help-38

[slim pawn]

because it is 10√2cm

[granite light]

you can find that side too

but actually no

yes

[slim pawn]

and the question is asking about 10√2cm

[granite light]

it is 10√2

coincidence

it is asking to consider the case where container is full to the brim

in other words the h√2 hypotenuse is equal to the segment you marked in purple

[slim pawn]

in this case

i gotcha now

so it is 20√2 right?

[granite light]

20/√2

because h√2 = 20cm -> h = (20cm)/(√2)

[slim pawn]

yes

[granite light]

now this doesn't look like 10√2

but actually

[slim pawn]

do we have to rationalise?

(20cm)/(√2)

i think we do have

because if we rationalize

it will be 20√2 / 2

[granite light]

you got it

[slim pawn]

let's go

[granite light]

well 20/2

is 10

[slim pawn]

yeah

perfect

[granite light]

so the other part asks you to find the volume

which you can do by finding the area of rectangle+area of triangle and multiplying that by 10cm

(because formula for the volume of a prism or something)

[slim pawn]

idk how to proove it

but i think the liquid will be the purple

so

the volume is 1000

[granite light]

no the liquid is green

the surface of the liquid has to be parallel to the ground

[granite light]

.

because gravity

[slim pawn]

the are of the triangle is 1000

10 * 10 * 10

[granite light]

why?

[slim pawn]

the yellow side is 10

no

it is 500

[granite light]

why?

[slim pawn]

we know that yellow, red and blue is 10

[granite light]

do you mean area or volume?

[slim pawn]

volume

[granite light]

ok

and we know angle between red and yellow is 90°

[slim pawn]

yes

[granite light]

if that wasn't the case we couldn't apply this formula

so what about the rectangle

[slim pawn]

we don't know the length of the missing side

the height

[granite light]

we know height + yellow = 20

[slim pawn]

so the height would be 10

[slim pawn]

not in this figure right

[granite light]

no

[slim pawn]

ok

[granite light]

but we are saying it's full all the way for part b

[slim pawn]

so the height is really 10

[granite light]

yes

[slim pawn]

so the volume is 1000

[granite light]

yes, and then the total volume is

[slim pawn]

the liquid?

[granite light]

yes

[slim pawn]

1000

isnt it?

[granite light]

no?

[slim pawn]

1500

[granite light]

why?

[slim pawn]

1000 + 500

[granite light]

yes

[slim pawn]

no

[granite light]

1500cm³

[slim pawn]

the triangle is not like this anymore

the triangle changed

[granite light]

it actually didn't

[slim pawn]

really?

[granite light]

it just moved up

[slim pawn]

no, in this case we added liquid

because the rectangle in bigger and the triangle is the same

so we have more liquid now

[granite light]

yes

but we calculated the volume for the big rectangle

(rectangular prism, technically)

[slim pawn]

i can see the answers too if you want

but i think we didnt added liquid

lemme see here

[granite light]

1500cm³ is correct

[slim pawn]

hmm

you are right

[granite light]

there is an easy way to see this actually

[slim pawn]

tell me

[granite light]

the total volume of 10x10x20 is what?

[slim pawn]

2000

[granite light]

we can split it into 2 equal cubes

[slim pawn]

1000

[granite light]

we can split each cube along the diagonal in half

[slim pawn]

500

[granite light]

split the top cube such that you remove the part which would overflow

so you have 3/4 of the 2000

[slim pawn]

1500

that is weird

[granite light]

that's the easy way to answer b

but this methods doesn't work at all for c

[slim pawn]

i think i understood now

so it really added liquid

i thought that we could move the arrow

[granite light]

well the amount of liquid is determined by where the arrow points

[slim pawn]

wait

[granite light]

idk how to explain this better

[slim pawn]

so the arrow continued in the same place?

[granite light]

in the new diagram it points to the edge

of the thing

[slim pawn]

understand

my biggest problem was probably because i didn't understand the question

anyways

thank you

have a great day

.close

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[cobalt cloak]

how to show that there is instataneous acceleleraton at t = 10 given v=0.5t for 0≤t≤10, v=0.25t^2-8t+40 for 10≤t≤20

[livid thunder]

find v at t=10 for both equations

[cobalt cloak]

and then

how do i justify it?

[livid thunder]

well if its velocity gives different answers at t=10 then the only way for that to be true is if it accelerated instantly

[cobalt cloak]

i don't need to differentiate it?

[livid thunder]

i mean u can if u want

[cobalt cloak]

if i do it this way how do i prove

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[leaden nacelle]

you can draw it

that proves that there is a vertical line in the velocity

meaning instant acceleration from one velocity to another

if that makes sense

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@cobalt cloak Has your question been resolved?

[cobalt cloak]

question doesn't let me draw though

😭

i get what you mean

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[wraith hinge]

hello

[wraith hinge]

i need help with calc 3 problem

i solved a

but i dont know how to solve b

<@&286206848099549185>

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[wraith hinge]

<@&286206848099549185>

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[unique minnow]

Well using the hint, can you give a vector from the equation of the plane that we know is always perpendicular to the plane?

