precalculus

[thick raptor]

@tardy olive calc is just to show there are no other solutions, not for solving it

You can see x=1,3 are solutions by plugging it in.

[tardy olive]

is that the only way to find solutions

just guessing numbers and putting them in

how do u know those are the only two solutions

nvm

but how do u know the solutions r 1 and 3 other than plugging them in

@thick raptor

[fickle moat]

how do you solve 4log (x-6) = 11

[frozen needle]

what does log mean for you?

[fickle moat]

?

[frozen needle]

when you write "log", what function does it refer to?

[fickle moat]

base of 10?

log_10

[frozen needle]

alright

[fickle moat]

ok wait

so log(x-6)=11/4

[frozen needle]

You can do 10^()

to get rid of the log

[fickle moat]

10 log(x-6) = 10 11/4

[frozen needle]

uhh

[fickle moat]

x-6 = 10 11/4

solve X?

[frozen needle]

yes

[fickle moat]

x = 6+10 11/4

[frozen needle]

yes

[thick raptor]

@tardy olive well you can look at the graph, but I don't have too much advice.

[hexed ermine]

Yes @tardy olive with these sorta problems there are not a set way to solving

[thick raptor]

well you can use Lambert W functions

but there's no elementary approach other than looking at graphs and kinda guessing if you want trivial points

[hexed ermine]

Ya

[fickle moat]

for #72

i know that i should use A= Pe^rt

than is the equation should set up as like this?

2 = 2500e^.0375t?

[hexed ermine]

Well set equal to double 2500

So 5000

[fickle moat]

can i as why

so you're saying

2x2500 =2500e^.0375t

[hexed ermine]

Yes

[fickle moat]

can i ask why

[hexed ermine]

Its asking the time required for double and triple the amount

Well your principle amount is 2500

So double 2500 is 5000

It wants you to find out the time needed to reach 5000 dollars

Where A is your ending amount

It wouldn't make sense to have your starting amount at 2 dollars

[fickle moat]

you have to solve for T but

whats the 1st step to write this this equation

5000 = 2500e^.0375t?

[hexed ermine]

Yes

[fickle moat]

one sec pls let me write it down

ok now

5000 = 2500e^.0375t divied by both side by 2500e

[hexed ermine]

No no

[fickle moat]

to cancel so i can have .0375t?

[hexed ermine]

Nooooo

5000=2500*e^0.375t

Divide both sides by 2500

2=e^0.375t

[fickle moat]

so you dont dived e with it

just the 2500 itself

[hexed ermine]

No

You cant divide with it

Its a base

[fickle moat]

OK

wait wait

e is stands for LN

so isnt it be ln_2 =ln_e^.0375?

@hexed ermine

[hexed ermine]

2=e^0.375t

[fickle moat]

i got that yea

[hexed ermine]

So ln(2)=0.375t

[fickle moat]

now i have to re-write it has log

?

[hexed ermine]

Huh?

Solve for t

ln(2)/0.375=t

[fickle moat]

.0375

yea[p

ok

[hexed ermine]

==ln(2)/0.375

[granite stirrup]

8×log(2)/3 = 1.84839248149319

[fickle moat]

sowhen they were given 2500 as inveset in account means i can put it in A and P?

[hexed ermine]

No put it as P

Your A is your ending amount

So if it asks for double it

You do double 2500

Or 5000

[fickle moat]

than why did we have to put 2500 into A and P

isnt it suppose to be than 2 = 2500e^

???

nevermind i red the wording wrong

double the invest 2500 = 5000 thats A

so for A we wrote in equation form as 2 x 2500 = 2500e^.0375t

thank you

[hexed ermine]

Ya

[fickle moat]

Ok so where do i begin?

since they are telling me who are under "x" inches tall is model by 64< equal to 78 . i can choose any number between it? so let say x = 75

do i have to plug that 75 into x

and solve the equation?

[patent beacon]

@fickle moat
The functions take a height → a percent of people who are shorter than that height.

The horizontal asymptotes are 0 and 100, corresponding to the fact that you can't have less than 0% of people, or more than 100% of people

[north bough]

why is sin^4(x) written like this for the power reducing function:

=tex sin^4\left(x\right)=\left(\frac{1-cos\left(2x\right)}{2}\right)^2

[north bough]

instead of

=tex sin^4\left(x\right)=\frac{\left(1-cos\left(2x\right)\right)}{2}

[north bough]

if sin^2(x) is written like this

=tex sin^2\left(x\right)=\left(\frac{1-cos\left(2x\right)}{2}\right)

[north bough]

oh wait nvm

[hexed ermine]

Square both sides

[north bough]

yeah lol Im bad

I didn't see that

[hexed ermine]

😂

[fickle moat]

Can someone tell me i solve this correct?
Its #49

#49

#49

[silk sequoia]

do you guys have any resources on determining the domain, vertical asymptotes, and x intercepts of log functions?

[limber bone]

well

domain of log is [0,inf)

no negatives

x intercepts are when y =0

just google if u want resources

find lim x go to inf for asym

[silk sequoia]

ohh ok

thank

[alpine portal]

Can someone help explain what I'm not understanding?

I tried .02x and thats not it either

Since its piecewise is it .02x+300

?

[hazy crown]

It would be 0.02(x-300)

Just check it, and if it's right I'll explain

[alpine portal]

Haha I think Im on my last try

before it locks me out 😦

[hazy crown]

Ooo

It should be 0.02(x-300)

What happens if it locks you out?

[alpine portal]

Cant try the problem anymore

sucks but I think it shows me the answer

Ill do it anyways

Its wrong

I think it was .02x+300

[hazy crown]

It doesn't show the answer?

