help-10

[winter cargo]

finally! 🥳

[brave bramble]

Unless I'm wrong and there is a double count, that's the number of committees that don't contain all 3 teams.

[winter cargo]

that was HARD

[winter cargo]

no you are right

[near pier]

Wow thank you

[brave bramble]

I'm not 100% confident lol

What's that as a decimal?

[winter cargo]

,calc 12091275/25621596

[warm shale]

Result:

0.47191732318315

[winter cargo]

almost 0.5

[brave bramble]

About half chance, neat

[winter cargo]

nice

[near pier]

noice

[winter cargo]

there should be a more elegant solution though 🤔

[near pier]

that would be solving it using the table

so the result seems to be correct

[winter cargo]

this is the simplified ans

[distant moth]

i thought about this too

[winter cargo]

hmmmmmmmmmmmmmmm

[distant moth]

but it didnt match the table way

as in the two answers are coming diff?

oh wait

they are the same!

I think the total- method is better

[winter cargo]

so we found the answer!

[distant moth]

But why is wolfram alpha so weird

[winter cargo]

(at least 0.6 sure)

[distant moth]

This was what it showed me when i tried the total- myself

Like is it a typing mistake

[near pier]

I don‘t know exactly how the total method works, but should it be 45 instead of 46 for the first one

[distant moth]

Oh wait maybe a counting mistake

Oh yes

It should be 36

[distant moth]

Basically p(at least one of each sport)= 1- p(F' union V' union H')

[near pier]

is it correct now?

[distant moth]

Yea

[near pier]

Ty all, I also ran a test using python to verify the results it is correct

.close

[obtuse pebble]

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[whole summit]

need help with this, it's calculus 1 ( Integration )

[tardy epoch]

What's your question

[whole summit]

So i understand the basic parts

so i know why he moved the lower part up and changed the sign of the power

but after that idk where the 1/2 came from

[timid silo]

Consider making a substitution of $u = x^2 -4x -1$

[warm shale]

♡A(lex)♡

[whole summit]

oh i can solve this in the u and du thing?

[timid silo]

yes

you can indeed solve it using that thing

[whole summit]

U substitution

but which is gonna be U and which is gonna be du

[timid silo]

well i already kinda told u

just take the derivative of that

[obtuse pebble]

@whole summit Has your question been resolved?

[obtuse pebble]

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[fickle badge]

Simple thing but it can't seem to get in my head, can someone explain this to me. Especially the quadratic part

[royal basin]

what's the goal? find the values of x and y?

[fickle badge]

Yes

[royal basin]

are you unable to even start, or do you get stuck somewhere midway?

[fickle badge]

Wait let me show where I'm stuck

I'm stuck at the quadratic part

[royal basin]

yes, show your work up to & including the point you got stuck at.

[fickle badge]

Ignore the last Line, well that's where I got stuck

[royal basin]

ok, i can tell you exactly where you went wrong already.

first off, these ^2's don't belong here

and second, (3-x)^2 ≠ 3^2 - x^2. exponents do NOT distribute over addition (nor subtraction).

[fickle badge]

Ohhhh

x^2 + (3-x)(3-x) = 5 then

[royal basin]

yeah

you could either recall the identity (a-b)^2 = a^2 - 2ab + b^2 or rederive it on the spot if you can't.

[fickle badge]

Alright noted

[fickle badge]

Can I get the rest?

[royal basin]

wdym "get the rest"?

[fickle badge]

Solution

Please

[supple granite]

you can't just ask for the solution

[fickle badge]

Mm

Alright I'll figure it out

[fickle badge]

Thanks ann

[royal basin]

keep going from here and tell us where you get stuck.

[obtuse pebble]

@fickle badge Has your question been resolved?

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[steel pasture]

$\lim_{x \to 0}\frac{\sin^2 x}{x}$

[warm shale]

Willow

[steel pasture]

$\frac{\sin x}{x} \cdot \sin$

[warm shale]

Willow

[royal basin]

sin(x), not just "sin"

[steel pasture]

where does the second x come from in the numerator here:

https://www.calcchat.com/book/Calculus-11e/1/3/67/

[royal basin]

$\sin^2(x)$ means $(\sin(x))^2$

[warm shale]

Ann

[royal basin]

$\sin x$ is not, and should not be treated as, the product of ``sin'' and $x$.

[warm shale]

Ann

[steel pasture]

oh. why is this?

what rule is this?

[royal basin]

it's just a notational convention.

[steel pasture]

anywhere I can read about it?

[royal basin]

no

[steel pasture]

or is it just something I have to know and memorize?

[royal basin]

there is not much to read about

[deft sun]

isn't sin a function?

[royal basin]

indeed it is.

[deft sun]

so of course it means nothing alone

[vast wren]

Do you know about the trigonometric squeeze/sandwich theorem

in order to eliminate the 0/0 uncertainty

[steel pasture]

we went over it in class today here:

but I don't understand what it means

[vast wren]

Basically

1 <= x <= 1

[steel pasture]

oh wait

[vast wren]

x has to be = 1

[steel pasture]

nevermind, the meaning is at the bottom

those three limit equivalences are the meaning of the squeeze theorum

but what does "open interval" mean in my teacher's definition above?

[royal basin]

do you know what an interval is?

and are you familiar with interval notation?

[vast wren]

R - {c}

buh

[steel pasture]

yes, interval notation vs set builder notation

[royal basin]

right, so you know what notations such as [a, b) or (a, b) or [a, b] mean?

