precalculus

[willow bear]

i mean when θ=0 you get (0, -R) which is exactly what it should be at the starting position

[viscid thistle]

lol i don't get it

why should that be the starting position

i mean i know you put it there

idk this is a difficult concept for me

[willow bear]

at the start the wheel has yet to rotate so the angle of rotation is 0

idk how else to explain it

[viscid thistle]

wait

you said the wheel starts out at (0, R)

not (0, -R)

[willow bear]

the center of the wheel starts out at (0,R), but the position of the yellow spot relative to the center is (0,-R) at the start

[viscid thistle]

where is (0,R) in your diagram

would it just be the rightmost point?

[willow bear]

it's not explicitly marked bc the starting position of the wheel isn't explicitly marked

yknow maybe i'm not really explaining this well

[viscid thistle]

@willow bear idk it's also very new to me and i am very dumb either way so it doesn't help

parametric equations new i mean

[willow bear]

what have i told you about self deprecation

[viscid thistle]

@willow bear well it's true tho if you look at like terrance tao he knew this stuff at like 10 years old

[willow bear]

don't compare yourself to terry tao lmao

i mean ig if you wanna wallow in self pity and nothing else then go right ahead

like there's obv gonna be giants better than you

or me

[viscid thistle]

why not?

compare i mean

[willow bear]

it's not a good use of your time

to like. compare yourself to one particular math giant and always end up concluding that you're nothing compared to them

[viscid thistle]

well

how do i get better at parametric equations

they are so overwhelming

do you have any resources

[willow bear]

not that i can think of rn

[viscid thistle]

no hope

[willow bear]

??

[viscid thistle]

@willow bear i don't understand this stuff at all idk why

it's so weird to me

[odd helm]

I’m not sure where I went wrong how should I set this problem up?

[odd helm]

<@&286206848099549185>

[fleet yew]

@odd helm looks right

[odd helm]

But when I do tan(15) I get a decimal and if I divide 18810 by a decimal it’s going to get bigger

[fleet yew]

Are you in degree mode

Also it should be a decimal

The horizontal distance IS greater than the vertical distance

Because the angle is less than 45

[uncut mulch]

what's the issue if it's big?

also remember to convert units

[fleet yew]

Lmao didnt even catch that

I hate when they try to pull that

[odd helm]

Ah I see now thank you

[odd helm]

Where did I go wrong?

[rigid sun]

your angle is on the wrong side

when you're walking and someone tells you to look at something, are you looking strait at the ground and then tilting your head up?

obviously not

so when someone says, "look up!" or "look down!" it obviously means that you were looking forward to begin with

[fleet yew]

Angle of depression is the angle below the horizontal

[cedar cosmos]

$5x+1$

[obsidian monolith]

charlestang06:

[viscid thistle]

Hi I need help with this easy problem, I have no idea why its wrong

[uncut mulch]

8* ?

show work if that didn't clear things up

[viscid thistle]

Ohhhhh

I meant 2

Omg i wrote it in paper 2 but on the website 8

Fuck it happens to me everytime

I am blind

Thanks a lot ^^

[scarlet sky]

Hello

I was wondering if someone could help me on ellipses and hyperbolas ?

<@&286206848099549185>

[uncut mulch]

!15m

#❓how-to-get-help

Ask the actual math question first. If its unanswered after 15m THEN you can ping helpers once

[regal gorge]

Can anyone recommend study sites for precalc

I’m confident inThe classroom but keeping up on it by myself is hard

So I’m looking for a good alternative support

[patent beacon]

http://tutorial.math.lamar.edu/Classes/Alg/Alg.aspx

Paul's is seriously one of the best. Everything below calc is there

[harsh cipher]

Question

I'm asked to sketch Cos^2x + 2cosx -1 = 0.

If i put in the following equation in ti 83 calculator

cos(x)cos(x)+2cos(x)-1

i get the same graph as putting in the -1 in a new equation

it means the same thing right?

[patent beacon]

Maybe you were asked to sketch
cos²(x) + 2cos(x) - 1?

Including the = 0 fixes the value of x

[scarlet sky]

Wait why did my picture get deleted

[harsh cipher]

yes that's the equation

ok let me put in y2 = 0

no I'm getting this weird graph that has a straight line pi to 3pi/2

[patent beacon]

,w graph cos(x)^2 + 2cos(x) - 1

[obsidian monolith]

Display results online and refine query

Requested by Kaynex#7520

[harsh cipher]

yea I have that weird straight also

maybe I'm right

weird straight line but I guess that's because of the 2x

2cosx omg can't type today

[harsh cipher]

@patent beacon thanks

[keen sigil]

could someone please explain how i'd do a question like this?

A rectangular hyperbola has centre (0,0) and vertices on the y-axis. If (5,7) and (10,k) are points on the graph of the hyperbola, determine the value of k.

[obsidian monolith]

Yugito Abyss:

Yugito Abyss:

[ripe wasp]

and

[obsidian monolith]

Yugito Abyss:

[willow bear]

no

first off, bad tex.

second, even disregarding that, these conclusions are not valid

not at the surface, anyway.

[ripe wasp]

I see, and i was gonna question if those were right or not

[willow bear]

ok well

TECHNICALLY, for THIS PARTICULAR EQUATION, these conclusions actually do happen to be correct. HOWEVER, stating them like this is, at best, unjustified and obscure.

[ripe wasp]

I am here to myself ask WHy this works

cause i don't understand

its in one of my class notes

and i so confused why teacher did that

[willow bear]

your teacher did that???

