help-27

[high knot]

so Rs = B^x+y

[fossil locust]

okay you keep forgetting your indices rules

$b^{x + y} = b^x \cdot b^y$

[woven radish]

south

[high knot]

yeah

[high knot]

isn't this the same as the one u did

oh wait nvm

mb

[fossil locust]

so you get RS = b^x * b^y

by your definition, just multiply both together

and then the rest is fine

[high knot]

yeah so where do we go from there

to get back to the og equation

so I wrote it out

[fossil locust]

$rs = b^{\log_b r + \log_b s}$

[woven radish]

south

[fossil locust]

ah wait so now you can take log base b of both sides

[high knot]

but rs = r + s doesn't make sense

[fossil locust]

no, you need to use laws of indices on $b^{\log_b r} \cdot b^{\log_b s}$

[woven radish]

south

[high knot]

right but b^logbr is basically js r right

[fossil locust]

yes

[high knot]

im wondering

from here

where do I go

[fossil locust]

yes, so this is where you go

remember you said, so we can say x = logbR and y = logbS

[high knot]

can I plug b^x instead of R for the original equation

and B^y for S

[fossil locust]

in where?

[high knot]

in the rule

[fossil locust]

no, that's what you want to prove

you can't use that until you have shown it from other statements

[high knot]

can I prove it like this

[fossil locust]

ah okay now we're talking

you still need to define your terms, so let x = logbR and y = logbS

actually let R = b^x and S = b^y is better

[high knot]

would this work

[fossil locust]

yes! now this works

[high knot]

bro..

[fossil locust]

<@&268886789983436800>

[high knot]

💀

[fossil locust]

also this formula is only valid for R, S > 0 of course, as the right hand side would be undefined otherwise

[high knot]

right but if I was asked to prove just based of the variables

I could use it right

[fossil locust]

oh wait it's trickier when the base can be negative

[fossil locust]

yes

my point is, the assumption that there exists an x such that x = log_b R is valid

[high knot]

oh we didn't get dat far in our content

but its unlikely it will show up on the test but I'll ask again if I come across it in my hw questions

but thank you

[fossil locust]

no worries!

[high knot]

.close

[devout snow]

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[long moss]

guys i cant solve a integral: 1/1+ey the whole *dy

[winter torrent]

let u = 1+ey

[long moss]

we dont study it like this

no wait

[winter torrent]

oooooooookay then how do you study it

[long moss]

yes u = 1+ey

I BRAINFARTED

too much math

[ancient sluice]

Multiply numerator and denominator with e^(-y)

[winter torrent]

that doesn't seem like a good idea

[ancient sluice]

Oh I didn't see, it's ey

[long moss]

how can dy to become d(ey+1)

this is my idea

[winter torrent]

find du/dy

[storm crater]

$\int \frac{1}{1} + ey$, surely

[woven radish]

M8327

[winter torrent]

then multiply both sides by dy

[solid osprey]

[winter torrent]

yes very good you've cracked the code

it's quite obvious that he meant 1/(1+ey)

[storm crater]

I think he might have meant e^y there.

[long moss]

yes yes

like this

[winter torrent]

ugh

can you try writing your question again

[storm crater]

You should put brackets and stuff though. If this was an exam, people will take away points.

[long moss]

im a first year pharmacy and this math stuff aint that important we just have it

[safe knoll]

arctan isnt 1/tan , btw

[winter torrent]

ok starting at the top, before all that fraction stuff, let u = 1 + e^y

[long moss]

i lit watched organic chemistry on that

[winter torrent]

does that say $\f1{\tan}\cdot x²$ at the bottom?

[woven radish]

hayley, who shakes the world

[long moss]

yes

[winter torrent]

do you know what tan is

(tg)

[long moss]

tangens

[winter torrent]

yes it's tangent (are you french?) and it's a function not a number

tan * x doesn't make sense any more than log * x

[long moss]

im from bulgaria

and i feel like you are talking chinise to me

idk i just watch the trigonomy function : tg^-1=arctg

[winter torrent]

yeah our notation sucks

[storm crater]

log*x makes more sense though, https://en.wikipedia.org/wiki/Iterated_logarithm?useskin=vector

[winter torrent]

$\tan\inv$ is the same thing as $\arctan$, which is not the same as $\f1\tan$

[woven radish]

hayley, who shakes the world

[storm crater]

I will now stop interfering here, good luck. 😉

[winter torrent]

this is extremely unhelpful here. if you want to feel smart about something like this put it in the lounge

[long moss]

how if its -1 it can go down right?

[winter torrent]

ordinarily that's what that means, yes. but in the case of trigonometric functions (sin, cos, tan), that symbol means their inverse

so because $\sin(\sfrac\pi6) = \sfrac12$, we can say $\arcsin(\sfrac12) = \sfrac\pi6$

[woven radish]

hayley, who shakes the world

[winter torrent]

which we can also write as $\sin\inv(\sfrac12) = \sfrac\pi6$

[woven radish]

hayley, who shakes the world

[ancient sluice]

I just thought that 😭

[winter torrent]

personally I prefer to just stick to arcsin, arctan, etc

so in your case the right hand side would just be arctg(x²) + C

[long moss]

okay

but how do we solve the integral

[winter torrent]

like this

your manipulations were valid but they actually made the problem harder than it originally was

starting from $\int\frac{e^y}{1+e^y}\dd{y}$, let $u = 1+e^y$

[woven radish]

hayley, who shakes the world

[long moss]

okay then what

put the ey into dy

[winter torrent]

then proceed as you normally would with substitution. I'm not going to do the whole problem for you.

find du/dy

[long moss]

so it become d(ey) and then d(ey+1)?

