precalculus

[split fossil]

Oh! sorry for 0.2 it is -13.98

for 0.1 it is -24

So what mode is your calculator inn now

[terse ravine]

Ok I'll get back to you when I have all the values

The calculator is in radian mode now

[split fossil]

Ping me if you need more help

[terse ravine]

Will do ^^

@split fossil This should be correct.

cymath gave me -28.499999 for the last value

My calculator in radian mode gave me -28.5

[split fossil]

No. your table is correct, but you can see that it is not approaching -29, it is approaching -28.5

[terse ravine]

Ok

[split fossil]

The 0.00001 should be a little less than -28.5, but the calculator is rounding

Just let me know if you need anymore help

[terse ravine]

I'm curious why we have to compute in radians.

I know now that calculus is always done in radian measure.

[split fossil]

Most higher math is in radians

Basically after you learn radians, you use radians

Unless told otherwise

[opaque olive]

you could use a graphing calculator so you can visualise whats going on

[split fossil]

@opaque olive Clearly that is not allowed, otherwise they wouldn't make you use a table

[opaque olive]

just for personal understanding?

[split fossil]

Oh, well yes

But its not the point of the question

[terse ravine]

x->4^+ is not 1 or DNE hmm

[split fossil]

It is 3

It says from the right hand side

[terse ravine]

OMG

I'm blind...

Is it 1,2 ?

@split fossil I see 2 points that DNE on -1

I should select DNE for part 3?

[split fossil]

Yes. If direction is not specified, and it is not continous it DNE

[terse ravine]

@split fossil This question is challenging since L is infinity here.

x is never equal to a and f(x) is never equal to L, only approaching a.

[split fossil]

Is this question saying that f(x)=infinity, or limit of f(x)=infinity

[terse ravine]

I think this is asking me to click on what is true and not false.

[split fossil]

Ok, 3 is false definitely, right

[terse ravine]

yes

The first appears to be true.

[split fossil]

Infinity means the limit doesn't exist, so 2 is false

5 is true

1 is debatable, but most likely true

4 is wrong, I think because f(x) will never be infinite, just a really big number

Do you agree @terse ravine ?

[terse ravine]

A number so big it has to be calculated in radians

Let me check

It seems me might be missing one based on the points I think 6/10.

[split fossil]

So we were right

[terse ravine]

The ans is correct when its green and 10/10 points.

[split fossil]

Yeah, I know :). But I was checking if those were our answers.

[terse ravine]

So is 4 true?

[split fossil]

Yes

Third is wrong I am sure

[split fossil]

Then remove 1

and 2

I am almost positive that we were right at first, but its all about how you phrase it I guess

[terse ravine]

Indeed.

[split fossil]

I don't like this question.

Very vague 😦

[fluid sail]

Uh I need some help

how can I tell if this is one to one

[split fossil]

Clarify please @fluid sail

[fluid sail]

lol this is the question

[patent beacon]

I like how the definition isn't there

A function f is 1 -to- 1 if no two elements in the domain of f correspond to the same element in the range of f . In other words, each x in the domain has exactly one image in the range.

Stolen from some online source @fluid sail

[fluid sail]

yea, I think I got it

I think if it has 2 y values that are the same it's not one to one

they got to be different

🤔

she never taught us this lmfao

[odd abyss]

I need help

if you have √7 (5√7√5 + 3√7√3)

how do you distribute

[viscid thistle]

what do you mean how?

√7 * 5 * √7 * √5 + √7 * 3 * √7 * √3

then you simplify

@umbral current are you sure that's precalc?

looks like an optimization problem

ok hang on then there's definitely a way to do it simply

[odd abyss]

I didnt ssee this

[viscid thistle]

There isn't any you have to deal with messy derivative

oh alright

that's what i thought but its in precalc

Though you're only concerned with the derivative's numerator since you have to equate numerator of derivative to 0

so why is he having to use derivatives

not really precalc-y

have you used derivatives?

then there's no way he can expect you to do this

if it hasnt been covered

whats the rest of the worksheet like?

did you do those ok

well if you have any questions or concerns slap em here

plug in one of the functions into the other and it should result in x

if it does they're inverses

looks like you are right they're not inverses

the inverse of f would be something like (1/x) - 7

nice job

yep

thats one way to prove it

find the inverse of g

yeep

f(f^-1(x)) = f^-1(f(x))

yessir

i dont see why not

f(x) = x is its own inverse because it fits that whole

f(f^-1(x)) = x

thing

idk if that actually counts or not

can you think of any other functions?

1/x

if you plug in 1/x to 1/x you get x

?

oh i see

how are you solving for the inverse?

nice

thats how i do it too

(x-3)/(x^2-x-6)

(x-3)/((x-3)(x+2))

VA at x=-2

but why is it a "hole" at x=3 and not a VA as well?

because at x =3 you get 0 in the denominator

When the zero and VA are the same, does that indicate a hole?

so its undefined or DNE or whatever

is VA vertical asymptote?

yeah

'

-2 doesnt give you 0 in the denominator

3 does

oh wait

oop it does

interesting

i didnt look at the whole thing my bad

i think it's because the zero in the num and zero in the denom line up

its giving a distinction between them?

hence a hole

as opposed to a VT

VA*

ah yea

0/0 is a special sort of form

same with infinity over infinity

also how do you get the inverse of that?

