help-10

[static beacon]

so just add them and simplify

[crisp lichen]

now i get it

[static beacon]

tell me what u get as your final answer

[crisp lichen]

ok

133

[static beacon]

r u sure thats the full question?

[crisp lichen]

yes it is

[static beacon]

x can literally be anything tho

[crisp lichen]

ik thank you for helping

can i friend u

[static beacon]

sure

[obtuse pebble]

@crisp lichen Has your question been resolved?

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[obtuse pebble]

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[solemn flax]

on the 3rd line where does the 2 vanish to?

[timid silo]

they divided 2 on both sides

[solemn flax]

why is it 2d/dx

[timid silo]

it should be 2dx/dtheta

[solemn flax]

nah its legit but where is the 2 from?

[timid silo]

chain rule via implicit differentiation

[solemn flax]

OH I see it now

[obtuse pebble]

@solemn flax Has your question been resolved?

[obtuse pebble]

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[exotic creek]

if 3 points ABC are in a straight line in 3d vector, what is true?

[exotic creek]

line AB and BC are parallel to each other?

[grizzled shore]

They are colinear so yes

AB, AC, BC are all parallel to each other

[exotic creek]

how would i show it in terms of ijk

[grizzled shore]

Well, suppose A ≠ B ≠ C as well

[exotic creek]

one of i j and k would equal to each other? right?

or like in a multiple of a number

[grizzled shore]

Look at AB, figure out which set of vectors lie on this line

Then show that every vector on AC is some point + the vectors of AB

In this case the “some vector” is the 0 vector as they are the same line

There’s no offset

[exotic creek]

I think it would be better to show the question

1 min

@grizzled shore

[grizzled shore]

Ok just find the point then

You know that every straight line is some point + a vector scaled

[exotic creek]

what do you mean?

[grizzled shore]

First think of any vector in 3D

From the origin to any point

[exotic creek]

i+j+k

[grizzled shore]

No that’s not any vector

ai + bj + ck

[exotic creek]

oh so if its a straight line it would be x(ai + bj + ck) ?

[grizzled shore]

Yeah

Now also realise that if I take some point

Then use it as a vector

[grizzled shore]

Then add this to the point

I can make any straight line in 3D

[grizzled shore]

This allows me to make any line that passes through 0,0,0

[grizzled shore]

This allows me to make any line that passes through p

[exotic creek]

i don't quite get it

if i use the question for an example, (4j+6k) = x(4i-3j+5k) ?

[grizzled shore]

Use 2 points to first determine the direction vector

Hold up I misread

We don’t even need to find the line

We just need to find the direction vector from A to B

Then extend that to C

And find the value of p

So now really it’s just asking 1) what is the vector from A to B

2. if adding a scaled amount of this vector from A to B reaches C, what is the value of C

[exotic creek]

i think im getting it

im gonna try that

[hearty flame]

Do you know how to represent a line in vector form?
Like r=a+(lambda)b

[exotic creek]

so i got $\vec{AB} = 4i+j+11k$ does that mean c is $\lambda (4i+j+11k)$ ?

[warm shale]

yomiko

[exotic creek]

not really no

[hearty flame]

Ohhokok, then frosst's method is good

[grizzled shore]

No it’ll be either point A or point B + some scaled version of vector A to B

Remember we want to still be on the line here

[grizzled shore]

This goes through the origin by λ = 0

(Not saying it can’t but that’s what it implies)

[exotic creek]

i don't really what ur saying here

[grizzled shore]

Ok we have the vector from A to B

We want to extend this direction vector from A until we hit C as well

We know this is possible because A, B and C are colinear

If we think about the vector from A to C

This vector is a scaled version of the vector from A to B

[exotic creek]

i can think of it like y=mx +c where c is position A and the gradient is the vector AB ?

[grizzled shore]

If I tell you in 2D that my line passes through point A and B

If I tell you 1 coordinate of point C you can find the other coordinate

Yes?