[steel pasture]

I think so, yes

[royal basin]

an open interval is an interval of the form (a, b).

i.e. it is an interval which does not include its endpoints

[steel pasture]

closed interval = [a,b] ?

and what about [a,b) ?

[deft sun]

it contains a but not b

[royal basin]

i've seen such intervals be called half-open

[steel pasture]

I mean, the name

ok

ty

.close

[obtuse pebble]

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[timid silo]

if you have the domain, then just check the behaviour of the function over the interval the domain lies in

like if the domain is something like (3, infinity), see what happens as the function approaches 3 and infinity

should be mainly intuition then

[zenith raft]

computing ranges for even just real domain and codomain functions can be hard

[timid silo]

again this goes back to my reasoning earlier

if you have the domain of (-inf, 2] then check the behaviour of the function as it approaches those endpoints

at x = 2, clearly you will have f(2) = 8

as x approaches minus infinity, the function will just grow without bounds in the negative direction because you have 8 - something huge

so ultimately (-inf, 8] is ur answer for the range

yeah i mean thats how i like to think of it

although like layla said, ranges can get reaaaaaaaaallly complicated

basically what i said but more rigorously i guess lmao

[warm canopy]

Yeah in general a function doesn't have to achieve all points between the values it takes at its end points

[timid silo]

well you are only approaching 2 from the left because your domain is (-inf, 2] first of all

also i use this just for the sake of intuition

the method here is more clear

[zenith raft]

zac

[timid silo]

lmaoo

a square root of whatever has to be greater or equal to 0

the rest of the steps is just them multiplying and adding things to it

you are trying to check the bounds of your function by constructing the inequality starting from something you know already

[zenith raft]

every question that asks you to compute a range requires different analysis facts about different functions

[timid silo]

^ this tbh

you just cant memorise with stuff like ranges lmaoo, you just have to use your intuition and logic behind the functions

[zenith raft]

lmaooo

[timid silo]

quite the analogy there

lmao

[trail cloak]

Sure

Remember the straight line slope

[timid silo]

oh god limit definition bs

[trail cloak]

$m=\frac{f(x_2) -f(x_1)}{x_2 -x_1}$

[warm shale]

VulcanOne

[trail cloak]

Here instead of f(x2) and f(x1)

No

I'm helping you connect the dots

Instead of f(x2) and f(x1), we use f(x+h) and f(x), and h is a really really small amount

$f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h) -f(x)}{h}$

[warm shale]

VulcanOne

[trail cloak]

The difference in x's is the really really small amount h

Find 4(x+h) -6 first

Then do 4(x+h)-6 -(4x-6)

Then apply the limit definition

Distribution

4x+4h -6

Now we find the difference

4x+4h-6 -4x+6

4x cancels -4x and 6 cancels -6

Yepp

Okay

Derivatives

Yeah

Then make your derivatives equal 0

Did you differentiate x^3 properly?

Set both your derivatives to 0 and solve for x and y

You will get 2 values for x and one value for y

You will replace the negative value of x and the y value you got once, and the positive value of x and the same y value another time in your function

What x and y values did you get?

Yes they are corrext

Sorry my mom wanted me to help with something

You will plug in

(-1,2) in f(x,y)

And

(1,2) in f(x,y)

You will get a minimum

And a maximum

When you plug your points in, you will see

One will be larger than the other

In terms of z value

Hmm

[timid silo]

what

just

take partial derivatives

and use the discrimnant

?

[timid silo]

yeah just use the det of the hessian

whats the problem exactly

[trail cloak]

https://cdn.discordapp.com/attachments/903430332324405288/1063418020665499648/image.png

[timid silo]

yeah just plug and chug i think

oh

okay what i said wasnt important lmao nvm

wait did i remove it

pain okay

linear approximation formula?

yeah sure but what u r saying is all vague

just tells u an approximation of the original point on the function innit

[obtuse pebble]

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[viral karma]

ABCD is a deltoid
length of AF = BF
DG = GC
AC = 18
BD = 24

it asks the length of FG

[trail cloak]

Deltoid means rhombus?

[viral karma]

no

kite? same thing I guess

[trail cloak]

Oh

[trail cloak]

Kite

Okay

Given that this is a kite

$AC \perp BD$

[warm shale]

VulcanOne

[trail cloak]

$BE = ED = \frac 1 2 BD$

[warm shale]

VulcanOne

[obtuse pebble]

@viral karma Has your question been resolved?

[winter cargo]

do you know what "midpoint theorem" is?

this question uses that idea

[viral karma]

yes I know that

can't figure out how to apply that here tho

[obtuse pebble]

@viral karma Has your question been resolved?

[obtuse pebble]

@viral karma Has your question been resolved?

[obtuse pebble]

@viral karma Has your question been resolved?

[earnest arrow]

Can somebody help me

sorry if i wrote it in a wrong channel

[graceful scaffold]

S= 1/2 (b.c.sinA)

[obtuse pebble]

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[dim egret]

Can someone help me prove that the one on the top is equal to the one on the bottom? I tried my best but I couldn't prove it.

[dim egret]

I need to simplify the first one to get the second one*

.close

[obtuse pebble]

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[delicate salmon]

Can someone help me to compute the limit with l'hopital's rule

[dark stirrup]

https://tenor.com/view/log-power-fire-burning-mad-gif-16474420

[timid silo]

https://cdn.discordapp.com/emojis/1063559604090454116.webp?size=40&quality=lossless

[dark stirrup]

I'll teach you the method I learned

That should get you there.

[obtuse pebble]

@delicate salmon Has your question been resolved?