[ripe wasp]

yeah

[willow bear]

excuse me what the fuck

i mean

can you like. take a picture or screenshot of your teacher's work

bc honestly that's not something i would even think to do even on a bad day

[ripe wasp]

Can't rlly its in my notes

and maybe its correct

i mean it does look absurd

but then again as you said it works out to be true

[willow bear]

i view it as an accident more than anything

[ripe wasp]

Well, i can neither confirm nor deny whether it was an accident or not

Oh and it is given that x lies in the range 0 to pie/2

[willow bear]

pi, not pie

[ripe wasp]

yeah my bad

pi

autocorrect

[ripe wasp]

Oh i understood why now

It is given that cos^6 x + sin^4 x = 1

[willow bear]

okay well

that's the exact opposite of what you had originally presented

[ripe wasp]

Hm actually i only presented the thing after this, so i was confused myself, sorry lol

[willow bear]

it'd help if you presented the problem in its entirety exactly as it was stated to you with all details included and everything written out verbatim and word-for-word

[ripe wasp]

I will do that from now on, thanks

[odd helm]

I’m not sure what I’m doing wrong I chose 15 and 13 as my answers because when I plugged them in for B I got a number greater than one so there can’t be any inverse sin so no solution

[uncut mulch]

,w solve 0 < 18sin(75deg)/x < 1

[obsidian monolith]

Display results online and refine query

Requested by ramonov#0921

[uncut mulch]

,w 9/2 * (sqrt(2) + sqrt(6))

[obsidian monolith]

Display results online and refine query

Requested by ramonov#0921

[uncut mulch]

those seem correct

[proud jetty]

https://gyazo.com/c80e231d59e3ac838974caa68e90cdb8.png

step 2: why isn't -3 multiplayed by -2

[uncut mulch]

because whoever did that is bad at math

[proud jetty]

00f that was on my schools website for answer key

how would u do it?

turing it into p(x)/q(x)

https://gyazo.com/ba4214160e40a2c2240b0d49e953f3fd.png

[harsh cipher]

Hello

Question

for 2sin^2x + sinx -1 = 0

What is the general solution?

I tried graping but my general solutions are

0.52+ 2pi n, 2.62+2pi n , 4.71 + 2pi n

answer says its pi n not 2pi n

[valid violet]

It's quadratic in s=sin x so you can write it as 2s²+s-1=0

[harsh cipher]

yes. how is the solution 2pi + n

because our b value is 1....

p = 2pi/b

[worn prairie]

using the quadratic formula, s = -1 or 1/2. So since arcsin(-1) = pi, i would guess that's why pi showed up as the answer

[harsh cipher]

ok

one more question

this question....if you divinde 0 by root 2

we are left with cos^2 x - cos x=0

to get cos x (cos x -1 ).

how can we have pi/4 and 7pi/4 as our answer

I only got pi/2 and 3pi/2

I mean how can we just create root 2 back in to our equation when solving for cos x -1=0

[stark ivy]

wait a minute here

factor out cosx from each function

[harsh cipher]

we're solving for the exact value here

[clever inlet]

You definitely did not divide correctly

[stark ivy]

you get cosx(root2cosx-1)

[harsh cipher]

ok..i tried to lose the root 2 first

[stark ivy]

thats where you went wrong

[clever inlet]

cos^2(x) - cos(x)/sqrt(2) = 0

If you divide through by sqrt(2)

[stark ivy]

yeah

[clever inlet]

With theta of course

Instead of x

[stark ivy]

easier to look at

anyways solve for each of the factors

[harsh cipher]

so you take out cos theta

[stark ivy]

sqrt2cosx-1=0

[harsh cipher]

from both

[stark ivy]

and cosx=0

[harsh cipher]

thank you both

[stark ivy]

you get pi/2 and 3pi/2 for the first factor and pi/4 and 7pi/4 for the other one

np

[viscid thistle]

Hello

How do you make a polynomial function based on a data table using a calc?

[lost mesa]

depends on the calculator

basically, you plug them into your calculator on a list, then plot a regression

[viscid thistle]

TI-83?

[lost mesa]

https://www.dummies.com/education/graphing-calculators/compute-a-regression-model-on-the-ti-83-plus/

[valid violet]

are you trying to find the best fit line or to find an exact n-1 degree polynomial that goes through n points?

I dont know if the ti83 can do best fit quadratic or cubic etc

[leaden stratus]

How can I verify this identity?

[willow bear]

what have you tried

[leaden stratus]

I tried to write that tan as sin/cos @willow bear

[willow bear]

uh huh

[leaden stratus]

Wdym

[willow bear]

i mean

okay

you did that

did it also occur to you to write cos(α - 3π/4) in the denominator as -cos(α + π/4)

and perhaps do a similar trick with the cosine on top except differently

[leaden stratus]

-cos (a + pi/4)?

Why?

[willow bear]

wrong parenthesis placement.

anyway

this should bring you closer to establishing that the right hand side is equal to sin(α + π/4)/cos(α + π/4).

[leaden stratus]

Yeah but why the denominator becomes -cos (a+pi/4)

[willow bear]

because i CHOSE to apply the identity cos(x-π) = -cos(x).

[leaden stratus]

Ohhh that's why

But -cos(x) isn't also cos (x+pi)?

But in that way it won't become pi/4 🤔

@willow bear

[willow bear]

But -cos(x) isn't also cos (x+pi)?
did i say it wasn't?

when someone says 3+4 = 7, do they in saying so deny that 2+5 = 7?

[leaden stratus]

No

I was just asking

[willow bear]

the way you worded it made it sound as if what i said made you believe cos(x+π) = -cos(x) was false

[leaden stratus]

Nono

Didn't mean that

[willow bear]

anyway, another identity that might come in handy here is cos(x + 3π/2) = sin(x)

[leaden stratus]

-sin(x) = cos(x+pi/2), right?

[willow bear]

er

yes

i screwed that up this time

[leaden stratus]

@willow bear now, the denominator is negative

And the numerator is positive

[willow bear]

what exactly do you have rn

[leaden stratus]

But how would I rewrite cos (a - 5/4pi)?