[winter torrent]

do Bulgarian keyboards not have the key on them?

[long moss]

yes

but its my first time typing math

ima right it down

on my paper

[winter torrent]

ok well when we write exponents we use ^ to denote them

ey means e times y
e^y means e to the power of y

anyway

[long moss]

yep i got it

ima right it down

[winter torrent]

find d(1+e^y)

[winter torrent]

what happened to the 1 + e^y on the left?

also, $\ln|e^{\psq}| = \psq$

[woven radish]

hayley, who shakes the world

[long moss]

what do you mean

can you point where

[winter torrent]

what happened here

[long moss]

integral: dU/U

[winter torrent]

yep

[winter torrent]

yep

what happened to the +1 though

[long moss]

aaaaaa

so my final answer should be: Y + 1 = e^arctgx^2 + C ,right?

[winter torrent]

hmm, no

$\ln|e^y+1| = \arctan(x^2)+C$ is where we were

[woven radish]

hayley, who shakes the world

[winter torrent]

and then your original idea of taking e^ both sides is correct here

we're trying to isolate that y

[long moss]

so

y= e^arctgx^2 + C -1

[safe knoll]

can i add smthing in here?

[winter torrent]

sure

[safe knoll]

idk if u noticed but

dontknow how they got arctanx^2

[winter torrent]

oh i didn't even really look at the right hand side

i should have done so more closely

[long moss]

well

so its

dx^2/1+x^2

[winter torrent]

doing the right hand side correctly will make the result significantly simpler

[long moss]

wait wait wait wait

ima right it again

[winter torrent]

you don't have that though

[long moss]

its U not U^2

[winter torrent]

you'll find that if u = 1 + x^2

[long moss]

yea yea

i think it sohuld look like that

[winter torrent]

yeah that looks much better

a note on style: $\ln e^y + 1$ will be read as $(\ln e^y) + 1$

[woven radish]

hayley, who shakes the world

[winter torrent]

if you want ln to be applied to the whole thing (as you do) then you should write it as $\ln(e^y+1)$

[woven radish]

hayley, who shakes the world

[long moss]

whats the differense

bc its Ln| U|

so its Ln|e^y|+1?

[winter torrent]

no

it's ln( |e^y + 1| )

have you seen notation like f(x) before?

for functions

[long moss]

seen

[winter torrent]

if i said f(x) = x^3 + 2, would you understand that and be able to calculate f(3)?

[long moss]

no

i lit just have to learn basic easy integrals and DI

like i said im not a math major

i dont get most of the things bc i dont study it as much as biology latin and chem

[winter torrent]

idk what DI is but okay

[long moss]

the thing that i showed

differential equations

yes

DE

[winter torrent]

ummmm sin, cos, log, tan -- these are all functions. they take an input and they give back an output.

they are usually written like log(100), and even though this kind of looks like multiplication it is not. it is the number 100 being fed into the machine called "log", which spits out a result (in this case that result is 2)

we can define our own functions, for example if i said g(x) = x^3 + 1, then i'm defining a machine called "g" which takes its input, cubes it, and then adds 1

so if i wanted to calculate g(2), then i'd follow those steps by replacing "x" with "2" and evaluating. in this case i'd get the result of 9, so we can say g(2) = 9

[long moss]

ughhhhhh

do i need to know that

[winter torrent]

tbh yes

you said pharmacy or someething right?

log shows up a lot there

in chemistry

[long moss]

its dif

and we never transform it

we just use log for equsiaons of pH and stuff

and we just keep it log

or Ln10

we never go beyond

and in practical pharmacy logs are never used

[winter torrent]

what class was this problem for? it was slightly more than a basic integral

[long moss]

math class

[winter torrent]

anyway the point i wanted to make with functions was that g(2) + 1 is not equal to g(2 + 1)

just like log(e^y + 1) is not the same as log(e^y) + 1

[long moss]

ahaaaa

i get it a little

btw can you check if these are correrct

the first two are like i woild get on the exam

[winter torrent]

the second one isn't right

the + c should show up as soon as the integral sign disappears

[long moss]

What’s the difference

e^C= C

Isn’t it

[winter torrent]

sure

but e^{x + c} is not e^x + c

[safe knoll]

also im pretty sure
$$e^{- \ln x} \neq x$$
using some log rules/properties :d
$$e^{- \ln x} = e^{\ln x^{-1}} = x^{-1}$$

[long moss]

Talked with my teacher about it

He said leave the minus be there

xD

[woven radish]

JustToPro

[safe knoll]

he said e^-lnx = x? is fine?

[long moss]

Yea it’s e^x*e^C

[winter torrent]

yes

sometimes written as k * e^x

[long moss]

Mhm

He said that minus ain’t that important

And that we can leave the answer like without transforming

So my 3 equsians are correct , right ?

[devout snow]

@long moss Has your question been resolved?

[devout snow]

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[worldly sonnet]

Guys what this means if lets say an eigen value has the multiplicity of 3 aka (L-6)^3 that means that the same eigen value would have 3 eigenvectors and the solution is a 3 dimensional subspace

[arctic field]

(L - 6)^3?

are you saying that as part of the char poly?

[torn vessel]

The number of eigenvectors is not always the same as the degree of the factor.
That’s why they specify “geometric” multiplicity.