[wraith hinge]

well i can know the normal to the plane

using the equation of it

but idk what to do after that

[wraith hinge]

.

[unique minnow]

Well we would want a point on the sphere where the vector from the center of the sphere to that point is parallel to the normal vector.

Can you give the center of the sphere?

[wraith hinge]

yes u have it

look i formed a line equations

equation *

using the normal vector as a direction vector

and the center as a poin on it

[unique minnow]

That's good.

[wraith hinge]

then i used

the parametric equation

in the eq of the plane

found t

then found x y z

does that work?

[unique minnow]

Yeah that's fine.

[wraith hinge]

very good

thank u for ur help

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[nova spire]

I don't understand what you mean by that

and I'm not sure why 6*2 gives you the second digit

sure last digit is 1

but what you're doing isn't really justified

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@wraith hinge Has your question been resolved?

[sudden mist]

Oh you could just do this with the binomial expansion

idk if the video does that

Maybe the method you did it in is a lot more effective, but i could do it by rewriting 61 as 6*10 + 1

To find the last two digits

we want to find the coef. a,b which are in the form a 10^1 + b 10^0

the binomial expansion lets us do this

by expanding the first two first of (6*10 + 1)^12

we get the two terms ${12 \choose 0}(6\cdot 10)^0 + {12 \choose 1}(6\cdot 10)^1$

[solid kiln]

Aslan

[sudden mist]

here the left term is just 1

while the right term is 720

so the first digit is 1

and the 2nd is 2

Since its the first two terms of the binomial expansion, and anything higher would be a bigger exponent in the (6*10) factor, thus a bigger digit and it does not contribute to the digits before it

I.e you sum 1 and 720

the one from above

then read off the two first digits

the 7 here is not correct for the third btw

you need to calculate another term for that

Remember what i wrote above

To read off the digits, we want to find a way to express the number in the form ... + a 10^1 + b 10^0

where a is the second digit and b the first

we could do this via the binonal expansion

But again, this might not be the best method

Did you read the above?

Go from the start and start there

What is the second digit in 721?

im only including the 7 here to show that this comes from the above calculation

but its irrelevant here

Where did you get?

Oh i see you rewrote it

Yeah so did you follow the part before it?

This does not give the second digit in general btw

yeah

So let's go trhough the details again

are you familar with the binomial expansion?

Of the binomial expansion?

uh idk expand (a+b)^n for fixed a,b and natural n

youll have to know how to use the binomial expansion formula

but idk if ur method requires that; so maybe useless to learn

We'll need it for my method

but thats assuming you want to understand it

if not, thats perfectly fine

Oh i just checked it and it seems theyre doing something very simliar with the binomial expansion

but not rewriting 61 as 6*10 + 1

in any case

youll have to understand the binomial expansion formula if you want to understand why this works

the binomial expansion formula?

well it depends on what level, do you want to understand why its true or just see the formula?

let's try

i'll first just tex the formula and work from there

Suppose $a,b$ are real numbers and $n$ is a natural number. Then we have that $(a+b)^n = {n\choose 0}a^0 b^{n-0} + \cdots + {n\choose n}a^n b^{n-n} $.

This is the formula

are you familiar with the notation used?

Hm

lets maybe try and rewrite it without sigma notation

[solid kiln]

Aslan

[sudden mist]

Does this look more familiar?

Uh thats fine, no worries

This one ${n\choose k}$?

[solid kiln]

Aslan

[sudden mist]

Are you famliar with factorial, like n! ?

like combinatorics in general

neat

so are you aware of permutations?

n! would be a kind pf permutation

so we could permutate a subset of objects from a bag of things, say we want to permute 3 from 6 objects

then we notice that in total theres 6! ways, but we have to divide away the number of way of not selecting 3 things, so (6-3)! number of things.

I.e the total number of ways to select 3 out of 6 objects (when order matters) is 6!/(6-3)! number of ways, so in notation that would be P(6,3). Again, in other words the number of permutations of 3 things out of 6

Okay so the bracket notation is basically permuations but without order

so ${n\choose k} = \frac{P(n,k)}{k!}$

[solid kiln]

Aslan

[sudden mist]

there we go

I.e this bracket notation is the total number of combinations

Where the total number of permutations (from $k$ elements out of $n$) as we saw was just $P(n,k)=\frac{n!}{(n-k)!}$

[solid kiln]

Aslan

[sudden mist]

Makes sense?

yup!

Why would you think that?

0! is infact defined

and its not zero

Its defined as 1

incase that was unclear

In any case, hopefully that notation is clear

[sudden mist]

Lets go back to this?

I'll try explaining why this is true

Sounds good?

Let me also just repost the image for clarity

Alright, so in order to get a term in (a+b)^n = (a+b)(a+b)....(a+b); then we must in each parenthesis choose either a or b.

Right?

