precalculus

[maiden igloo]

okay let me write it out

[steel venture]

prefect

[maiden igloo]

Like that ?

Excuse my hand writing

[steel venture]

awesome

ok do you know logarithms

[maiden igloo]

a little

i know

logging it

removes the

variable from the exponent

like pulls it down ?

[steel venture]

sort of

but thats not what we're looking for here

[maiden igloo]

aight

[fleet yew]

you can simplify that a lot

(a^b)^c = a^(bc)

[daring yarrow]

u dont need logs to solve this do you?

[maiden igloo]

i don't think so

[daring yarrow]

just equal exponent rule thing

[maiden igloo]

Now that we have it equal the same base

do i do

...

lost

[sour inlet]

i'm completely lost maximo, still can't figure out

[fleet yew]

$(2^2)^{x^2} = 2^{2x^2}$

[obsidian monolith]

AMD:

[maiden igloo]

uh

okay

so that's how it should be written

[fleet yew]

uh huh

then you have

[maiden igloo]

got it

\

[fleet yew]

$2^{2x^2} \times 2^{5x} = 2^{3}$

[obsidian monolith]

AMD:

[maiden igloo]

why did we switch it

from the 8 being on the left side

?

[fleet yew]

doesn't matter lol

[maiden igloo]

ahh so it's your preference ?

[fleet yew]

yeah sure

[maiden igloo]

aight

[fleet yew]

now what can you do

if you have 2^a times 2^b

how do you simplify that

[maiden igloo]

2a * 2b

that would

4ab

?

wait no

wait yes

[fleet yew]

no

let me give you an example using real numbers

what is 2^2 * 2^3

[maiden igloo]

oh

4 * 8

= 32

?

[fleet yew]

and what is 32?

in terms of powers of 2?

[maiden igloo]

and even num

um

it is a ....

[fleet yew]

32 = 2^5

[maiden igloo]

oh

okay

make sense

[fleet yew]

and 5 = 3 + 2

so we can just add the exponents together

and put it over the same base

2^2 * 2^3 = 2^5

see

[maiden igloo]

care to throw up that wonderful text

[fleet yew]

let's just look at the left hand side of the equation now.

[maiden igloo]

yeah

starting from 8

🙂

[fleet yew]

$2^{2x^2} \times 2^{5x} $

[obsidian monolith]

AMD:

[maiden igloo]

wait okay

i see

okay so starting from here

[fleet yew]

$ = 2^{2x^2+5x}$

[obsidian monolith]

AMD:

[fleet yew]

because you add the exponents together

[maiden igloo]

ahh we can combine

okay

and we can do this because we share the same base

is that correct

[fleet yew]

yes

and now

$ 2^3= 2^{2x^2+5x}$

[obsidian monolith]

AMD:

[maiden igloo]

now

we

multiply

2 by the whole exponent

?

[tardy ridge]

the bases are the same

[maiden igloo]

oh

so can i set to zero

[tardy ridge]

so if 2^5=2^x what is x

[maiden igloo]

0 = 2^2x2+5x - 2^3

[tardy ridge]

(that x has nothing to do with the question just an example)

[maiden igloo]

oh rop

rip

um

do i just solve now

2x^2+5x ?

[tardy ridge]

$ 2^5=2^x$

[obsidian monolith]

Fishraider:

[tardy ridge]

what is x

[maiden igloo]

5

?

[tardy ridge]

YES

!!

[maiden igloo]

im laughing

oh man im so dumb

okay

[tardy ridge]

you have

$ 2^3= 2^{2x^2+5x}$

[obsidian monolith]

Fishraider:

[tardy ridge]

so what is 3 equal to

[maiden igloo]

8

wait

[tardy ridge]

lol

[maiden igloo]

😦 lost me

[tardy ridge]

ok so

$2^x=2^n$

[obsidian monolith]

Fishraider:

[tardy ridge]

what is x in respect to n

infact we can do this

$b^x=b^n$

[obsidian monolith]

Fishraider:

[maiden igloo]

x=n

?

[fleet yew]

yes

[tardy ridge]

YES!!

[fleet yew]

but only over the reals

[daring yarrow]

cmon amd

[tardy ridge]

so if you try that with your equation

[maiden igloo]

x = 3

[daring yarrow]

no nono

[tardy ridge]

no he's comparing to orginal equation

let's see what he makes n equal to

[daring yarrow]

oh

yesyesyes

[tardy ridge]

that's what I think s/he is doing anyway

[maiden igloo]

based on the examples you gave me I am assuming whatver my x is on one side it's reflected on the other

but i feel like that isn't right

[tardy ridge]

x and n are variables in my equation

they can represent a whole expression=

[daring yarrow]

$b^(n)=b^(5+2)$

[obsidian monolith]

piece:

[daring yarrow]

wtf

[tardy ridge]

piece what

lmao

[daring yarrow]

idk how this bot works

wanted to look smart

[tardy ridge]

I think you use the swiggly one

[daring yarrow]

eh just nvm me i'll lurk

[tardy ridge]

ok it might be easier to explain if you use log, do you know what log is @maiden igloo ?

