precalculus

[mystic rune]

now go and practise

[deft flume]

thanks for the wise words

i will do my best^^

[south rivet]

[royal gull]

,rotate

[valid vector]

Can someone help me solve for the variable in these four exercises?

I'm confused as to how to do it.

I've got all the log properties, but I'm still confused.

Please tag me

[forest geode]

log(3z) = 2 => 3z = 10^2

3z=100, z= 33.333..3

@valid vector

[valid vector]

Yeah, I managed to do that one. The others I'm still missing, though. Thanks, tho!

[forest geode]

The other ones are the same

instead of 3z write sqrt(x-8)

so sqrt(x-8) = 10^5

just a huge silly number

x = 10^10 - 8

[viscid thistle]

i hate math

[kind pier]

I need help factoring something simple

(1/x) - 1 / x-1

[fringe stream]

is it supposed to be $\frac1{x} - \frac1{x - 1}$?

[obsidian monolith]

.jun:

[slow wharf]

What's the differebce between lo g division and synthetic division?

They achieve the same thing as far as I know.

[dull tartan]

The method of solving

Synthetic division can be used to factor and solve polynomials

Watch a video on long division and synthetic division. So long as there are no fractions, synthetic can be very clean and easy

Synthetic substitution can also be used with the rational zero theroem

[short sorrel]

Synthetic division is really just a special case application of long division

For specific polynomials

[dull tartan]

Yes

The divisor has to be (x - p)

For synthetic

[short sorrel]

Like how the long division you learn in elementary school is really just polynomial long division where your variable is an implied 10 instead

Ie long dividing 475 is like long dividing 4(10^2) + 7(10) + 5

[dull tartan]

@slow wharf

[slow wharf]

Yes

Oh, I see

Because synthetic division seems simpler

[dull tartan]

And is multipurpose

[slow wharf]

Multipurpose?

I'd use it for like integrals

If you know what I mean

[dull tartan]

Sometimes you'll get a polynomial with the factor x+2

If that polynomial is x^2 + 4x + 4

And they say to factor

-2 | 1 4 4

(Im on mobile, sorry)

( x + 2 ) ( x +2 ) is that factored

If the remainder is zero, it is a factor of the polynomial

This can ve done with more complicated polynomials

[slow wharf]

Can the same thing be achieved with both syntheti division and long division

[dull tartan]

Tbh I'm not sure

But synthetic is def the way to go if permissible

Mayne @short sorrel can shed some light

[short sorrel]

Basically yeah, synthetic division is used when it works (division by x-a) mostly because of convenience

Can't think of any cases where synthetic division is effective where long division wouldn't be

[timber forum]

Anyone here know what is the limit+inf of 1^x

[patent beacon]

1

[final rover]

Isn't 1^x an undetermined form ?

Since $1^x=e^{x ln(1)}$

[obsidian monolith]

TSM64CM:

[fading token]

And ln(1) = 0... What's your point?

[mild warren]

1^x is only indeterminate in an infinite limit

1^x for any real given value should just be one

So it's just 1=1 lol

[final rover]

But... he was talking in an infinite limit.

And "0*infinity" is an undetermined form.

[barren hedge]

Eh, that ln(1) is a constant 0, so a bit different. You can think of it as 0+0+⋯(∞ times) in this case. Now if it wasn't a constant 0 (like 1/x as x→∞) you would need to show more work (like in the indeterminate case of x·1/x).

[thick raptor]

Analogy:
0/0 is indeterminate, but lim{x->0} 0/x is very clearly 0.
Similarly, 1^infinity is indeterminate, but lim{x->infinity} 1^x is very clearly 1.

[spring thunder]

@final rover so that you may see the answers one day

[final rover]

I think I took the problem when the 1 was from something that converges to 1... That's why I couldn't understand... Well, I think. But if the 1 is constant, then yes, 1^x when x goes to the infinity is 1.

[patent beacon]

The "very large real number" analogy works for me. It only fails at something like (1 + 1/x)ˣ

[final rover]

That's what I was thinking

[winged lava]

yo im lost what do i do next

[willow bear]

what's the question?

[rigid beacon]

^

also it should be r^2 on the left hand side

since it's x^2 and y^2

[willow bear]

yes

[winged lava]

Change the rectangular equation into a polar equation

[rigid beacon]

oh

[willow bear]

well you just did that then, did you not?

[winged lava]

Thats only if its x^2 + y^2 right?

i have to isolate R

[rigid beacon]

oh fuck that lol

oh

[willow bear]

divide out by r

[rigid beacon]

well r^2

on the left side

[willow bear]

to get $r(\cos^2(\theta) - \sin^2(\theta)) = \cos(\theta)$

[obsidian monolith]

Ann:

[rigid beacon]

(it should be r^2?)

nvm

[willow bear]

i divided both sides by r

[rigid beacon]

you divided on the right

yea

[winged lava]

o

[willow bear]

this can, if you want, be rewritten as $r \cos(2\theta) = \cos(\theta)$

[obsidian monolith]

Ann:

[winged lava]

we didnt learn that identity 💦

[rigid beacon]

wot

[willow bear]

did you not? fair enough, that's optional then

[rigid beacon]

you never learned cos(a+b) and sin(a+b)?

[winged lava]

wel we did

That one ^

But not the r one

[rigid beacon]

it's the same thing

[winged lava]

uwot

[rigid beacon]

yea

it's still sin and cos

just cause you're in polar doesn't mean trig is different

sin is sin, cos is cos

[willow bear]

But not the r one
??

