precalculus

[harsh cipher]

okay, thanks Ann, as always

[viscid thistle]

Quick question

Like for trig powers

Can somebody show me how I reduce cos^4(x)

[willow bear]

wdym reduce

[viscid thistle]

Write in terms of cosx

@willow bear

I’m stuck

[willow bear]

,rcw

[willow bear]

,,,,,,, what even

[viscid thistle]

Write the cos^4 x in terms of cosx

I uhh

Got confused

[willow bear]

"in terms of cos(x)" is vague

[viscid thistle]

That’s all it says

[willow bear]

do you have the exact problem statement

[viscid thistle]

That’s it verbatim

[willow bear]

wow ok like

[viscid thistle]

I have like half a solution

[willow bear]

okay the smartass answer would be "(cos(x))^4

[viscid thistle]

From class

[willow bear]

i mean ok like... you can use the formula $\cos^2(t) = \frac12(1 + \cos(2t))$

[obsidian monolith]

Ann:

[willow bear]

twice

[viscid thistle]

Lemme try

1 sec

Wait what’s that formula

Half angle

[willow bear]

yes

[viscid thistle]

Uhh

Then

That’s the right one

@willow bear

[willow bear]

hngh

neck: broken

anyway now use the same formula on cos^2(2x)

[viscid thistle]

0.o

Ok 1 sec

Then?

@willow bear

[willow bear]

now simplify the fuck out of this

[viscid thistle]

@willow bear I’m actually stuck now

[willow bear]

ugh

please

stop breaking my neck

,rcw

[viscid thistle]

My back hurts

[willow bear]

i mean...

wdym you're stuck lol

you're done

[viscid thistle]

No

She said u can simplify more

In class I think

[willow bear]

no, you're done.

i'd like to hear exactly what your teacher said

[hard hornet]

Maybe possibly decompose it into sin x and cos x only

[viscid thistle]

Like u can pull out something I know the fraction is supposed to be 1/16

[willow bear]

how do you know that

bc like... it's not

[viscid thistle]

She showed us the answer

But I only wrote down some of it

And I lost the paper

:,(( tomorrows test gonna suck ass

[viscid thistle]

So I’m right?

I swear she fucking showed us something different

[uncut mulch]

as mentioned earlier, the way the question was presented was extremely weird

and we have nfi what it actually wants

[viscid thistle]

it wants it in the for of cosx

There other problems which ask for it in cosx or sinx (different questions tho)

I remember there being a tan3x express in terms of tanx

@uncut mulch what did u use to solve it

Like what program

[uncut mulch]

as mentioned earlier cos^4(x) is already in terms of cos(x) so the question doesn't really make sense

and what you reached is in terms of cos(ax), cos(bx)

[viscid thistle]

Tomorrow’s test gonna be fun

What did u use to solve that

[uncut mulch]

wolfram

[viscid thistle]

Ty

I can figure out the tan problem now

Is it alternate representation or alternate forms?

@uncut mulch

Where is alternate forms

Wtf is a derivative

Is that the slope

[rigid sun]

Sure

[rigid beacon]

The derivative of some function f(x) can notated as f'(x)

At a given x value c

f'(c) = the slope of f(x) at x = c

[obtuse stirrup]

uh ignore the align environment stuff, its just latex formatting

how can I simplify this

does it just want me to solve n(5)+(2n(n-1))/2 >165165

[willow bear]

=

but yes

[obtuse stirrup]

So the answer should be n = 405 because it has to be an integer

@willow bear thanks

[mystic oasis]

so what did i do wrong here? why is the limit (n -> infinity) = 1?

[serene heath]

who said it was 1

[willow bear]

[mystic oasis]

its in the solutions my teacher gave me and wolfram alpha said it too but i dont have the pro version to see the steps :(

[serene heath]

WA said it was 1?

[mystic oasis]

yes

[serene heath]

doubt

[willow bear]

are you sure that's a minus sign and not a plus sign between the roots

[mystic oasis]

ill check

[willow bear]

[serene heath]

thatll make sense

should be a plus then

thats how u get 1

[mystic oasis]

ah yes

that was the initial term

by expanding with the same but instead of minus a plus

[willow bear]

$(a-b)^2 \neq a^2 - b^2$

[mystic oasis]

can someone explain why you would expand with a plus instead of a minus

[obsidian monolith]

Ann:

[mystic oasis]

oh

i see

yes

thanks lol

[mystic oasis]

where can i find a summary on the ]0;∞) form of writing

and ]2;4[ for exmaple

[stuck lark]

https://en.wikipedia.org/wiki/Interval_(mathematics)#Notations_for_intervals

[mystic oasis]

thanks i lack the vocabulary and its frustrating to google

[viscid thistle]

7 = x^2 + 12x + 34

How do I solve this?

[clever inlet]

What have you tried?

[viscid thistle]

I've subtracted 7 to 34 so...
0= x^2+12+27 and now and factoring

[clever inlet]

Yeah

That works

[viscid thistle]

Oh, amazing

[clever inlet]

So it factors into (x + 9)(x + 3) = 0

[viscid thistle]

What about here?