[daring yarrow]

2^(x²+2x+1) = 2^-1

idk if this example could help but this is what got one of my friends to understand how it works

[unique hill]

$2^{x²+2x+1} = 2^{-1}$

there

[tardy ridge]

good job

that -1 part

[daring yarrow]

still messed up

[obsidian monolith]

Nightingale:

[tardy ridge]

wait what happens if I react to it

o

[daring yarrow]

and you know that
$b^n=b^x$

[obsidian monolith]

piece:

[daring yarrow]

when n=x

[tardy ridge]

I think subi is long gone

[daring yarrow]

rip

this selfstudy stuff is pretty tough sometimes

[maiden igloo]

Nah i am

here

gf stepped in and helped

[tardy ridge]

do you know what log is

[maiden igloo]

i do know

[tardy ridge]

just log 2 both sides

[maiden igloo]

but i probably not sure how to use correctly

I got the answer

[tardy ridge]

nice

[maiden igloo]

I realized that

it was

3 2x^2+5x

I kept thinking we had to do something with the bases

so i got confused

But i understand now that the example you showed me

$2^x=2^n$

[obsidian monolith]

Subi:

[maiden igloo]

is saying

x = n

which are my exponent values

so set

x = n whatever that is and in my case was

[tardy ridge]

yes

[maiden igloo]

$ 2^3= 2^{2x^2+5x}$

[tardy ridge]

so what's the answer you got

[obsidian monolith]

Subi:

[maiden igloo]

-3 , 1/2

[daring yarrow]

[tardy ridge]

nice

[daring yarrow]

goodjob

[maiden igloo]

but dinner timw

ill be back

i appreciate you guys

a lot

im trying to get better

[tardy ridge]

wait

one thing

[daring yarrow]

dont give up and you'll be fine

[tardy ridge]

if the bases are different but they can turn into the same you can do that

like 4^x = 2^2x

[daring yarrow]

$4^x=2^{2x}$

[obsidian monolith]

piece:

[daring yarrow]

ez

[tardy ridge]

ok piece do want to try a problem relating to this? It's not too difficult but I'll just put this here cause i came up with it like a minute ago

$b^{2k}=(-b)^{k^2+1}$

[obsidian monolith]

Fishraider:

[daring yarrow]

its 4am so cant guarantee ill be able to solve

[tardy ridge]

logs aren't gonna work now

wth

[daring yarrow]

lemme get up and grab a piece of paper

[unique hill]

you gonna solve for k or b?

[daring yarrow]

k

[unique hill]

2 unknowns

[tardy ridge]

k because b can be anything

yes that's true

but piece you should go to sleep

[daring yarrow]

what are the solutions

@tardy ridge

ima just say there arent any

[tardy ridge]

if k^2+1 is even then there is

[daring yarrow]

i tried squaring both sides but idk

@tardy ridge how many solutions?

[tardy ridge]

2

k=1 and k=s1

-1

look if k^2+1 is even then -b doesn't matter

[daring yarrow]

figured but idk how to get there without just thinking

[tardy ridge]

if you square (-2)^even then it's 2^even

[daring yarrow]

can u actually calculate this

or is it just thinking

[unique hill]

but you dont know if it's odd or even

[tardy ridge]

@unique hill if k^2+1 is even, k=1 k=-1 but -1 doesn't work so k=1

and k^2+1 is indeed even because 1+1 is even

the equation basically becomes b^2=(-b)^2

when k=1

[maiden igloo]

I have two more question , the first one I just wanna confirm if im understanding the problem correctly

which is this

Information is given about a polynomial f whose coefficients are real numbers.
Find the remaining zeros of f .
Degree 5; zeros: -4, i, 5 − 3i

I said the remaining zeroes were -4 , i ,-i , 5-3i , 5+3i

[cedar pawn]

Can u send a picture

[maiden igloo]

Yes sir

[cedar pawn]

Oki

[maiden furnace]

https://gyazo.com/792bebda44f194b849a18039fd28cc76

how do i get the period of this?