Yeah but without order now, so the bracket notation

Notice that the terms a^k b^n-k occurs as many times as we can choose a in k, out of the n parenthesis's, i.e exactly n choose k, (the bracket notation)

So each term a^k b^n-k has a corresponding coefficient n choose k

and thats pretty much it

Yeah sorry i should of made that clear, here k is the term we're talking about, so k = 0 is the first term and k = n is the last term

i could of added that back in again

let me do that quickly

Suppose $a,b$ are real numbers and $n$ is a natural number. Then we have that $$(a+b)^n = {n\choose 0}a^0 b^{n-0} + \cdots + {n\choose k}a^k b^{n-k} + \cdots + {n\choose n}a^n b^{n-n}. $$

Looks rather messy though

thats why i prefer the sigma sum notation for stuff like this

[solid kiln]

Aslan

[sudden mist]

So first off a^k b^(n-k), do you understand this part?

the terms

Yeah

do you understand how we got there from the line above?

So think of it like this way, when you multiply each parenthesis, you naturally do so by distributing the a's and b's depending on which parenthesis youre working with, right?

Say in the first parenthesis you picked a, then you have a binary choice what to multiply this with, either a or b in the next parenthesis.

Suppose you only pick b afterwards

then you get a term of the form a^1 b^(n-1), right?

Now its just the n parenthesis from (a+b)^n = (a+b)(a+b)...(a+b)

i.e the start

Take (a+b)(a+b) for example, you could start from the left and say first pick a, and then multiply it with either a or b in the next parenthesis, right?

so you either get a^2 or ab

or in general, a^2 b^(2-2) or a^1 b^(2-1)

right?

Notice that this is just of the form a^k b^(n-k)

im trying to rewrite it in the more general setting, therye the same still

b^(2-2) = 1

just to motivate where a^k b^(n-k) comes from

its basically a sequence of yes and no's for selecting a or b in the multiplication

Do you notice how when doing the multiplication for (a+b)(a+b)...(a+b) starting from the leftmost parenthis and going to each one to the right

we're asking

do you want to multiply a or b?

Yes, but theyre seperate terms

we want to get a hold of every term separately

hence the formula above

not to mention that some of these will occur more than others!

for example

a^n

only occurs once

and so does b^n

and infact these are the only two terms in the expansion that occurs exactly once

do you see how?

Yup

and why is that the only way to do so?

notice how multiplication is commutative

the order does not matter!

while yes there are n! ways of multiplying all the a's

it doesnt matter

as all the a's are the same and multiplication commutes and we only pick a each time

so we'll only get one such term

so i guess this smoothly transitions to how we find how many times the other terms occurs, since you seem to agree that all the terms are of the from a^k b^(n-k) now?

yeah

so going back to (a+b)(a+b)

notice how theres another way of getting ab

by picking b instead at first

and then a

you're picking them from different parenthesis

so the first ab was by multiplying the a from the first parenthesis, and the b from the 2nd. While the second ab we get by multiplying b from the first parenthesis (a different element, not the same as a) and then the a from the 2nd.

Notice how these are genuinely two different ways of getting ab

Comparing it to a^2, we can only do so once

doesnt matter

and the reason here is really because we're picking a and b differently in each parenthesis, thats why that counts as a different way

youve already used up the a from the 1st parenthesis

from the first ab

if that makes sense!

So this leads up the sentence "that the terms a^k b^n-k occurs as many times as we can choose a in k, out of the n parenthesis's, i.e exactly n choose k"

We can make use of the notion of n choose k

since this is what we're dealing with

we're dealing with the amount of combinations, selecting an object out of a bigger collection, where order doesnt matter

For example, how many ways are there of selecting a in 2 out the n parenthesis's?

This hardly relates to it at this point in time

but you need the formula for it

which you said you didnt understand

and wanted an explanation

so heres my attempt of one!

Well, its quite hard to go trhough a few weeks of combinatorics or whatever in just a few sentences, so if the explantion is not coming through its totally understandable; hopefully if you still care i think a video or two about the binomial expansion is better

They might be more concrete

Yeah i could tell, i feel mean for not stopping

also its 4 am atm lmao

I hope you got something out of this atleast

hopefully it'll come easier when you decide to watch videos about it maybe

or ask about it here

maybe someone else can chime in, if u decide to keep the channel open

The first part being?

Oh i see

well again, so what i was getting at before was that in either case youll need to be comfortable with the binomial expansion, in order for me to explain this

since the method basically is just the binomial expansion

he does every so slightly differently

but it still uses that

Yeah, im just using the formula is all

And the fact that the formula writes it almost in the form a_1 10^0 + a_2 10^1 + ... etc

so we can just read of the cofficents a_1, and a_2 to see what the first and 2nd digits are

and since these are just the first two terms of the expansion this is rather easy

[sudden mist]

by writing the number in a form like 6*10 + 1 for example instead of 61

but the person in the video does it slightly differently for some reason, using x instead of 10

which is not clear why

i alteast explain why that will work if i use 10, it makes sense atleast that way

maybe the confusion here is that youre not aware what the purpose of rewriting the number as a_1 10^0 + a_2 10^1 + ... does?

So you do get it?

[sudden mist]

Since im taking this almost for granted btw

the rest is just the binomial formula

[acoustic idol]

Too

Yo

Can anyone help me out with this question

[sudden mist]

If you read #❓how-to-get-help someone kind enough will

@wraith hinge Im gonna sleep now, so goodluck!