(Given C is on the same line as AB)

[exotic creek]

yeah

[grizzled shore]

Ok now consider an arrow from point A to B

We can scale this arrow (stretch or squeeze) in the direction of the line

One of these scaling will get us from A to C

The combination of all the possible scaled versions gives us every point on the line

In other words, let v be the vector from A to B, then A + kv where k ∈ ℝ is the set of all points on the line AB

[exotic creek]

yeah

[grizzled shore]

So then

We’ve found v, A is given

[grizzled shore]

We just need to find p in the point C such that it satisfies this

Such that it satisfies A + kv

[exotic creek]

so i can say c is $4i-3j+5k + \lambda (4i+j+11k)$?

[warm shale]

yomiko

[exotic creek]

point A + how much longer c is compared to AB

[grizzled shore]

I don’t think you did vector AB correctly

[grizzled shore]

But that’s the idea yes

[exotic creek]

wait, how is it wrong?

A(4i-3j+5k) B(0i+4j+6k)

[grizzled shore]

(4i - 3j + 5k) + (ai + bj + ck) = (0i + 4j + 6k)

a = -4, b = 7, c = 1

[exotic creek]

don't you just add them together?

[grizzled shore]

No

[exotic creek]

oh you do AO +OB

right?

[grizzled shore]

What’s AO

Think about it geometrically

[exotic creek]

$\vec{AO}$

[grizzled shore]

We are at point A

[warm shale]

yomiko

[grizzled shore]

We want to get to point B

We need to move toward point B

[grizzled shore]

This is what this equation is saying

We start at A

[exotic creek]

$\vec{AO} +\vec{OB} = \vec{AB}$

[warm shale]

yomiko

[grizzled shore]

Plus some vector

We get to B

[grizzled shore]

I don’t think so

[exotic creek]

yh its right

[hearty flame]

Though this is correct 4i-3j+5k is OA and not AO

[exotic creek]

yh just flip the signs

[hearty flame]

Yess

[hearty flame]

Then this would change

[exotic creek]

the mark scheme have done it in a way simpler method

but im not sure what they meant by "uses two of the three vectors in such a way as to find the value"

[hearty flame]

You're doing the same thing, frosst is explaining everystep that's why it could seem that it's longer

[hearty flame]

They are just comparing the coefficients of i, j and k to find p

[exotic creek]

if i compare AB and AC, i would get p+3 = 7 so p = 10?

[hearty flame]

No no

You gotta make sure that i and k are the same before you directly compare j

[exotic creek]

oh so (p+3)/5 = 7

riiiiight

[hearty flame]

Yeah

@grizzled shore I'm sorry to have overtaken your conversation. I just wanted to help out a bit

[grizzled shore]

All good I’m not particularly skilled at line geometry if I’m honest

[hearty flame]

Hope i was able to help out

[exotic creek]

yess

thanks

.close

[obtuse pebble]

Channel closed

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[delicate pond]

can chat gpt help me with stuff?

[delicate pond]

Squares:
A quadrilateral is a four-part figure with equal sides and four right angles.
The perimeter of a square is the sum of the lengths of all sides, which can be calculated by multiplying the length of one side by 4.
The area of ​​a square is calculated by multiplying the length of one side by itself (squared).
Formula for perimeter: P = 4s
Formula for area: A = s^2

Triangles:
A triangle is a three-part figure with three sides and three corners.
The perimeter of a triangle is the sum of the lengths of all sides.
The area of ​​a triangle can be calculated by multiplying the height of the triangle by half the width.
Formula for perimeter: P = a + b + c
Formula for area: A = 1/2bh

Circles:
A circle is a closed curve where all points on the surface are equidistant from the center.
The circumference of a circle is equal to the product of the diameter and pi (π), where pi is a mathematical constant with the value 3.14.
The area of ​​a circle can be calculated by multiplying the radius of the circle by itself, and then multiplying the result by pi.
Formula for circumference: P = 2πr or P = πd
Formula for area: A = πr^2

is thise right?

[uneven palm]

chat gpt is not made for doing math

[delicate pond]

yeah i kinda realised that

but i need fast help

and yeah im just stressing

[uneven palm]

what do you need help with?