[stuck lark]

,rccw

[clever inlet]

Theres a common factor between the two terms

[clever inlet]

What is it?

[clever inlet]

I mean

Theres a common factor between the two terms

Just factor that out

[viscid thistle]

I'm bad with that

[clever inlet]

?

[viscid thistle]

Like, factoring

Idk, the common factor is 1?

[clever inlet]

You have an x in both terms don't you?

[viscid thistle]

I have an x in the # 10

[clever inlet]

And you have an x in the first term

[viscid thistle]

Ah yeah

[clever inlet]

Factor that out

[viscid thistle]

The x^2?????!!!!!

[clever inlet]

Factor out x

[viscid thistle]

Idk how to factor that!

[clever inlet]

x multiplied by what gives x^2?

[viscid thistle]

By it self...

[clever inlet]

Yes

x multiplied by what gives -10x?

[viscid thistle]

-10

[clever inlet]

Yes

So

x(x - 10) = 0

[viscid thistle]

Can I do it like this?

[clever inlet]

Well, considering it's wrong

No

[viscid thistle]

[clever inlet]

This would imply x = 1 is a solution

Which it clearly isn't

[viscid thistle]

What about this?

[clever inlet]

Ok, if you want to do it the long way

What two numbers add to -10 and multiply to 0?

[viscid thistle]

0 × -10= 0
0+ (-10)= -10

[clever inlet]

So 0 and -10

(x - 0)(x - 10) = 0

[viscid thistle]

So my factor tree would look like this?

[clever inlet]

Those are not factors

Those are what sums to -10

But either way, you should learn how to factor common factors in general

[viscid thistle]

Yes...

[clever inlet]

Just rote learning methods is not useful if you don't understand why they work

[viscid thistle]

:// I'll ask my teacher but now I really need to get this done

Can you help me here?

Not 3, but 2

-26 doesn't have more possibilities

[clever inlet]

Why subtract 15

Is that not a 16

[viscid thistle]

That what I did here

[clever inlet]

Is that not a 16 on the LHS?

[viscid thistle]

No, 15

[clever inlet]

Then it's not factorable over rational numbers

[viscid thistle]

What should I do then? No solution?

[clever inlet]

Quadratic formula or complete the square

It would have rational solutions if the LHS was 16

[clever inlet]

Then it has no rational solutions

[viscid thistle]

Noice!

I'll leave space just in case my teacher wants me to do something with it

[clever inlet]

Ok

[viscid thistle]

I didn't do algebra 2, that's why I struggle with simple stuff

[clever inlet]

You should probably revise algebra 2

[viscid thistle]

I'll talk with my teacher about what we can do

Stock again

[clever inlet]

Has no rational solutions

Either use the quadratic formula, or complete the square

[viscid thistle]

I fell like he would want me to do something

Yeah

Can you teach me?

Or should I watch a video?

Or should I wait for him to teach me :/?

[clever inlet]

You should learn it now

[viscid thistle]

I watched a video and I don't know how to continue

The exercise wants me to give a result in order pairs

[clever inlet]

That was quick

Good work

[viscid thistle]

Yeah, but I don't know why am doing what am doing xD

I just followed the template

[clever inlet]

It's completing the square

You want to get to a point where you have

(something)^2 = numbers

So you can square root both sides to solve

[viscid thistle]

Oh, okay

[viscid thistle]

Is this okay?

[sturdy haven]

How do I find p given a conic equation of a parabola

[clever inlet]

Example?

[sturdy haven]

(x-3)^2=4(y-5) so how do i know what p is

[clever inlet]

Do you have any more information?

[sturdy haven]

What if it looks like this

How do i find p

[clever inlet]

Well the right hand side is just

4(1)(y-5)

So 1

[sturdy haven]

Uhhh

What

Look in a regular equation it is just a=1/4p

But idk about this kind

[clever inlet]

?

[sturdy haven]

In an equation like a(x-h)^2+k

It is easy to find p

What is the formula for a conic equation

[clever inlet]

Well

If you have

y = a(x - h)^2 + k

Then you get

[sturdy haven]

a=1/4p

[clever inlet]

4(1/4a)(y - k) = (x-h)^2

Which is consistent with what I already said

[sturdy haven]

Wait so it is refused

[clever inlet]

??

[sturdy haven]

Reversed

It was 1/4p now it is 1/4a

[clever inlet]

a = 1/(4p) <=> p = 1/(4a) no?

[sturdy haven]

I am so confused omg

[clever inlet]

Parentheses for clarity

[sturdy haven]

So answer me this

two questions

In a conic equation of a parabola

WHERE IS a

AND WHAT DOES P EQUAL

[clever inlet]

Stop with the caps

It's literally just algebra

Which I have already shown

Scroll up

[fluid shore]

earlten, have you done your basics first

you were learning about differentiation just about a week ago

[slate tree]

how do i solve for a secant of something

such as

secant(theta)=12/4

[clever inlet]

Well

What can we rewrite secant(theta) as?