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

swap x's to y's and solve for y

you end up factoring y out, right?

yeah

let me try it real quick since I should know how

alrighty

lmao I get to y(x+5)=3x-2

then I keep making it worse from there

how are you supposed to go about that

(-2/y+5)+3 ?

that doesn't work nvm

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

thats what i got

took me a while idk why

its straightforward

(3y - 2)/(y + 5) = x

3y - 2 = x(y + 5)

3y - 2 = xy + 5x

3y - xy = 5x + 2

y(3 - x) = 5x + 2

y = (5x + 2) / (3 - x)

yours is the same answer @umbral current

its just multiplied by -1

extremely large brained

@viscid thistle @umbral current (5x+2)/(3-x)

yup!

oh

yeah

good job

the factoring is just weird

like you have to shuffle the stuff around

yeah you that xy term

looks scary

my hand writing is terrible

so mine always looks like spaghetti

[willow bear]

cancel out the factors that were causing the hole

[smoky needle]

I’m stuck whether the interval would be (-infinity, infinity) or [0, infinity)

[viscid thistle]

its supposed to be open interval so i dont think you can use an inclusive [

i think its (0, infinity)

[smoky needle]

Ohh got it, so not [ ?

[viscid thistle]

Increasing/decreasing intervals usually are marked with ()

[smoky needle]

Thank you!

[viscid thistle]

it says open interval so yea

[lime kestrel]

$\frac{x^2}{x^2 + 1}$ is equivalent to $1-\frac{1}{1 + x}$

[obsidian monolith]

humble human bean:

humble human bean:

[lime bolt]

lol u still messed it up

[viscid thistle]

denominator is x^2+1

[viscid thistle]

help

I personally got
8 < t <= 37/7

[blazing parrot]

show us your working

[uncut mulch]

integer values

what integers are between 8 and 37/7

[viscid thistle]

OOOh

Frick thanks

[unkempt magnet]

why is x = 0 a critical value?

[blissful ridge]

Because LHS cannot be less than Zero

[unkempt magnet]

what if for example x=0.5

then we would have sqrt(4)

wouldnt the answer to that be +2 or -2

[blissful ridge]

No

Negative values don't come out of sqaure roots

[unkempt magnet]

oh yh

thx

@blissful ridge sorry sstill confused

wouldnt small negative values still satisfy the inequality

as long as its greater than -4

for example x=-2 that would give sqrt(6) which is greater than -2

[willow bear]

sqrt(2 * (-2) + 3) is undefined

but what you seem to be getting at is that for values of x in [-3/2, 0] the inequality is guaranteed

[shut sinew]

x=-1 is defined

[unkempt magnet]

oh shit

sorry my brain broke lol

hold on

no it didnt

kind of

i didnt notice something

the video im watching and the notes associated to it have different examples

my bad i didnt notice

im working with sqrt(2x+8) > x

[willow bear]

oh

ok so the restriction set here will be [-4, +∞)

[unkempt magnet]

yep

[willow bear]

same idea: for x in [-4, 0) the inequality is guaranteed

since then you have sqrt(2x+8) ≥ 0 > x

[unkempt magnet]

ok so i get that sqrt(2x+8) will always be >= 0

so because thats the LHS

that means the RHS also has to be >= 0?

[willow bear]

no

it means that the set of points at which RHS is negative automatically becomes part of your solution

[unkempt magnet]

how can RHS be negative

wait nvm its an inequality

but why is x=0 a critical point

@willow bear

[willow bear]

idk, reads as bad/obscure terminology to me

[unkempt magnet]

oh

i thought you find critical points by solving it as an equation

so sqrt(2x+8) = x

[willow bear]

i mean heres the thing

you want to square both sides

to get rid of the sqrt

but you can't just do that blindly

you can only do it if both sides are positive

[unkempt magnet]

yeah since sqrt gives positive output

you would only take the positive solution

as a critical point?

[willow bear]

no, that is not what i'm talking about!

no! no! no!!!

[sand harbor]

[unkempt magnet]

sorry

[willow bear]

like

$a < b \implies a^2 < b^2$

[obsidian monolith]

Ann:

[willow bear]

we WANT to do this

do you understand that much

[unkempt magnet]

not really

[sand harbor]

Performing the same thing on both sides make equations unchanged😎

[willow bear]

we have a square root

we want to get rid of the square root

square roots are undone by squaring

therefore we want to square both sides

[unkempt magnet]

yes

[willow bear]

do you understand what ive said so far

y/n

[unkempt magnet]

yes

[willow bear]

ok

great

so, we want to square both sides, but we cannot do it blindly

going from a < b to a^2 < b^2 is only valid when a and b are both positive

[unkempt magnet]

this will make the whole thing positive but x could be negative?

[willow bear]

no, if either side is negative the inequality just breaks

[unkempt magnet]

yh

[willow bear]

we don't want it to break

[unkempt magnet]

you would flip the sign

[willow bear]

no.

not what i mean.

[unkempt magnet]

oh

[willow bear]

it would BREAK, like

if you took -5 < 5 and squared both sides

[unkempt magnet]

yh

[willow bear]

you'd get 25 on both sides

but you can't write 25 < 25

that's just false

worse still, if you took -7 < 5 you'd get 49 < 25 which is even more BS

[unkempt magnet]

yes

[willow bear]

going from a < b to a^2 < b^2 is only valid when a and b are both positive

so going back to our original inequality, sqrt(2x+8) > x

sqrt(2x+8) is positive always when defined, so there's no issue with that

but x also has to be positive

in order for the squaring thing to work

understand? y/n

[unkempt magnet]

y

[willow bear]

ok so like

we've covered that part, more or less.

or will cover.

but we still need to deal with the other case, when x ≤ 0.