[uneven palm]

this is just a bunch of facts (maybe I didn't read it all)

[delicate pond]

il translate everything

[tardy epoch]

start all over and just show your work along with the original question

[marsh geyser]

ChatGPT just predict the next word. It can be good at saying some facts but it can also lie or make mistakes.

[delicate pond]

Area/perimeter of squares, triangles and circles
Trapezium
Pythagoras
Properties of triangles
Formality and congruence

[marsh geyser]

What's the question?

[tardy epoch]

this is just a bunch of topics. help channels are meant for actual problems

[delicate pond]

its a test

tommorow

we can have a rule book kinda

[tardy epoch]

go through your homework then

[delicate pond]

book with all the formels and stuff

we dont have homework

[marsh geyser]

This is how this works. This channel can be used to make 1 question. The question must be specific and then, someone will help you out with that question.

[delicate pond]

where can i find like

the ebst sites for these things

[tardy epoch]

then google "khan academy" and each one of your topics

[delicate pond]

where i can just write down as much

[delicate pond]

tysm

[tardy epoch]

or just "practice problems" and your topic

[delicate pond]

you guys are such good help

[delicate pond]

dont have time

i only have time to stay up at night and write down as much

so in the test i can bassicly look up any problems in my work book

[tardy epoch]

can't help you then

[tardy epoch]

.

[delicate pond]

alright

thank you guys tho

.close

[obtuse pebble]

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[verbal cloud]

How do I convert undistorted planar coordinates (traditional x, y) into Equirectangular projection planar coordinates? I.e x, y -> distorted (x, y) that will project nicely onto a sphere

[obtuse pebble]

@verbal cloud Has your question been resolved?

Channel closed

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Remember:
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[boreal cipher]

SIUUUUUUUU

[boreal cipher]

@teal turret did u answer my questions

@teal turret

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[boreal cipher]

Nah bro

[obtuse pebble]

@boreal cipher Has your question been resolved?

[boreal cipher]

No

[obtuse pebble]

@boreal cipher Has your question been resolved?

[obtuse pebble]

@boreal cipher Has your question been resolved?

[boreal cipher]

Well ok no one can help me so yeah

[obtuse pebble]

Channel closed

Closed by @boreal cipher

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[obtuse pebble]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[simple burrow]

halp

[tacit gate]

have you tried anything about the problem @simple burrow

[simple burrow]

i mean its a really open ended quesiton, right?

question*

[tacit gate]

it is definitely not

you don't wake up and know the answer to it, you see patterns that go with perfect squares, what numbers they end with and observing it finding the patterns and what happens when you divide with any number gives you the answer

[simple burrow]

alright

so in tthis case, the remainder of 0 and 1 is a pattern?

after being divided by 4?

[obtuse pebble]

@simple burrow Has your question been resolved?

[obtuse pebble]

@simple burrow Has your question been resolved?

[obtuse pebble]

@simple burrow Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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• Show your work, and if possible, explain where you are stuck.
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[potent spire]

Hey im working on trig and i need help with this problem

[potent spire]

this was my work

the sin was supposed to be 30, what was wrong

[viral blade]

Why are you taking the sin of 60?

[potent spire]

tbh idk

I got 60 bc of the triangle thingy

I'm really rusty of geometry

[viral blade]

but you want to find the vertical leg right

[potent spire]

Yes

[viral blade]

and the hypotenuse is 1

[potent spire]

yes

[viral blade]

you're just doing sohcahtoa wrong then

[potent spire]

oh

I honestly dont even know how to apply soh cah toa correctly anymore, I just know oposite hypotenuse and adjacent, i forgot all of this stuff

[viral blade]

sine of an angle is opposite over hypotenuse, right

[potent spire]

yea

[viral blade]

What's the opposite of the 60° angle in the triangle

[potent spire]

uh OB, or cos(30degrees)

i think

[viral blade]

OB, right

Which is not the side you want to be solving for

[potent spire]

oh

[viral blade]

both cos(30) and sin(60) solved for OB

That's why they came out the same

[potent spire]

i just looked at this chart for sin cos tan for angles in trig but idk how they are even derived

I just feel completely lost with this problem

[viral blade]