[alpine portal]

Hold on

[hazy crown]

Sorry man

[alpine portal]

Not letting me

its okay I got 1 last question for homework

If i get this right ill have 39/40 questions correct

Lol But I still want to know the correct answer

Yeah I think .02x+300 wouldve been right because at that piece of the function the rule is no longer that it starts at 2 dollars. In that part the rule is that it starts at 300 dollars

[hazy crown]

yea prob

Sorry once again

[alpine portal]

Its cool not like I failed an exam, I'm learning

[blazing raven]

hi

sup? 😃

[alpine portal]

Hey my question is up there

My final answer Is .02x+300 but It locked me out so I will never know

[blazing raven]

38+0.02(x-300) ... 32 + 0.02x ?

Like okay. What is the amount paid if x = 300?

Then how do you represent "amount of Kwh after 300"?

😃

[alpine portal]

Ohh I was supposed to plug in 300 to the first function

[blazing raven]

😃

[alpine portal]

So its .02x+38

right?

haha

[blazing raven]

OH wait ...
Then how do you represent "amount of Kwh after 300"?

[alpine portal]

thats how

.02x+38

[blazing raven]

um ... looking for x - 300 😃

[alpine portal]

before 300 is .12x+2

[blazing raven]

like if I use 350 kWh ... how many of those are at the 0.02 rate? 50 not all 350 😃

[alpine portal]

ooh ok that makes sense

[blazing raven]

so yeah
38+0.02(x-300)
= 38+0.02x-6
= 32 + 0.02x

😃

[alpine portal]

I see now

[blazing raven]

yay

It is a tricky problem.

That's good.

[alpine portal]

Thank you so how would I go about looking at these problems in the future?

So far listing every single variable in the word problem helps

[blazing raven]

Ooh, good question. In this problem, the x's at the "elbow" points have y values that match. So if it is sensible to assume that then it will help.

Um, that's probably the best advice I can give. Um, another important idea was the "x - 300" part.

[alpine portal]

Yeah I think we went over it once in class

[blazing raven]

You are only being charged 0.02 for each additional KwH past 300.

But yeah that's how this problem panned out.

[alpine portal]

But, I was trying to do it only based on the information because If I used the graphs then they could've taken out that part of the problem

Can you explain how you knew right away that it was x-300?

[blazing raven]

this is true. My final piece of advice (which hasn't been done) is check your work. what should the answer be at 100, 200, 300, 350, 500 kwh

Anything that is for "additional" stuff ironically uses subtraction 😃

so 3 free bags at the airport and additional bags are $50 means...

for x > 3, 50(x - 3) is the extra cost

(it is basically a ... you've done enough problems thing)

[alpine portal]

So any problem that has an excess of something I should be thinking that the excess is minus the "not excess"

[blazing raven]

Well minus the minimum where the excess kicks in usually. But I'm not going to say that works every time.

[alpine portal]

haha thanks I will try to understand this more now

[blazing raven]

yup 😃 yw and anytime

[alpine portal]

excess minus excess min

[blazing raven]

sure .... most of the time 😃

[alpine portal]

Did I develop a new theory?

jk

[blazing raven]

it's fun stuff this math when ya get the hang of it.

😃

[alpine portal]

So in this case the key word was "usage over"

[blazing raven]

yes!

[alpine portal]

alright thanks for helping, do you have any books you recommend for understanding math in a different perspective or books on the language of math turned into English haha?

[blazing raven]

Not that I know of but a book filled with word problem secrets would totally help everyone.

You are welcome.

[viscid thistle]

Please help in question 10

[shell salmon]

Hint: cos/sin = cot

[fringe stream]

*cot

[shell salmon]

Yup

[tardy olive]

Is this true?

Nvm it is true

[snow estuary]

Can someone help me with this worksheet, I keep getting answers that are close but not correct

Also, was I supposed to ask this in a different chat??

[shell salmon]

multiply whole equations to get common denom. then add/subtract. then cross multiply and solve for x

[fickle moat]

A=Pe^rt

now i have a question since they are asking me a interest earned

the equation should be looked like

A = 7900e^(.08x15) ?

<@&286206848099549185>

[viscid thistle]

hi fren

[fickle moat]

hi

[viscid thistle]

thats correct

[fickle moat]

the answer is 18,328.92

but i am getting like

26k

where did i mess up

[viscid thistle]

=pup (7900(e^(0.08*15)))

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@languid wind

[fickle moat]

is this mean the teacher put wrong answer?

@viscid thistle

[viscid thistle]

I don't know how they'd get an answer of 18.328.92

[fickle moat]

wait one other question is do i have to use

A= P(1+r/n)^(n x t)

?

[viscid thistle]

what is r?

[fickle moat]

r =8%

[viscid thistle]

what is n, what is t

[fickle moat]

t=15

i couldnt find n

[fickle moat]

r =8% t=15 p=7900

[viscid thistle]

This problem you have clearly states continuous compounding

[fickle moat]

it says compunded continuously for 15 yrs

that means i use a =per

t

where is this 18k coming from though..

[viscid thistle]

I would discuss this with your teacher/professor and in worst case scenario they will show you the correct method (if theirs was wrong you're ahead of the game)

[fickle moat]

i get the idea

but.. i feel like they input the wrong answer >.>

i have a question
t=1
37000 should be K?
92500 = p?

<@&286206848099549185>

[fickle moat]

i get i have to use A=Pe^rt

P = 4000 , T = 200 what is this interest rat for this one?

or do i have to use differernt format

or is this mean i have to solve for R =?

[hexed ermine]

A=Pe^rt

Plug in your 3 knowns

A=8mil, P=4000, t=200

8000000=4000*e^(200r)

Solve for r

[grizzled solstice]

I need help ! How to show that both of them equal to each other

[grizzled solstice]

Anyone ? TT

[subtle narwhal]

hey can someone help identify what these type of problems are called?