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[ancient edge]

My work to solve the SA however when I plugged in the values, I get differing answers

[unique minnow]

I've not completely diagnosed it but the inside the of the square root should be 1 + 4u^2, not 1+2u^2.

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[nimble grove]

is there a way to simplify this before I make the limit goes to 0?

[spiral kettle]

this step that you did to simplify the numerator was not valid in the first place (you can’t just bring the thing you’re multiplying by inside the tangent)

[nimble grove]

What am i supposed to do then..

[nimble grove]

Can i eliminate sin and tan before i put in y+5 to x?

[spiral kettle]

that is certainly something that you can try. if you write tan as sin(the stuff inside)/cos(the stuff inside) you might end up with something that you can work with. also is this for a class or just standardized exam practice? if it’s just to prep for standardized exams then we can do some funny tricks

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[arctic saddle]

use the circle theorem to help find the values of the pronumerals (pls help thank you :))

[worldly lagoon]

Isn't that 125

115

[arctic saddle]

HOW DID YOU GET THERE??

[worldly lagoon]

The angle at the circumference is the half the angle at the diameter

[arctic saddle]

Wow you're an actual wizard

[arctic saddle]

could you possibly dumb it down for me? TwT

[worldly lagoon]

Basically of there's a circle and an angle in the middle

With straight lines to any point on the edge of the circle

The angle formed by those lines will be 1/2 the angle in the middle

Sorry I'm not too good at explaining things hope that helps

[arctic saddle]

so does it matter if the straight lines are equal or does it just have to meet at the edge of the circle??

[worldly lagoon]

Nope

Just had to meet at the edge

[arctic saddle]

OHHHHHH

[worldly lagoon]

Basically try remember the shape in that question and if you ever see it again you know what to do

[arctic saddle]

is the other lines for distraction btw??

[worldly lagoon]

The other line also has the angle 115

Cause it's the same shape

[arctic saddle]

ohhhh okayy

tysm man!!

[worldly lagoon]

You're welcome

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[arctic saddle]

Sorry to ask another question but how do I prove that these triangles are similar?

[arctic saddle]

This one too sorry

[supple saffron]

So proving triangle similar u need minimum 2 angle the same

,rotate

[prisma stone]

just place the angles

[supple saffron]

in this case u can see BAE = DAC

[supple saffron]

bro they haven't study trigonometry i think

[arctic saddle]

Ya not very good at trig sorry 😭

[prisma stone]

they don't have to'

35+55 = 90
180 - 90 = 90

[supple saffron]

u dont have to

[supple saffron]

u were asked for x value right?

[arctic saddle]

Yes

[supple saffron]

also angle AEB = ADC because they are paralel

[arctic saddle]

Oohhh

[supple saffron]

use similarity

[arctic saddle]

How do I do that?? Sorry

[supple saffron]

so see triangle ABD and DBC

are they similar?

if yes show me

[arctic saddle]

Yes because we use AAA

BAE = CAD

[supple saffron]

bro for similarity we dont use same sides

we use sides for congruent which the triangle the same

[arctic saddle]

Ohh

[supple saffron]

now with that u can do this

[arctic saddle]

AAA

[supple saffron]

$\frac{BD}{BC} = \frac{AB}{BD}$

[solid kiln]

Newt

[supple saffron]

plug every value

you got

$\frac{10}{7} = \frac{x}{10}$

[solid kiln]

Newt

[supple saffron]

$x = \frac{10 \times 10}{7}$

[solid kiln]

Newt

[supple saffron]

@arctic saddle here you go

[arctic saddle]

OHHHH

[arctic saddle]

Btw is this all I need to prove they're similar??

[supple saffron]

also angle AEB = ADC because they are paralel and
in this case u can see angle BAE = DAC

[arctic saddle]

Using that how do I figure out x?

[arctic saddle]

Ohhh okay that makes sense

[supple saffron]

have u study comparison in similarity

[arctic saddle]

I don't think so

But it might be on the test and I couldn't find any info on how to do it in the power points 😭😭

Is it the same as ratio??

[supple saffron]

power of points?

[sage jacinth]

?? Only 1 triangle

[arctic saddle]

PowerPoint as in the presentations

[arctic saddle]

It's similar triangles so there's 2

[sage jacinth]

Sorry I'm unable to understand your question,can u send that again

[arctic saddle]

Basically I have to prove similarity and find x

I know how to prove it's similar but idk how to find x

[sage jacinth]

Ohh

[supple saffron]

Bruh i thought power of point in mathematics lol

[arctic saddle]

Sorry bout that😭😭

[supple saffron]

nah it's okat

[sage jacinth]

Angle A is common &
'DC' is parallel to 'EB'
So angle(AEB)=(ADC)
Hence ∆ADC~∆AEB

[arctic saddle]

So how do I find x??

