precalculus

[viscid thistle]

are u the daughter of shakespere

[willow bear]

no, Shakespeare lived way too long ago.

[viscid thistle]

ok ur english is hard

[vestal thorn]

Am I doing this correct yall?

[willow bear]

no

[vestal thorn]

rip

[willow bear]

there are multiple issues with this

the derivative of sin(x) is not 1-sin(x), for one.

[vestal thorn]

Ah wow wtf am I doing I messed up cos

Wait no

[willow bear]

and then you had some alge-bruh moments

[vestal thorn]

Ah ye I see nvm lol

[heady lotus]

if u hate something u will hate called by that name
I like noodles
Doesnt mean i would like being called noodles
Your logic is flawed

[viscid thistle]

vice versa may not be true

[heady lotus]

Vice versa isnt true for many true statements
All congruent triangles are similar is always true
Vice versa isnt

[viscid thistle]

yes that means my logic isnt flaw

[heady lotus]

Your logic doesnt hold either way

[viscid thistle]

give examples either way then

You should continue this thing in a different channel

[heady lotus]

Yes

[viscid thistle]

.purge

[trim hemlock]

[viscid thistle]

why are u sweaty ?

[vestal thorn]

Aight then is it correct now?

mb I should also probably be asking in the other channel but ye

[heady lotus]

The - sign shouldnt remain
It's vu'-uv', remember
You probably did vu'+uv'

@vestal thorn

[vestal thorn]

Ah yeah my bad I see my mistake I did uv’ - vu’

[heady lotus]

👍

[vestal thorn]

Thanks :0

🙂 *

[heady lotus]

Np

[oak arch]

How does one go about simplifying this?

Seems simple enough to me

[trim hemlock]

what base is the log?

if its base 10 then the answer would be xlog(e)

[oak arch]

Base is not given

So we should assume by default its 10?

Could you please help me understand how it simplifies to xlog(e)

[trim hemlock]

do you know this identity

[oak arch]

Nope

Oh but with that example I understand now

Thanks

Actually nvm

I dont understand

[trim hemlock]

ok lets start proving it

[obsidian monolith]

waler:

waler:

waler:

waler:

[trim hemlock]

so that leaves us with xb

[obsidian monolith]

waler:

waler:

[oak arch]

When we have log_a(a^xb), and log_a, and a^xb cancel each other out, how comewe are still left with xb

[trim hemlock]

only the base cancels out

the power is still there

[oak arch]

Ah ofc

[trim hemlock]

so now just use that same rule for log(e^x)

[oak arch]

Why do you say "but x = log_a(y)

[trim hemlock]

because y = a^x

[oak arch]

oh yes

[trim hemlock]

so take the log_a on both side

[oak arch]

log both sides

[trim hemlock]

yes

exactly

[oak arch]

Awesome, I understand it now! Thanks a lot ❤️

[trim hemlock]

np glad could help

[flint pulsar]

hi i was just wondering for part ii, where it's asking to find the percentage increase, would you just differentiate the equation to get the answer?

[willow bear]

no

youre not asked for instantaneous weight increase, but weight increase per day

[flint pulsar]

hm okay, so how would you go about solving it? do you need to differentiate at all in this case?

[willow bear]

no

notice that each time t increases by 1 (marking the passing of one day), the weight goes up by a factor of 10^0.05

that makes sense to you, right

[flint pulsar]

yup bc t = 1 is 1 day that has gone by

[willow bear]

yeah ok so

,calc 10^0.05

[obsidian monolith]

Result:

1.122018454302

[flint pulsar]

ohhh

12% increase

[willow bear]

12.202%

[flint pulsar]

ohh that makes complete sense now, thank you for clearing that up!

[fallow solstice]

saying ||FUCK|| is now banned (WARNING PRESS SPOILER AT OWN RISK)

[viscid thistle]

What

How do u spoiler smth

[shadow plaza]

||x|| will spoiler 'x'

[lament fiber]

How do I find the domain of f(x)= sqrt(sin x-cos x)?

in [0,2pi]

without using the graphs of the functions?

[viscid thistle]

which functions?

sin and cos?

[lament fiber]

yes

one sec

[viscid thistle]

it's easier if you draw them though

[lament fiber]

I can do tan x>=1
in the first and third quadrant
in the second and fourth quadrant I can't divide
or I can

yeah, but I'm curious what the method would be if you don't use a graph

actually
I think I have a way

[shadow plaza]

@lament fiber just look for the smaller intervals in [0,2π] in which sinx≥cosx

[lament fiber]

I can do tan x>=1
or tan x<=1 in Q2

I don't know if it's a correct way to do it idk

[shadow plaza]

Is my approach not helpful, Obi?

[lament fiber]

it is

I'm just asking if doing what I said is correct

[shadow plaza]

No.

You want an explanation?

[lament fiber]

Sure

(If it's about dividing by cos x
the sign of the inequality will change when cos x<0, which is in Q2 )

but I think my approach is wrong in some other way

[obsidian monolith]

Sidharth:

[viscid thistle]

||x|| will spoiler 'x'
@shadow plaza thanks

[lament fiber]

abs(|x|) is the way to go, then?

why would we want to do that

[obsidian monolith]

Sidharth:

[lament fiber]

ok

I follow

[shadow plaza]

Now notice that there are points in the third quadrant for which tan(x)≥1 but cosx≤0 for all x in third quadrant.

Therefore f(x) is undefined

[lament fiber]

I'm confused

[shadow plaza]

So you cannot base your answer on tanx without some extra restrictions

[lament fiber]

okay, let me read what you said fully
thanks for putting up with me

[shadow plaza]

okayy

[lament fiber]

but how does this relate to me saying
sin x- cos x must be >=0
which means sin x>=cos x
and I can divide by cos x without reversing the inequality when cos x>0 and reversing when cos x<0
to get x when tan x>=1 in Q1,Q4 and x when tan x<=1 in Q2,Q3

how does this relate to changing sqrt(sin x-cos x) to sqrt(cos x * (tan x-1))?