[worldly sonnet]

yea

[arctic field]

well that's not going to be equal to the geom multiplicity

there's no single "multiplicity"

[worldly sonnet]

thats equal to algebric multiplicty

?

[arctic field]

yes

[worldly sonnet]

the number of eigenvectors

[arctic field]

they're not equal

[worldly sonnet]

okay

[arctic field]

no

[arctic field]

this is not equal to the alg multiplicity

[worldly sonnet]

okay so number of eigenvectors is not related to any multiplicity?

[arctic field]

no it's the geometric multiplicity

[worldly sonnet]

sorry?

[arctic field]

you probably shouldn't say number of eigenvectors

[worldly sonnet]

right okay

so what is geometric multiplicty

[arctic field]

there's always infinitely many over an infinite field

[worldly sonnet]

and algebric multiplicity

[arctic field]

you talk about dimensionality

[worldly sonnet]

oh oh I see yeah I also meant the span thingy

[arctic field]

geometric multiplicity is the number of lin indep eigenvectors

algebraic multiplicity is the degree in the char poly

[worldly sonnet]

okay right

perfect thank you

.close

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[small badge]

is a linear function denoted as f(x) = ax or f(x) = ax + b

[iron kindle]

in what sense linear?

[wicked coral]

#1330206112569491473

[iron kindle]

!occupied

[devout snow]

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).

[wicked coral]

Pls help

[ancient sluice]

A linear function, means the highest degree of a term in the equation won't exceed 1

[tardy edge]

usually y = ax + b

so that x = -b/a when y = 0

[small badge]

so

[iron kindle]

but could also mean, f(ax+by)=af(x)+bf(y) right?

[small badge]

which one

[ancient sluice]

Ah yes

[tardy edge]

i've been taught it's ax + b

since by the logic

for example a = 1

f(x) = x

that is a linear function

[ancient sluice]

Yes it is !

[onyx tusk]

b=/= 0?

[tardy edge]

but how would you write the form for something such as f(x) = x+2

[ancient sluice]

y= a, is linear too !

[tardy edge]

yes, it is

[small badge]

f(x) = ax is a linear function
f(x) = ax + b is an affine function

[tardy edge]

it's linear in all cases where the highest power is 1

[small badge]

thats how i was taught

[onyx tusk]

i thought the first one would be linear too

[tardy edge]

all are linear

[iron kindle]

that is also correct

[tardy edge]

both are linear

[onyx tusk]

right

[iron kindle]

but generally for high school maths ax+b is seen as linear

[tardy edge]

ax is just another form of ax + b when b = 0

and when theres no vertical shift

[small badge]

so like

linear function

just means it has a curve in the form of a line?

[ancient sluice]

Yeah !

[small badge]

ok so

constant, linear, affine

js all can be called linear

?

[ancient sluice]

Yes, in a wider sense yes !

[small badge]

okay

got it

thanks

.close

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[devout snow]

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[woven yacht]

I am so lost on this problem and don't know what to do.

[iron kindle]

do you know sohcahtoa?

[woven yacht]

I will do pythag and fnid the other side rq

[woven yacht]

hm?

[tardy edge]

cos = A/H

[iron kindle]

,tex .sohcahtoa

[woven radish]

Bonk

[tardy edge]

SOH-CAH-TOA

S = O/H
C = A/H
T = O/A

[woven yacht]

ohh

A is adjacent?

alright let me do that really quick

i got sqrt 26/sqrt 79 is that right?

[ancient sluice]

It is

[woven yacht]

alright

ok ty for the help

much appriciated

.close

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[cedar sparrow]

The polygon P is formed by taking a square of side length
2 and attaching an equilateral triangle of side length 2 to each of the sides. Can someone please help me show that this can be broken and rearranged into a square?

[uneven kernel]

I can text here for my math problems?

[cedar sparrow]

I started by breaking the square and the 4 triangles on its sides and then breaking the square into two triangles

and I’m trying to use this process https://en.wikipedia.org/wiki/Wallace–Bolyai–Gerwien_theorem

<@&286206848099549185>

[devout snow]

@cedar sparrow Has your question been resolved?

[finite briar]

prove that the total area is a perfect square

[cedar sparrow]

I actually need to perform the dissections and show the formation

[finite briar]

thats why i sent it as a strike..

[cedar sparrow]

ok

[cedar sparrow]

<@&286206848099549185>

[rancid cedar]

We need to prove that this complex shape can be rearranged into a single square.

Area of the Square: 2 * 2 = 4 square units
Area of an Equilateral Triangle: (√3/4) * side² = (√3/4) * 2² = √3 square units
Total Area of the Figure: 4 (square) + 4 * √3 (triangles) = 4 + 4√3 square units
Finding the Side Length of the Equivalent Square
Let the side length of the equivalent square be 's'.
Area of the equivalent square = s²
Therefore, s² = 4 + 4√3
s = √(4 + 4√3)

[cedar sparrow]

that’s not the question tho

[devout snow]

@cedar sparrow Has your question been resolved?

[potent tusk]

I cant solve my problem as well so ill just stare at your problem

[cedar sparrow]

it shouldn’t be too difficult maybe I’m just making some stupid choices when breaking it apart

it gets to a rectangle with dimensions 1 and area(P) but I’m stuck after that

[devout snow]

@cedar sparrow Has your question been resolved?

[devout snow]

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[sage pagoda]

What does it mean?

[iron kindle]

what does what mean?

whcih part dont you understand

[sage pagoda]

What the question is asking.

