help-19

[odd edge]

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[hardy bluff]

How to draw this trig function this is my attempt but I got confused in the middle and the other pic is the answer key

[hardy bluff]

Also what was that rule about horizontal stretch or vertical stretch being the reciprocal or something

[worldly rune]

does it say -20 to 100?

[hardy bluff]

No

Where did u get that from

[worldly rune]

$-20\leq x \leq 100$

[clever fjord]

edwardborn

[hardy bluff]

No

[worldly rune]

nvm i see the x axis now

it's p

[hardy bluff]

^^^^^^^

[hardy bluff]

^^^^^

[worldly rune]

so you know how f(x) looks like

[hardy bluff]

Yes they already gave it to us as u can see in the question

[worldly rune]

it has a maximum at x = 0 and its first minimum when x = 2p

f(x) = cos(bx)

[hardy bluff]

I know

[worldly rune]

where do you think cos(bx/2) will have its first minimum?

[hardy bluff]

I already drew its minimum

[worldly rune]

that's -3cos(bx/2), not cos(bx/2)

or well some of it

[hardy bluff]

Yes because that’s the question

[worldly rune]

i see, well have a good day then

[hardy bluff]

Do you understand the question I am asking

[hardy bluff]

^^^^^

[hardy bluff]

^^^^^

<@&286206848099549185>

<@&286206848099549185>

[odd edge]

@hardy bluff Has your question been resolved?

[hardy bluff]

<@&286206848099549185>

[odd edge]

@hardy bluff Has your question been resolved?

[hardy bluff]

<@&286206848099549185>

Ain’t no way nobody knows how to draw this

Alr whatever I’ll figure it out myself

.close

[odd edge]

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[grizzled granite]

How is i^i a real number?

[gritty oar]

Are you familiar with polar representation of complex numbers, in particular, i = e^(ipi/2)

[grizzled granite]

yes

[gritty oar]

Perfect, exponentiate both sides by i

you get e^(-pi/2)

[grizzled granite]

oh

[gritty oar]

[grizzled granite]

.closed

.close

[odd edge]

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[grizzled granite]

.close

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[grave seal]

alr gang

[grave seal]

8 : 2 (2+2) = 1

8 : 2 (2 + 2)
8 : 2 (4)
8 : 2 x 4
8 : 8
= 1

right?

[ivory dock]

[grave seal]

I got a c in maths

its 1

[mint mirage]

By order of operations, no

[grave seal]

right?

[ivory dock]

This is illegal notation

I'm calling the cops

[mint mirage]

If you had
8 : (2 (2+2))
Then that equals 1

[grave seal]

well doesn’t it go pharanthases > exponents > multiplication > division > addition > subtraction?

[ivory dock]

it's ambiguous, notation like this is illegal

use fractions

like a good boy

[mint mirage]

PEMDAS
Multiplication and division is done with whatever comes first from left to right

Since 8 : 2 comes first, you do that first, in 8 : 2 (4)

[ivory dock]

PEMDAS wasn't universal, people didn't agree on it everywhere, europe did one thing, americas another

it's illegal

[grave seal]

so its 16

[mint mirage]

Yes, by order of operations, it results in 16

[mint mirage]

,calc 8 / 2 (2+2)

[clever fjord]

Result:

16

[ivory dock]

this question doesn't have a correct solution, it depends on the convention of the time it was written in and the intentions of it's author

I swear I read old math books in which 4 / 2 + 3 = 4/5

[fierce kindle]

why wouldn't it have a correct solution

I think it makes sense

[mint mirage]

First, it's been recent with these types of problems. Yes the questions are ambiguous so it depends on what the author meant. But if you read, I stated that by order of operations, it's 16

[mint mirage]

Lack of parentheses

[grave seal]

isnt that just 5?

[ivory dock]

Yeah, but you're assuming the order of operations to be the correct one

[ivory dock]

It didn't have any

[mint mirage]

Order of operations is the norm now

[ivory dock]

The book just used that convention

[fierce kindle]

[mint mirage]

Most people don't deviate from it

[ivory dock]

The book went by : everything left is the top of the fraction, everything right is the bottom of the fraction

[ivory dock]

Tell that to the russians of 18th century

[grave seal]

,calc 4 : 2 + 3

[clever fjord]

Result:

[4, 5]

[grave seal]

oh

[mint mirage]

Don't use the :

Use /

[grave seal]

,calc 4/2 + 3

[clever fjord]

Result:

5

[grave seal]

the european sign

[ivory dock]

It's just bad convention

to use :

or not include brackets

[grave seal]

nuh uh?

: is superior

[ivory dock]

: is for nerds

[mint mirage]

If you use : then use parentheses as well

[grave seal]

ur a nerd

[mint mirage]

To denote what is in the numerator vs denominator

[fierce kindle]

this is a server for nerds

I'm proud to be called a nerd

[ivory dock]

#usebrackets

#deathtothecolon

[grave seal]

local Players = game:GetService("Players")

'PlayerAdded' event
Players.PlayerAdded:Connect(function(player)

—username is "Specter"
    if player.Name == "Specter" then
       
        print(player.Name .. ", ur a nerd")

        — the player (optional)
        local ChatService = game:GetService("Chat")
        ChatService:Chat(player.Character or player, "ur a nerd")
    end
end)

type shi

[fierce kindle]

no way bro just created a roblox script calling me a nerd

-- print to the output console says everything 😂

[grave seal]

okay sir

old code I reused

to print messages

[fierce kindle]

username is "Specter" will break the script

[grave seal]

from one of my roblox games

probably

[fierce kindle]

also the player (optional)

[grave seal]

wait

theres supposed to be

—

[fierce kindle]

there's a bunch of comments without their dashes

[grave seal]

I ripped some of my old code out in a hurry

no hate given

[fierce kindle]

I love roblox

[grave seal]

please

[grave seal]

for real

lua is fun

[ivory dock]

omg you guys really are nerds

I'm out

[grave seal]

lua is easy bro

[fierce kindle]

you should learn python it's very similar

[grave seal]

isnt c++ more similar?

just with integers?

instead of local stuff

[fierce kindle]

wait u fr?