Mostly people just memorize the 2 special triangles

45-45-90 and 30-60-90

[potent spire]

let me draw them real quick

[viral blade]

45-45-90 is the easy one because it's just isosceles

[potent spire]

ya 45 is the same length

[viral blade]

You can derive all the trig ratios just using Pythagorean theorem to find the hypotenuse

[potent spire]

I don't know how to do that

[viral blade]

stuff like sin (45) I mean

or do you not know Pythagorean theorem

[potent spire]

yea a^2+b^2=c^2

and c is the hypotenunse

[viral blade]

yeah

so just make any 45-45-90 triangle with any side lengths you want and find the opp/hyp of a 45° angle

[potent spire]

okey

I still feel really confused on the given cos(330) and the sin(330)

What I was doing was just going 330 and marking the ray

then i just made a triangle

but it feels like I was doing this all wrong. Im trying to self learn trig without a textbook and am just thinking about picking up a textbook

[viral blade]

Khan academy is good too

[potent spire]

I've been trying khan academy

I was able to go through algebra 1 and most of 2 with little to no problems

but at the end of algebra 2 they started with trig and I feel like I hit a brick wall

[viral blade]

You got the cos right

sin you got the sign right

Better than half the people in the trig class I TA would have done

[potent spire]

I started trig concepts like 2 days ago

😦

[viral blade]

Seems like you're doing pretty good then

You just solved for the wrong side length is all

[potent spire]

okay let me redraw what I made

Okay I made this so far

ah I forgot to label the triangle

A is bottom right

this is what I mentally understand so far

[viral blade]

right

let's start with OB ig

compared to angle O, is OB adjacent, opposite, or hypotenuse

[potent spire]

adjacent

[viral blade]

right

actually maybe this isn't necessary

we have a 30-60-90 triangle

[potent spire]

I don't have the values memorized

is that somthing I should do

[viral blade]

Probably

[potent spire]

how would I even do that

[viral blade]

Can you reflect A about B

[potent spire]

uhmm

[viral blade]

and draw another triangle

[potent spire]

with 30-60-90

[worn yoke]

you should memorize the values for the triangles of 30-60-90 degrees and 45-45-90 degrees with hypotenuse 1

[potent spire]

im not sure

[viral blade]

just make a mirror triangle

keep the old one

[potent spire]

oh

[viral blade]

but act like OB is a mirror

[potent spire]

okay

like this?

[viral blade]

ye but keep the old one

put them back to back

[potent spire]

oh so make like a larger triangle

[viral blade]

yeah

[potent spire]

oh shoot

now its 60

[viral blade]

yeah all the angles are 60

angles in a triangle need to add up to 180 right

[potent spire]

yea

[viral blade]

so anyways

since all the angles are the same what do we know about the sides

[potent spire]

they are all the same length

[viral blade]

yeah

so if OA is 1, what's AB

[potent spire]

hmm im not sure

i want to say 1/2

but

not sure

[viral blade]

you reflected the triangle right

[potent spire]

oh true

[viral blade]

so AB is exactly half of the side

so yeah 1/2

And then you can use Pythagorean theorem to find OB

So that's how you could derive the side lengths

[potent spire]

i forgot how to apply the pythagoream theorem to find lengths

[viral blade]

can you set up the equation?

[potent spire]

yea,

(1/2)^2+x^2=1^2

[viral blade]

yeah

then try to isolate x^2

to isolate x

[potent spire]

ok

[viral blade]

Yeah

[potent spire]

sqrt(3)/2 looks familiar

[viral blade]

Yeah that's the (correct) length you already got for OB

Now you have a length for AB too though

[potent spire]

wouldnt AB just be 1/2

[viral blade]

Yeah

[potent spire]

OH WOW

now If I look at unit circle

I know that cos(330) will be +sqrt(3)/2 and sin(330) will be -1/2

I learned unit circle yesterday pretty cool stufff

[viral blade]

ye nice

[potent spire]

honestly I just forgot a lot of geomerty tbh

rules of triangles and whatnot

[viral blade]

oof yeah that'll happen

[potent spire]

Its been like 8 years since I took geometry

so its all so rusty, I was thinking about self learning geometry again

but not really too exicted for it, maybe integrating it with something else

becuase out of all the maths geometry wasn't my fav,

[viral blade]

Ye not mine either

[potent spire]

I mostly like algebra and abstract ideas, I want to learn calculus soon

anyways thanks for the help eggnote, hopefully I start to make sense of these a bit more

also one last question

[viral blade]

you're welcome

[viral blade]

sure

[wintry mountain]

Help

[potent spire]

oh sorry, ill close channel for you vibe slayer

[graceful spruce]

hi im new how do i get help?