[slate tree]

ik that cos is the reciprical

[clever inlet]

So 1/cos(theta) = 12/4 right?

[slate tree]

yes

[clever inlet]

Take reciprocal of both sides

[slate tree]

yeah and solve

[clever inlet]

cos(theta) = 4/12 = 1/3

[viscid thistle]

What’s cooking?

[slate tree]

thanks for the help much appreciated

[viscid thistle]

mmmM yes trigonometric functions...

[long pond]

Trig is boring

[slate tree]

how would i do cosecant(pie/3) without a calculator

[clever inlet]

Well

[long pond]

*pi

[clever inlet]

What can you rewrite cosecant as?

[slate tree]

1/sin

but i need to solve in terms of radians i believe

[clever inlet]

Yes

1/sin(pi/3)

[uncut mulch]

pi not pie

[clever inlet]

What's sin(pi/3)?

[slate tree]

sorry typo

dont know they want us to solve without calc

[clever inlet]

It has an exact value

[uncut mulch]

^ special angles

[long pond]

Just use the approximation sin(x)\approx x

[clever inlet]

I believe you can derive this by cutting down the perpendicular of an equilateral triangle

[uncut mulch]

and/or do you know how to convert radians to degreens

[slate tree]

i still dont erally know

how would i do cos(pi/4)

without a calculator

[clever inlet]

You should learn your special angles

[long pond]

Just memorize

or better yet,

picture a unit circle in your head

[slate tree]

the answer comes out to be square root (2) / 2

we havent gotten that far

but is it possible to do in terms of 2pi

[clever inlet]

?

What do you mean?

[slate tree]

it says to evaluate cos(pi/4) without a calculator and i have no idea how to do it

the answer comes out to be Rooot(2)/

[clever inlet]

You need to learn your special angles

[slate tree]

2

is that hte only way

[clever inlet]

These are special angles

[slate tree]

alright

[clever inlet]

You need to learn them somehow

[slate tree]

alright

[clever inlet]

Either through deriving or memorising

[slate tree]

were talking about the unit cirle tomorrow so guess ill learn about it then

thanks for the help

[clever inlet]

np

[uncut mulch]

you can derive them from isosceles and equilateral triangles

and that pi (radians) = 180°

[slate tree]

thanks

[flint kite]

x|log2x|3log2(x-4)+1
list three points
X= 1,2,4

[clever inlet]

🤔

[uncut mulch]

don't double post

and don't dm me

[flint kite]

sorry bro

[uncut mulch]

and that doesn't make much sense

[long pond]

That’s a warnable offense

read the rule

s

[flint kite]

i just wanna sleep

@uncut mulch its a graph question

[uncut mulch]

so y= ?

[flint kite]

yea

[uncut mulch]

what about those x values?

[long pond]

Why not just plug in the points

[uncut mulch]

are those where they want the points?

[flint kite]

yea

X|logbase|logequation

[uncut mulch]

do you have a pic?

that notation is horrid

[long pond]

^

[flint kite]

no

[long pond]

Let me try to format

[uncut mulch]

log_(base) arg

[long pond]

$y=x|\log_2(x)|\cdot 3\log_2(x-4) +1$

?

[obsidian monolith]

IAmJon:

[long pond]

This what you mean?

[flint kite]

no

my teacher wants it in a specific way

[uncut mulch]

write is as least ambiguously as possible

[long pond]

Don’t just say “no”

[uncut mulch]

^ say how its different

[long pond]

tell me how to fix it then, so we are on the same page

[slate tree]

how do i find sec(theta) and csc (theta) if cot (theta) = -4/3 and cos(theta) < 0

[uncut mulch]

determine which quadrant theta is in

[slate tree]

what do you mean

oh would it be 4th

[flint kite]

@long pond lemme just try to do this myslef because this stuff is kind of on a test tmr

[uncut mulch]

no not 4th

[slate tree]

i mean 2 or 3

[uncut mulch]

$\text{use log_(base) (arg) to represent } \log_{base} (arg)$

[flint kite]

sorry and thanks for ur efforts

[long pond]

Ok

[slate tree]

it would be the 2 or 3rd right

cause those arnt cos

[long pond]

My teacher would have a brain aneurysm if i wrote that notation on a test, just saying

[slate tree]

then what do i do

@uncut mulch

[uncut mulch]

well cos is negative in 2rd or 3rd,
where is cot negative?

[slate tree]

i dont really know

[uncut mulch]

do you know what cot is?

[slate tree]

recip of tan

[uncut mulch]

and where is tan negative?

[slate tree]

2 and 4

so would it be the overlap

[uncut mulch]

would cot be neg when tan is neg?

[slate tree]

yes indeed

so then what do i do

[uncut mulch]

so which quadrant are you in?

[slate tree]

2

[uncut mulch]

ok good.