[iron kindle]

you need to find the minimum of $(3^x)^2-4x+8$

[woven radish]

Bonk

[sage pagoda]

Like the lowest possible amount?

[iron kindle]

yes

[sage pagoda]

The whole thing, other than the 3 is a exponent btw.

[iron kindle]

have you had derivatives?

[iron kindle]

wdym?

!original

[devout snow]

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.

[sage pagoda]

Everything is a exponent except the 3 which is the base

[iron kindle]

but its not in brackets

wdym

[sage pagoda]

Yeah, I did

1 moment

[iron kindle]

that is very different than what you showed first

anyway

to get the minimum

you need to get the minimum of x^2-4x+8

[sage pagoda]

Can't x just be 0?

[iron kindle]

it can be

but thats not the minimum

$f(0)=3^{0^2-4\cdot 0+8}=3^8$

[woven radish]

Bonk

[iron kindle]

if x is 1 oyu already get a smaller number

$f(1)=3^{1^2-4\cdot 1+8}=3^5$

[woven radish]

Bonk

[iron kindle]

which is smaller

have you had derivatives?

[sage pagoda]

No I have not.

[iron kindle]

what do you know about parabolas

[sage pagoda]

I know the quadratic formula.

[iron kindle]

do you know how to find the minimum of a parabola?

[sage pagoda]

meaning?

[iron kindle]

,w graph x^2-4x+8

[woven radish]

Results provided by WolframAlpha

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

Requested by .bonk__

[iron kindle]

thats the minimum

the lowest point of the graph

[sage pagoda]

Okay, I'm following

[iron kindle]

youve never had to find an extremum?

[sage pagoda]

no? what grade do u learn this in

[iron kindle]

idk, grade 9 perhaps?

what grade are you in

[sage pagoda]

8

[faint gorge]

you certainly know how to find the vertex of that parabola right

[sage pagoda]

Haven't learnt that stuff yet unfortunately.

[faint gorge]

do you know how to complete the square?

[sage pagoda]

wdym

[iron kindle]

,tex .cts

[woven radish]

Bonk

[faint gorge]

you can also find the x coordinate of the vertex using x = -b/(2a)

what you are essentially looking for

[devout snow]

@sage pagoda Has your question been resolved?

[strange nimbus]

OK, so the quadratic formula has two parts. You have the -b/(2a) part and then the square root over 2a, right?

The -b/(2a) part tells you where the minimum or maximum of the parabola will be.

So, with x^2 - 4x + 8, you have a = 1, b = -4, c = 8.

What's -b/(2a) with this parabola?

[devout snow]

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[stark shuttle]

Hello

[stark shuttle]

$\int \sqrt{4-\left(x-3\right)^2}dx$

[woven radish]

Chaewon

[stark shuttle]

Is there a way to solve this integral without using trig sub?

[wind mason]

dont think so.

[crisp tapir]

there is a formula for this, but i believe it has been derived using trig substitutionn

[wind mason]

unless you know the integral for $\int\sqrt{4-x^2}dx$, in which you can perform u-sub

[woven radish]

;(

[wind mason]

but that involves trig

[stark shuttle]

One of the students that I'm tutoring showed me this question, but I know fs that they didn't cover trig sub yet

[potent tusk]

those who know this is probably a circle

[crisp tapir]

its a semicircle and there are no limits

[potent tusk]

the top half

[crisp tapir]

oh well i dont know how you would solve it without trig, but it would be interesting

[stark oxide]

it can be done via IBP

[potent tusk]

if there is it will make me a very happy person

[potent tusk]

what do you make f and g' or whatever

[stark oxide]

u=sqrt term
dv=dt

[crisp tapir]

there will be a x^2/sqrt(a^2-x^2) term right? how will you handle that

[stark oxide]

+a^2-a^2 in numerator

[crisp tapir]

Oh so write it in terms of I and divide by 2?

[stark oxide]

yup

[crisp tapir]

but you will still get a^2/sqrt(a^2-x^2) whose integration is in sin inverse

but cool method

[potent tusk]

im pretty sure thats a standard integral so i guess you are supposed to know it

[stark oxide]

that can also be done without trig sub

[devout snow]

@stark shuttle Has your question been resolved?

[slender mirage]

solve geometrically ._>

that a semicircle with radius "2" at (3, 0)

[potent tusk]

It is an indefinite integral so its a little more annoying than that

[slender mirage]

? but we know how to evaluate area of a segment of a circle

[potent tusk]

wait wait how

[slender mirage]

isn't that just area of sector - area of triangle?

[devout snow]

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[potent tusk]

ive never seen this formula..

oh yeah i remember it now

[devout snow]

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[smoky wedge]

What am I doing wrong?

969 can’t be right because that’s more than what we originally started with

This is the formula I’m using

[supple knot]

,calc 200 * exp(log(0.15)/4.25*48)

[woven radish]

Result:

9.901653967522e-8

[smoky wedge]

Huhhh

[supple knot]

Maybe you entered into the calculator wrong

[smoky wedge]

math will be the end of me

I have the same problem with the next one

I’ll send my work rq

Wait maybe I got it

Is 5.16 hours right for the second part of the question?

[supple knot]

Show work

[devout snow]

@smoky wedge Has your question been resolved?

[supple knot]

Where do you solve for t

[devout snow]

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[smoky wedge]

.reopen

[devout snow]

✅

[potent tusk]

it took me a while to actually comprehend this but this is a really cool method

[potent tusk]

sorry for speaking in this chat

[devout snow]

@smoky wedge Has your question been resolved?

