precalculus

[hoary valley]

@stuck lark I'm not sure how to get rid of -4.9

[stuck lark]

-4.9 is multiplied by t^2, what's the inverse operation of multiplication?

[hoary valley]

Dividing?

[stuck lark]

sure

[hoary valley]

So ?

I divide everything by -4.9 ?

[stuck lark]

sure

[hoary valley]

There is no way I can solve this now..

I don't know 2 numbers multiplied equals to -1.63

[stuck lark]

what oscillatingEquilibrium posted is useful here

[hoary valley]

@gentle vigil Can you tell me what's (-b) relate to ?

[gentle vigil]

$ax^2+bx+c=0$

[obsidian monolith]

oscillatingEquilibrium:

[gentle vigil]

Is your equations

$\Delta =b^2-4ac$

[obsidian monolith]

oscillatingEquilibrium:

[gentle vigil]

Put a,b,c values from your equation into the equation I sent you

[patent beacon]

It's a quadratic in disguise

Let e^x = u

[tidal rain]

Hey how do i go about simplifying this

[patent beacon]

@tidal rain
Common factor the top and bottom

@queen sundial
Np. Did it work?

[trail badge]

Hum

x<=In(5)

Not >=

I think

[patent beacon]

u² - 3u - 10 ≤ 0
(u + 2)(u - 5) ≤ 0
-2 ≤ u ≤ 5

-2 ≤ e^x ≤ 5
But of course, e^x is always greater than 0

[trail badge]

Really?

Ohh nvm

Yeah >0

[leaden canyon]

If in a geometric sequence g_1=3 and g_5=243/16 how do i find the konstant quotient?

[proud sparrow]

??

geometric progression?

are you finding the common ratio?

@leaden canyon

[leaden canyon]

Yes! Sry common ration is the correct phrase! Sry 😅

[patent beacon]

A geometric sequence follows
t(n) = ak^n

@leaden canyon

Plug in your two points, can you get k?

[leaden canyon]

Kaynex Ty i got it now

[green zenith]

i need some help equating exponents

i have this problem 9^(x)-(6)(3)^(x)+8=0

what i did was change 3^x to y, so i have 3y-6y+8=0.

then i got y = 8/3

then i change y back to 3^x

solve x with logarithms and i found x=0.893

[spring thunder]

yeah there's a lil problem with your 9^x -> 3y

[green zenith]

is that incorrect?

[spring thunder]

yeah

3y would just be 3*3^x ie 3^(x+1)

now think about this, i can rewrite $9^x$ as $(3^2)^x$ right?

[obsidian monolith]

emeric75:

[green zenith]

ahhh

thats where i went wrong

i thought multiplying 3^x and 3^x would result to 9^x

[spring thunder]

well that's the case indeed

3^x * 3^x = 9^x

[green zenith]

okay then. thanks for the correction.

[rigid sun]

[spring thunder]

[atomic bone]

what property are they using to get from the first one to the second?

[serene heath]

Compound angle formula

[atomic bone]

thank you

[green zenith]

i need some help solving this equation (3^(x)-3^(-x))/(3^(x)+3^(-x))=1/4

[serene heath]

Multiply top n bottom of lhs by 3^x

[green zenith]

wow

that is magical

thank you

[green zenith]

is there any answer to this equation log6(x+2) + log6(x-3) = 1?

the x's are 0 and 1

[proud sparrow]

what can you do with logs?

when you add them?

[green zenith]

i multiply them. the product rule.

well, i actually alreardy have the solution

with the x's i have, there are no answer to the equation

[stuck lark]

i multiply them. the product rule.
could you write down what you mean? and how it applies to the problem?

[green zenith]

log6((x+2)(x-3))=1

[stuck lark]

sure, nice, keep going

[proud sparrow]

yeah I found 1 solution

there is, keep going

[green zenith]

okay

so i have log6(x^2-x-6)=1

then x^2-x-6=6^1

is this good so far?

[stuck lark]

yes

[green zenith]

ah i see my problem

im bad

[proud sparrow]

yay you fixed your solution

[green zenith]

then x^2-x-6-6=0

yeah i have bad eyes

then x^2-x-12=0

(x-4)(x-3)=0

ahh

there it is

[stuck lark]

factoring seems a bit off

[green zenith]

ah yes

its (x-4)(x+3)

[proud sparrow]

so what are the solutions?

[green zenith]

well -3 isnt a solution because log (-3-3) and log (-3+2) aren't possible

so its 4 because it outputs positives within the logs

[stuck lark]

very naisu!

[green zenith]

thanks

thanks guys

[tidal rain]

@patent beacon cool and what are these types of problems called?