[languid nebula]

Hi guys

I want to write a math sentence using the correct signs but Im stuck with this one I don't know why

Let n belong to integers (neg/pos). If n>1 and n isn't a prime number, then there is P a prime number that divides n and p<=sqrt(n)

Is it correct to write,
$\forall n > 1, \exists P , n not prime\Rightarrow P|n and P\leq \sqrt{n} $

[hot frost]

Could I get some help real quick on these?

1. Adding two functions: what does it mean?

2. How do you compute the domain of the sum, difference, or product of two functions?

3. How do you compute the domain of the quotient of two functions?

4. In your own words, what does it mean to compose two functions?

[sour hemlock]

@hot frost very ambiguous questions, can I have some context here. Is this some homework? Class assignment?

[vast trellis]

Hello. So i’m doing some homework online and I come across this question.

So I know that because there’s 2 exponents for x and y and the coefficients are the same, making it a circle. And it’s x-y making it a hyperbola. But how do I know if the graph will look like C or D?

[hexed ermine]

You mean B and D

x^2=4+y^2

x^2-y^2=4

Since the x and y are subtracting itll form a hyperbola

If it were x^2+y^2=4 it would be a circle

So with it being a hyperbola it would be either B Or D

Now since its x^2-y^2 the hyperbola will open to the sides

If it were y^2-x^2 the hyperbola would open up and down

So you can say that the answer is D

[vast trellis]

Thank u so much

[vast trellis]

So I got this question wrong and i’m a bit confused. I thought an ellipse was when is x^2 + y^2 and a circle was an ellipse

[rocky bison]

No

Ellipse = Oval

[frozen needle]

iirc oval is an actual thing of its own

[spring thunder]

[frozen needle]

not much is still a little bit :p

[spring thunder]

😛

[pulsar lynx]

im struggling to find the domain of (3/x - 5/x+5)

[viscid thistle]

The domain is everywhere where the denominator is not zero

@pulsar lynx it's the very similar to the one you've already found

(Same)

[viscid thistle]

guys, what discontinuities does 1/x^2 have????

[devout osprey]

It has one, at 0

[viscid thistle]

yea, but is there another disconiuity?

cuz i make a table for the limit and the y values are -10, -100, -1000, limit, 1000, 100, 10

[empty pecan]

no

[devout osprey]

I don't think so.

Oh I think you're not operating the signs

You know, (-10)(-10)=100

So when you approach from the left side, the signs become positive as well because (-x)^2=(-x)(-x)=x^2 😄

[viscid thistle]

wait sorry

yeah

its 10, 100, 1000, limit, 1000, 100, 10

[devout osprey]

Here's the graph, if it helps.

[viscid thistle]

ah

so infinite discontinuity?

[empty pecan]

ye

[viscid thistle]

kk thx

[devout osprey]

Yes, it's an asymptote 😄

npnp

[viscid thistle]

ye

thanks

[stark trench]

yo

can anyone help

[late mauve]

sure

[stark trench]

I need help with kinematics

high school level stuff

can u help with that?

@late mauve

well rip

[late mauve]

mm

maybe, shoot

[stark trench]

A police car 150km/h is trying to catch a speeder 1km ahead who is driving at 130km/h. How many meters should the police drive to catch him?

idk where to start with this

[late mauve]

y_p= 150*t

that'S the equation for the police car

y_p being in km

t in hours

[stark trench]

ok so d=150t and d=130t

[late mauve]

nope because the theif is 1km ahead

so his equation is

y_thief = 130*t + 1

now find the intersection of these two equations

[stark trench]

ah I see

[late mauve]

so at t=0 you get y_thief = 1, meaning he is already at 1km

[stark trench]

now good old systems

[late mauve]

exactly!

[stark trench]

ic ic

thanks

[late mauve]

no problem!

[viscid thistle]

???

on both, set f(x)=0 and solve for x, then do the reverse.. set x=0 and solve for f(x)

[stark trench]

the answer I got was 7500m @late mauve

is that right?

[viscid thistle]

@stark trench yes

[stark trench]

merci mon ami

[viscid thistle]

@viscid thistle wait those r just 0s

i have to describe end behavior in terms of limits

i got 5A, but what would 5B be?

try the same way

let me look at the question hold on

do you know how to take derivatives?

you can use L'Hôpital's Rule

or you can reason like this, if we take the limit as x -> inf in both directions, we see that the expression converges to the same number.

if we take the limit of the top and bottom, or ignore everything else except the main factors, we arrive at -6x^2)

(-6x^2)/(2x^2)

=-3

so that f(x) converges to -3 as x approaches infinity by L'Hôpital's Rule

conversely the same is true as x approcahes negative infinity

kk thx

L'Hospitals rule states that if we take x-> inf and we get inf/inf or 0/0, then we are allowed to take the derivative of the top and bottom until we arrive at a easy limit to compute

for example. (2x)/(5x) since both top and bottom go to infinity as x goes to infinity, we can take the derivative and arrive at 2/5

wait 2.5?

2/5?

what does that have to do w -3

no 2/5 is the example I just made

ohh

i get it

thanks!

;3you're answer is -3

:3

[stark trench]

boiz need help

"An object is thrown from ground into air with a velocity of 30 m/s at an angle of 35.0° to the horizontal. What is the range of this object?"

@viscid thistle

<@&286206848099549185>

anyone on?

rip

[fringe stream]

what have you tried

[viscid thistle]

im bad at physics

http://www.problemsphysics.com/mechanics/projectile/projectile_solution.html#Solution_to_Problem_1

this may help

@stark trench

[nocturne lake]

May be more algebra, but I cant seem to get this one.
I was asked to simplify cos(u)sec(u)
I expanded like this: cos(u)*(1/cos(u)) then simplified to 1 but that was incorrect?