[trail cloak]

🪵

[delicate salmon]

I know, i am stuck here, can't get a whole number after replacing the x in the derivatives

[dark stirrup]

I don't get what's happening here

[delicate salmon]

i tried getting derivatives of both with l hopital rule, i switchec hem into a fraction so i indeterminate form

[dark stirrup]

This step looks good

But i have no idea what this is

[delicate salmon]

that's the doble fraction after i got the derrivatives

the result is the end one

that i circled

Am i doing good or"

[dark stirrup]

Nah I don't think so

Also, it would help if it weren't cut off

I'm missing context

[delicate salmon]

Can you guide me to the first steps :/

[dark stirrup]

Take a better picture so I can see the cutoff parts

[delicate salmon]

I solved it' i have messed up the deivative of cot(-pix) ..

Thanks tho xD

.close

[obtuse pebble]

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[wooden cipher]

it needs to go through the center

[obtuse pebble]

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[wooden cipher]

??

[undone sparrow]

A sphere it's center (2,-1,-2) it's radius 3 is placed on the placed 2x+6y-3z+k=0

[undone sparrow]

Sorry I thought my connection was screwed up because I didn't see the bot pin the message then noticed that there was a forum so I was about to post it there and delete the question here

[wooden cipher]

you need to get a new channel then

[undone sparrow]

Oh

[obtuse pebble]

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[trim thistle]

Hello, in a calculus exam with a very strict teacher I got a point decrease because in the exam I wrote something along the lines of x * -3 = .... which lead to a grade decrease. I got a "notation error point decrease" because according to the teacher its a mistake to not use parantheses in this situation so I have to write it like x * (-3) = .... And, yes I am aware that using parantheses is conventional, I just happened to forget because of time. However, I wanted to ask whether it is true that is actually a formal mistake in mathematical notation since I dont see how the parantheses actually change anything besides it looking more clear

[trail cloak]

It can look confusing

[trim thistle]

sorry, i know its a stupid question for this server

[trail cloak]

$x \cdot -2$

[warm shale]

VulcanOne

[trail cloak]

What if you forgot that dot/the multiplication sign

Is it still x*(-2)?

Also

Easier thing to do

Just write -2x

Clearer and easier and more conventional

[trim thistle]

yes i know and i do but my question is more

now that i already this time wrote it like that

is it a mistake?

or a notational error?

[trail cloak]

It can be acceptable in some cases

But generally more confusing to deal with

Notationally it's not wrong

It's just awkward

[trim thistle]

so uhm

[knotty crow]

I think it's relative and depend on a teacher, you can try asking, it won't do any harm

[trail cloak]

Not sure tho

[trim thistle]

i did

[trail cloak]

I would agree

[trim thistle]

he said no i will have your point deducted anyways because you wrote it in a non conventional way

[trail cloak]

Personally I wouldn't deduct marks if I see something like that but I would make sure that I teach the proper and most accepted notation before hand

[trim thistle]

but im more confused on your answers than i was before the answers: relative and acceptable in some cases? I thought math was like.... definitive right or wrong

and if its not wrong then no point deduction

[fervent nacelle]

That seems so unnecessarily pedantic

[trail cloak]

Definitive right and wrong when you deal with problems and their solutions

However, representing solutions can be relative and depending on conventions

[fervent nacelle]

Like yeah its technically x * (-1) *3 but... Jesus christ, a decrease in score for that?

[trail cloak]

Yeah it's weird

I personally won't deduct anything on that

But it's just super harsh

[knotty crow]

you know... imagine you are a teacher and you see someone did the right notation hundred times and this one time he/she did it wrong, I mean teacher can presume this person knows what he/she does

[trim thistle]

wait so it is a mistake?

[trail cloak]

It's not

[fervent nacelle]

Eh.... one could argue -3 is a number like any other

[trail cloak]

Notationally it's acceptable

[fervent nacelle]

Definitely not a mistake
A conventional error at best

[trail cloak]

And you put a multiplication sign before the negative sign so you made it clearer

[trim thistle]

well oh well

the teacher was pretty pisssed off for me even asking actually

but oh well rip ill just try to not do any non conventional notations next time

like he also deduced me a point for forgetting "dx" once but that I can actually understand cause its an actual mistake

and RIP my A+

thanks guys for clearing up the confusion

[trail cloak]

Hope the teacher leans up and gives you an A+

[trim thistle]

thanks

[obtuse pebble]

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[keen spoke]

How to use the sin cose rule when the complete c and complete b is not given

[static beacon]

cosine rule

a^2 = b^2 + c^2 -2bc cos(A)

you can than find all the angles in the upper triangle

than doin 180 - u can find most angles in the quadrilaterial

[keen spoke]

Ok

[keen spoke]

I found the all angles

[static beacon]

in the triangle?

[keen spoke]

Upper one

ADE

[static beacon]

ok

what'd u get for it

and find the othr angles within the upper triangle

[keen spoke]

Now what

[static beacon]

Ya

Now u can work out the other angles

By doing 180- the angle

Coz it’s on a straight line

[keen spoke]

Ok

[obtuse pebble]

@keen spoke Has your question been resolved?

[obtuse pebble]

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[coarse condor]

x^2/4 + (4/2√ 7)^2 = 1^2

[coarse condor]

so far ive done

ive solved up to

[zenith raft]

this is not written very clearly

and is it really supposed to be 1^2?