[patent beacon]

I've seen the term "complex fraction" before

[trim fable]

hey i need help

with this

@patent beacon will u help me? 😛

[formal patrol]

@trim fable what are the 4 roots

[trim fable]

hiii

theres a root at 2,-1,0,3

[formal patrol]

only one of those is right

[trim fable]

there

[formal patrol]

still only one

[trim fable]

1,3,-2,-1

[formal patrol]

are you just guessing

[trim fable]

uh ill actually do it one sec

[formal patrol]

why waste people's time

[trim fable]

-2,1,3,-1

that

well tbh i was actually doing it but u said 1 was right

so i kept doing other stuff so ye

idk

sorry

[formal patrol]

that's still wrong

[trim fable]

im not trying to waste ur time

well then idk

which is why i asked 😛

lol

[formal patrol]

what do you think "double root" might mean

[trim fable]

it means theres a bounce at 3

exponent of 2

[formal patrol]

okay but how does that translate to when you list the roots

[patent beacon]

Try instead to list them as factors

[trim fable]

that one of them would be 3

[formal patrol]

double root

[trim fable]

exponent of 2

[formal patrol]

😐

[trim fable]

is a double root

on the 3

[patent beacon]

In a way, two of the roots are 3

[trim fable]

yes

which is why the power would be 2

and there would be a bounce

at that point

[patent beacon]

That may or may not be ridiculous depending on how you're thinking of it

[trim fable]

ik what double root is tho..

lol

i just told u guys

[uncut mulch]

the factor would be to the power of, two but when listing double roots,
you would write the 3 twice

[patent beacon]

It means there's a factor of (x - 3)²

[trim fable]

yes

thats what i was saying

exponent of 2 at 3

[uncut mulch]

also why do you think x=-1 is a root of f(x)=0?

[patent beacon]

There are four factors, since it's a quartic

[trim fable]

ye

i listed 0 before too

[uncut mulch]

is f(0) equal to zero?

[trim fable]

-1 would be a point to (x+1)

so two points so far are (x-3)^2(x+1)

[patent beacon]

It might be easiest to graph this function to get the info you need

[trim fable]

true..

😛

[uncut mulch]

x= -2 , 1 are roots from the inequality
x = 3 is a double root which is given

but f(-1) = 96 which isn't 0. so why was it included in your list?

[trim fable]

oh coz he kept on saying i was wrong

so i said 0

[uncut mulch]

0 for what?

[trim fable]

the points would be -2,-1,3,1

and ik i said that at first

[formal patrol]

that's still wrong

[uncut mulch]

but f(-1) = 96 which isn't 0. so why was it included in your list?

[trim fable]

-1 is a pointtttttttttttt

when x= -1 the value of y is 96

[uncut mulch]

do you understand that from that, x=-1 is NOT a root to f(x)=0?

[trim fable]

u know..

[uncut mulch]

yes,
so what are your 4 roots to f(x)=0

[trim fable]

?

i think i messed up the bounce lol

there so

-2,1 and 3,3

those are the points

[uncut mulch]

*x coodinates

[trim fable]

ye

my bad

(x+2)(x-1)(x-3)^2

is the equation

[uncut mulch]

not quite,
you would need to include a scaling factor

and use the information of f(-1) = 96 to determine its value

[formal patrol]

would* ?

[uncut mulch]

lol, fixed

[trim fable]

the equation is

f(x)=-3(x-1)(x+2)(x-3)^2

there

ik

and ye thats the answer

[uncut mulch]

yep

[trim fable]

tbh that wasn't hard hmm

i should have began with a sketch my bad

i have another question

i honestly have no clue so ye

wait

is a coz it will always be positive?

coz positive exponents?

[uncut mulch]

not because they're positive

or be more clear

[trim fable]

ye coz the number will always be greater than 0 coz

the number will always be a positive number

[uncut mulch]

can you explain why they'll be positive more clearly?

[trim fable]

umm..

ok

like if u were to have a negative number for x

if u solve with a positive exponent the number will always be positive

and same with positive ones so

if u solve that

the value will always be

greater than0

making the inequality false

[uncut mulch]

is (-5)^5 positive?

[trim fable]

what about for b?

..

i meant to say

an even number

sorry

i wasn't clear 😛

[uncut mulch]

yes that's it

[trim fable]

😄

heheh

sorry

what about b 😛

[uncut mulch]

its positive because the exponent is even

[trim fable]

i actually dont have an idea for that one

ye!!

but for b

how would ik?

[uncut mulch]

similar idea. move everything to one side first

[trim fable]

ok so

-4x^12 -7x^6 + -2x^2 -10

there

so now what

[uncut mulch]

might be easier to see if you moved everything to the other side

[trim fable]

ok

just switch the signs^^

[uncut mulch]

inequality gets flipped too

[trim fable]

yes

ik

😛

[uncut mulch]

were you able to get the answer?

[trim fable]

well

how does that have a solution of -infinty <x < +infinity

do i solve this?

[uncut mulch]

what inequality did you reach?

[trim fable]

idk how i would

oh

4x^12 + 7x^6 +2x^2 +10 >0

so same thing

the exponents are even

and its true?

[uncut mulch]

true for what values of x?

[trim fable]

all

and its

idk

[uncut mulch]

how would it be represented as a set?

