2(\theta)(1 + \tan^2(\theta)) = 1$ and $\tan(\theta) = -\frac{3}{4}$?

[obsidian monolith]

Ann:

[vernal spindle]

would i divide 1+tan^2(x) to get cos^2(x) alone?

[willow bear]

if you wish

yes, you can get $\cos^2(\theta) = \frac{1}{1 + \tan^2(\theta)}$

[obsidian monolith]

Ann:

[willow bear]

can you calculate the value of cos^2(θ) now just to make sure you're on the right track computation wise

actually do you wanna continue with my assistance or without

[vernal spindle]

Could I little plug it in and get 1/(1+(-3/4)^2)

[willow bear]

that is in fact precisely what i would expect you to do here

are you able to simplify that fraction afterwards?

[vernal spindle]

cos^2(x)=16/25 ?

[willow bear]

bravo

can you now find cos(x) and sin(x), minding their signs as per the quadrant info you're given?

[vernal spindle]

sin(x)=3/5

cos(x)=4/5

[willow bear]

minding their signs as per the quadrant info you're given

remember x is in quadrant 2

[vernal spindle]

cos(x)=-4/5

[willow bear]

there we go

can you now do the one final step of putting it all together

to find the value of 2sin(x)cos(x)

[vernal spindle]

16/25?

[willow bear]

check your multiplication again

$2 \times \frac{3}{5} \times \paren{-\frac{4}{5}} \neq \frac{16}{25}$

[obsidian monolith]

Ann:

[vernal spindle]

-16/25

[willow bear]

check your multiplication again

$2 \times \frac{3}{5} \times \paren{-\frac{4}{5}} \neq -\frac{16}{25}$ either

[obsidian monolith]

Ann:

[vernal spindle]

oh, -24/25

[willow bear]

don't rush it

yes okay good

as a bonus, doing parts b and c is way easier now since you don't have to repeat the work of finding sin(x) and cos(x) individually

[vernal spindle]

I really don't get where the part came from before we plugged it into the given equation

[willow bear]

uhh

what

what part

can you please be more specific in what you're talking about rn bc i'm at a loss

[vernal spindle]

so what you want now is a way to extract the values of sin(x) and cos(x)

how we extracted it

do you understand how to find the value of $\cos^2(\theta)$ from the equations $\cos^2(\theta)(1 + \tan^2(\theta)) = 1$ and $\tan(\theta) = -\frac{3}{4}$?

[obsidian monolith]

kingbluesapphire:

[willow bear]

didn't you do just that tho

you found cos^2(θ), and from that you found cos(θ) and sin(θ)

[vernal spindle]

yeah but i don't get where it came from and when to know to do that

[willow bear]

where what came from

the idea of getting cos^2(θ)?

in this type of problems (i.e. "<trig function>(θ) = <value>, θ is in <quadrant>, find the value of <other trig function>(θ)")

in general a good first step is to find both sin(θ) and cos(θ)

bc once you have those, you can find just about anything you want

[vernal spindle]

So for any of these problems to find sin and cos, i can start with that?

[willow bear]

it doesn't hurt to

so yeah, you can start with that

generally the thing to keep in mind is the pythagorean identity

with some algebraic manipulation you can whip it into the form you need

i.e. a form which involves whatever quantity you actually know

'course, it helps to not be rusty with algebra

bc some things become way easier to see if you know your way around it well

[vernal spindle]

is that these?

[willow bear]

sure

i'll prob sound pretentious but perhaps it helps the algebraically inept to memorize these as three separate things

[vernal spindle]

but these are all 'pythagorean identities' right

and b) cos(2x) is 16/25?

[willow bear]

but these are all 'pythagorean identities' right
as far as i am concerned these identities are the same thing in slightly different forms

and no, cos(2x) is not 16/25

[vernal spindle]

7/25*

[willow bear]

that's more like it

[vernal spindle]

andd tan(2x)=24/7

[willow bear]

nope

[vernal spindle]

-24/7

[willow bear]

mind your arithmetic

you're making really stupid mistakes that you would really regret on a test

it sounds like you're rushing it

which is like the last thing you should do

[vernal spindle]

i cant find my mistake i did it twice

[willow bear]

alright your first mistake is that your thetas are sorely misshapen

they look more like φs

the line should be horizontal

[vernal spindle]

😦

[willow bear]

anyway, your answer of -24/7 is correct

i'm just remarking on the fact that it takes me saying "no that's wrong" for you to produce the right answer

which let's just say is not ideal

[vernal spindle]

I keep over looking the negatives

[willow bear]

that's exactly what i'm trying to get you to not do

[vernal spindle]

I'm going to sleep now, I'll work on d,e,f tomorrow, and hopefully understand it for my midterm Thursday!

[pale bison]

<@&268886789983436800>

nice

[stuck lark]

[pale bison]

you missed the fun

[stuck lark]

[mild sparrow]

hey not sure if this is the right place

log25(x)-log0,2(x)=3
can someone help me get the x
first I changed the base of log0,2(x) to log25(x)/log25(0,2)
and then idk what to do

[fleet yew]

Lemme write this out

$\log_{25}(x)-\log_{0,2}(x)=3$

[willow bear]

\log

[obsidian monolith]

AMD:

[fleet yew]

Latex be like

\\\\

[stark trellis]

Practice 2 a and b

[versed basalt]

differentiate the function and put it equal to the slope of the line that is 3 and you get the value of x now put that x in equation of curve so you get y and now you have x and y pair so put both in equation of the line

@stark trellis makes sense ?