[coarse condor]

yeah

[zenith raft]

lol ok weird

[coarse condor]

lemme see if i can write it bettwe

[zenith raft]

it's also good to take a picture of the whole problem if possible 🙂

[coarse condor]

alr

[fierce lagoon]

Holy shit Dr Sins is also my mom's personal trainer!!

[fierce lagoon]

I would not try to rationalize

Lemme see

Uh

[zenith raft]

you can rewrite this as x^2 = something

[fierce lagoon]

The second to last line should be = 7

Actually hold on

No

Subtract 4/7 from 1

Then multiply by 4

[zenith raft]

oh i see that's you were trying to do loll

[coarse condor]

which'd be the better solution then

so x^2 = 12/7

?

I got a final answer of 2√ 21/7

<@&286206848099549185>

[lilac finch]

sir big fan

why u wan't help

[coarse condor]

i gotta know if its right

and i need help on a second problem

<@&286206848099549185>

[tardy epoch]

plug your answer into your original equation to verify

[coarse condor]

yeah i did that i just need help on something else now

@tardy epoch

[tardy epoch]

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

[coarse condor]

thats the problem

i dony know where to start

i know i should probably simplify firsy

but how would i simpligfy √6x^4

[tardy epoch]

$(ax)^b = a^b x^b$

[warm shale]

riemann

[tardy epoch]

$\sqrt{y} = y^{1/2}$

[warm shale]

riemann

[obtuse pebble]

@coarse condor Has your question been resolved?

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[coarse condor]

@tardy epoch i dont get it

[coarse condor]

i get this i think

.reopen

[obtuse pebble]

✅

[coarse condor]

√6 * √x^4 + 4√5 = 5√6 * √x^4 + 4√5 = 56 * √6 * √x^4 + 4√5 = 5x^2

simplifies down to

√6 * X + 4√6 = 5√6 * X

<@&286206848099549185>

[long dove]

You can start to find the value of x now?

[coarse condor]

but how would i do that part

i tried to do that at first

but i did something wrong

idk what

[long dove]

$\sqrt{6} x + 4 \sqrt{6} = 5 \sqrt{6} x$

[warm shale]

Welf

[long dove]

How bout now, can you see it?

[coarse condor]

so would it go to 4√6 = 4√6x

?

[long dove]

yes

[coarse condor]

so x is plus or minus 1

[long dove]

wait I think something missed

[coarse condor]

oh

[long dove]

$\sqrt{6x^{4}}=\sqrt{6} x^{2}$

[warm shale]

Welf

[coarse condor]

oh right

i forgot about the exponents

[long dove]

so, I guess now we have

[coarse condor]

didnt i simpligy it down?

what happened to the other two digits

oh ok

so whats the next move

i assume you simplify them further

[long dove]

If this is the one you get then

[long dove]

I think it just have the value of 1 since the degree of x is 1

[coarse condor]

oh

alright

so its solved then?

[long dove]

I just tried it

so

$\sqrt{6 x^{4}} + \sqrt{96} = \sqrt{150 x^{2}} \ \
\sqrt{6} x^{2} +4 \sqrt{6} = 5 \sqrt{6} x$

[warm shale]

Welf

[long dove]

the you can simplify and get the quadratic equation

[coarse condor]

would 6x^2 become 6x

[long dove]

nope

By my knowledge this far, it can't

[coarse condor]

oh no youre right

lemme see the quadratic equation

√6x^2 - 5√6x + 4√6 = 0

[long dove]

simplify it

[coarse condor]

how?

[long dove]

how bout this illustration

$a(x^{2}+x+c)=ax^{2}+ax+ac$

[warm shale]

Welf

[coarse condor]

ok

lemme see if i can get it

[long dove]

now equation:

$(\sqrt{6}) x^{2}-(\sqrt{6}) 5x + (\sqrt{6}) 4 = 0$

[coarse condor]

howd you get to that

[long dove]

umm ._. ?

owh I think I wrote it bad

[warm shale]

Welf

[coarse condor]

ok i see the equation

but im struggling to simplify it down

[long dove]

start by seeing the √6 as 'a' in the illustration

[long dove]

the illustration

well, since we already in the quadratic equation, you can find the solution by the quadratic formula

[coarse condor]

im trying to but im messing up and idk where

[long dove]

no worry

so

$(\sqrt{6}) x^{2}-(\sqrt{6}) 5x + (\sqrt{6}) 4 = 0 \
\sqrt{6} (x^{2}-5x+4) = 0 \ \$
divide both side with $\sqrt{6}$
$\ \ \frac{1} { \sqrt{6} } \times \sqrt{6} (x^{2}-5x+4) = \frac{1} {\sqrt{6}} \times 0 \
1 (x^{2}-5x+4) = 0 \
x^{2}-5x+4 = 0$

[warm shale]

Welf

[long dove]

there you go

[coarse condor]

i see

x=1 or x=4

thanks bro

.close

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[lavish swift]

Hi

[daring rock]

Hi, this channel is yours now. You can go ahead and post your question

[lavish swift]

I am puzzled on how the author manipulated (c) to (d)

[knotty crow]

Have you tried their hints?

[lavish swift]

Yes

[green epoch]

Sin and cos are complementary so arcsin+arccos is always π/2

[lavish swift]

I do not see how that helps set up the integral

My work so far

[obtuse pebble]

@lavish swift Has your question been resolved?

[lavish swift]

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[past hull]

hi

[past hull]

I need help writing a constraint line

basically the constraint limit is 500 calories, donuts count for 50 and coffee count for 0

would the line in that case be

50d + c = 500

?

[tardy epoch]

you left out the zero for calories of coffee

[past hull]

50d + 0c =500 so just 50d = 500, since we can consume infinitely many coffee without going over the calorie constraint?