[potent spire]

.close

[obtuse pebble]

Channel closed

Closed by @potent spire

Use .reopen if this was a mistake.

[viral blade]

u have right of way

[wintry mountain]

I have already but I I cant get any help

[viral blade]

They need to get their own channels

[potent spire]

i already closed it

[viral blade]

They'll regret not doing that anyways

can .reopen if you want

[potent spire]

.reopen

[obtuse pebble]

✅

[potent spire]

okay, back to the circle

so this 330 degree I derived from cos and sin being 330 degrees, is that logic okay? becuase im kinda doubtfull on that

[viral blade]

That's exactly what that means

[potent spire]

okay good

[viral blade]

x and y coordinate of a point 330° around the unit circle

[potent spire]

all the problems ive had so far have been equal degrees

so in the future I will learn more about that

[viral blade]

ah yeah otherwise you'd have to make 2 triangles

[potent spire]

thats for the future

[viral blade]

I mean it's not harder to find cos(330) than it is to find both cos(330) and sin(330)

[potent spire]

oh thats good to hear

[viral blade]

So if they're not equal degrees it would just be two different problems

[viral blade]

It's definitional though, right

[potent spire]

what is?

[viral blade]

You could just find both cosine and sine anyways, and then only input the cosine value

[potent spire]

oh i see

Also do you have any recommendations for memorizing the special angles

30-60-90 and 45-45-90

for cos sin and tan

or would I eventually remember them with time

[viral blade]

sin(0, 30, 45, 60, 90)

goes √0/2, √1/2, √2/2, √3/2, √4/2

[potent spire]

kinda like radians

[viral blade]

cos does that in reverse

[potent spire]

so cos(0, 30, 45, 60, 90)

[viral blade]

sin(15) is like (√6-√2)/4, definitely not memorized

so that's why 30 45 60

[potent spire]

i see

isnt cos(0) = 1?

[viral blade]

yeah

[potent spire]

oh sqrt4/2

[viral blade]

It goes in reverse for cos

[potent spire]

hahahah

[viral blade]

Yeah lol

3 of the 5 simplify

[potent spire]

I need to memorize that

[viral blade]

It's just a funny trick

[potent spire]

sqrt0/2 sqrt 1/2 sqrt2/2 sqrt3/2 sqrt4/2

again thanks for the help

[viral blade]

np

tan you can always find by dividing but it goes 0, 1/√3, 1, √3, (undefined)

[potent spire]

uh oh

[viral blade]

tan(90) = sin(90)/cos(90) = 1/0 lol

[potent spire]

dang i really need to memorize those, seems so powerfull

once I take trig in like a year or so in college again i'll hopefully memorize this

okay im going to close the channel now I don't want to take up any more time 🙂

.close

[obtuse pebble]

Channel closed

Closed by @potent spire

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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.

[safe lodge]

Weird, glitches. Every time I do this problem, in part b, the calculation for me comes out negative, and I can't do negative inverse cosine. What might I be doing wrong?

[viral blade]

,w arccos(-.7)

[warm shale]

Results provided by WolframAlpha

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

Requested by Eggnote#0205

[viral blade]

are you getting -44.9?

Or no

-45.1

er no that doesn't make sense either

what are you getting?

,w plot arccos(x)

[warm shale]

Results provided by WolframAlpha

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Requested by Eggnote#0205

[viral blade]

-135.1°?

[safe lodge]

whoops sorry, I am getting an undefined answer if I put it into inverse cos, but -.7078781513

that's what I get before putting it into inverse cos, but I get an undefined answer. So not even a negative answer.