BUT it turns out

that part is actually trivial

since EVERY admissible value of x in that range will be a solution

[unkempt magnet]

btw

could the first case be x >= 0 and the other x < 0

[willow bear]

sure

you could examine zero separately from both, even

[unkempt magnet]

since EVERY admissible value of x in that range will be a solution
@willow bear and why is this the case

[willow bear]

sqrt(2x+8) > 0 > x

[unkempt magnet]

sorry im still confused

the x>0 case actually

so when x > 0 you can square both sides

then you get a quadratic inequality

which becomes x^2 - 2x - 8 < 0

i think this solves to (-2,4)

for the x<0 case

[willow bear]

hhh

[unkempt magnet]

i guess (-4,0) feels too trivial to explain

i feel like im just looking at the LHS with x<0

oh i think it should be x>=0 because you can square when both sides are non-negative

oh so that must mean

(-2,4) -> [0,4)

and then with x<0 we have [-4,0)

OH

so

since sqrt(2x+8) > 0

and we are looking at x < 0

sqrt(2x+8) > 0 > x satisfies the inequality

so we can use -4 all the way up to (but not including) 0

but alternatively could you write

sqrt(2x+8) >= 0 so then sqrt(2x+8) >= 0 > x

[sand harbor]

You guys still on this

[unkempt magnet]

but anyways we have [0,4) and [-4,0) which combines to [-4,0)

nope done

[old rune]

Does someone know this

[viscid thistle]

okay

i'm assuming you are asking for help and not for answers

because it it's for answers this is not the place

anyways, what have you tried so far?

@old rune

[wind holly]

I've got a question

and its probably a stupid one given that the answer is right in front of me

but what does find positive and negative intervals even mean?

[terse ravine]

Yeah some times the answers in our face, but our eyes don't see it.

I'm misunderstanding something here.

@patent beacon can you assist me on this one Kaynex?

[patent beacon]

The small + means "from the right"

[terse ravine]

It looks like its headed towards positive infinity

[patent beacon]

Yus

[terse ravine]

but its wrong

I entered oo and it didn't like the answer

that's why I'm confused.

[patent beacon]

That's because it doesn't think that you deserve the mark

Math questions are very judgemental

If you give it a lot of food it will like you more

[terse ravine]

[patent beacon]

The limit should be inf. There must be a typo somewhere

[terse ravine]

I probably typed in 00 instead of oo like a noob.

I got it now.

[viscid thistle]

can anyone help me with question d?

I've already tried to do it, but my answer is way off

[snow quiver]

any ideas on b)? using the binomial theorem

[blissful ridge]

Applying the binomial theorem as it is will be a pain
So as suggested first try to factorise the given quadratic

[snow quiver]

I got that, what do I do next? it's still one term

[remote veldt]

@snow quiver What you can do is note that $\left((2x + 3)(x - 1)\right)^8 = (2x + 3)^8 \cdot (x - 1)^8$

[obsidian monolith]

Nicholas:

[remote veldt]

do you see what you could do from there? Think of all the ways you can make x when you multiply those two things, and you should b able to find the coefficient

@viscid thistle the best way to do this is probably to count the number of arrangements that include both tim and gwen, and then just subtract

[rancid crest]

anyone alive

i need help with like the simplest stuff

reeeee

metaL

@slender river

0-0

[slender river]

why ping me

[lapis egret]

can anyone help me with some algebraic tests to check for symmetry with respect to both axes and the origin questions?

[bleak lance]

May someone please help me with dividing an interval into subintervals?

[fiery wren]

how do you evaluate d

[flat maple]

Anyone know how to solve for y?

Oh nvm nvm got it

[fiery wren]

how do you find the range of a function without graphing

[willow bear]

in general? take the equation f(x) = k and find which values of k make it have a solution

those values of k will make up your range

f here is your function, of course

[rare zephyr]

Question's about finding coordinates of point q. How tf did I got x=24

elp

[uncut mulch]

text is too small/faint to read
also what's the original question

[rare zephyr]

Find coordinate of q

question also said p is the midpoint of A and B

[uncut mulch]

text is too small/faint to read
what are A,B etc

like do you have a legible pic of the problem

[rare zephyr]

Translation :
A flower garden is located on a hill at point (-1 , 30). The resting station is located at the peak at point (9 , 50), while the gas station is located at the hillside at point (20 , 22). A market is to be built such that its distant from the flower garden and resting station are the same. The market also must be at minimum distance with the gas station. Find the coordinates for the market to be built at.

My interpretation :

q is where the market is supposed to be

pq should be a locus moving in a straight line but its always equidistant from A and B

so that's where I got q

my diagram looks weird I know

back to the problem Im faving I somehow got the x of q is 24

[willow bear]

wait ok your naming is weird

[rare zephyr]

A is the garden
B is the resting station(?)
C is the gas station
q is the market

[willow bear]

ok now it makes sense

,rcw

[willow bear]

A = (9,50)
B = (-1,30)
P = (4, 40)

so far so good

[rare zephyr]

that bottom left is cq

[willow bear]

ok so AB has a slope of... 2

so PQ has a slope of -1/2

[rare zephyr]

yep

[willow bear]

ok so $PQ$ has equation $y = -\frac12x + 42$

[obsidian monolith]

Ann:

[willow bear]

and $CQ$ has equation $y = 2x-18$

[obsidian monolith]

Ann:

[willow bear]

so $-\frac12x + 42 = 2x - 18$ for the $x$ coordinate of $Q$

[obsidian monolith]

Ann:

[willow bear]

yeah so you get x=24

your diagram's just not to scale

[rare zephyr]

but... something seems... wrong

[rare zephyr]

ohhhhhhhhhhhhhhhhhhhhhhhhhhh

i played myself for this whole 2 hours

whatever thank you for reassuring my confusion

❤️

[earnest coral]

Can someone help me?

I know it’s negative exponent, and both the numerator & denominator are odd

Is the b/c greater than or less than 1??

[split fossil]

@earnest coral I think you have covered all the options, and the absolute value of b/c is greater than 1.

[earnest coral]

ohh

[split fossil]

Are you good?

[earnest coral]

uhh I think so

[split fossil]

Good 🙂

[earnest coral]

so it's greater than 1?