[trim fable]

why does it go from - to +

what do u mean

[uncut mulch]

misread the format they had

-inf < x < inf
represents all real x

[trim fable]

ok

uh so

OH

COZ IT WILL

WORK FOR BOTH

ok makes sense

thanks!!

oh i had 1 other question i have an answer but i wanna ask something

so for e and f

well starting with e

i got x<= 2 or x>=3

but the answer is

-3/2<=x or x>=3

and im confused on how they got the fraction

so am I wrong?

[uncut mulch]

yeh

[trim fable]

yeh for what

coz the roots i got were -2,-1, and 3

[uncut mulch]

you should start by factoring the two terms

[trim fable]

which 2?

like what do u mean

[uncut mulch]

factorise (x-3)(x+1) + (x-3)(x+2)

[trim fable]

how tho

[uncut mulch]

do they have a common factor?

[trim fable]

no

well 1

[uncut mulch]

which is:

[trim fable]

1

[uncut mulch]

(x-3)(x+1) + (x-3)(x+2)

[trim fable]

oh thats what u meant

lol

[uncut mulch]

since its a sum of quadratics, it usually won't have the exact same roots as the individual quadratics

so, were you able to factorise it?

[trim fable]

hmm u know

ill ask my teacher this instead

but like

even if u dont factor it

is my answer wrong

[uncut mulch]

it'll be the same as how you factorise
ax + bx

[trim fable]

or would that count as right

x(a+b)

[uncut mulch]

yep and apply the exact same method here

[trim fable]

so

(x-3) ((x+1)+(x+2))

[uncut mulch]

yeh

[trim fable]

ye

but like

[uncut mulch]

=0

[trim fable]

ye

but like

how does that give u -3/2?

[uncut mulch]

and then simplify the (x+1)+(x+2)

[trim fable]

expand?

3x+2

oh makes sense..

😮

[uncut mulch]

wait

[trim fable]

why did they do that tho

[uncut mulch]

not 3x+2

[trim fable]

oops

LOL

i got excited to get 3/2 sorry

[uncut mulch]

no

[trim fable]

really?

what

[uncut mulch]

x + 1 + x + 2 is just 2x + 3

[trim fable]

oh true

LOL

OHHH

i was multiplyingggg

ooooops

didn't see the + lol

hhehe

[uncut mulch]

and now you have
(x-3) (2x + 3) >= 0
and i believe you know how to solve that

[trim fable]

ye

[uncut mulch]

same idea with f)

[trim fable]

3, -3/2

oh alright

[uncut mulch]

^remember the inequalities though

[trim fable]

ok

hey

is it alright if i ask u one last thing?

@uncut mulch

[uncut mulch]

yeh sure

[trim fable]

this when i put it into a graphing calc

im getting this

but the answer key looks like this

why?

like what am i doing wrong?

[uncut mulch]

only the appropriate part of the graph is being used

[trim fable]

oh ok

thanks

for the help

😄

😛

@uncut mulch

u still there?

[uncut mulch]

yeh

[trim fable]

one thing

its a question from this morning i forgot to ask

9c/5+32 >c

so when u solve that and i checked it

why did the answer change the 9c/5 to 4/5?

[uncut mulch]

$\frac{9c}{5} +32 > c$

[obsidian monolith]

ramonov:

[trim fable]

so then 9c/5>-32

they did 4c/5>-32

why?

the answer is -40 btw

[uncut mulch]

you mean -40?

[trim fable]

ye

[uncut mulch]

you left something out in the 1st line

[trim fable]

?

[uncut mulch]

uh i mean first line

[trim fable]

?

[uncut mulch]

9c/5 **-c ** >-32

$\frac{9c}{5} -\frac{5c}{5} > -32$

[obsidian monolith]

ramonov:

[trim fable]

why?

[uncut mulch]

so they could subtract the fractions

alternatively they could've multiplied by 5 first

[trim fable]

where did the 5c/5 come from tho

[uncut mulch]

5c/5 = c

[trim fable]

oh i should know that?

[uncut mulch]

adding fractions 101

c = c * a/a

common denominator etc

[trim fable]

ik but like

i was confused coz

nvm

im kinda confused like

why subtract the c?

oh right..

nvm..

lol

😛

[uncut mulch]

there was a c originally on the right

[trim fable]

ye

heheh

so u multiplied by 5

to get rid of it lol

i just made the c vanish for some reason

well thank u!!!!!

[uncut mulch]

np

[trim fable]

hey

wait how old are u?

@uncut mulch

[uncut mulch]

that's rude

[trim fable]

how? ;-;

[uncut mulch]

never ask people their age

[trim fable]

im sorry 😦

sorry :c

[uncut mulch]

nah, just don't really want to disclose that info

[trim fable]

ok 😛

i was asking coz u know how to do this all so

i was assuming u were older and ye

which is why i asked

coz ur smart lol 😛

[uncut mulch]

there are plenty of people here that are better than me

[trim fable]

well thats alright!