[trail cloak]

Yes if coffee is 0 calories

[past hull]

ok thanks

.close

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[bronze folio]

can someone help me with this question? thanks a lot !!

[pastel meadow]

so first, do you know how to graph the equality y = 5?

[bronze folio]

yes !! i think so

[royal basin]

then do it and show us

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[frosty spoke]

A friend asked me, and I'm a little lost: is there a general method for solving systems of equations like:
x^2 + y^2 = a
x^3 + y^3 = b

or even things like:
x^2 + y^2 + z^2 = a
x^3 + y^3 + z^3 = b
x + y + z = c

[gleaming ridge]

These are called Diophantine equations (assuming you're restricting the solution spaces to integers) and there are no general method for solving such equations

[frosty spoke]

I hate discrete mathematics, so I'd never ask a question about integers (gross!)

what about in the continuous case?

[gleaming ridge]

I am not sure about the continuous case cause for one, you'll always be able to find approximate (numerical) solutions

Just plot the graph (ofc only works upto 2-3 variables)

[frosty spoke]

hmm well clearly some gradient based methods might work, but no analytical solutions?

,w solve x^2 + y^2 = a and x^3 + y^3 = b

[warm shale]

Results provided by WolframAlpha

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Requested by Saccharine#3780

[frosty spoke]

OH YIKES

what even is that

[gleaming ridge]

Looks like a differential form but never seen that # symbol

[frosty spoke]

it's some mathematica syntax I think

looks like it's finding roots of a bunch of polynomials

.close

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[scarlet talon]

Hello, I'd like someone to explain how to do this question.

[timid silo]

What makes a function differentiable at a point

[scarlet talon]

if it's continuous

[timid silo]

That's it?

[scarlet talon]

the left side derivative needs to be same as right side derivative?

[tardy epoch]

Yes

[timid silo]

Yes

Now check the derivatives of the left and right sides at x = 2

[scarlet talon]

left side = 1

right side = 2ax = 4a

[timid silo]

Where did the a go

[scarlet talon]

4a my bad

[timid silo]

Alright so both sides need to equal, so 4a = 1 right?

So what is a

[scarlet talon]

a = 1/4

ohhhh

and then do i substitute the a value into ax^2 = x + b?

[timid silo]

Yeah what do u get for b

[scarlet talon]

b = -1

[timid silo]

Seems right

[scarlet talon]

Thanks for your. help

got it now

[timid silo]

Anytime!

[scarlet talon]

!close

[timid silo]

Type .close

[scarlet talon]

kk

.close

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[fluid snow]

Does this always hold?

If f''(x0) = 0 and f'''(x0) > 0, then f is concave down for (-infty, x0) and concave up for (x0, infty)
If f''(x0) = 0 and f'''(x0) < 0, then f is concave up for (-infty, x0) and concave down for (x0, infty)

[stable rain]

hm

concave down is f''<0

we can pick

f=x^4-x^3

[fluid snow]

Yes

[stable rain]

wait hm

[fluid snow]

We can reduce this to if f(x0) = 0 and f'(x0) > 0, then f is negative for (-infty, x0) and positive for (x0, infty)

so it should be right

Right?

[stable rain]

er

ok um

for one add continuity

lemme think abt the rest lol

[stable rain]

i dont think so?

only if f is a line

aka y=mx+c

c=0

[fluid snow]

f(x0) = 0 and f'(x0) > 0 implies that f is negative for (-infty, x0) and positive for (x0, infty), the inverse is not true, yes

Since we have x^3 for example, where f'(x0) = 0

But the statement itself should hold, right?

[obtuse pebble]

@fluid snow Has your question been resolved?

[obtuse pebble]

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[stable rain]

.eh more like there exists some delta>0 such that f(x0+delta) is positive and f(x0-delta) is negative

[obtuse pebble]

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[fluid snow]

Yes, if x0 is not the only point where f is 0

[stable rain]

nothing abt x0 being the only point where f is 0 is stated

[fluid snow]

Well, if x0 is not the only point where f'' is 0, then this won't hold, since there may be another inflection point

If $f''(a) = f''(x_0) = f''(b) = 0$ where $a < x_0 < b$ and $f'''(x_0) > 0$, then $f$ is concave down for $(a, x_0)$ and concave up for $(x_0, b)$. \\
If $f''(a) = f''(x_0) = f''(b) = 0$ where $a < x_0 < b$ and $f'''(x_0) < 0$, then $f$ is concave up for $(a, x_0)$ and concave down for $(x_0, b)$.

[obtuse pebble]

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[mossy timber]

Fairly simple question, but does this vector notation make any sense? https://i.imgur.com/geLHfuM.png

The x-portion of the vector has an unbound value (b_x = 3_x+9) and I don't remember ever dealing with vectors wherein one of the components (x-component in this case) has an x-component and an undeclared component

[prime pebble]

pretty sure this is 9.9

[haughty coyote]

When you use English so often you forget your language writes decimals differently

Yeah it's clearly 9.9

[mossy timber]

You're probably correct 🤦‍♂️

.close

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[keen spoke]

I did

[static beacon]

Ok now u know angles in a quad literal add up to 360 so u can find the last angle ACB

[keen spoke]

Ok

[keen spoke]

Angle ACB I'd equall to 78.14

Now what should I do

[static beacon]

Now u got the 3 angles of a big triangle

and one side

U can use the sine rule

a/sin(A) = b/sin(B)

[static beacon]

It’s 77.64

[keen spoke]

Ok

[obtuse pebble]

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[distant moth]

can anyone check option D

i think its wrong

[rigid lintel]

why do you think so?