[grave seal]

yeah

[fierce kindle]

no no no

[grave seal]

wait

no

python is more similar

[fierce kindle]

if lua was similar to c++ then python would be similar to c++ too, both arent

[grave seal]

yeah

lua is easy asf tho

learnt it in a few weeks

c++ is pretty easy too tho ngl

.close

[odd edge]

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[odd edge]

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Remember:
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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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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[light jungle]

solving ofr shsaded area

[sly juniper]

!status

[odd edge]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[light jungle]

1

[light jungle]

im lost

[sly juniper]

connect the end points touching the circle to the center

[light jungle]

that crreates the center angle right

its 120?

[sly juniper]

like that

[light jungle]

ya

[sly juniper]

and the shaded area = area of a kite - area of an arc

[light jungle]

how do I find the radius?

[sly juniper]

you can find angles of a kite

and you know 2 of the sides are = 12

[light jungle]

so would I make triangles and od soh ca toa?

[sly juniper]

that should be enough to find r

[sly juniper]

yeah

you can cut it to 30-60-90 triangle

[light jungle]

did I do something wrong I got this

[sly juniper]

im think 120 is on the other side

[light jungle]

wouldnt it be 120 onb both sides

cuz the toal is 360?

[sly juniper]

i doesnt need to be 120

120 on one side, 60 on other

[light jungle]

so then what does the kite look like

[sly juniper]

wait lemme check

[light jungle]

yeah im stuck

[sly juniper]

i gotta go but to not gonna hung you

the are of a kite = 1/2 * (h1+h2)* a

h1 is height of a 12/12/alpha triangle and h2 is height of r/r/120 triangle

[light jungle]

yeah I dont get it at all 💀

wtv

[sly juniper]

try to work out with this info

bye

[light jungle]

ima just copy the answer

[sly juniper]

ok

[light jungle]

.close

[odd edge]

Channel closed

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[odd edge]

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Remember:
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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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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[mystic saffron]

I need help

[barren coral]

alright. with what?

[mystic saffron]

Its geometry hw and test corrections

I need to be taught because I have a test tuesday

[barren coral]

if you can, could you send some pictures of it?

[mystic saffron]

I don’t know what else to do

The corrections? Yes

[barren coral]

particularly parts you need help on

[mystic saffron]

Okay

This is one problem

I got it wrong

As you can tell

[barren coral]

ohh

okay

let me try to rotate it rq

so

this is something called the two secants/two tangents theorem

whenever two lines intersect outside of a circle and both touch the circle, the angle they intersect at is equal to the angle of the larger arc of the circle (QP in this case) minus the angle of the smaller arc (QR)

if it seems random, I can try to get you an explanation for it

or

sorry

its that divided by two

[mystic saffron]

Can you ft me so i can write it on paper?

Its corrections so i have to show how i got the answer

[barren coral]

alright

one second

[mystic saffron]

okie

.close

[odd edge]

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[odd edge]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[idle bridge]

find all x on the interval -2pi =< x =< 3pi such that cos (x) + 1 = 0

[idle bridge]

i moved 1 to right side and got cod

-1/1 -1= x 1 = r

now idk what the " find all x on the interval -2pi =< x =< 3pi " is asking

[odd edge]

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).

[crimson hazel]

sketch a y=cos(x) graph and see where y = -1 and try determine from that

[idle bridge]

can i do it without graphing

[crimson hazel]

yes

[idle bridge]

the way my teacher does it is with the triangles

so i found out that the position of cos (x) + 1 = 0

is on the negative x-axis,

now idk what the " find all x on the interval -2pi =< x =< 3pi " is asking

[crimson hazel]

there are multiple solutions with cos(x) = -1 and so it wants you to find all the possible x values

and the cos wave values repeats every 2pi

so you can add or subtract 2pi from your answer to get other solutions

[crimson hazel]

for -1 and 1

[idle bridge]

its just pi?

thats the start right

[crimson hazel]

yes

[idle bridge]

so i add 2 pi

so

pi, 3pi , 5pi

[crimson hazel]

yes except its asking for solutions within the interval of -2pi and 3pi

so 5pi wouldnt be one of your solutions

[idle bridge]

yes thats what im confused about

wait

so it cant go more than 3 pi ur saying from the adding of 2 pi from pi

[crimson hazel]

yes

because then it will be out of bounds

[idle bridge]

ok we are getting somehwere

i see

[crimson hazel]

you can also subtract 2pi

to get more solutions

[idle bridge]

so basically

[crimson hazel]

i.e -pi, -3pi

[idle bridge]

the answer would be just -pi and 3pi

?

[crimson hazel]

and pi

[idle bridge]

oh yes

okay i understand

i got another question similar to it that i got wrong

find the exact values of angle over [ -4pi, 0 ]

sec = - 2sqrt3 / 3

so far i know that

sec would have to be

r over x

so -3 here must be the x?