[viral imp]

,, a) x>= \frac{\log(5)-\log(2)}{\log(2)+\log(5)} ---> b) x>= \log(\frac{5}{2})

[obsidian monolith]

Umma.Gumma:

[viral imp]

can someone enlighten me on how to get from a to b?

I mean, are these two inequalities even ...equal?

[heady jewel]

log(a)+log(b)=log(ab)

and a^-1=1/a

these two are enough

[rigid sun]

[viral imp]

not sure I'm seeing it

[heady jewel]

a=5

b=2

log(a^{-1})=-log(a)

now?

[rigid sun]

Just use this

log(a)+log(b)=log(ab) log(a)-log(b)=log(a/b)

Don’t even use the other one

Now do what it says using that formula

[viral imp]

,, \log(10)

[obsidian monolith]

Umma.Gumma:

[viral imp]

this is so funny

I worked around this for like 15 minutes without seeing it

gotta love math

[rigid sun]

Just do it

[viral imp]

thanks guys

[rigid sun]

[viral imp]

yeah done done

thanks!

[rigid sun]

😫

[viral imp]

what?

[heady jewel]

LOL

[hoary valley]

What's this ?

[heady jewel]

the logarithm of a number

with the base of the log as 'e'

a constant

@hoary valley understandable?

[hoary valley]

@heady jewel yes, can you rewrite the equation

[heady jewel]

in terms of y?

or x

[hoary valley]

As a function equation

[heady jewel]

ok

y(x)=ln(x)

[hoary valley]

Oh that's n

[heady jewel]

lol

did you think lax

"lax"

[hoary valley]

I didn't know what that letter was

[heady jewel]

ok

[hoary valley]

Shouldn't it be lg x ?

[heady jewel]

so $\log_e{x}$=$\ln(x)$

@hoary valley its just abbreviations so it dosent matter ya know

[obsidian monolith]

Lionel:

[hoary valley]

Awesome man, thanks ❤

[heady jewel]

🍻

[hoary valley]

I call it the logarithmic function or what @heady jewel ?

[heady jewel]

ya

[hoary valley]

It's the inverse of the exponential function ?

[heady jewel]

yes

ln is

ok?

to specify log _ e

you just write ln

if $a^b=c$

[obsidian monolith]

Lionel:

[hoary valley]

ln x is an inverse of e^x

[heady jewel]

yes

take e^ln(x)

youll see

ok so $a^b=x$ implies $\log_a{x}=b$

[obsidian monolith]

Lionel:

[heady jewel]

thats the definition of the logartihm

its kinda just notation

@hoary valley

understood?

[viral imp]

,, \frac{9^x-1}{3^x-4}<=0

[obsidian monolith]

Umma.Gumma:

[viral imp]

I basically solved this inequality, but I have a general question

I get to the point where I have to define the graph in order to solve the inequality, and the two points are 1 and \frac{log4}{log3}

assuming I have no calculator

is there a reasonable way to determine log4/log3 and understand whether it is < or > than 1?

[hoary valley]

does this shrink or stretch the domain ?

[patent beacon]

Stretches the graph, compared to f(x)

Horizontal stretch by 2

[hoary valley]

Ok I just wanted to make sure thanks for responding.

[novel dirge]

Hello people, do you know of any good resource where I can see how to solve rot functions?

by root functions I meah functions like y= (x-2)/root(x+2)

or something like that

will have that in an exam this week, but our teacher never explained how to do them

[patent beacon]

@hoary valley
That's quite a bit different. It's not easy to relate f|x| to f(x)

@novel dirge
What do you mean "solve"?

[hoary valley]

@patent beacon Ok If I asked you, what's the equation for the graph of y= |x| after shifting it to the left 5 units, ?

[patent beacon]

g = |x + 5|

In general, f(x + 5) is f(x), but shifted 5 units to the left.

That means, whereever there's an x, replace it with x + 5

[hoary valley]

@patent beacon y=|x| +5 this means the range starts from +5 right ?

[patent beacon]

The range of y = |x| + 5 is [5, inf)

[hoary valley]

Yeah got it

Thanks a lot Kayn

[patent beacon]

I think you're trying to ask about the transformation. You're shifting up 5 units

Np, feel free to ask if you need anything else

That's a vertical compression

y = 1/5 |x|
If it helps to see it like that

[novel dirge]

@patent beacon by solve I mean find its x and y interceptions, its domain, sign, asymptotes and plot a graph

[patent beacon]

If you're used to these terms, it shouldn't be too much to apply them onto any function

x and y intercept are setting y or x to zero, and solving

Do you have an example? We can go through it

[novel dirge]

Currently not since I am already in bed

I will post again tomorrow

With the example

[hoary valley]

Are these the same ? both are exponential functions ?

[patent beacon]

Yes, but with different bases

They are not the same function, but behave similarly

[hoary valley]

Yes gotcha

[hoary valley]

What's the name of this function ?