[sage jacinth]

If 2 triangles are similer,then ratio of their side is constant
ie (10-x)/10 = (3/8)

[arctic saddle]

Ohhh I seee

So if I get the ratio

[sage jacinth]

U get the equation to solve for x

[sage jacinth]

Which I have given

[arctic saddle]

3/8=2.67

That's the ratio

[supple saffron]

huh

[arctic saddle]

Is that not right?? 😭😭

[supple saffron]

$\frac{10-x}{10} = \frac{3}{8}$

[solid kiln]

Newt

[sage jacinth]

Solve
(10-x)/10 =(3/8)
& Calc x

[supple saffron]

the equation was like this

use the method to multiply diagnoally idk what the name my english bad

[arctic saddle]

Ohhh

[sage jacinth]

U are right but u don't need to divide like that , that's waste of time

[arctic saddle]

Wait how would you do it??

[supple saffron]

btw 3/8 isn't 2.67

it is 8/3

,calc 3/8

[solid kiln]

Result:

0.375

[supple saffron]

but it is useless

[sage jacinth]

The ratio of sides of similar triangle is equal to each other

@arctic saddle i think you first need to study some properties of similar triangle

[arctic saddle]

Okk

[sage jacinth]

Your profile animation is good and unique

I like that

[arctic saddle]

Thank you I brainrot on val

[supple saffron]

bro my pfp is empty idh any money i am broke

[arctic saddle]

Mines free

[supple saffron]

😭

[supple saffron]

how

[arctic saddle]

It's from a quest my guy!!

[sage jacinth]

❤️

[sage jacinth]

Ohh

[arctic saddle]

I think all I did was join call and it gave me a quest

And then I accept it and just had to run val till it reach 100%

[sage jacinth]

I don't know much about this😅

[arctic saddle]

At least yk math

Which is more useful

[sage jacinth]

No-one can say he know maths

Know a little bit

[arctic saddle]

More than me lmaoo

[sage jacinth]

@arctic saddle u are most useful as u r edible, mushroom 🍄😉

[supple saffron]

i dont know math

math is to complex to know perhaps

[arctic saddle]

Yes I am

[arctic saddle]

Trig hard

[sage jacinth]

Friends

[arctic saddle]

Y'all are philosophical 🫡

[supple saffron]

idk Trigonometry

i am still grad 9

[sage jacinth]

Trigonometry is easiest,i think

[arctic saddle]

HUH

WHAT COUNTRY

THIS IS YR 10 EXTENTION

[supple saffron]

depends

[arctic saddle]

WHAT

[supple saffron]

no i study math comp that why

but since it math comp i haven't study trig

[arctic saddle]

That's actually insane 😭😭

[sage jacinth]

Yes it depends on practice

[supple saffron]

no

[arctic saddle]

I thought y'all were in uni or sum😭😭

[sage jacinth]

What

[arctic saddle]

Idk man y'all are just smart like that 😭

[sage jacinth]

Ya we r smart, but little less than u

As u r mashroom

[arctic saddle]

Insane

Also off topic but do you guys have tips on what to put on a maths notes page??

[supple saffron]

but in my curriculum it was 9 grade though

[sage jacinth]

We'll talk about it later on dm,as I have to go,bye

[arctic saddle]

Byee

[supple saffron]

wdym?

[arctic saddle]

Diff country probs

[supple saffron]

yeah

[sage jacinth]

Very nice as u r trying to improve u in maths
All the best

[arctic saddle]

Like cuz in my school we're allowed to take a notes page for the calc section

[supple saffron]

calc is calculus?

[arctic saddle]

Calculator section

We're not learning calculas yet

[supple saffron]

depend on how u study anyways

[arctic saddle]

Should I put examples to help me??

[sage jacinth]

Yes

[arctic saddle]

okii thanks guyss

[supple saffron]

that will help

[supple saffron]

Can you do .close?

[arctic saddle]

yup

thank you for the help

.close

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[vernal fern]

how can i find the exact value of sin(-165)?

[vernal fern]

i need to use one of those half angle identities right

[split chasm]

not necessarily

you could so stuff like compound angle identities and oddness

[cold dune]

sin(-165) = -sin(15) = -sin(45-30)

[vernal fern]

what is this?

why?

[split chasm]

you could so stuff like compound angle identities and oddness
(and supplementary identity)

its more convenient than the half angle identity which gets you nested roots in many cases

[cold dune]

for why sin(-165) = -sin(15) you can use the unit circle, or prove it like this:
sin(-165) = -sin(165) = -sin(180-15) = -sin(15)

[vernal fern]

so just a second, can we go step by step?

first thing I have to do to solve for sin(-165)

[cold dune]

sure

[cold dune]

a general solution would be using the supplementary angle identities to get to angles which you can calculate their exact value (30, 45 60 etc), and then using compound angle identities to evaluate it

let's do that step by step

[vernal fern]

okay

[cold dune]

you have sin(-165)

do you agree that it's equal to -sin(165)

[vernal fern]

yes

[cold dune]

alright, so we have -sin(165) = -sin(180-15)

this is where we can use supplementary angle identities

do u know them?