I figure this is where my poor knowledge factors in

[shadow plaza]

1 is also greater than 0 i.e 1>0 but you cannot divide both sides by 0,can you?

[lament fiber]

yeah

[willow bear]

@Okka#4940 please do not give people false ideas. cussing is not against the rules of this server.

ah they left

[lament fiber]

lmao

[shadow plaza]

Similarly, let sinx=0.7 and cosx=-0.3(purely hypothetical values solely for the purpose of explanation). Therefore, 0.7>-0.3 but 0.7/(-0.3) is not greater than 1, is it?

[lament fiber]

okay, so I consider when cos x=0
or do I just give up on this approach altogether because it's fundamentally wrong

oh

Similarly, let sinx=0.7 and cosx=-0.3(purely hypothetical values solely for the purpose of explanation). Therefore, 0.7>-0.3 but 0.7/(-0.3) is not greater than 1, is it?
yeah I know this, dividing by a negative value on both sides reverses the inequality

I mentioned that

[shadow plaza]

The point is if sinx≥cosx then tanx≥1 only when cosx>0

[lament fiber]

yes

I mentioned this?

[shadow plaza]

So this doesn't hold in third Quadrant

[lament fiber]

so in Q2 and Q3 I'll have to solve for x with tan x<=1 ?

I mentioned this

[shadow plaza]

Ohh. I am sorry. I didn't see that

[lament fiber]

so is the approach fine, then?

[shadow plaza]

I am not sure that it is

[lament fiber]

so here
the domain is found by solving
cos x>=0 and tan x>=1 and get the intersection of the sets?

[shadow plaza]

Okay let's try to analyse this using f(x)

[lament fiber]

okay

[shadow plaza]

https://cdn.discordapp.com/attachments/363224154469826562/767036062300110848/736188748492177408.png

We know that this is true

The value of f(x) is defined only when:
A) cosx≥0 and tanx-1≥0
B) cosx<0 and tanx-1<0

makes sense?

[lament fiber]

not B)

for a value under the radical you can't have it negative right?
for real functions

[viscid thistle]

this could be done graphically also right ? sinx>=cosx

[lament fiber]

yeah

I wanted to know how to do it without the help of the graph

[viscid thistle]

ohh okayy

[shadow plaza]

Okay, so let cosx= -a then √cosx=√-a=√a×i

Similarly we'll obtain another i from √(tanx-1)

And since i²=-1

[lament fiber]

yeah

[shadow plaza]

Therefore the value of f(x) is defined

Okay?

[lament fiber]

wait

but don't I have to consider only real values?

[shadow plaza]

Ofc

[lament fiber]

oh ok

nvm

[shadow plaza]

That is the definition of domain, is it not?

[lament fiber]

yeah

[shadow plaza]

Okay, so you follow so far?

[lament fiber]

yes

[shadow plaza]

Okay so now can you obtain the respective intervals satisfying those two conditions?

[lament fiber]

yeah

[shadow plaza]

The union of those two sets of values will be your domain

[lament fiber]

thanks a lot for answering

[shadow plaza]

Anytime

[lament fiber]

I never considered the part with i^2 in any problem
I'll keep it in mind now

I don't know if this is a stupid question, it probably is, but how do I get better at this stuff

I guess I should start thinking more instead of mechanically going for stuff like sin x>=cos x => tan x<=1 or tan x>=1 etc 🤷

[shadow plaza]

You should pick a book(that you trust is good for this sort of stuff) and read it. Don't cram things up but try to make sense of them. Once you step up to a point where things seem more logical and you don't feel the need to remember anything is the point where you'll start to get better at this stuff

[lament fiber]

Any suggestions?
I guess I should try to get through high school Math first, or is it not needed?

[shadow plaza]

Ofc, you'll also need some practice to feel confident enough to tackle problems

Aah. I am not very good at suggesting books. You can ask in #book-recommendations

[lament fiber]

Yeah I've been actively trying to do this for the last few months

okay, thanks a lot for all the help!

[shadow plaza]

yw

[viscid thistle]

not sure how to go about applying exponents -2/3 to 125/8, can anyone help? Thanks

This is how I attempted to solve the problem

[lament fiber]

so the fraction gets flipped

(8/125)^2/3

keep in mind that the bracket means both 125 and 8 get the ^-2

(8/125)^2/3=

[(8/25)^2]^1/3
because (a^x)^y= a^(x*y)

simplifying you get 2^(3*2/3) for the top

and similarly 5^(3*2/3)

so it's (2/5)^2

in summary
what you did is right but you need to do what you did to 125 to 8 as well

[viscid thistle]

oh i see

thanks man

[quaint mason]

i am confused on what I am supposed to do

[sick steppe]

make a log function that models the data

[viscid thistle]

[sand harbor]

LOL

[patent lance]

Hey there

I cant figure out this problem

if anyone could like explain how i would do it

[sick steppe]

do you know graph transformations?

[patent lance]

I dont know what that means

but if it means like

how a graph changes depending on how the equation is affected

then I usually do

[quaint mason]

@sick steppe wym by making a log function from that data

[patent lance]

Im not sure what type of problem this is though if that makes sense

[sick steppe]

It's a graph change based on the equation frosty

[pastel cloud]

Halp

[patent lance]

Ah ok

[quaint mason]

u mean the lnReg?

[sick steppe]

@pastel cloud remainder theorem

@patent lance g(x) is a translated version of f 1 unit in a direction

[patent lance]

Yeah

I know that part but

[sick steppe]

do you know what direction?

[patent lance]

my first thought

was down

because -

then left

because -

but Im not sure how to figure that specific thing out

[quaint mason]

y=a+b(lnx)?

[sick steppe]

If it were down, it'd be f(x)-1

[patent lance]

so itd be to left?