[devout snow]

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[thick orbit]

Hi guys I’m stuck with 1b

I’m confused where to get x-x^2/2

I’m thinking of just x^2/2 using geometric interpretation

[lethal pollen]

Is this the answer keys answer or something?

[thick orbit]

yea

it is

[lethal pollen]

Probably the trapezoid formula

[thick orbit]

could u help me ?

[lethal pollen]

That's not quite it though

[thick orbit]

that is not equal to the key answer

[lethal pollen]

Oh mb

Yea I know

[tame palm]

By geometric interpretation, I would hazard a guess and say they mean calculating the area of a rectangle and the area of a triangle.

[thick orbit]

can you help me ?

im confused

[lethal pollen]

1/2*x*(-x+1+1)

I'll break it down ig

So for 0<=x<=1

f(x)=-x+1 yes?

[thick orbit]

no is not

it is 1-x

[lethal pollen]

Which is equal to -x+1

[tame palm]

How was the geometric interpretation explained to you in class?

[lethal pollen]

I just prefer it in this form

[thick orbit]

make the integral formula by calculating the area by using the variable

[lethal pollen]

Alright then

Maybe we should enlarge this diagram

Can you draw the triangle a bit bigger

[tame palm]

So converting a Reimann Sum to an integral?

[thick orbit]

not using the reimann sum

just making the formula of the area

[lethal pollen]

So here we have a shoddy diagram

[tame palm]

Well, one thing to note is that F(x) is defined as the integral from 0 to x, so anything less than zero can be disregarded.

[lethal pollen]

We see that a trapezium is formed for 0<=x<=1

This trapezium has a base of length 1 and another which we are going to determine

But would you like to attempt to express the base in terms of x

Before I reveal the answer

[thick orbit]

ohh yea

let me try

Is it like this right?

[lethal pollen]

Ye

[thick orbit]

thank you so much

appreciate it !!1

.close

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[devout snow]

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[hollow glacier]

Am I correct?

[hollow glacier]

<@&286206848099549185>

[past juniper]

Mate I want to help you but I am curious if that sign is t or + sign

And also you need to square both sides so it becomes 4h^2 instead of 2h^2

[vital pilot]

there a fucking shortage of helpers or wtf...?

[devout snow]

@hollow glacier Has your question been resolved?

[hollow glacier]

• sign

[devout snow]

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[warped mesa]

Can someone please explain how (a+b) became (a+b+4)?

[unreal stone]

Cuz they factored out the (a-b) from both equations

U left with (a-b)(a + b -4)

[mystic scarab]

It doesn't, if you look carefully. Otherwise the 4(a - b) would still be there

[unreal stone]

😭🙏

[devout snow]

@warped mesa Has your question been resolved?

[warped mesa]

oh 💀

this is embarassing im literally in my final year of high school

oh well

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[restive river]

i have two questions, the other one is short

[woven radish]

villegas_bozo

[restive river]

Second question, if both definitions are logically equivalent, where is the unidirectional step of proof 2 "encoded" in proof 1 of the previous example?

[devout snow]

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[gilded snow]

I'm stuck on how to find the area of the intsection

I know that the radii of the middle circle is. 5 tho

[dark tundra]

I would draw the line connecting the intersections of the two larger circles

Actually it might help to assign names to those intersection points

Maybe A and B

Then draw PA, PB, QA, QB, and AB

[dark tundra]

Try sketching the above and think of what you need to do next

[gilded snow]

Idk should I find the area

[dark tundra]

What area

[gilded snow]

Of the diamond

[dark tundra]

Try drawing the figure with the points and lines I suggested first

Well, not really

For now, I think it would be best to temporarily block out the middle circle

Because obviously the area you want is the intersection area minus the small circle, right?

[gilded snow]

Yea

[dark tundra]

So getting the area of the small circle is easy, you can probably do that in your sleep

[gilded snow]

Yea

[dark tundra]

It’s the intersection area that’s not as obvious

[gilded snow]

What should I do nect

[dark tundra]

My hint is that perhaps to get the area of the intersection, it would be better to divide it into two shaded regions

[gilded snow]

Wait so then it would just be the area of a semicircle, or no bcuz it's not a proper semicircke

[dark tundra]

No it’s not a semicircle

It’s formally called a segment (confusing I know)

[gilded snow]

Oh now what

I don't know what to do cuz my teacher hast taught this yt

And google says something about sin cos tan

[dark tundra]

Kinda looks like this, yeah?

[gilded snow]

Yeah

[dark tundra]

So to get the area of the segment

It’s the area of the sector (the round part of the circle enclosed by two radii) minus the area of the triangle enclosed by the radii and the chord between them

Do you know how to get the area of a sector of a circle?

[gilded snow]

Nope

[dark tundra]

Well you know the area of a circle, right?

So a sector is a fraction of the area of a full circle

[gilded snow]

Ok then how to get the fraction

[dark tundra]

Based on the central angle of the sector

so basically we need to find angle APB

What can you observe about triangles APQ and BPQ?

[gilded snow]

It's equiangular

[gilded snow]

Will it be 120

[dark tundra]

Yes!

[gilded snow]

Then what do we do w the angle

[dark tundra]

so yeah area of sector = $\frac{\theta}{360}\pi r^2$ (angle in degrees)

[woven radish]

CST (please ping when replying)

[dark tundra]

next, area of a triangle with two sides and the included angle, do you know that one?