[vernal fern]

um

no

[cold dune]

ahh i see
so there are a couple identities which you can use when you have sin(180±x) or cos(180±x), to make them simpler, the one you're gonna use here is
sin(180-x) = sin(x)

you can prove it just by doing that on the unit circle

https://imgur.com/mlQfmsf

the other ones are
cos(-x) = cos(x)
cos(180-x) = -cos(x)
cos(180+x) = -cos(x)
sin(180-x) = sin(x)
sin(180+x) = -sin(x)

[cold dune]

try using this one to change -sin(180-15)

[vernal fern]

wait so decimal form -0.99

[cold dune]

we're still dealing with angles though

where'd you get -0.99 from?

[vernal fern]

i just used a calculator to check

[cold dune]

oh

I meant try using the identity to change -sin(180-15)

we'll get to the exact value in a bit

[vernal fern]

oh wait you did

which identity though?

[cold dune]

sin(180-x) = sin(x)

try using this^ to change -sin(180-x)

[vernal fern]

im so sorry i dont understand

[cold dune]

it's ok let's break it down

you know the unit circle though right?

[vernal fern]

yes yes i do

love the unit circle

[cold dune]

frr

ok so

imagine you start from 0 on the unit circle, and move an imaginary amount, say x degrees on it

[vernal fern]

yes alright

[cold dune]

it'll look like something like this

[cold dune]

https://imgur.com/fFYIBHP

now about the point on the circle we're pointing at, do we agree that it's y coordinate is equal to sin(x)?

[vernal fern]

understanding..

[cold dune]

for example if we move 30 degrees on the circle, we'll get to a point where its y coordinate is sin(30), or 1/2 to be exact

[vernal fern]

okay

[cold dune]

now imagine if we start from 0, then move 180 degrees on the circle, and then move back x degrees

something like this

https://imgur.com/CMDsldu

we see that if we move x degrees or 180-x degrees on the circle, we get the same y coordinate (in other words, we get the same sine)

[vernal fern]

wait why are we doing that though

[cold dune]

that proves sin(180-x) = sin(x)

so in your question we have -sin(180-15)

following the same logic, we can say -sin(180-15) = -sin(15)

[vernal fern]

oh

so we move the thing that isnt sine to get the same sine?

[cold dune]

all we did was change sin(-165) to -sin(15), to make it simpler to solve the problem

do you know about compound angle identities though?

[vernal fern]

also no

but i have google!

[cold dune]

nice!

in short

sin(x+y) = sin(x).cos(y) + sin(y).cos(x)
cos(x+y) = cos(x).cos(y) - sin(x).sin(y)

as u can see, we can write -sin(15) as -sin(45-30)

[vernal fern]

oh oh oh

[cold dune]

imagine x = 45 and y = -30, and use this formula to evaluate it

[vernal fern]

okay thank you very much i have to run to work i will do this on my break and come back in a couple hours

thank you very much!!!

[cold dune]

cheers! good luck

[vernal fern]

so sorry i cant see it thru rn time management where

.close

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[drifting storm]

guys can anyone pls help guide on how to solve this

[lilac flame]

you can start by simplifying what's inside the modulus

[drifting storm]

i tried this way but i think it’s wrong

[drifting storm]

do i do that by removing fractions?

[lilac flame]

yes

commond denominator

[drifting storm]

oooh

[lilac flame]

this is almost correct

you should plug in the solutions you found

to check if they're correct

because for example the left side is only true if what's inside the modulus is positive

[drifting storm]

ooohhhh

ohhhh ure rightt

i got it correct alreadyy

thank you so muchh

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[lilac flame]

np

[drifting storm]

.reopen

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✅

[drifting storm]

btw guys, is |21/5| = -21/5?

[nova spire]

no

|21/5| is 21/5

since 21/5 > 0

|x| always outputs the positive (non-negative) value

[drifting storm]

ohhh wait yes ure right, absolute value should give non negative results

so do i conclude that this equation only has one solution?

[drifting turtle]

solve 2x+7 = 0 first to note when the absolute value function changes the sign of 3x

which is -7/2

now for each case x >= -7/2 and x < -7/2

for the first case the RHS is just 3x, for second case RHS is -3x

then solve for each case

reject when the answer contradicts the assumption (for the first case l, x>= -7/2)

[drifting storm]

ooohh okk thanksss

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[drifting turtle]

np

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[prime lynx]

Yeah sure

[nimble stone]

i would say or rather than and, but sure, both are okay

[prime lynx]

Although change the and to or

[nimble stone]

and means it has to satisfy both of them to use that expression

which doesnt work

or means it can satisfy either or both

that much is fine, yeah

no real difference really, one just takes less time to write

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[untold phoenix]

Hey I need help with this

[untold phoenix]

This is the drawing I’ve setup

Now Im a little confused with finding the top of my Z, because Z = to squareroot of 9-y^2 seems right

But the thing is why is that the case? Its a cylinder extending infinitely into X no?

[prime sierra]

the plane y=3x prevents x from getting to infinity.

and y is limited to 0<=y<=3

[untold phoenix]

Oh yeah ur right, I got confused because my drawing skills suck LOL

That black shape is my new drawing, that’s the solid right?

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[deep solstice]

if 100% is 124
.... %? 279

[pseudo zodiac]

100/100 * x = 124
=> x = 124

y/100 * 124 = 279
y = 27900/124

[deep solstice]

225%?