[idle dust]

yeah i cant find the 2r one either

[winged lava]

hold on lemme process this rq xdd

oh i see

[rigid beacon]

oh

[winged lava]

you factored out r first

derp

[rigid beacon]

oh yea

[idle dust]

nani

[winged lava]

but then how would I isolate it on one side

[idle dust]

im beyond confused

[winged lava]

i think r is just 0

[willow bear]

@viscid thistle no spam.

[rigid beacon]

@winged lava lemme write out the work 1 sec

[winged lava]

👌

[rigid beacon]

you see what I did?

@winged lava

[winged lava]

uh hold on my data is kinda slow

[rigid beacon]

oof

is aight

[winged lava]

wait

omg

im a fool

[rigid beacon]

that happens to all of us lol

[winged lava]

It really was r^2 oof

Tyty

[rigid beacon]

(i been saying t hat from the beginning but ok)

[winged lava]

no I thought you meant like

x^2-y^2=r^2 xdd

[rigid beacon]

oh lmao

[ruby otter]

some one help me with wavy curves

how do i draw wavy curves for functions having mod function in it

[serene heath]

post a question

[ruby otter]

ya im finding difficulties to tex it

thats why its taking time

[spring thunder]

tex is usually harder than drawing things tbh

[serene heath]

there should be a || thing on your keyboard

bottom left

[ruby otter]

tex it
(2|x| -2) / ( x^2 -2x )

[serene heath]

that looks rough

[ruby otter]

tex that i have got the rest

[spring thunder]

$\frac{2|x|-2}{x^2-2x}$ so this

[obsidian monolith]

emeric75:

[spring thunder]

?

[serene heath]

best thing with thses

is to look at 2 different cases

when x>0

and when x<0

[ruby otter]

f(x) = ({x}^{2} - 2|x|)(2|x|-2)-9 ($\frac{2|x|-2}{x^2-2x}$)

[obsidian monolith]

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

[serene heath]

[ruby otter]

-_-

f(x) = $({x}^{2} - 2|x|)(2|x|-2)-9 ($\frac{2|x|-2}{x^2-2x}$)$

[obsidian monolith]

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

[spring thunder]

$f(x) = ({x}^{2} - 2|x|)(2|x|-2)-9 (\frac{2|x|-2}{x^2-2x})$

[obsidian monolith]

emeric75:

[spring thunder]

englobe every math smh

[ruby otter]

thanx bruh

how do i put wavy curve for it

wait

[obsidian monolith]

Radical Ninja:

[ruby otter]

this is the correct question

[rigid beacon]

what are you trying to do?

[ruby otter]

wavy curve this

indeed my doubt is big

but if i had to explain it will take years to tex it

so better u can say the answer i will clarify my doubt

[spring thunder]

so you wanna graph this ? oof

[rigid beacon]

plug and chug lol

this shit is dumb

[ruby otter]

not graph

just wavy curve

its easy

[rigid beacon]

[spring thunder]

wym?

the variations of the curve?

[ruby otter]

wait first u know what is wavy curve right??

or u know it by some other term??

[spring thunder]

explain wavy curve then ig

[ruby otter]

Wavy curve

• number line rule
• sign scheme for rational functions

[spring thunder]

ahh

[ruby otter]

these are alternatives names

[spring thunder]

so you just wanna find the sign of the function ?

[ruby otter]

exactly

where it is +ve where it is -ve and zero

[spring thunder]

okok

https://cdn.discordapp.com/attachments/363224154469826562/551481660374253598/271974653977690112.png

[ruby otter]

dont the simplifications

i give the terms wait

(2|x|-2) [ ( |x| -1 ) ^2 +2 ] [ ( |x| +1 ) ( |x| -3 ]

numerator

x^2 -2|x| is Dr.

[obsidian monolith]

Radical Ninja:

[ruby otter]

$f(x) =(\frac{(2|x|-2) [ ( |x| -1 ) ^2 +2 ] [ ( |x| +1 ) ( |x| -3) ]}{x^2 -2|x|})$

[spring thunder]

ok yeah seems gud

[obsidian monolith]

Radical Ninja:

[ruby otter]

my only prob is denominator

[spring thunder]

yeah there's a little trick you can use

the fact that |x|² = x²

[ruby otter]

-__-

[rigid beacon]

it's true tho

[ruby otter]

even then i cant handle |x|

[spring thunder]

x^2 - 2|x| = |x| (|x|-2)

[ruby otter]

i did it though

[spring thunder]

it's positive iff |x|>0 and |x|-2>0 (you can't have |x| < 0, so you don't have to consider the case when the two factors are negative)

ie when |x|>2

[ruby otter]

no bruh

what are you doing

it is supposed to be |x| , |x| -2 =0 not >

[spring thunder]

it's even easier then

you get |x| = 0 or |x|-2 = 0

[ruby otter]

ya i thought the same but while plotting the signs in the graph i made the mistake

can u just try it once pls

[spring thunder]

try what once ?

[ruby otter]

sign scheme

solve the question

[spring thunder]

and if you're just doing +,-,+,- for the signs of x^2 - 2|x|, one advice : never trust absolute value

that's why i'm just solving the inequation manually 😎

[ruby otter]

ya do it'

i will wait

the graph comes to be symmetric

the graph of mod is always symmetric

but when i graph it
it is not symmetric

[spring thunder]

when i graph on wolfram it is symmetric

[ruby otter]

ya i know that

what did u get??

do u get smtng symmetric ?