[uncut mulch]

don't need to list all 5

[maiden igloo]

you talking to me

[uncut mulch]

the remaining (2) zeroes in Q11 would just be the:
-i and 5 + 3i

[unique hill]

he's doing prev question

[maiden igloo]

Okay so i figured it was those for question 11

it just seem so simple

so thought it was wong

wrong*

I got 12 btw

the demon question

I skipped cause I was struggling is this one

[uncut mulch]

that info layout is horrid

[maiden igloo]

THANK YOU

do you see the lonely i

in the corner

[uncut mulch]

3 is a zero with multiplicity of 2
i is also a zero

[cedar pawn]

(2+i)2

[maiden igloo]

but why is I not listed with the zeroes

[cedar pawn]

(2+i)^2

[maiden igloo]

i thought it would be

[cedar pawn]

[maiden igloo]

(x-3)^2

[uncut mulch]

they separated the list with ;

[cedar pawn]

[maiden igloo]

Can someone walk me through or point me to resources to explain this better

[cedar pawn]

Khan academy

[uncut mulch]

it uses the idea as the previous questions

[maiden igloo]

The degree should reflect the amount of zeroes ... is this correct

[uncut mulch]

conjugate root theorem

[maiden igloo]

gotta look that up

[fleet yew]

if a+bi is a root, then a-bi is a root, assuming the polynomial has real coefficients

[uncut mulch]

its pretty much exactly the same thing as what you just applied

[fleet yew]

because (a+bi)(a-bi) = a^2 + b^2 so it must have the conjugate so the imaginary parts can cancel out

[unique hill]

i see the word imaginary...

is this precalculus?

[fleet yew]

yeah

[unique hill]

wait im pretty sure complex numbers are not precalculus

[cedar pawn]

Ain't it prealgebra

[unique hill]

for my curriculum, calculus is taught before complex numbers

[maiden igloo]

Complex nums is College algebra

But I don’t recall these theorems

I’ll openly admit I didn’t try as hard as I should have in the past and I’m obviously paying for it now , but I’m willing to go through that now

Cause I need this to be where I wanna be

[unique hill]

i havent touched on complex numbers

[maiden igloo]

What math are you in

[unique hill]

as in?

[maiden igloo]

so

[fleet yew]

complex numbers are like algebra 1 wtf

[maiden igloo]

chill man

take it easy on us

[fleet yew]

like how else do you solve x^2+1=0

[maiden igloo]

uh

oh

[fleet yew]

no i'm just saying how do you do calculus without knowing what i is

[maiden igloo]

x^2 = -1

oh square root of negative

give ...

i know it's like i or something

i think

[fleet yew]

+-i

[maiden igloo]

So

I am confused how conjugate root theorem is used here

[unique hill]

there's no theorem

it's like x^2=1

x=+-1

but with i

[fleet yew]

conjugate root theorem says

if $(x - (a + bi))$ is a root of a polynomial with real coefficients, then $(x - (a - bi))$ must also be a root of the polynomial

[obsidian monolith]

AMD:

[unique hill]

@fleet yew in response to ur comment, ik what i is but havent done anything with it

all ik is 1/i = -i

and stuff i did in calculus has no use of complex numbers

[willow bear]

@fleet yew factor not root

[maiden igloo]

I am referring to my question

I thought Degrees were connected to the zeros

for exmaple

Degree 3

zeroes 1, 2 , 1+i

sorry

Degree 4 zeroes 1, 2,1+i

i would assume it's the conjugate paiur

pair*

But here

the Degree is 3

[patent beacon]

If you count multiplicity, then there's exactly as many zeroes as the degree

[maiden igloo]

Hmm

First time dealing with mulitplicity in this regard

[harsh cipher]

Good morning Ann

[maiden igloo]

I am aware of it for determining if

[patent beacon]

If the polynomial has real coefficients, then 1-i is a root as well

[maiden igloo]

it cross x or bounces

So confused 😭

so

is x-3 not a zero here ?

[patent beacon]

Oh, but you said it's degree 3. So, the polynomial must not have real coefficients

[maiden igloo]

Here

Form a polynomial f (x) with real coefficients having the given degree and zeros. Show your work! Degree 4; zeros: 3, multiplicity 2; i

Oooooooo

wth

I kept reading it like this

Degree: 4
Zeroes:3
Multiplicity:2

i

So multiplicity is apart of the zeroes

[patent beacon]

A zero also has a multiplicity, yeah

[maiden igloo]

is i not apart of the zero ?

[harsh cipher]

question

(2)^1-x = (5)^x

solve exactly for x then evaluate your solution rounding to 3 decimals

why cant I log both sides and perform simple algebra?

