help-23

[sand oar]

so um

this is x

[unreal fiber]

<@&268886789983436800>

[sand oar]

f(x)

[unreal fiber]

yes

[sand oar]

I need to put it in the f(x+h)

[unreal fiber]

correct

[sand oar]

do I just put a +h in there

[harsh shoal]

Tyvm

[sand oar]

+h

like this

[unreal fiber]

not really

you substitute x+h in for x

[sand oar]

then I put it to every x I see

[unreal fiber]

sure

[sand oar]

something like this?

[unreal fiber]

yeah

[sand oar]

can I even simplfy this?

[unreal fiber]

unsure...

[sand oar]

what should I do next

[verbal oak]

Don't functions like x^nsin(1/x) always have an oscillatory type discontinuity?

[unreal fiber]

I think derivative at a point definition would helps us more than just the h → 0 one

[sand oar]

[unreal fiber]

the function itself is continuous but the derivative may not be continuous at every point..?

I am really unsure though

[verbal oak]

I mean we have defined the function at x=0. So that shouldn't be a problem.

[unreal fiber]

also is oscillatory type of discontinuity even a thing

[sand oar]

then what should I do

[verbal oak]

I am pretty sure it is.
But I am not sure if it can be removed. My best guess is that we can not diffrentiate this function at x=0.

[unreal fiber]

the derivative of f(x) at point a: $f'(a) = \lim_{x → a} \frac{f(x) - f(a)}{x - a}$

[flat frigate]

biggboy

[unreal fiber]

not for all r yes, but I think it can be for some

like r = 2... I think

[verbal oak]

Maybe right hand derivative= left hand derivative definition might work.

But then again I don't think we can remove oscillatory type discontinuity.

Another easier definition of derivability to use in this case can be
limit of f'(0+)= limit of f'(0-).

[unreal fiber]

the graph of xⁿsin(1/x) seems to be bounded by xⁿ and -xⁿ for all positive integer n bigger than 1

so I think that "oscillatory discontinuity" disappears(?)

[sand oar]

guys I think I should ask someone else

but thanks anyways

[unreal fiber]

sure

[sand oar]

.close

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[faint saffron]

the ans has to be 50 cm^2 and (50pi-100)cm^2 but i dont know where to start

[junior smelt]

See you finally changed the days

For b, you can work out what the unshaded parts are pretty easily

[faint saffron]

hehe yess

[junior smelt]

Then use that and the area of the square to get the area of the shaded

[faint saffron]

but i have no idea how to do that

[junior smelt]

You have a square right, and inside it is a quarter-circle of radius the same as the square’s?

[faint saffron]

yess

a

wait

do i have to find the area of arcs

like 90/360 multiply 10^2

and then

i forgot the english word

[junior smelt]

Don’t forget the pi

[faint saffron]

ohh yess

and then like area of square - area of 2 arcs right?

[junior smelt]

For one of the unshaded you could do area of square - area of that quarter circle

Then you know that bit of the unshaded are the same so double that

Then remove the unshaded total area from that of the square

[faint saffron]

omg yess

i can see it now

and how about a?

it is a bit harder than b

seems to me

i mean

[junior smelt]

A tiny bit but not by much

Similar idea to the last one, the unshaded area is easy to work out

[faint saffron]

maybe hm i think i can find the area of the semi circle and it will be 25pi/2

but what about another unshaded area

[junior smelt]

Create like “sub-squares” to see it easier

[faint saffron]

okay a sec

[junior smelt]

Like this

You should have it from there

[faint saffron]

aaa

okaaaay

i will try it now

thank you very much chartbit!! u r cool!

.close

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[lean otter]

is this correct ?

[safe radish]

@lean otter Has your question been resolved?

[waxen patio]

nope

$g\left(f\left(h\left(x\right)\right)\right)=g\left(f\left(\dfrac{-2}{x}\right)\right)=g\left(\dfrac{4}{x^2}\right)=\dfrac{4}{x^2}-1$

[flat frigate]

SAKUUN

[waxen patio]

@lean otter

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[tepid torrent]

How would I finish this: Tan(53.2)/1 = 14/x

[quasi bison]

are you asked to find the value of x?

[frank glen]

You're solving for x?

[tepid torrent]

Sorry, let me show u the full problem

It's a trig question- I got it down to Tan(53.2) = 14/x

and I know you're supposed to cross multiply it

I just don't know how to finish that

[unreal fiber]

how did you got that from the diagram... and what does x represent in the diagram

[tepid torrent]

omg wrong question

here's the right one

[unreal fiber]

is your x the length GH

[tepid torrent]

yes

[unreal fiber]

then you got your fraction wrong

[tepid torrent]

oh wait no

x is G

[unreal fiber]

angle G?

[tepid torrent]

yes

[unreal fiber]

wuh

[tepid torrent]

14 is adjacent G is opposite

so wait it'd be

[unreal fiber]

G is an angle though

[tepid torrent]

oh

omg

wait so can you just help me with the whole problem

[unreal fiber]

firstly, calm dowm

[tepid torrent]

okok sorry

[unreal fiber]

hmm

tan(53.2 degrees) is not a very nice value but workable I guess

so what would tan(53.2 degrees) would represent as a ratio of sides on your diagram

[tepid torrent]

well wait

are u sure it's tan

I THINK I JUST NEEDA RESTART THE PROBLEM

[unreal fiber]

I guess cosine works better lol

I just went with your setup

alr

starting fresh

[tepid torrent]

okok yes

[unreal fiber]

GJ would be the hypotenuse of your triangle, yes?

[tepid torrent]

mhm

[unreal fiber]

and the length HJ would be the adjacent length respect to the angle J

so we need the trigonometric function that relates both the hypotenuse and the adjacent side length

which is the cosine function

[tepid torrent]

cosine!

[unreal fiber]

yes!

[tepid torrent]

yay!

[unreal fiber]

so our cosine value of the angle J would be the ratio 14/GJ, right?

[tepid torrent]

mhm

[unreal fiber]

so solving for GJ would get us with

[tepid torrent]

how do I solve for GJ

[unreal fiber]

you have $\cos(53.2\deg) = \frac{14}{|GJ|}$

[flat frigate]

biggboy

[tepid torrent]

yes

can I replace GJ with x?

[unreal fiber]

multiplying both sides by length GJ: $|GJ| \cdot \cos(53.2 \deg) = 14$

[flat frigate]

biggboy

[unreal fiber]

you don't really need to but you can

dividing both sides by cos(53.2 deg): $|GJ| = \frac{14}{\cos(53.2 \deg)}$

[flat frigate]

biggboy

[unreal fiber]

you can just plug that in a calculator and round it the way your question suggests

[tepid torrent]

so 14/cos(53.2) ?

23.4?

[unreal fiber]

yeah

[tepid torrent]

thank u so much!

u explained it so well

.close

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[dark terrace]

Hello im trying to to do numerical analysis.

I did 2 equation called d2_1 and d2_2 where d2_1 is the finite difference approximation of f'(xi) and d2_2 FDA of f''(xi) both is in the second order. I want to study the error of | d2_1 - f'(1) | and | d2_2 - f''(1) | in python with a graph where h reduces where f(x) = e^-x^2 and xi = 1.

[dark terrace]

this is the graph i get

Does this look reasonable?

here is the code

[plucky elk]

shouldn't your error decrease with decreasing step size

can you type out your function and its derivatives in latex

[safe radish]

@dark terrace Has your question been resolved?

[dark terrace]

tbh I don't know

[grizzled stirrup]

theres some weird things computers do with floating point calculations

[dark terrace]

So you mean it has to do something with the code so my calculation are not wrong?

[grizzled stirrup]

im not sure i didnt look at the code

[plucky elk]

you're missing a factor of h/2 multiplying the -2e^(-x^2) term

use $a(b+c) = ab + ac$ with $a=h/2$

[flat frigate]

riemann

[plucky elk]

oh maybe it looks right in the code

[dark terrace]

So based on what you said earlier i rewrote the function d2_1(h) so it look like this now:

now the graph look like this

@plucky elk

[plucky elk]

why is there an x and a 1 here?

[dark terrace]

what do you mean?

[plucky elk]

this is your f''(x). the x in 4x^2 should be the same as the x in the two exponentials

[plucky elk]

also looks like the parentheses don't match. you have -4x^2 multiplying both exponentials

[dark terrace]

but when x = 1 isn't e^-x^2 just e^-1 which is the same 1/e^1

[plucky elk]

that's not your issue

[plucky elk]

read the screenshot more carefully

oh you only pass in x=1

well that's completely inconsistent.

[plucky elk]

either re-write your function to have x or don't at all.

[plucky elk]

fix this and plot again

[dark terrace]

I had x first but when I was trying to debug the code and put 1 to check if there were different but there was no difference then i left it like that

[plucky elk]

yea that's bad

[dark terrace]

I changed the 1:s to x but the graph is the same

[plucky elk]

.

[dark terrace]

Ok i c

[dark terrace]

Now i get this when i fixed the parentheses

[plucky elk]

the blue curve looks like it makes more sense.

[dark terrace]

Ok im gonna check the other function in my code tomorrow tp fix it

[plucky elk]

[dark terrace]

Tnx for help

[plucky elk]

.close

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[vital parrot]

I feel like such an idiot but please help me figure out where I'm going wrong

If p = -4, p squared is -4 x -4 which is 16. 2p would be 2 x -4 which is -8, plus four brings it up to -4. Therefore, 16 minus -4 equals 20 due to a double negative yet somehow, the answer came out to 28. How is that even possible?

[vital parrot]

I'm just going through practice stuff so I'll get similar questions to this til I pass so I just want to figure out how to properly do this shit

[stoic dune]

(-4)² - 2(-4) + 4

Use brackets if you're having trouble

Then that's
(16) - 2(-4) + 4
= 16 + 8 + 4

[vital parrot]

So its not a subtraction at all

Okay, so basically each one of these is its separate formula?

[stoic dune]

I'm choosing to evaluate them separately then combine them after, sure

Any term seem weird? Want to go over any of it?

[vital parrot]

I'm gonna work through another similar problem and see if I get stuck

Give me a minute

I just don't get it

Ahh wait a sec

I misinterpreted the whole thing

It makes more sense when I write it like a sentence

Oh wait my answer is still wrong

[stoic dune]

A few things that catch people with these:

• Squaring a negative number turns it into a positive number

• Subtracting a negative number is the same as adding a positive number

[vital parrot]

I dunno how much more I can break this down. I went r^2 is 9, -2 times -3 is 6, plus four equals 10, therefore its -1

This is driving me nuts

[stoic dune]

Okay, so your terms become 9, 6, 4. That's correct!

Those add to become 19.

[vital parrot]

But wait a second, the root at the end is a positive, so wouldn't it be subtraction?

[stoic dune]

Root at the end? I'm not sure what that refers to

[vital parrot]

Because -2 x -3 is a positive, plus a positive 4 we're left with a positive 10

[stoic dune]

Right! Don't forget about your 9 that came before

[vital parrot]

Yes, so we subtract a positive 10 from a positive 9, we're left with a negative 1, that's what I don't get

[stoic dune]

You used your - when you multiplied -2(-3)

[vital parrot]

Ahh

I keep misreading the problem when its really simple, I guess

[stoic dune]

(-3)² - 2(-3) + 4
9 + 6 + 4

You can also think of it as such:
9 - (-6) + 4

[vital parrot]

So lets work through this again then, so positive 2 times negative -3 is -6 plus +4 is a negative -2

At least that's what I'm coming to

[stoic dune]

So note as well that the - only applies to the term to its immediate right

Your explanation is trying to do:
-(6 + 4)

But you really want to do
-6 + 4

[vital parrot]

So okay, its a negative -2 plus a four giving us 2

So 11 in total?

Wait no

So it'd be 9 minus 11

9 minus 2

But that leads me to my answer of 7

I just don't get it!

[stoic dune]

Not sure where -2 + 4 happens

[vital parrot]

Oh

I messed up

-2 is the end result

Not -2 + 4

So 9 minus -2 is 11

[stoic dune]

Doing this in sentences is mixing you up

[vital parrot]

God I hope this is finally the correct answer

THE ANSWER IS 19?

HOW?

Oh my god I'm gonna have to come back to this later

.close

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[rustic oracle]

Quick Question: I would like to know how much sin cos and tangent is in calculus - higher level math due to my adamant distain towards it due to not learning it well enough.

[final halo]

yeah you're really gonna wanna learn to like it more

[rustic oracle]

f my life

i guess

thanks

.close

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[fiery socket]

∀a∈Z, a=kq+r, k∈z+ ∧ 0<r<q

[fiery socket]

Does this notation make sense? I'm trying to say "For all/every integer a, a can be rewritten as kq+r, where k is a positive integer and r is between 0 and q.

If it doesn't, what would be the correct version?

[plucky elk]

yes it makes sense

[fiery socket]

ok ty! It's my first time learning math notations and I'm just practicing with random theorems

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[iron bolt]

how do i start this?

[stray socket]

Start with the sketch

,tex \dintarea

[flat frigate]

Umbraleviathan

[safe radish]

@iron bolt Has your question been resolved?

[safe radish]

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[frozen harbor]

Can someone explain what this means

Ive been staring at this and i just have no idea what im looking at

In part b

[worthy hemlock]

Do you know what f(3) means?

[frozen harbor]

You plug in 3 into f(x)

[worthy hemlock]

$f(f^{-1}(x))$ is the same logic

[flat frigate]

dldh06

[worthy hemlock]

Take $f^{-1}(x)$ plug it into f(x)

[flat frigate]

dldh06

[frozen harbor]

So I can plug in x-9 / 2 into 2x + 9

So 2(x-9 / 2) + 9

[worthy hemlock]

Yes, and the reason for the many options, it wants you to select the one that is in that specific form that you just gave

[worthy hemlock]

Like this

[frozen harbor]

Oh

It'll be the first one

[worthy hemlock]

Yes

[frozen harbor]

Now it's asking me the same thing but plug the other one into the other other one

Is there an actual use for this or is it just asking me things

[worthy hemlock]

There is

[worthy hemlock]

Notice how if you simplified choice a, after plugging that in, it results in x

[frozen harbor]

What does that mean then

If it gives me x

[worthy hemlock]

It means that the f^-1(x) and f(x) are inverses

Like say if I gave you f(x) = x + 2 and g(x) = 3x, and I asked to prove if those were inverses of each other. If you did f(g(x)) and it resulted x and did g(f(x)) and that also resulted in x then you know that f(x) and g(x) are inverses

[worthy hemlock]

That example, I just made up two random functions that weren't inverses but the idea is still the same if you were given something more complex

[frozen harbor]

That makes sense

Plugging them into each other and getting x as both results will tell me that they are inverses of each other

[worthy hemlock]

Yep exactly

[frozen harbor]

Thank you so much

[worthy hemlock]

.close if you are done with the channel

[frozen harbor]

.close

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[tardy copper]

how would I do a problem like this lol

.close

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[graceful urchin]

Hi I need help finding the velocity from a limit at t seconds

[graceful urchin]

im given the limit as x approaches s(a)-s(t)/a-t

s(t)= -4.9t^2+300

[stray socket]

x appaorches that fraction?

For what?

[graceful urchin]

for the velocity

[stray socket]

x approaches a?

[graceful urchin]

s(t) is the position equation

a=7.825

which is the time

[stray socket]

Just compute s'(a)

[graceful urchin]

then I get something wrong

[stray socket]

I wouldn't bother using the limit definition for derivatives unless you're being forced to

[graceful urchin]

cus i checked it by taking the derivative

I have to cus im in ab calc

[stray socket]

It should be the same

[graceful urchin]

we havent learned it yet technically

i end up getting 300 for some reason

do you think if I just wrote the right answer the teacher would ask for my work

cus i wrote the first part

[stray socket]

$\limit{\frac{(-4.9a^2 + 300) - (-4.9t^2 + 300)}{a-t}}ta$

[flat frigate]

Umbraleviathan

[stray socket]

This isn't so bad ngl

Just simplify

And then factor a-t from the numerator

[graceful urchin]

so i get 300.027-4.9^2+300

over a-t

over 7.825-t

?

ok thanks I got it

.close

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[lapis shadow]

try joining E & A, O & A, O & B

[faint saffron]

Okay i will try it now

[wintry condor]

Use this

[cunning cloak]

btw, if an arc is 80 deg, then isnt the angle on the opposite side equal to exactly half of that?

so would angle ADC be 40 deg?

[wintry condor]

That's only true if the radius of the circle is 1, isn't it?

[wintry condor]

Also don't you mean ADB?

[faint saffron]

I did a bit different

But i still dont see anything

[lapis shadow]

im thinking 80deg = AOB?

[wintry condor]

I dont think that will work

[lapis shadow]

what abt angle EAD

[faint saffron]

I will try this

[lapis shadow]

that and ECD are the same

[faint saffron]

Ead is 60?

Wait

30

[lapis shadow]

EAD = 30

then u have a triangle

[faint saffron]

Okay i will do this

[wintry condor]

There's a lot of ways to do this ig

But this looks simpler

[lapis shadow]

yes another way of doing it

[faint saffron]

Is this correct?

[lapis shadow]

yes

[faint saffron]

Ooo i could also do thisss

[lapis shadow]

but u cud have just said 30+40 = x

[wintry condor]

Both are good ways 👍

[lapis shadow]

coz exterior angle = sum of interior opp angles

[wintry condor]

You did a good job here

[faint saffron]

Yepp, thank you silversoldier and jay!!!!

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[bold ferry]

what is a?

[light shoal]

well what does the mean value theorem say?

[bold ferry]

f'(c) = f(2) - f(0) over 2-0

wait...

is it a not 0

ohhhhhh

[light shoal]

yea should be a

[bold ferry]

it's just saying that whole thing is > 0

I see

[light shoal]

yea that will help because you'll probably have multiple answers and you want the positive one

so $$f'(c) = \frac{f(2) - f(a)}{2 - a}$$

[flat frigate]

Bungo

[light shoal]

you know c, so you can calculate f'(c)

and you know a formula for f(a)

try plugging all of that in

see if you can solve for a

[bold ferry]

I see

I got 1.424

correct?

okay yeah i have the key

ty for the help

.close

[safe radish]

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[fickle trail]

wtf

[hardy lion]

I love that i can see the original problem

[dull sequoia]

Me too

[dull sequoia]

What about it

[fickle trail]

Look at the answer for range

[dull sequoia]

It looks like we are trying to find domain and range of this expression?

[fickle trail]

As per the verified account https://quizlet.com/explanations/textbook-solutions/thomas-calculus-early-transcendentals-14th-edition-9780134439020

Meanwhile...

[dull sequoia]

How do you even get -inf

t² has range >= 0

[fickle trail]

very small negative value

[dull sequoia]

Oh

Close to 16

Ah yeah forgot about that

[fickle trail]

limit towards 4 for t, or limit towards Infinity?

[dull sequoia]

Yeah so the range is everything but 0

[dull sequoia]

4

Something close to 4 will be very big or very negative

[fickle trail]

And a huge number for t would be positive, right

[dull sequoia]

Would be close to 0

Make t really big

[fickle trail]

right

[dull sequoia]

t² -16 is really big

[fickle trail]

So anyways

[dull sequoia]

2/really big = close to 0

[fickle trail]

Which is the correct answer?

for Range

[dull sequoia]

Think the first one

[fickle trail]

because this textbook f'ing stinks

[trim swan]

No, (-inf, -1/8] U (0, inf) is correct

[dull sequoia]

Really

[radiant crypt]

looking at it in GeoGebra, the second is correct

[fickle trail]

Meanwhile, their verified Quizlet page is full of crap

[trim swan]

The denominator has a minimum value of -16

[dull sequoia]

Oh

Yeah sorry

[fickle trail]

I still don't see it either

I would have went with the Quizlet answer

[trim swan]

So, the function cannot equal -1/16, for example, since that is equivalent to 2/(-32), but the denominator cannot ever have a value of -32

[trim swan]

This explanation is right, if a little bit hard to decipher

[dull sequoia]

Ok sorry my break is over sorry for confusing you :c

[trim swan]

@fickle trail Consider the denominator alone: t^2-16

Do you agree that it cannot be less than -16, for any value of t?

[fickle trail]

yes, denominator cannot be 0

that was discovered in the domain

[trim swan]

What I mean is, there is no value of t such that t^2-16 < -16

[fickle trail]

what if t is 0?

0 - 16 = -16

[trim swan]

Right

That's the minimum value of the denominator

There is no t such that t^2 - 16 < -16

[fickle trail]

oh I see what you are saying

OK yeah

I agree

[trim swan]

And that minimum occurs at t=0, like you said

[fickle trail]

right

only getting 0 or positives from an even power

[trim swan]

Now, in the domain, you specified that t cannot be -4 or 4

[fickle trail]

Right

[trim swan]

To be clear, t CAN equal 0 (0 is in the domiain), but t=0 is a special point, because it gives the minimum value for our denominator

So, it might help to look at four separate intervals:

-inf to -4
-4 to 0
0 to 4
4 to inf

Does this make sense so far?

[fickle trail]

Yes

[trim swan]

So consider the first interval -inf to -4

How does the function behave on this interval?

What does it do as t approaches -inf
and what does it do as t approaches -4 from the left?

[fickle trail]

-Inf would make the denominator a huge number

towards -4 from the left would make the denominator close to 0

[trim swan]

Right, so what about the entire function? (not just the denominator)

[fickle trail]

close to 0?

[trim swan]

Yep, but still positive, right?

2/(big positive number) = small positive number

[fickle trail]

yes, for -Inf

[trim swan]

yep

[fickle trail]

for -4 lemme think

[trim swan]

-4 from the left specifically

[fickle trail]

yeah

well, if we end up getting 2/0.0001

that would be a big number

[trim swan]

yep

in fact, it's approaching infinity

[fickle trail]

right

so approaches 0 and approaches infinity

[trim swan]

So on that interval, our range is (0, inf)

right

[fickle trail]

yes

[trim swan]

perfect

The right-most interval is similar

t from 4 to inf

[fickle trail]

And I would just stop there with my answer lol

what else am I supposed to test?

[trim swan]

So far, we only considered t between -inf and -4

But, now consider the interval from 4 to inf

I think you can convince yourself it has basically the same behavior

[fickle trail]

so really, I could just test the first 2, or the last 2

no need to test all 4?

[trim swan]

hm

In this case, that's true, but not all functions are so symmetrical

[fickle trail]

and it's symmetrical because of the leading term power?

x^2

if it were odd I would want to test all intervals maybe

or understand when the intervals change due to multiplicity

[trim swan]

This is a good strategy

[fickle trail]

OK

[trim swan]

If you had both even and odd degree terms in the denominator, it would be much less predictable

[fickle trail]

so if I know this interval is true, -4 to 0 must be false

[trim swan]

not sure what you mean by true/false

[fickle trail]

yeah

not sure either

lets continue

-4 from the right

[trim swan]

okay

[fickle trail]

approaching -4 from the right would be a very small denominator

we can say approaches 0

[trim swan]

the denominator

So what about the function as a whole?

[fickle trail]

very big positive number

approaches positive infinity

[fickle trail]

for this

[trim swan]

Actually, if we're approaching -4 from the right, the denominator will be a very small negative number

[fickle trail]

oh right,

we are counting down back to 0 on the number line, for the negatives, when approaching from the right side

keep forgetting this

[trim swan]

(-3.999)^2 - 16 < 0, for example

[fickle trail]

yeah

[trim swan]

So this will approach -inf, not +inf

[fickle trail]

I was imagining 4.00001 because it's to the right of 4, and x^2 is positive.. but we havent squared yet

[trim swan]

right, exactly

[fickle trail]

alright

[trim swan]

So then, what about as x goes to 0?

from the left

[fickle trail]

close to -16

[trim swan]

so tell me about the function as a whole

[fickle trail]

2/-16

-1/8

but it includes -1/8

why?

[trim swan]

Because t=0 is in the domain

t=-4 was not

2/(t^2-16) is perfectly well-defined for t=0

We plug in 0, we get -1/8

Nothing weird

[fickle trail]

so basically to find all range you just need to test the limits on all sides of the domain intervals

left and right sides, if domain permits it

like if domain says (-Inf, 1) you could test from the left side of 1... 0.999999

but no point in testing from the right side of 1 ... 1.000000001

it's not included in the domain to begin with

[trim swan]

Yeah, I mean

If you test a value that's not in the domain, you should find that the function is undefined there

Otherwise, it would be in the domain

[fickle trail]

interesting

OK

so domain is EZ to find

range takes a bit more time

[trim swan]

Yes

[fickle trail]

and i guess rational functions can be the trickiest for range?

need to be careful you dont skip over domain intervals

but finding domain first seems like a good strategy for finding range

and understanding limits

[trim swan]

As far as the kinds of functions you'll study in a course like this, yeah

[fickle trail]

OK

[trim swan]

You can make all kinds of weird functions that do all kinds of weird things though lol

[fickle trail]

question about range.. if domain is (-Inf, Inf) is it guaranteed range will be the same? regardless of the function?

[trim swan]

No, definitely not

For example f(x) = x^2

[fickle trail]

OK lol

so generally domain has more numbers in a set than range does

generally speaking

or is it like comparing apples with oranges

[trim swan]

"""""yes"""""

[fickle trail]

can go either way

[fickle trail]

yes for which? haha

[trim swan]

this

[fickle trail]

ty

[trim swan]

Your intuition is correct, but when speaking of the size of infinite sets, things get a little weird

[fickle trail]

https://tenor.com/view/thanks-thank-you-gif-20870035

I think I'm finally starting to get it

[trim swan]

🎉

Awesome 👍

[fickle trail]

.close

[safe radish]

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[safe radish]

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[lean otter]

i have a big problem

[lean otter]

ok

draw a diagram

.

wait i got this far

to save time for u

k

k

seems to me u hav found the solution ; where is the workings

[lean otter]

.

depending if i put it in degree or rad mode

i get different answers

put degre

so here i wnt it in degree since 36 is degre??/

....

ye

when do i want rad mode

if the angle r in radians

oh ok ty

ty for helping me

k

.close

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[brittle prairie]

I'm not quite sure if I have done part a correct

[dull sequoia]

,w solve 0=(1/20)*(18x-3x^2)

[flat frigate]

Results provided by WolframAlpha

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

Requested by Frosst#1111

[dull sequoia]

Yeah part a is right

[brittle prairie]

Okay, could you help me on part b ii

[safe radish]

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[lean otter]

can someone explain

how would i get the numbers of theta and beta

[vocal portal]

trigonometric identity probably

[lean otter]

oh ok let me check

[vocal portal]

more like angle change formulas

pi/2 is 90degree

so theta = 90degree - beta

so sec theta - csc 90 - theta would be:...

0

cot(theta)/tan(90deg-theta) = 1

and last again 0

[lean otter]

i dont really get it sorry

[vocal portal]

ok

FCK

I dunno how it is in english

ok

[vocal portal]

reduction formulas

you got them?

[lean otter]

no i dnot think so

i have trig identitiy i think

[vocal portal]

https://trigonometrywithereza.wordpress.com/reduction/

below table you got them

so for example

[lean otter]

yea

i have that

[vocal portal]

cos(90deg+x) = -sin(x)

[lean otter]

all students take calculus

is what the teacher said

[hexed cedar]

You can get expressions for theta and beta using the first expression they gave

[hexed cedar]

so for example if you want beta make beta the subject

[lean otter]

??

[hexed cedar]

okay so a simple way to understand this is

[lean otter]

theta = frac

• b

?

[hexed cedar]

do you know the answer to Sin(pi/2 - x) @lean otter

[lean otter]

yea with calculator

[hexed cedar]

No without calc

[lean otter]

no i dont then

[hexed cedar]

no problem it’s really simple don’t worry

[hexed cedar]

so when ever you get something like pi/2 - something or p/2 + something for the trigonometric ratio. remember that the sign will change

sin will change to cos
so
Sin(pi/2-x) = cos x

cos(pi/2-10) = sin10

and so on

[lean otter]

oh ok i see now

[hexed cedar]

but remember to see which quadrant the angle is in

[hexed cedar]

@lean otter can you tell me which quadrant this in ?

[lean otter]

idk if its 0

then no quad between 1 and 4

?

[hexed cedar]

pi/2 is basically 90

[lean otter]

sorry

[hexed cedar]

so 90-10 is in ?

[hexed cedar]

you can use this diagram for reference

[lean otter]

1

[hexed cedar]

yes

and that’s why it’s positive

so if it was something like Cos(pi-10) the answer will be -Cos10 (here the cos doesn’t change to sin because it’s pi not pi/2 minus or plus something)

And the minus is because in the second quadrant only sin is positive

[lean otter]

oh ok

[hexed cedar]

if you didn’t understand let me know. it’s alright

[lean otter]

let me see

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

i dont understand sorry

i dont think any of that is in the homework or textbook so far

.close

[safe radish]

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[stray axle]

just a quick question but i was wondering, why dont we also make a vector QR?

[zinc token]

you can too

equation of plane is not unique

[stray axle]

but we need only 2 vectors to make a vector form equation?

is that how it works

[zinc token]

yes

but ofc they have to be not parallel

[stray axle]

alright thats cool, so is it in general you need 1 vector less than the dimension? so for (x,y) you need only 1 vector and (x,y,z) you need 2 and (x,y,z,a) you need 3?

[safe radish]

@stray axle Has your question been resolved?

[zinc token]

hmm I can't speak to that

for the x, y case yeah

[stray axle]

.close

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[rapid lantern]

im trying to understand how that became (1/cosx) * (1/sinx)

[thin bridge]

did you understand the first step?

[rapid lantern]

yes

common denominator

[thin bridge]

do you know what the numerator:
$$\sin^2(x) + \cos^2(x)$$
simplifies to?

[flat frigate]

ℝamonov

[rapid lantern]

umm

no

i would assume it's 1?

[thin bridge]

you shouldn't be assuming that

it is unequivocally 1

it's the pythagorean trig identity

[rapid lantern]

so the x is 45 degrees?

i can understand now thinking that its 45 degrees

oh nvm

it works for any angle

now that i think about it

not just 45

ooooooooh

thank you so much

.close

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[pliant badger]

Consider (O; e1 ;e2) , the vectors u(-1;0) and v = 2e1 + e2

The following propositions can be classified as:

[pliant badger]

A) both false
B) p is false and q is true
C) q is false and p is true
D) both true

I been struggling with understanding the q one

I can easily understand how p works

not the q one

and what'd make sense for me would be p being true and q false

but would like an explanation if possible as I'm trynna learn 😄

[final halo]

Have you tried actually calculating the things?

[pliant badger]

yeah I did try but I'm losing myself in the "q" one

because i don't know if I can convert the u + v into their norms like that

[final halo]

What do you mean convert u+v into their norms

[pliant badger]

|u| = square root (c-a)^2 + (d-b)^2

for example

[final halo]

What are a,b,c,d?

[pliant badger]

the coordinates of the point

[final halo]

These vectors have two coordinates so you only need two letters

Anyway you should explicitly find the vector 2(u+v) and then compute its norm

Then you should compute the norm of u+v and then times it by two

And see if you get the same result

[pliant badger]

alright I'll try that

[safe radish]

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[weary grotto]

sorry let me re add the images i missed up the extension

[safe radish]

@weary grotto Has your question been resolved?

[weary grotto]

<@&286206848099549185>

[plucky elk]

#old-network for CS server

some formulas here at the bottom
https://en.wikipedia.org/wiki/B-tree

[weary grotto]

i will try my luck there thank you

.close

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[upbeat swan]

Do you know how to convert units?

[worldly bay]

no

[upbeat swan]

Then why are you doing this exercise in the first place

1km²=1,000,000m²

Use this

[worldly bay]

ah ok

[safe radish]

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[vital wing]

I am lost in an exercise about similar triangles

[vital wing]

I have to justify if they are similar triangles or not

I was able to do a) and b) but I'm really out of ideas in the c)

i don't know where to even begin

[safe radish]

@vital wing Has your question been resolved?

[hasty wagon]

for c) they should be guiding you to use
SAS

try to find if the ratios are the same

like,is
||9/4.5 = 7/3||?

[vital wing]

yeah i tried that but

even if 9/4.5 doesnt equal 7/3 it doesnt really justify it I think

because the homological side of 9 or 7 could be the side which value I don't know

I don't know if homological is the correct name but it's called like that in spanish

homologous*

that's the word

[hasty wagon]

good reasoning

and you need to check also if
9/3 = 7/4.5

if both are not equal, then it's good

or we would say we cannot prove they are similar

[vital wing]

yes i think so too

[hasty wagon]

although, for later on , when we learn more about sine cosine, we can prove it's impossible to be similar

so, i think that should be it

[safe radish]

@vital wing Has your question been resolved?

[safe radish]

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[pseudo sentinel]

Hi, I need help with this:

[pseudo sentinel]

Write the equation of the line passing through the origin O and tangent to the curve of equation y = -ln 2x

The problem must be solved with derivatives and the result is
y = (-2/e)x

[safe radish]

@pseudo sentinel Has your question been resolved?

[pseudo sentinel]

<@&286206848099549185>

[tulip saddle]

ok so

what do we know about the line?

@pseudo sentinel

specifically, what two points do we know it must go through? hint: one of them has no constants, and isn't really a specific point that we know of rn, but both points are necessarily traveled through as stated by the problem

[pseudo sentinel]

I called the tangent line t and solved the system; the result comes out but I didn't understand the relationship between m and x

in a few words, we know two pieces of information about m (i.e. the relations placed in the system) and by uniting them we obtain the abscissa x of the tangent point, subsequently substituting the value of x in the derivative function we obtain the angular coefficient of the tangent

[tulip saddle]

huh this is a much better way of solving that I was going to do

though I'm not sure how you did it

[pseudo sentinel]

I had to think about it for quite some time but finally I was able to do it on my own

[tulip saddle]

good job!

[pseudo sentinel]

thanks anyway!

[marble imp]

I believe u did somewhat same… cool

[safe radish]

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[safe radish]

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[steady loom]

Can someone help me with this problem?

I dont understand how to find the solution to this ODE

I think its a first order linear ODE

so I should solve by method of integration factor, but I get stuck when trying to deal with h(t)

[marsh walrus]

try splitting it?

find a solution valid for each piece of h as a domain

then check that you are able to join the halfs

[steady loom]

splitting it?
As in solve for y' + y = 0 and y' + y = 1

[marsh walrus]

yea

[steady loom]

how does putting it back together work

wouldn't I end up with 2 different solutions

[safe radish]

@steady loom Has your question been resolved?

[steady loom]

@marsh walrus these are the solutions I got

what did you mean by does it make sense to put it back together?

<@&286206848099549185>

[marsh walrus]

sorry, ill be home in a bit

[steady loom]

oh ok nw appreciate it

[steady loom]

also it has a hint to solve it as if it wasn't discontinuous

which is why I didnt try separating it at first

[safe radish]

@steady loom Has your question been resolved?

[safe radish]

@steady loom Has your question been resolved?

[safe radish]

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[lean otter]

I do not understand how to get the final answer once cos is solved

[lean otter]

uhm

<@&268886789983436800>

[harsh shoal]

ty

[lean otter]

<@&286206848099549185>

[hybrid wedge]

Do you know the general solution of $\cos{x}=\cos{α}$

[flat frigate]

Anagh

[light shoal]

your reasoning in the screenshot seems fine, what is the issue?

[hybrid wedge]

@lean otter

[lean otter]

its a worked out question i just do not understand what is happeing after gettin cosx = 1/2 and cos x = 1

[light shoal]

well what angle x has cos x = 1?

there's only one such angle in the interval [0,2pi)

[lean otter]

i dont know what that is

[light shoal]

i'm not sure how you can solve this problem if you don't know the cosine function...

[hybrid wedge]

Bro first you need to learn trigo

[light shoal]

probably you should take a step back and review it

[hybrid wedge]

How to get a general solution

[lean otter]

so what should i look up

[hybrid wedge]

Then how to get solution when θ lies in some intervals

[hybrid wedge]

How to solve Trigonometric eqns

[lean otter]

ok ty

[safe radish]

@lean otter Has your question been resolved?

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[worn lark]

how do you derive the cartesian coordinates of an icosahedron? like this \
$(\pm1,\pm\varphi,0), (0,\pm1,\pm\varphi), (\pm\varphi,0,\pm1)$

[flat frigate]

🤓

[worn lark]

dodecahedron also

[worn lark]

*the vertices

[wooden ivy]

is this a real question of a goofy ahh question

what course is this

[worn lark]

not an assignment

[wooden ivy]

[worn lark]

just me wondering how are these coordinates derived

[safe radish]

@worn lark Has your question been resolved?

[safe radish]

@worn lark Has your question been resolved?

[safe radish]

@worn lark Has your question been resolved?

[rare folio]

HAHAHAH

[rare folio]

Lmao

[rare folio]

🤓

[worn lark]

<@&286206848099549185>

[lean otter]

[worn lark]

this is a serious question

[safe radish]

@worn lark Has your question been resolved?

[woven swan]

i dont get why everyone is laughing at this

what do you mean by derive @worn lark

the coords u gave dont give u a regular icosahedron anyway

[worn lark]

it does

according to wikipedia

theyre the vertices

[woven swan]

o ye u might be right

actually ye

what do u want to do tho

[worn lark]

how do you find these coordinates

im not asking for proof

[woven swan]

hmm okay

[worn lark]

like, you cant just give me the formula of the gamma function and expect me to know how did you find it

i need to know what method did the first person that calculated these coordinates use

[misty bay]

hmm well it depends on how you define an icosahedron

but like if you just draw out the shape and say its angles are something

it's in general possible to compute arbitrary rotations of arbitrary vectors

and then you can just calculate the points one by one

obviously this is not very elegant or satisfying I guess

[toxic stratus]

if you take two points of the icosahedron to be placed at the poles of a sphere, you can compute the co-latitudinal angle using spherical coordinates

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[worn lark]

.close

[safe radish]

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[brave narwhal]

hi guys, does anybody know how to get the inverse function of this function with normal procedure (without drawing)?

[lapis shadow]

what do u mean drawing

graphing?

[brave narwhal]

yup

[lapis shadow]

okay u can set y = 2x/(x^2 + 1)

and solve for x

[brave narwhal]

yess

but there's 2

[lapis shadow]

so it has no inverse

[brave narwhal]

seriously???

it's pretty similar to sine or cos, and i believe they do have inverse

[lapis shadow]

it has an inverse if u restrict the codomain

[lapis shadow]

sines and cosines are also only invertible once u restrict the codomains

[brave narwhal]

codomain is the y that comes after any x is inserted right?

[wintry condor]

This isn't a bijective function, so it has no inverse

[lapis shadow]

no, the values of y that actually do come out when u insert values of x are called the range

[lapis shadow]

the bigger set that all these values of y reside in, is called the codomain

[brave narwhal]

oh yeah, understood

[lapis shadow]

when u write $f : A\to B$

[flat frigate]

SilverSoldier

[lapis shadow]

A is the domain, B is the codomain

[brave narwhal]

yep the function is not bijective, cuz its not injective

[brave narwhal]

yep thanks

[lapis shadow]

neither is it surjective

[brave narwhal]

why?

[lapis shadow]

not everything in the codomain has a preimage

[brave narwhal]

but its the case if the codomaim is all Real numbers right?

if we restricht the codomain it can be surjectibe right?

[lapis shadow]

but no this is never equal to something like 2

[lapis shadow]

yeah if we restrict the codomain to [-1, 1]

is it surjective, but not injective

[brave narwhal]

yeah not injective cuz at some points y has 2 x values

[lapis shadow]

if we further restrict the codomain to like [0, 1], then it is both surjective and injective and it will have an inverse

[lapis shadow]

yeah