[sick steppe]

cause f(x) are y values

[patent lance]

Yeah

that makes sense

[sick steppe]

no, horizontal translations are backwards, x-1 means 1 right

[pastel cloud]

if you have (x-1) what value of x will make it zero?

[sick steppe]

^

[patent lance]

1

[sick steppe]

right, so the x=0 point gets moved to x=1

[patent lance]

so to the right?

dude

i feel so dumb

[pastel cloud]

lol don't

[sick steppe]

oh trust me horizontal transformations period are hard when learning them

[pastel cloud]

it's learning new things

[sick steppe]

cause everything is backwards to vertical

[patent lance]

im not even sure

what type of equation

the graph is

looks like absolute value

but its like

wonky

[pastel cloud]

it's not it's a piece wise function. Not just one function.

[patent lance]

ah

[pastel cloud]

all that matters though is all x values are shifted in that example

[patent lance]

usually i use demos

to figure these out\

[pastel cloud]

if it was f(x-1) + 1 then all x values shift right by 1 and all y values would shift up by 1

[patent lance]

why would y values shift up

oh

yes

yes

[pastel cloud]

the constant

u know

[patent lance]

so this?

[pastel cloud]

yes

[patent lance]

i got half credit

on it

whatever ty for your help realy appreciated

[pastel cloud]

half credit?

[patent lance]

.6/1

[pastel cloud]

it should be a full mark

[patent lance]

its fine ill just take it

Oh

maybe its because I put too many points

that was it, next problem was extremely similar

[pastel cloud]

@sick steppe do you think u can help me with my question?

[sick steppe]

yeye

[sick steppe]

yeah, use remainder theorem

[pastel cloud]

I know how to divide polynomials, and I also know how to use synthetic division.. but im not sure how to work backwards

[sick steppe]

do you know remainder theorem?

[pastel cloud]

if there's a remainder than x+3 is not a factor / zero

[sick steppe]

yes or no lol

[pastel cloud]

i don't think so

[sick steppe]

If I divide a polynomial P(x) by x-a, P(a) will be the remainder

[pastel cloud]

ok so

3?

I would sub in 3 for x?

[sick steppe]

sub in -3, since you want the root value

P(-3)=5 by Remainder Theorem

[pastel cloud]

hmmmm

ok

let me try this

2=K ?

[sick steppe]

ye

[pastel cloud]

I don't think I want to know the questions you have problems with lol

What level are you at?

[sick steppe]

1st year uni

[pastel cloud]

Good stuff.

[potent rover]

If at some point the derivative of a function is 0. And the double derivative is also 0, will that point be a point of maxima or minima or will it be none of it?

I am experiencing this problem with the above question

Answer given in (3)2

But double derivative is also 0 at that point

Pls help

[sick steppe]

if first derivative = 0, then it's a horizontal tangent

if 2nd = 0 then it has no concavity

so it'll be a saddle point

(Think of x^3 at x=0 for what a saddle point looks like)

[patent lance]

ok so

I got this question right

but it took wayy too long

and i needed demos

desmos

so i feel like ive been doing it wrong

what is the proper way to go about that question

sorry for asking so many questions

[potent rover]

so it'll be a saddle point
@sick steppe Thank you

[quaint mason]

idk how to start, do i plug this into calculator and use lnReg in Calc?

[vapid mica]

damn u guys are really struggling huh

[sick steppe]

@pastel cloud same thing with the last one i helped you with.. except you have to solve 2 equations for 2 unknowns

[quaint mason]

yeah, do u have any ideas? @vapid mica

[pastel cloud]

@sick steppe I looked at the solution for it and I still couldn't understand it

It's really disheartening

I could just be tired.. had a long day yesterday and been doing math for 5 hours today.

[elder charm]

$$p(x) = (x+1)(x+a) + 4$$
$$p(x) = (x-5)(x+b) + 4$$

• that's what being divisible with remainder means. a and b are some constants. That means:
  $$
  p(x) = x^2 + (a+1)x + a + 4 = x^2 + (b-5)x -5b + 4
  $$
  Two polynomials can only be equal when all their coefficients are, so we have a system of 2 equations for 2 variables:
  $$
  a+1 = b-5
  $$
  $$
  a+ 4 = -5b + 4
  $$
  By solving them for either of the variables, we can obtain p by substituting into the polynomial equalities above. So:
  $a = -5b$, $-5b+1 = b-5 \implies b = 1$, therefore $p(x) = (x-5)(x+1) + 4 = x^2 -4x -1$

@pastel cloud

[obsidian monolith]

ConfusedReptile:

[pastel cloud]

I tried reading it a few times I'm just not understanding completely.

So for the first equation you subbed in (x+1) and 4 then solved?

which gave you x^2 + (a+1)x + a + 4?

[patent lance]

in f[g(x)]

what do the sqaure brackets

[pastel cloud]

f of g(x)

it's the same as f(g(x))

example if g(x) = 2

you would then input it into f(x) = f(2) and then find the answer.

g(x) = 1 + x

f(x) = 3x

What is f[g(x)]? if x = 1

g(1) = 1 + 1 = 2
f(2) = 3(2) = 6

therefor, f[g(1)] = 6

It's a composite function @patent lance

Two functions combined.

[viscid thistle]

This chat

Will save my life

Thank you for the person who made this discord

[patent lance]

@pastel cloud ty got conifused when it wasnt normal parentheses

[patent beacon]

No, if your life is threatened we cannot save it for you

[pastel cloud]

if his parents will kill him if he gets a bad grade in math.... lol

[viscid thistle]

^

[patent lance]

f(g(0))

in that

would you find f of g but input x as 0

[patent beacon]

Take 0, put it into g. Then whatever you get, put that into f

Or instead yeah, there's a single function we call f o g that is just the combination of those two

[tribal vapor]

can someone help with this?