[split fossil]

Yes, but the exponent is negative, so it is the absolute value that is greater than onne

[earnest coral]

what instance would it be less than 1

[split fossil]

wdym?

[earnest coral]

how would a graph that has an absolute value of b/c less than 1 look like

???

[split fossil]

Oh. Very similar, just wwarped slightly

You can try graphing it. I tried it and found that the original function was x^-5/3. You could try x^-3/5

[earnest coral]

ohh

alright

[violet quail]

A practical question. Let's say I input 2000 money (any currency) every year into a bank account, and I get 3% in return each year, how much money will I have after 11 years?

How do I calculate this for n =11?

let me edit so it's readable..

[blissful ridge]

Where did you get that equation?

[violet quail]

i see a pattern

thats all

[viscid thistle]

Bring a_n to the left side

Run a summation from n=0 to 10

On both sides

[blissful ridge]

If you try to see more
You'll see a better pattern

[viscid thistle]

Left is telescopic and right is a GP

You'll have a_11 - a_0 on left and a gp sum on right

[violet quail]

GP?

[viscid thistle]

Geometric progression

[violet quail]

ah

hmm, still dont know what the expression should be

i mean, all i see is recursion

[blissful ridge]

If a_n represents the money after 1 year
Then shouldn't a_0 be 2000

[violet quail]

yeah´, you're right

alternatively

[blissful ridge]

And this looks like compound interest to me

So a_1=a_0+a_0(0.03)^1

[violet quail]

I solved a similar problem a year ago, and now im coming up short

[fiery wren]

how do you determine the net change of a function?

[viscid thistle]

Wdym

[fiery wren]

i got it actually, thanks

f(b) - f(a)

[terse ravine]

@split fossil any Idea?

I think the ans is -2.

<@&286206848099549185>

[crude bough]

yo

when we are dealing with rate of change questions, we always think of differentiating

so first try differentiating the function

then try subbing in x = -2 into the differentiated function

that should be the answer

[terse ravine]

Oh that was too easy.

The ans is -2

[crude bough]

wait, do you know calculus

[terse ravine]

I was so used to harder ones I forgot about subbing in the variable first when the denominator does not equal zero.

If the quest had asked y(t) = 2 / (s-2) at the point t = -2 I would have to make t slightly more negative to around 1.9999.

[crude bough]

wait, was the answer -2 for the earlier question

[terse ravine]

Yes.

[crude bough]

give me a sec

[terse ravine]

ok

[crude bough]

uhh, i think you got very lucky for this type of question

because u aren't meant to directly sub x = -2 into the function

do you like f'(x)

like derivative

[terse ravine]

I have no choice but to love calculus.

I'm going to take a break since I've been doing math problems since this afternoon.

[crude bough]

$f'(x) = \frac{2}{(x+1)^2}$

[obsidian monolith]

KDen:

[crude bough]

this is what you are meant to sub x = -2 into

[terse ravine]

Ok I'm listening

[crude bough]

but you got very lucky since it is the same answer as if you did f(2)

[terse ravine]

[noble bison]

Quick question, I understand how to do logs (for the most part except when you get to e^x etc), but what are these instructions asking? It confuses me : "Using laws of logarithms, write the expression below using sums and/or differences of logarithmic expressions which do not contain the logarithms of products, quotients, or powers."

i have log((x+9)/x^58) but i have no idea how I am suppose to enter it given the constraints

[onyx wharf]

two laws of logarithms are being used:
log(a/b) = log(a) - log(b)
log(a^b) = b*log(a)

so you can try those and see how to get rid of the products, quotients, and powers

[ornate wolf]

any hints?

[willow bear]

write 1/x as (1/x^2) * x

[viscid thistle]

$\lim_{x\to 0}\frac{f(x)}{x^2}x$ as Ann said

[obsidian monolith]

HoboSas:

[ornate wolf]

thank you

[odd abyss]

Do I simplify sqrt (35) and sqrt of (14) to be

7 [5(sqrt(5))(sqrt(7)) + 2(sqrt(2))(sqrt(7))]

[uncut mulch]

,rotate

[uncut mulch]

the stuff inside the parentheses can be expressed like that,
HOWEVER, that doesn't meant that your sqrt(7) on the left can suddenly turn into a 7

@odd abyss

[odd abyss]

Oops

Meant to write sqrt(7)

So is this the first step in multiplying?

[uncut mulch]

$=\sqrt{7}(5\sqrt{5}\sqrt{7} + 2\sqrt{2}\sqrt{7})$

[obsidian monolith]

ramonov:

[uncut mulch]

would be an acceptable first step

[viscid thistle]

iirc you asked this problem before

i thought you got the answer then

[odd abyss]

Ok so I know that we distribute the sqrt of 7 to each side of the +

How do I write it after that

Do the sqrt of 7s cancel out?

[uncut mulch]

wdym by cancel...

[odd abyss]

5 sqrt(5) + 2 sqrt (2)

As answer

[uncut mulch]

no

are you saying that sqrt(7)*sqrt(7) = 1?

[sand harbor]

No but √a * √a = a

[odd abyss]

No it’s 7 right

[uncut mulch]

yes, in which case it should be clear that

5 sqrt(5) + 2 sqrt (2)
would be wrong

[odd abyss]

35sqrt(5) + 14 sqrt(2)

?

[uncut mulch]

that's better

[odd abyss]

As answer?

[uncut mulch]

yep

[odd abyss]

Ok

So everything just flipped

[uncut mulch]

wdym by flipped

[odd abyss]

Look it’s flipped

The radical and the number outside

[uncut mulch]

flipped is an extremely inappropriate term here

[odd abyss]

It’s flip flopped

Is this correct

[uncut mulch]

no

[odd abyss]

Sorry

X^2*

?