[spring thunder]

for the signs also ye

[ruby otter]

pic?

[spring thunder]

dem my phones out of battery

[ruby otter]

.........

[spring thunder]

is ok just lemme charge it 2min

dem i didn't think it was so blurry

[ruby otter]

i think the phone is high

[ruby otter]

-_- i want the net graph

the final product

[spring thunder]

the bottom one is the final

[ruby otter]

shit

i didnt even notice that

ya its crct how did u do it?

[spring thunder]

just taking the sign of each factor in the product and combining them

(the 5 middle lines are my intermediary steps basically)

[ruby otter]

no how did u combine in final step

but multiplying??

[spring thunder]

so yeah use the fact that even number of negative factors => the whole thing is positive

odd number of negative factors => the whole thing is negative

[ruby otter]

How did he put it directly

i think he took on side and mirror imaged it

what do u think??

[spring thunder]

idk what the fuck the person did but w/e

[ruby otter]

w/e??

[spring thunder]

whatever

(now that i think about it, w/e could also be interpreted as weekend)

[ruby otter]

just listen

i was working with another prob

where i got |x| >A and i suddenly put
-A > X> A 😂

took some time to realize my mistake

[spring thunder]

ewf

[ruby otter]

ewf??

[spring thunder]

a mix of eww and oof

[ruby otter]

u wanna try some interesting prob??

@spring thunder

[spring thunder]

idk

show it

[ruby otter]

5n there are 2 choose one

differentiation or domain

[spring thunder]

both

(let's start with domain)

[ruby otter]

okie

wait

[spring thunder]

(wait i don't even have problems at hand )

[ruby otter]

how do i put alpha

and beta

[spring thunder]

in tex?

\alpha \beta

[ruby otter]

$ if ((\alpha +1 )(\beta +1) + (\alpha -1 )(\beta +1))a + (\alpha -1)(\beta-1) =0$

[obsidian monolith]

Radical Ninja:

[ruby otter]

$ and (\alpha +1 )(\beta +1)a - (\alpha -1 )(\beta -1) =0 $

[obsidian monolith]

Radical Ninja:

[ruby otter]

Also , let $ A = { \frac{\alpha+1}{\alpha -1}} ,{\frac{\beta+1}{\beta -1}} and
B ={ \frac{2\alpha}{\alpha +1}},{\frac{2\beta}{\beta +1}} $

[obsidian monolith]

Radical Ninja:

[ruby otter]

If A intersection B not equal to NULL then find all the permissible values for the para meter a

u der??

@spring thunder

[spring thunder]

@ruby otter tell my internets they've been really bad

[ruby otter]

5n can u understand the prob

[spring thunder]

i 'understand' the prob

approx. 0 ideas come to me atm tho

+i gtg have dinner

[ruby otter]

lol i never understood the pro ever since

so if u can help me im happy

@spring thunder bruh u keep trying ping me when u get the answer im off now u want the 2nd prob??

its a easy one

[spring thunder]

well just post it (i'll be off for quite a bit also)

[ruby otter]

ok its in #calculus

$ if ((\alpha +1 )(\beta +1) + (\alpha -1 )(\beta +1))a + (\alpha -1)(\beta-1) =0$\
$ and (\alpha +1 )(\beta +1)a - (\alpha -1 )(\beta -1) =0 $\
Also , let $ A = {{ \frac{\alpha+1}{\alpha -1}}} ,{{\frac{\beta+1}{\beta -1}}} $ and $
B ={ \frac{2\alpha}{\alpha +1}},{\frac{2\beta}{\beta +1}} $\
If A intersection B not equal to NULL then find all the permissible values for the parameter a

[obsidian monolith]

Radical Ninja:

[gaunt gate]

does anyone here know what sec2theta is equivalent to?

I know that the sin2theta is equal to 2sin^2cos^2

[short sorrel]

are you given a similar identity for cos? @gaunt gate

[gaunt gate]

what do you mean namington?

[short sorrel]

do you have an identity for cos(2theta)?

ah, I missed that you were already answered in #geometry-and-trigonometry @gaunt gate

apologies

[gaunt gate]

oh no worries, I really appreciate that you tried to help

[full garden]

guys please I need some help with grade 11 functions

how do I find the x and y intercept for this one

please help

[vale pewter]

remember the definition of x- and y- intercept?

o wait that's not a function .-.

[spark gyro]

is that a lowercase L

[vale pewter]

......

o.O

jesus

ur right

[full garden]

yes that is L

[spark gyro]

ew ew ew

[full garden]

is the t = y and L= x?

[obsidian monolith]

tubular:

[spark gyro]

probably

looks like a pendulum question

[full garden]

yes exactly it is

i just do not know how to graph this

[spark gyro]

t would be the period and l the length

and the usual pendulum thing is figuring out the period given the length

so the period is the dependent variable aka the vertical axis

[full garden]

yes tubular but it is asking me to graph for Earth, Mars, and the moon

[spark gyro]

???