[pseudo zodiac]

yup

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[snow night]

I need help

[snow night]

A guy asked 50 dollars from his mom and dad so he could buy a shirt, he now has 100 dollars. He then went outside to buy a shirt, he said "Hey, how much is shirt?" The seller said "100 dollars" he then said "can I get a discount?" The seller "said sure I'll sell it to you for 97$" he now has 3 dollars that he will give it to both his parent's and the extra 1 dollar for him. He then gave 1 dollar to both his parents, since he gave the 1 dollar back the parents gave him 50 each in total. 49+49 = 98$ + 1 from his extra change = 99$. Where did the extra 1 go$

[solid kiln]

𝗖𝗼𝗿𝗺𝗮𝗰

[empty sluice]

Wait. If he gives 1 dollar each to his parents, he has 1 dollar with him. Then his parents give him 50 each. So won't he have 1+50+50=101? Or did I get the question wrong?

[snow night]

His parents gave him 50 each

Then he bought shirt for 97

He now has 3

He gave 1 dollar to his mom and dad

He kept 1 dollar

[empty sluice]

Right

[snow night]

Then

Since he gave 1 to his parents

The parents made 1 dollar

So like

Wait

It's like this

[empty sluice]

They don't give him 49, they give him 50

!original

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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.

[snow night]

It was just told

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[wind cloak]

So after he gave 1 dollar to each of his parents, we can see this new situation like the both the parents gave him 49 dollars each.
So he has total 98 dollars. From these 98 dollars, he bought a shirt for 97 dollars and kept one for himself

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[strange imp]

In the following system of equations, y can be expressed as m/n, where m and n are positive integers and relatively prime. What is the value of m + n?

[strange imp]

(A)30 B(14) C(27) D(12) E(24)

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[static girder]

Javier draws a quadrilateral. The first and second angles are equal. The third angle is 45° more than the first angle. The fourth angle is half the size of the first angle.

(a) Let the size of the first angle be x. Write down the sizes of the other three angles in terms of x.

(b) Form an equation in x and solve for y x to find the size of each angle of the quadrilateral.

I have done a which is
1, 2 = x
3 = x + 45°
4 = x ÷ 2.

I'm still stuck on b because I don't know what equation to make.

[mild oxide]

Yo

[static girder]

Hi

[mild oxide]

1st is x and 2nd is x 3rd is x+45 and 4th is x/2

Fr

[static girder]

Yes I've done that

[mild oxide]

Now see

Do u know sum of all angles of quadrilateral is 360 degree

Enjoy

[static girder]

Ok ty, I'll see what I can do

How do I close this

[mild oxide]

.close

[static girder]

ok tysm

.close

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[mild oxide]

Welcome

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[hollow arrow]

quadratic equation?

[lavish sequoia]

Yh

I put 2.

Which is wrong

[hollow arrow]

Isn't it supposed to be 0?

[lavish sequoia]

yh

[hollow arrow]

the discriminant is less than 0

[lavish sequoia]

yh

[hollow arrow]

So there are no real solutions

[lavish sequoia]

oh

okay

tyyyy

[hollow arrow]

if the discriminant = 0

[lavish sequoia]

yep

[hollow arrow]

then there is one real solution

[lavish sequoia]

yh

[hollow arrow]

Okay I think you got it

[lavish sequoia]

thnx

[lavish sequoia]

yeah

[hollow arrow]

np

[lavish sequoia]

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[empty sluice]

What are you supposed to find?

[supple copper]

i mean

i just guessed

[vernal briar]

Yeah guessing is easy here

[supple copper]

pick 3 integers that multiply to 4

[elder glade]

I know but how do u do it with equations

[supple copper]

you don't

maybe you do actually im not sure

there's some weird cubic factorisation methods i've not studied

in general it's really ass though

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✅

[supple copper]

i wouldn't

i just said, i wouldn't

i'd look at it

try to guess

if it doesn't work i'd give up

lol

i mean i suppose you can substitute things in and see if it works

solve for alpha, sub into beta, etc etc

that's a pain though

idk it sounds terribel

there's a cubic formula if you want to take a look

no one really uses it though

[sudden mist]

IIRC you could also try to exploit some kind of symmetry to avoid solving a cubic

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[fathom flicker]

Is the answer of question 17 D?

[chilly bobcat]

yes

[fathom flicker]

thanks!

.close

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[steady silo]

hhelpme

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[pale tulip]

I had a attempt at the question

not sure how I’m doing

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[marsh forum]

Can I have this verified

[marsh forum]

I can tex it if required

[grim sparrow]

do you mind elaborating a bit when you say “by definition”?

[chilly bobcat]

hmm, for sums like these a more formal answer would be better

you can start with calling a variable x, where x belongs to A

[worldly wing]

What does U_f represent?