[crimson hazel]

sec(x) = -2sqrt3 / 3?

id prefer to change it to cos by doing the reciprocal as to make it easier which gives cosx = -3/2sqrt3 which gives x=pi/6 as your first solution

but obviously that is out of bounds so you have to figure out the other possible solutions

[idle bridge]

howd u know pi/6 is out of bounds

[crimson hazel]

because it is greater than 0

[idle bridge]

can u explain what is this first

find the exact values of angle over [ -4pi, 0 ]

because i see another problem like it but its [0, 2pi]

i got that problem right but idk why

[crimson hazel]

[-4pi, 0] means you dont accept answers which are either greater than 0 or less than -4pi

so for example -pi would be an acceptable answer

however pi wouldnt be

as it is greater than 0

[crimson hazel]

-5pi wouldnt be accepted as it is less than -4pi

[idle bridge]

it means that the answers must be all negative then cause it cant be 0?

[crimson hazel]

it can be equal to 0 but it cannot be greater than 0

[idle bridge]

okay

so lets try to solve

how i interpretted it first

sec = -2 sqrt3 / 3

r = 2 sqrt 3 (hypotenuse)

x = -3 (adjacent)

is that correct?

[crimson hazel]

honestly don't know that method

hold on

oh

right

ok

yes that is

[idle bridge]

so now i look for its triangle

that has those

[crimson hazel]

yes

[idle bridge]

ah but wait idk a triangle with 2 sqrt 3/ 3

but im assuming this triangle would be in quadrant 2 or 3 then cause x = -3

[crimson hazel]

quadrant 3

well

yeah both

for solutions

[idle bridge]

but since it has to be -

its in quadrant

3

LETS FKING GO

[crimson hazel]

is this non-calculator btw?

[idle bridge]

but how do i figure out its exact value from r being 2 sqrt 3/ 3

[idle bridge]

yes no calc for everything

[crimson hazel]

ok lovely

[idle bridge]

this for my pre calc, ive never needed a calculator

because my teacher does not give problems that require calc

[crimson hazel]

do you just memorise a ton of triangles then 😁

[idle bridge]

i know unit circle

and the 3 right triangles

[crimson hazel]

if it wasnt negative the angle would be 5pi/6 but as it is and it is sec(x) it is symmetric and so x=-5pi/6

[idle bridge]

howd u know it was 5pi/6

[idle bridge]

okay ? i see - sqrt 3 / 2

[crimson hazel]

secx = -2sqrt3 / 3 => cosx = -3/2sqrt3 = -sqrt3/2

alternatively you could use 7pi/6

and because of the symmetry of the sec(x) function about the y axis, you can state that x=-5pi/6 and x=-7pi/6

[idle bridge]

ok, i understand this problem better. i appreciate you

.close

[odd edge]

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[quaint fern]

https://i.imgur.com/w99bPJq.png Trying to determine if this is injective, surjective, or bijective. My thought is that since there's an infinite number of primes (where each one will always have an input), and only one natural number will always point to a prime, it's bijective.
Image

[steady tide]

!sseei

[odd edge]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

@quaint fern Has your question been resolved?

[odd edge]

@quaint fern Has your question been resolved?

[fresh ruin]

yea it's bijective, fastest way to check is that the inverse obviously works perfectly fine

[odd edge]

@quaint fern Has your question been resolved?

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[odd edge]

Available help channel!

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Remember:
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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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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[radiant chasm]

Trying to understand why does the discriminant of the derivative have to be positive.

[radiant chasm]

for the inverse of g

[sleek moth]

If g^(-1) doesn't exist then you know that g is not strictly monotonic meaning the derivative can't be all positive or all negative

If a>0 then this means that the discriminant is positive (meaning there are two real roots)

If a<0 then this means that the discriminant is well actually once again positive (meaning there are two real roots)

[radiant chasm]

what does a function being monotinic mean if I may ask?

monotonic*

[sleek moth]

https://en.wikipedia.org/wiki/Monotonic_function

[frigid canopy]

it's either increasing or decreasing

only

[radiant chasm]

oh ok got it

[sleek moth]

Basically a differentiable function is invertible if and only if the derivative is all positive or all negative, except for countably many points

or something like that

[olive needle]

I'm going to go out on a limb and say there was a typo and that b^2 - 3ac should have been b^2 - 4ac.

[radiant chasm]

nah its 3 ac

anyways I didnt get it exactly

some of the terms went over my head

but thank you for the help

.close

[odd edge]

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Remember:
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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!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fierce kindle]

Suppose that x is in [1,5), and y=(3-x)/2, prove that y is in [1,4)
where do I begin with proving this?

[fierce kindle]

I tried this but that got me nowhere

[nimble blaze]

did you copy down the question correctly? because that isn't true

[fierce kindle]

oh wait

Suppose that x is in [1,5), and y=((3-x)/2)+2, prove that y is in [1,4)

[nimble blaze]

-1 + 3 isn''t 1

[fierce kindle]

sorry my bad, I fixed it now

[fierce kindle]

@nimble blaze but that still doesn't lead me to what I have to prove

[nimble blaze]

any value in this interval are also in [1,4)

[fierce kindle]

3.1 - 3.9 would be bigger than 3 and smaller than 4 though

[nimble blaze]

you didn't add the 2 to the -1

1 < y <= 3
any value of y satisfying that inequality will also satisfy
1 <= y < 4

[fierce kindle]

yea I wrote -1 on accident, ig I should slow down a bit

[nimble blaze]

e.g. y could be 2
which is in [1,4) so its fine

[fierce kindle]

but y could also be 3.5 or 1 which doesn't satisfy 1 < y <= 3

[nimble blaze]

implication is one way

doesn't necessarily have to work the other way

consider a question like
show that x is between 1 and 10 billion
x - 2 = 3

[fierce kindle]

wouldn't it still be wrong because of 1 < and 1 <=?