[slate tree]

i dont really know what to do from there

heres the question again if you need it

v

how do i find sec(theta) and csc (theta) if cot (theta) = -4/3 and cos(theta) < 0

[uncut mulch]

you can draw the triangle in quadrant 2

and figure out the ratios from there

[slate tree]

would sec be -5/3

and csc be 5/4

[uncut mulch]

re-check your ratios

[slate tree]

isnt sec hyp /x

and csc hyp/y

[uncut mulch]

yes

but from cot(theta) = (-4)/3, which is which

[slate tree]

oh i see

x=4

y=3

[uncut mulch]

well x = -4

[slate tree]

yeah

[fluid kayak]

How does one go about solving cos3x = 1

Interval 0<=x<2π

[clever inlet]

I would adjust the interval first

[fluid kayak]

Ok thatd be between 0 to 6π then right?

[clever inlet]

Yeah

Good

You can inverse cos both sides

But like

You should know the cos of what gives 1

[fluid kayak]

Ye 0

Er

[clever inlet]

Yep

So 3x = 0 right?

[fluid kayak]

Yea

[clever inlet]

And so x = 0 is a solution

When do more solutions occur?

Wait

Don't divide by 3 yet

[fluid kayak]

Oh okay

[clever inlet]

3x = 0

When's the next solution?

When 3x = ?

[fluid kayak]

Would that be 2π

[clever inlet]

Yep

Keep going

[fluid kayak]

Then 4π?

[clever inlet]

Yes

And one more

[fluid kayak]

5π?

[clever inlet]

Not quite

You can keep adding 2pi for more solutions

[fluid kayak]

Ah I see

Alright I get it now, thanks

[clever inlet]

[trim fable]

hi

@fluid shore hey u there? 😛

coz

i have a question

what does it want me to do

well

[fluid shore]

Uh show the full statement of the problem

[trim fable]

im comparing too variables

oh its an assignment

thats what a part of it is

so

I have that so this is being done on a data software and

what do i even do

write about the mean and median?

what can u even say about it?

[fluid shore]

Hmm I can't say much about this cos I've not really done that sort of thing before. Uh take this to #probability-statistics and you might get more help there from others who know about this more than I do.

[trim fable]

ok

nah ill finish it tom

its late so

[winged cloak]

They say that to calculate the derivate of F in x you just have to know the derivate of in 0...why is it so?

[hard hornet]

ur gonna have to give a lil more information

[willow bear]

what's f

[winged cloak]

So we have the function f(x) =a^x

And we want to derivate ir

It

F is the function

And we use the average velocity between x and x+h

[stuck lark]

is french your first language?

[winged cloak]

And then we use the limit of h->0 and we get
f'(x) = a^x f'(0)

@stuck lark yep

[stuck lark]

ok ann speaks french, so perhaps you two can converse in french so she can decipher what you're doing

[winged cloak]

Who s Ann?

[stuck lark]

discount Tuong

[winged cloak]

Oh i see

Thanks

[willow bear]

moi

[winged cloak]

Salut jpx ptetre t envoyer la feuille ?

[willow bear]

envoie la ici

[winged cloak]

Att jvais la refaire

[willow bear]

,rccw

[willow bear]

ok

alors qu'est-ce qui te trouble exactement

[winged cloak]

J ai compris le calcul

Mais jvois pas comment on est passé de la vitesse moyenne à f'(x)

[willow bear]

la variation moyenne

[winged cloak]

Oui c ça variation pardon

[willow bear]

la limite de V(x, x+h) lorsque h tend vers 0 c'est la DÉFINITION de f'(x)

littéralement

[winged cloak]

Oui hahaha et pk on dit que il suffit de connaitre f'(0)?

[willow bear]

parce qu'on a exprimé f'(x) pour tous les points x en termes de f'(0)

si on connaît f'(0) on connaît f' partout

en utilisant la formule obtenue dans ta photo

[winged cloak]

OK merci de ton temps

[mystic oasis]

why is that 3/4?

[rigid beacon]

@mystic oasis you see how when you factor it

You can cancel a factor from the top and bottom of the fraction?

[mystic oasis]

ohh

[rigid beacon]

So you cancel that out

[mystic oasis]

i see

🤦‍♂️

[rigid beacon]

And then directly sub in

[mystic oasis]

thanks bro

sometimes im blind

[hardy cave]

@viscid thistle I watched this to learn completing the square

https://www.youtube.com/watch?v=C206SNAXDGE

[deft peak]

<@&286206848099549185> I need to know how to calculate the bi-weekly payment for a car.

Does anyone know?

I'll tell you everything you need to know but I just can't figure out the formula.

[viscid thistle]

Do you know how to calculate the weekly payment for a car? @deft peak

Also next time send your question then ping helpers after 15m

[deft peak]

Oh I apologize.

And no I do not.

what formula should I use?

The total cost of the car is $28,585. The interest rate is 1.99%/a bi-weekly for 4 years.

<@&286206848099549185> Not trying to spam. Class ends in 20 mins 😛

[viscid thistle]

@deft peak if you're in class get off your phone and ask your teacher

[plain lake]

i'll help you after class ends if you wnt

want

[deft peak]

I'd ask my teacher if it wasn't a sub.