[steel venture]

before anything

is it

(2)^(1-x) = (5)^x

or

(2)^(1) - x = (5)^x

[harsh cipher]

opps

[steel venture]

i.e. $2^{1-x}=5^x$

[obsidian monolith]

Maximo:

[harsh cipher]

its that one

[steel venture]

ok

[harsh cipher]

I forgot parentheses

[steel venture]

they aren't the same base

[harsh cipher]

i hope Ann is not watching

yes they are not

[steel venture]

lmfao

[harsh cipher]

so I need to log both sides

hahahaha

[steel venture]

the problem is you can't because theyre not the same base

what can be done though is using one of the log rules

$log(a^b)=blog(a)

$log(a^b)=blog(a)#

$log(a^b)=blog(a)
$

fuck

[obsidian monolith]

Maximo:

[harsh cipher]

power rule?

[steel venture]

yep

[harsh cipher]

okay

do I apply the power rule to both sides of the equation?

I just solved the previous problem

using the same rule

(3)^(2x+5) = 54

[steel venture]

54 = 3*18

so you can use log base 3

[harsh cipher]

I did it different way

(log 54 )/ (log3) = 2x+5

also that did not make sense when you said 3*18

[steel venture]

ok well its applicable to both problems i think

wait it didn't make sense

why

[harsh cipher]

because 3^(2x+5)= 3*18

basically what you're telling me is that

I can equate (2x+5) to 18?

[steel venture]

no

what i was implying

is that you can log base 3 to get

$2x+5=1 + log_3(18)$

[obsidian monolith]

Maximo:

[fleet yew]

RHS simplifies to 3 + log_3(2)

[steel venture]

oh

[willow bear]

\log

[steel venture]

thanks ann

[harsh cipher]

oh hi there

[steel venture]

ill remember for next time

[maiden igloo]

last question before i go to bed. Dealing with these real world problems as such , I get confused with what data goes where

Between 12:00 PM and 1:00 PM, cars arrive at a bank’s drive-thru at the rate of 12 cars per hour (0.2 car per minute). The following formula from statistics can be used to determine the probability that a car will arrive within t minutes of 12:00 PM.

F (t) = 1 − e −0.2t

Determine the probability that a car will arrive within 10 minutes of 12:00 PM. Round to two decimal places as needed.

[harsh cipher]

I can be patient let's help subi first

[fleet yew]

No lets finish your question

[harsh cipher]

okay sure

[maiden igloo]

no

finish your stuff first

[harsh cipher]

okay thank you

[viscid thistle]

is e the constant?

or variable

[harsh cipher]

so my question (2)^(1-x) = (5)^(x)

[viscid thistle]

@maiden igloo

[maiden igloo]

yeah it is

[viscid thistle]

oh constant

[maiden igloo]

I believe this is an expotential growth function

[viscid thistle]

yes

[maiden igloo]

i always get them confused

between

decay

growth

[willow bear]

(2)^(1-x) = (5)^(x)
three of these four pairs of parentheses are redundant

[steel venture]

ann you're very nitpicky haha

[viscid thistle]

is it not simply F(10)? @maiden igloo

[maiden igloo]

uh

[viscid thistle]

oh

[maiden igloo]

Idk

[viscid thistle]

within

[maiden igloo]

I am trying to understand how to break this down

[steel venture]

mirrion go to a questions channel ill help you

[viscid thistle]

hmm

[harsh cipher]

which channel

[fleet yew]

#help-2

[steel venture]

any one not occupied

[harsh cipher]

okay thank you see you there

[maiden igloo]

do i put 10 in for F(T)

[viscid thistle]

It would be negative

[maiden igloo]

and put 10 where t is

[viscid thistle]

No

so 0.2 cars / min

[maiden igloo]

correct

[viscid thistle]

Are you sure F(t) is written correctly

[maiden igloo]

let me send pic

[viscid thistle]

ok

[viscid thistle]

so thats

different

now please tell me

i have never taken probability

but its between 0 and 1?

yes

ok

yes it should just be F(10)

[maiden igloo]

okay

So

[viscid thistle]

but its saying within 10 mins

yes

[maiden igloo]

1 - e^-0.2^10

?

[viscid thistle]

F(1)

*F(10)

[maiden igloo]

let me work it out

[viscid thistle]

86.4%

[distant rampart]

How do you graph a 3d plane
Like
Its easy to plot the points by just finding intercepts

But i dont know how to draw the plane

[steel venture]

https://www.geogebra.org/3d?lang=es

plot in what you need and copy it

its your best bet

[maiden igloo]

i got the same

@viscid thistle

[distant rampart]

Hmm i thanks
I tried that already i still couldnt get it
What im thinking of right now is just make up line equations in two dimensions then just somehow piece them together

[viscid thistle]

Alrigh

t

[maiden igloo]

Is there a way for me to donate to you guys

[viscid thistle]

Never donate

Just express thanks using English

[maiden igloo]

THANK YOU !