[nimble blaze]

values slightly above 1 will be in the given interval

[fierce kindle]

as my final answer would be that y is in (1,3] which wouldn't fully prove y is in [1,4) as I still have to prove that y can also be 1, right?

[nimble blaze]

no

you don't need to prove that y can attain very single value in that interval

just that the values will be in that interval

show that x is between 1 and 10 billion
x - 2 = 3
just like how you don't need to show that x can be any number between 1 and 10 billion

[fierce kindle]

ohh wait I figured it out

[nimble blaze]

x = 5 only
and that number is between 1 and 10 billion

[fierce kindle]

if a<b then a<=b

[nimble blaze]

which is completely fine

[fierce kindle]

because a<b satisfies a<=b, as a<=b means a<b OR a=b

if either one is correct then a<=b

I forgot about that

[odd edge]

@fierce kindle Has your question been resolved?

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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[severe oxide]

How to find the area between those 2?

[nimble blaze]

what have you tried

[lusty nova]

try plotting

[severe oxide]

y = x - 2, y = sqrt(x), y = -sqrt(x)

Then I tried to solve integral from 1 to 4 of (sqrt(x) - (x - 2))

Then I tried to solve -integral from 1 to 4 of (-sqrt(x) - (x - 2))

And finally to add them up

[nimble blaze]

did you do a sketch/graph

[glass wedge]

Are you sure the limits of integration are right

[glass wedge]

Sooo integrating from 1 to 4 you dont get the entire area

You still have some area from 0 to 1

[severe oxide]

Oh boy, that's it

Thank you

[nimble blaze]

personally i'd integrate wrt y

,rccw

[nimble blaze]

and do it with a single integral

[odd edge]

@severe oxide Has your question been resolved?

[last jackal]

can someone help me in stastic?? please?? I really need help:'D

[severe oxide]

How to find with dx?

[nimble blaze]

the idea of subtracting areas, is fine, but you'd need to set up you integrals and bounds properly

sqrt(x) - (x-2) from 0 to 4 will give the area of this region

and you'd want to subtract

which would be
-sqrt(x) - (x-2) from 0 to 1

[severe oxide]

Oh boy, right

Thank you, now I get it

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[sharp urchin]

can someone explain me the sentence of phytagoras

[lusty nova]

$hyp^2 = opp^2 + adj^2$

[clever fjord]

e=mc^2

[tepid pelican]

opposite and adjacent are quite weird terms to use here ngl

[sharp urchin]

thah is not how I learned it

[lusty nova]

opposite and adjacent are the sides that are not the longest side

[sharp urchin]

its a²+b²=c²

[somber jackal]

you can call the sides whatever you want

but (one side)^2 + (other side)^2 = hypotenuse^2 is always true

for a right triangle

[lusty nova]

they probably labelled the sides a,b,c but it is still the same thing

c is the hypotenuse

[odd edge]

@sharp urchin Has your question been resolved?

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[visual hazel]

Hello, can you help me to demonstrate this property please?

[grim cradle]

Take an element of A \cup B then use assumption to show that the element is also in C \cup D

[fathom dock]

yes, if x is an element of A it is an element of C. same goes for B and D. and remember this:

[visual hazel]

But there I would be starting from the thesis, isn't that incorrect?

[grim cradle]

What?

[odd edge]

@visual hazel Has your question been resolved?

[visual hazel]

I mean that in the way you said you are starting from the conclusion of the conditional and I think I remember that this is not correct.

[grim cradle]

No I’m not?

[visual hazel]

a

[grim cradle]

Yes?

[visual hazel]

what you mentioned is this, and this is the conclusion

[grim cradle]

The conclusion is A \cup B subset of C \cup D

Not anything about A \cup B

Why can’t I take an element of that?

[visual hazel]

but that is in the conclusion

i don't understand

[grim cradle]

No its not

The conclusion is not $x \in A \cup B$

Which is all I’m doing

[clever fjord]

ScapeProf

[visual hazel]

I find it a bit counter-intuitive, but it's okay

[visual hazel]

what would you do after that?

[grim cradle]

.

[visual hazel]

let me see if I understand you, you would be looking for the antecedent to be equal to the consequent, then it would be demonstrated that

?

[grim cradle]

What?

If you want to show X subset of Y take an arbitary element of X and show that element is also in Y

Hence every element in X is also in Y and by definition X is a subset of Y

[visual hazel]

And what would you do with that? I just don't understand the way you're thinking of solving it.

I'm confused

[mystic saffron]

Whata bout proving that all elements in A u B are in C u D

Exactly like he said

Prove that:

$\forall x \in A \cup B, x \in C \cup D$

[clever fjord]

Jadεn

[mystic saffron]

If you can prove that, we conclude that:

$A \cup B \subseteq C \cup D$

[clever fjord]

Jadεn

[grim cradle]

I wrote what to do with it?

[mystic saffron]

$\text{because } A \subseteq C \rightarrow \forall x \in A, x \in C \rightarrow \forall x \in A, x \in C \cup D$

[clever fjord]

Jadεn

[mystic saffron]

Also

$\text{because } B \subseteq D \rightarrow \forall x \in B, x \in D \rightarrow \forall x \in B, x \in C \cup D$

[clever fjord]

Jadεn

[mystic saffron]

And from these two statements we can get:

$\forall x \in A \cup B, x \in C \cup D$

[clever fjord]

Jadεn

[mystic saffron]

$\text{Therefore } A \subseteq C \land B \subseteq D \rightarrow A \cup B \subseteq C \cup D$

[clever fjord]

Jadεn

[mystic saffron]

@visual hazel

[visual hazel]

hmm

[mystic saffron]

So...

