help-39

[merry python]

yeah, and if the curve is continuous, it must cross the x-axis at some point

between -1 and 1

its called the intermediate value theorem, or IVT for short, which Denascite said

[amber basin]

the graph is shown like this

that means uh, f(-1) < x < f(1)?

and that f(c) = x

[merry python]

suppose we have the function f(x) = x, because f(-1) < 0 and f(1) > 0 it must cross the x-axis, and there must be a solution to f(x) = 0 for -1 < x < 1

[amber basin]

ahh i see

i do have another question tho

find a and b, that f(x) has a derivative on every real element

[merry python]

that means that you have to find a and b such that both x^3 - 5 and ax^2 + bx have the same value and derivative at x = 1

if they don't have the same value, the derivative wouldn't be defined, and if they don't have the same derivative the derivative also wouldn't be defined at x = 1

[amber basin]

so i should derive both ax^2 + bx and x^3 - 5?

[merry python]

yes

[amber basin]

then it should be
2ax + b
3x^2

what next

[merry python]

pretty sure the verb is differentiating btw, but it doesn't matter

[merry python]

correct

set them equal

and also set ax^2 + bx equal to x^3 - 5

[amber basin]

2ax + b = 3x^2?

[merry python]

ye

for x = 1 btw

[amber basin]

why it should be 1 tho

[amber basin]

a + b = -4

[amber basin]

2a + b = 3

[merry python]

because that's the "splitting point" in the function definition

[amber basin]

ah so, a = 7

[merry python]

yup

and b?

[amber basin]

and that b = -11

[merry python]

that's correct

[amber basin]

i still dont get this tho

[merry python]

well the function has one definition for x <= 1 and another definition for x > 1, one is thus the point at which the definition "switches"

[amber basin]

what if we change x<=5 and x >2 for both function, what's the splitting point then

[merry python]

then the function isn't defined for 5 < x <= 2

[amber basin]

hence we could use 3 or 4 to be the point?

[merry python]

this is what it looks like in desmos

[merry python]

there wouldn't be any values of a and b for which the derivative is defined for every real element

because there is no point but a range

[amber basin]

owh right

[amber basin]

i see what you mean by splitting point

[merry python]

nice

[placid thicket]

Hi guys, can u give me an advice about which is the best program to do graphs and stuff?

[placid thicket]

programs like this

[amber basin]

say that x <= 3 and x > 6, then the function isnt defined for x>3 and x<=6?

[cursive fulcrum]

That’s desmos

[placid thicket]

k thank u

[merry python]

yeah

[amber basin]

3 < x < =6

[merry python]

yes

[amber basin]

so, uhm, will there be any values for a and b?

[merry python]

no, because the derivative at x = 4, for example, will never be defined no matter what a and b you choose

4 isn't part of the domain of the function

[amber basin]

where did u get x=4 from?

[merry python]

its just a random number between 3 and 6

5 would've worked as well

or 4.238382

or 6

[amber basin]

soo, there will be no splitting point to be inserted into the f(x) function?

[merry python]

note that splitting point isn't a formal name for it

[amber basin]

ehh, what is then

[merry python]

I don't know if there is a formal name for it

You could ask your teacher

[amber basin]

my brain

thanks for the help tho! @merry python , ill try to process everything said here

[merry python]

okay

.close

[amber basin]

.close

[pearl pond]

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[amber basin]

gotchu

[pearl pond]

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[frigid oriole]

So I'm trying to wrap my mind around this Trig question and I'm not entirely sure if I'm going about it right.
I know its in the 1st quadrant so my values will end up positive and I attempted to draw a triangle with cos(x) =1/4. with that I pulled my adjacent and hypotenuse length to enter into cos(x/2) and made cos(1/4/2) which is cos(1/8) I think . Im stuck now thinking im going in the wrong direction. Do I make another circle cos(1/8)? idk

[clear plank]

cos(2x)=cos²x-sin²x=2cos²x-1

cos(x)=2cos²(x/2)-1

cos²(x/2)=(1+cosx)/2

[frigid oriole]

Ty I have those from my professors lecture notes but I'm confused as to how to apply them here

[clear plank]

cos(a+b)=cosacosb-sinasinb

every other formula derives from this one

[thorn crag]

hey can someone help with precalc?

[frigid oriole]

check out the how to get help channel this one is currently occupied

[thorn crag]

oh alr sorry

[frigid oriole]

np gl

[frigid oriole]

was I on the right path with my triangle drawing and assuming cos(1/8) was what I needed?

I'd show you a pic but my handwriting is garbage

[clear plank]

nah

triangle cant do much here

[frigid oriole]

darn Im still grasping at trig concepts so im resorting to other methods that have worked on other questions

[clear plank]

you can try

but you will see that you cant use it

[frigid oriole]

im trying not to cross a line and ask you too much but if I can where do I get started? Which formula do I use and how do I input my information?

again sorry if thats asking too much

[clear plank]

this

[frigid oriole]

seems obvious to use the half angle but

oh

hm

[clear plank]

just change the a,bs

for example cos(a+a)=cosacosa-sinasina

[frigid oriole]

thats very confusing to me. I just used something similar when I had to use that formula but it was two triangles

oh

[clear plank]

which is cos(2a)=cos²a-sin²a

[frigid oriole]

Im new to this channel not sure how to report this

referring to the troll

[lapis mango]

This channel is occupied

[frigid oriole]

im going over what you've written btw..

[frigid oriole]

this dude tryna get banned or something

[lapis mango]

yes

<@&268886789983436800> This person @bold ravine is spamming

[clear plank]

🥶

[frigid oriole]

i blocked him and it deleted all his stuff nice

[opal lantern]

ty

[frigid oriole]

is there a way to keep this up and not let it time out while I try and work this out?

[lapis mango]

just respond no when the bot pings you

[frigid oriole]

ok ty

[lapis mango]

if the channel accidentally gets closed, you can always use . reopen

[clear plank]

the way to get this formula

[frigid oriole]

man you drew that for me? you sweetheart

ima stare at this for a min

[clear plank]

but you dont really wanna draw this every single time to remember the formula

so thats why ppl are telling you to memorize it

[frigid oriole]

lol no but the visual helps

just curious you a teacher/student?

[clear plank]

the distance between the two points = cosine rule

im a student

[frigid oriole]

I get what you've given me but am unsure how to put my information into this formula and derive a start

[clear plank]

you mean to get the answer?

or to get the formula

[frigid oriole]

lol well the answer is hte goal but Im trying to understand how to put my formula together with the question. I'm dangerously close to asking you for too much help

Like i'd like the answer but I want to know how and why we got there. My math teacher friend was helping me earlier but he had to run. Wolframalpha is nice cause I can get some context as I work toward the answer but this one is not possible to enter into wolframalpha. At least I think I cant

[clear plank]

this is probably all of the formulas i think

but theres too many things to memorize

[lapis mango]

Weird. I always visualise the unit circle in my head to recall all my formulas

[clear plank]

if youre just trying to memorize all of it

but if you think about it you can just change a,b to get them

using cos(pi/2-a)=sin(a)

you can change cosine into sine

and you can put -b instead of b

and tangent is just sine/cosine

so theres just really 1 formula to memorize which is cos(a+b)

i mean its your choice

so i cant really say anything

but im just telling you how i memorized them

[frigid oriole]

I appreciate it I'll def take that to heart.

[clear plank]

let me know if you cant solve the question

[frigid oriole]

oh ty my girl let me sign into her Chegg which I didnt know was a thing and I found a question thats almost exactly the same so Im about to work through it

I've been on this for like 40 minutes now and am basically wasting time but I want to get it. The exams my professor gives are timed so I cant look things up on the fly so I really gotta get the concepts down

here lemme show you this similar question i found on chegg

ignore 1

oh lemme show the question lol my b

2. Given that cos 𝑥 = − 2/3 , and 𝑥 is in quadrant III, find the exact value of cos 𝑥/2

almost identical to mine

[frigid oriole]

yeah I basically copy my half angle formula my professor gave me and work it down. Not having the square on it through me for a loop

this dude threw the whole right side into a square root

[frigid oriole]

is there a way to write formulas the same way coding discords allow for special typing?

my answer

i just struggled with understanding how he simplfied a couple things but thats just me sucking at basic stuff.

[frigid oriole]

thanks for the help pimp im off to watch the dolphins

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[smoky ingot]

hello, I have to find out what the limit is

[lapis mango]

Multiply and divide by (x+2) for a start

[smoky ingot]

one second

you mean like this

[midnight haven]

so

[midnight haven]

add 3 and remove 3

[smoky ingot]

you mean like this

[midnight haven]

no

hold on

@smoky ingot

this is the method

[smoky ingot]

aight

mind if I ask what's the point of those paranthesis

[midnight haven]

am just lazy to rewrite it

[smoky ingot]

so basically you applied that

and the square root disappeared

cuz of the formula

[midnight haven]

yes

is it clear now?

now i need someone to help me with my limit lo

l

[smoky ingot]

ye thanks

.close

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[patent yew]

can someone help me finding derivative

[patent yew]

i did chain rule

but then i can't figure out how to simplify it for dy/dx

[midnight haven]

drrivative of?

[patent yew]

the problem above

implicit differentiation

[plain blade]

Wrt what ?:

[patent yew]

d/dx

[plain blade]

Show ur progress

[patent yew]

after this

i cant figure out how to simplify

[light helm]

expand, separate terms with/without dy/dx

[patent yew]

but how

u get a fraction

[light helm]

so?

what's wrong with fractions

[plain blade]

Bring all the dy/dx to one side

[light helm]

the main focus is isolating dy/dx
and if there are fractions, so be it

do simplification to make it look nicer afterwards if possible

[patent yew]

ik that but how do u expand (x^2 dy/dx +2xy)(2√xy)

without making it

[light helm]

basic algebra/distribution

the product of p and q can be expressed as pq or (p)(q)
you don't have to explicitly perform excessive exponent/radical laws

this is not that different to solving
a(b + ck) = d + ek
for k

[patent yew]

so would

be acceptable

[plain blade]

Yep

[patent yew]

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[tough ivy]

How do I go about simplifying this?

[hollow cobalt]

According to the distributive property, this can be expanded into (a + 2) * b + (a + 2) * 2

Then you need to apply it again

[midnight haven]

ab+2a+2b+4

[tough ivy]

I've never done a question like this before, or used the distributive property in a way like this... so i'm a bit lost

[lapis mango]

would have have any problem with applying the distributive property to x(b+2)?

[tough ivy]

I would just multiply x with everything in the bracket

So xb + 2x

[lapis mango]

Good. Now treat (a+2) like one term. Same as how you did for x

[tough ivy]

I get something completely wrong..

[lapis mango]

What did you get?

[tough ivy]

Oh my

Thank you haha

I get ab+2b+2a+4

[lapis mango]

That's great!

[tough ivy]

Thank you

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[primal bluff]

can someone explain to me this linear algebra theorm: For each b in R^m, the equation Ax=b has a solution.

[sharp quest]

what about it?

[primal bluff]

I dont get it

I dont get the For each b in R^m

[sharp quest]

pick any vector in R^m, call it b

there is an x vector such that Ax = b

[primal bluff]

but how is that a theorm

theorem

[tropic saddle]

well so far its not even true. you need extra assumptions on A

[primal bluff]

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[naive musk]

How do I solve this using elimination? I've been trying and failing to do so. I know what the answer is but I don't know how to get there using elimination.
i) 4x-3=3y
ii) 25+5x=-2y

[honest bough]

It's certainly not the best for elimination

What have you tried?

[naive musk]

multiplying the coefficient of either y or x so that one of em matches

[honest bough]

And what have you done to that end?

[naive musk]

or rather it doesnt match, but it's opposite of the other coefficient. Soooo let's say I do:

i) -5x+3.75=-3.75y
ii) 25+5x=-2y

Now if I add everything together, I get 28.75=-5.75y

ugh now that i wrote it out ofc i got the right answer T_T

i musta been subtracting 3.75 from 25 or something cause I was getting 21.25 before

and then unable to divide the number on the left by the y coefficient I was just like "????"

but that's all correct right? y=5?

[honest bough]

Should be

[naive musk]

cool, thanks. my bad! it's like when ya ask where something is and only once ya've asked does it show up lol

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[pearl pond]

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[ornate cloak]

how do i prove continuity on a closed interval, for example sqrt (4-x^2) on [-2,2]

[honest bough]

Show its continuous on (-2, 2), and show its continuous from the right at -2 and from the left at 2

[pearl pond]

@ornate cloak Has your question been resolved?

[ornate cloak]

can you give an example

i'm not really sure what that means

[honest bough]

What have you tried for your function

[ornate cloak]

i just found f(-2) and f(2)

[honest bough]

Where is the function discontinuous at all

[ornate cloak]

when x is less than -2 and greater than 2

[honest bough]

Ignore the domain restriction

When is √(4 - x²) discontinuous

[ornate cloak]

when the inside is negatiev

[honest bough]

I'm high nvm

So the only place it could be discontinuous is on the boundaries, right?

So we only need to check those

What's the right hand limit as x goes to -2?

[ornate cloak]

wdym on the boundaries

[honest bough]

It's discontinuous at -2 and 2

Which is at our bounds

[ornate cloak]

doesn't it exist at -2 and 2 though

[honest bough]

So since we're looking at the restricted interval [-2, 2], we only need to check at x = -2 and x = 2

[honest bough]

Continuous ≠ defined

[ornate cloak]

oh

so we're checking if it is discontinuous there

[honest bough]

Yep. Normally we'd check the two sided limit, but since it's on the bounds of our interval, we just check the one sided limits

[honest bough]

.

[ornate cloak]

0

[honest bough]

Is that equal to f(-2)?

[ornate cloak]

yeah

[honest bough]

So it's continuous at -2

What's the left hand limit as x -> 2?

[ornate cloak]

also 0

[honest bough]

Is that f(2)?

[ornate cloak]

yeah

[honest bough]

Tada

You're done

[ornate cloak]

oh

i see

thanks

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[honest basalt]

hey i need help with some trigonometry problems that just are not making sense to me

[plush bramble]

Use double angle formula

[honest basalt]

could you show me how to work it out?

idk if ive used double angle formula before

[plush bramble]

Oh shoot you don't even need it

Just recognize that 2theta is a special angle

And solve for 2theta. Then divide by 2 to get theta

[honest basalt]

so i find the two values of cos that are equal to sqrt3/2?

kinda working through this one and this is how my textbook is setting it up for me to solve

do i then take those and divide them by 2?

[old geyser]

[old geyser]

Yeah

But then you have to find values of k, after dividing by 2, that fits within the answer range

[honest basalt]

so for instance (pi/12)+(-1)pi?

would i just throw that into a calculator or what

[old geyser]

You shouldn't need a calculator to simplify $\frac{\pi}{12} - \pi$

[jolly parrot]

Umbraleviathan

[pearl pond]

@honest basalt Has your question been resolved?

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[lyric nacelle]

I wanna know if this is right, i’m not sure if i rounded correctly 😭😭

[pearl pond]

@lyric nacelle Has your question been resolved?

[pearl pond]

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[elfin gust]

i can show you what i got up untill but im having a bit of difficulty after that

[ember bay]

Send a picture of your work

[elfin gust]

i think im doing something wrong

because you would have to take out x twice

[ember bay]

x(x⁴-5x²+4)=0 is a good idea.
Now you should introduce another variable to make this equation a second degree equation. Try to substitute t=x², then try to solve for t first

[elfin gust]

to make it a second degree we keep factoring using synthetic division

thats how we were taught

[light helm]

you're missing a 0 for the x term in your synthetic division (between the -5 and 4)

[elfin gust]

yes i see that lol thanks

that basically solves the whole thing for me

thank you very much

!

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[rancid wraith]

Please help

[pearl pond]

Please don't occupy multiple help channels.

[bronze oyster]

I know if i wanna calculate the possible ways to arrange x amount of objects is x!
to arrange x objects with repeat is
x^(amount of digits or cheaters or the maximum possible amount)
but what about repeated letters and different amount of possible letters?

like "a" and "b" for 2 different places is 6

aa
bb
ab
ba
a
b

is there's a general formula?

[plush bramble]

This channel's occupied. Read #❓how-to-get-help and delete your messages

[bronze oyster]

Thx i don't think i need help anymore, i thought about it maybe it's

if anyone interested

m is the maximum amount of digits (i dunno what to call them English isn't my mother language,)

[plush bramble]

Stop spamming other people's channels

Open up your own and delete your messages in this channel

[bronze oyster]

srry i didn't know it was occupied

[rancid wraith]

.close

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[runic hazel]

how do i divide numbers with decimals, for example:
256/3,20

[runic hazel]

sorry for the stupid question, i just never learned

currently im studying about circles and sometimes we have to do big numbers divided by 3,14

[spice fable]

ok so the main concept is that $$3.20 = \frac{320}{100}$$

[jolly parrot]

ohNoiAmHere

[spice fable]

so its essentially the same as classical division but at the end you adjust the decimal

so like in your example

$$\frac{256}{3.20} = \frac{256}{320/100} = \frac{256}{320}\cdot 100$$

[jolly parrot]

ohNoiAmHere

[runic hazel]

hm ok

wouldit also work to do something like $$\frac{256}{1} / \frac{320}{100}$$

[jolly parrot]

chaves

[spice fable]

yes

[runic hazel]

but that seems a bit more difficult so maybe doing what u said is btter

lemme try to solve it

[spice fable]

theyre essentially the same if you look at the second step i have

[runic hazel]

true, but then how would i do 256/320?

simplifying?

[spice fable]

long division

any sort of division

or fraction

depends on what kind of answer you want

[runic hazel]

i am really bad at division omg

[spice fable]

you could simplify the fraction if you want a fraction for an answer

but since youre working with pi id assume a decimal is better

usually, this is done with a calculator btw

cause long division is tedious

[runic hazel]

true, my teacher accepts fractions as well

i replied to the wrong thing sorry

but also, we cant use calculators

[spice fable]

oh i see

if you can use fractions

its nicer i think

[runic hazel]

the test probably wont have very big numbers dividing

at least thats what he said

[spice fable]

$$\frac{256}{320} = \frac{2^8}{2^6\cdot 5}$$

[jolly parrot]

ohNoiAmHere

[runic hazel]

but i just wanna know because its btter to know

[spice fable]

you would factor it to reduce the fraction

[runic hazel]

i see

[spice fable]

yea its always good to know

[runic hazel]

i think i got it now, thx

.close

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[hardy valley]

Is this the right way to start out or should I use power reducing ID first ?

Doing a trig substation integral if it wasn’t obvious

[fierce oxide]

Let u1 = x^2 , du1 = 2xdx
Simplify

Let u1 = sin(t)
du1 = cos(t)dt

Simplify

[hardy valley]

Ahh so regular u sub first then proceed ?

[fierce oxide]

Use half angle

[hardy valley]

Ok true that will make it fit my x= asin(theta) form right ?

Doing the u sub first

[fierce oxide]

Yeah

[hardy valley]

Right on gonna try to work it from there - thank you much!

[pearl pond]

@hardy valley Has your question been resolved?

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[void harbor]

I’m stuck on how to start Q2 part 2

[pearl pond]

@void harbor Has your question been resolved?

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[midnight haven]

As I have seen this a lot, what does sin0 mean? I have been wondering what it means for a long time. Any help is appreciated, thank you!

[plain blade]

0 ?

[hollow cobalt]

Just refer to the definition of sin that you have

[midnight haven]

Can you answer this at least, what is sin used for and what does it mean?

[plain blade]

Which class are u in ?

[midnight haven]

8th

[hollow cobalt]

You can just google instead of waiting for someone to answer here

[midnight haven]

ok... I thought someone would explain it to me here better than in google, alright...

[plain blade]

For now just think of it as something that would be used in triangles and the no. In front of the function is the angle of the triangle

[midnight haven]

ok

[pulsar lark]

You will know once you learn about the unit circle

[pearl pond]

@midnight haven Has your question been resolved?

[vestal tapir]

sin(x) is your height if you walk x radians counterclockwise from that spot on the right

the full circle is 2pi, so sin(0) = sin(pi)

= 0

[pearl pond]

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[weak surge]

For 4/7x^2 is the 4 considered degree 1?

Numerator degree 1, denominator degree 2?

Don’t know how else to write that clearly for numerator degree equals denominator degree

[rotund talon]

no 0 ig cuz 4 = 4x^0

definitely not 1 though

[weak surge]

Degree is only referring to a term with a variable?

Constant is always degree 0?

I thought anything degree 0 always equals 1

Oh I see it’s x^0

Interesting

So we need to compare like terms to make this observation

Comparing 4 with 7x^2 is not fair

But 4x^0 with 7x^2 is a fair comparison for degrees?

Otherwise it could be 4/7x and both are degree 1.. that would be incorrect

[weak surge]

It would be degree 0 for numerator and degree 1 for denominator, right?

[weak surge]

Btw what is “ig”?

[rotund talon]

"I guess" ig

[weak surge]

OK

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[fickle finch]

i dont really understand how to approach this

[buoyant panther]

firstly find f'(x) using definition

[fickle finch]

-3 right

[plush bramble]

yes

[fickle finch]

then is it just (f'(-3+0)- f'(-3))/0 = 0?

i feel like im getting dumber by the day if i couldnt get that right away

[plush bramble]

yup

that's right

,w second derivative 2-3x

[buoyant panther]

it's because expression (value) doesn't depend on h

[jolly parrot]

Results provided by WolframAlpha

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

Requested by riemann#2529

[fickle finch]

alright thanks for the guidance!

.close

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[hard locust]

Ok back to asking
" Isn't sin opp/hyp? Why is the answer for sin sqr5/5?

[pearl pond]

@hard locust Has your question been resolved?

[hard locust]

<@&286206848099549185>

[hard locust]

<@&286206848099549185>

[spiral mantle]

What's the problem here?

[hard locust]

Isn't sin opp/hyp? Why is the answer for sin sqr5/5?

[spiral mantle]

Because the length of hypotenuse is wrong, it's not -1.

It cannot be

Why all the "lengths" are negative in your triangle?

and also, $-1^2 \neq (-1)^2$

[jolly parrot]

Robin S.

[spiral mantle]

How did you find those lengths though

[hard locust]

I corrected it so bow its just one

[hard locust]

I use tan

[spiral mantle]

Well, do you think $(\sqrt{5})^2 + 2^2 = 1^2$

[jolly parrot]

Robin S.

[spiral mantle]

Is it correct?

[hard locust]

I dont think so? I don't get 5 in the question itself I get 5 by using the hypotenuse which is 1 and the opposite which is 2

[spiral mantle]

Well, something is wrong with lengths obviously.

How did you find those 2 and square root 5 with Hypotenuse?

If hypotenuse is 1, the other ones must be fractional number

[spiral mantle]

Do you know about areas?

That negative sign is not for length

[hard locust]

I am a bit confused on the graph tho, how do I know which one is negative?

Cuz -1/2 is the problem

Do I make both 1 and 2 negative?

Or just one

[spiral mantle]

Also $tan = \frac{opposite}{adjacent}$

[jolly parrot]

Robin S.

[hard locust]

Oh

I get it now

I think

[spiral mantle]

You should learn about areas then, your area is (+,+) atm, which means your tan must be positive on that side of Coordinate System.

It's not really well but if you say x axis is cos and y axis is sin, then we know about $tanx = \frac{sinx}{cosx}$ and $cotx = \frac{cosx}{sinx}$

[jolly parrot]

Robin S.

[hard locust]

This better?

I do get it now tho

[spiral mantle]

Yes, but you should consider that, lengths will never be negative, that minus sign is just specifies quadrant.

So your triangle must be on Quadrant 4 or Quadrant 2

[hard locust]

It's 2

[spiral mantle]

It doesn't really matter

Alright

Oh

Yes

It's on Quadrant 2

[hard locust]

The negative seems to have to go too 2 tho since the answers that are negative all involve 2

[spiral mantle]

Do you know what signs sinx and cosx will get in quadrant 2?

[hard locust]

Wdym?

[spiral mantle]

I mean, if your angle on quadrant2, what signs x axis and y axis will get?

It says (-, +) on Quadrant 2.

[hard locust]

Yeah so is 1/2 (-1,2) or (-2,1)?

[spiral mantle]

Yes but your x value is negative.

[hard locust]

Yes that's why I'm seeing whether the numerator or the denominator is the x-axis

[spiral mantle]

Did you fix your triangle?

[hard locust]

Wdym? Besides the negative thing I don't know how it's wrong?

[spiral mantle]

It's not wrong, that -1 should be 1

Did you fix that?

[spiral mantle]

Alright, it's good now

Now you should write Trigonometric Values, but be careful with those signs.

[hard locust]

So the denominator is the x-axis?

[spiral mantle]

It's not denominator, it's cos value

Well, it's not cos value though since your not on unit circle

[hard locust]

Then how do I determine whether 1 or 2 is negative?

[spiral mantle]

x axis is negative

and y is positive

[hard locust]

How do I tell which on is the x-axis

That's why I was asking about the denominator of 1/2

[spiral mantle]

the left side of origin is always negative for x axis, and below is negative for y axis

[hard locust]

K wrote it down

[spiral mantle]

Your triangle is still wrong.

Because $1 + 2^2 \neq 5^2$

[jolly parrot]

Robin S.

[spiral mantle]

You should find the hypotenuse with pythagorean theorem.

[pearl pond]

@hard locust Has your question been resolved?

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[hot niche]

What value of x makes line l parallel to line m?

[hot niche]

I'm having trouble because we didn't cover this type of angle pair in class, but it's on the homework anyways.

There is corresponding angles theorem, but this isn't a corresponding angle because it's on opposite sides of the transversal?

So how would you set up the equation and why?

[light helm]

use a combo of multiple theorems

[old geyser]

105 + (3x - 18) = 180

External angle theorem, and then it's part of a 180° angle

So yeah

[light helm]

e.g you can combine with angle sum on a line / linear pair

[old geyser]

Linear pair

That's what it's called

[hot niche]

Wait what, external angle theorem?

Like alternate exterior angle theorem?

[old geyser]

Yeah

[hot niche]

How does that work for this if one of them is interior

[old geyser]

The interior + exterior = 180

They're a linear pair

Linear pair's angles always add up to 180

[hot niche]

Is there a theorem or something that states this?? The only thing we covered that adds up to 180 was the SSI angle theorem

or consecutive side angle theorem I think

[old geyser]

Definition of a linear pair

It's a postulate

[hot niche]

So you just assume that these are parallel, which let's you set up 105+3x-18?

=180

Cause these aren't even on the same side of the transversal??

I thought it would have been a vertical angle

let alone a linear pair

[light helm]

based on the wording of your question technically you'd apply the converse theorem

those angle have to satisfy a certain property for those lines to be parallel

[hot niche]

So is converse of corresponding angle theorem what states this?

I just lost my notes that I had littearly 5 minutes ago 💀😭

Nvm we good

So a converse of a corresponding angle is angles that lie on the opposite side of the transverse and in opposite positions?

The only thing I think of to do without that is looking at 105

Saying oh this is a linear pair, then 180-105

75 degrees

OOHHH

I see

alt exterior angle creates the linear pair

Alright boys I'm good

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[misty saffron]

Why isn't e^x equal to x * e^(x-1) ?

[night urchin]

why would it be

think of an example that u can figure out

is 2^3 equal to 3 * 2^2

[misty saffron]

oh sry wait I forgot an important part

I ment

why isn't the deravative of e^x equal that

[night urchin]

I mean, because it is not

but I think you are confusing power rule

power rule is when x is the base

and is raised to a constant exponent

here we have a constant base

raised to the x

[misty saffron]

ahhhh i think my teacher skipped over something important ;-;

ty thanks my question is answered

[night urchin]

np

[misty saffron]

.clos

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[night urchin]

when is this function not continuous

[midnight haven]

x = -1

I think

[night urchin]

well, that was kind of a trick question, it is continuous everywhere it is defined

polynomial functions and rational functions (division of 2 polynomials) are very well behaved, basically when is this function not defined

when you need to divide by zero

when does that happen

[midnight haven]

x=1

[night urchin]

not just 1

[midnight haven]

1, and -1

[night urchin]

yeah

[midnight haven]

how would I write that in interval notation tho

[night urchin]

so it is not continuous at x = -1 and x = 1, and continuous everywhere else

how would you write that indeed

start from (-infinity, something) as your first interval

[midnight haven]

(-inf, -1) (-1,1) (1,inf)

is that it

[night urchin]

yes

[midnight haven]

ty

[night urchin]

np

[midnight haven]

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[meager grotto]

Hi I need help with geometrical problems

[hazy crypt]

just send

[meager grotto]

this 3 problems

annoying homeworks

[meager grotto]

Sent

[hazy crypt]

what is the question.

?

[meager grotto]

It's not a question but a problem

[hazy crypt]

what are you trying to accomplish

[meager grotto]

I'm trying to accomplish on how to measure and how this thing works

I was absent in school for a week

My arm was broken*

So I really need to answer this till lunch time

[meager grotto]

Will you be able to help me?

[hazy crypt]

sorry i still don't understand what is trying to be done here

[meager grotto]

.close

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[prime loom]

hello im not sure how to study this on youtube i dont really know this subject name was wondering if i can get any assistance, not really asking for help with the problem itself tho

[prime loom]

just wanna know what to search up

all i know is complex numbers but i get other similar problems that isnt exactly this template

[untold atlas]

looks like a standard imaginary numbers problem, just factor out an "i" from the negative sqrts, and evaluate like normal

[prime loom]

so rooted -8 is just 8i then im guessing

[untold atlas]

it's just sqrt(8) * i

the 8 is still under the sqrt, mind you

[prime loom]

hmm alright ill try that

[prime loom]

going to try but i'll still hope someone can tell me the name of this subject still, all i know is this is related to complex numbers i dont really know the specific subject so i cant study it on my own without that information

[lyric lintel]

Complex Numbers: Real and Imaginary parts of a complex number, complex plane, terminology and notation, and complex algebra

https://youtu.be/SP-YJe7Vldo

I first learned it in HS algebra II if that's what you're asking

[prime loom]

oh thanks

i'll look into this, really appreciate it

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[chrome ferry]

i understand left side where he divided by n^2

but the right hand side with hopitals rule

i keep doing lohpital until i cant anymore?\

is that the right way to think of it

[sharp quest]

yeah, basically

if you do l'hopital's rule and get another indeterminate form, you can do l'hopital's rule again.

[chrome ferry]

got it

ty

[sharp quest]

happens all the time with polynomials like that.

[chrome ferry]

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[rocky willow]

I am finding it hard to identify if a statement is a theorem or an axiom. For example is: a+(-a)=0 for every integer; an axiom or a theorem. Another example would be is the statement: On every plane there are atleast 3 non collinear points, and there exist points outside of that plane; an axiom or a theorem? And how would i identify them.

[wind wigeon]

theorems are proofs that follow from axioms

[sharp quest]

is there a reason you want to distinguish them?
Axioms are statements taken to be true without proof.
Theorems are statements that are proven true, given a set of axioms.

Both of those statements are (probabily) theorems.
The first is arguably a definition depending ont he context.

[rocky willow]

a professor is being very keen on us identifying axioms vs theorem statements. but from the way she taught i absolutely got nothing out of the lecture.

is prooving a statement or trying to proove it the only way to distinguish them?

[sharp quest]

More or less, yeah.
Axioms are just true.

Theorems are proven from axioms (and other theorems proved from axioms)

[rocky willow]

Ok ill try and solve the rest of them and see if i run into any problems

tysm

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[gaunt steeple]

i have two questions:
how do i calculate 2.5x(5V3/2)
and
how do i calculate 0.5x(12/√3)x6?

[gaunt steeple]

please ping me if someone answers

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[finite crown]

https://gyazo.com/f5485eb57067414daa4ebbc920a057fc

[finite crown]

Hey ive like totally forgot

what any of that means

anyone gimme a little boost

[wind wigeon]

absolute is asking the distance from the origin

arg is the angle with the positive real axis

[finite crown]

anticlockwise is positive?

for the theta

[wind wigeon]

yes

heres my shitty drawing

[finite crown]

what is the modulus

thats r?

[wind wigeon]

yes

[finite crown]

so 9^2 + 6^2 and the root of that

would be the modulus

[wind wigeon]

yes

[finite crown]

thanks

<3

[wind wigeon]

np

[finite crown]

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[copper stratus]

what is the value of x

[minor vortex]

The triangle has a total value of 180°.

If one corner has 114° the others have a combined value of (180-114)°.

3x = 180°-114°

[copper stratus]

so 180°-114° is the answer?

[coral fiber]

you have to subract first

[copper stratus]

so 66?

[coral fiber]

66+2x+x=180

[minor vortex]

?

[copper stratus]

my options for the question is 38, 57, 76, and 85

[coral fiber]

multiply?

[minor vortex]

None of them is correct

[abstract locust]

uh yes?

38 is correct

[coral fiber]

what

[abstract locust]

3x = 114

[coral fiber]

180-114= 66

and your finding the inside of the triangle

[minor vortex]

Oh, sorry

I was wrong.

[coral fiber]

66+2x+x=180

[abstract locust]

you don't even have to do that

[coral fiber]

add like-terms then isolate x

[abstract locust]

you can just use extenral angle formula

[copper stratus]

ok so 38

[abstract locust]

max do you actually understand it

[coral fiber]

this is wrong?

[copper stratus]

not really tbh

[abstract locust]

how is it wrong

it's the same as what you wrote

[coral fiber]

why would you divide 114

[abstract locust]

just quicker

[abstract locust]

?

[abstract locust]

angles on a straight line add to 180

so the missing angle would be 180-114 = 66

angles in a triangle add up to 180

so 66 + x + 2x = 180

3x = 114

[coral fiber]

oh sry

[copper stratus]

so it is 38 then or no

[coral fiber]

yep

[pearl pond]

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[abstract narwhal]

So im struggling on how to tackle this problem i graphed the first function on the left but im not really understanding the concepts of continuity

[pearl pond]

@abstract narwhal Has your question been resolved?

[pearl pond]

@abstract narwhal Has your question been resolved?

[abstract narwhal]

<@&286206848099549185>

[next salmon]

So with piecewise functions they only apply within the domain specified. Here, for values of x greater than 2 you must only graph y=0. (Rather than the linear equation for x=all real numbers) Then you see if the function is continuous, meaning it has no steps, no cusps, and no undefined points.

[abstract narwhal]

@next salmon i graphed y=0 so does that mean that for the piecewise function on the left is countious at x=5?

[next salmon]

Well, I think I see what you did. You should graph y=0 only at and after x>/=2

But a function is continuous for any point which is does not violate the criteria above.

More or less smooth lines with no gaps.

[abstract narwhal]

Im kind of confused is there anyway you could show me what the graph should look like?

[next salmon]

yeah, give me a second. I have to get on my laptop.

in any piecewise function, a domain is specified on which the function holds true. Here, it's 2x-10 for x<2, and for every other value y=0.

so at 2 on the x-axis, you stop following the first function and switch to the second.

(and x=2 is included in the latter function)

[abstract narwhal]

So i have that graphed now what? Do i set them equal to each other and solve or?

[next salmon]

nope. you're looking to see whether the graph is continuous. Do you know what the criteria for a function to be continuous are?

[abstract narwhal]

I think so in order for a graph to be continues there cant be any jumps skips and then something else

[next salmon]

yep, and i think cusps are the last thing you're thinking of. so do you see any values of x where the function isnt continuous on this graph?

[abstract narwhal]

No I dont unless im mistaken and then i might need to look and see what the defintions for jumps and steps are again

[next salmon]

look at it this way:

is the graph one, unbroken line?

[abstract narwhal]

The 2x-10 graph? No

[next salmon]

the entire graph. piecewise functions are one function

but yes. it's not just one line, unbroken. so it cannot be continuous.

so now, we need to find the values of which it is continuous, or simpler, where it isnt continuous.

do you see where the only x-value is where there is a jump?

[abstract narwhal]

Ahh

[next salmon]

so at that x value, if you see it, that's where the function is not continuous. so every other number than that is continuous on this function.

[abstract narwhal]

Wait the answer to this problem is gonna be in interval notation isnt it?

[next salmon]

yep

[abstract narwhal]

Uh

[next salmon]

you have to use union (U)

[abstract narwhal]

(-oo,5) u (5,oo)?

[next salmon]

first of all, where is x not continuous? what point? Just so I don't accidentally give away the answer lol

[next salmon]

not 5...

where is the jump?

on the graph i posted

[abstract narwhal]

Oh 2

[next salmon]

yep

[next salmon]

so its this answer, but instead of 5, put 2

[abstract narwhal]

I see i keep graphing the first function to the point where the intersected and thought the intersection meant something but it doesnt

[next salmon]

nope, the two functions don't exist outside of the domain specified

anyways, good job, hope I was able to help you understand it all a bit better 👍

[abstract narwhal]

.close

[pearl pond]

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[abstract narwhal]

.reopen

[pearl pond]

✅

[abstract narwhal]

@next salmon real quick question What about the second graph?

[next salmon]

the one right of it on this paper? #4?

[abstract narwhal]

Yeah i should use my graphing calculator to figure that 1 out yes?

[next salmon]

if you're allowed to, you could yes. but i think this one would be helpful to do without it, will make your skills with polynomials stronger.

factor the numerator, and find the zeroes and their multiplicity

then graph it with the asymptotes as well

once youve graphed it lmk

and show if youre not confident in it ofc

[abstract narwhal]

Alright ill do that

[abstract narwhal]

Sorry it took me longer i might really wanna go over how to factor again that knowledge is gone

[next salmon]

no problem, so now you can graph it as the numerator's zeroes are the function's zeroes and the denominator's are the asymptotes.

so there is one point where x is undefined (per the definition of an asymptote)

[abstract narwhal]

So im trying to use my graphing calculator because we can indeed use it on the test but i need help adjusting my window

[next salmon]

not sure how to do that, sorry. depends on the calculator but im sure theres some resources online

[abstract narwhal]

Give me a second

Ill respond in 30 something came up

[abstract narwhal]

@next salmon sorry do you mind show me the graph on you're side so i can figure that out

[pearl pond]

@abstract narwhal Has your question been resolved?

[pearl pond]

@abstract narwhal Has your question been resolved?

[pearl pond]

@abstract narwhal Has your question been resolved?

[pearl pond]

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[granite dome]

Someone explain me the second step of the second image

The question is the ex 2 of image1

[smoky musk]

Is Lt limit?

[granite dome]

Yes

@smoky musk

[tiny shore]

they used 2 things

• addition formula for sin(x+y)
• log property log(a) - log(b) = log(a/b)

ah i see, you meant the other step, take a look at the taylor expansions of sin & cos

upon simplifying the fraction you have
$$\cos(h/a) + \cot(x/a) \sin(h/a)$$
you can insert the taylor expansion for both $\cos(h/a)$ and $\sin(h/a)$

[jolly parrot]

nvx

[tiny shore]

,w taylor sin(h/a)

[jolly parrot]

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[tiny shore]

,w taylor cos(h/a)

[jolly parrot]

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[pearl pond]

@granite dome Has your question been resolved?

[smoky musk]

Taylor expansion? This early?

[tiny shore]

well they used it

[pearl pond]

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[pearl pond]

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[fervent briar]

Can someone help me with number 1

[buoyant panther]

try to use geometric series

[fervent briar]

I can’t figure out how

[buoyant panther]

look at this

[fervent briar]

The 2n-1 is confusing me

[old marsh]

break up the numerator

[fervent briar]

How

[buoyant panther]

$$\frac{2^{2n-1}}{5^n}=\frac{2^{2n} \cdot \frac{1}{2}}{5^n}=\frac{\frac{1}{2} \cdot 4^n}{5^n}=\frac{1}{2} \cdot \Big(\frac{4}{5}\Big)^n$$

e

[old marsh]

shdnt it be 2^-1

[jolly parrot]

Modus

[old marsh]

there u go

[buoyant panther]

now? can you determine common ratio?

[fervent briar]

Ya I see, thx

How did you get 4^n

Is 2^(2n)

4^n
?

[buoyant panther]

2^(2n)=(2^(2))^n = 4^n

[fervent briar]

!close

[pearl pond]

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[drowsy tangle]

?

[drowsy tangle]

i graoh this and its just a straight line x=0

why is that?

[snow skiff]

um it most definitely should not be

plug in p(1) and you'll see it's not 0

did you plug the equation into desmos?

[snow skiff]

evaluate p(1)

p(2) etc.

also, does your teacher allow you to use desmos to graph this?

intuitively, $p(t)$ is of the form $\frac{\alpha t}{t+1}$ where $\alpha$ is simply some constant.

[jolly parrot]

DerpZ

[snow skiff]

So you can set $\alpha$ to be something smaller like 3 and observe the shape of the graph

[jolly parrot]

DerpZ

[drowsy tangle]

am i typing it in wrong

[snow skiff]

nope, it's right. but evaluate $p(1)$ and tell me what you get

[jolly parrot]

DerpZ

[snow skiff]

which is?

[snow skiff]

ok, I mean evaluate it without desmos

[drowsy tangle]

3000/2?

[snow skiff]

which is?

[drowsy tangle]

1500

[snow skiff]

can you see why that graph looks like a straight line now?

[drowsy tangle]

the population is decreasing

[snow skiff]

^^

[snow skiff]

^^

[snow skiff]

at the normal desmos scale this is very large

[drowsy tangle]

lol i dont get it so what happening

problem wise

[snow skiff]

the 3000 is just a scale factor

you could set it to anything you like and the graph will have the same shape

[drowsy tangle]

yeah

[snow skiff]

I don't see the problem then

[drowsy tangle]

i see that but idk how to answer the question

[snow skiff]

a? or b

[drowsy tangle]

b

the graph is just straight up

[snow skiff]

Observe t/(t+1)

which I just explained the reason it goes straight up is because desmos at that scale does things like that

so plot a graph with a similar shape

[drowsy tangle]

ok but can u just help me answer my question one the population i get that no matter what i plufg in is lower than 3000

how can i say what happening to the population

[snow skiff]

^

Plot this graph

You'll have your answer

[drowsy tangle]

i dont know what that meanssss lol

plot on my graph

[snow skiff]

p(t) = 3000t/(t+1)

this graph = t/(t+1)

[snow skiff]

good

now what do you see as t tends to infinity

[drowsy tangle]

so 3000 is my asymptote?

[snow skiff]

very good