[willow bear]

as cos(α + π/4 - 3π/2) perhaps

[leaden stratus]

But 3/2pi is an associated angle

I would do an associated angle of an associated angle O_O

[willow bear]

,,,,what

[leaden stratus]

I didn't understand this exercise

Thanks anyways, Ann

[willow bear]

alpha + beta + 30° = 180°

[sturdy haven]

I got 310

[viscid thistle]

hmm

10*2=20

20*2=40

40*2=80

80*2=160

160*2=?

[sturdy haven]

I used geometric series

[viscid thistle]

why

it doesn't accumulate

it doubles

[sturdy haven]

But it is asking how much money you will have after 5 months

[viscid thistle]

so what

why does that make it a series?

[sturdy haven]

You are adding up all the money for a total

[viscid thistle]

it never says that

it says your money doubles

which i assume means multiplying by 2 every month

[sturdy haven]

In problems involving money how do you know when to use series or explicit formula?

[astral mountain]

does 1/sqrt(x) have a vertical asymptoot ?

[viscid thistle]

yes

[fleet yew]

Yeah

Think about it

Where is the denominator 0?

[harsh cipher]

Hello

Question

2cosx -3 = 0

there is no solution for the above equation?

agree?

whereas cosx +1 = 0

x = pi

agree?

[lost mesa]

agree for both

[harsh cipher]

thank you

[viscid thistle]

For 2cos(x)-3=0, you would have cos(x)=3/2 which is greater than 1, therefore, outside the domain of cos(x). This means there are no solutions to the equation. For cos(x)+1=0, cos(x)=-1, x=pi is your solution in (0,2pi) but you're general is pi+2pi(n)

[harsh cipher]

yes and in degrees it would be 180+ 360(n)

[viscid thistle]

yeah

[lost mesa]

i mean, i just simplified cos(pi) = -1

-1 + 1 = 0

therefore true

[tender knoll]

Hello!can someone please help me with this question

This is what i got and i just dont know how to get rid of the x....

[stuck lark]

https://i.gyazo.com/c9c07dc86610129a01d64375ee7fd621.png

try expanding the right side here, collect x terms on the same side, factor out x

[tender knoll]

Thats where i get stuck on, i thought if i expand (1-x)ln4 it becomes division

[stuck lark]

a(b+c)=ab+ac

[tender knoll]

Ok leme try again

I DID ITT

Yasss thank u so much 🙏 ~!!!

[stuck lark]

very naisu

[tender knoll]

I didnt know u can do that

I thought i had to put it in divison mode right away

[stuck lark]

i actually don't get how (1-x)ln(4) led to ln(4)/(xln(4))

[fleet yew]

No it makes sense

[stuck lark]

[tender knoll]

O.O

Man im so glad i joined this discord >:)

[harsh cipher]

Hi

2Cotx + root 12 = 0

cot x = -root 3

I couldn't figure this one out on my own

[harsh cipher]

nm figured it out

1/tanx = adj/opp

• root 3 in quad 2 and 4

[tender knoll]

Nicenice

[harsh cipher]

it's naisu~

thanks anyways

[sturdy haven]

If there is a question that says the population of rabbits on an island is increasing at 4% and you start with one thousand

Then in the first year do you start with one thousand or 1040

[willow bear]

the population is 1000 at the beginning of the first year and 1040 by the end

[sturdy haven]

So how would you find the fifteenth year

[willow bear]

what do you mean by "find the fifteenth year"

do you have the problem you're looking at, EXACTLY AS IT IS STATED

[sturdy haven]

The population

[willow bear]

VERBATIM and WORD-FOR-WORD

since apparently using one of this pair of synonyms fails to get the point across most of the time

do you have the exact problem statement?

[sturdy haven]

Yes

[willow bear]

then post it

[sturdy haven]

The population of rabbits on an island increases annually at a rate of 4%. If the starting population is 1000, what will be the population in the 15th year.

[willow bear]

okay so this is kind of ambiguous but i would assume they mean "if the starting population is 1000, then what will the population be 15 years later"

i.e. 1000 * 1.04^15

[sturdy haven]

Thanks

Also this lol

[willow bear]

well, there is a quick and easy way to sum all the numbers from 1 to n

[sturdy haven]

N/2(a1+an)

[willow bear]

yknow what fuck it i'm too tired to tell you that capitalization matters and also that you should be more careful about notation in general

[sturdy haven]

Okay

Sorry

I just have an exam today and Im trying to understand this

[willow bear]

you're asked to solve n(n+1)/2 = 465 for n.

[willow bear]

meh

it's a quadratic equation

[viscid thistle]

How would you do letter A

I’m so lost on this unit lmao

[willow bear]

you're told that there is no phase shift. so the equation for the temperature can be written down as $T(t) = T_0 + A \sin(kt)$, where $A, k$ and $T_0$ are parameters to be determined from the information you have

[obsidian monolith]

Ann:

[willow bear]

you're told that the period of this function is 16, from which you can find k, and that the low is 37° while the high is 40.4°, so you should know the amplitude (A) and the vertical shift (T_0)

[viscid thistle]

Someone help me with this, I can’t find the answer

I don’t understand this

[weak fractal]

Can someone help me with this questiion

[valid violet]

@weak fractal rectangles have equal opposite sides

and all those expressions can be simplified by factoring and cancelling things

[weak fractal]

@valid violet ty

[viscid thistle]

,rotate

[proud jetty]

o.O

[viscid thistle]

can you make an equation with that data

[proud jetty]

idk probably

[viscid thistle]

well

compute $f(3)$

[obsidian monolith]

mop:

[proud jetty]

im kinda confused?

[viscid thistle]

so $f(x)=\frac{2-k}{5x+k}$

[obsidian monolith]

mop:

[viscid thistle]

then $f(3)=\frac{2-k}{5(3)+k}$

[obsidian monolith]

mop:

[viscid thistle]

right?

[proud jetty]

yea I did that much

[viscid thistle]

so what can we say about f(3)

it's equal to what

[proud jetty]

15?