You get it?

[visual hazel]

I'm trying, but no

[mystic saffron]

$\text{because } A \subseteq C \rightarrow \forall x \in A, x \in C \rightarrow \forall x \in A, x \in C \cup D \text{ and because } B \subseteq D \rightarrow \forall x \in B, x \in D \rightarrow \forall x \in B, x \in C \cup D \text{ We can conclude that } \forall x \in A \cup B, x \in C \cup D \text{ Therefore } A \subseteq C \land B \subseteq D \rightarrow A \cup B \subseteq C \cup D$

[clever fjord]

Jadεn

[mystic saffron]

Here I summed it up

@visual hazel do you get it now?

If you don't

tell me that hard part

[visual hazel]

why?

[mystic saffron]

Because C exists inside C u D, therefore all x in C is in C u D

And since all x in A is in C

It applies to all x in A that they are in C u D

[visual hazel]

you're taking that from the conclusion

[grim cradle]

\begin{align*}
x &\in A \cup B \
&\subseteq C \cup B \
&\subseteq C \cup D
\end{align*}

Since A subset C and B subset D from assumption

[clever fjord]

ScapeProf

[visual hazel]

it makes me dizzy because when I was demonstrating the validity of valid reasoning I started with the premises and tried to reach the conclusion by means of inference rules, here it is different?

[grim cradle]

What?

You have some assumptions

Let A,B,C,D be sets such that A subset C and B subset D

[visual hazel]

sorry if I'm not clear, I'm using a translator

[visual hazel]

assumptions is not the hypothesis of the conditional?

[grim cradle]

What?

[visual hazel]

this

[grim cradle]

What about it?

[visual hazel]

that's assumed to be true and it's all about getting to the right thing, wasn't it?

[grim cradle]

Whats wrong about the proof in your mind?

You don’t think if we show every element of X is in Y then X subset of Y?

Isn’t that litterally definiton of subset?

[visual hazel]

where is the comma, wouldn't the conditional go? at least I have that definition of inclusion

[visual hazel]

yes

[mystic saffron]

The comma means imply in a sense

[visual hazel]

ok

[mystic saffron]

Here it says for all x in A that condition(x in C) muxt hold true

[visual hazel]

I think I understand a little better now

[mystic saffron]

Alright

[visual hazel]

So here, if I understand correctly, A is included in C when it is satisfied that if x belongs to A then x belongs to C. Since x belongs to C, x can belong to the union of C with any other set, in this case you took union with D because we need to get to that.

[mystic saffron]

Exactly

@visual hazel you're getting it now

[visual hazel]

So, from the hypothesis you were able to infer this and you were able to join both propositions and then apply the definition of inclusion to reach the conclusion?

I would like to know what strategy you take to deal with these types of exercises.

[mystic saffron]

I mean, since both of them are subsets of the set (C u D), their union is also a subset

$A \subseteq (C \cup D) \land B \subseteq (C \cup D) \Rightarrow A \cup B \subseteq C \cup D$

[clever fjord]

Jadεn

[visual hazel]

I understand, thanks

[mystic saffron]

Np

:3

[visual hazel]

I want to see one more exercise and that's it, this one is a bit more complex than the previous one, what strategy would you use?

[mystic saffron]

Let's prove that this is only valid for B subset of A, meaning that if B weren't a subset of A it won't hold

Assume that B isn't a subset of A, and try to prove it by contradiction

@visual hazel do you have any idea of how to start solving?

[visual hazel]

You mean assume that the conclusion is false and try to get to the hypothesis being false? That is, if the conclusion is false, the hypothesis must be false for the property to be valid.

[visual hazel]

By the way, how would the last step be justified? Because I am asked to justify each step, is there any property or definition that says this?

[mystic saffron]

Look, it can be prooven but it is trivial

just think about it

if A is in a room and B is in the same room, both of them are in the same exact room

A in C u D and B in C u D therefore A u B in C u D

[visual hazel]

Sure, it makes sense, but I was asking maybe for a brief and more formal justification, for example we use this property, this definition, etc.

[mystic saffron]

I don't have that much experience with set theory

but it can be proven

but it may take some time ro figure out

one second

[visual hazel]

ok

[mystic saffron]

I don't really know

Sometimes trivial stuff are very hard to prove

Someone more experienced with set theory can help us

<@&286206848099549185> Someone experienced with set theory here please

@visual hazel I don't think we should prove it

Just go with it

and let's do the other question

can you send it again?