It's a study period for exams

[viscid thistle]

You have peers to help you ;-;

[plain lake]

ask people in your class then

[viscid thistle]

After class I'm down to help

[deft peak]

Ok. Thanks.

Do you guys think I could get some help with the question like later tonight?

What time zone are you guys on?

Either way I'll be around at 7:30ish EST.

I appreciate the coming help!

[green zenith]

I need help understanding this question

[patent beacon]

Same problem but with a picture

Important is the 2D interpretation to the left

@green zenith

[green zenith]

Okay ill take a look

I still dont undwrstand how to put this as a function

I tried relating my volume with the surface area

[viscid thistle]

A square has its length increased by 4 meters and its width decreased by 5 meters. The resulting rectangle has an area of 132 meters. What is the original perimeter?

[patent beacon]

Np. So the picture shows a 2" cut out, but you actually want an x" cut out @green zenith

The height of the box is x". What's the bottom side length?

[trim fable]

hi

remember me 😛

[patent beacon]

Yeah, hi!

[trim fable]

heyy!!

[patent beacon]

If you call the bottom side length y, you can solve for it in terms of x.

[trim fable]

oh so i have a question 😛

[green zenith]

Idk. I dont have any value for any side

All i know is 2 of the 3 sides are equal

[trim fable]

did u ever do a data statistics cpt?

im almost done but

im confused about some parts

[patent beacon]

So call the height x, call the bottom side length y. Can you express the surface area of the box in terms of x and y?

[green zenith]

I could try

S=4xy+2x^2

Is that correct?

[patent beacon]

I think we're mixing up x and y, I was thinking
S = 4xy + y²

[green zenith]

Oh yeah

[patent beacon]

Each side is xy, the bottom is y²

[green zenith]

I definitely did

[patent beacon]

But we know S. Your question states S = 9

Wait hold up I think I'm mixing up what the question wants. I think the question means x is the bottom side

[green zenith]

I think the x is the height

Since its a square bottom

It is not specified anyway

[patent beacon]

The question would be easier if it were x = bottom side length, so I think that's what they want oop

If that's the interpretation, then:
S = 4xy + x²

But that also means:
9 = 4xy + x²

[green zenith]

Wait

Why is S=9?

[patent beacon]

Finally, we can solve for y:
y = (9 - x²) / 4x

[green zenith]

9 is my volume

[patent beacon]

OH MB that's volume

Wow I'm really screwing with you here. Okay okay let's go back

[green zenith]

It is fine

[patent beacon]

Complete restart. Let x be the bottom side length. Let y be the height:
V = x²y
S = 4xy + x²

Cool with both of those?

[green zenith]

Yeah

[patent beacon]

Now V = 9. Solving for y in the first equation, we got:
y = 9/x²

Plugging that into the bottom:
S = 4x(9/x²) + x²

And simplifying:
S = 36/x + x²

[green zenith]

Wait hold on

Shouldnt it be 4xy+2x^2?

[patent beacon]

It's open top. That is, it has no lid

[green zenith]

Oooooh

So it has no surface

Got it

So we only 5 sides

Well i think thats it

Thank you for helping!

[trim fable]

@patent beacon 😛

u there?

Or no 😮

[serene heath]

what's the question

[trim fable]

oh

would i have to explain those?

is that what its asking

its part of the rubric for my assignment

[twilit shadow]

May I have some help please?
This is my problem:

cotX=sin2X/1-cos2X and I have to verify that it is true

Here’s what I have to work with

And this

[serene heath]

that's a lot of repeated identities

anyway what have you tried so far

[twilit shadow]

Yes it is

Well, I tried making cotX into cosX/sinX but that didn’t get me anywhere

I have other work that basically just took me back to where I started but I’ve already erased that

[serene heath]

that's a good first step

now try multiplying top and bottom of that fraction by 2sinx

[twilit shadow]

The bottom becomes 2sin^2X, correct?

[serene heath]

indeed

[twilit shadow]

I can make sin2X on the other side into 2sinXcosX

Should I do that?

Oh yeah that’s the other side

[serene heath]

stick to working with one side

[twilit shadow]

Okay

This is what I have now

Sorry if it’s terrible quality

Can I reduce the twos on the left?

To just get rid of them

Oh yeah and then I can make sin^2X into 1-cos^2X

Oh but that’s not the same as the other side

Poop

[serene heath]

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

[twilit shadow]

Oh woah

Okay

[uncut mulch]

best not to structure it like that

do LHS =
=...
=...
=RHS when you actually reach it

[twilit shadow]

Thank you very much!

I got it now

[twilit shadow]

Here’s another problem I have

I’ve been working at it for a while

I’ve got the sinA/2 and cosA/2 made into the radical things

And I’ve tried many things, like separating the fractions, dividing by a negative to try and get something I can work with, multiplying top and bottom by sin or cos, and nothing’s been working

I even tried multiplying out the square

It all led to dead ends

[patent beacon]

@twilit shadow
Let's say you square the right side. What does it look like after?