[zenith jacinth]

is it possible to find the height using 2 angles and 1 side?

[quaint mason]

Yes

ASA theorem

[cobalt storm]

#11 I need help with

I don’t understand the theory

[worthy summit]

x+5/x+1 is less than or equal to x-2/x-1

[patent beacon]

Yes that's correct

[earnest jungle]

$\frac{x+5}{x+1} \leq \frac{x-2}{x-1}$

[obsidian monolith]

LifeSource:

[patent beacon]

(x + 5)/(x + 1) - (x - 2)/(x - 1) ≤ 0

Then, combine them into one fraction

Cross-multiply to make the denom the same

Note that this is the exact same process as
5/2 - 5/3

Just, with x in it

[worthy summit]

5(-2)-3/(-2-1)(-2+1)

[patent beacon]

Where'd your x go haha

First fraction, multiply by (x-1)/(x-1)

[earnest jungle]

$\frac{(x+5)(x-1)}{x^2 - 1} - \frac{(x-2)(x+1)}{x^2 - 1} \leq 0$

[obsidian monolith]

LifeSource:

[patent beacon]

Yeah that's the end goal

[earnest jungle]

$\frac{(x+5)(x-1)-(x-2)(x+1)}{x^2 - 1}\leq 0$

[obsidian monolith]

LifeSource:

[earnest jungle]

$\frac{x^2 + 4x - 5 - (x^2 - x - 2)}{x^2 - 1}\leq 0$

[obsidian monolith]

LifeSource:

[earnest jungle]

$\frac{5x - 3}{x^2 - 1}\leq 0$

[obsidian monolith]

LifeSource:

[charred frigate]

i feel stupid that i cant do this
solving the inequality 3+20/x >=0

$3+\frac{20}{x}\geq 0$

[obsidian monolith]

nix:

[full pagoda]

isolate x

what's the first step?

[charred frigate]

i got 1/x >=-3/20

[full pagoda]

not wrong but how you flipped the entire thing is odd lol

[charred frigate]

but i tried flipping everything and switching the inequality sign but that doesnt work apparently lol

[full pagoda]

i'm not sure why you have everything flipped

walk yourself through it

step #1

[charred frigate]

lol tutoring a student in it so gimme a min

[full pagoda]

gotcha haha

[charred frigate]

ok so im just nervous about multiplying x because it can be negative or positive

do we need to split it up into cases

[uncut mulch]

there are multiple approaches.

taking cases is one of them

[full pagoda]

i would set up a line and find where each solution is positive or negative

honestly not too sure what that's called, it's been a while

[charred frigate]

same

[full pagoda]

i'll draw out what i mean real quick

[charred frigate]

i havent had to do something like this in years

[uncut mulch]

alternatively you can multiply both sides by x^2 (which is known to be non-negative)
so that you don't have to worry about the direction of the inequality

[uncut mulch]

and solve the resultant quadratic ineq using curve/wave method

[full pagoda]

this is what my approach would be

then i'd take a number in between each of those to find out when it is positive/negative

and that's what you get for answers

so it'd be (-infinity, -20/3) U (0, infinity)

because those solutions are positive, which satisfy the original equation

of being greater than or equal to zero

@charred frigate does that make sense? i can explain if you'd like

[charred frigate]

so basically identify critical points and test?

[full pagoda]

yes

[charred frigate]

thats definitely a lot smarter than trying to finagle it algebraically

[full pagoda]

haha yeah it's a bit easier

[quaint mason]

i get the concept but i just dont understand on how to solve it.

[viscid thistle]

I have no idea what the question is asking me

😢

How am i supposed to read this table?

[iron acorn]

first, start with finding g(1)

g(1) basically means g(x) when x=1

when x=1, g(x)=0

the question is asking for f(g(1)), and now that we know g(1)=0

we have to find out what is f(0)

and f(0) is 4

so f(g(1))=4

[willow bear]

@viscid thistle do you know what function composition is?

if not, it might do you some good to read up on it

oh god you multiposted

[swift jasper]

Hi guys, my math teacher told me to prove the Cauchy-Buniakovski-Schwarz inequality using mathematical induction. I solved P(1) P(2) and P(3) correctly and I also wrote P(K) and P(K+1), but I am having some difficulties proving P(K+1). Any help will be much apprecitated. I have also attached P(K) and P(K+1)

[viscid thistle]

Guys for the second one I'm getting the answer to be 3430, while the textbook says that the answer is 4606

my methodology for it would be to make two cases, one where the husband served and one where the wife served

and just add them both at the end

using this method you get 8C4 x 7C5, when the husband is serving

and when the wife is serving you get 8C5 x 7C4

but when you add these two you get 3430

Nice name

[tropic wyvern]

@viscid thistle you forgot to account for the case where both husband and wife did not serve

the best way to solve this is to find the number of committees where both husband and wife are serving and subtract that from the total

you'll find that you do get 4606

[viscid thistle]

Ahhh, thanks a ton mate that makes sense

[blissful ridge]

The flaw in your above Calculation was you didn't consider the case neither of them served

[lament fiber]

What’s the range of sin(cos x)?
My textbook says [-sin 1, sin 1] which as I understand comes from [sin(-1), sin 1]
but if we take f(x)= sin x and g(x)= cos x
then q(x) = f o g = sin(cos x)
wouldn’t that mean the range of q(x) is the range of f(x) that is [-1,1] ?

I understand that since -1<= cos x <= 1 you’re only getting the set of values of sin x for x belonging to [-1,1] so where did I go wrong in my approach with the composition of functions

[olive zephyr]

the range of q(x) is the range of f(x) but in a restricted domain (specifically of the range of g(x))

You can think of it as sin(x) being 1 at pi/2 radians but cos(x) only outputting values up to 1 (<pi/2)

[lament fiber]

yeah the domain of f(x) must be a superset of the codomain of g(x) right

oh yeah that’s basically what I said at the end

You can think of it as sin(x) being 1 at pi/2 radians but cos(x) only outputting values up to 1 (<pi/2)
@olive zephyr can you please explain

[olive zephyr]

If you think of cos(x) assigning each real number to another real number

[lament fiber]

doesn’t cos x have the range [-1,1]

[olive zephyr]

yes

[lament fiber]

that doesn’t line up with what you said earlier though (“You can think of it as...”)