[visual hazel]

ok

[mystic saffron]

Okay

let me think about it

[visual hazel]

✅

[mystic saffron]

I assumed that B is not a subset of A, and found out the left hand side is only true if B is the empty set

But the empty set is a subset of all sets

Therefore it must be a subset

Let $B \not \subseteq A$

[clever fjord]

Jadεn

[mystic saffron]

$(A - B) \cup (A \cap B)=((A - B) \cup A) \cap ((A-B) \cup B)$

[clever fjord]

Jadεn

[mystic saffron]

Which is an identity

$(A - B) \cup A = (A \cup A) - (B \cup A)$

[clever fjord]

Jadεn

[mystic saffron]

which is equal to $A - (B \cup A)$

[clever fjord]

Jadεn

[mystic saffron]

But see that it's equal to the empty set cause all the elements of A are in A u B, and we're removing them from A

So, we get: $(A - B) \cup (A \cap B) = \varnothing \cap ... = \varnothing$

[clever fjord]

Jadεn

[mystic saffron]

$\forall A, \varnothing \subseteq A$

[clever fjord]

Jadεn

[mystic saffron]

And since here $B = \varnothing, B \subseteq A$

[clever fjord]

Jadεn

[mystic saffron]

But as we assumed, $B \not \subseteq A$

[clever fjord]

Jadεn

[mystic saffron]

And we get a contradiction

Therefore B must be a subset of A

@visual hazel

[visual hazel]

I'm reading

[mystic saffron]

Okay

[visual hazel]

I think you were very clear

[mystic saffron]

So, you got it all?

[mystic saffron]

This is the only one Idk how to prove

[visual hazel]

yes

thanks

[mystic saffron]

:>

[mystic saffron]

Don't worry

[visual hazel]

np

[mystic saffron]

:3

Have a great one

Bye

[visual hazel]

🙂

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[limpid aspen]

Basically just write a^2 + 1 for example, if the limit was a^2 + 1

Oh sorry, I misread the question. That is strange phrasing. I misread it as "write the limit as a function of a", but it seems to be asking for values of a as functions of a.

Apologies

[grim cradle]

Then what are you asking? It means it should be a function of a

by noticing a^2 behaves differently for a>1, a=1 and a<1

Clearly your answer depends on value of a

so its a function of a?

[odd edge]

@mystic saffron Has your question been resolved?

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[amber pecan]

Id just swap a and b here right?

since b is the angle given

I think thats right but i cant remember when 2 angles are used

@dusk warren

I g2g so guess ill ask again later

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[stark comet]

Can someone help me know how I got this wrong? If standard deviation = max-min/4 -> 100-40/4=15. Why would the correct answer be 10 here?

[odd edge]

@stark comet Has your question been resolved?

[stark comet]

I've tried it as many was i could think of and can't make 10. It's not urgent i get an answer since i'm just reviewing for an exam, but i'm also wondering if I really made a mistake or if theres an error in the homework program : ( Is there a certain way i had to factor in the mean?

<@&286206848099549185>

[summer hollow]

Considering the distribution is bell -shaped, it will be approximately normally distributed, and around 99.7% of values fall within 3 standard deviations of the mean

so you can see that within the range, you will have 6 standard deviations approximately , so 60/6 = 10

[stark comet]

thank you for explaining to me! i'm kinda mad it was as simple as that, but i have no idea how curves really work in stats so i didnt know 😭

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[torn pond]

need help with number 3 and 4

[narrow cove]

There are two approaches to 3, either calculate the voltage across the resistor or calculate the total resistance and hence current in the circuit and go from there

As for number 4, P = VI

[torn pond]

voltage times I and I isamps?

or current

[narrow cove]

Same thing, current is measured in amps

But yes, I is current

[torn pond]

but im not sure on what to do for the seperate current

[narrow cove]

Have you calculated the total current through the circuit or not yet?

[torn pond]

8/80

because the combined restisiance of 45 and 90 is 30

add 50 and 30

so 8/80

or .1

i think

[narrow cove]

Ye that's right

And from KCL you know that the current through both your branches adds to your total current

[torn pond]

yeah 1/45 and 1/90 =.1

but what about the 50 ohms at first what do i do about that?

is it 1/45 + 1/90 =.1?

[narrow cove]

No

[narrow cove]

yes XD

[torn pond]

woopsy

so do i just do 1/45 = .1- 1/90?

so is .08 repeating the current going through that resistor?

[narrow cove]

No, use the fact that the PD across the branches is the same

V=IR

So $I_1R_1 = I_2R_2$ and $I_1 + I_2 = 0.1$

[clever fjord]

TayBee

[narrow cove]

2 eqs 2 unknowns solve

[narrow cove]

If you use KCL you don't need to consider it other than for calculating the total current in the circuit

[torn pond]

8?

also whgats KCL?

[narrow cove]

The alternative is to consider the voltage drop across the 50 ohm resistor and hence calculate the voltage in the branches

[narrow cove]

Oh soz

Kirchoff's Current Law

Just the thing that says the total current entering a junction of a circuit is equal to the total current leaving

Hence I can say I1 + I2 = 0.1

[narrow cove]

I also do not know what this number is lol

[torn pond]

2/15?

[torn pond]

im ;lost as to how i use this?

[narrow cove]

So R1 = 45 and R2 = 90 yes?

[torn pond]

yes?

i fucking hope so

[narrow cove]

$45I_1 = 90I_2 \ I_1 = 2I_2$

[clever fjord]

TayBee

[torn pond]

so 2 i1 = i2

[narrow cove]

So now using I1 + I2 = 0.1 you can say 2I2 + I2 = 0.1 -> 3I2 = 0.1

No, I1 = 2I2

[torn pond]

im lost again

[narrow cove]

Do you understand this

All I've done is substitute in the values of R1 and R2

And then divided by 45

[torn pond]

ok

so 3i2 = .1

[narrow cove]

Ye, so i2 = ?

[torn pond]

and since we can divide .1 by 3

i2 = .1/3

[narrow cove]

Ye, or 1/30 because decimals in fractions scare me

[torn pond]

90=1/30

[narrow cove]

So if i1 = 2i2, and i2 = 1/30, what is i1?

[torn pond]

so would i1 be 2/30?