[viscid thistle]

it's in the q

[proud jetty]

go on?

[uncut mulch]

the graph passes through (3 , -2/19)

what is the value of the function when x = 3?

[proud jetty]

the y value?

[uncut mulch]

\frac{numerator}{denominator}

[proud jetty]

ah ic lmafo

but yea how do you solve for it when u have two unknown tho?

[uncut mulch]

what do you mean 2 unknowns?

as your bad tex implied earlier, f(3) = -2/19

$f(3)=\frac{2-k}{5(3)+k} = -\frac{2}{19}$

[obsidian monolith]

ramonov:

[uncut mulch]

right?

[proud jetty]

yes

but like in this case the values of k is same right?

which happens to be 4

;O

tf i was solving it like u slove for x

[uncut mulch]

i mean its the same variable

so it would have the same value

[valid violet]

When you use the same variable for than once in the same expression then they're all.the same variable

More 4than

[proud jetty]

ic i couldn't solve cuz I didn't know how i would slove for two x

[valid violet]

And you can solve for k that would work if you didn't see the solution by inspection

(x+3)/(2x-1)=19

You can have things like that

[proud jetty]

icic

how about this

[uncut mulch]

,rotate

[proud jetty]

12

[uncut mulch]

cross multiply

[proud jetty]

dam

that still exists

jk thx thx

[fleet yew]

with the stipulation that x cannot be -2 or -9

[harsh cipher]

Question if anyone is awake

I cannot solve this problem.

I almost did but went wrong somewhere towards the end.

Let me explain where I got stuck.

So I use the conjucate for the left equation and arrive at

Sec^2x -1/ Sin^2x (secx-1)

why am i not getting 1-cosx in the denominator

[uncut mulch]

consider: factorising the numerator

[harsh cipher]

ok let me try that

[uncut mulch]

of the original

[harsh cipher]

to tan^2x?

I tried that let me redo it

[uncut mulch]

uh to be more specific, take out a factor of sec(x)

[harsh cipher]

is that even possible

Sec^2x - 1?

[uncut mulch]

$\sec(x) \cdot \br{\frac{\sec(x) + 1}{\sec(x)}} = , ?$

[obsidian monolith]

ramonov:

[harsh cipher]

secx+1

[uncut mulch]

that's the original
simplify the stuff in the parentheses

to get the factorisation

[harsh cipher]

2secx

[uncut mulch]

how are you getting 2sec(x)?

[harsh cipher]

tbh I dont know

[uncut mulch]

$\frac{\sec(x)}{\sec(x)} + \frac{1}{\sec(x)} = \ ?$

[obsidian monolith]

ramonov:

[harsh cipher]

1+secx/sec x

[uncut mulch]

that's not really what i'm expecting

$\frac{\sec(x)}{\sec(x)} = \ ? $

[obsidian monolith]

ramonov:

[harsh cipher]

1

[uncut mulch]

$\frac{1}{\sec(x)} = \ ?$

[obsidian monolith]

ramonov:

[harsh cipher]

csc x

i mean

[uncut mulch]

no

[harsh cipher]

1/secx

[uncut mulch]

i mean that's exactly what's written...

can you express it with something simpler?

[harsh cipher]

cosx

[uncut mulch]

yes

so sec(x) + 1 = sec(x) * ( 1 + cos(x)) right?

[harsh cipher]

yes

[uncut mulch]

and that's the hard part done

applying 1-2 simple identities should get you what you need

LHS = $\frac{\sec(x)(1 + \cos(x))}{\sin^2(x)}$

[obsidian monolith]

ramonov:

[harsh cipher]

LHS should be Sec^2x -1.....

[uncut mulch]

i'm providing an alternate method

that's less tedious

[harsh cipher]

okay

[uncut mulch]

and then consider:
sin^2(x) + cos^2(x) = 1

[harsh cipher]

ok

Hmm anyways

Thank you @uncut mulch

[uncut mulch]

did you manage to get it?

[harsh cipher]

no I did not

but it's almost 3:40 am

I need sleep.

Thanks!

[uncut mulch]

ok i'll put stuff in spoilers for when you wake up

[harsh cipher]

tyty!

[uncut mulch]

|| sin^2(x) + cos^2(x) = 1|| → ||sin^2(x) = ?||
consider: ||a^2 - b^2||

[warped swift]

How do you substitute ln3 into 3e^-x

and get 3e^ ln1/3

?

[uncut mulch]

$-\ln(3) = \ln(3^{-1})$

[obsidian monolith]

ramonov:

[winged cloak]

What does it mean if E is in symmetrical relation with V?

[willow bear]

context?

[winged cloak]

E is the edges of a graph

And v the vertices

[willow bear]

no, can you show like

the text where you saw that phrase

and maybe check if it's defined elsewhere

[winged cloak]

It s a definition I'm learning about graphs

@willow bear

[willow bear]

on donne la définition de «E est une relation symétrique» en bas

c'est pas «E est en relation symétrique avec V» comme suggèrerait la façon dont tu l'as dit en ang

[winged cloak]

Donc en gros v1,v2 c est la mm chose que v2, v1 ?

[willow bear]

en gros, on fait pas de distinction entre une arête de v1 à v2 et une arête de v2 à v1

[winged cloak]

Et une partie de VxV ça voudrait dire quoi alors ?

[willow bear]

une... partie... c'est à dire un sous-ensemble

[winged cloak]

Donc c possible pr un graphe orienté pour un x on est plusieurs v ?

[willow bear]

mon dieu tes messages sont beaucoup trop difficiles à lire avec tout ce langage sms

[winged cloak]

Pardon 😥

[willow bear]

ça veut dire quoi « pour un x » ?

il n'y a pas de x

[winged cloak]

On a un sommet x qui fait partie de l ensemble V

[willow bear]

oui

[winged cloak]

Oui j aurais mieux fait de l appeler v

[willow bear]

ok on a un sommet

et ?