[twilit shadow]

I’ll go through that and show you

That’s what I got

[uncut mulch]

that was definitely not what was recommended

square the rhs immediately

don't do any extra stuff

[twilit shadow]

Square the what?

Sorry, I’m very dumb and already have a headache

[uncut mulch]

$\left(\sin(\frac A2) - \cos(\frac A2)\right)^2 = \ ?$

[obsidian monolith]

ramonov:

[twilit shadow]

I have to square that first?

[clever inlet]

You'll get a much nicer result

[uncut mulch]

you can start from whichever side is more convenient

[twilit shadow]

Okay

[uncut mulch]

you can also start from the lhs but it isn't as obvious which identites to apply

[twilit shadow]

Okay I’ve done that

Now do I make them into the radicle things?

[uncut mulch]

no

[twilit shadow]

I don’t really know what to call that

[uncut mulch]

forget about the radicals for this question

what do you have after squaring

[clever inlet]

There's a special identity

[uncut mulch]

don't overthink it

[twilit shadow]

Sin^2 + cos^2?

[uncut mulch]

just that?

[twilit shadow]

And -2sincos

[uncut mulch]

don't leave out the arguments

[twilit shadow]

I didn’t I just don’t want to type them

Oh it’s all over 2

[uncut mulch]

huh?

[twilit shadow]

Wait no

Nevermind

[uncut mulch]

this is exactly why you should type them in

especially when the quesion involves both A/2 and A

[twilit shadow]

sin^2(A/2)-2sin(A/2)cos(A/2)+cos^2(A/2)

[uncut mulch]

ok. now what identites can you apply?

[twilit shadow]

I can do sin^2(A/2)+cos^2(A/2)=1
And
sin2(A/2)=2sin(A/2)cos(A/2)

I think the second one is good to use

[uncut mulch]

re-check what you typed for the second one

[twilit shadow]

Was that wrong?

[uncut mulch]

the left side of it

[twilit shadow]

Yes that becomes just sinA, right?

[uncut mulch]

not becomes

supposed to be

sin2(A/2) is wtf notation

[twilit shadow]

Oh

[uncut mulch]

sin2 is not a function

[twilit shadow]

Okay

[uncut mulch]

you would've at least needed
sin(2 (A/2)) = sin(A)

[twilit shadow]

Well yeah that’s how I was thinking

[uncut mulch]

but yes, apply those 2 identites

[twilit shadow]

Oh hey now it’s the same as the left side

Thank you very much!

You relieved a ton of stress

This is why I love math, because it just works. Even when I just can’t get it and it stresses me out, I love the end result when it’s finally solved

And now there’s only one more to do for my homework

If I have cotA-1/cotA+1 I can make that into tanA+1/tanA-1, right?

[uncut mulch]

that doesn't look right,
can you put parentheses where appropriate?

,w (cot x-1)/(cot x+1 ) = (tan x+1)/(tan x-1)

[obsidian monolith]

Display results online and refine query

Requested by ramonov#0921

[lethal oracle]

Anyone there to help with something?

[clever inlet]

?

[lethal oracle]

I need to sketch this

How do I figure out the vertices and foci

[clever inlet]

Is this supposed to be an equation?

=1?

[lethal oracle]

Ya sorry

[clever inlet]

Well

What shape is the curve firstly

[lethal oracle]

It’s an ellipse

[clever inlet]

Yes

Compared to a circle

[lethal oracle]

I know how to figure the vertices and foci our when the center is 0,0

[clever inlet]

How much is it stretched vertically?

[lethal oracle]

That’s what I’m confused about

Is it 2

[clever inlet]

Yes

So you can go 2 units up from the centre: (-3, 3)

And 2 units down from the centre: (-3, -1)

[lethal oracle]

Where did you get the center (-3,3)

[clever inlet]

It's not

The center is (-3, 1)

[lethal oracle]

You typed (-3,3)

[clever inlet]

Yes

If you go 2 units up from the center

[lethal oracle]

Oh I see what you mean

So would this be correct

[clever inlet]

Looks about correct

Label your points though

[lethal oracle]

How do I find the foci

I need to use c^2=a^2-b^2

Right?

[clever inlet]

Sure

[lethal oracle]

And do I add that to the x values of my two vertices?

[clever inlet]

From the center

Not the vertices

[lethal oracle]

So would these be my two foci then

[clever inlet]

Hmm

How did you get 3sqrt(2)?

[lethal oracle]

Sorry 2sqrt(3)

?

c^2=16-4

[clever inlet]

Yeah, that seems fine

[lethal oracle]

Alright

Thanks so much for the help

[clever inlet]

np

[viscid thistle]

I need help

STAT

I have been trying to understand this all night

And I have a quiz tommorw

problem:

e^2x - e^x -6 = 0

If anyone could help before I fall asleep of sleep deprivation that would be nice

thats not the only problem btw

[uncut mulch]

assuming that first term is e^(2x)
that is a quadratic in e^x

[undone pawn]

i. e, to solve, substitute e^x=y and proceed

[uncut mulch]

although not necessary

but if its your first few times dealing with these, it will help

[viscid thistle]

Number 4 is the problem

[undone pawn]

ok ramonov just told you what to do

[uncut mulch]

do you understand the comments we made?