[patent beacon]

Looks right to me

[limpid plover]

Same

=pup cosx×secx

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@limpid plover

[nocturne lake]

Hmm. Must be an issue with the question. Thanks!

[limpid plover]

Try inputting one if it's some online bullcrap

[nocturne lake]

cos(u)*(1/cos(u)) worked. I hate this thing.

[patent beacon]

Wat

[nocturne lake]

Yeah, the question was Simplify and write the trigonometric expression in terms of sine and cosine.
I guess they really meant that.

[limpid plover]

Heh

[nocturne lake]

Hey, so I ran into another one. I got 1/(sin(x)cos(x) but the site says its sin(x)cos(x)

Oh, I went from addition to multiplication in the first few steps. My bad.

[hexed ermine]

Yep

[royal gust]

=tex 9^x+2 = 27^-x

[granite stirrup]

Rendering failed. Check your code. You can edit your existing message if needed.

[spring thunder]

$$ 9^x+2 = 27^{-x}$$

[spring thunder]

brackets are life

[royal gust]

ohh thanks

[spring thunder]

so you wanna solve this eq?

[royal gust]

yes please

$$9^{x+2} = 27^{-x}$$

[spring thunder]

ah dem

that's even easier

so we can notice that 9 and 27 are powers of 3

ie 9 = 3^2 and 27 = 3^3

[royal gust]

so we put them in prents right?

$$9^{(x+2)} = 27^{(-x)}$$

[spring thunder]

ah you can if you want

but that's not compulsory

but can you rewrite the equation using the two equalities i showed you above ? (like that we'll have the same base raised to some powers on both sides)

[royal gust]

ok

so its $$9 = 3^{2} 27=3^{3}$$

[royal gust]

what about when we put logs in front of them

[spring thunder]

well you could take logs first, but i'm just doing some work before so that it will be 100% obvious

so then $$9^{x+2} = 27^{-x} \iff (3^2)^{x+2} = (3^3)^{-x}$$

[royal gust]

oh I see now

sorry I had such hard time finding that in notes here it looked familar

now we distribute them?

2x+4

-3x

[spring thunder]

yus

[royal gust]

$$3^{2x+4} = 3^{-3x}$$

[royal gust]

and we some how cancel out both 3s?

[spring thunder]

what's your level of math?

my explanation may depend on that

[royal gust]

I like to break things down to know where things go

each step is important

im in precalculus now

[spring thunder]

one explanation is just : log_3 both sides, but there are more interesting approaches

like saying the function x->3^x is strictly increasing on its domain, therefore you can't have two x-values associated to the same y value

which justifies the equivalence $$3^{2x+4} = 3^{-3x} \iff 2x+4 = -3x$$

[spring thunder]

(theres the notion of bijection behind that idea, but that's definitely not precal dope)

[royal gust]

yes I see it then we cancel out 4

2x = -1

am I doing this right?

im getting 0.5 now

or just leave it as

-1/2?

[spring thunder]

we all prefer -1/2

[royal gust]

thats great

thanks alot for your help

[spring thunder]

👌

[royal gust]

oh wait wait

is that an x by the 3 I see?

I cannot tell from the improper rending of the bot

[spring thunder]

yes

[royal gust]

wow

[spring thunder]

(just click on the image you'll see it)

[royal gust]

that means we might be wrong then

[spring thunder]

wait -1/2

you screwed up

sorry

[royal gust]

no not me

its the bot

[spring thunder]

yes you

my thing is right

you screwed up the equation after

[royal gust]

your equation is good but the bot threw us off

[spring thunder]

$$2x-4 = -3x \iff 5x = 4$$

[spring thunder]

:/

[royal gust]

yes we are on track

just that last part

that looks good now

5x = 4

so now its 4/5

2x-4 ? isnt it 2x+4?

[spring thunder]

demit yes

@royal gust

[royal gust]

great

so it would be -4/5 then

[winter spoke]

A certain radioactive isotope has a half-life of approximately 1050 years. How many years to the nearest year would be required for a given amount of this isotope to decay to 70% of that amount?

Can someone help me with this

[viscid thistle]

this is actually a physics question.

There is something called exponential decay/first order decay. It is called first order decay because the following expression is obtained from a first order differential equation.
N = Ae^(-kt)

where A is initial amount of substance
N is the current amount of substance
k is decay constant
t is time

Solving for t, eqn will be t = -[ln(N/A)]/k

Plug in the values now

[winter spoke]

thank you

Determine the half-life of the substance that decays from A0 to A in time t. Round to the nearest tenth of aunit.27.A0 = 11.3, A = 8.5, t = 33 minutes

Ao=11.3, A=8.5, t=33minutes

[viscid thistle]

another direct question. Same procedure but find k instead.

half life is 0.693/k as arbitrary as it sounds

[winter spoke]

thank you

[alpine portal]

I have no Idea what its asking... Isnt the height of the graph x=2?

Im confused

[safe maple]

a

[alpine portal]

oooh my goood

really that simple

fock

thanks @safe maple

[safe maple]

np

[viscid thistle]

i have a question too

=pup lim (sinx/x)^(sinx/x - sinx) as x approaches 0 + lim 1/(1 - x) as x approaches 1

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@woeful pivot

[viscid thistle]

The answer is coming to be - infinity

and when i calculated it i got 1

and secondly - infinity is even not in the options of the answer

so can someone pls confirm the answer

@alpine portal can u help me

[alpine portal]

@viscid thistle sorry I haven't learnt that yet.