[winged cloak]

Et donc si on a un sommet v et les arêtes qui font partie de VxV ça veut dire qu on peut avoir plusieurs arêtes Pr un même sommet ?

[willow bear]

« un sommet V » putain non

y a une différence entre v et V

[winged cloak]

Oui j ai changé

[willow bear]

V majuscule c'est l'ensemble de TOUS les sommets

v minuscule désigne habituellement un des éléments de V, c'est à dire un sommet

VxV c'est l'ensemble de toutes des couples de sommets

couples ordonnées

[winged cloak]

Oui je sais je sais ds le cas d un orienté

Mais est ce qu'on peut dire la même chose Pr un non orienté en parlant des E ?

Je sais que ma question a l air un peu stupide mais J essaie de comprendre

Non OK g compris merci de ta patience👌

[tender knoll]

Hello can someone help mewith this

I dont get how to get x fromthis

I tried to get everything to base 5 so i can just make the exponents as the equation but its not working with -100

And in the answer sheet it says: 1/2

[uncut mulch]

a^(b+c) = a^b*a^c

[tender knoll]

I domt get it

This is my work so far but i think its not right

[maiden blade]

Doing ln(-100) and using log laws is not permitted

[uncut mulch]

log does work like that

[tender knoll]

Ohh

[maiden blade]

Oh wait

That's not even

[tender knoll]

Then i can only chnge base to get answer?

[uncut mulch]

log isn't linear
consider applying what I posted above

[tender knoll]

Not sure how i can get -100 to be 5^__

Ok leme try again

Wihtout added ln

[uncut mulch]

consider simplifying the left side first

[tender knoll]

Would it be this?

[uncut mulch]

that's a good start

[tender knoll]

I thiught that only works wihth adding ln

Isnt that not 5^(4x+1)

[uncut mulch]

log(a) + log(b) \neq log(a + b)

[tender knoll]

So i do need to add log or something?

[uncut mulch]

not yet

and probably won't need to for this q

continue simplifyingbthe left side

[tender knoll]

What indont get is how 5^(4x+1) can equal to 5*5^(4x)

[uncut mulch]

a^(b+c) = a^b*a^c

[tender knoll]

So this kind of question is ok to make + into * without adding ln or log?

Cuz in class we always added them

[uncut mulch]

added what ?
wdym turning + into *?

[tender knoll]

The 4x+1

Into *

I get that if thee was ln infront i’d do that

But i never knew u can do that without it

[uncut mulch]

• doesn't "turn" into *

[tender knoll]

Oki so if the exponenet is polynomial of a number

I can just change it into that form alwqys?

Not using ln ?

[uncut mulch]

the:

a^(b+c) = a^b*a^c
exponent law is applied to get
5^(4x + 1) = 5^(4x) * 5

[tender knoll]

Okii

[uncut mulch]

and you have a decent knowledge of that

[tender knoll]

What if it was 25^(4x+1)?

[uncut mulch]

just because it's a question supposedly in the log section doesn't mean. you should chuck log at it directly

[tender knoll]

Would i turn that 25 into a 5^2(4x+1)

Oh okii

[uncut mulch]

parentheses

5^( 2(4x+1))

[tender knoll]

Ah

Then i’d have to distribute 2 to 4x+1?

And get the new exponent terms?

[uncut mulch]

if you want tto

[tender knoll]

Then 25^(4x+1) turn into 5^(8x+2)?

[uncut mulch]

yes.

[tender knoll]

Ah icic

When i see an exponent with number outside of term idk what to do, but now i know we can just distribute like normal right?

[uncut mulch]

can you show what you're doing by continuing with the original question

[tender knoll]

Yes one sec!! Trying

[uncut mulch]

nope.

[tender knoll]

Am i allow yup knew i was wrong

[uncut mulch]

you just implied that
5^(4x+1) = 5^(8x)

[tender knoll]

Oh

[uncut mulch]

note that both your terms have 5^(4x) in them

[tender knoll]

Then where do i put that 5^1

Then can i use a vqriqble to represent that number

[uncut mulch]

$5^{4x} -5 \cdot 5^{4x} = -100$

[obsidian monolith]

ramonov:

[uncut mulch]

you can use a substitution if you want
(but not necessary)

[tender knoll]

Oh

Oh soafter i change 5^(4x+1) to (5^4x) and (5^1)

I should just treat those like its own body?

[uncut mulch]

weird wording

[tender knoll]

And cuz its multi i can move that 5 over?

Yea i dont really know the terminology

Oh

[uncut mulch]

u*5 = 5u
is applying the commutative property of multiplication

[uncut mulch]

yeh that works

[tender knoll]

Ooo yess

This took me like 1 hour o_o..

Thanks so much 🙏!!

[uncut mulch]

np

[tender knoll]

Is it illegal if log(-#)?

It always has to be positive?

It says to find the domain for this function

[uncut mulch]

when dealing with real numbers, it(#) needs to be greater than 0

[tender knoll]

Idk y im not getting that

I have the numbers correct but it doesn’t make sense

[uncut mulch]

are there any other situations when
(x+5)/(x-7) is positive?

[tender knoll]

It doesn’t say if it is pos or neg

[uncut mulch]

also that isn't a union

& isn't the same as U

you need to consider 2 cases

[tender knoll]

Oh yea i just put that there instead of writing and

[uncut mulch]

firstly what are the intersections of
(-5, inf) AND (7,inf)?

[tender knoll]

Umm

Leme draw

[uncut mulch]

not U

[tender knoll]

Ah thats totally wrong

Icic

But the answer key says U

(-inf,-5) U (7,inf)

[uncut mulch]

not for what you have atm
we'll get to that

[tender knoll]

Ok

[uncut mulch]

for this case it should be: $(-5,\infty) \cap (7, \infty)$

[tender knoll]

I just do t get which direction i need to set < or >

[obsidian monolith]

ramonov:

[willow bear]

ram

are you sure log_10( (x+5)/(x-7) ) is undefined for values like x=-100

bc that's what you seem to be saying

[uncut mulch]

"it" was referring the the # which was meant to represent the argument

[tender knoll]

The case u mean by my numebr line graph?