Instead of 2x^2?

[uncut mulch]

can you rewrite the whole thing

[odd abyss]

x^2 (sqrt(9x))

[uncut mulch]

is the original denominator sqrt(2x) or sqrt(2)x

whoops

[odd abyss]

2x

All under the radical

[uncut mulch]

then no

[odd abyss]

Is this the correct answer

[uncut mulch]

no

you're not applying multiplication laws properly

[odd abyss]

Woops

What about now

[uncut mulch]

good, keep simplifying

[odd abyss]

3x^3?

[uncut mulch]

yes

[odd abyss]

Ok thanks

[odd abyss]

Is this just 15

[uncut mulch]

no

[odd abyss]

I got it

If I had to find the function value

And I got sqrt(0)

Do I write f(3) = 0 or doesn’t exist

[viscid thistle]

why is the inequality x^2 - 4x-21 < 0 in f (x) < 0 ?? why not f(x)=0 or >

[half star]

no particular reason, apparently you're simply being asked to find all "x" that verify that

f(x)=x^2-4x-21 has two real roots so it should have points that verify f(x)=0 and f(x)>0 as well

[elder charm]

you can solve 0 > f(x) if you like that inequality better 😅

[viscid thistle]

@half star thank you 🙏

[floral basin]

Consider the functions f(x)=−(x2)+6x and g(x)=x2−9x+1. Which of the following is true?

f(-3)>g(-3)

f(-2)<g(-2)

f(0)=g(0)

f(4)<g(4)

help

[elder charm]

did you miss some powers there?

[viscid thistle]

I believe x2 is x^2

[floral basin]

tis

My bad

[elder charm]

just evaluate the polynomials at these points.

[viscid thistle]

f(x)=x^3 instant rate of change is 3 at x=-1 // finding the equation of the tangent line to graph y = f(x) at x cord = -1 .. can someone help me understand how i would get y+1 = 3 (x +1)

[elder charm]

Is that supposed to be a question?

[viscid thistle]

yea sorry just edited 😅

[uncut mulch]

what's the y coordinate at x=-1

and apply point slope formula

[viscid thistle]

what is the y cord at x=-1?

[uncut mulch]

that's what i'm asking you

[viscid thistle]

3?

[uncut mulch]

no

y=f(x)=x^3

f(-1) = ?

[viscid thistle]

1?

im not sure..

[uncut mulch]

replace the x in x^3 with -1

[viscid thistle]

@short sorrel

What answer do you get when you solve for x

I’ll tell you what I got after I hear yours so I don’t bother your work

[short sorrel]

there's no closed form solution

an approximation is x = 1.307776829...

[ancient stirrup]

@viscid thistle let $x=t^2$ and solve it for t

[obsidian monolith]

majasaro:

[terse ravine]

I've been doing Math for 8 hours straight.

Break time 🙂

[viscid thistle]

I’ll try that @ancient stirrup

[drowsy juniper]

How to do this?

[blissful ridge]

Did you make the diagram

[willow bear]

@drowsy juniper what have you done so far & where are you stuck?

& are you willing to cooperate or did you expect someone to just do the problem for you?

[viscid thistle]

Haha the kid posted it in multiple channels so he seems a bit serious about it

[willow bear]

well they haven't replied to me so i don't know for sure

maybe they chose to run away after seeing the problem not get solved in an instant

[viscid thistle]

Hmm maybe

He hasn't replied anywhere

[willow bear]

do you know for sure zeroxyl goes by he/him pronouns?

[viscid thistle]

No I just assumed which ig is wrong now that you pointed it

mb

[viscid thistle]

I've been doing Math for 8 hours straight.
@terse ravine Only 8?

[willow bear]

bruh

[drowsy juniper]

Who know how to solve this ? HEHE

@terse ravine Only 8?
@viscid thistle Same 8 hours of solving :))

[willow bear]

ok so like first off

[drowsy juniper]

I gets the formula in my first Question sorry about that

[willow bear]

can we PLEASE not spread the extremely toxic mindset of "working for 3478528934 hours straight is good"

like seriously

8 hours straight is an easy way to burn yourself out

[viscid thistle]

Haha I was just messing around, ofc I didnt mean it like that

[willow bear]

YEAH? AND HOW WAS ANYONE MEANT TO KNOW?

[viscid thistle]

Maybe shouldve put /s at the end my bad

[willow bear]

yes

k whatever

@drowsy juniper have you made any progress on this question on your own?

[drowsy juniper]

W8 let me screenshot my work on this

[willow bear]

ok

so...?

[drowsy juniper]

TanA = Sqrt (3) Cot B= -sqrt(2) = sqrt (3) sqrt (2) = sqrt (6) Then Ievaluate the angle of tan and cot Im not sure about this Im only focusing on my reviewer but I think my answer is incorrect.

[willow bear]

okay first off

i thought you said screenshot

not type out in plaintext and garble it

Cot B= -sqrt(2) = sqrt (3) sqrt (2)
for starters, what even is this

and second, you didn't even give an answer

[viscid thistle]

8 hours straight is an easy way to burn yourself out
@willow bear Not really tbh, for some people doing mathematics 8 hours straight is normal. Not unhealthy at all

[willow bear]

idk 8 hours straight w/o breaks is like an express ticket to burnout from my pov

i don't know anyone who actually does that

[viscid thistle]

So like..PhDs do it ?

Haven't heard of school kids do that

Possible but haven't heard. Even college students 8hrs seems to be ah..too long for a stretch

[half star]

it's pretty boring at a school or freshman maths level, but it could be fun if you can understand and study many different topics. PhD people get in there because they love the idea of researching something hours straight (with some rest obvs since they're humans)

[viscid thistle]

Hmm makes sense ^^

[viscid thistle]

it's pretty boring at a school or freshman maths level
But if you are doing Olympiad mathematics it's extremely interesting and fun

Hmm

[viscid thistle]

For graduate/PhD mathematicians 8+ hours a day is normal.