[viscid thistle]

ok

np

[vast trellis]

Hi. How do I do the Gauss Jordan method

[fading token]

You add the two equations in such a way that you get an equation with just one variable

Solve for that variable, and use that answer to solve for the other variable

[gritty blaze]

Matrix method ?

[floral tide]

I dont get why people take pictures off their screen when you can screenshot on the computer and then paste it which is faster and gives better quality

pro tip: ShareX

[fading token]

Maybe they're using a public use computer

[slender river]

you mean pro tip for windows users: snipping tool

[viscid thistle]

Can anyone explain simpler steps to Law of Sines?

My teacher is making the entire class do a self teaching Unit (Flipped Unit) so she barely teaches the class...

Ping me if you can be an solution/aide to my problem.

Theres no simple steps to law of sines @viscid thistle

Its just one equation

and also i understand how bad self study is

When it comes to solving the oblique triangle I mean.

my geometry teacher doesnt do much

o idk then

rip me

she just sent a series of barcodes

soz cant help u with that

it's alright

yeah my geometry teacher is the same

I prefer my Algebra II Teacher.

They actually did something.

tru same

I think I'll go back to tutoring.

if you really need help with something ask your teacher

thats what im doing this year

[alpine portal]

Is this question asking me to put any number as h except for 0?

So can I put h= 1?

Thats the answer I got for that

in the pic

and it marked it wrong....

[alpine portal]

In the pic its simplified

ok lemme redo

@viscid thistle howd you get rid of 8/h?

Oh yeah

Lol thanks

[viscid thistle]

I think that question was abt calculating differncial

[alpine portal]

Calculating slope of secant

[viscid thistle]

ok

[tardy olive]

What is the difference between something being undefined and something being not in the domain of a function?

[fading token]

Can you give an example?

[fallen cloud]

well say you have a function that can only be applied to integers like the factorial function, it is undefined for noninteger values. But something not in a domain means that you can't plug in a certain value in a function or you'll get some complex number. EX: the domain of sqrt(x) for real output y is x >= 0;

factorial is only defined as such: x! = x * (x-1)!, 0! = 1. you can't input fractions or such. it's just not defined

[patent beacon]

The domain is all points where the x value is defined.

sin(0) / 0 is undefined.
The domain of sin(x) / x does not include x = 0

[tardy olive]

so by saying something is undefined, I am saying that it is not in the domain of the function?

for example, is log(-3) undefined

[hexed ermine]

Yeah -3 is not in the domain

[tardy olive]

but is it undefined?

[fallen cloud]

no

[hexed ermine]

Yes in the reals

[fallen cloud]

if you split it up it would be log(3) + log(i), which we dont know, but at least we have a concise idea of what it is

rather if you input a fraction or such into a function that can only be calculated with integers, then its undefined

[hexed ermine]

You mean log(3)+2log(i)

[fallen cloud]

ye

lel

[patent beacon]

But if you're working in the reals, ln is only defined on positive numbers

[tardy olive]

so if I say log(-3) in undefined instead of saying it is not in the domain, (working with real numbers), the two statements are equivalent?

[patent beacon]

Which, funnily enough, is another way to say "the domain is positive numbers"

Yeah, I would say those two things are equivalent

[fallen cloud]

:\

[hexed ermine]

But it's cool to note that log(-n)= log(n)+2e^((pi/2)i) :v

[fallen cloud]

don't have the e there, then youd be correct

log(-n) = log(n) + pi * i

[hexed ermine]

well + e^(ipi)

[patent beacon]

z = re^(iθ)
log(z) = log(r) + iθ

Note the complex logarithm is multivalued. log(-1) = πi, but also log(-1) = 3πi

[fallen cloud]

ye, but we'll just have fun with the simplest value for now 😃

[tardy olive]

Ay guys sorry to bother but can u help me on this one

[fallen cloud]

ooo

[tardy olive]

I have a general idea

[fallen cloud]

derivatives >:) or do we not use that yet?

[tardy olive]

But the (x-4)^2 is troubling

[hexed ermine]

What do you have so far?

[tardy olive]

no calculus

okay, so I reasoned that the bottom term is positive no matter what

so y-4 should be as negative as possible

[fallen cloud]

the top would be xx - 2x - 7,

[tardy olive]

except if we just go and look for the most negative y value, we might come up with a x value that makes the denominator really big

making the entire expression only slightly negative

if that makes any sense lol

[fallen cloud]

you do limits?

[tardy olive]

nope

i mean i know limits but i need to solve it without limits

[fallen cloud]

what if we just set the top to 0 and solve for x that way?

[tardy olive]

that might not be the most negative value it can achieve though

[fallen cloud]

like xx - 2x - 7 = 0, use quadratic?

[tardy olive]

because there could be some negative value which is smaller than 0

like, x^2-2x-7, there is some x value that will make it negative

[fallen cloud]

then... how about take the average of the roots of 'xx-2x-7' and plug it in to the top an bottom

[tardy olive]

why would we do that?

[fallen cloud]

well, maybe test it?

[tardy olive]

kk

1

the average of the roots in 1

if my arithmetic is correct ofc

[wise umbra]

Hii! I was wanting some help w these exponential functions.

32^-2x-3 = 64^-2x

I know that we do the root and for this it would be 2, but idk how to like write that out.

and the other one is

(1/4)^3x-1=64^-x-3

[glass venture]

could you please use some more parenthesis? I can't tell what's supposed to be in the exponent

[wise umbra]

ofc sorry!

Problem 1: 32^(-2x-3) = 62^(-2x)

Problem 2: (1/4)^(3x-1) = 64^(-x-3)

[glass venture]

so by the way, are you familiar with logs yet or no?