[viscid thistle]

Okay so make a variable that represents e^x and plug it in and work it out?

[lilac pier]

not necessary, but if you want, you can use e^x = y to make things easier for you

[viscid thistle]

Okay

[lilac pier]

once you solve for y, you'll be able to solve for x using y=e^x

[viscid thistle]

I'm guessing negative values are no solutions

For y values going into e^x

@lilac pier

[uncut mulch]

well no real ones

which are the ones they want

[viscid thistle]

What do u mean sorry

[lilac pier]

assuming the question is asking for real solutions, yeah you'll ignore negative values for y

[viscid thistle]

Okay thank you

[uncut mulch]

(you'll learn about that at higher math, don't worry too much about it now)

what did you end up getting?

[viscid thistle]

So wait I find that e^x = 3 is true but where do I go from there

[uncut mulch]

ln both sides

[viscid thistle]

What

[languid crane]

ln = the natural log base e

[uncut mulch]

oh noes

[viscid thistle]

So ln 3

[uncut mulch]

yes, x = ln(3)

[viscid thistle]

Which is approx 1.10

[uncut mulch]

just leave it as ln(3)

unless they ask you for an approximation

[viscid thistle]

okay

Having problems with this one

Number 3

[uncut mulch]

try expressing them as exponents with the same base

[viscid thistle]

How would I go about doing that

[lilac pier]

know your tables!

[viscid thistle]

Wait I'm just confused what ut saying is all

Sorry

[uncut mulch]

note that 9 and 27 are both powers of a certain number

[viscid thistle]

Both of 3

[uncut mulch]

yes

27 can be expressed as 3^(?)

[viscid thistle]

3

Lol

So write it as that

For that side I assume

[uncut mulch]

9 can be expressed as 3^2
hence 1/9 can be expressed as 3^(?)

[viscid thistle]

-2

[uncut mulch]

yes

[viscid thistle]

Ouhh

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

[uncut mulch]

parentheses

around the (x+3)

[viscid thistle]

so do you times the exponets inside the parentheses and out side like the (-2) * (x+3) ?

[uncut mulch]

yes

[viscid thistle]

okay I got a answer

-6/5

as x

[uncut mulch]

seems ok

[viscid thistle]

Number 8 holy shit I have no clue on . The only thing I know to start with is say that e^2x is = y

Past that this is the hardest question

And I don't know

maybe factor

vbut how

[undone pawn]

you dont know to solve quadratics?

[viscid thistle]

I do

[undone pawn]

this is same as the first question you asked except it's e^2x instead of e^x

then solve the quadratic lol

[viscid thistle]

okay LOL

[uncut mulch]

these all seem to involve tricks with exponents

[viscid thistle]

When you divide exponents does it just minus

Idk if If I. Explaining fully

am*

[uncut mulch]

show the expression show there's no confusion

[undone pawn]

$\frac{a^b} {a^c} = a^{b-c} $

[obsidian monolith]

FlynnXD:

[undone pawn]

this?

[viscid thistle]

yeah I guess lol

[undone pawn]

wdym you guess

do you not know what your own doubt is?

[willow bear]

"im not a math person"

[fluid shore]

Sometimes, people find it hard to express their own doubts about something if that thing is particularly confusing.

[viscid thistle]

@undone pawn https://cdn.discordapp.com/attachments/363224154469826562/667644866511110165/JPEG_20200117_022130.jpg
in problem 8 u said I could use quadratic equation
However I am not sure about my answer...

I plugged in

a = 3 b = -9 and c = -15

then I assumed u square it but I am not sure really keep in mind it is 2:38 am for me so i am not as confident in my math abilities or anything for that matter so I am sorry lol

[uncut mulch]

what exaclty are you squaring?

[viscid thistle]

I thought it had to do with the e^2x if that even makes sense

IDk really I am reallyt ired

Or do u just do it normally

[uncut mulch]

ah, but you shouldn't be doing any more squaring

you could also divide everything by 3 first to work with smaller numbers

plugging it into the QF will give you the solutions to
e^(2x)

and similar to be before, log both sides

(also remove the negative solution)

[winged ether]

how do i do (2-2i)(1+i)

[clever inlet]

what are you trying to do?

simplify?

[winged ether]

ya

[clever inlet]

just expand like normal

[winged ether]

multiplying complex numbers i guess

cuz the answer sheet says it equals 4 but i dont understand why

[clever inlet]

well

2(1 - i)(1 + i)

we can rewrite as that right?

[winged ether]

oh wait

im really dumb

lol

thx tho

[clever inlet]

np

[green zenith]

Im so confused with fig 1.67 here

It says in the book that sin a=12/180 and sin b=12/62.

How??

Am i to assume that if angle a have the opposite side value of 12, the hypotenuse is 180?