[gilded snow]

Nope huhu

[dark tundra]

$\frac{1}{2}ab\sin C$

[woven radish]

CST (please ping when replying)

[dark tundra]

basically it's just bh/2 but adapted

[gilded snow]

Ok how do we do that

[dark tundra]

ab are the sides and C is the included angle

what do you think would be the most convenient choices

[gilded snow]

Idkkk

[dark tundra]

in this figure

you see the triangle, right?

[gilded snow]

Yep

[dark tundra]

so you basically have two choices for the pair of sides

either the two radii

or one radius and the chord

which do you think is better?

[gilded snow]

Two radi

[dark tundra]

yes!

so it's the two radii and the included angle, which is the central angle of 120 again

[dark tundra]

so now take the area of the sector

[dark tundra]

area of the triangle

subtract them, and come back to me

[gilded snow]

What does sine do

[dark tundra]

wait you don't know sine? the trig operation?

[gilded snow]

Nope

[dark tundra]

if you don't mind me asking, what grade or year level are you in

[gilded snow]

7

[dark tundra]

hmmmm...

[dark tundra]

Basically, the best way I can describe sine is that it's a ratio between two sides of a right triangle, like so

So because you haven't encountered sine yet, I'll just give you $\sin(120)=\frac{\sqrt{3}}{2}$

[woven radish]

CST (please ping when replying)

[gilded snow]

Ok thanks ima compute the area

[gilded snow]

So it's gonna Abe 1/3 pi - √3/4

[dark tundra]

360 degrees

theta is the central angle, in this case 120

[gilded snow]

Ok thank you

So is the area just 1/3 pi - √3/4

[dark tundra]

so yes that's the area of the sector

but note that's just half

because there's the other side

[gilded snow]

So x2 minus pi 0.5 squared?

[dark tundra]

and then there's the circle we ignored until now

ye

[gilded snow]

Thank you

[dark tundra]

that should be it

[gilded snow]

Ats it right?

Thank you

[dark tundra]

no problem!

[devout snow]

@gilded snow Has your question been resolved?

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[thin brook]

hello can someone help me with this non calculator question?

[winged bluff]

by the binomial expansion theoreom the coeffecient of x in (x+1)^n is simply n,and since summing up the coeffecients of x is the same as just summing up regular numbers the question is basically just asking
"you have to add the first what numbers to get 210?"
and you can do that pretty easily by basic guessing and checking
try 1+2+3...+10 = 55
try 1+2+3...+20 = 210

so n = 20

does that explanation make sense?

[devout snow]

@thin brook Has your question been resolved?

[thin brook]

is there a way to not just check and guess tho? my tests only give one mark for correct answers but

they want like method marks

like there's six marks total in this question for example

[winged bluff]

ohh youre doing gcse/igcse right?

[thin brook]

nah I'm doing IB

[winged bluff]

oh ok

[thin brook]

sorry for the inconvience

[winged bluff]

anyways so until the part we're weve transformed the question into adding numbers to get 210 we already have full steps

we just need full steps for finding the numbers that add up to 210

[winged bluff]

its fine dont worry

[winged bluff]

the only way i can think of for doing that is abit overcomplicated tbh but i think its good enough

[thin brook]

okie dokie thank you!

[winged bluff]

theres a formula(i think it can be derived from arithmetic series) that states the sum of numbers from 1 to n is n(n+1)/2 for even n and n(n-1)/2 + n for odd n
lets try both cases
n(n+1)/2 = 210
n(n+1) = 420
n^2 + n = 420
n^2 + n - 420 = 0
solve the quadratic to get n = 20 or n = -21, n cant be negative here so n = 20
if n is odd then
n(n-1) / 2 + n = 420
n^2 /2 - n/2 + n = 420
n^2 /2 + n/2 - 420 = 0
in this case both roots end up being irrational so n can never be odd
so the only possiblity is n = 20

i feel like you dont really need to account for the case where n is odd if youve already found an even n that works but i included it just in case that want that much rigor

keep in mind however i definetely feel like im overcomplicating this, so itd be better if you looked at the markscheme and saw what method they went with in case it isnt this one/if they wont accept this one

[devout snow]

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[devout snow]

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[nimble zodiac]

how do i solve question 29c?

my only working out for that question is figuring out where the velocity is 0

and getting a value of t for it

i never get these right 😭

i’m only asking for an easy way to solve these types of questions

[lunar harbor]

Recall that the distance traveled by a particle on the interval $[a,b]$ is given by $\int^{b}_{a} |v(t)| \dd{t}$.

[woven radish]

Civil Service Pigeon

[nimble zodiac]

yes

and it needs to be split up right?

I can only solve this using a calculator and not by hand

idk how to do it by hand and i’d love to learn how

[lunar harbor]

Oh you need help calculating the actual integral

You should’ve said that

$\int^{4}_{0} |2-4\sin^2 t| \dd{t}$

[woven radish]

Civil Service Pigeon

[lunar harbor]

We’re calculating this, right?

I skimmed over it kinda quickly, hence why I’m asking.

[nimble zodiac]

yes this

[lunar harbor]

Alright

[nimble zodiac]

sorry about that

[lunar harbor]

You were on a good start with finding where the function is zero

b/c we are going to split it, as you also said

So where does the velocity change sign?

[nimble zodiac]

at pi/4 right?

it goes from positive to negative

if i’m not mistaken

wait

[lunar harbor]

Correct, but it also changes sign two more times on the interval.