[olive zephyr]

basically, sin(x) is 1 at the principal value of pi/2

which is greater than 1

[lament fiber]

oh

[olive zephyr]

so my cos(x) will have to output pi/2 for q(x) = sin(pi/2)=1

[lament fiber]

yeah makes sense now

[olive zephyr]

great!

[lament fiber]

I understood that part(I stated that already) but how do I proceed to find the range from there?

I understand that since -1<= cos x <= 1 you’re only getting the set of values of sin x for x belonging to [-1,1] so where did I go wrong in my approach with the composition of functions
this was what I said

[olive zephyr]

You are now inputting [-1,1]

into sin(x)

[lament fiber]

which I then remembered you need to ensure that codomain of g(x) is a subset of the domain of f(x) as you put it

[olive zephyr]

and the range of that is [-sin(1),sin(1)]

[lament fiber]

so you need to find the maximum and minimum values of sin x for x in [-1,1]

yeah but how do I get there

finding the max and min
without calc

[olive zephyr]

You can look at the graph

[lament fiber]

yeah I guess

thanks for answering!

[olive zephyr]

welcome!

[pseudo sonnet]

https://i.imgur.com/b3wVfAx.png

how do I get a total flight time equation

do I solve the first equation for the variable t?

[lime bolt]

use s=ut +1/2(at^2)

in the vertical direction, so s=0, and then u can easily find t

[pseudo sonnet]

i got it

thanks though

[robust sky]

any suggestions to go forth without a calc?

[patent beacon]

You just wouldn't simplify the number I suppose. But you can express it

Why 1-r?

[robust sky]

thought it was exp. decay, would it just be r?

[patent beacon]

I'm not exactly sure, really does depend on the class haha. I would suspect arⁿ

But if your teacher has made it clear that 1-r should work instead, I can't trump the teacher

Anyway, divide the bottom line by the top one, that eliminates a

[fiery wren]

is this right?

[uncut mulch]

partially

[fiery wren]

oh, and including csc?

[uncut mulch]

then yes

[fiery wren]

thank you

sorry, last question.

i'm starting to doubt my answers lol

[uncut mulch]

it's a bad question

[vapid mica]

does anyone know how to tell if its cos or sin from the graph if horizontal translations are present?

[viscid thistle]

Yeah prob

[vapid mica]

i just need to know how you would know the difference

[viscid thistle]

Usually cosine starts at the top and goes down and sine starts from the bttom

That's how I figure it out but I may be wrong

In your case, pretty sure it would be a sine function

You could also test those in desmos

[vapid mica]

yeh ik that but the graph translates horizontally

i dont wanna do that

im gonna be screwed when we have in person tests

wait im dumb nvm there isnt a horizontal translation

its sin yeh ur right

[viscid thistle]

Exactly lolll

Vertical doesn't change anything

But yeah

[vapid mica]

it flips doe right

so its negative

[viscid thistle]

Uh kinda forgot trig functions lol

[vapid mica]

yeh

[viscid thistle]

That might help you

But ya I think it's a sine function

[vapid mica]

yeh it was

[viscid thistle]

Was it negative?

[vapid mica]

naw

i compared to a regular sin graph

and it didnt flip

just moved up

[viscid thistle]

kk bet

[naive prism]

how do u do this
i got the equation is (x-7)^2+(y-8)^2=r^2
and that the other equation is y= -3/5x + 19
but like what do i do from there

[sick steppe]

@naive prism r would be the perpendicular distance between the centre of the circle to the tangent point

[mossy tiger]

How did i do this wrong. I did long division. (x^2+8x+7)/(x-1)

I got x + 7 as the quotient

[uncut mulch]

looks like you messed up your division

[honest depot]

hello, im starting precalculus and was doing trig identities. Can anyone help me solve this, and perhaps give the steps?

tanX + cotX

the goal is to simplify using trig identities

[full pagoda]

what do tan and cot break down into?

[honest depot]

tan should break into sinx/cosx

and cot to cosx/sinx

[full pagoda]

correct

now you want a common denominator to combine the fractions

[honest depot]

so then itd be (sinx/cosx) + (cosx/sinx)?

[full pagoda]

yes

[honest depot]

this is what i have: (sinxcosx/cosxsinx)+ (cosxsinx/sinxcosx)

i feel that thats wrong

is it?

[full pagoda]

yes

[honest depot]

what would it be

[full pagoda]

what is the least common multiple of the fractions (sinx/cosx) and (cosx/sinx)?

[honest depot]

im not sure

[full pagoda]

if you want to combine both fractions, you need to get a common denominator

what is cos(x) missing that sin(x) has?

(not a trick question)

[honest depot]

one sec

sin

[full pagoda]

yes

and what is sin(x) missing that cos(x) has?

[honest depot]

cos

[full pagoda]

yes

so to get a common denominator...

[honest depot]

so then the denominator is cosxsinx?

[full pagoda]

yes

[honest depot]

but then id have to multiply the numerators as well

let me do that real quick

[sick steppe]

You were close with your original answer, just slightly off

[honest depot]

ok so heres what i have

sin^2x+cos^2x/sinxcosx

[full pagoda]

correct so far

now what does the numerator equal?