[eternal lotus]

o_O

[full garden]

sorry to interput you

continue please

[spark gyro]

anyway this is just a rescaled sqrt so the only places where it will intersect the t or l axes would be t = l = 0

[full garden]

yes how do i solve for the L

i dont know how to solve for L

I think T=0

but I dont know the L point

[spark gyro]

do they mean solve for l in terms of t, or solve for l given a specific value of t

[full garden]

i dont even know man the teacher just told me to graph it but I have no idea how to do that

please just tell me how I should graph this funcigon

@spark gyro please help me bro

[spark gyro]

to graph a functio means to draw a picture of it

try plugging in some values for l and typing it into your calculator

and then draw little dots on a graph for each (t,l)

and take a guess at what the function is meant to look like

i recommend l = 0, 1, 2, 3, 4

maybe also l = 1/2 if you wanna look real snazzy

[full garden]

thank you so much bro

you are the best bro

I really appreciate your help @spark gyro

[full garden]

guys

what does it mean when it asks to find the initial value of this?

[eternal lotus]

That is a velocity vs time function right

[full garden]

@eternal lotus the question says (V) is thousand of dollars , of a certain car after (T) years

[eternal lotus]

Right

So initial value t=0

V(0)=...

[full garden]

ohh

so 35/0+3?

[eternal lotus]

Yes

[full garden]

so initial value means like the start value? Stephen

[eternal lotus]

Yeah basically

initial means start

[full garden]

@eternal lotus thank you so much bro

sorry for the dumb questions man I am not at your level

thank you

[ruby otter]

https://cdn.discordapp.com/attachments/363224154469826562/551511536741515304/271974653977690112.png

<@&286206848099549185>

A , B are sets i cant put {} around them in tex

[willow bear]

you can

\{ and \}

[ruby otter]

-_-

[obsidian monolith]

Radical Ninja:

$ if  ((\alpha +1 )(\beta +1) + (\alpha -1 )(\beta +1))a  + (\alpha -1)(\beta-1) =0$\\
$ and  (\alpha +1 )(\beta +1)a - (\alpha -1 )(\beta -1) =0 $\\
Also , let $ A = \{{ \frac{\alpha+1}{\alpha -1}} ,{{\frac{\beta+1}{\beta -1}\}} $ and  $
  B =\{{ \frac{2\alpha}{\alpha +1}},{\frac{2\beta}{\beta +1}} \}$\\
If A intersection B  not equal to NULL then find all the permissible values for the parameter a
```Compile error! Output:

! Missing } inserted.
<inserted text>
}
l.13 ...lpha -1}} ,{{\frac{\beta+1}{\beta -1}}} $
and $
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.

[ruby otter]

its fine now some one help

[full garden]

hello guys

is y=a(b)^x the exponential growth function?

@ruby otter what grade is this math

[ruby otter]

you asking me??

[full garden]

ya it looks pretty hard

[compact tendon]

@full garden
1-This is called the exponential function.

[ruby otter]

what are a,b??

have i talk to u before??

[compact tendon]

Me?

[ruby otter]

are u pinging me in random??

[full garden]

@compact tendon there was a question asking about a rumor that has an initial value of of five. and each person that knows the rumor tells two more students the day after they hear about it

for the first day is it y=5(2)^1

[compact tendon]

You must be taking differential equations

[full garden]

does that mean for day two y=5(2)^2 = 100?

@compact tendon yes exponential functions

[compact tendon]

In my opinion you are correct.

According to what I know.

[full garden]

@compact tendon yes it looks rational but in the answers it says in day two there will be 20 students that know

[compact tendon]

I forgot how to deal with these.

[full garden]

i dont even know where that came from

[compact tendon]

5*(2)^2

This is 20

[full garden]

its 100?

[ruby otter]

wait

[compact tendon]

Not 100 😂 😂 😂

[ruby otter]

lets start with n members

[full garden]

oh

[ruby otter]

next day 2n will get to know the rumor

so in total 3n

where does exp func comes here

lemme think

[compact tendon]

It is growing exponentially

According to the question probably.

[full garden]

but isn

t

2*5= 10 10^2=100

?

[ruby otter]

its kind (3^n ) m

n- no of days

[full garden]

ya

[ruby otter]

m - number students initially

[full garden]

oh yeah

[ruby otter]

thats the answer substitute acc

[full garden]

thank you bro

[ruby otter]

👍

[full garden]

👍

[compact tendon]

@ruby otter concerning your problem

[ruby otter]

my prob??

that parameter a ??

[compact tendon]

Alpha and beta

Yes

[ruby otter]

ya u got the answer/

[compact tendon]

I will give you a sketch

I did not really do it because there is a lot of work

You solve for one of either alpha or beta

[ruby otter]

there are 3 variables

thats why i find it hard to solve

[compact tendon]

a is a variable too?

Now this is fun

Please wait

[ruby otter]

ya in here a is treated variable actually its a parameter but since we dont the value

[compact tendon]

Is it arbitrary?

I mean, is it any real number?

[ruby otter]

it is not mentioned in the question

[compact tendon]

We assume that all of them are real

So that we do not restrict ourselves

[ruby otter]

ok

[compact tendon]

But beyond that...

[ruby otter]

u dont have to restrict from real numbers u have to restrict from complex nos.

[thick raptor]

what question

[ruby otter]

https://cdn.discordapp.com/attachments/363224154469826562/551690185495281684/271974653977690112.png

[thick raptor]

:l

set the elements equal

and solve the system of equations

[ruby otter]

how??

[thick raptor]

$$\frac{\alpha+1}{\alpha-1}=\frac{2\alpha}{\alpha+1}$$

[obsidian monolith]

Simple_Art:

[thick raptor]

Add this to the equations and see if there's a solution

and repeat this for each element of A and B

[ruby otter]

former or latter?

[thick raptor]

?

[ruby otter]

can u explain first one or two steps

i will continue from there??

[thick raptor]

Well start by solving for α from that equation

[ruby otter]

how it is 3 variable?? isnt it?

[thick raptor]

?

[ruby otter]

ohh ok got it

[thick raptor]

[ruby otter]

wait how do u use that emoji??

[thick raptor]

[ruby otter]

how ??

i get -1 , -1/3

[thick raptor]

Certainly α = -1 can't be right

[ruby otter]

it can be why not??

[thick raptor]

actually neither of them look right

how'd you solve for α?

[ruby otter]

QE

[ruby otter]

PLOM u dead??

[compact tendon]

alpha cannot be one

Niether beta

Sorry for being late mate.

[ruby otter]

no prob

thats not a issue i cant force people to solve or help

take ur time

[compact tendon]

Neither -1

One question

Are you sure A is a set and is not an ordered pair?

[ruby otter]

it is a ordered pair

[compact tendon]

Wait, I am confident that you know that an ordered pair is like points of functions

[ruby otter]

well know something i said its ordered pair

i can be wrong coz it s not mentioned in question

[compact tendon]

If it is really an ordered pair...

a is a constant then.

Otherwise it would be absurd to figure out.

But wait....

Intersection...

This is indeed a set...

[ruby otter]

is it??

coz it is mention intersection

[compact tendon]

How would two ordered pairs intersect???

There is no such thing called intersection between two ordered pairs.

Are you familiar with the Cartesian product?

Or should I say, the External direct product of sets.

You are probably not.

This is not in precalculus after all.

But there are infinitely many solutions for alpha and beta

[ruby otter]

-_-

say what is cartesian prod

[compact tendon]

Alpha is -0.5a for any a not equal to 2 or -2

Cartesian product is a product of two sets

For which they form points

Like functions and their images.

Suppose R is the set of real numbers.

[ruby otter]

i know ut

*it

[compact tendon]

R x R is the xy plane

Well you do not need it at the moment

I found the solution

Something came up and I was sort of busy

alpha=-beta =-0.5a for any real number a except 2 and -2

This it what satisfies both the equation

And the intersection of sets not to be null

[ruby otter]

hmm

then