[chilly bobcat]

all the elements in the family of sets F

if F is {{1, 2}, {2, 3}} then U F is {1, 2, 3}

[worldly wing]

How is it different than F itself then

[worldly wing]

Ohhhh okay

[sudden mist]

It’s just the union

[worldly wing]

Oh then it’s just the union of all the sets in F

Yeah

[sudden mist]

It has a precise definition in this case, which the exercise is all about

[marsh forum]

One minute, have some bank work to finish, will be back in 5

So I basically have to define the unions of multiple sets here

[sudden mist]

You have to define? Isn’t it already given

[marsh forum]

It is

So what do I even write then

[sudden mist]

Try and unpack what you have and what the goal is

For example unpack what it means for a set to be a subset of another, since after all this is what you’ll need to show

[marsh forum]

$x \in A_i$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[marsh forum]

Something like this right

[sudden mist]

What is this for?

[marsh forum]

One minute

Trying to construct the definition of union

[sudden mist]

Why are you constructing it? It should already be given

In your textbook or whatever it may be

[chilly bobcat]

so basically, $x \in A$, and\
$A \subseteq \cup F \implies x \in \cup F$

i dont know if there's any special theorem to make it clearer so this is the best I can do

[sudden mist]

??

[chilly bobcat]

what happened? it looks quite right to me

[sudden mist]

A subset U F is what we want to show

This looks confusing

[chilly bobcat]

they have given that A is a subset of F

[chilly bobcat]

i know

im trying to make sense of it too

[sudden mist]

No they’re given A is an element of F

[chilly bobcat]

ahhh my bad

[marsh forum]

So my proof is wrong?

[chilly bobcat]

better now?

[sudden mist]

I mean in any case I don’t see how this is helping

[sudden mist]

You just need to clarify it, or else I could write every proof of my own as “proof by definition” and be done

[chilly bobcat]

not really, just that I'm not sure if they'd accept a worded solution

[marsh forum]

So I need to write the definition of union essentially right?

[sudden mist]

No that’s just gibberish to me

[sudden mist]

Yeah just unpack what everything says

First off start slow

You’re essentially showing a set is a subset of another set

What does that mean? Start like that and peel the layers slowly

[chilly bobcat]

x is an element of A
A is a subset of UF
so, x belongs to UF as well

that's essentially what it means

[sudden mist]

Your second line is literally what we want to show

[chilly bobcat]

yes that's where I'm stuck

[marsh forum]

So this essentially means that $x \in A \implies x \in B$

[chilly bobcat]

I've never done questions using UF

[chilly bobcat]

if A is a subset of B or the other way round

[sudden mist]

Well if you can’t help, then don’t? We’re trying to help someone else here

[solid kiln]

ƒ(Why am. I here)=I don't Know

[sudden mist]

Yeah for arbitrary x

And in our case B is?

[marsh forum]

so F

wait

UF

[sudden mist]

Yeah

[marsh forum]

$\bigcup F$

sorry

[solid kiln]

ƒ(Why am. I here)=I don't Know

[sudden mist]

It’s fine don’t worry

So in your textbook, do you have a definition for a union over a family of sets?

[marsh forum]

so as every arbitrary element x that belongs to A belongs to $\bigcup F$ we have that $ A \subseteq \bigcup F$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[sudden mist]

That’s what we’re trying to convince ourselves of in the moment yes

But we haven’t shown it yet

[marsh forum]

what do you mean

[sudden mist]

We want to show that A is a subset of union F

[marsh forum]

yes

[sudden mist]

This is just restating the definition

It doesn’t mean it’s necessarily true

So convince me!

[marsh forum]

so you want me to define union now right?

[sudden mist]

I could do it for you, the reason we need to is to unpack what the notation is saying; or else it could really mean anything

So x is an element of union F iff there exists some element B in F such that x is an element of B

Does your textbook state something similar?

[marsh forum]

for union my book defined it as follows

$A \cup B = {x| x \in A or x \in B }$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[sudden mist]

That’s a union between two sets, we’re talking about over a family of sets

Is there one for a family of sets?

[marsh forum]

let me check

$\bigcup F = {x | \exists A \in F(x \in A)}$

[solid kiln]

ƒ(Why am. I here)=I don't Know

[sudden mist]

Yup this is the exact same

Note that the letter A here isn’t necessarily the same set so it’s best to use a different letter like I did

So if we pick an element in the union it satisfies the latter condition which you have written

[sudden mist]

I.e this

[marsh forum]

hmm, yeah

[sudden mist]

But notice that by assumption we have that x is in A

so there does indeed exist an element in F (pick B=A)

Such that x is in B

[marsh forum]

oh , that's it?

[sudden mist]

Yup!

[marsh forum]

wow

thanks

I've really got to learn to use defns

[sudden mist]

Np

[marsh forum]

I'm used to using intutution

[sudden mist]

Yeah that’s what I realized too

[marsh forum]

sadge

[sudden mist]

Definitions are really important

[marsh forum]

yeah, thanks

[sudden mist]

Use intuition to make a strategy

But use definitions to actually prove it

[marsh forum]

got it

thanks

.close

[trim jolt]

Channel closed

Closed by @marsh forum

Use .reopen if this was a mistake.

[trim jolt]

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[spring bolt]

If four sides of parallelogram sides touches a circle like this then it will be

[spring bolt]

Square?

[trim jolt]

@spring bolt Has your question been resolved?