[honest depot]

1

[full pagoda]

there ya go

so simplify it further

[honest depot]

so its 1/sinxcosx which becomes cscxsecx

wow

omg

[sick steppe]

Ye

[honest depot]

thank you very much

this was a very uplifting experience ngl

[sick steppe]

You did the most repeated/used trig identity question there is🥳

[honest depot]

lol

i get the other ones fast, the fraction ones mess me up

[full pagoda]

yeah, trig identities can suck a lot

[sick steppe]

But legit it is a good question: you do the 3 common trickd

[full pagoda]

i always go back to one place: if you can't think of anything to do with it, break it down into sin and cos

[sick steppe]

Sin and cos everything, common denominator, pythagorean and reciprocal

[patent lance]

so my teacher just did the thing on the right

and i was a bit confused

she factored out a negatibve

from -x^2-x+12

nvm i get it

[mossy tiger]

How do I find it if I can't identify the x-int?

[honest depot]

thats not centered

[mossy tiger]

so there isn't a way to solve for it?

[honest depot]

this is a weird one- the two x intercepts should be equidistant from the y intercept, but they are not

[mossy tiger]

what if there aren't x-intercepts

[honest depot]

so do you see the two points where the line crosses over a coordinate exactly

its (-1,-5) and (2,0)

[mossy tiger]

how do you know that\

[honest depot]

thats how you know what the amplitude is- in cot and tan graphs there is no height, so the A value determines how above or below the line will cross a point from its center

look at the graph again, the line crosses over (-1,-5) and (2,0)

so yeah that graph is really weird

they should both be +/- 3 away or +/- 1 away- it can't be both

if one of the trig identities is --> 1+tan^2x=sec^2 , can it be use like this? 1+tanx=secx

is anyone still up

wait

hm

im at this step of the problen

this is the question

so the lcm is secx^2-1

so its like

cos(x)(secX-1) + (cosX)(secx+1)/(sec^2x-1)

(sec^2x-1) simplifies to tan^2x

can anyone help after this step? i now have cos(x)(secX-1) + (cosX)(secx+1)/(tan^2x)

my final asnwer is 2cos^3x/sin^2x

if anyone can confirm or reject that that'd be great

@strong ermine what do you think, sorry for the ping

[strong ermine]

no idea sorry

[honest depot]

lol

[strong ermine]

just browsing through channels

[honest depot]

i really hate the one with fractions ughh

[strong ermine]

while procrasinating

[honest depot]

well it was nice to meet you

[strong ermine]

i hope you get help

try going to questions tab and asking this

and after like 15 min if no one responds

ping helpers

[honest depot]

thank you

[viscid thistle]

how do I decompose this?

this is what i did so far, is everything correct? I'm not very confident in decomposition and still practicing it.

[strong ermine]

yeah it looks good to me

@viscid thistle

[uncut mulch]

the degree of the numerator is higher than the denominator so you should probably do some long division first

[gilded brook]

I've gotten this problem wrong 3 times now I'm using the Law of sines for it and I'm either doing it wrong or messing up somewhere

What am I suppose to be doing?

[astral mantle]

what answers are you getting?

[gilded brook]

It changes each time but for my last one I got

[astral mantle]

i would use the same significant figures for the answers

and b is close but not correct

wait this is a different question than the last one

wtf

[gilded brook]

Significant figures?

[full pagoda]

@gilded brook you're given tthe measure of angle a, the side length a, and the side length b

what can you figure out with these?

[gilded brook]

So a/sinA = b/sinB = c/sinC

So I can figure out Angle B

[full pagoda]

correct

so, let's do that

if you were to isolate B, what would you get?

[gilded brook]

47.34026985

And

132.65973014

[full pagoda]

not what i got

hm

what did you do to solve for B?

[gilded brook]

Simplfied left side
Then got sinB alone by times it both sides
dividing 25sqrt2 both sides
arcsining the right side

[full pagoda]

25sqrt2...?

where did that come from?

[gilded brook]

from 1/sin(45)

to csc(45)

[full pagoda]

why did you do 1/sin(45)

[gilded brook]

Separate fractions

[full pagoda]

what

[gilded brook]

so its 25/sin45 then I did 25/1 and 1/sin45

(25/sin45)(1/sin45)

=25/sin45

[full pagoda]

where are you getting a 45 degree angle from

[gilded brook]

Thats angle a

[full pagoda]

angle a is 50

[gilded brook]

thats old angle a I'm solving picture above it

[full pagoda]

ah

give me a sec then

[gilded brook]

kk

[full pagoda]

i got 47.34 degrees for angle B

okay cool

so did you

[gilded brook]

Yeah

Is the standard rounding to the second decimal?

If it doesn't say what to round it to

[full pagoda]

i normally do that, yeah

if you really want, you can use significant figures

[gilded brook]

Idk what that is

[full pagoda]

but it seems you haven't learned that yet since you questioned it before

yeah

have you taken chemistry yet?

[gilded brook]

Nah

[full pagoda]

you'll learn then

for now, don't worry about it

[gilded brook]

I won't be taking chem anytime

[full pagoda]

ah

lol

[gilded brook]

But now with angle B and angle A I can get angle C

[full pagoda]

132.65973014
@gilded brook is this what you got for C?

[gilded brook]

Angle C? or c

[full pagoda]

angle C

[gilded brook]

Angle C I got 87.7

[full pagoda]

correct

[gilded brook]

for a+b+c=180

[full pagoda]

yes

so now you're just missing side c

[gilded brook]

and I can get by doing 25/sin45=c/sin87.7

[full pagoda]

yes

[lean tusk]

how do I get the range for this function?

[gilded brook]

Which is 35.33

[full pagoda]

👍

[gilded brook]

Says it's all wrong

😦

[full pagoda]

oh shoot

been a while since i've done law of sines

forgot about amibiguous triangles

damn that's my bad i led you in the wrong direction

[gilded brook]

Its all good, I got 2 more attempts for this same problem then it'll have to switch

Do you know the steps for the one obtuse angle?

[full pagoda]

if i remember correctly, when you're solving for angle B, you just take that number and subtract it from 180

[gilded brook]

Do I subtract angle A with it?