[sour inlet]

Guys I'm supposed to use a calculator to solve for theta on the interval [-90deg,90deg]
for Sin(theta) = 3/4

I did arcsin(.75) and got 48.5903 deg

[unique hill]

48.6 (rounded) degrees is one of the solution

[sour inlet]

and I also got 131.4096 deg from subtracting 48.6 from 180 deg

since sin i positive in q2 right?

[unique hill]

yep

[sour inlet]

but i don't understand what it means

from a negative 90 deg

to a positive 90 deg

[unique hill]

from a neg, you go clockwise along the ASTC circle

so from 0 to 90 degrees, u go anticlockwise
but from 0 to -90 degrees, u go clockwise

[sour inlet]

One sec trying to process

[maiden igloo]

Good ole trig

I have that class tomorrow

Morning

[sour inlet]

what would that be

-41.4 deg?

[maiden igloo]

+369

360

I think would give coterminal angle

That is the same

One of the smart guys can verify

[unique hill]

not +360

x is between 90 and -90 degrees

[maiden igloo]

Rip so 90 ?

90 + 41.6 ?

41.4*

[unique hill]

no 90+41.6 exceeds 90

so it's not in the range of x

[patent beacon]

There's only one solution in that range, the one your calc gave you

[sour inlet]

hmm i see

[maiden igloo]

💀

Wait

[sour inlet]

thanks guys

[maiden igloo]

90 -41.4

[unique hill]

that's arcsin 3/4

[maiden igloo]

Would that be the coterminal

Ahh okay I’m dumb

Got it

Thanks for all the help guys

I hope this discord is around for the next 2 years help me through college

Goodnight

[sour inlet]

gn

[sour inlet]

Last question for the day if anyone is still here

how would you go about getting the equivalent of 6sinxcos(5x)

would it be

3sin(6x)+3sin(4x)

?

[willow bear]

"the equivalent"?

[unique hill]

are you talking about double angles?

[sour inlet]

that's all it asks

[unique hill]

ah

do u know the product-to-sum formula?

[willow bear]

oh

well

[sour inlet]

you mean like cos(a+b) = cosAcosB-sinAsinB

?

[willow bear]

no that's not what nightingale means

[unique hill]

that's not product to sum

those are compound angles

formula

[willow bear]

$\sin(a) \cos(b) = \frac12 (\sin(a+b) + \sin(a-b))$

[obsidian monolith]

Ann:

[unique hill]

thx for typing it in texit

[sour inlet]

oo

1/2(6sin(a+b) +sin (a - b))

what would a and b

be in this case?

[unique hill]

a and b are the respective angles

[sour inlet]

how would you get angles from something like this

doesn't look like I have much to work with, not much that i can see anyways

[unique hill]

first focus on the ones inside the brackets

ignore the 1/2 for now

so just $sin(a+b)+sin(a-b)$

[sour inlet]

sin(5x+x)

[obsidian monolith]

Nightingale:

[sour inlet]

?

[unique hill]

a and b are the matching angles

so compare them with $sinacosb$

[obsidian monolith]

Nightingale:

[unique hill]

my bad

with the question

[sour inlet]

all I can think of is
6sin(X+5X) + sin (X-5x)

6sin(6x) + sin(-4X)

[unique hill]

your working out is correct

but u might want to

[sour inlet]

6sin(6x) - sin(4X)?

[unique hill]

let a=5x and b=x

that's favourable bc you dont get a negative inside the bracket

[sour inlet]

ahhh

6sin(6x) + sin(4x)

then the 1/2?

[unique hill]

half it

remember the parenthesis

it's 6 times the WHOLE thing

[sour inlet]

6sin(6x)+6sin(4x)

1/2

3sin(6x) +3sin(4x)?

[unique hill]

minus in the middle

you're using a different formula now

becuz a and b are switched

[sour inlet]

o ok

so to avoid negative in bracket

formula was different and sign was switched

?

[unique hill]

$2cosAsinB=sin(A+B)-sin(A-B)$

uhh

why's there no parenthesis

[sour inlet]

where

o

[obsidian monolith]

Nightingale:

[unique hill]

no for my one

ok nice

[sour inlet]

and when did this fo rmula come into place

on the 1/2

?

[unique hill]

that is just 2 multiplied by the whole thing

ok now, u know that a=5x and b=x

you have everything u need to know

sub into $2cosAsinB=sin(A+B)-sin(A-B)$

[obsidian monolith]

Nightingale:

[sour inlet]

2(cos5X*sinX) = sin(5X+X) - sin(5X-X) ?

wait what happened to the 3

[unique hill]

yes and simplify

simplify first

[sour inlet]

2cos5XsinX=sin(6X)-sin(4x)

?