[narrow cove]

Exactly

Alternative method for you because I feel like that might have imploded your brain

Kirchoff's second law states that the sum of voltages in a closed loop is equal to zero

[torn pond]

i remember using 1/8 =1/45 +1/90

change it doing 1/45 to 2/90

[narrow cove]

In other words, the amount of voltage your battery supplies is the amount of voltage consumed by the circuit

[narrow cove]

Well this breaks the universe as 1/45 + 1/90 = 1/30

[torn pond]

oh woops

it was 1/ 45 pluis 1/90 = 3/90

[narrow cove]

That would make more sense

[torn pond]

my previous way of doing is staryting to show flaws in my head

[narrow cove]

Yeah effectively the formal KCL method we just did proves that current splits in parallel in relation to your resistances

[torn pond]

i won tlie i forgot how to deal with total resit

resirt

resist

[narrow cove]

It's chill

Thinking about voltage is sometimes easier (it made my brain hurt less in my first year of engineering at least)

Remember we worked out the total current in the circuit

[torn pond]

would you minding herlping me a little longer for the next questiuons?

[narrow cove]

That also means we can work out the voltage drop across that first resistor very easily

[narrow cove]

About to sleep but by all means paste them I'll have a brief look or pass the baton on lol

[torn pond]

luckily its on the same pagfe

[narrow cove]

Big win

Just quickly though

We worked out the current as 0.1

We know that V = IR

So the PD across your first 50 ohm resistor is 0.1 * 50 = 5

That means it's used 5 of your 8V supplied

So there are 3V remaining

The PD across parallel branches is all the same, so 3V goes into every branch (one of the several benefits of parallel circuits in engineering XD)

So knowing the PD across both your branches is 3V, we can say that I = V/R

So in branch 1, I = 3/45 = 1/15, and in branch 2, I = 3/90 = 1/30

Different method, same results

Both methods have their place, that method is probably faster provided you're comfortable with the concept of voltage drops as you go round a circuit

Anyway, number 4 you wanted help with, P = VI (Power = Voltage x Current)

[torn pond]

would number 4 be 10?

[narrow cove]

Rearrange for current, I = P/V, you have P = 1200, V = 120

[narrow cove]

[torn pond]

than would 5 be 350 ohms resist?

6=.2

7=1/75?

oh no its volt lost

7=15 vlots lost?

[narrow cove]

How did you get to 15

[torn pond]

.2 times 75

.2 being the current

[narrow cove]

No, total current splits between parallel branches remember

The voltage is the only thing equal across parallel branches

[torn pond]

oh so .1 times 75

7.5

[narrow cove]

Nope, because current doesn't split evenly unless the resistance in branches is even, it is inversely proportional to resistance XD

I think with where you're at it's easier to work out voltage drops across everything else and work out how much you have left for the parallel bit

So

How much voltage is lost across the 125 and 175 ohm resistors

[torn pond]

25, 35

[narrow cove]

Exactly

So

How much do you have left over from your supply

[torn pond]

10

[narrow cove]

Precisely, so that is how much is left for your parallel bit

And every branch gets that

[torn pond]

so 10 volts lost?

[narrow cove]

Ye

And you can use that to calculate the current for the next Q

As V = 10, and you have R

nvm I cant read

The next Q is the 175 ohm resistor

So you've already answered that lmao

If you needed to calculate the current in the branches you'd just use I = V/R with V = 10 XD

[torn pond]

wait i did?

when?

[narrow cove]

You calculated the current a while back as 0.2

And remember Kirchoff's Current Law

Current entering a junction is the same as current leaving the junction

[torn pond]

35 ogmas

[narrow cove]

And since that 175 is not part of a parallel bit

The 'full' current is flowing through it

[torn pond]

ohms*

[narrow cove]

What's 35 ohms

[torn pond]

175 right?

i might be brain famaged

[narrow cove]

You've just told me the current flowing through a 175 ohm resistor is 35 ohms, you can tell me what is wrong with that answer lmao

[torn pond]

lmao

ohmsd is resirt not current

[narrow cove]

No XD Voltage = Current x Resistance

Resistance is the measure of how much a component resists the flow of electrons

more resistance = less current

And vice versa

[torn pond]

35 volts

[narrow cove]

That would be voltage

[torn pond]

fuk

[narrow cove]

The Q wants the current if memory serves

Amps

How many amps

[torn pond]

.2

[narrow cove]

yes

Job done

That's your answer for that one

Same as the total current in the circuit as the answer to Q6

[torn pond]

so wairt 7 = 10 volts lost right?

[narrow cove]

Yee

[torn pond]

ok

so how would 9 work then?

[narrow cove]

What is the unit of power?

[torn pond]

kW?

kW/Hr?

[narrow cove]

Close, it's Watts (W)

kW is kilowatts

1kW = 1000W

[torn pond]

ah

[narrow cove]

A kWh represents the energy delivered by one kW (1000W) of power over an hour

[torn pond]

so if 95 watts divided by 115 volts

=19/23

then what

[narrow cove]

It's how you are charged for energy, so for example if you have a 2kW heater running for 3 hours, and energy costs $1/kWh (which it probably will when WW3 begins), the cost would be 3x2 = $6

[narrow cove]

No

because we already have power

Power = Voltage x Current

If they gave you the Voltage and the Current, then you would multiply them together to find the power

But they've already given us P

So the voltage is irrelevant to the question

[torn pond]

so 115 times 95

[narrow cove]

Still no

[torn pond]

wha?

[narrow cove]

A 95 Watt TV will always draw 95 Watts if it is available to it, the only thing that can vary is the voltage and the current that deliver that 95 watts

[torn pond]

ok so

95 watts an hour?