[distant moth]

wait

These were done in class and we are dealing with p(ab)

[rigid lintel]

ab meaning the event that A and B both happen right?

A intersect B

[distant moth]

Yes

[rigid lintel]

does that say max(0, P(A) + P(B) -1)?

[distant moth]

And if p(ab)<= min(p(a),p(b))
I can think of cases where p(a) or p(b) are less than p(a)+p(b) so wouldn't option D be including wrong answers too?

[distant moth]

Yes

[next reef]

Can't that thing be boiled down to 0 <= P(A intersection B) <= 1 ?

[sage geode]

I'd prove the first inequality this way
$$P(A\cup{B}) \le 1$$
$$P(A) + P(B) - P(A\cap{B}) \le 1$$
$$P(A) + P(B) - 1 \le P(A\cap{B})$$

[warm shale]

A Lonely Bean

[next reef]

I think that's what Gijs thought of saying

[distant moth]

I have seen that, im asking abt this

[sage geode]

Oh the second inequality can be proven in a similar way

I don't think min and max are needed

$$0 \le P(A\cup{B})$$
$$0 \le P(A) + P(B) - P(A\cap{B})$$
$$P(A\cap{B}) \le P(A) + P(B)$$

Now I shall depart because I don't do probability

[warm shale]

A Lonely Bean

[distant moth]

Im saying that if p(ab)<= max (p(a),p(b)) but option d is saying p(ab)<= p(a)+p(b), but there are cases i can think of where max (p(a),p(b)<p(a)+p(b) so wouldnt option D be including some wrong answers?

[rigid lintel]

p(ab) != max (p(a),p(b)

not the same

[distant moth]

I know💀

[rigid lintel]

i dont get what youre asking

this is easy to prove

this is a bit more difficult

[rigid lintel]

but still this has been shown many times in this chat

[royal basin]

indicator variables

or just draw a venn diagram

[rigid lintel]

proof by venn diagram

you sound like my professor lol

[royal basin]

P(A) + P(B) - 1 counts A ∩ B once, A \ B and B \ A zero times, and (A ∪ B)' -1 times.

that's less than or equal to only counting A ∩ B once.

[distant moth]

The extra part was useful in one question, i just have it incase i ever need it

[distant moth]

Okay wait

@rigid lintel

[obtuse pebble]

@distant moth Has your question been resolved?

[obtuse pebble]

@distant moth Has your question been resolved?

[distant moth]

Is there an easier way to prove what ive proved?

[obtuse pebble]

@distant moth Has your question been resolved?

[distant moth]

.close

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[wide swift]

what is a vertex?

[sage geode]

Just a point

[wide swift]

??

[sage geode]

Not really sure if there's a formal definition

[wide swift]

what is vertex?

[sharp pecan]

for graphs it seldom is used for an extremum

[wide swift]

@sharp pecan I am in 7th

i think is graph is for higher classes

[sharp pecan]

then show us where you encountered your vertex

[fierce lagoon]

A vertex is any local/global maximum or minimum im pretty sure

A "turning point"

Vertices also refer to corners of a shape or 3D object

[wide swift]

Here it is

[fierce lagoon]

_ _

[sage geode]

In this case A B C D E F G and H are vertices

[sharp pecan]

also ABCD is a vertex for a 2d body example in 3d space

for task 4 you have to find a 3d body

[wide swift]

And the 3rd one?

[fierce lagoon]

Pretty sure they're only referring to corners

[sharp pecan]

are just areas

[sharp pecan]

isnt the 4th task finding the pyramid formed by the faces?

[fierce lagoon]

No it's asking you to identify the corner made by those sides

[wide swift]

3rd one??

[fierce lagoon]

The fourth is asking you to find the corner formed from ABCD, ADHE, and CDHG

[wide swift]

I fount that oe

what about 3rd one

[fierce lagoon]

The third is asking the opposite: find the sides that form corner A

[wide swift]

oh

But like its says 3 faces?

Does that mean all 3 faces those are not using the vertex a?

[fierce lagoon]

_ _

[wide swift]

huh?

[fierce lagoon]

The faces gotta form vertex A

It would make literally zero sense to identify the faces that have nothing to do with A

Think about corners of a room

They're (typically) formed with 3 faces/planes/sides

[wide swift]

And perpendicular means 90 degree agnle formed

?

[fierce lagoon]

Well yeah

The angle doesn't matter here

It's the vertex/corner, A, itself that matters

You're just identifying what sides/planes/faces forms vertex A

[wide swift]

Ok

[tall tusk]

oh man

So we back in the mine

[wide swift]

what?

its my room

[fierce lagoon]

Got our pickaxe swinging from side to sode

Side side to side

[tall tusk]

This task, a grueling one

[fierce lagoon]

I forgot the lyrics shit fuck piss

Side side to side

[tall tusk]

LMAO

[wide swift]

aren't we here to study?

[fierce lagoon]

Eh sometimes we goof around a bit

Have you found the planes that make A

[wide swift]

Is 8 one possible?

Sir ?? U there??

[fierce lagoon]

Uh, if the planes refer to faces

Well ... they all cannot lie on one plane

So one point could just not be on the plane I'm pretty sure

[wide swift]

Ok

Are faculty online over night?

?

[teal turret]

Faculty xd

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[obtuse pebble]

@wide swift Has your question been resolved?