Like straight without breaks ?

It's just like full time job for them

Also research problems are really hard, it takes hours, even days to make some progress with one

Hmmmm

[burnt sonnet]

Azazeel breaks are allowed 🙂

[viscid thistle]

8 hours straight damn

ive never done anything for 8 hours straight, I get diminishing returns after an hour of productivity unless I take a break and do another task

[somber yew]

I can sleep for 8 hours straight if that counts

[viscid thistle]

I have insomnia so sleeping even 2 hours is too much

[somber yew]

Ouch, take care! I often despise myself for oversleeping but I guess those who can't get much of it have it worse.

[viscid thistle]

It's bad, I suddenly wake up in the middle of the night and wonder who or what the fuck am I

[somber yew]

Have you consulted a professional? This sounds terrible.

[gilded brook]

I got one of these wrong, how do I figure out which one is wrong?
https://gyazo.com/7e1fe787befa3e2988dfcaf1e2635020

[cursive rock]

how do i use

can someone help me use synthetic division to solve this

i dnt know how to write my answer in polynomial format

i have -5, 5, -5, 5, -5

[willow bear]

the last number is the remainder, is it not?

thus -5x^3 + 5x^2 - 5x + 5 + 5/(x+1)

[cursive rock]

how did you get that?

[glad turret]

Help me with point slope formula

[uncut mulch]

help how?

[glad turret]

I need help understanding it

My school didnt go over it in algebra 2

[elder charm]

I mean, slope is just difference in y divided by difference in x.
that's what this implies

[glad turret]

Oh, ok

[fiery wren]

am i missing anything

[elder charm]

I don't think you are. Why?

[fiery wren]

just making sure

it's asking me to select all that applies

and that's the only answer choice that i feel is true

or maybe im missing something

[glad pasture]

could someone show me how to solve this i got to (2x-5)ln(4) = (x+4)ln(3) but i think the x cancels out

[elder charm]

$$
4^{2x - 5} = 3^{x+4}
$$$$
(2x-5) \ln(4) = (x+4) \ln(3)
$$

[obsidian monolith]

ConfusedReptile:

[elder charm]

$$
x(2 \ln(4) - \ln(3)) = 4 \ln(3) + 5 \ln(4)
$$

[obsidian monolith]

ConfusedReptile:

[elder charm]

so they most certainly don't cancel out.

[glad pasture]

how do you do it if you could show me

[elder charm]

Do what?

[glad pasture]

like the problem

[elder charm]

I mean, I basically just solved it above

the only step left is to divide by the paranteses on the left to obtain x = <something>.

[glad pasture]

what did u do to get it to one side of the equation

i thought u had to divided one side to the other

[bold maple]

does anyone know a good video to explain 6.2, the unit circle with co tan, co secant, etc

[stuck raft]

Hi, could anyone work through these questions with me?

[split fossil]

Do you know quadratic formula?

[stuck raft]

12. b2 Square root b2- 4a all divided by 2a? I think

[split fossil]

Close

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

[obsidian monolith]

Ionic:

[stuck raft]

Yes

[split fossil]

Good. Put in a=1,b=1,and c=-1 for first problem

[stuck raft]

$x=\frac{-1\pm\sqrt{1^2-4x1x-1}}{2x1}$

$x=\frac{-1\pm\sqrt{1^2-4×1×-1}}{2×1}$

[obsidian monolith]

2433red:

[stuck raft]

@split fossil like this?

[split fossil]

Yes

[stuck raft]

$x=\frac{-1\pm\sqrt{5}}{2}$

[obsidian monolith]

2433red:

[split fossil]

Yes

And you can split that into two solutions

[stuck raft]

@split fossil

Is this the final answer?

[split fossil]

Yes

[stuck raft]

Is there a way to do this with completing the square?

[split fossil]

Yes, technically the quadratic formula is completing the square

If you didn't know that, you should check out a proof of the formula, because it is actually quite cool

And simple

[stuck raft]

@split fossil

Its different

?

[split fossil]

Solve for x

And also it should have been divided by a at the start so it is incorrect

Khan Academy has a good proof

I've got to go

[stuck raft]

Yes, technically the quadratic formula is completing the square
@split fossil so if i get a question like this I just put the quadratic formula in its simplist form? Non calc

I've got to go
@split fossil OK sure

<@&286206848099549185>

[terse ravine]

I ended up at -9m/-9 which canceled both -9 and left me with m.

[hoary adder]

-3m-9=-9, m=0

[terse ravine]

-9m/-9 = 0/-9

m = 0/-9

m= 0

[hoary adder]

where did the image go

smh

[dreamy spade]

hi, this is a question from grade 12 precal class, i believe i can ask for help for this question as this is for a review hand in worksheet, this is a question frm transformations of graphs to reciprocal graph using invariant points and asymptotes, i have not came across anything similar to this question before so i have doubts if i did this right or not, i m sry if it's difficult to understand, tysm ^_^

[hoary adder]

,rccw

[dreamy spade]

<@&286206848099549185>

[umbral cloud]

Hey guys

I have some questions about implication in math....
Can anyone help me?

[viscid thistle]

Rule 1, #❓how-to-get-help (just post it and don't ask if you can get help)

[umbral cloud]

To prove something while answering a question... Should I use => or <=>?

Sometime i get confused about them

Between them XD

Especially when doing an exam question

And thanks in advance

?

[patent beacon]

@umbral cloud
A → B can be read as "A if B"
It is important to know that → doesn't work backwards. B being true doesn't say anything about A.