[patent beacon]

y

It's like a polynomial but it isn't since the power isn't an integer

[hoary valley]

@patent beacon If it's not a polynomial nor exponential, what can we call it ?

[noble halo]

Just checking to see if what i did was right

[viscid thistle]

👍

[hexed bolt]

Is expanding an ln function the same way as expanding a log function?

[willow bear]

idk what you mean by "expanding"\ but $\ln$ is just $\log_e$

[obsidian monolith]

Ann:

[hexed bolt]

ok

[glacial island]

my calculator isnt graphing parametric equations correctly

t-84, changed the mode to para

desmos isnt working as well?

[stuck lark]

$3(\sin(t))^3$

[obsidian monolith]

RokettoJanpu:

[glacial island]

thank you, i plugged it in, i think im still doing something wrong?

i think it suppose to look something like

[vernal rapids]

Um

In desmos you make a parametric equation by defining pairs in terms of t

(x(t),y(t))
Where x(t) and y(t) are the stuff you wrote above

(3sin(t)^3,3cos(t)^3)

[viscid thistle]

Hello!

I don't know what does shrink/ stretch a function

I don't know how that would look

[flint river]

well, what could make it go "higher"?

[chilly delta]

could someone help me with this stuff?

I don't understand

[dark fable]

In sum and difference formulas:
Since tan(a+b) = sin(a+b) / cos(a+b)
Is tan(a-b) = cos(a-b) / sin (a-b)?

[green zenith]

i need help finding an equation for something

bacteria initially contains 1500 and doubles every half hour

im not familiar with finding equations

but i did try to find one

i made up this one (InitialAmount)(4t)

where t is in hours

[viscid thistle]

Yeah no.

So every half hour you double your amount correct?

[green zenith]

yes

[viscid thistle]

So let's just say we have 1 bacteria and it can spread by itself.

It first starts with 1, then after 30 mins, 2, then 4, then 8, then 16 etc.

Makes sense?

[green zenith]

ahhhh

yes

[viscid thistle]

So you have (Initial Amount)(Doubler for ever half hour)

[green zenith]

now i understand

[viscid thistle]

So your function is 1500 * ?

What's the ? mark.

[green zenith]

hang on let me figure it out

[viscid thistle]

Kk.

[green zenith]

wait

is it 1500(2)^(2t)?

@viscid thistle

(InitialAmount)(4t) means that the initial amount increases by the initial amount every quarter of an hour

it took me almost an hour to get that, how embarassing

[viscid thistle]

@green zenith Nope, but you're on the right track and it's very close.

[green zenith]

wait, that wasnt the right answer?\

[viscid thistle]

Nope.

Here, after 30 minutes, you know 1500 doubles to 3000 right?

,w 1500(2^{4})

[obsidian monolith]

Display results online and refine query

Requested by leviosa#5599

[viscid thistle]

Is this right?

[green zenith]

i dont understand the equation

whats 4

[viscid thistle]

Oh I meant 2.

Sorry.

,w 1500(2^{1*2})

[obsidian monolith]

Display results online and refine query

Requested by leviosa#5599

[viscid thistle]

Here's your equation. After 30 minutes it's supposed to be 3000.

[green zenith]

uhuh

[viscid thistle]

Wait, is t half hours or hours.

[green zenith]

t is an hour

[viscid thistle]

Oh then you're right.

I thought it was by half hours.

[green zenith]

yes!

[viscid thistle]

Mb.

[green zenith]

no probs

[hoary valley]

How to solve this? I don't know how to get rid of the number on the left of x^2

[green zenith]

at least it was confirmed

thanks homes

[viscid thistle]

Mhm.

Np.

$(2x-6)(2x-6)=2(x-3)2(x-3)=4(x-3)^2$

[obsidian monolith]

leviosa:

[viscid thistle]

@hoary valley

how to get rid of the number on the left of x^2

Other than multiplying 4 on both sides, this is the closest you can get.

[hoary valley]

@viscid thistle I solved it, and got the wrong answer... I got -3 and 3

The right answer is only 3

[viscid thistle]

$(2x-6)^2 = 0 \ 4(x-3)^2=0 \ x=3$

[obsidian monolith]

leviosa:

[viscid thistle]

FYI, what you solved was what x can't equal to.

Plug in -3 and it exists.

@hoary valley

[noble isle]

i kinda forget this domain range stuff can someone help

[noble isle]

-_-

[harsh cipher]

on the third question

in pink, the video explains that 4x(x+3) (X^2 -3x+9)

in the part (X^2-3x+9). He explains different sign after X^2, which is -3x

I don't get why it's using -3x+9.

difference of cubes = (x-y)(x^2+xy+y^2)

it has to +3x - 9 not -3x.