[unique hill]

yes

very good

wait

it's 2cos(5x)sinx for LHS

there's no extra 2

[sour inlet]

oh right cus we multiplied first in the paranthesis

[unique hill]

so now what's your LHS and RHS

[sour inlet]

2cos5XsinX
sin(6X)-sin(4x)

I'm not confused by the question, what do you mean what's on my left hand side and right hand side.

[unique hill]

there's an equal sign in between them

cuz you subbed in a and b into the formula which has an equal sign in the middle

so LHS=2cos(5x)sinx and RHS=sin(6x)-sin(4x)

[sour inlet]

right

[unique hill]

you follow me?

that eqn above

[sour inlet]

Well it's the first time I've seen something like that, but yes we have a LHS and RHS

and they're set to equal each other

[unique hill]

now what's the question asking?

[sour inlet]

Which of the sums is equivalent to 6sinxcos(5x)

my best guess would be 3sin(6x)-3sin(4x)

[unique hill]

yes that's correct

dont guess btw

[sour inlet]

but our RHS is completely different

ever since we dropped the 3 and moved onto a new formula

so i'm confused

[unique hill]

since u found out 2cos(5x)sinx, u can times 3 on the other side since the question is asking for 6cos(5x)sinx

[sour inlet]

ohhh

[unique hill]

so you times 3 for RHS

which makes 3sin(6x)-3sin(4x)

which should be correct

[sour inlet]

and lhs 6sinxcos(5x)

i see

[unique hill]

wait wha

no that's LHS

[sour inlet]

typo

[unique hill]

okok

[sour inlet]

thank you very much for your time and patience

appreciate it

[unique hill]

no worries

[daring yarrow]

lim(f(x)) = -inifinity
(x -->+inifnity)
i have to prove this using the epsilon gamma method but i uh

yeah i dont really get it

[tardy ridge]

What is the question

[daring yarrow]

wait nvm they're being as vague as ever in the book

just want me to write it in epsilon-delta form

aight then

[maiden igloo]

good morning

[harsh cipher]

good morning

[steel venture]

,rotate

[sturdy haven]

Help with number four PLease

[sturdy haven]

IM SO OCNGUSED

[steel venture]

try different points to see how it looks

[sturdy haven]

uhhh what

Thats not

I mean how do I space it out

And what intervals

And when use period

[steel venture]

ok the center is zero

and you have all that room on the right to get to 2 pi

so space it as you will but utilize the room wisely

[sturdy haven]

So i spaced it out an it is PI OVER FIVE

WHAT

ALSO THE PERIOD IS FOUR PI

what does that mean

how is im so confused right now

[steel venture]

ok

[sturdy haven]

because listen if spaced OUT evenly

[steel venture]

lets take it slow

[sturdy haven]

SO

ok

[steel venture]

count the lines to the right of the origin

vertical lines

[sturdy haven]

21

[steel venture]

ok

now you need to seperate that in 2

but oh no! its odd!

[sturdy haven]

o

[steel venture]

so move one back

and write 2pi

[sturdy haven]

what the

WHAT

[steel venture]

and then count 10 back

and write pi

[sturdy haven]

oh

wait what

so just leave EMpty space that tje end

[steel venture]

yes

no one is going to die from that

dw its safe

[sturdy haven]

I Wrote two pie now what

[steel venture]

move ten lines to the left

and write pi

[sturdy haven]

ok infucekd up

its only 20

so i wrote two pied at the end

And wrote pie in centrez

ok no waht

[steel venture]

then just earase and rewrite it

[sturdy haven]

i did

now what to do

[steel venture]

set the top of the graph to be equal to 1

and the bottom to -1

[sturdy haven]

ok

but whar if period is four PIE

[steel venture]

it doesn't matter

now all you have to do is plot points

[sturdy haven]

oh shet i fucked up

[steel venture]

until you can fill in the lines

[sturdy haven]

FREQUENCY IS ONE HAD

FUCK

the fricken frequency is one half!!!