A ⇔ B (or A iff B) can be read as "A if and only if B"
This works forward and backwards. That is, A and B always have the same truthity, even if you use B to get A.

[umbral cloud]

Okay... May be this example will help me..
"prove that for n € N...
((2+3i)^n) +((2-3i)^n)) € R"
Without knowing how to solve it....
Should I conclude by...
=> ((2+3i)^n) +((2-3i)^n)) € R or <=> ((2+3i)^n) +((2-3i)^n)) € R?

And thanks for the response 😊

[sudden bay]

so im doing piecewise functions. I have (x+1)^2, x > 3

do i multiply the exponents or leave them and just go graph

and if so how do i graph x^2+1

[viscid thistle]

did you just do a freshman's dream

$(a+b)^2\neq a^2+b^2$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

@sudden bay

[sudden bay]

?

[viscid thistle]

i mean if you expand that correctly

it'd be easier to graph

?
you did (x+1)^2=x^2+1

which is wrong

[sudden bay]

So what do i do instead?

[viscid thistle]

identities

$(a+b)^2=a^2+2ab+b^2$

[obsidian monolith]

Al𝟛dium:

[sudden bay]

oh god

[viscid thistle]

are you able to follow this

[sudden bay]

yes

sort of

[viscid thistle]

so what would you get after expanding correctly

take your time if you want

[sudden bay]

x^2 + 2x+1 ?

[viscid thistle]

sort of
let me know if there's any doubt wandering in your head rn

[sudden bay]

well idk why the 2 is there

[viscid thistle]

x^2 + 2x+1 ?
yeah

that's just an identity

very known one

you can prove it

look

[sudden bay]

My math teacher was never very good so he skimmed this stuff

[sand harbor]

LOL

[sudden bay]

So why does that happen?

[viscid thistle]

hold up i'm writing the proof

should not take more than one 1 min

i'm doing it with colors and stuff so that you can see it better

[sudden bay]

ok

[viscid thistle]

$$(a+b)^2=({\color{yellow}{a}}+{\color{red}{b}})({\color{green}{a}}+{\color{blue}{b}})=\overbrace{{\color{yellow}{a}}{\color{green}{a}}}^{a^2}+\overbrace{{\color{yellow}{a}}{\color{blue}{b}}+{\color{red}{b}}{\color{green}{a}}}^{ab+ba=ab+ab=2ab}+\overbrace{{\color{red}{b}}{\color{blue}{b}}}^{b^2}=a^2+2ab+b^2$$

[obsidian monolith]

Al𝟛dium:

[sand harbor]

😎

[sudden bay]

I get it

[viscid thistle]

alright

so y=x^2+2x+1

do you know how to plot it?

[sudden bay]

it would look like a U shape right

[viscid thistle]

yes

[sudden bay]

alright yea

[viscid thistle]

because it's a parabola with a>0

[sudden bay]

i get the idea behind it but i need to cram to remember this for my test

[viscid thistle]

well, all i can say from experience, the more you practice, the more it will stick in your head

[sudden bay]

I have this as part of a unit that im doing because they never properly reviewed this

i'm basically having to teach myself precalc because last year in 8th grade my teacher just said " well tough for you but we're not doing this"

[viscid thistle]

that's bad, on any case, i'm always here at this server if you need any more help or anybody else will do as well.

[sudden bay]

alright thanks man

if i'm graphing this as a peicewise function do i write the parabola until x > 3

[viscid thistle]

like, wdym by graphing this a piecewise function? is it something like
$f(x)=\begin{cases} \text{something} & \text{something} \ x^2+2x+1 & x>3 \end{cases}$

[obsidian monolith]

Al𝟛dium:

[sudden bay]

yes

[viscid thistle]

then yes

the parabola should only be draw from 3 to the right

[sudden bay]

ah

i understand

so the leftmost side of the parabola should be past 3

[viscid thistle]

like i haven't looked how the parabola will look but

so the leftmost side of the parabola should be past 3
?

[sudden bay]

so would the side of the parabola be past 3 on the graph

because it has to be more than 3 so

i was thinking that the x cannot go past 3

[viscid thistle]

i mean like yeah, from 3 to the left is graphed the other thing they told you about the piecewise function (which of you didn't told me)

,w plot x^2+2x+1 from 3 to 10

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Al𝟛dium#4084

[viscid thistle]

this is from 3 to the right

[sudden bay]

yes

That is what i meant

[viscid thistle]

then the rest is simply up to what they tell you about the rest

[sudden bay]

ok

[terse ravine]

Hmm not sure where I went wrong since the algebra is correct.

[viscid thistle]

can i see what you did on the algebra

[terse ravine]

x= -4

mx-11 = x^2+7x-7

-11m-11=(-4)^2+7(-4)-7

-11m-11=-19

[uncut mulch]

-11m?

[viscid thistle]

^

[terse ravine]

+11 +11 on both sides

-11m = -8

[uncut mulch]

-11m?

[terse ravine]

-11m/-11 = -8/-11

[uncut mulch]

where's the -11m coming from

[terse ravine]

m= -8/-11 = m= 8/11

[viscid thistle]

what?

[uncut mulch]

this prompts are strongly implying that you made a booboo at the start already

[viscid thistle]

you are not answering us

[terse ravine]

mx-11 = x^2+7x-7, when x= -4

[viscid thistle]

and the left part?

[uncut mulch]

that part is ok

[terse ravine]

how do I rearrange mx-11?

-11m-11?

[uncut mulch]

how are you getting -11m

[viscid thistle]

when x=-4

[uncut mulch]

when replacing the x in mx with -4

[viscid thistle]

and not

when x=-11

[terse ravine]

m(-4)11=(-4)^2+7(-4)-7 better?