Oo

Do i actually need to test this before coming with solution

[uncut mulch]

there are two cases when the inside is positive,

1. when both the numerator and denominator are positive
2. when both the numerator and denominator are negative

currently you are working on 1.

and what are the common set of values when you draw (-5, inf) and (7,inf) on the number line

[viscid thistle]

Yo

[tender knoll]

Yeey

Thanks!!!

I keep forgetting bout the negatives so i didnt test it

[uncut mulch]

that works

[uncut mulch]

where are you stuck?

[viscid thistle]

30

[uncut mulch]

what have you tried?

[tender knoll]

Ramonov

Are u bust atm @uncut mulch

Busy*

[violet quail]

how do i solve for x?

[harsh cipher]

Question

I used 9pi/12 and 4pi/12 to simplify and arrived at the answer

root3+1/ 2 root2

Answer key says root3 -1 for the numerator

hmmmm

question is for a)

question is, is it negative sin or positive sin because its 3pi/4. Reference angle is pi/4

sin is positive in quadrant 1 and 2

[remote veldt]

@violet quail x = 0 is clearly a solution. Consider the graphs of ln(1 - x) and x - x^2/2
To the left of x = 0, ln(x) is strictly decreasing and x - x^2/2 is strictly increasing. See what you can draw from that - is it possible for them to intersect to the left of x = 0, if one function is going down, and the other is going up?

ln(1 - x) is not defined for x >= 1. So, the other place you need to check is (0, 1). You'll notice that ln(1 - x) is still decreasing in this interval, while x - x^2/2 is increasing. Same question - if they're equal at x = 0, one graph is going down, and the other graph is going up, is it possible for them to intersect again?

Once you address those cases, you have a solution, and you have whether or not there are any other solutions

@harsh cipher are you sure you used the formula correctly? yes, sin is positive in Q2, but keep in mind, the formula has a subtraction sign

[violet quail]

can't i just solve it algebraically somwhow?

[harsh cipher]

Yes I did

Because it has a negative sign

We can change the signs later when writing the answer.

(1/root)(1/2) - (1/root2)(root3/2)

Same as root3-1/2 root2

[remote veldt]

@violet quail short answer? not at the level you're working at

[viscid thistle]

Yooo

What’s going on

What does symmetrical about the pole mean

[harsh cipher]

Hi question.

For question b)

we need to write the question as tan(a-b).

[tender knoll]

Ooo leme also try this

I need my notes...

[pale bison]

yes that's how you can apply the sum and differences formula

[harsh cipher]

okay I'm trying my new method if I can solve as tan(pi/6 - 3pi/4)

it's prob the same as writing alpha and beta as pi/4 and pi/3

i mean the answer should be the same but I just couldn't solve with ratios that were used for pi/6 and 3pi/4

[pale bison]

break it up into sine and cosine

[harsh cipher]

well the video lesson had instructions to solve using trig ratios

and to simplify and use rationalizing fractions to further simplify

[tender knoll]

Not sure but this is what i got

[harsh cipher]

cool

[tender knoll]

U dont have answer key?

[harsh cipher]

i do

answer is 2+root 3

I

[tender knoll]

2+root3?

Oki leme see if i got it right ..

I’m not getting the same answer rip

[harsh cipher]

i didn't use -pi/4 and -pi/3

did you use conjucate

[tender knoll]

Yup

[harsh cipher]

oh

[tender knoll]

And its cancelling out the rad 3s

When i add them up

[harsh cipher]

this problem took me 1hr+

LOL

i got it now though

https://www.symbolab.com/solver/factor-calculator/tan\left(\frac{-7pi}{12}\right)

[tender knoll]

Oh wow

Man im just like u hahah i take 1hr+ cuz im always stuck on 1 question

[harsh cipher]

hahaha

[harsh cipher]

Question

Question 5a)

my answer was 0

answer key is (2-2root10)/9

because I'm sure 2root10/9 - 2root10/9 - 0

oh no i got it wrong

[viscid thistle]

@harsh cipher you still need help?

@harsh cipher you use the pythagorean identity to get sin^2 A + cos^2 A = 1 and sin^2 B + cos^2 B = 1. then you can solve for sin B to get sin B = sqrt(1 - cos^2 B). and you know cos B, which is 2/3, so sin B = sqrt(1 - 4/9) = sqrt(5)/3.
doing the same with cos A, you get cos A = sqrt(1 - sin^2 A) and you know sin A = 1/3, so cos A = sqrt(8)/3.
using the angle addition identity, you have sin(A - B) = sin A cos B - sin B cos A, substituting in known values you get sin(A - B) = 1/3 * 2/3 - sqrt(5)/3 * sqrt(8)/3 = 2/9 - sqrt(5)/3 * sqrt(8)/3 = 2/9 - sqrt(40)/9 = (2 - 2sqrt(10))/9.

i might've made a mistake since i typed it out on the computer without a piece of paper to check my work but hopefully that's correct

see if you can do the second one from there.

[viscid thistle]

Can somebody help me. Book answer is 2sin(a). But whatever I try, simply cant get there

[willow bear]

consider that tan(a/2) = sin(a)/(1 + cos(a))

[viral trail]

Limit

[serene heath]

?

[viral trail]

I can figure it out by using limit h ->0 of f(x+π/2 + h), but how can I find the limit without changing to variable h?

I can't simplify 1 - sinx

[serene heath]

what is it asking for

theres no question there

[viral trail]

I want to find the limit of the equation as x -> π/2 without using variable h

[willow bear]

do you mean without using the substitution h = x - π/2

i mean ig you could like... write 1 - sin(x) as 2sin^2(x/2 - π/4)

kinda meh tho

[harsh cipher]

@viscid thistle no I figured it out. Thank you!