[full pagoda]

no, you're given angle A

[gilded brook]

So I would just 180-47.34 and that would be angle B?

[full pagoda]

i believe so

it's been a while, but i'm pretty sure that's how you do it

[gilded brook]

and then you can find angle C The normal way?

[full pagoda]

yes

[gilded brook]

and you can find c the normal way?

[full pagoda]

yes

this all depends on whether angle B is correct, which i think i remember

[gilded brook]

So I got angle C Right

But I got c wrong

[full pagoda]

what did you do for c?

[gilded brook]

angle C = 2.34

and I got c=0.082

I did the same thing as before but instead of 87.7 I used 2.34

[full pagoda]

i got c = 1.44

[gilded brook]

By doing 25/sin45=c/sin2.34

?

[full pagoda]

i do it the other way around, but as long as you keep it consistent, it works either way

[gilded brook]

Weird let me try again.

[gilded brook]

Now I'm getting 25.4

c= 25.4

[full pagoda]

how are you isolating c

?

[gilded brook]

Same way you just showed

[full pagoda]

are you sure?

[gilded brook]

Yeah

[full pagoda]

So you're typing into your calculator: 25sin(2.34)/sin(45)

and it's giving you 25.4?

[gilded brook]

Yeah

[full pagoda]

that's uhh

odd

[gilded brook]

Tried it on my hand calculator, and two website calculator

All 3 are giving me 25.4

[full pagoda]

am i going insane

can you take a picture?

[gilded brook]

Heres Mathaway

https://gyazo.com/7c358e7c1c4928fce32c02d3f1849226

Heres symbo

https://gyazo.com/3297821ab1839edc9cdf2a8953be52d7

And I cant take a pic of hand calc

But its a TI-84

[full pagoda]

ah

you're in radian mode aren't ya

[gilded brook]

Maybe? I'm not sure

[full pagoda]

click "mode" next to the blue 2nd button on your calculator

and go down to the 4th set of options

where it says radian and degree

switch to degree

[gilded brook]

Yeah I see that

Alright now I get 1455.41650302

[full pagoda]

that's a few decimals off

[gilded brook]

But should I answer in radians since it's a side and not an angle?

[full pagoda]

no, radians are still dealing with angles

an angle measurement can be converted from degrees to radians by multiplying it by pi/180

[gilded brook]

1455.40254551

Thats with the exact thing

[full pagoda]

the decimal is off

[gilded brook]

Still?

[full pagoda]

yeah

[gilded brook]

That was just by putting in 25sin2.34/sin45

[gilded brook]

Hmm

I got it

[full pagoda]

lol good

[gilded brook]

Yeah Idk what I was doing differently

[full pagoda]

haha

[gilded brook]

I just turned the calc off and then back on

And it gave that number

[full pagoda]

nice

[gilded brook]

Alright man thanks for the help!

[full pagoda]

no problemo

[lean tusk]

how can I get the range of this function?

[full pagoda]

also, side note: whenever you see the degrees symbol, make sure your calculator is in degrees

[gilded brook]

Will do!

[full pagoda]

how can I get the range of this function?
@lean tusk do you know what the range is for a normal csc function?

[lean tusk]

yeah, (-oo,-1] U [1,oo)

[full pagoda]

correct

now apply the vertical transformation to that

[full pagoda]

the vertical translation is down 2

wait is it telling you the answer or is that your answer

[lean tusk]

oh wait I got it

[full pagoda]

alright

[lean tusk]

that was my answer

wait nvm I don't got it

this is the expected answer

[full pagoda]

ah, there is another vertical transformation

[lean tusk]

oh yeah

thanks!

[gilded brook]

Can I get help with this final part?

At the end I do c/sin21=243.9/sin115.92659176

[gilded brook]

I figured it out was using rounded 21 instead of exact value

[viscid thistle]

is that the trig section in pre cal???

@gilded brook

[lean tusk]

how do I find the domain for this function

I know the parent function's domain is (2n+1)pi/2

[blissful ridge]

You have to find the values of x wherever sec(2x+4) is not defined

[lean tusk]

yeah I understand that

but I don't understand how to find those values

[gilded brook]

is that the trig section in pre cal???
@IcyKing#4126 no this is pre calc

[hasty nest]

@lean tusk is this what you meant when you typed out the parent function’s domain or is it something dif? I can’t tell

Bc we can use this to help with that last question

[lean tusk]

yeah that's right

I figured it out though thanks

[hasty nest]

Oh ok

Glad you figured it out!

[tough onyx]

help

[lean tusk]

actually @hasty nest would you mind running that by me to make sure I got the right way to solve it?

[hasty nest]

Yeah so basically what I did was I just set the domain of the parent function equal to the inside of the parenthesis (so in this case 2x+4) and then I solved for x which gave me the domain @lean tusk

Did you use a dif method?

[lean tusk]

I found a workaround of sorts

thanks!

[hasty nest]

Ah ok

Also if you ever wanna double check then doing a quick sketch of the graph can help sometimes

np

[tame wedge]

Is there any way to find an interval for 'a' for which the polynomial: 42x^{5}-36x^{4}-20/ax^{3}+36x^{2}-18x-4 has some number of solutions?

Say I wanted to know for what values of 'a' there were 3 solutions

[willow bear]

what is the equation you are analyzing? $$42x^5 - 36x^4 - \frac{20}{a} x^3 + 36x^2 - 18x - 4 = 0?$$

[obsidian monolith]

Ann:

[halcyon edge]

Anyone free?

[viscid thistle]

So you don't know what the greatest integer function is?

(Which says it in the name)

Also this isn't a question, you just stated f

[halcyon edge]

I mean literally this was the question

Im confused maybe im dumb

[viscid thistle]

Where is your doubt at?

So you don't know what the greatest integer function is?