I'm definitely not getting 135.1 like the answer indicates I should be getting.

.close

[obtuse pebble]

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[obtuse pebble]

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[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@astral narwhal Has your question been resolved?

[obtuse pebble]

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[dense drift]

Need some help with this logic question.
"A and B are mountain climbing. On their way back A says "Only i got to the top!" and B says "That's not true, we both got to the top". Knowing that both are lying, it's certainly true that:
A) if B got to the top, he got there alone
B) B got to the top
C) both got to the top
D)A got to the top
E) neither got to the top"
I think the answer is E since the negation of the B statement would be " We both didn't get to the top". But im not sure what the negation of the A statement would be, any help pls?

[obtuse pebble]

@dense drift Has your question been resolved?

[rich plume]

It is clear that a alone didn't go to the top
Also both a and b didn't go to the top
So its sure that a didn't go to the top (either alone or with b)

We're left with option A B and E

Your statement may/ may not be true since Both A and B didn't go to the top
Either may hav

[dense drift]

i think A makes the most sense

[drifting wraith]

yes

[rich plume]

Yeah, I feel the same

[drifting wraith]

they each just named one outcome, so 2 are left

and A describes them

[rich plume]

A seems the most appropriate one because it doesn't say B went went to the top, It says 'if'

[dense drift]

perfect, thank you guys!

[dark stirrup]

P1: (A got to top) and (B did not get to top)
P2: (A got to top) and (B get to top)
~P1: (A did not get to top) or (B got to top)
~P2: (A did not get to top) or (B did not get to top)
~P1 and ~P2: A did not get to top

[obtuse pebble]

@dense drift Has your question been resolved?

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[earnest linden]

Question : find all polynomial P ∈ R[X] which ∀t ∈ R, P(cos(t)) = cos(P(t))

[earnest linden]

I tried to consider the degree of the polynomial :

if deg(P) = -inf then P doesn't work

if deg(P) = 0 , P = a only work with a the only solution for cos(x) = x

if deg(P) = 1 only P = X work

but if deg(P) >= 2 I don't know

then I saw that there is some congruence within let' say P(0) et P(2*pi) then I don't know what to do with that

[jolly ginkgo]

I may be wrong but I don't think there may be any two degree polynomial satisfying that for all t in R

[earnest linden]

unfortunately I made a mistake

[jolly ginkgo]

What?

[earnest linden]

I just have : cos(P(x+t+2π))=cos(P(x+t)) with x an antecedent of 2kpi with k enought big

[earnest linden]

the fact that P(t + 2pi ) = P(t) is false (or I didn't prouve it )

[proud edge]

limit at infinity

[earnest linden]

where ?

[proud edge]

nevermind 😔

[obtuse pebble]

@earnest linden Has your question been resolved?

[obtuse pebble]

@earnest linden Has your question been resolved?

[obtuse pebble]

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[balmy sequoia]

useless

[balmy sequoia]

explain

[earnest elk]

Explain what?

What are you having trouble with?

[rich plume]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[balmy sequoia]

1

[rich plume]

Separate them all

[balmy sequoia]

what

[west sierra]

Think about where to begin like this, $$(abcd)^{1/4} = a^{1/4} * b^{1/4} * c^{1/4} * d^{1/4}$$

[balmy sequoia]

how do i do that

[rich plume]

$\sqrt[4]{abc}=\sqrt[4]{a}\cdot\sqrt[4]{b}\cdot\sqrt[4]{c}$

[warm shale]

B-eard

[royal basin]

{} to make exponents not fuck up

$a^{1/4}$

[warm shale]

Ann

[balmy sequoia]

how do i put 1/4 into 8/3

[rich plume]

Multiply

[royal basin]

apply the law $(x^p)^q = x^{p \cdot q}$ on all of these

[warm shale]

Ann

[royal basin]

$(c^{8/3})^{1/4} = c^{\frac{8}{3} \cdot \frac{1}{4}}$

etc.

[warm shale]

Ann

Calamity

[balmy sequoia]

so it would be 8/12?

[royal basin]

also just as a reminder:

!volunteers

[obtuse pebble]

Helpers are just people volunteering their time to help you. Be polite.