-_-

[stuck lark]

it's flipped for sum of cubes

[harsh cipher]

sigh I don't get it

can you explain please? 😦

[stuck lark]

https://36gu5d4dxary1824ba1o7kkq6uc-wpengine.netdna-ssl.com/help/wp-content/uploads/sites/2/2014/06/Sum-and-difference-of-cubes.png

[harsh cipher]

oh because when we simplified it it's sum of cubes

Perfect you're awesome.!

Thank you

[stuck lark]

np

[uncut mulch]

bad notation inserting those parentheses like that in the first one

[harsh cipher]

He doesn't explain it's sum of cubes, "he just goes different sign this time".

Then how should I insert parentheses when asking on discord?

[uncut mulch]

you're trying to represent the grouping of terms
however,
(x^3 - 5x^2)(-9x + 45) implies (x^3 - 5x^2) * (-9x + 45)

[harsh cipher]

ah I see.

I'll imply that I'm trying to factor binomial

factoring a binomial

[uncut mulch]

either put (x^3 - 5x^2) + (-9x + 45)
in a separate step or jump to
x^2(x - 5) -9(x-5) immediately

[harsh cipher]

Thanks for that

[pseudo sonnet]

https://i.imgur.com/E3B8fEF.png

can someone help me with this

<@&286206848099549185>

[pseudo sonnet]

this is as far as i got im confused

[willow bear]

you plugged each equation into itself of course you're gonna get a tautology

[pseudo sonnet]

oh

was i suppose to substitute my S term equations into the other systems

[willow bear]

tbh

subtract the two equations

you're gonna be able to do some factoring

[pseudo sonnet]

$rs-t + t-st = 1$

[obsidian monolith]

Vici:

[pseudo sonnet]

like that?

[willow bear]

no, you messed up

typo'd

$rs - r + t - st = 1$

[obsidian monolith]

Ann:

[willow bear]

see to it that this can be factored to $(r-t)(s-1) = 1$

[obsidian monolith]

Ann:

[pseudo sonnet]

OH

ok so

I got S = 2

and

r = t + 1

[willow bear]

yup

[pseudo sonnet]

now I have to find r?

r = t + 1

r = 4 + 1 = 5

are these all my possible ordered triplets?

[willow bear]

there is only one

(5,2,4)

no others

[pseudo sonnet]

these are the only values that will make the system of two equations true?

from the original question

[willow bear]

the two equations*

[pseudo sonnet]

you're right

thank you! 🙂 this one made more sense

[hoary valley]

How are these two different ?

[serene heath]

What makes u say they are

[hoary valley]

@serene heath they have different domains.

[blazing parrot]

are you sure?

[hoary valley]

@blazing parrot Why they have different domains?

[blazing parrot]

yea

[acoustic laurel]

@hoary valley You know that rule where stuff that get multiplied together inside a square root can have the root distributed to each thing multiplied? It also applies to division.

I personally don’t really see how its domain is different

The denominator will be both zero in x=-4

Same with x+4=0

sqrt(x+4)=0 will yield x=-4

[uncut mulch]

that only applies when the insides of the square roots are positive

[acoustic laurel]

Ahh

Darn, good to know

Is there a name for that rule

[novel dirge]

Can someone help me with the 3rd and 4th one

With the roots

How to get the asymptotes

And the flow, ai am not sure how to translate it

[novel dirge]

If we get a functio that has a square root which will be an imaginary number for certain parts of the domain, do we modify the domin to exclude those parts and what do we use to get the vertical asymptote?

[serene heath]

,rotate

[serene heath]

Why did u stop

And what does $0^-$ mean

[obsidian monolith]

the one n only:

[serene heath]

@novel dirge ?

[novel dirge]

So

My teacher told us that when we look for the vertical asymptotes we have to look for break points or domain edges

In his case, it should be 0 but since we will get an imaginary number for every x between - 2 amd 2, i took those 2 to be my vertical asymptotes

Noe I have to prove that

We have to do that by having lim x->n- and n+

Where n+ is approaching n from the right, like n. 0001 and n- i n. 0001

But we also have to get it to be + or - infinity to be a vertical asymptote

But now I am getting this root(0-)/ - 4

Whicj would be root(-0.000001)/-4

And now I am not sure if this will be - inf or if there is no VA

[novel dirge]

another thing I have problems with is how to get asymptotes when we have a root in the denominator

[ripe dust]

https://cdn.discordapp.com/attachments/578249966716321814/633404041963241492/unknown.png

idk if it belongs in calculus but

(or precalculus) but do you necessarily need to use lambert w function in order to find the answer?

[viscid thistle]

how to do you begin to solve this 3^2x + 3^x - 6 = 0

[serene heath]

Let u =3^x @viscid thistle

@ripe dust think u can do it using logs

[viscid thistle]

thanks

The half-life of silicon-32 is 710 years. If 10 grams is present now, how much will be in 800 years

Very confused on how to set this up

this section is related to logs and exp

[scenic heart]

Can someone help me with this word problem? Grade 12 Advanced Functions (I think it's precalc but idk. Canada)

[patent beacon]

Define variables. What are some things you wish you knew?