[wise umbra]

yeah but i dont really understand them

[glass venture]

ok well I can talk about those after I give you a way of doing this without them. They're simple but teachers often manage to explain them badly (as with most high school level math).

ok so to do these without logs, the key thing to notice is that the bases happen to be powers of 2

in the first problem, 32 = 2^5, 64=2^6

so I'm going to substitute those in and use some standard exponent properties

[wise umbra]

yeah i did see those 2 powers i was just so confused on how to write them

because i know that base rule thingy

[glass venture]

=tex 32^{-2x-3} = (2^5)^{-2x-3}

[wise umbra]

OHH ok thats how you write it

[glass venture]

so yeah do you think you could do some more from here then?

[wise umbra]

yes :)

thanks a bunch 💕

[glass venture]

np

btw, for the second one, you'll use 1/4 = 2^(-2) . That's basically as complicated as this type of thing can get

[wise umbra]

oh thanks! so we just leave the 1/4 alone lol.

[glass venture]

well you should replace it with 2^(-2)

[wise umbra]

OH i see

[glass venture]

the general idea is that if you have a^x = a^y, then x=y

so you want to get everything to the same base always

[wise umbra]

yeah i didnt notice you were saying 2^(-2) was = 1/4 i thought u were saying the whole equation

thanks xD

[glass venture]

np

so anyway, for logs, the definition is pretty simple

suppose you know this:

=tex a^x = y

[wise umbra]

mhm.

[glass venture]

then we define the log y in base a to be x

and we write this:

=tex \log_a y = x

[glass venture]

so this is a different way of writing a^x = y

[wise umbra]

OK SO THATS WHY exponentials and logs are related

[glass venture]

yeah, the log is the inverse of the exponential basically

log_a is a function that's defined so that the following is true:

=tex \log_a a^x = x

[glass venture]

log_a y is the number you raise a to in order to get y

so it's reversing exponentation

[wise umbra]

so would the 2 a's cancel eachother out?

[glass venture]

ehh sort of but don't think of it like canceling

so let's say we fix a

[wise umbra]

ok

[glass venture]

we can think of the function which take a number to the number a^x

log_a is the function which reverses this process

it takes things of the form a^x back to x

[wise umbra]

hmm that makes more sense

[glass venture]

except usually you're not given something in the form a^x. Like you might be asked "what is log_2 8"

do you think you can figure out what log_2 8 is?

[wise umbra]

ill try

64?

XD

[glass venture]

how'd you get that?

[wise umbra]

because 8 squared is 64

[glass venture]

well I'd probably just write 8^2 if I wanted you to find 8^2 :p

[wise umbra]

xD

[glass venture]

so log_2 8 means "the number x for which 2^x = 8"

[wise umbra]

OH

which would be 3 in that case :3

[glass venture]

yeah exactly

so it's a bit to swallow and think through, but once you get some practice with it it won't be too bad

it's like with square roots, you have to think "what squared is 64"

and with logs you think "what power do I raise this to"

[wise umbra]

i really can't thank you enough

you explain much better than my teacher does in a hour lesson

maybe i wont fail this test

[glass venture]

well one on one is a little easier lol

so btw, you may have seen some log properties

these all follow from exponent properties, and they're easy to prove if you've really swallowed the definition

[wise umbra]

yes i have a sheet for some of the properties

[glass venture]

so like for example, you learned this:

=tex \log_a x + \log_a y = \log_a xy

[wise umbra]

YES

because the a is the same

thats why you can do that with just adding

[glass venture]

this comes from the fact that (c^m)(c^n) = c^(m+n)

let me just prove this formally so you can get an idea of how this sort of thing goes.

[viscid thistle]

Actually that's a 2 liner

[glass venture]

which is why I'm doing it lol

[viscid thistle]

😛

[glass venture]

Let

=tex \log_a x = A, \log_a y = B

[wise umbra]

xD

[glass venture]

Then by definition,

=tex a^A = x, a^B = y

[glass venture]

This implies

=tex xy = a^A a^B = a^{A+B}

[glass venture]

and thus

=tex \log_a xy = A+B = \log_a x + \log_a y

[wise umbra]

im not gonna lie thats so much easier on my brain now

[viscid thistle]

yea

[wise umbra]

like it makes so much more sense

[glass venture]

protip: everything in math makes sense. don't accept stuff which don't makes sense. this is how you get good at math

[viscid thistle]

Just knowing how exponents work, you'll understand root, log, etc.

[glass venture]

anyway if you have the time I encourage you to try and derive the other log rules with this type of strategy

[wise umbra]

yeah, i have like 5 pages of packet to do and a test so ima be up late studying

[viscid thistle]

Actually, instead of straight-up not accepting the fact, you can analyze what's wrong in that so that you won't come to conclusions just like that.

[wise umbra]

mhm.

any tips you rec for like last min studying or future tests and such.

[viscid thistle]

Current education system exposes students to applications of math, but actual math involves relating stuff and THAT is more important.

last min studying, huh

[glass venture]

my main tip is avoid doing stuff last minute lol

[viscid thistle]

You could pay attention to lectures during class.

Otherwise, 0 way to do it at the last min

[wise umbra]

ik. i do pay attention just sometimes i cant really make those conneections.

[glass venture]

so some people get away with cramming more than others, but it eventually falls apart for everyone

[wise umbra]

like the lesson is easy but the sheet is just so hard

[viscid thistle]

They aren't arbitrary.

Trust me. You haven't jumped to spatial math yet

[wise umbra]

do i wanna know what that is xD

[glass venture]

so I would recommend in general to focus effort on making sure everything you do makes sense

[wise umbra]

lol sorry if i sound so immature im a sophmore in hs so unprepared for college.