[full garden]

guys what did i do wrong here

this question says patient has 1 mill on Monday

[serene heath]

idek whats happenning there

[full garden]

this is the question

I answered a and b

but Idk c

[serene heath]

can you walk me thorugh exactly what u did for c?

because i really cant tell what u did

[royal gull]

lmfao

you cant cancel like that

[full garden]

yes so I had the equation y=1000000(1/x)^t

[serene heath]

it seems you cancelled the 2 with 10000 and somehow brought the x expoenetdown

[full garden]

oh

[serene heath]

but thats not how exponents work

[full garden]

@serene heath I just dont know how to solve question C

could you please help me

[serene heath]

sure

[full garden]

i think I have to substitute something in but I don't know what it is

[serene heath]

$1950=1000000 \cdot \left(\frac{1}{2} \right )^{x}$

[obsidian monolith]

lemon catto:

[full garden]

wait how did you do that

[serene heath]

you wanna solve for x

[royal gull]

thats what you wrote

[full garden]

ya but how did you convert it like that

[royal gull]

in your pic as well

[full garden]

it looks nice

[royal gull]

LaTeX

[full garden]

yes lemon catto keep going please

[royal gull]

do you know logarithms?

[full garden]

ohhh I didnt know that

[serene heath]

have you come across logarithms before ?

[full garden]

no

I dont even know what that is

[royal gull]

so I guess you could just divide by two and see waht you get

[serene heath]

well let's isolate that 1/2 term

so dividing both sides a by million

[full garden]

yes sir

[serene heath]

but not sure how you're gonna solve it without logs

,w 1950/1000000

[obsidian monolith]

Display results online and refine query

Requested by lemon catto#7133

[royal gull]

yeah I'd say you are supposed to check what you get after dividing by 2, it won't be taht much

[serene heath]

hmm

[serene heath]

does this look familiar $\log_{2}(x)$?

[full garden]

guys is this correct?

[obsidian monolith]

lemon catto:

[full garden]

no @serene heath I hade never done this before

[serene heath]

is that photo maths lmao

[full garden]

ya is it not good?

[royal gull]

no

you wrote t/2

[serene heath]

lol

[royal gull]

not (1/2)^t

[serene heath]

its solving a different thing

[full garden]

lemon catto is the only way to solve this is by using the inverse of exponential?

[barren hedge]

Unless you want to make approximations using the regular exponential, but like why.

[royal gull]

lmao I said how you can solve it before, just read it

if you didnt have logs yet

[full garden]

omg okay thanks guys

i really appreciate it

[vestal plaza]

Can anyone help with Quadratic Functions?

[patent beacon]

#info Assume yes, no need to ask to ask

[idle dust]

p = 2a +2b

p = 2a+ 2a +14

206 = 4a+14

192=4a

a=48

feet

b = 55 feet

area = ba

area = 48x55

= 2640 feet^2

@vestal plaza

._.

.-.

xd

[vestal plaza]

let me check

where you get the 55?

[idle dust]

1 is 7 longer

[vestal plaza]

oh

[idle dust]

lol

a is 48, then b is 48+7

it says in the question

is it right?

[vestal plaza]

yea

what about this one? is it the same way?

[idle dust]

oh

hi

sory

oh

ok a is height and b is width

so a = a and b = a+3

by the pythagorean theorem, a^2+(b+3)^2=26^2

oops i mean a+3^2

which becomes a^2+b^2+6b+9= 676

bruh

i keep writing b instead of a

but these bs are all a

sorry

let me rewrite it

a^2+a^2+6a+9=676

ok now it becomes

2a^2+6a+9=676

now we can make 1 side 0 to make it easier to use factoring techniques or other things...

ok now we do minus 676 on both sides of the equation

so 2a^2+6a-667=0

now this is a little complicated, so lets just use the quadratic formula

which is too complicated to write

let me just tell the answer

16.823482201808694

ok anyway

it says round to 1 decimal place

so 16.8

@vestal plaza

[echo lynx]

sum1 help

[idle dust]

uhh are we supposed to solve theta

[vestal plaza]

I see

[echo lynx]

no make LS = RS

by only simplifying LS

using trig identities

[idle dust]

lol

ok multiply both sides by sintehta+cos theta

now it becomes

the left is 1+2sinthetacostheta

the right is sin2theta+1

now minus 1 on both sides

2sinthetacostehta=sin2theta

bruh

ok

i mean

;-

;

[vestal plaza]

What about this one?

[idle dust]

girly cat boy

do u undrstand

cuz im just spilling the answers for u

i dont wanna just do ur homework without u being able to do it usrself after

[vestal plaza]

yea, I had another discord explain it to me also in the same time I asked this one

[idle dust]

;-;

[rigid beacon]

_>

[idle dust]

k

[rigid beacon]

@idle dust you been cheated on

[idle dust]

frik

[vestal plaza]

I got a math tutor anyways

[idle dust]

BOI

ok kms

[rigid beacon]

B R U H

[vestal plaza]

I am super bad at word problems and my teacher goes too fast

[rigid beacon]

Why ask here then

then like

[idle dust]

oh

[rigid beacon]

ask them to slow down

[vestal plaza]

because I only go to him on tuesdays not weekends and its only 1 hour

[rigid beacon]

you're paying them to teach you, if you don't understand then you're paying for nothing

[idle dust]

are u in 9th or 10th grade?

[vestal plaza]

thing is he gives homework of stuff he doesnt go into detail in class

I am in college

[idle dust]

wait

WHAT

[rigid beacon]

Yo chill

[idle dust]

ok

[rigid beacon]

different people learn math at different times

[echo lynx]

^

[idle dust]

yeash

[vestal plaza]

I havent done math for 6 years

[rigid beacon]

not everyone does calc in HS

[rigid beacon]

in fact a minority of kids do calc in HS

[idle dust]

dont mind me

[idle dust]

yesh

[echo lynx]

how can this be done

[idle dust]

i dont get the point of this

[echo lynx]

with a one side solve

[rigid beacon]

1 sec

[spring thunder]

use the fact that 1 = cos^2 + sin^2

[echo lynx]

my teacher wants us to simplify one side only

[spring thunder]

done

[rigid beacon]

wait hold up

[idle dust]

but u can simplify both sides

[spring thunder]

(yeah you can tbh)

[idle dust]

this is just and identity multiplied by another identity