[narrow peak]

tangent=derivative

normal is product of 2 gradients=to-1

$m_1\cdot m_2=-1$

[obsidian monolith]

Hmm:

[narrow peak]

where m1 and m2 are perpendicular to each other

[stark trellis]

yea i got the m1 m2 thing

i was thinking more like 3=4x-5, then 8=4x

x=4

sub in eqution

[narrow peak]

can u send a clearer pic

i cant see anything

@stark trellis

[stark trellis]

okay

[stark trellis]

@narrow peak

[steel venture]

@stark trellis

need help?

[stark trellis]

erm nvm i got the answer

i just got the x coordinate and y coordinate and subbed it back

[novel dirge]

Can someone help me with probability of random events(not sure if that is the right translation)?

if I have 3 events, A,B and C

how can I write that only one of them turned out to be true?

or all of them

write it like for group operations

[steel venture]

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

[obsidian monolith]

Buncho Maximos:

[novel dirge]

not sure if I am using the correct words for this since I have to translate

[steel venture]

something like that?

[novel dirge]

yes

that's it

[steel venture]

if they are all exclusive then you can just write $P(A)$

[obsidian monolith]

Buncho Maximos:

[novel dirge]

I understand the first one

but not the last one

[steel venture]

if they are exclusive it means

if A happens

then B and C can't

so if they are exclusive

[tardy ridge]

so if you are a boy you can't be a girl

[steel venture]

the probability of A means that B and C don't happen

[tardy ridge]

actually that is debatable

[novel dirge]

you really chose an unfortunate example

[tardy ridge]

if you are 9 years old you cannot be 10 or 11 years old

[steel venture]

a race between a b and c

the probability of a winning

is the same as a winning and b losing and c losing

[novel dirge]

@steel venture but what is the difference between that and the first one?

[steel venture]

for the second one

[novel dirge]

aren't they the same

[steel venture]

the events have to be exclusive

[novel dirge]

oh

and the first one means that the other ones can happen, but just haven't?

[steel venture]

the first one is saying the probability that A happens, but B and C don't

regardless of if they are exclusive or not

[novel dirge]

and what about if all 3 happen

[steel venture]

so the top will work for exclusive and inclusive

P(A\cap B\cap C)

[novel dirge]

or at least one or two of the 3 happen?

[steel venture]

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

[obsidian monolith]

Buncho Maximos:

[steel venture]

that is all 3

for at least 2 it gets a little longer

you can write out $P(A\cap B\cap C') + P(A\cap C\cap B') + P(B\cap C\cap A') + P(A\cap B\cap C)$

wait

[novel dirge]

ok

[obsidian monolith]

Buncho Maximos:

[novel dirge]

and that means that only 2 happened?

but what if I know which 2? does it stay the same?

and how can I figure this out on my own?

I really want to stop constantly having to ask questions

[steel venture]

what i sent is for at least 2

$P(A)$ is the probability of A happening

[obsidian monolith]

Buncho Maximos:

[steel venture]

$P(A')$ is the probability of A not happening

[obsidian monolith]

Buncho Maximos:

[steel venture]

which can also be written as $1-P(A)$

[obsidian monolith]

Buncho Maximos:

[steel venture]

$P(A\cap B)$ is the probability of A and B happening

[obsidian monolith]

Buncho Maximos:

[steel venture]

and with those you'll probably be able to write what you want

[novel dirge]

what is this called?

these formulas and symbols

[tardy ridge]

interesection

[steel venture]

probability notation?

im not sure

[novel dirge]

the U and the rest

[slender sun]

set builder notation right?

[tardy ridge]

U is union, n is intersection

[steel venture]

the $\cap$ is intersection

[obsidian monolith]

Buncho Maximos:

[slender sun]

or maybe I'm thinking of something else

[steel venture]

but in probability we can just say it means "and"

[slender sun]

nvm I am thinking of something else lol

[novel dirge]

found it

[steel venture]

hunter its similar

[novel dirge]

so when I have 2 of the 3

[steel venture]

ah wait

no

i was talking to hunter

continue mr pancake

[novel dirge]

is it enough to just say AnB or do I have to say AnBnC' or AnBuC ?

or do the other 2 just mean that C can't happen

but the first one doesn't care about C

[steel venture]

if we say $P(A\cap B)$

[obsidian monolith]

Buncho Maximos:

[steel venture]

there is no mention of C

so it doesn't matter if C happens or not

[slender sun]

this is set builder notation lol

[steel venture]

this gives you the probability of A and B happening

[novel dirge]

I think I get it

I'll try solving some problems on my own now and ask when I get stuck again

thanks for the help

[steel venture]

👍

[novel dirge]

if at least one happens it is just P(A)?

[steel venture]

no

that is the probability of A happening

[novel dirge]

so I have to write it like P(A)UP(B)UP(C)?

[steel venture]

$P(A) + P(B) + P(C)$ works

[obsidian monolith]

Buncho Maximos:

[steel venture]

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

[obsidian monolith]

Buncho Maximos:

[novel dirge]

what does the + mean?

[steel venture]

so you'd be excluding a few probabilities with the notation you mentioned

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

[obsidian monolith]

Buncho Maximos:

[novel dirge]

and how is my notation wrong?

[steel venture]

the + just means you add them

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

[obsidian monolith]

Buncho Maximos:

[novel dirge]

...

fml

we didn't do any of these in school

and my teacher gave us a pdf of definitions without anz examples

and told us we have to do it for homework

[steel venture]

ouch