[glass venture]

avoid memorizing procedures

[wise umbra]

yeah ofc.

[glass venture]

try to see why rules and procedures work

[viscid thistle]

math which deals with spaces. They have so many different terminologies which confuses me sometimes

[glass venture]

if you can do that you'll be way ahead of most people

but after that, just do practice problems until you stop making silly mistakes

[viscid thistle]

I still haven't found even ONE satisfactory video/site which explains why curl indicates rotation

[wise umbra]

yeah i really like math its honestly fav subject its just getting it is my problem

[glass venture]

don't be afraid to go slowly and write lots of things at first

@viscid thistle did you see the 3b1b video on that?

[viscid thistle]

Is there one?

[glass venture]

yeah I'll try to find it

I thought it was good

[viscid thistle]

ok.

[glass venture]

https://www.youtube.com/watch?v=rB83DpBJQsE

[viscid thistle]

oh that

But, i don't think that he explained why that happens

[wise umbra]

geez that looks crazy

[viscid thistle]

He said "it's possible to prove that cross product is curl"

I think, towards the end of the vid or something

[glass venture]

@wise umbra everyone who gets that had to go through the same path you're going through now, so don't worry about it

@viscid thistle yeah it's been a while since I watched it, I just remember enjoying it

[wise umbra]

what do you rec me take next year for math

thanks u guys btw u have helped me a ton :o

[glass venture]

I mean you can follow you're schools curriculum and you'll be fine. If you like it, try to be on the most advanced track. If you really like it, read up ahead, maybe try to skip a grade (you could take a class in the summer to do this maybe).

[viscid thistle]

Everyone is gonna fail at one point or the other. The actual thing that separates people who are good at math from those who are bad is: people who are good NEVER give up and never get discouraged by the dip

[wise umbra]

the dip

[glass venture]

when you get to college there are usually placement exams, so even if your school doesn't have a very advanced track, you can get ahead on your own

[wise umbra]

what is that

[viscid thistle]

Dip means failures.

[wise umbra]

i mean ik that

lol

[viscid thistle]

Just a metaphoric way of saying that

[wise umbra]

just saying i thought u meant a time that u hit dip

like COUGh junior year

[viscid thistle]

Dip typically occurs in any year b/w 13-18

Giving up is the worst thing you could do at that time

[wise umbra]

15 rn

[glass venture]

but yeah everyone has to deal with some failures. Also 90% of the time spent on learning math involves not totally understanding something, so don't get discouraged by that either

[wise umbra]

honestly me.

[glass venture]

lots of people aren't comfortable with not knowing what to do and all, so just don't get discouraged by that

[wise umbra]

im just worried that i wont be able to make it well in ap cal

[glass venture]

if you get the basics well you'll do fine

[viscid thistle]

^

[wise umbra]

ok :)

[viscid thistle]

I'll be honest rn. I am not satisfied by my math performance this time. I am not giving up that easily.

practice and interest is most important imo

[glass venture]

and remember, if you don't like your teacher, or whatever, it's on you to get past that. Read the book, hang out on the discord and ask questions, and you'll be able to do well for sure.

[wise umbra]

yeah. thanks so much! you made my day ik it sounds weird but thanks

[viscid thistle]

Good to hear!

[glass venture]

lol it's really no problem, I'm glad we could help 😃

[wise umbra]

ofc thanks c;

[viscid thistle]

One more tip: Help others in solving problems

Them asking doubts can make you think more

[wise umbra]

i do help my friends sometimes

true

anyways you all have a nice day or night

thanks~

[viscid thistle]

np

good luck in math

[wise umbra]

thanks same to you

So one other question, I'll probably have a lot in the next few hours, xD. Just to make sure im right, if im trying to find x for the expression below, it would be x=8.5 right?

=tex \log_5 x + \log_5 (x+1) = \log_5 20

[glass venture]

I think there should actually be two solutions

and I don't think either is 8.5

[wise umbra]

OOF

how would you solve it?

[glass venture]

ok so first I would use one of the log rules

[viscid thistle]

x^2 + x - 20 = 0

[glass venture]

the sum of logs is the log of the product

then yeah it becomes that

[wise umbra]

oh so we always set it to 0?

[glass venture]

not exactly

[viscid thistle]

I guess, x = 5 and x = -4

[glass venture]

ok so I mentioned this before:

[hexed ermine]

For finding quadratic solutions yes

[glass venture]

=tex log_a(a^x) = x

[glass venture]

this is also true:

=tex a^{\log_a(x)} = x

[wise umbra]

WAIT No I did what you guys were saying i just messed up the problems!

[glass venture]

if you think carefully about the definition this is obvious

[wise umbra]

sorry if i wasted your time. xD

[viscid thistle]

But, the bigger concept is negative numbers should not be inside logarithms

[wise umbra]

i put them out of order omg.

Oh i defenitely know from expirence xD

[glass venture]

it's fine no worries

[wise umbra]

my little ti-84 goes like nurp

=tex \log_5 (x+1) +

[wise umbra]

oops not done yet ack

=tex \log_5 (x+1) - \log_5 (x-1) = 2

[wise umbra]

Wouldnt x=0

Just checking btw c;

[hexed ermine]

You know your log properties?

[wise umbra]

i have a sheet with them and sorta understand them but applying them is a whole nother story

[hexed ermine]

Okay well since the two logs are subtracting the argument itself is dividing

So we have log_5((x+1)/(x-1))=2

[wise umbra]

Oh yeah i forget about all this dividing ahhh

[hexed ermine]

Now you know how log_b(a)=c is the same as b^c=a?

[wise umbra]

yes

[hexed ermine]

Okay so we have 5^2=(x+1)/(x-1)

Or 25=(x+1)/(x-1)

I'm sure you can do the rest.