[steel venture]

the period is 4 pi

you are correct

don't worry

[sturdy haven]

but the frequences

[steel venture]

the question didn't ask for that

all its asking is for you to plot the graph in the given interval haha

[sturdy haven]

now the whole grzphis is fucked up

the frequency is one hafe

[steel venture]

plot the point for theta=pi

and theta = 2pi

[sturdy haven]

i can’t

[steel venture]

and theta = pi/2

[sturdy haven]

ghe graph

ok

wait

so

for two pie

Is just cos pie

so

thats neg one

how on gods green earth do i do 3pie over Two

[steel venture]

unit circle

[sturdy haven]

no

its just one and negaitve e oneo

is coffect

or NO

[steel venture]

looks fine

[sturdy haven]

if i fail i will blzme you but i thinkz you did it right sonidk

BRUH

HOW THE FUCK

[steel venture]

😗

[sturdy haven]

!rotates

!rotate

[steel venture]

,rotate 270

[sturdy haven]

!rotation

[steel venture]

dawg i taught you how to do it

[sturdy haven]

nomber feiv

but this oens hard

its amplificationals and uhh FREQUENCY

[steel venture]

you just need to plot points until ou find the shape

[sturdy haven]

listen my frenh

how is y POINTS

[steel venture]

listen my guy i got some sleep to catch up on

[sturdy haven]

LIKE ONE AND NEGATIVE ONE

[steel venture]

think about it

plot points and figure out what the maximum and minimum y values can be

[sturdy haven]

i like you

[steel venture]

😄

[molten forge]

How do you do this? It's in an online worksheet

They didn't explain how

this too

[viscid thistle]

@sturdy haven still need help?

[sturdy haven]

YES

hwo do i do this

@viscid thistle

[viscid thistle]

Ok lets move to delta questions

#help-4

[maiden furnace]

https://gyazo.com/ee8cf80df3eb355b8bb5dc7b1081cd18

how would i go about to solve this

[stuck lark]

think in reverse, think of an angle x between 0 & pi such that cos(x)=-1/2

[tardy ridge]

two values I think?

[unique hill]

one angle

between 0 and pi

[tardy ridge]

where are the restrictions

there should be 2

[unique hill]

there's 2 if restriction is 0<x<2pi

so depends

[tardy ridge]

yeah

[dim basin]

can someone quick help me with this?

[tardy ridge]

1 thing is that theta is in the 4th quadarant

[dim basin]

yep

[tardy ridge]

are you supposed to use a formula or something for this?

what's stopping you from just saying theta=-80.53

[dim basin]

tbh we did this before corona/spring break and its been 3 weeks since we have done this so i completly forgot how lol

oh

[tardy ridge]

and just plugging the value in

[dim basin]

waity

idk im in the wrong chat btw

i should be in early university

sry

[unique hill]

i think you're fine

[dim basin]

ok

[glad vale]

The perimeter of a stadium is an ellipse with major axis 640 feet and minor axis 549 feet. Find the distance between the foci of this ellipse.

giw di u di this?

how do i do this lol sorry

[maiden furnace]

ok i need someone to walk me through this one i managed to solve my other problem

https://gyazo.com/be3623fac73704ab4d6b05d86359fbf7

so i know the answers isnt 7pi/6 because of the restriction of inverse cos being between 0 <= x <= pi

so what do i do next?

[tardy ridge]

well cos (pi+theta)=-cos(theta)

so cos 7pi/6 is -cos pi/6

and you can calculate that

also the answer would be 7pi/6 and another value if there wasn't that restriction, don't think 7pi/6 isn't right though

[maiden furnace]

well back of the book we have answer and it says the answer is 5pi/6

and going through some website it says to use the refence angle

[tardy ridge]

it is 5pi /6

but 7pi/6 is also an answer if the restriction wasn't there

do you know cos (pi+theta)=-cos(theta)

[maiden furnace]

no i dont know thecos (pi+theta)=-cos(theta)

[tardy ridge]

you could use that and change cos(7pi/6) into -cos(pi/6)

oh it's because of the unit circle

do you know the unit circle

[maiden furnace]

yup

[tardy ridge]

do you see how for quadarants 2 and 3 cosine is negative

because it's adj / hypo

[maiden furnace]

wait

is that just a refence angle?

[tardy ridge]

and adj is negative

what?

[maiden furnace]

is that just a reference angle?

[tardy ridge]

theta?

[maiden furnace]

well u get theta bar i believe

unless im just confusing my slef

[tardy ridge]

theta is a variable angle

if that's what you mean

theta is usally a variable referring to angles

and cos pi/6 is cos 30 which is sqrt3/2

that's also on the unit circle

since we are taking the arcos of -sqrt3/2 it's in the 2 or 3rd quadarant

but by the restriction it can only be the second one

which would be 180-30 degrees

150 degrees

did you know that cos (180-theta) also equals -cos theta?

[maiden furnace]

nope i didnt know that

[tardy ridge]

well it'll be hard to explain

[maiden furnace]

well looks liek this math teacher missed out on alot of information in that case

[unique hill]

have u learnt ASTC?

[maiden furnace]

yeah

[unique hill]

pretty sure ASTC covers that

[slate oracle]

Can anyone help with this..?\

[tardy ridge]

draw a triangle

[gaunt bay]

cant figure out how to start this

if someone can pm me. i know that i should use tan but i keep getting a negative number and not a big enough number

[unique hill]

looks like the channel's taken

@slate oracle draw a diagram of the building and the shadow

[slate oracle]

@unique hill

Sorry if it's messy

[unique hill]

nice aight

now what is the question asking?