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[rare scroll]

In trapezium ABCD, AB||DA and DC=2AB. EF drawn parallel to AB cuts AD in F

and BC in E such that ECBE​=43​. Diagonal DB intersects EF at G. Prove that 7FE=10 AB.

[rare scroll]

can anyone provide a solution

[vestal thorn]

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

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[timid silo]

$\frac{405}{180}$

[warm shale]

ED

[timid silo]

can someone send me a video or explain how to simplify fractions like this

without a calculator

[tired skiff]

find the great common factor

[thick gyro]

@timid silo you can see that 405 is a multiple of 3. You so it's 180. You can divide both by 3 to get 135/60, 135 is also a multiple of 3, so you can reduce it further to 45/20, you can see these are now multiples of 5, so it's now 9/4

Can't simplify more than that

You can start small and just keep simplifying over and over when you spot new factors

[stable peak]

multiples of 5 end in 0 or 5 and they both are multiples of 5 so divide both sides by 5 it becomes 81/36 and if u add all the digits of a number and its a multiple of 3 then the number is a multiple of 3. for example in 81, 8+1 is 9 and 9 is a multiple of 3 so 81 is a multiple of 3 so divide both by 3 it becomes 27/12 then u can see again theyre multiples of 3 so divide by 3 it becomes 9/4

u just gotta find the hcf highest common factor cause the way i did it was long and slow but if u notice that the hcf is 45 i think so then divide both by 45 and it becomes 9/4

[timid silo]

how do i know 405 is a multiple of 3

oh

if i add 4 plus 5 its 9 so 405 is a multiple of 3

?

ty i didnt know that

thank u for helping

.close

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[rose zenith]

Does this process work ?

[late stump]

,rotate 90

[rose zenith]

My bad heres upright

[late stump]

do you know the sum/difference of cubes formula

[rose zenith]

I do not

Did I mess up it to much lol

[late stump]

okay well it’s $a^3+b^3=(a+b)(a^2-ab+b^2)$

[warm shale]

light

[rose zenith]

Ohh interesting

[late stump]

do you know pythagorean identity for trig ratios? @rose zenith

[rose zenith]

Yea

Which ones would good?

[late stump]

cool so you know $1=\sin^2x+\cos^2x$

[warm shale]

light

[rose zenith]

Yea

[late stump]

so if you replace the one on the RS with that

you get the sum of cubes formula

[rose zenith]

But that one you need an addition in the middle

Did I mess up my algebra

I can't see where I can use it here

[late stump]

in your initial problem
$$(\sin^3x)+(\cos^3x)=(\sin x+\cos x)(\fbox{1}-\sin x\cos x)$$

[warm shale]

light

[rose zenith]

Wait where did I do this?

[late stump]

sorry?

you didnt do that

[rose zenith]

This

[late stump]

this this initial problem youre asked to prove

[rose zenith]

Oh yea yea

[late stump]

so if you replace 1 with cos^2 x + sin^2 x

you get $$(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)$$

[warm shale]

light

[rose zenith]

Ohhh so I should go from there

Bet

[late stump]

and now that you have this, you can use this formula to prove that LS = RS

[rose zenith]

Bet bet

Ty bro

I think I have it now

.close

[obtuse pebble]

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[obtuse pebble]

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[brave edge]

NO CALCULATOR ALLOWED!

Find Ea

My work so far:
308-293 = 15 in the numerator

Denominator = 293 × 308.
(200+93)×(300+8)
Apply FOIL:

1. 200×300 = 60 000 EZ (cancel the 0s and add them back later for easier time)
2. 200 × 8 = 1600
3. 93 × 300 = 90× 300+3×300 = 27000 + 900 = 27 900
4. 93 × 8 = 90 × 8 + 3×8 = 720+24=744

GRAND FINALE:
60 000 + 1 600 + 27 900 + 744 = 90 244

15/90 244

2.303 × 25/3
2303/1000 × 25/3 (cancel 1000 and 25)
2303/40 × 1/3
2303/120

SO FAR: (Ea)/(2303/120) × (15)/(90244)

(15Ea)/((2303/120)×90244/1)

(15Ea)/(2303×90244/120)

Keep change flip
15Ea/1 × 120/2303×90244

1 800Ea/2303×90 244

1.8 × 10^3Ea/(2.303×10^3)(90.244×10^3)

No fucking clue where to go. This was my method. Very long. Tedious. I tried every shortcut I knew. Now I'm stuck. How the FUCK do I find Ea? I'm going to BEAT something with a CHAIR extremely fucking hard.

[brave edge]

God this is fucking messy

This isn't my problem, a friend's problem. I tried to help but I got stuck

And I'm a stubborn goat

[obtuse pebble]

@brave edge Has your question been resolved?

[obtuse pebble]

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[halcyon sail]

Is there a symbol that represents "any non-zero real number" ?

[halcyon sail]

.close

[obtuse pebble]

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[brave edge]

BRUH

BRUH

.reopen

[obtuse pebble]

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[timid silo]

APPARENTLY

IT IS NOT

0.915

[timid silo]

But

-2.9518

How.

And why like

Also this is log stuff

[late stump]

,calc 0.0899-1+0.8435-(-3+0.8751)-(3+0.1065)

[warm shale]

Result:

-1.0482

[late stump]

well apparently its not either of those

[timid silo]

This is how he did it

What are these fat steps in the middle

.