[nimble zodiac]

no sorry

other way around?

also at 3pi/4 right?

[lunar harbor]

No you were right the first time

sin (t)<1/sqrt(2) to the left of pi/4 and sin(t)>1/sqrt(2) to the right of pi/4

[nimble zodiac]

and then back to 9pi/4?

[lunar harbor]

9pi/4 isn’t in the interval

[nimble zodiac]

oh

you’re right

too far off

[lunar harbor]

$\sin t=\pm \frac{1}{\sqrt 2}$

[woven radish]

Civil Service Pigeon

[lunar harbor]

You neglected the negative case

[nimble zodiac]

oh true

i forgot about that

so

what about 5pi/4?

[lunar harbor]

Yeah

[nimble zodiac]

and 7pi/4?

[lunar harbor]

Outside the interval.

So how would we split the integral given all of this?

[nimble zodiac]

well

it’s positive between 0 and 3pi/4 right?

or is that incorrect

[lunar harbor]

,w 2-4sin^2 (t), t=5pi/8

[woven radish]

Results provided by WolframAlpha

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Requested by civil_service_pigeon

[nimble zodiac]

?

i think i got it

[lunar harbor]

.

5pi/8 is between 0 and 3pi/4

Hence 2-4sin^2 (t) is not always positive on the interval

[nimble zodiac]

so many numbers

[nimble zodiac]

so is this incorrect?

[lunar harbor]

Yes

[nimble zodiac]

what do i do

[lunar harbor]

You found that the zeros are at pi/4, 3pi/4, and 5pi/4

So figure out how the sign changes from 0 to pi/4, pi/4 to 3pi/4, 3pi/4 to 5pi/4, and 5pi/4 to 4

[eternal siren]

CAN SOMEONE HELP ME PLEASEEEEEE

how to solve for trapezoids and kitesssss☹️☹️☹️

[lunar harbor]

You may find it easier to rewrite the function using the cosine double angle formula

Up to you

[nimble zodiac]

That looks like a lot of integrals

[lunar harbor]

k

[devout snow]

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).

[devout snow]

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).

[eternal siren]

OH MY BAD IM NEW HERE

[nimble zodiac]

what is that?

[lunar harbor]

$\cos(2t)=1-2\sin^2 (t)$

[woven radish]

Civil Service Pigeon

[nimble zodiac]

omg

finally

i got the right answer

i spent a whole hour on this unbelievable

[lunar harbor]

You didn’t find the a antiderivative by hand ):

,w 6-sin(8 rad)

[woven radish]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by civil_service_pigeon

[nimble zodiac]

i’ve already spent too long on this question 😭

the antiderivative i wrote on the previous page i think

sin2t

[lunar harbor]

🤔

Ok fine

[nimble zodiac]

i confirmed with my calculator too

my goodness

so tedious

but i’m really grateful for your help

thanks a lot

.close

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[crisp fjord]

Heya, still got nothing. Like I have no idea how we get to this limit being 0.

[crisp fjord]

I might rewrite the second log, but that's about it

[slender mirage]

the right sin(1/2 log(1 + 1/(x² + 1))) is just sin(ln(√2)) nvm

you're concerned about the first sin

[crisp fjord]

but then there is still log with inf in there

as in there is 1/x^2 etc

[slender mirage]

no wait

let me type =_=

[crisp fjord]

this problem sucks

[slender mirage]

$\frac{\sin \left(\frac{1}{2} \log\left(1 + \frac{x^2}{x^2 + 1}\right)\right)}{x\sqrt{x}}$

this expression will bless you

[woven radish]

Arya

[slender mirage]

the log is ln(1 + y) as y -> 0

and sin is sin(z²) as z -> 0

at the end of all criss-cross apple sauce you should be left with √x/2 on this side

since the other sine is a finite, 0 * finite = 0

[crisp fjord]

Ty for all the help - this seems very useful, though I now still need to do something about the denominator (? idk, the bottom is what I mean - the x^3/2)

.close

[devout snow]

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[finite briar]

u could apply 2sinAsinB here ig

[crisp fjord]

tried and did, though I am still left with the x^(3/2) and cant get rid of it

[devout snow]

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[winter dome]

Help

[winter dome]

Where did the -2 come from when there isn’t a -2 in the first highlighted integral? I know that the -0.5 is there to balance the integral

The green highlights are mine and I don’t get it

[slender mirage]

1 was written as -0.5 * -2, idk what got you thinking -0.5 is there to balance the integral

Also they sent the -2 there to use 4 - x² = u sub, so that -2x = du

[knotty sage]

feels like a generalized u-sub

[winter dome]

The person didn’t sub

So I don’t get it

Why -2?

Does that also mean we could use either 3, 10 and so on?

Please look at the core 2 bit and tell me

[knotty sage]

try u-subing it and you'll see why

[winter dome]

K

Nvm I should’ve slowed down the video from x2 to x1.5

Because I missed that part

[knotty sage]

lol i felt that

[winter dome]

Which confused me

Thanks @knotty sage

.close

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[sand monolith]

can someone help??

[storm crater]

!status

[devout snow]

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

[storm crater]

@sand monolith

[sand monolith]

1

[storm crater]

I see no task to do.

[sand monolith]

i know

[storm crater]

It's just a description, right?

[sand monolith]

i just need some explaination

cause I really dodn't know where to start

@storm crater

[storm crater]

Does (1) make sense?

[sand monolith]

yep

i think it makes sense

hellp???