I dont even know if it is the same angle with the angle on Chicago

[vapid plaza]

well angle a is the same as the angle at the very top (the Chicago one)

If the seemingly parallel lines are really parallel cuz it didn’t specify

Oh wait it did (in the question)

So the two angles are corresponding angles

Same for angle b and the other one

[heady jewel]

vector algebra pset is what you should search

[grizzled totem]

I'm doing matrices:

I've completed the first 2

I want to know whether the WZ means to multiply

W x Z

[rigid beacon]

Yea typically 2 variables next to each other means multiply

XY = X•Y = X*Y = X × Y

[grizzled totem]

I did some research regarding the T symbol shown in question (iii)

T means transpose

Matrice V has 2 rows and 3 columns

Would it look like this after being transposed: -1 0
5 3
2 1

[stuck lark]

yes

[grizzled totem]

cheers

So for this

I'd need to multipy Matrice U by the transposed matrice V which is now this

[stuck lark]

yes

[safe kettle]

alright so

what's the domain and range of this parametric equation

how do i know what is is

what it is

i thought it's domain would be all real numbers from [-5, 3], but when i graphed it on desmos it was just excluding all real numbers from (-3, 1)

[prime prawn]

how did you get -5 to 3

[safe kettle]

cuz if you plug in pi/3 into 2sec(t) - 1 you get 2(2) - 1 = 3

[prime prawn]

and then

[safe kettle]

right?

and for the other one you plug in 2pi/3 into 2sec(t) -1

to get 2(-2) - 1 = -5 ??

i dont get it

[prime prawn]

2pi/3?

or pi/3

[safe kettle]

both

to get the domain right?

the max?

and min?

[prime prawn]

sorry i dont get your thought process, i can tell you mine though

find the range of sec(t) and when its undefined (from -1 to 1). so 2sec(t)-1 is undefined from 2(-1)-1 to 2(1)-1, therefore the domain (where x is defined) is all x but x from -3 to 1

now tanx has no range restrictions so y =4tant+3 has range of all real numbers

[safe kettle]

where did you get -1 to 1 from

im confsued

[prime prawn]

whats the range of secx

[safe kettle]

everything excluding -1 to 1?

[prime prawn]

yeah so whats the range of 2secx-1

[safe kettle]

ahh

i think

i understand

[prime prawn]

well tell me your understanding cause you dont wanna get this wrong in the future

[safe kettle]

wait

why are we getting the range of secx

[prime prawn]

so we can get the range of 2secx-1

if im confusing you, i mean the range of x = 2sect-1

x is the range of 2sect-1

but its the domain of the whole function

if it makes sense

[safe kettle]

i thought we had to plug in values on t

to find the range of 2sec(t)-1

[prime prawn]

you cant just plug in random values to find the range of 2sec(t)-1

[safe kettle]

am i stupid or something

but why is sec(x) excluding -1 to 1

[prime prawn]

hmm what is sex(x) equal to

1/cos(x) right?

surely you know the range of cosx

[safe kettle]

from -1 - 1

-1 to 1

[prime prawn]

so lets see

if cos(x) = -1, then sex(x) = -1

if cos(x) = 1, then sex(x) = 1

[safe kettle]

yea?

so why is it restricted

[prime prawn]

if cos(x) is 0, then secx is undefined

so if cosx is from (-1,0) then secx is from -1 to negative infinity

if cosx is from (0,1) then secx is from 1 to positive infinity

so overall, the range of secx is all x excluding (-1, 1)

[safe kettle]

how did we get negative infinity

i forgot the basics

oh

i understand

[safe kettle]

k thx bro

[viscid thistle]

boom

[zenith bough]

is e^-x = some kind of ln or log

i tried to find zeros for this problem and not know where to start

[stuck lark]

e^(-x) is an exponential

solving $2xe^{-x}-x^2e^{-x}=0$, one may start by factoring

[obsidian monolith]

RokettoJanpu:

[spare hull]

$2e^{-x}=xe^{-x}$

[obsidian monolith]

Angelium:

[spare hull]

$2=x$

[obsidian monolith]

Angelium:

[spare hull]

wow that was easy lmao

[blazing parrot]

you forgot x=0

[fallen thorn]

@spare hull you shouldn't divide by x... You should factor, otherwise you will lose solutions.

[spare hull]

yes i am aware

[viscid thistle]

let e^-x = c
2xc - x^2c = xc(2-x)=0
x=0 and x=2

[elfin cradle]

hi, so I don't understand ambiguous cases in ssa triangles

i've watched many videos, asked friends, read about it

i'm still really troubled on the cases where there are 2 triangles

i just completely don't understand

[patent beacon]

Depending on the info you have, there can be two triangles that satisfy your constraints

There are two different solutions to sin(θ) = a for some a. So the sin law can give two triangles

@elfin cradle

[verbal leaf]

Can someone help me understand how (-f(x))+(-g(x)) is -(f+g)(x)

I’m really tired but it’s not clicking

[willow bear]

$(-a) + (-b) = -(a+b)$

[obsidian monolith]

Ann:

[verbal leaf]

Ooooh right