@willow bear Bonjour

[viral trail]

Thank you

[blazing raven]

Just for practice / thought (algebra II / Trig / limits / other)...

Which TWO of the following graphs are the same:

f(x) = sin(x)
f(x) = (x sin(x))/x
f(x) = ((x+1) sin(x))/(x+1)
f(x) = ((x^2+2x+1) sin(x))/(x^2+2x+1)

(inspired by an AMC problem)

[uneven dawn]

I dont even know what to search online to find a tutorial?

if anyone even knows what this type of problem would be called that would help a ton. on my exam in a couple hours and i have no clue what it is

[dim jungle]

uh

are u given anything else cold stew

cuz this is part (c)

[viscid thistle]

I have to solve this vector application and i just don't understand where to even start tbh

[patent beacon]

5 miles west:
(-5, 0)

4 miles southwest:
(-4/√2, -4/√2)

Add the vectors together to get the vector displaying where she is now:
(-5 - 4/√2, -4/√2)

That much make sense? @viscid thistle

This approximates to (-7.828, -2.828)

[viscid thistle]

do you add those two numbers up ???

[shy briar]

herro

[viscid thistle]

hola

[shy briar]

If a scientist mixed 10% saline solution with 60% saline solution to get 25 gallons of 40% saline solution, how many gallons of 10% and 60% solutions were mixed?

😦

me no understand how to do

[viscid thistle]

sorry i'm dumb asf so im not helpful 😅

[shy briar]

the basic rule for doing this is you need to have atleast two variables and have it like (Let: x=(something) and y= (something))

but its too hard

😡

[patent beacon]

@shy briar
Heyo. Let x be the gallons of 10%, let y be the gallons of 60%. You can make two equations with x and y.

[shy briar]

thanks

i already got someone else to solve it

tho

i got another equation tho

An investor who dabbles in real estate invested 1.1 million dollars into two land investments. On the first investment, Swan Peak, her return was a 110% increase on the money she invested. On the second investment, Riverside Community, she earned 50% over what she invested. If she earned $1 million in profits, how much did she invest in each of the land deals?

[fleet yew]

@shy briar still need help with this?

[harsh cipher]

Hi

Question.

Expand double angle identity

6tan 4x

[stuck lark]

$\trig$

[obsidian monolith]

RokettoJanpu:

[tawny nacelle]

how did u do that ?

[stuck lark]

custom command

[tawny nacelle]

oo

fancy

[harsh cipher]

this is so lame.

no where in the lesson they taught about double angle identity for tangent

for sum and difference they did.

not for double angle identities.....

wtf?

[harsh cipher]

can someone check my work please

1/2-cos^2 3x

write as single trig identity

i got 1/4 cos 6x

answer key says -1/2 cos6x

oh nm

[viscid thistle]

why is that image so large lol

you can't even zoom in without making one identity the size of your whole screen, jeez

[patent beacon]

@stuck lark
Can I make a custom command like that? Where can I learn?

[willow bear]

learn latex

new commands are created with the command \newcommand shockingly enough

[pale bison]

\newcommand{\newcommand}{nou}

[willow bear]

that wouldn't work

you'd have to use \renewcommand to override an existing command definition

[pale bison]

[patent beacon]

I guess I'll Google the details, then

[pale bison]

like, if you do \newcommand{\RR}{\mathbb{R}}
then whenever you type /RR, it's gonna get replaced with \mathbb{R}

[harsh cipher]

hi

what does determine non-permissible value fors theta given 0=<theta<2pi

for

[viscid thistle]

Could anyone explain how she got this out of 5 in the fraction?

[willow bear]

0.8 = 4/5

[viscid thistle]

Yes but where did she get the 0.8 from

It's 4 times more likely

[willow bear]

P(H) = 4P(T)

P(H) + P(T) = 1

do these equations make sense to you

[harsh cipher]

Hi Ann

🙂

[viscid thistle]

yes

it does

but i still don't know how it's out of 5 chances.

[willow bear]

you have these equations

can you solve for P(H) and P(T)

i want a "yes", a "no", or a "let me try"

[viscid thistle]

no

i was trying

wouldn't it just be TTTTH

[willow bear]

okay, can you show your attempt then?

[viscid thistle]

and that's where the 5 comes from

[willow bear]

no look

look

[viscid thistle]

i am

[willow bear]

$\begin{cases} x = 4y \ x + y = 1 \end{cases}$

[obsidian monolith]

Ann:

[willow bear]

would you be able to solve this system of equations for x and y

it's really not rocket science. i'm sure you can do it. just don't overthink it.

[viscid thistle]

yeah i can, but i think it's simpler than that. it's a probability equation and i know that it has to equal to 1. knowing hte sample size in this example would be {TTTTH}, that's where the 5 would come from

correct?

[willow bear]

man your wording

but i mean

how's this system of equations not as simple as it gets

like you can just

susbtitute the first eq into the second

get 4y + y = 1

or 5y = 1

and there's your five that you were confused about the origins of

[viscid thistle]

lol i know. i wanted to make a picturable sample size. it's easier to solve probabilities when u can envision it.

for me, at least.

[undone pawn]

oh

[viscid thistle]

I'm suck at how she got 33 and 11

I figured out the example I posted on top, but I can't quite get to a destination here..

[willow bear]

P(2) = 2P(1)
P(3) = 3P(2) = 6P(1)
P(4) = 4P(3) = 24P(1)

P(1) + 2P(1) + 6P(1) + 24P(1) = 1

33P(1) = 1

it really is just that

@viscid thistle

[viscid thistle]

Oh...

Right

I'm stupid

System of equations it is

Like you said

at the beginning.