[viscid thistle]

m(-4)***-***11

[uncut mulch]

wrong in a different way now

missing sign between the two terms on the left

[terse ravine]

m(-4)-11 = (-4)^2+7(-4)-7

[uncut mulch]

looks ok now

[terse ravine]

Ok let me erase all my noob mistakes on my paper.

[viscid thistle]

you should get to the answer if you don't do any more mistakes (the answer you should get: ||m=2||)

[terse ravine]

m*-4-11=16+7*-4-7

-4m-11=16+7*-4-7

-4m-11=16-28-7

-4m-11=-19

+11 +11 both sides

-4m=-8

[viscid thistle]

so far so good

[terse ravine]

m=-8/-4

8/4= 2

m=2

[viscid thistle]

yeah correct

[sudden bay]

how would i evaluate a piecewise function using f(2) with 2/3x -5, x =< -4 and 4x+2, x>4

I dont plug 2 into X right?

[viscid thistle]

wait

x=<-4 and then you move to x>4?

what's in between

[sudden bay]

well its equal to or less than

[viscid thistle]

you sure it's not a typo

[sudden bay]

idk how to write it via keyboard

[viscid thistle]

?

[sudden bay]

the question says 2/3 x -5, x equal to or less than -4

and then 4x + 2, x > 4

[viscid thistle]

$f(x)=\begin{cases} \frac23 x-5 & x\leq -4 \ 4x+2 & x>4\end{cases}$?

[obsidian monolith]

Al𝟛dium:

[sudden bay]

yes

in this f(2)

[viscid thistle]

are you 100% sure there is no typo

[sudden bay]

yea

ill send a screenshot

[viscid thistle]

bc what you are asking me rn does not have any sense

okay

[sudden bay]

we're supposed to show our work but i dont get what i do with f(2)

[viscid thistle]

exactly

[sudden bay]

im watching a video that says to plug 2 into X

[viscid thistle]

do you know what happens with the function between -4 and 4

from what you are given right there

[sudden bay]

no?

i dont know

[viscid thistle]

exactly

there is no function between those

so you can't plug x=2 to anything

therefore the result

[sudden bay]

because if I do x > 4 or x is less than or equal to -4 wont be correct?

[viscid thistle]

if you mean that if you plug x=2 into any of the parts of the function where x>4 or x=<-4 and get a value no that's non sense because 2 is not on the x>4 and neither on x=<-4

which means that the function does not exist from -4 to 4

so you can't plug x=2 into anything

hence the result there

[sudden bay]

im still not sure if i understand. Is the reason why it is ∅ because I cannot plug 2 into any X without getting a result that does not correlate with x=< -4

[viscid thistle]

okay hold up

[sudden bay]

i dont understand this at all

wait

i think i get it

i get it

thank you

[viscid thistle]

@sudden bay you understood it?

i was gonna plot it

and showed you how there is no such thing as f(2)

let me know if you still want it

[sudden bay]

no its good.

i understand how to do it now

originally i didnt even understand what it was asking now i do

[viscid thistle]

oh lol

okay glad you understood, yw

[terse ravine]

Oh I think its this instead..

@viscid thistle Is this correct?

(x-2)^2=0

sqrt both sides and add +2

wouldn't adding +2 on the left only remove the ^2?

[uncut mulch]

what was the point of square rooting to you

[viscid thistle]

^

[terse ravine]

to cancel out the ^2

[viscid thistle]

You don't "cancel" like that on any case if you strictly wanna follow that

[terse ravine]

ok

[viscid thistle]

$\sqrt{t²}=\abs{t}$

[obsidian monolith]

Al𝟛dium:

[terse ravine]

sqrt2^2=2

[viscid thistle]

And ends up with |x-2|=0, which is x-2=±0 and no matter what x=2. The point of sqrting both sides loses it's sense

So

Instead

Use your logic

The only way that a power equals 0 is if the base is 0

Hence x-2=0 so x=2

Do you understand this logic? @terse ravine

The only way that a power equals 0 is if the base is 0

[terse ravine]

x-2^2=0 is the base correct?

[viscid thistle]

No, the base of a power like a^7 is a, likewise the base of (x-2)² is just x-2

[terse ravine]

x-2 is the base and ^2 is the power

[viscid thistle]

Yep

[terse ravine]

Since the base = 0 the power = 0

[viscid thistle]

Not exactly, if the base = 0, no matter what the power is (excluding ≤ 0 as power) the outcome will be 0:
0^3=0
0^1763=0

Since the base = 0 the power = 0
If the power is 0 and the base is 0 we have 0^0 which is undefined

[terse ravine]

ok

[grim stirrup]

Can I plz get help with this problem

[elder charm]

hint: think about what inverse functions mean, and try to rewrite that expression in English, and you'll probably realise it's very simple then.

For example:
$$
\sin(\sin^{-1}(2x))
$$

[obsidian monolith]

ConfusedReptile:

[elder charm]

"The sine of such an angle that its sine is 2x". Doesn't sound like a hard task now, does it?

[valid estuary]

Hey could anyone help me out for solving for theta on this problem?

I think I've got the right idea for finding some answers but extra answers are what confuse me

[willow bear]

what have you tried

what part of the problem are you struggling with? the underlying mathematics or the communication?

[vague grotto]

can anyone help me, how do i solve 2y²+8=x+8y its a parabola

[blissful ridge]

Solve the Quadratic

[uncut mulch]

solve for what

[vague grotto]

since this one is y²=4px right?

[shadow plaza]

since this one is y²=4px right?
@vague grotto it isn't but it can be converted to

$(y-m)²=kx$ where $k,m$ are constants

[obsidian monolith]

Sidharth:

[vague grotto]

how will it be converted

[uncut mulch]

completing the square

[tulip lynx]

pls