[slate oracle]

For the measure(in feet) of the shadow and top of building.

[unique hill]

angle between the end of shadow and vertical side of building from the person

[slate oracle]

Yes.

[unique hill]

but yes

now what would be the appropriate trig ratio to use here?

[slate oracle]

Tangent..?

[unique hill]

that's correct

[slate oracle]

So then, tantheta= 100/230.. correct?

[unique hill]

yes

[slate oracle]

And then we would move tan to the other side making the equation theta=tan^(-1)100/230

Right?

[unique hill]

yep

[slate oracle]

So then, we would find the degree.. which is.. ~23.

Right..?

Would be estimate the degree or not?

[unique hill]

if it's rounded to one decimal place, it's 23.5 degrees

[slate oracle]

Yeah yeah, 23.498...

So then, what would be do afterwards?

[unique hill]

it says to the nearest degree

[slate oracle]

Oh, sorry.

[unique hill]

round to nearest degree

[slate oracle]

So it's 25 degrees..?

[unique hill]

so that would be 23 degrees

nono the 49856.... after the decimal place rounds down

[slate oracle]

You said it's 23.5 which is 24.

[unique hill]

no that's if its rounded to one decimal place

[slate oracle]

Okok

So what would we do now that we know the degree is 23?

[unique hill]

that's it, you've found the angle

[slate oracle]

I'm stuck here..

WHat?

[unique hill]

yeah that's the answer

[slate oracle]

oH.

Thanks for helping, Night

[unique hill]

np

[slate oracle]

Help..?

I drew this up.

[unique hill]

just make sure the 47 degree angle is clear

[slate oracle]

Oh ok

Can you verify my steps?

[unique hill]

your diagram looks correct

[slate oracle]

I would use sin, which leads it to being sin35=x/59 which would make x=(sin35)(59). Then x would be 33.84 or ~34 meters.

Ye?

[unique hill]

uh not sine

[slate oracle]

Then what..?

[unique hill]

are u aware of the trig ratios?

for sine, cos and tan?

[slate oracle]

sohcahtoa?

[unique hill]

yes

sine is opp/hyp

but 59 is none of those

[slate oracle]

cosine?

[unique hill]

ok first, dont guess

[slate oracle]

ok

[unique hill]

what's the trig ratio for cosine?

[slate oracle]

adj/hyp

And since we're finding for oppo, that's why I chose cos..

[unique hill]

where's the 'opp' in cos?

you said that it's adj/hyp

[slate oracle]

Uh. I don't know.

[unique hill]

ok you're trying to find the height of the building right? excluding the radio tower

[slate oracle]

Yeah, which would be considered the opposite.

[unique hill]

mhm, and you are given the distance of the person to the building, which is adjacent

[slate oracle]

Yeah

[unique hill]

you need to find the opposite and you are given the angle and adjacent

so what trig ratios would u use?

[slate oracle]

We would use tangent because that's the ratio with oppo and adj?

oppo/adj*

[unique hill]

yes correct

so what does tan 35 equal?

[slate oracle]

It's 0.46..

7*

[unique hill]

uh okay, first tan 35 = opp/adjacent correct?

[slate oracle]

so (tan35)(59) is 41.31 or ~41

wait

MB

[unique hill]

you follow?

[slate oracle]

Yeah

Wait.

[unique hill]

the tan ratio is opp/adj, and you are given the adj

[slate oracle]

Yeah but not the oppo.

[unique hill]

$tan{35}=\frac{height of building}{59}$

[obsidian monolith]

Nightingale:

[slate oracle]

Yeah

[unique hill]

ignore the spacing

do u get that?

[slate oracle]

Yes.

Then we would seperate the heightofbuilding

[unique hill]

yes, you make the height the subject

[slate oracle]

So multiply everything by 59

[viscid thistle]

hey does anybody know a simple way to solve this problem?

[unique hill]

@viscid thistle this channel's taken

[slate oracle]

Making the equation (tan35)(59)

[viscid thistle]

?

[gaunt bay]

@viscid thistle im waiting for someone to help me next

[slate oracle]

(tan35)(59)=~41

[unique hill]

@slate oracle yes, that's correct

yep looks good

[slate oracle]

So then what would be doo

[unique hill]

you have just found the height of the building, excluding the height of radio tower

[slate oracle]

Subtraction?

So do we find the second part now?

[unique hill]

the second part is find the height of building and the radio tower together

[slate oracle]

Ok