[scenic heart]

I wish I knew how many students there are. So let x - 50 rep the number of students buying and let x rep the number of students

If my let statements are correct, where am I supposed to next?

[hoary valley]

How come this is a function ? I see it as 2 functions..

[merry sphinx]

It is a piecewise function

You are somewhat right, but it is still one function

[lean steppe]

Anyone know how to insert [[x]] into a ti nspire cx?

[hoary valley]

It's separated though

[merry sphinx]

This is a domain separation

Search up graphing a piecewise function to understand it better

[lean steppe]

My teacher told the class how to put it into the ti84

Not the nspire

My teacher doesnt need me to understand the concept, at least for now, she needs me to able to graph it

[patent beacon]

@scenic heart
Another is the cost of the textbooks. Let C be the cost of all of them. There are now a few relations that can be made

[scenic heart]

I need 3 let statements?

[patent beacon]

Why not?

[scenic heart]

Idk my previous questions only required 2

[patent beacon]

There will be two equations here, you have that bit right

[scenic heart]

"let 50x rep the cost payed by students of the yearbooks"?

50 cents per student

[patent beacon]

You'd want to focus on equations that have an = sign

[scenic heart]

I don't understand

Like "rational function + rational function = number + rational function"?

[patent beacon]

Like C = 2x + 1

Except not that cuz that makes no sense for the question

[scenic heart]

How am I supposed to find a numerator for each fraction?

[patent beacon]

Oh wait, C is given, my bad

[scenic heart]

3300 = /x-50 + /50x

[patent beacon]

A better "let" would have been "let c be the cost each student pays when they all are buying books"

cx = 3300
(c + 0.5)(x - 50) = 3300

[scenic heart]

Where are the rational functions?

[patent beacon]

c = 3300/x

And so
(3300/x + 0.5)(x - 50) = 3300

[scenic heart]

I'm sorry but I'm very confused and do not understand. The other questions where distance time and speed questions then I suddenly have this.

[hoary valley]

How come this is not an Even function ??

[merry sphinx]

Definition of an even function?

[hoary valley]

I replace X with (-x) and see if it comes out negative or positive.. if positive then even..

[merry sphinx]

No

[hoary valley]

Help

[modern nymph]

1/1+1?

[merry sphinx]

A function is even if f(x)=f(-x)

[hoary valley]

1/2

[merry sphinx]

$f(x)=\frac{x}{x+1}\neq f(-x)=\frac{-x}{-x+1}$

[modern nymph]

Lol

[obsidian monolith]

pocofrosty12:

[hoary valley]

So here If I put a negative it comes out the same

So it's Even ?

[merry sphinx]

Are the 2 sides the same?

[hoary valley]

Yes

[merry sphinx]

Visually speaking, a function is even if it is symmetric about the y axis

Also how are the two sides equal

Cause they are not

[hoary valley]

what u mean..

(-) cancels (-)

[merry sphinx]

$\frac{x}{x+1} and \frac{-1(x)}{-1(x-1)}$

[obsidian monolith]

pocofrosty12:

[merry sphinx]

Remember you need to factor out the negative from the entire bottom quantity

[hoary valley]

@merry sphinx

[merry sphinx]

Yes and?

Look at my latex above

[hoary valley]

The negative cancel man..

In ur latex, u randomly changed everything...

[merry sphinx]

You need to factor out the negative from -x+1 which turns it into -1(x-1)

-1 cancels out on top and bottom

[hoary valley]

We're supposed to replace (x) with (-x)

[merry sphinx]

In the initial function

[hoary valley]

So it's Even.

[merry sphinx]

no....

[hoary valley]

I make perfect sense.

[merry sphinx]

try graphing it online to see if it is even

[pale kettle]

Here, think about something

What's 2 + 1

[hoary valley]

I don't want to graph it, because it's not graphed in my book..

@pale kettle 3

[pale kettle]

What's -2 + 1

[hoary valley]

-1

[sleek pawn]

11

[pale kettle]

Isn't that aproblem?

Don't you want that to be -3?

[hoary valley]

No

What you mean

[uncut mulch]

$\frac{2}{2+1} \neq \frac{-2}{-2 + 1}$

[obsidian monolith]

ramonov:

[hoary valley]

@uncut mulch I have to sum it first ?

[uncut mulch]

you can't just do 'stuff' to a single term in the denominator

if you can't clearly see whether they're equal, either try to make the numerator or denominator the same first

this was done by multiplying by -1/-1

[hoary valley]

@uncut mulch I can't cancel the signs, until I add -2 + 1 first ?