[rigid beacon]

yea

but

[echo lynx]

ik but he says he wants us only to do one side for now

[rigid beacon]

teacher wants 1 side

so

we doing 1 side

[echo lynx]

: )

[idle dust]

oK just carry us spamakin

[rigid beacon]

I hate trig identities gimme a sec

[open apex]

Interesting comversation

[spring thunder]

$$\frac{1+2\sin(\theta)\sin(\theta)}{\sin(\theta)+\cos(\theta)} = \frac{\sin^2(\theta)+\cos^2(\theta)+2\sin(\theta)\cos(\theta)}{\sin(\theta)+\cos(\theta)}$$ $$ \frac{1+2\sin(\theta)\sin(\theta)}{\sin(\theta)+\cos(\theta)} = \frac{(\sin(\theta)+\cos(\theta))^2}{\sin(\theta)+\cos(\theta)}$$

[obsidian monolith]

emeric75:

[idle dust]

oh my

so much skills

[echo lynx]

o damn

[rigid beacon]

how the fuck people so fast with LaTeX

[echo lynx]

thx

[spring thunder]

i'm not honorable for nothing boi

[rigid beacon]

but yea that's the answer I got as well

just like on pencil and paper and slower

[spring thunder]

and copy pasta for ever

[idle dust]

i feel like a dum fuk

[rigid beacon]

I hate teachers being like "wOrK WItH oNe SiDe!!11!!1!!!1!!!1!"

like there's no point

[idle dust]

baguette

i dont even know why

[spring thunder]

i'm a baguette

🥖 arise

[idle dust]

bruh

i cant get anothe letter

0bam4 is gon3

play crab rave

[echo lynx]

How can I study for my trig test tmrw

It is on reciprocal trig ratios, trig ratios, ambiguous cases, 3d mapping problems, and pythagorean trig identities.

[barren hedge]

If it's tomorrow, work to memorize them like you would for a History test. If for an end of the course exam, work to understand the derivations/proofs.

[echo lynx]

ok ty

Prove that L.S. = R.S. while solving only one side

[rigid beacon]

common denominators would be my first step

[echo lynx]

cos(theta) / sec(theta) = cos^2(theta)

so i reduced that so far

ugh how do i do this

[rigid beacon]

@echo lynx cos / sec = cos^2

cause you have cos/ 1/cos

[echo lynx]

ik

[rigid beacon]

oh you have cos^2 = all that

[echo lynx]

this is the farthest ive got:

[rigid beacon]

man why can't we work with both sides ;-;

[echo lynx]

cos^2(theta) cot(theta) over cot(theta) - sin(theta) over cot(theta)

and im wondering that too lol

and sin(theta) over cot(theta) = tan(theta) over csc(theta)

[rigid beacon]

oh that's ez

[echo lynx]

but how do you combine that with the first term to get the R.S.

[rigid beacon]

oh eai

*wait

[echo lynx]

i got itttt

turn the first term into cos^2(theta) and the second term into tan(theta) over csc(theta)

then get a common denominator and combine the two numerators

im just confused on what my thought process should be to do those types of questions easier and quicker

[barren hedge]

From (1) to (2) I got a common denominator, and from (2) to (3) I used the definitions of secant and cotangent. From there, hopefully you see how we get to the RHS of the equation.

[rigid beacon]

oh

I think we both just oversimplified

and tried to condense stuff from there

[barren hedge]

You guys lost me very quickly from what you were saying, I can hardly read text equations like "cos/1/cos" as well as you guys xD

[swift glacier]

f(-4) would be undefined, right?

[barren hedge]

Nope

[obsidian monolith]

𝓗𝑒𝓃𝓇𝓎 𝓒𝒶𝓈𝓉𝓁𝑒:

[soft zephyr]

can someone help me with my hw

[barren hedge]

Gotta post the questions my dude.

[soft zephyr]

there is a lot

[barren hedge]

One at a time then.

So what angle of Sine gives 0?

[soft zephyr]

how do i figure that out?

[barren hedge]

Memorizing your unit circle, lol.

Atleast until Taylor series, and you're a computer.

[soft zephyr]

what is taylor series

[barren hedge]

Oh nothing you need to worry about until Calculus 2.

[soft zephyr]

is sin of pi 0?

[barren hedge]

Please memorize this entirely, as there's no reasonable way around it. Oh, and yes, sin(π)=0

[soft zephyr]

you have any tips to remember it >

?

[severe verge]

its pretty symmetric

you only really need the top right quadrant

the rest you can just get by reflecting it over y axis or x axis

there's this memorization mnemonic "SOH CAH TOA"

[soft zephyr]

i yes i use that

[severe verge]

sin, cos, tan are functions that relate the angle of a right triangle to the side lengths of it

what you do is you think of every right triangle as part of the unit circle

1 point at the origin, connected to 1 point on the circle

that's the hypotenuse

if (x,y) is the point on the circle, then (x,0) is the last vertex
(maybe this is really hard to imagine and I should draw a picture)

[barren hedge]

Maybe the Desmos preset graph of the unit circle could help.

Hm, they don't seem to describe thing all too well. They throw you in quite abruptly. If you want to know what I'm talking about, https://www.desmos.com/calculator/qmzx2skkzy

Please do not look at the last 4 lines, they do not serve any other purpose than to form the image.

[severe verge]

the unit circle sets Hypotenuse = 1, so that
sinθ = Opposite/1 = Opposite = y coordinate
cosθ = Adjacent/1 = Adjacent = x coordinate

[soft zephyr]

y/r x/r y/x

what am i doing wrong here?

[viscid thistle]

inverse sin is defined where?

@soft zephyr

which quadrants

[soft zephyr]

that is all the question gies

gives

[viscid thistle]

what is the domain of sin^-1

sin^-1 = arcsin

you need an angle either in the first or 4th quadrant

where does sin(x) = -rt(2)/2?

there's 2 places

but the inverse sine function only tells us ONE of those

it occurs in the 4th quadrant

[soft zephyr]

so should it be 7pi/4?

[viscid thistle]

no

-pi/4

definition is tricky with these bad bois

[barren hedge]

Essentially, going forward 7π/4 is the same as going backwards (-π/4)

[short sorrel]

yeah i didnt make an active effort to "memorize" the unit circle

i just knew how to construct the 30-60-90 and 45-45-90 triangles

(and radian equivalents)

I "memorized" the unit circle sometime in calculus, purely naturally/accidentally

but an understanding of how the unit circle is constructed is enough to do it yourself

[soft zephyr]

i need help with this

[barren hedge]

Recall that sin⁻¹(-√{3}/2) is -π/3