@storm crater bro please😖

help me

[supple knot]

No one can help if there's no question

[supple knot]

Read this

[supple knot]

Start what

[storm crater]

I think he might want some extra explanations but I don't know where to start with that either.

I think I get the idea of it now but it is difficult to say in words.

Maybe you try making some drawings yourself

Think about what the derivatives mean graphically

Another hint

the x - (x dot t) t thing removes the part of x that is parallel to t.

[devout snow]

@sand monolith Has your question been resolved?

[white stirrup]

I just wanna confirm if this is right I’m asked the evaluate the expression

[supple knot]

! occupied

[devout snow]

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).

[white stirrup]

Gotchu

[devout snow]

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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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[supple knot]

,rotate

[reef crest]

hello

<@&286206848099549185>

[supple knot]

Did you learn any equations about growth

[reef crest]

honestly i’m just tryna graduate

i just need these to get done

[supple knot]

That doesn't quite answer my question

[reef crest]

lol okay

[reef crest]

sure

[supple knot]

Show them

[reef crest]

What u mean?

.close

[devout snow]

Channel closed

Closed by @reef crest

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[strange vale]

I have to classify the systems

[strange vale]

I dont really understand how I should do

[nova dove]

this is a guess but I think they are asking you to answer what the dimensions of the matrix B should be

You understand that matrix have dimension $n\times m$ for some n and m

[woven radish]

龙傲天

[nova dove]

where n refers to the rows and m refers to the columns

[strange vale]

yea but in the solutions they tell me that i. its impossible

[nova dove]

and you know that matrix multiplcation with $n\times m$ and $m \times k$ results in a $n\times k$ matrix

[woven radish]

龙傲天

[strange vale]

yea

[nova dove]

so in i. you have Bx = c right

what's are the dimensions of x and c

[strange vale]

x = 4x1

c = 3x1

[nova dove]

so if it's always $n\times m$ times $m\times p$ is $n\times p$

[woven radish]

龙傲天

[nova dove]

what are n, m, p here?

c has $n\times p$ so n = 3, p = 1

[strange vale]

oh ok

but the m is not equal

cause in x the m = 1

and in c the number of lines is 3

and to multiplay two matrix AxB CxD the B and C need to be equal

the number of columns of the first matrix and the number of lines of the second matrix

[nova dove]

numbers of COLUMNS of the first matrix and number of ROWS in the second matrix

rows x columns

oh wait

no you're right I mixed them up

no wait I'm right

so supposed $B$ is a $m\times n$ matrix

[woven radish]

龙傲天

[strange vale]

yea

B is M_3x4

and car(B) = 2

that means the last one row is empty

[nova dove]

we know that $x$ is a $4\times1$ matrix ($n\times p$) so $n=4, p=1$
we know that $c$ is a $3\times1$ matrix ($m\times p$) so $m=3, p=1$
that means that $B$ is a $3\times4$ matrix

BUT remember that for matrix multiplcation it needes to be the case that the number of COLUMNS of the first matrix IS EQUAL TO the number of ROWS of the second matrix, the number of columns of $B$ is $4$, the numbers of rows of $x$ is $4$

these DO match

thereforee B exists

[strange vale]

ohh nice

ok thx

.close

[devout snow]

Channel closed

Closed by @strange vale

Use .reopen if this was a mistake.

[nova dove]

@strange vale apologies I made a typo they are actually equal

[strange vale]

.reopen

[devout snow]

✅

[woven radish]

龙傲天

[strange vale]

wait

B shouldnt be 4 x 1?

[nova dove]

B should be 3 x 4

since x is 4 x 1

and c is 3 x 1

the 4 of B matches with the 4 of x
the 3 of B matches with the 3 of c

clear?

[strange vale]

but 3 is from the rows

imagine

[nova dove]

Are the (a), (b) parts of the question relevant for (c) ?

[nova dove]

Yes, the rows of c

[strange vale]

hm I think no because the (a) its about |A| and (b) about A^-1

[strange vale]

exactly

[nova dove]

m = 3, n = 4, p=1

[strange vale]

so 5 x 3 multipy for 3 x 4 its 5 x 4

[nova dove]

Yes

[strange vale]

and 4 x 1 multply 3 x 1 its 4 x 1

[nova dove]

No because 4 x 1 multiple 3 x 1 is not possible

since m = 4, n = 1, n = 3, p = 1

n can't be both 1 and 3

[strange vale]

ohh ok

I think I got it

[nova dove]

so it's ALWAYS of the form
m x n multiply n x p is m x p

the n's have to match up

they cannot be different

[strange vale]

ok got it

do u understand portuguese?

[nova dove]

I do not

[strange vale]

ohh okk

Letb= 1 2 3 T. Considering thatB´ is a ladder matrix and car(B)=2,
The matrix[B|b] is also in a matrix and has to dry([B|b])=3. So, how
car(B)=car([B|b]), the system is impossible.

thats the answer

[nova dove]

Can you post the b and c parts of the question

Is B given anywhere?

I think I misinterpreted the original question since it was in portugeese

[devout snow]

@strange vale Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[restive river]

How do they get 4y^2? I’m confused by this example

[lunar harbor]

$(y^3+2y^2)-(y^3-2y^2)=y^3+2y^2-y^3+2y^2=2y^2+2y^2=4y^2$

[woven radish]

Civil Service Pigeon

[deft trout]

for y^3-2y^2 you flip the sign under the long division

that's why you're adding it to get y^4

and that's why it cancells out the y^3

[restive river]

I can’t tell if I’m on the right track