[late stump]

that step before the log is definitely not right

how did -2+0.9518 turn into 2.9518

[timid silo]

Ehy cant i putallofit in the calc

[tardy epoch]

Probably missing parentheses

[timid silo]

You do it

[timid silo]

You wont get the same ans as this one

[fast turtle]

-1,0482

[timid silo]

My bookiswrong

Ugh

[tardy epoch]

Nah

[timid silo]

Wateva im done

[timid silo]

Good choice

[timid silo]

hi

[late stump]

huh

[timid silo]

hi

[timid silo]

Howdy

[late stump]

was there something you needed help with specifically?

[timid silo]

no

Just a simple howdy

Rejected

.close

[obtuse pebble]

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[fast turtle]

Do I have mistakes please tell me if I’m somewhere wrong/missing

[fast turtle]

.close

[obtuse pebble]

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[vestal thorn]

you can't distribute exponentiation over subtraction

ie this

binomial multiplication (FOIL)

exponentiation is multiplication

$(\sqrt{u}-\sqrt{7-u})^2=(\sqrt{u}-\sqrt{7-u})\cdot(\sqrt{u}-\sqrt{7-u})$

[warm shale]

baro | awake

[vestal thorn]

does that help?

not that i know of!

it's not atrocious, and it gets rid of some of your square roots

yes

because $(\sqrt{u}-\sqrt{7-u})\cdot(\sqrt{u}-\sqrt{7-u})\neq u-(7-u)$

[warm shale]

baro | awake

[vestal thorn]

it's kind of hard for me to tell what you're saying

sorry!

[nocturne minnow]

Squares roots can not be split across addition or subtraction

$\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$

[warm shale]

dldh06

[nocturne minnow]

One of the only options you have is to distribute

No

Because exponents don't get distribute across addition or subtraction

[vestal thorn]

exponentiation does distribute over multiplication though

so yes for this one

[nocturne minnow]

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

[warm shale]

dldh06

[nocturne minnow]

That's a no, there's a minus sign

[nocturne minnow]

Same logic to this

[vestal thorn]

it should read 4(√u*√(7-u))^2, i'm assuming from the original problem

$(\sqrt{a}\cdot\sqrt{b})^2=(\sqrt{a})^2\cdot(\sqrt{b})^2=a\cdot b$

[nocturne minnow]

Were you meaning $(\sqrt{u}\sqrt{7} - u)^2$ or $(\sqrt{u}\sqrt{7 - u})^2$?

[warm shale]

dldh06

baro | awake

[nocturne minnow]

That's when you can do it

[vestal thorn]

i think we are confusing you more than we are helping at this point

[nocturne minnow]

Exponents cannot be split across addition or subtraction, only across multiplication and division

You can simplfy this still $(\sqrt{7-u})^2$

[warm shale]

dldh06

[nocturne minnow]

Same process with the $\sqrt{u}$

[warm shale]

dldh06

[nocturne minnow]

Yes, if that minus was a multiplication sign, that is possible to do

Yes

Because exponents represent repeated multiplication, like a^2 = a * a

Tell me this, is
$(\sqrt{4} + \sqrt{9})^2$ equal to $(4+ 9)$?

[warm shale]

dldh06

[nocturne minnow]

Well do the math

Order of operations

No

You can find square root 4 and 9 first

[nocturne minnow]

.

And order of operations, you are failing to do that

PEMDAS/BODMAS

So in (2 + 3)^2, if parentheses are first, what do you get?

[nocturne minnow]

Now answer this

Do you see why your original question you had was wrong?

[obtuse pebble]

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[obtuse pebble]

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[wise hedge]

Is the fraction 2/3 repeating or non-repeating when converted to it's decimal equivalent?

[nocturne minnow]

Use a calculator and see

[wise hedge]

But a calculator is only going to show so many decimal places just to be practical.

[nocturne minnow]

Or you can google it

[wise hedge]

🙂 Yes I could but I thought we came to these #help channels for questions we have about Math to get help from people who know.

That's your answer to my question, is Google it?

[nocturne minnow]

I mean if it's something you can easily google, why waste people's time?

[wise hedge]

Everything can be Googled so why even have help channels at all?

Why even have a Discord application?

[nocturne minnow]

For instances when people have done their research and fail to accomplish to find anything

[wise hedge]

What??? You sound like you don't know the answer to my question so you pushed it off to Google. That's all you had to say, man.

[nocturne minnow]

I'm just saying, that you can easily google "is 2/3 a repeating decimal" and it'll give you the answer

[static beacon]

@wise hedge

[nocturne minnow]

But yes, it's repeating

[static beacon]

What is the decimal of 2/3

[wise hedge]

Well, it's anything that start with .6

And repeats...

[static beacon]

Oh

So it does repeat?

There u go

[wise hedge]

But if it repeats doesn't that imply non-terminating?

And if it's non-terminating doesn't that imply irrational?

[nocturne minnow]

No

[nocturne minnow]

And yes, that is what non terminating means

[static beacon]

Bro we talked about this yesterday

[wise hedge]

I think another person said it does.

[nocturne minnow]

a rational number is one that can be written as the ratio of two integers

[wise hedge]

Yes, I get that part.

[static beacon]

If it’s repeating and is non terminating, it can be expressed as a pattern

[wise hedge]

But I don't like conflicting definitions, I like consistency, if there is no consistency then I get confused.

[static beacon]

Because in your definition your using words to confuse you

[timid silo]

irrational is non terminating without any periodicity in its digits

[wise hedge]

Repeating implies non-terminating but non-terminating does not imply repeating. I think this is the mistaken assumption I've been making. It doesn't work both ways.

[static beacon]

Irrational is simply a number that can’t be expressed z/y where z and y are both intergers