[halcyon edge]

This is the solution

Like why negative 1 there

[viscid thistle]

Yeah so you didn't post the question

As i said

The question is probably about studying continuity

Am i right?

[halcyon edge]

Yes u are

Like why we replace x with negative 1

[viscid thistle]

Because of floor function/greatest integer function

,w plot floor(x)

[obsidian monolith]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Al𝟛dium#4084

[halcyon edge]

Ohh nice i see

[viscid thistle]

Do you see how it rounds to the integer -1 in this case

[halcyon edge]

Actually im new to this concept will google about it more

They never told us about it in the class

[viscid thistle]

That's weird

Be sure to check it

[halcyon edge]

Yes thanks a lot

[copper breach]

hey can someone help me with a problem

[sick steppe]

@copper breach we can help if you post the question

[celest gulch]

Hey so far I have this for my equation

shouldnt the x^2 cancel out?

my graph still touchs y=3

[neon escarp]

quick question: for an even function with a jump discontinuity, does it need to be on both sides?

[sick steppe]

@neon escarp what's "it"

[neon escarp]

the jump discontiniuty

[sick steppe]

a jump discontinuity is vertical, so it's 1 dimensional

Im not entirely sure what you're asking

[neon escarp]

yeah but for an even function doesnt it need to be symettrical

[sick steppe]

Even functions are symmetric about the y axis

reflectionally symmetric*

[neon escarp]

so if theres a jump at x=-3

would there be one at x=3

[sick steppe]

Yes there would have to be one

[worthy summit]

alright soo

I need some hel

help*

[sick steppe]

ok with what

[worthy summit]

difference quotient of 4^x

$$4^x

[obsidian monolith]

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

[sick steppe]

[f(x+h)-f(x)]/h

just plug in the function into that

[worthy summit]

hold on

$$frac{4^x+h-4^x}{h}

I think I did that right

maybe not

[obsidian monolith]

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

[worthy summit]

ooof

[sick steppe]

$frac{4^(x+h)-4^x}{h}$

[obsidian monolith]

moshill1:

[sick steppe]

rip

anyway

[worthy summit]

alright well we know what we are trying to say

[sick steppe]

yeah

[worthy summit]

so I know how to set up but

I don't know how we get the answer which is 4^x(4^h-1)/h

the steps to get there illude me

[sick steppe]

do you know power/exponent laws?

[worthy summit]

yes

which is what is confusing

unless im jsut missin something

[sick steppe]

what's the one where you multiply 2 things with the same base but different exponents

(a^b)(a^c)

[worthy summit]

you just add the exponenets

[sick steppe]

yeah

so... if i have 4^(x+h)

what is that equal to?

[lime bolt]

4^x * 4^h

[worthy summit]

so if you subtract 4^(x+h)-4^x

[sick steppe]

@lime bolt that question wasnt for you

[lime bolt]

[sick steppe]

@worthy summit you agree 4^(x+h) = 4^x * 4^h right?

[worthy summit]

yes

I got that far

lets go coms

[sick steppe]

so you have [4^x * 4^h - 4^x], what can you do?

[worthy summit]

wehre do you get the 4^x * 4^h from? I thought you would be adding those two terms

[sick steppe]

4^(x+h) = 4^x * 4^h

[worthy summit]

no it's 4^x + 4^h

[viscid thistle]

Why are you doing limits

[worthy summit]

wait

[sick steppe]

no, multiply things with the same base means add the exponents

[worthy summit]

ahh yeah

ok that's where im getting confused

[sick steppe]

so do you see now?

[worthy summit]

can you go math voice coms?

[sick steppe]

no

[worthy summit]

ok

[lean tusk]

can someone walk me through how I'd get the domain for this function? I've been getting -3/4 + 3πk/8 but the answer is -3/4 + πk/8

[worthy summit]

okay so I know that 4^x * 4^h

[sick steppe]

yeah

[worthy summit]

so know we have 4^x * 4^h - 4^x all over h

[sick steppe]

yep

[worthy summit]

ok

so

from there we would combine terms

[sick steppe]

combine terms?

[worthy summit]

oh man

im having a bad day with this crap

cuold you walk me through the rest of it

[sick steppe]

factor out the 4^x and you're done

[worthy summit]

wdym

[sick steppe]

you want to show 4^x*4^h-4^x = 4^x[4^h-1]

[worthy summit]

right

ohhhh

itj ust clicked

okay

thankyou duhhh

[sick steppe]

ye np

[worthy summit]

I was getting confused with where the -4^x was placed

not realising that if you just put like this -4^x(4^x * 4^h)

you can factor it out from there

and boom

@lean tusk I don't think I can help you, but someone else may be able to

[lean tusk]

I'll repost the question so people can see it 🙂

[worthy summit]

Thank you @sick steppe

[lean tusk]

can someone walk me through how I'd get the domain for this function? I've been getting -3/4 + 3πk/8 but the answer is -3/4 + πk/8

https://cdn.discordapp.com/attachments/363224154469826562/767826325108949022/unknown.png

the way I've tried solving this is by setting 4x+3 = π/2 + πk (the parent function's domain) and solve for x, but that leaves me with the wrong answer

[echo pelican]

@lean tusk

So you need to consider where secant is undefined

It happens at pi/2 and 3pi/2

Actually hold up

I think the solution they gave you may not be right

You get it by setting 4x+3 = kPi/2

But this is only right if you take k to be an odd number

Since secant is undefined at odd numbers of pi/2

[shadow plaza]

no. That is only true when k is even

[echo pelican]

Checking for secant to be undefined this would be at pi/2 or 3pi/2

[shadow plaza]

That too a multiple of 4 only

[echo pelican]

So that would be odd

Going on it would also be undefined at 5pi/2, 7pi/2 etc

[shadow plaza]

The solution says πk/8. Let's say k=3, then how is sec(3π/8) undefined?

[echo pelican]

You are saying x would equal that