[compact tendon]

Use Sin(A+B) formula

There is nothing odd with this

[old marten]

<@&286206848099549185> i need help please quick

[tiny verge]

!15m

!15

mhmm

okay, don't ping the helpers before 15 mins of you askig you question

and post the question before requesting for help

[old marten]

im sorry i havent used the server before

[tiny verge]

good idea to read the rules before using the server

anyway, post your question

[willow bear]

what do you need help with?

[old marten]

so Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. f(x) = (x-9/x+5) and g(x) = -(5x-9/x-1)

[tiny verge]

right

[old marten]

ive tried a bunch of different stats to solve it but none of them are lining up

[willow bear]

uh huh... and what's giving you trouble there?

[tiny verge]

wdym diff stats?

[old marten]

i plugged in each value into the other equation and i cant seem to line up the numbers right i guess

[willow bear]

the instructions couldn't be more straightforward.

can you show your work?

[old marten]

like i made g(x)=-5(x-4/x+5)-9)/(x-9)/x+5)-1 then i got g(f(x))= 14x +25/14x

so in plugged that into f(x) to get

[willow bear]

...do you have this written out on paper?

it's really hard to read in plaintext

[old marten]

yes, ill send a pic

[willow bear]

do that

[old marten]

Right to left

Its not finished on this paper but that’s as far as I got until it stopped making sense

[willow bear]

okay just wait a minute.

is $g(x)$ meant to be $-\frac{5x-9}{x-1}$, or $\frac{-5x-9}{x-1}$?

[obsidian monolith]

Ann:

[old marten]

Its supposed to be the second option

[willow bear]

ok.

so then, how do you go from $\frac{-5\left(\frac{x-9}{x+5}\right) - 9}{\left(\frac{x-9}{x+5}\right) - 1}$ to $\frac{14x+25}{14x} $?

[obsidian monolith]

Ann:

[old marten]

My brother put them through a calculator and he said that’s what it should be

[hexed ermine]

Lol

[willow bear]

...........

are you not able to do the simplification yourself

[old marten]

He said he knew this stuff and he was my first source

[willow bear]

if he puts shit like this through a calculator... i would honestly not trust him much

[hexed ermine]

Hahaha

[short sorrel]

meh its kinda tedious - mind, putting it into a calculator might be even more tedious

[old marten]

Well when I multiply the -5( (x-9)/(x+5) )how do I properly simplify that?

[short sorrel]

distributive property

[willow bear]

you should, first and foremost, get rid of the fraction nesting.

[short sorrel]

but yeah

id knock out those fractions first

[willow bear]

because nested fractions are a great source of algebraic fuckups

[short sorrel]

make your life easier

[willow bear]

this can be done by multiplying the numerator and denominator of the big fraction by (x+5)

giving $\frac{-5(x-9) - 9(x+5)}{(x-9) - (x+5)}$

[obsidian monolith]

Ann:

Namington:

[short sorrel]

in case you're unclear what's happening

(x+5)/(x+5) = 1, so this is algebraically valid

and yeah, it gives what ann said

can you go from there?

[old marten]

i understand so far but im not sure how to go to the next step

im sorry

[short sorrel]

distribute the -5, -9, and -1 into the binomials

$\frac{-5(x-9) - 9(x+5)}{(x-9) - (x+5)} \
= \frac{-5(x) - 5(-9) - 9(x) - 9(5)}{(x-9) -(x) -(5)}$

[obsidian monolith]

Namington:

[short sorrel]

if you want it spelled out deliberately

[old marten]

so would i plug that into the other equation for x in f(g(x)?

bc this is what we get from g(f(x)) right?

im sorry i just havent done this in a long time and its just making me very nervous

[short sorrel]

first simplify the above

[old marten]

ok

[short sorrel]

what function do you get?

[old marten]

-14?

g(f(x)=-14?

[short sorrel]

...not quite

[old marten]

oh shit

sorry my brain only recognized the bottom half

lemme try again

[short sorrel]

$\frac{-5(x) - 5(-9) - 9(x) - 9(5)}{(x-9) -(x) -(5)} \
= \frac{-5x -9x + 45 - 45}{x-x -9-5}$

[obsidian monolith]

Namington:

[short sorrel]

of course, 45 - 45 = 0, and x-x = 0

[old marten]

so we would be left with
-5x-9x/-9-5?

[short sorrel]

so we get $\frac{-5x-9x}{-9-5} = \frac{-14x}{-14} \
= x$

[obsidian monolith]

Namington:

[old marten]

ok that makes sense

[short sorrel]

one note: theres a restriction on the domain here

ie, one value x cannot be

what value is that?

[old marten]

well would the two -14 's cancel out?

[short sorrel]

yeah, -14/-14 = 1

and 1x = x

so g(f(x)) = x

buuuut with a restriction

[old marten]

ok so then if g(f(x))=x then would f(g(x)) just be (x-9)/x+5) still?

[short sorrel]

can you try to find the restriction first?

look back at the steps we did

where we were messing about with the x

[old marten]

am i getting caught on the wrong detail?

[short sorrel]

is there some assumption we made at some point?

nah, its a minor thing

but worth mentioning

[old marten]

um let me look

[short sorrel]

hint: we multiplied by $\frac{x+5}{x+5}$

[obsidian monolith]

Namington:

[short sorrel]

and said it was equal to 1

is that always true?

[old marten]

no, x could equal something other than 0 or -5

i think

[short sorrel]

other way around

anything divided by itself is 1

so that's always true

...anything except 0, that is

0/0 is not 1, it's undefined

[old marten]

oh so it can equal anyuthing but those values?

[short sorrel]

what value of x makes the denominator equal to 0?

in other words, solve x + 5 = 0 for x

[old marten]

-5

[short sorrel]

yes

so in our simplification process

we assumed $x \neq -5$

[obsidian monolith]

Namington:

[old marten]

so the restriction would be -5?

[short sorrel]

if x = -5, then we cant do what we did

the restriction is $x \neq -5$, yes

[obsidian monolith]

Namington:

[old marten]

ok, that makes sense

[short sorrel]

in other words, g(f(x)) = x for every value of x

except -5

it's a "small" detail but it is worth noting

anyway, as for f(g(x))

$\frac{\left(\frac{-5x-9}{x-1}\right)-9}{\left(\frac{-5x-9}{x-1}\right)+5}$

wait hold on

whered the stuff in brackets come from

[old marten]

from my miscalculations

[obsidian monolith]

Namington: