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[charred birch]

when it asks to consider the function do i just plug each funtion into each roman numeral function? so each problem would have two solutions or??

[charred birch]

thats my question #1 as for question #2 , in this picture do i just find the piecewise of absolute value and plug into the whole function?

[torn jolt]

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@charred birch Has your question been resolved?

[charred birch]

Q1

Consider the functions
(i) 𝑓(𝑥) = 3𝑥 − 5𝑥'2 and (ii) 𝑓(𝑥) = 𝑥/x+3
For each of the functions find and simplify the following:
(a) 𝑓(𝑥 + ℎ)
(b) 𝑓(𝑥 + ℎ) − 𝑓(𝑥)
(c) 𝑓(𝑥 + ℎ) − 𝑓(𝑥)/ h

Q2/
3. Write the functions as piecewise functions and simplify.
(i) 𝑓(𝑥) = |5𝑥 − 7|
(ii) 𝑓(𝑥) = |5𝑥| / x
(iii) 𝑓(𝑥) = 5 − 𝑥/ |𝑥 − 5|

(iv) 𝑓(𝑥) = |𝑥 + 2| / 𝑥'2 + 𝑥 − 2

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[stone trench]

where did this came from

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[warm harbor]

do you have photo reference? of actual equation

ok let me try to graph and see

this is what symbolab got

ahh so i see why its -inf, im confused tbh. where does it say DNE? sorry. maybe we need moer helpers in here

ahhh i see. hmmm. that is confusing. im so sorry. im not really sure myself

<@&286206848099549185>

i almost actually feel like that dot isn't that necessary for actually finding the limit

there are many points with y at zero so thats why that specific point also kind of confuses me.

[pulsar sigil]

👋

please show me the original question

tbh i don;t see why it does not exist

,w plot (x +3) * e^- x at x = -∞

from this graph it discontinues at x = -3

but as x increases y tends to 0

[glossy valve]

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Requested by ishigamisenku_

[pulsar sigil]

lol

yes

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@tulip lantern Has your question been resolved?

[hollow spire]

Lemme take a look

While there is an asymptote at that location, he is showing what the point would be at that location if there wasn’t an asymptote

Both Symbolab and Desmos does not visually show asymptotes and the same thing with most calculators

I’m going to base my explanations off the graph my Ti-89 gives me

I don’t trust wolfram, it’s not the best resource to go to when it comes to math, it is ok for simpler questions but not something as complex as this

There’s a lot of gaps in the coding of wolfram so it often cannot compute graphical equations correctly

The problem there is desmos cannot compute a missing point in x=inf, y=inf because the point is not defined, inf is considered an unknown variable in that sense

It can graph x=inf, y=inf but it cannot put an asymptote there

Could you restate the math question so I can read it clearly? I scrolled up to f(x)=(x+3)e^-x

what is it you are trying to find

Ok lemme rephrase my question. Is the equation asking you to graph the equation? From the information you are providing it seems like you may be getting caught up in complexities that are made to distract you

Don’t use wolfram

There is no true way of graphing something by hand perfectly

So if I remember correctly

Lim as x approaches infinity, you have to find the actual limit

what I want you to do is

Find Lim as x approaches infinity for (x+3)e^-x

Try doing that. I have to go sadly, I have to go to work

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[warm abyss]

What have you tried?

[trail harbor]

Ive rewritten sin(x/2) as equal to cos(x/2 -90) so that i can use the cos(A-B) compound formula but I realise that only simplifies to sin(x/2)

[warm abyss]

How about applying double angle formula for cos(x)?

[trail harbor]

wouldn't I need cos(2x) to be able to use 1-2sin^2 (x) to then rearrange to get 0 = 2sin^2 (x) + sin(x/2) -1

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[dull seal]

Show your work

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@hybrid quartz Has your question been resolved?

[inland trellis]

Try subbing in the values and see which one makes sense

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[solar arch]

I need help in #help-13 please

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[solar arch]

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[torn jolt]

how to solve these kind of questions

[thick hedge]

two ways

1 is to use l'hopital's rule

the other is to factorise

[hot herald]

note that the denominator → 0 as x→-2
for the limit to exist for this expression, the numerator would also need to approach 0 as x→-2

[silent wagon]

making it an indeterminate form, no?

[silent wagon]

🤔
is this even taught that early

[torn jolt]

0/0 exist ?

[torn jolt]

using factorise

[hot herald]

0/0 is an indeterminate form (and a limit could exist)

but getting something (non-zero)/0, the limit does not exist

[silent wagon]

wait, you can conclude that? 👀

[torn jolt]

so we Equalize x^2+kx-6 = 0 then substitute x = -2 to find k ?
(-2)^2 +(-2)k - 6 = 0
4-2k-6=0
-2k=2
k=-1 ?

[hot herald]

yes

[torn jolt]

thank u

[silent wagon]

and because you had to make the function 0/0, can you find the limit given k unassisted?

[torn jolt]

didnt understand

what u mean

[silent wagon]

can you find h now, now that you had k?

[torn jolt]

yes
h = 1

[silent wagon]

sure...
wait.

[silent wagon]

this question asked for a positive integer k ;-;

[torn jolt]

oh

so there no solution for k ?

[silent wagon]

I think 🤔

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@torn jolt Has your question been resolved?

[hot herald]

didn't see the N+
question is somewhat flawed

were there any other instructions above this?

[torn jolt]

no
I think he meant +6, not -6

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[granite torrent]

is that a 1 over ln(10) ?

[lime jackal]

lol

[granite torrent]

your teacher appears to be incorrect

you can take the natural logarithm of both sides (but this is overkill)

because if you define logarithms in the complex plane, taking ln(-infinity) does give you positive infinity (+ an imaginary constant but this is presumably unimportant)

...

well that process technically works... your teacher just applied it incorrectly it appears

1/ln10 is less than 1, taking that to the power of infinity would give you 0

this limit is not an indeterminate form so you can do that and comfortably reach a correct answer

it might make more sense to put it like this

ln((x - 4)/(1 - x)) = ln(3/(x-1) - 1)

and the limit as x approaches 1 from the positive side of 3/(x-1) - 1 is most definitely positive infinity

sorry for not noticing but your teacher was wrong here as well

you would have ln((-3)/(-0)) because you're approaching 1 from the positive end (i.e. x - 1 < 0)

this approaches positive infinity not negative infinity

that would be correct if (1/ln10) is greater than 1

which it is not

oh is that a product

the ln((x - 4)/(1 - x)) was written slightly higher than the 1/ln(10) so it looked like exponentiation

uh...

lol 😅

yeah then infinity would be correct if it was a product

no

whenever that is the case, the logarithm would be undefined in the real plane

no because the logarithm of a negative number is always undefined

might have been because he found it incorrectly

(x - 4)/(1 - x) approaches positive infinity from the positive side

the limit as x approaches 1 from the positive side of (x - 4)/(1 - x) is not negative infinity

the limit is rewritten as ln(inf) not ln(-inf)

if it was -inf, the entire limit would be undefined

because 1 - x is always less than 0 when x > 1

therefore the limit would be more accurately written as (-3/-0), not (-3/0)

😅

that certainly is interesting 🤔

wolfram is thinking in the complex plane

ln(-inf) does not mean inf when you try to evaluate limits

in fact, this entire function is actually undefined when x < 1

yes

(for clarification, ln(-inf) would more accurately be written as inf + ipi)

would appear so

ah rip

welcome 🙂

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[restive marten]

question on natural domain of functions

[restive marten]

i understand in questions such as 1/x-3, the natural domain for x is {x ∈ R: x ≠ 3}, but in an equation like f (x) = 2x + 3, why isn't the natural domain -3/2?

[onyx glen]

are you sure you did not mistype that?

f (x) = 2x + 3, why isn't the natural domain -3/2

[restive marten]

yeh?

[onyx glen]

are you asking why the natural domain isn't a one-point set consisting of the number -3/2 ONLY?

or what

[restive marten]

no sorry i mean

why can't it equal -3/2

i understand the natural domain is infinite

[onyx glen]

why can't what equal -3/2...

[restive marten]

f (x) = 2x + 3, this

sorry excuse my lack of terminology, im new to this concept

[torn jolt]

are you saying why can't the domain have the element -3/2

[onyx glen]

this function can receive the number -3/2 as input...

[restive marten]

yeh

[onyx glen]

f(-3/2) is defined...

[restive marten]

i think

ok but in 1/x-3, x can't equal 3

[torn jolt]

well, it can that would just give you 0 with which there's no problem

[granite torrent]

this function is undefined at x = 3

[restive marten]

is that because of the additional 3 at the end

[onyx glen]

you mean $\frac{1}{x-3}$?

[glossy valve]

Ann

[restive marten]

yessss thank you

[onyx glen]

that's properly written as 1/(x-3), just fyi.

[restive marten]

i had no idea how to do that

[onyx glen]

and the reason x ≠ 3 in that

is that if x WERE 3

then you'd run into a division by zero

division by zero is not allowed

understand?

[restive marten]

uh

yeh

ok

[restive marten]

oh alr

ohhh so if it was for example f(x) = 2x

its got infinite domain

even -

0

[onyx glen]

"infinite domain" is bad wording

its domain is all real numbers

or just R for short

[restive marten]

but if its this on it can't be 3 cus of the rule yo mentioned

[restive marten]

right sorry real numbers

[onyx glen]

like a fancy R because mathematicians like giving fancy symbols to important objects

$\mathbb{R}$

[glossy valve]

Ann

[restive marten]

yeh haha that thing

[onyx glen]

there are various ways of writing it by hand

yeah so

the three big things you wanna watch out for

are

• division by zero
• sqrt of a negative number [or more generally roots of even degree; however, odd roots (3rd, 5th, 7th etc.) are safe!]
• logarithm of 0 or a negative

[restive marten]

sorry sqrt

whats that mean?

[onyx glen]

square root

[restive marten]

oh thanks

[onyx glen]

square root

[restive marten]

hahaa right

[onyx glen]

you'll encounter this abbreviation a lot

it's standard in basically all programming languages

[restive marten]

oh alright

thank you lots

i had one more question

ik i can solve domain by hand, but what does it mean by solving range graphically?

like calculater?

[onyx glen]

it means graphing the function (by whatever means) and looking at what values it covers on the y-axis

[restive marten]

ohh

ok

thank you sm

ok i think im good

[restive marten]

anything else i should look out for?

[onyx glen]

?

[restive marten]

like trick questions

allg thank you for the help 🙏

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[chrome jolt]

how can I solve this

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[coral vortex]

Could someone help me with question 7

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[torn jolt]

so

a) part

A is a point then
$\ R_Q(A) = T_p(R_O(A)\
2Q-A = T_p(-A))\
(2A-2P)-A = -A+p\
2A = 3p$

[glossy valve]

Gary

[torn jolt]

i don't know

what they mean by translation there

like by what factor

is that translating?

im confused

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[swift fulcrum]

Hello

[swift fulcrum]

I'm wondering why this is represented in a piece-wise function

[oak ruin]

Because f(x,y) is discontinuous at the origin otherwise.

[empty sapphire]

they're just making it a continuous function

[swift fulcrum]

Why do we want to make it continuous?

[umbral dome]

a function must be continuous to be differentiable

[spice orchid]

the more general point is they just want the function to be defined everywhere. you can just say f(x,y) = (x^3+y^4)/(x^? + x^2) for all x,y since then it wouldnt be defined at (0,0) (you cant divide by zero)

[swift fulcrum]

Can't we also differentiate functions like 1/x?

1/x isn't continuous

[next lantern]

I need help

[swift fulcrum]

Then would f_x(0, 0) and f_y(0, 0) equal 0?

[empty sapphire]

#❓how-to-get-help

[oak ruin]

They probably want it to be differentiable at the origin too

[oak ruin]

I would use the limit definition to calculate that

[spice orchid]

yeah it will not be in general

y = x is 0 at x=0 but its derivative is 1 everywhere

[swift fulcrum]

$\lim _{\left(x,y\right)\to \left(0,0\right)}\left(\frac{x^3+y^3}{x^2+y^2}\right)=0$

Like this?

[glossy valve]

LE SSERAFIM

[oak ruin]

No

not at all

$\lim_{h \rightarrow 0} \frac{f(x+h, y) - f(x,y)}{h}$

[glossy valve]

Ari

[swift fulcrum]

Ohhh ic

Looks like pain in the ass calculation

I understand now thanks guys

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[oak ruin]

It won't be because x and y are both 0

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[swift fulcrum]

Hello

[swift fulcrum]

This is what my prof wrote in class

And I think they are missing "- f(a, b)" in the numerator

Am I right?

[oak ruin]

Yes

[swift fulcrum]

🤦

Thank you

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[torn jolt]

so for c)

i split it into cases

case A) only the unit digit varies
case B) only the unit AND the ten digits vary
case C) all of them vary

so A by counting gets you 6 numbers

B gets you 6(3) numbers

C gets you 6(3)(2) numbers

is my logic right

[spice orchid]

check by calculating how many are less than 330

[torn jolt]

wait i think its fucked because like

3(6)(5)

should be it for the C

then like fixing the hundred position

you have 3(3)(5) for B

if im not mistaken

and for A

6 again

,calc 3(6)(5) + 3(3)(5) + 6

[glossy valve]

Result:

141

[torn jolt]

uhh

like

thats going to be a bit hard

so we cant have 0 for the hundred digit

so like

2(6)(5)

and then also

3*2

and also

no hm

if the hundred digit is 4,5,6 then the tens is 6 and the unit is 5

so 3(6)(5)

if the hundred is 3 then the tens is 3 choices and the unit is 5

so ,

,calc 3(6)(5) + 3(5)

[glossy valve]

Result:

105

[torn jolt]

ok i tyhink that sounds legit

.clsoe

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[modest rain]

How does this function look like δ(t)?

[modest rain]

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[full prawn]

I was working on this problem . And a) part is very easy: $C_A \times C_B$ , $C_A + C_B - C_A \times C_B$, $1 - C_A $

However I'm very much so stuck on b). The result is very obvious but I can't think of rigorous reasoning.

[glossy valve]

NZzska

[kindred grove]

well you have to define yourself a set A (which will certainly depend on the function f) such that the equality works

if you look at the definition of C_A it shouldn't be too hard to see what A would fit

proving an existence statement by providing an object that satisfies the property is completely fine

[full forum]

@full prawn Has your question been resolved?

[full prawn]

sorry I'm helping someone else on other channel and trying to see if it makes sense

I'll reply soon

lol

[kindred grove]

aight

[full prawn]

I'm a little confused still.

So assuming set $A$ exists, where $C_A$ satisfies $f = C_A$. Then $f(x)$ will be either 1 or 0 and this completes the proof?

[glossy valve]

NZzska

[kindred grove]

no you're not assuming it exists

you have to find one yourself

[full prawn]

Ok so I need to define how A depends on f?

[kindred grove]

yea

[full prawn]

I don't understand...

[kindred grove]

alright let's try something else

if you had to explain what C_A does in english, what would you say ?

[full prawn]

It's equal to 1 when x is in set A and 0 otherwise

[kindred grove]

ok yea

so it tells you what elements are in set A essentially

[full prawn]

Oh

[kindred grove]

if the function's value is 1, then the element is in the set

otherwise it's not

[full prawn]

Yeah makes total sense

[kindred grove]

so now if I gave you a function $f(x) = \begin{cases} 1 & \text{ if } x > 0 \ 0 &\text{otherwise}\end{cases}$

[glossy valve]

aPlatypus

[kindred grove]

could you find a set A which has this C_A?

[full prawn]

well of course then $A = {x: x>0}$

[glossy valve]

NZzska

[kindred grove]

yea exactly

[full prawn]

but you used some form of f, we don't know when it's 1 and when its 0

[kindred grove]

now the f is unspecified in the question

but you can use its output still

you can ask for f(x)

for all x's

[full prawn]

Is this as obscure as...

if $f(x) = \begin{cases} 1 & \text{ if } x = \text{some set} &\text{otherwise}\end{cases}$
and we call "some set" A. so A exist where f(x) = C_A.

[kindred grove]

if f(x) = 1, what can you say relating x and the set A we're looking for ?

it's pretty much that

[glossy valve]

NZzska
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[kindred grove]

not very obscure

$A = \{ x : f(x) = 1 \}$

[glossy valve]

aPlatypus

[kindred grove]

x should be in the set if f(x) = 1

[full prawn]

I don't know if I start to understand what you're trying to say or I'm just getting used to it as truth, which is scary.

I understand the reasoning, but it feels like repeating same statement twice.

Ok so yet again:

[kindred grove]

well essentially you're going the reverse way as C_A, so it's gonna be very similar

C_A takes a set A and gives you a function telling you which elements are in A : set -> function

that question you have asks you from a function to find a corresponding set A : function -> set

[full prawn]

if function f(x) can be equal to 0 or 1 we can write it as f(x) = C_A =

1 for {x: f(x) = 1}
0 for R - {x: f(x) = 1} (everything else)

and we can see that here A = {x: f(x) = 1} so it exists and we can write f(x) as C_A

[kindred grove]

but yeah the point of that exercise is that the set and the function give the same information

you can compute the characteristic function (that's the name of the C_A) and get back to the set without any problem

it's two ways of talking about the same thing

[full prawn]

So asking if x in A and computing x is the same thing. It probably goes further than functions of 1s and 0s

[kindred grove]

asking if x in A and computing C_Ax
yes

well 1 and 0 is just a choice

where 1 signifies, I'm in the set

and 0 I'm not

[full prawn]

I'm still deeply unsatisfied with the "solution" it's "yeah we can write f(x) as C_A if we call A={x: f(x) = 1}"

But if that's it I'll accept it

[kindred grove]

unfortunately it is

just let it sink in for a while

gtg :/

[full prawn]

@kindred grove no problem, thanks for your effort. I mean this is very obvious result, but for some reason it doesn't sit right with me. At least c part looks very easy. Thanks again

.close

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[chrome jolt]

I have a function f:C-->C f(x+yi)=3-2x+y-(x+2y+2)i

I am trying to prove that the function is injective so I took arbitrary values x+yi and x'+y'i and assumed f(x+yi)=f(x'+y'i)

so we have 3-2x+y-(x+2y+2)i = 3-2x'+y'-(x'+2y'+2)i if and only if their real numbers are equal and their imaginary numbers are equal

so 3-2x+y = 3-2x'+y' AND x+2y+2 = x'+2y'+2, so far so good? this is confusing me because of the minus sign, do I multiply it into the imaginary bracket or not?

[chrome jolt]

in other words, what calculations do I go with

3-2x+y = 3-2x'+y' AND x+2y+2 = x'+2y'+2

OR

3-2x+y = 3-2x'+y' AND -x-2y-2 = -x'-2y'-2

[full forum]

@chrome jolt Has your question been resolved?

[chrome jolt]

,close

.close

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[rare dock]

<@&268886789983436800>

[full forum]

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as this action is irreversible, and this channel may abruptly lock.

[spiral vigil]

lmao

[full forum]

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[torn jolt]

Find the exact value of $\sin(\cos^{-1}\frac{3}{5})$

[glossy valve]

Matt

[buoyant pewter]

a hint is, use trig idenity

[empty sapphire]

so cos^-1 (3/5) means that we'll get an angle back. The fact that we have 3/5 tells us we have a special triangle: a 3-4-5 triangle

we know the angles of a 3-4-5 triangle. Use that information to get the angle, then apply that angle to sin

[torn jolt]

im not really familiar with special triangles but i am with trig identities

[torn jolt]

how would i use a trig identity for this

.close

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[solar arch]

How do I find the exact value of sin 75 using sum identities?

[hard veldt]

sin(45 + 30) = sin(75), so you can use the sum identity on that

[solar arch]

So I can split up 75 to whatever I want?

sin (69 + 1)
sin (50 + 25)

etc..

They all work?

[hard veldt]

well you would want it in something you know, and you (should) know sin45 and sin30

[quasi bronze]

Use degree to radian formlua

[hard veldt]

oh wait is it in rad or deg

[solar arch]

degrees

[hard veldt]

i assumed it was deg mb

[hard veldt]

oh ok

then its fine

just do sin(45 + 30)

[quasi bronze]

that will work to

[solar arch]

Okay thx

[hard veldt]

that should work, but you would need a calulator to evaluate

[solar arch]

I got the anwser hbankst

[solar arch]

How do I solve sec(-15)
degrees

[hard veldt]

sec(-15) is 1/cos(-15)

cos(-15) is cos(15), because cos(-x) = cos(x)

so we can simplify sec(-15) to 1/cos(15)

[hard veldt]

do you know half angle or angle subtraction formulas?

[solar arch]

Can i also change -15 to 345?

[hard veldt]

if you want to, but personally, I like smaller numbers

[solar arch]

Is this because the reference angle is always positive?

How do I then solve cos(345) because I need it in exact value

[hard veldt]

use angle sum

[solar arch]

So what would I turn it into?

[hard veldt]

cos(300 + 45)

[solar arch]

How do I know what it's supposed to turn into?

[hard veldt]

just guess values, and check if you can evaluate them

you can evaluate every multiple of 45, and every multiple of 30

without a calculator, that is

[solar arch]

Got it than ks

[hard veldt]

Remember to take reciprocal, because we are evaluation sec, not cos

[full forum]

@solar arch Has your question been resolved?

[full forum]

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[vivid grove]

Hello guys can somebody explain the procedures on this?

The picture on the left was is in a youtube channel, while on the right is when I try to solve it

[silent wagon]

!xy

[full forum]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[silent wagon]

(just for best measures)

[vivid grove]

I dont understand how did it got solved

This is the original problem

!xy

[full forum]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

@vivid grove Has your question been resolved?

[vivid grove]

nvm i already got the answer

[full forum]

@vivid grove Has your question been resolved?

[full forum]

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[torn jolt]

hello can someone pks help

[torn jolt]

!show

[full forum]

Show your work, and if possible, explain where you are stuck.

[torn jolt]

ok

show the original problem as well

I don’t have any work

All I need is part B

[full forum]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[torn jolt]

Actually no I don’t understand either

[lime ether]

i just helped u?

[torn jolt]

Yeah how tho

I said I’m confused

And u left

[lime ether]

u still had questions?

[torn jolt]

U never answered my first one

[dim compass]

Well that was my work

I wrote !help apparently, never mind

[lime ether]

what didn’t u understand

[torn jolt]

The steps u have me

Gabe*

They didn’t work

[lime ether]

we’re adding so we want a common denominator

[torn jolt]

Yeah I tried that

[lime ether]

what went wrong

[torn jolt]

I ended with 6(y-4)(y+4)

[lime ether]

the common denominator should be 6*(y-4)^2 * (y+4)

[dim compass]

We do LCM of denominator when we want a common denominator

[lime ether]

ur missing the y-4

that would just be the left denominator

the right has two (y-4)

[torn jolt]

But where did the square come from

[lime ether]

the right denominator

[torn jolt]

Oh nvm

Yeah

Ok

But then the numerator

Didn’t make sense for me

[lime ether]

y^2-16=(y-4)(y+4)

notice how there’s only one y-4

[torn jolt]

Yeah I realize that now

[lime ether]

so we need to multiply the whole fraction by (y-4)/(y-4)

and the right fraction has a 2 in the denominator but the left has a 6

so the get it to be 6

we multiply top and bottom by 3/3

additionally the right doesn’t have a (y+4) which the left does

[dim compass]

We manipulate the expression to have common denominator

[torn jolt]

So far I did allat

[lime ether]

so multiply by (y+4)/(y+4)

did u multiply top and bottom by what i said?

then just distribute

the numerator

and add

[lime ether]

combing like terms

ur work is hard to follow

[torn jolt]

Sorry

[lime ether]

u should get 21y(y+4)

[torn jolt]

It’s LIEK specialy for me

[lime ether]

for the right

[torn jolt]

LIEK a curse

[torn jolt]

IDID

[dim compass]

What is liek

[torn jolt]

I DID

[lime ether]

=21y^2+84y

[torn jolt]

Misspelled like

[dim compass]

Ok

[lime ether]

the left is y^2-4y

[torn jolt]

Yes

[lime ether]

then just add

since they have a common denominator

[torn jolt]

Ok lemme try

I got 22y^2+80y

Holy shit then u divide by two from the top and bottom

[sharp surge]

waht

[torn jolt]

Then factor Y

[torn jolt]

Yo wsp why u here

[sharp surge]

im looking

i saw a math qs i got ocnused in too

CONFUSED

[torn jolt]

Yeah but like I’m still trying to confirm

[lime ether]

r u looking for help

cloud

[torn jolt]

I think I got it right

[sharp surge]

mmm

[lime ether]

^

yes

[sharp surge]

kinda on the same situation w batata

[lime ether]

open a new help channel

[torn jolt]

Yippee

[sharp surge]

nono its fine

[torn jolt]

I think it’s a rule

[sharp surge]

i can just look over here anyway 😭

WAHT

[torn jolt]

Or smthn

[sharp surge]

THATS A RULE?

[torn jolt]

Idk

[lime ether]

yea

[sharp surge]

damn

[lime ether]

occupied!

or

wait

[torn jolt]

See ya later

[lime ether]

!occupied

[full forum]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[torn jolt]

My last questions

Of the day

Cause honestly I’m getting tired

[lime ether]

oh wait

this is a quiz?

[torn jolt]

More like a practice quiz

I swear I’m not cheating

[sharp surge]

mm

LMDOSOA

[lime ether]

y does it say practice quiz

<@&268886789983436800>

what do u guys think

[sharp surge]

huh

[torn jolt]

Cause we are practicing for a quiz

[sharp surge]

yeah

its not hte same thing tbh

[lime ether]

not practice it says online quiz

[calm nacelle]

If it's practice then it seems okay

[sharp surge]

diff numbers and such

[lime ether]

it says online quiz

[torn jolt]

Quiz is tomorrow

[sharp surge]

because its being done online??

[spiral vigil]

i think this is clearly practice, but also stop posting your name onlinr

[sharp surge]

our system of quizzes are quite different

[torn jolt]

Oh shit I poseyed her name

[sharp surge]

omg

BROBRO

DELETE IT

[torn jolt]

Yeah no I’m cropping hold on

[sharp surge]

good

[lime ether]

do y’all know each other

[sharp surge]

uhm

wdymm no

🧍‍♀️

[lime ether]

cloud and batata

yea

[sharp surge]

nuh uh

[lime ether]

yes u do

[torn jolt]

Bro u can’t lie

[sharp surge]

HOW DO YOU KNOW

🧍‍♀️

gagi joke lang

hsdha

joke joke dw

[torn jolt]

We u already said OUR school system lol

[sharp surge]

wdymmm

🐴

wait waht

we you

[torn jolt]

Aight now I actually need help

[lime ether]

yea

[sharp surge]

staker

stalker

💢

[lime ether]

also y would u be chillin here

[torn jolt]

So shoo shoo

[lime ether]

u caught me

[sharp surge]

eY?

im looking over too man

cmon

[lime ether]

batatA

[sharp surge]

it sthe same matieral honestly

[torn jolt]

Yeah we have the same material

[lime ether]

what do u think the first step is

[torn jolt]

Finding the LCD?

[lime ether]

yup

[torn jolt]

They both have 30 as a common multiple

[lime ether]

yes

what about the x terms

[torn jolt]

Left side has x-3

And right has x+1

[lime ether]

so

…

[torn jolt]

So I multiply left side by 3 and x+1

[lime ether]

yup

[torn jolt]

And the right side by 2 and x-3

[lime ether]

yup

[sharp surge]

ohh

[torn jolt]

Yakhi this gonna take me a year

Hold on

[sharp surge]

lmao

i think i kinda get it la

my teach kinda gives us a quicker way 😭

[lime ether]

blow out denominator?

[torn jolt]

Huh?

[lime ether]

u don’t want to do that here because N/M

[lime ether]

i’m guessing that’s the method

where u multiply everything by the common denominator

it does the same thing

but

if u do that

the right hand side

wont be N/M

so don’t do that

just do it my way

[sharp surge]

wait batata is the quiz mostly like denominator related?

like N/M

or not

[torn jolt]

Yeah I think it will be

[lime ether]

cloud who’s better

[sharp surge]

ahh okay

what

[lime ether]

u or batata

[torn jolt]

So pray cause we are finished

[sharp surge]

im clealry betetr

HDAHSDHA

kidding

hes better

[torn jolt]

Nah id win

[sharp surge]

yeah

im lowkey mediocre

[lime ether]

unfortunate

[lime ether]

true

[sharp surge]

xd

[lime ether]

i’m kidding

[sharp surge]

ik its alright lmoa

[torn jolt]

HAHAHAH

[lime ether]

highkey*

💀

[sharp surge]

oo kitty cat

[lime ether]

my fault

[torn jolt]

One sec

I can cancel the numeration and denomination

[sharp surge]

oh hey i got the 30 part

but no clue abt the others

[torn jolt]

Bro I got 1764 💀

[lime ether]

i didn’t do it

[sharp surge]

i got -360...

thing is

2-3 man

its -1

then after you have to multiply it no?

i guess im just guessing..

[lime ether]

(5x-1)*2(x-3)

[sharp surge]

eh?

[lime ether]

(3x+1)*3(x+1)

=(5x^2-16x+3)*2

[sharp surge]

ahh i think our methods are quite different

same answer tho

[lime ether]

(3x^2+4x+2)*3

so

[sharp surge]

shucks

[lime ether]

10x^2-32x+6

and

[sharp surge]

even the exam time is gonna drain

isnirfs

[lime ether]

9x^2+12x+6

subtracting the 10x^2 from the 9x^2

u get -x^2+44x

so (-x^2+44x)/(30)(x+2)(x-3)(x+1)

if x=2

N=-x^2+44x

=-4+88

=84

how tf did u get 1764

[torn jolt]

In actually tweaking

[lime ether]

no cap

[sharp surge]

DAWG

ITS 81

NOT 84

IM LAGGIN BRO

[lime ether]

fr?

[sharp surge]

FR

[lime ether]

lemme redo my arithmetic

[sharp surge]

bro

[torn jolt]

BOMBOCLAT

[sharp surge]

HUHU WERE SO SCREWED

jk im not

[lime ether]

i was doing it in my head

[sharp surge]

im dropping

xddsdadas

[torn jolt]

Hey it’s only one question

[sharp surge]

bro

[lime ether]

was it fr 81?

[sharp surge]

thers 27 questions..

YES

idk abt hte shapes ones

[lime ether]

it’s an practice quiz tho…

[sharp surge]

i might ask friends

[torn jolt]

I bet they’re gonna bring 20 of this question In The real exam

Cause our school hates us dawg

[sharp surge]

bro

chem was already enough for me..

8 qs

[torn jolt]

Ok lets forget this qs

[sharp surge]

got me on a chokehold

[torn jolt]

This don’t make sense either

[sharp surge]

TSTOP

IDK EITHER

IM SAD AND SCARED RN

[torn jolt]

Yeah no so far three question I can’t understand

But my maths teacher is gonna do this tomorrow

So hopefully he does it well

[sharp surge]

aw shucked

wait no

isnt ur math teacher good

[torn jolt]

Quite the opposite

[sharp surge]

mines lowkey just becoming the next solar panel

💀

[torn jolt]

What?

[sharp surge]

ahh yes taht guys

[sharp surge]

hes bald in context 😭

[torn jolt]

Oh Mr A….

[scenic thistle]

happy martin luther king junior day

[sharp surge]

HAHSHA YES

ah yes

[torn jolt]

Less go

[sharp surge]

martin luther king junior

[sharp surge]

mr K no?

[scenic thistle]

yes

[sharp surge]

manman

[torn jolt]

He’s good he teaches well

[sharp surge]

i heard he just sucks

[torn jolt]

Yeah

[sharp surge]

💀

[sharp surge]

kinda if you have krip in your class..

so loud

hdsaudhsu

[torn jolt]

Not just that

He actually teaches well

[sharp surge]

AND AL

BRO AL

:(

HE DOES

ITS JUST THE PPL IN MY CLASS THAT MAKE IT WORSE LA

[torn jolt]

Damn

Anyways where did knief go

[sharp surge]

RIGHT

wehres knife

knifeee

@lime ether

we need you bro

[torn jolt]

Ur our only hope

[sharp surge]

oh i actuall love that emoji wahtt

hm

omg

[torn jolt]

This is just becoming dms

Imma close this off and befriend knief cause he was cool

[sharp surge]

oo i wnat too aswell

[torn jolt]

Talk in dms

.close

[full forum]

Channel closed

Closed by @coarse crater

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[torn jolt]

countably infinite for a set X is f: N -> X right

[torn jolt]

and countably finite for a set X is f: [n] -> X right

the difference is one is a subset of the natural numbers and one is the entirety of natural numbers

since N is infinite, but is countable, if you find a bijection from N to some set X, it is then countably infinite

but since [n] is countable and finite, then a bijection onto a set X must be countably finite right

[full forum]

@torn jolt Has your question been resolved?

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[tropic wedge]

This should be done using the law of cosines but I got no idea how to apply it

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@tropic wedge Has your question been resolved?

[inner shale]

is it required to solve by law of cosines? because vector solution is quite easy

[tropic wedge]

well it's an exercise in a subchapter of law of sine and cosine

so idk if it matters that much

[tropic wedge]

what is the vector solution for this?

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@tropic wedge Has your question been resolved?

[tropic wedge]

<@&286206848099549185>

[wanton cedar]

yes

what do you need help with?

[tropic wedge]

with this exercise

[tropic wedge]

one above

[tropic wedge]

.close

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[rocky smelt]

aa

[rocky smelt]

surface area 😭

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@rocky smelt Has your question been resolved?

[gritty rose]

https://www.cuemath.com/measurement/surface-area-of-square-pyramid/

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[torn jolt]

Ik this is not math

But it’s ab balancing chemical reactions

What the hell should I start with?

Ik that I should leave O and H for last

But the rest

you cannot balance that PO4 like that 😭😭

THEN HOW

THEY BOTH HAVE CAPITAL LETTERS 😭

https://discord.gg/eexdsFw

chill out sweetie

lmao whyyy

why in math server

[undone vector]

It is pretty much math the chemistry is very elementary

[pulsar sigil]

[torn jolt]

Then tell me how to do it if it’s math math is like understandable but that is just a bunch of weird letters

[torn jolt]

it creates a bad precedent. Thats the issue

[gritty rose]

slopes be slippery

[torn jolt]

but anyways people can help with the traction here

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@torn jolt Has your question been resolved?

[undone vector]

You want both sides to have equal amounts of N,H,P,O,Pb,and N.
We can rewrite this in math terms as a(3N+12H+P+4O)+b(Pb+4N+12O)=c(3Pb+4P+16O)+d(N+4H+N+3O) where a,b,c,d are positive integers

You can create a system of equations from this and solve it from there

each variable (a,b,c,d) is correlated to one molecule

btw you cannot put the balancing numbers in the middle of a molecule like you did with with (NH4)3PO4

[torn jolt]

Wait I’m a bit confused

How should I solve the equation with that many variables? I’m not on that level yet 😭

[undone vector]

You will just need to do a little guess and checking. Make a guess for one of the equations and use that guess to solve the other equations and if everything comes out as an integer then you are good.

You kinda already did that by writing 3 infront of Pb(NO3)4 to get 3Pb(NO3)4. You probably did this because you saw there was 3Pb s in the first molecule on the right side. This guess is actually correct, and by plugging in these values to the other equations you will be able to get the answer

[torn jolt]

OMG I DID IT

I THINK

AYUAAHHAHA

[undone vector]

yeah thats correct

[torn jolt]

Thank youuu

Took me like half an hour though lmao

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[waxen flicker]

Hi!

I'm trying to 'solve' the following integral:
$$
\int \frac{3 \cos(x)+9 \sin(x)+11}{6 \cos(x)+3 \sin(x)+7} , dx
$$

Wolfram did it, and so it seems at least possible.. Tho my attempts didn't give anything: \

• I tried the common technique of completing the numerator to cancel it out with the denominator, but it didn't become simpler. \
• Also, the expression that Wolfram has found involves $\cos(\frac{x}{2})$, so I also tried to put it somewhere but unsuccessfully. 😦

[glossy valve]

Sweet Tea 🧋

[waxen flicker]

,w integrate (3cos(x)+9sin(x)+11)/(6cos(x)+3sin(x)+7) dx

[glossy valve]

Results provided by WolframAlpha

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

Requested by sweet.tea3

[waxen flicker]

Tho I think I will still leave this here:

\begin{align*}
\int \frac{3 \cos(x)+9 \sin(x)+11}{6 \cos(x)+3 \sin(x)+7} , dx = \cancelto{x}{\int \frac{6 \cos(x) + 3 \sin(x) + 7}{6 \cos(x)+3 \sin(x)+7} , dx} + \int \frac{-3 \cos(x) + 6 \sin(x) + 4}{6 \cos(x)+3 \sin(x)+7} , dx
\end{align*}

Since Wolfram's expression also includes the $x$ term

[glossy valve]

Sweet Tea 🧋

[waxen flicker]

Ok, one more insight: by playing around with Wolfram I think the last expression should be decomposed into the following two:

\begin{align*}
\int \frac{-3 \cos(x) + 6 \sin(x)}{6 \cos(x)+3 \sin(x)+7} , dx + \int \frac{
4}{6 \cos(x)+3 \sin(x)+7} , dx
\end{align*}

[glossy valve]

Sweet Tea 🧋

[waxen flicker]

Because if we try just the cos:

,w integrate (-3cos(x))/(6cos(x)+3sin(x)+7) dx

[glossy valve]

Results provided by WolframAlpha

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

Requested by sweet.tea3

[waxen flicker]

it gives nonsense

[wise gate]

And here the first integral is $\log(6\cos(x)+3\sin(x)+7)$

[waxen flicker]

wait, what

[glossy valve]

d

[waxen flicker]

on the website the answer was different lmao

[wise gate]

Or -log(6cos(x)+3sin(x)+7) maybe

[waxen flicker]

hm, why?

[waxen flicker]

I mean, yea, they look very similar: except that the coefficients are swapped 😦

[wise gate]

Because the numerator is the derivative of the denominator

[brisk obsidian]

You won't like how Wolfram solved it.

[waxen flicker]

ohhhhhhhhhhhhhhhh

[wise gate]

Also @waxen flicker you can always use the t=tan(x/2) substitution

[idle kelp]

My instinct is to use a weirstrauss substitution and partial fractions, but I am definitely too lazy to try it myself

What they said

[waxen flicker]

as far the w-sub: we definetly didn't learn that yet, so I would like to practice other methods

[waxen flicker]

hmm, and we didn't learn that either – only the u-sub in its most general form:

$$
\int f(x) , d(g(x)) = f(x)g(x) - \int g(x) , d(f(x))
$$

[glossy valve]

Sweet Tea 🧋

[waxen flicker]

tho was that abt the first ingeral or the second? (after I've split the initial one into two)

(mb u suggested an alternative solution to the first, idk, so just to be sure)

[wise gate]

The second, but it actually applies to any integral with sums, products and quotients of sin(x) and cos(x)

[umbral dome]

for reference this is usually called integration by parts

[waxen flicker]

ouch, yea, I didn't have much sleep xD

I meant the following:
$$
\int f(x) , dx = F(x)+C \implies \int f(g(x)) , d(g(x)) = F(g(x))+C
$$

[glossy valve]

Sweet Tea 🧋

[wise gate]

Then use this, with g(x)=tan(x/2)

[waxen flicker]

hmm, lemme try. It's new to me

tho why with $\tan (\frac{x}{2})$ and not just $\tan (x)$?

[glossy valve]

Sweet Tea 🧋

[waxen flicker]

or am I missing smth

[wise gate]

Well, if you use tan(x), you obtain an easier integral, but it might work, and it might not work as well. There is a rule to know when it works but I always forget it

However tan(x/2) always works

[wise gate]

At the cost of obtaining something that might be more complicated

[waxen flicker]

ok, so, well, we need to factor out $\frac{1}{\cos^2{x/2}}$ from that, right?

[glossy valve]

Sweet Tea 🧋

[waxen flicker]

1/2 times that technically

if I'm getting the idea of it right

[wise gate]

Oh wait no

[wise gate]

This is wrong

The inverse of that

[waxen flicker]

u mean arctan or 1/tan?

[wise gate]

I mean g^-1(x)=tan(x/2)
g(x) = 2 arctan(x)

And then prove and use these

Or the first two

[waxen flicker]

hmmm, so we get

$$
\frac{4 (1 + \tan^2(x/2))}{13 + 6 \tan(x/2) + \tan^2 (x/2)}
$$

[glossy valve]

Sweet Tea 🧋

[waxen flicker]

but $g(x)=2\arctan(x)$.... We cant do the u-sub..

[glossy valve]

Sweet Tea 🧋

[wise gate]

Why not?

[waxen flicker]

hmm, I've watched a video about it and I think I seem to get it now

We hence have the following:
\begin{align*}
\int \frac{
4}{6 \cos(x)+3 \sin(x)+7} , dx = \int \frac{4 \cancel{(1 + t^2)}}{13 + 6 t + t^2} \cdot \frac{2}{\cancel{t^2+1}} , dt = 8 \int \frac{1}{t^2+6t+13} , dt
\end{align*}

[glossy valve]

Sweet Tea 🧋

[waxen flicker]

What to do about the quadratic tho? It doesn't have real roots

so no partial fractions

[wise gate]

Well

[waxen flicker]

wait, lemme try to play a bit with it

mb something comes up

[wise gate]

I will also try to remember how this works