Thank you

[harsh cipher]

I have a question

finding non-permisssble values for the below equation.

sec x/1-sinx

1/cosx cannot equal 0

sinx cannot equal 1

[willow bear]

did you mean $\frac{\sec(x)}{1-\sin(x)}$

[obsidian monolith]

Ann:

[harsh cipher]

yes

[willow bear]

what did i say on multiple occasions about parentheses

anyway, it's cos(x) that cannot equal 0

1/cos(x) can never equal 0 anyway

[harsh cipher]

oh yes you're right about the numerator

can you please repeat what you said about parentheses. I'll make a note

[willow bear]

they are required when writing fractions in plaintext, because order of operations is a thing no matter how much you might wanna forget about it

[obsidian monolith]

Ann:

[harsh cipher]

correct

[willow bear]

yeah so

[obsidian monolith]

Ann:

[harsh cipher]

ok so I must write (secx/1)/(1-sinx)?

[willow bear]

no, you don't need to write explicit 1 denominators!

sec(x)/(1 - sin(x)) is fine!

[harsh cipher]

ah okay

let me explain my thought process for the question.

for the numerator cos x cannot equal 0. So pi/3 and 3pi/2.

pi/2

and the denominator 1-sinx I got pi/2

[willow bear]

are you asked to do this on [0,2pi] or on the entire number line

[harsh cipher]

Yes

[willow bear]

......

WHAT

bruh

[harsh cipher]

0<equal to theta and less than 2pi

[willow bear]

so [0,2pi], then.

but

[harsh cipher]

yes

[willow bear]

you don't respond to a "this or that" question with "yes"

it's just

bruh

[harsh cipher]

yes I didnt carefully read your question

[willow bear]

okay so [0, 2pi] then

[harsh cipher]

yes

[willow bear]

yeah so. ok

so your final answer is...

[harsh cipher]

pi/2, 3pi/2 for the numerator. pi/2 for the denominator.

[willow bear]

no

[harsh cipher]

okay

[willow bear]

your final answer

should be

"the expression is undefined for precisely the following values of x:"

followed by the list of all such values

[harsh cipher]

yes answer key shows x cannot equal pi/2 and 3pi/2

[willow bear]

did i ask you to peek at the answer key

no i did not

[harsh cipher]

no

you did not

[willow bear]

indeed

[harsh cipher]

so what did I do wrong?

[willow bear]

you failed to connect the dots between "the numerator breaks for π/2 and for 3π/2, while the denominator doesn't break but becomes zero for π/2" to "the expression is undefined at π/2 and at 3π/2"

[harsh cipher]

Thank you

[viscid thistle]

@willow bear what do I do from here?

[willow bear]

you have fucked up already

sin(2a) is not 2sin(a)

[viscid thistle]

Sin(2a) + 2sin(a) = 2sin(2a)

I think. And 1+cos(a) = 2cos(a/2)^2

I can maybe cancel out the twos.

[willow bear]

Sin(2a) + 2sin(a)

no!

no! no! no! no!

[pale bison]

[willow bear]

sin(2a) ≠ 2sin(a) !!!

sin(2a) ≠ 2sin(a) !!!

sin(2a) ≠ 2sin(a) !!!

sin(2a) ≠ 2sin(a) !!!

sin(2a) ≠ 2sin(a) !!!

[pale bison]

"true" for smol a

[willow bear]

@viscid thistle

how do you read my message saying in as direct a manner as possible that sin(2a) and 2sin(a) aren't the same thing and the immediately afterward pretend they ARE the same

[pale bison]

they didn't read your message duh

[viscid thistle]

It wasn't an offence to you :)

Just maybe not that good at math

[willow bear]

it's not math it's reading comprehension

how are the words "is not" so hard to understand

i was under the impression they weren't

[viscid thistle]

I dont appreciate getting talked to like that.

[willow bear]

ok but let's get back to the point i was trying to make

sin(2a) and 2sin(a) are not the same

like. yknow

sin(2a) = 2sin(a)cos(a), if you wanted to have an identity to apply here.

[viscid thistle]

Oh, i see. I cant just add them together

[harsh cipher]

Question

if cos theta = 2/5 and theta is in quadrant 4

find tan 2 theta

[willow bear]

for the love of god please

[harsh cipher]

my final answer is -2root21/19

oh no

lol

[willow bear]

and here too

[harsh cipher]

cos(x) = 2/5. Find tan(2x)

is that right?

[willow bear]

thank you

[harsh cipher]

my final answer is ((-2 root21)/17)

[willow bear]

,,,,,,

-2sqrt(21)/17?

[harsh cipher]

(-2root21)/19)

yes

[willow bear]

what's your denom. is it 17 or is it 19

[harsh cipher]

19

-2sqrt(21)/19

[willow bear]

your answer does not match mine

show your work with all steps written out clearly

[harsh cipher]

did you get sqrt4(21)/17?

[willow bear]

no, i did not

[harsh cipher]

well that's the answer in the answer key

I looked because I had to check my incorrect answer

😛

[willow bear]

the correct answer is tan(2x) = 4sqrt(21)/19

[harsh cipher]

hmmm I got it incorrect somewhere in the end again

because I did manage to write -4sqrt(21)/38

I might've got the (-4) part incorrect as well

well I know the answer is incorrect for now.

[uncut mulch]

should be /17 right?

[willow bear]

fuck

you're right

[harsh cipher]

lol the answer key is right?

[hard hornet]

F for Ann

[uncut mulch]

depends on what the key actually says

[harsh cipher]

answer key says.......

[uncut mulch]

because: sqrt4(21)/17 isn't and it seems like you're misrepresenting it

[harsh cipher]

4sqrt(21)/17

[uncut mulch]

yes, that would be correct

[harsh cipher]

omg Ann disappointed everyone..

now...question is how the hell did the denom become 17,,,

[uncut mulch]

as mentioned earlier, show your steps

[harsh cipher]

Well I first used pythagoras to find tan(x).