[narrow cove]

Getting close, 95 Watts means 95 joules of energy per second

So it's not 95 watts per hour

But in an hour, the TV uses 95 watt-hours (Wh)

That's equivalent to 95 joules per second for an hour (so actually, the energy used is 95x60x60)

Now, the price is given in kWh, so we should convert to that first

What is 95 watts in kW

[torn pond]

.095?

[narrow cove]

Good!

So 0.095kW, for 2 hours, is how many kWh

[torn pond]

.095 times 2 =.19

[narrow cove]

Exactly, so this TV has used 0.19kWh

[torn pond]

1.52 cents of electricity?

[narrow cove]

Yup!

[torn pond]

ok thank god

thanks for the help

[narrow cove]

So next time your parents complain at you for watching a movie on the TV, you can tell them you'll pay them a few cents for the privilege

Np 🫡

[torn pond]

i would not have finished this otherwise

[narrow cove]

Recite the difference between voltage current and resistance to yourself before you sleep, V = IR is the fundamental law of electricity, and once you've done that, have a read of Kirchoff's Current and Kirchoff's Voltage Laws, they're the next fundamental two laws for circuits XD

Haha, well congrats, you are now free of electronics for the night

[torn pond]

volt = current times resist?

[narrow cove]

Ye

[torn pond]

ok thx then

ill make sure to rember

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[mystic saffron]

um

[mystic saffron]

so

our range (set of ouput values) is:
f (x) > root 2.5
and
our domain (set of input values) is:
x > 3/2

And so,
The f^-1 (x)domain is > root 2.5 (as in f(x), our range was x > root 2.5)
And the f^-1 (x) range is >3/2 (as in f(x), our domain was..x>3/2)

[uncut remnant]

how is that your range?

i get something else

oh mb

[mystic saffron]

uhm lemme show u

[uncut remnant]

yes i agree

[uncut remnant]

yeah what youve written is correct

[mystic saffron]

okay

Shuriii

[mystic saffron]

Can u check that again Shuuuri

Or someone else

.close

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[vague spoke]

.starr

.start

can anyone explain how to use cosign rule when the triangle has 3 sides? (qns asking to find the smallest angle)

[royal herald]

smallest angle is opposite the smallest side

make sure the angle ur solving for is across the smallest side

then u will get the right answer

@vague spoke

[odd edge]

@vague spoke Has your question been resolved?

[vague spoke]

how do i do that

the 3 sides are 8cm, 12cm and 9cm so i did 12^2 = 9^2+8^2-2(9)(8)cosA

[royal herald]

no

the smallest side is 8 cm

so u want the 8 cm on the left side of the equation

[vague spoke]

ohh

[royal herald]

to get the smallest angle A

[vague spoke]

so 8^2 = 12^2+9^2-2(9)(12)cosA

so 64 = 9cosA

@royal herald

[royal herald]

yea looks right

,w cosine rule

hold on

,calc 64 - 144 - 81

[clever fjord]

Result:

-161

[royal herald]

wait howd u get 64

[vague spoke]

so do i move the 9 to the left?

8^2?

[royal herald]

ok but what happened to the 12^2 + 9^2

[vague spoke]

12^2+9^2-2(9)(12)

or can i not do that

[royal herald]

do what

i dont understand what u did

[vague spoke]

this

[royal herald]

what did u simplify that to

u cant rlly simplify that cuz the 2(9)(12) is multiplied to the cosA

[vague spoke]

ohhh

but isnt it just 9cosA?

wait nvm

so its 64 = 225 - 216cosA

[royal herald]

,w 144+81

,calc 2912

[clever fjord]

Result:

216

[royal herald]

yes

[vague spoke]

now what

[royal herald]

simplify

[vague spoke]

i move 225 to the left?

[vague spoke]

wait im stuck

[royal herald]

why

[vague spoke]

can i just swap them over to become positive?

[royal herald]

idk what u mean by swap them over

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[vague spoke]

nvm help me on this step

.reopen

[odd edge]

✅

[royal herald]

a = b - cd

solve for d

[vague spoke]

a-b+c = d???

[royal herald]

no

you need to review your fundamental algebraic operations

theres no point in attempting problems like these if you dont have those down

<@&268886789983436800>

[warped grove]

don't feed the trolls

[vague spoke]

64-225 = -216cosA ?

[royal herald]

yea thats the first step

[vague spoke]

-161 = -216cosA so then -161/-216?

[royal herald]

yes

[vague spoke]

so if i move to cos over does it become cos-1

[royal herald]

dont say things like "swap" "move over"

use words like add, subtract, multiply, divide

exponentiate

i cant understand otherwise

[vague spoke]

ok

[royal herald]

what did u get after this

-161 = -216cosA

[vague spoke]

161/216 = cosA

[royal herald]

yes

now what will u do

[vague spoke]

idk

cos-1 161/216 ?

[royal herald]

by cos-1, do u mean

[vague spoke]

cos^-1

[royal herald]

ok

yes

thats correct

[vague spoke]

ohh

alright

thx

[royal herald]

good job

review ur algebra again and u should be solid for these problems

[vague spoke]

ok

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@vague spoke Has your question been resolved?

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[dusky locust]

how would you do this question?

[odd edge]

@dusky locust Has your question been resolved?

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lavish frost]

not exactly sure how to solve b

[lavish frost]

average acceleration is v(tF) - v(tI)/tF - tI

but instantaneous is v'(t)

or is a(t) isntead of v(t)

[cold sage]

youre right in both first two lines