[uncut mulch]

if you want to divide by -1, you have to divide all those terms by -1 and that includes that 1 on the right

so it could be 2/(2-1) = 2

we used an integer example since you could add them first directly

[hoary valley]

@uncut mulch Thanks I understand it now. It's like 5+2/2 <<< I can't divide until I add them together.. so it will be 7/2

[uncut mulch]

$\frac{-x}{-x+1}\cdot \frac{-1}{-1} = \frac{x}{x-1} \neq \frac{x}{x+1}$

[obsidian monolith]

ramonov:

[hoary valley]

Omg Thanks ❤ @uncut mulch

[hard knot]

I'm not to sure how I should go about solving part c

[glacial island]

With unique examples of your own let’s discuss the number of solutions possible (0, 1, 2, 3, 4) to a system of equations that include a conic section

i have no idea what this means on my hw

oh wait i think i just make up an equation that is a parabola/ellipse/hyperbola, then solve for # of solutions

[pseudo sonnet]

why do i have the same phase shift and period?

[willow bear]

bc you're phase-shifted by exactly one period

[rigid sun]

wdym

lol wtf

[willow bear]

cos(t - 2π) = cos(t)

your phase shift may just as well be zero

[pseudo sonnet]

what?

isnt period

2pi/b

and phase shift you set equal to 0

[willow bear]

the period of cos itself is 2π

[rigid sun]

boi

[willow bear]

cos(t - 2π) = cos(t)

follows directly

from the fucking

DEFINITION

[rigid sun]

you're both saying the same thing

[willow bear]

of a periodic function

[rigid sun]

there is no remarkable discovery here

[willow bear]

yes

but vici is in overthink mode

[rigid sun]

ik vici its just random luck

the authors of your problems gave you a freebee

[pseudo sonnet]

dude what

im so confused as fuck

how do i have the same phase shift and period?

[rigid sun]

thats just what it is

why is that even an issue for you?

[pseudo sonnet]

how can my function start and end at the same point?

[willow bear]

wat

look

your phase shift

for all intents and purposes

might as well be ZERO

but also why are you putting π/3 there twice regardless

[rigid sun]

^

the function doesn't end btw

and it never starts either

you take the entire graph

and shift it by its own period

you'll just get the same graph you started with

don't view the trig graphs as just one period

[willow bear]

"bUt PeRiOd Is TwO pI oVeR b"

[pseudo sonnet]

im just trying to graph one period

[rigid sun]

ok

[pseudo sonnet]

and isnt my phase shift the beginning

and period the end

[rigid sun]

no beginnings or end

[pseudo sonnet]

thats why im confused why i have pi over 3 as the same values for both

[willow bear]

there is no beginning or end

[pseudo sonnet]

for example here are my notes

im trying to create the same number line but for the other problem

[willow bear]

if you want to graph one period's worth of your function

you can start at zero and end at the period

or start at the phase shift and end at the phase shift + the period

FUCKING HELL WHY ARE MY MESSAGES NOT SENDING

AAAAAA

there we go

[rigid sun]

yeah

the reason you're able to have period be your end point is because you're STARTING AT 0

period is only the change in x

so for instance, you could graph a sin function, by starting at 69.69

you'd just end the graph at 69.69+2pi

in this case, your phase shift move the point a distance the same as the change in x (period)

so it looks like they're right on top of each other

here's a better formula

phase shift+period=ending x value

[pseudo sonnet]

since im graphing one period

isnt the period just the ending value

i know it goes forever

but im just saying for one period

[rigid sun]

if you do that you have to start at 0

[pseudo sonnet]

yeah

thats why i set my phase shift equal to 0

then i solved for x

[rigid sun]

do whatever you want

[pseudo sonnet]

and thats what i got for my phase shift

[rigid sun]

nice

[pseudo sonnet]

wdym nice??

im trying to undert

understand

[rigid sun]

you have all the things you need

your phase shift if your start point

your period is your delta x, which can be either positive or negative

you can start at 0 or end at 2pi/3

whichever one floats ur boat

[pseudo sonnet]

so starting at pi/3

and ending at pi/3

is normal?

[rigid sun]

thats not whats happening

[willow bear]

no!

[pseudo sonnet]

oh wait!

[willow bear]

your phase shift is

Z

E

R

O

[pseudo sonnet]

do i just add 2pi to pi/3

my phase shift is pi/3

[rigid sun]

no

your period is pi/3

not 2pi

[pseudo sonnet]

ok

how come my phase shift is zero

dont i set the inside arguments equal to 0

of my fun tion

function*

[rigid sun]

uhhh

its zero because

[rigid sun]

phase shift=period in this case

remember how adding or subtracting 2pi to sin or cos functions resulted in no change?

same thing here

except your period is pi/3

like i said earlier, use this formula to make graphs quick af

phase shift= start point

period = change in x

phase shift + period= end point x value

just do that and you'll be golden forever

[glad shadow]

How do you answer b?

<@&286206848099549185>

[pseudo sonnet]

so were my phase shift and period right

[dim minnow]

Eccentricity is 1/2, so b²/a² = 3/4

So, b²=3a²/4

Equation of ellipse would be x²/a² + 4y²/3a²=1

Put the value of x & y

You'll get the values you a², & then b²

[viscid thistle]

Help plz

[serene heath]

What have u tried