[wise umbra]

yes, thanks c;

[hexed ermine]

No problem

[rare zephyr]

Which two lines should I start first?

[thick raptor]

#❓how-to-get-help
Wait a minimum of 15 minutes after posting your problem before pinging helpers.

[rare zephyr]

Aite

[hexed ermine]

@rare zephyr Okay

So for that top line, we have 2y=x+8

Or y=x/2+4

That means that for that line, the slope is 1/2

Since the line QR meets that meet orthogonally, the slope of QR is the opposite reciprocal

So -2

Now we have y=-2x+b

So to find b, since the y intercept is Q itself, you plug in 0 in for x in 2y=x+8 to find y which is 4

So now we have QRs line being y=-2x+4

Now notice with that line, (k,0) is on the line as well so plugging in k for x and 0 for y will satisfy

Plugging in them both, you get 0=-2k+4

Solve for k, gets you 2

All in all your final answer for k is k=2

@rare zephyr

[main geyser]

=tex \log_5 (x+1) - \log_5 (x-1) = 2

[elfin night]

=tex \log_5\left(\frac{x+1}{x-1}\right)=2 \ \frac{x+1}{x-1}=25 \ x+1=25\left(x-1\right) \ x+1=25x-25 \ 26=24x \ \frac{26}{24}=\frac{13}{12}=x

[solid reef]

#3?

[solid reef]

that helps a ton

thanks for the time

[old stag]

hey can anyone help me with a question

[viscid thistle]

question?

Let me try solving that.

[old stag]

the area of a circle is 48 πcm^2 - What is the exact length of the square

I know the answer

but I cnt figure out

how

• what is th elength of the diagonal of the squar

and its in a circle

[viscid thistle]

Is square inscribed within the circle or circle within square

Oh.

[old stag]

and they want the perimeter also

[viscid thistle]

Then the answer is just to find the radius of the circle.

sqrt(48)

4sqrt(3)

So,

Since square is symmetrical

[old stag]

they have the length as: 8(sqrt)3 and perimiter as: 16(sqrt)6

[viscid thistle]

wait wot

[old stag]

its a mixed radical

[viscid thistle]

length in the sense side length?

You asked for diagonal length, right?

[old stag]

length of the diagonal

[viscid thistle]

then it's radius x 2

4sqrt(3) x 2

8sqrt(3)

Anyways, since square is symmetrical, diagonal^2 = 2*side^2

This means side^2 = 32*3
side = 4sqrt(6)

perimeter = 4*side

16sqrt(6)

[old stag]

im trying to figure it out, i just dont get it

[viscid thistle]

Lemme sum stuff up because I think that my writing was a bit confusing.

Now, a square is inscribed within a circle. Your question asks you to find the diagonal length of the square. So, now since it is inscribed, it's safe to say that diagonal length is 2 times the radius of the circle. Find the radius using area formula.

Since square is symmetrical, by pythagoras theorem, (diagonal acts as hypotenuse, sides act as base and height) diagonal^2 = 2 x side^2.

Perimeter of the square is 4 x side.

[old stag]

what is the area formula

[viscid thistle]

πr^2

where r is radius

btw this question should be in algebra I think so.

[old stag]

its early pre-calc

im studying like 8 hours a day paced through

so I hope 2 get good, and help others ;p

thanx for tht

saved me time

[viscid thistle]

no probs

[old stag]

I been away from this stuff for years

wont give up this time

;v

[viscid thistle]

Good to hear!

[old stag]

how did you determine the radius

what values for the division

[viscid thistle]

It is 48 pi cm^2

not 48 pi^2

Read the question properly please. (;

[old stag]

but dont you leave out the cm

why would I include that

[viscid thistle]

Actually, it's our teachers asking to leave us so. As a matter of fact, when dividing, it should be:

sqrt((48 pi cm^2)/pi)

Now notice that both "pi"s get cancelled.

sqrt of cm^2 is cm

sqrt of 48 is 4sqrt(3)

It's how it SHOULD work but our teachers never explain why we are not including that

You'll get your answers as soon as you learn about homogeneity principle.

[old stag]

I think you are wrong dude

somewhere

thts not right

it just dosent work

your reasoning is ok

but the answer is wrong

[viscid thistle]

I got the answer

[old stag]

no

you did not

[fallen swallow]

I tried solving it but I'm unsure of the answer. Is the answer : the limit doesn't exist?

[viscid thistle]

Yeah

Actually, wait.

Limit doesn't exist

It's because it might be useful to separate out the function into parts

lim 1, lim (1/x)

lim 1 is clearly 1

lim (1/x) doesn't exist

[fallen swallow]

🆗

Thanks.

This means f(x) = |-x| + 3?

[viscid thistle]

Actually, I'm unable to understand your question.

[fallen swallow]

Hold on.

To know the limit, I must first know what ƒ(x) is. Problem is, I don't know what ƒ(x) is.

[viscid thistle]

f(x) is given...

It's wise if you check Left and right hand limits for this question.

[wicked token]

f(x) is a piece wise function

[vast trellis]

Question when multiplying two matrices with different sizes, how do you know the size of product matrix? Example: 3x2 and 2x3

[civic plaza]

M x N matrix multiplied on the right by N x P matrix will be M x P

[patent beacon]

@vast trellis
I go with this

Also shows the size requirement

[vast trellis]

Thank you

[wise umbra]

Hello! I am working on logs and such. If im expanding this log:

=tex \log _2\left(\frac{x+1}{y^2\sqrt{z}}\right)

[wise umbra]

thanks

lemme get my answer lol

[wise umbra]

owo

Thanks so much!

lemme check if thats what i got

ok yay it is!

thanks! 😂