[balmy sequoia]

i just realised i clicked on the wrong test

[royal basin]

the what

[balmy sequoia]

the wrong test

ive been doing the wrong one the entire time

[west sierra]

Atleast your plan to cheat the test failed

[balmy sequoia]

cheat

wtf

this is previous tests

what do you mean

i did the 2020 one instead of 2022

[west sierra]

Oh its previous test then that's good

[balmy sequoia]

what is this

ive fried my brain

[royal basin]

this is an expression

[balmy sequoia]

colecting like terms?

[royal basin]

if you have fried your brain you should take a break

[balmy sequoia]

i will be back with some water

ok i have water

explain everything

i mean

help me understand this problem better

[royal basin]

you shouldn't do math when your brain is fried.

you're not going to get anywhere.

[balmy sequoia]

im jokig

[royal basin]

"my brain is fried" is not a joking matter

[balmy sequoia]

its a collecting thing

i have to collect the like terms

[royal basin]

i mean, yes, sure, to simplify 3 sqrt(5) - 4 sqrt(2) + sqrt(5) - 3 sqrt(2) pretty much the only thing you can do is collect like terms

[balmy sequoia]

right

so it would be 4√5-7√2

why is my last year test much easier

this question is easier than 2020 test

i take it back

what is this

[sage dagger]

do you know how to expand a(b+c)

[balmy sequoia]

i assume not

[sage dagger]

Do you know how to expand in general

[balmy sequoia]

as in?

like

b^5

[sage dagger]

Do you know what expanding is

[balmy sequoia]

bxbxbxbxb

[sage dagger]

no

[balmy sequoia]

then no

[sage dagger]

If i gave you 2(a+b) do you know how to expand

[balmy sequoia]

no

[sage dagger]

It means you multiply the term on the outside to everything inside the brackets

[balmy sequoia]

ok so it would be 2a +2b?

[sage dagger]

Yes

[balmy sequoia]

isnt that like the same as (a+b)^2

[sage dagger]

no

those are different

[balmy sequoia]

oh wait yes

bcus that would be a^2 + b^2

ok so i jst do

[balmy sequoia]

this

[sage dagger]

Wrong

incorrect

[balmy sequoia]

um ok

[sage dagger]

Why do you think (a+b)^2 is a^2*b^2

[balmy sequoia]

index law 4

[sage dagger]

That’s not how index law works with this

[balmy sequoia]

this topic is about indicies and surds

[sage dagger]

That’s still not how the index law works

(a+b)^2 = (a+b)(a+b)

[balmy sequoia]

ok

then

back to the question

[sage dagger]

Are you studying from the Australian curriculum by any chance

[balmy sequoia]

yes

[sage dagger]

I knew that sounded familiar

[balmy sequoia]

ok

now back to the question?

[sage dagger]

You know how expanding works now. Apply the same logic

[balmy sequoia]

with roots

how

[sage dagger]

Treat sqrts like any other variable. You can only perform the same operation with them if it’s the same term

[balmy sequoia]

that doesnt help

[sage dagger]

Multiply sqrt(2) to everything in the inside

[balmy sequoia]

wait

[west sierra]

It's same as a(b+c)= ab + ac

[balmy sequoia]

so it would be √10+√6

or no?

[sage dagger]

No

[balmy sequoia]

hmm

im still confused

im looking at the answer rn

[west sierra]

The √6 part is correct though

[sage dagger]

sqrt(2) • 5 does not equal root 10

[balmy sequoia]

and they got 5√2 + √6

nit sure how

[sage dagger]

if it was sqrt(2) • sqrt(5) then you would get root 10

But that’s not the case

[west sierra]

If it was sqrt(2) • sqrt(5) then it would have been √10

[sage dagger]

Think of it like this

[balmy sequoia]

oh wait

[sage dagger]

That’s what I just said

[balmy sequoia]

is it because its not the same it has to go on the outside

or like infront

or whatever

[sage dagger]

5 • 1sqrt(2)

Look at it like that

[balmy sequoia]

ok i srt of get it

this be the next one

[sage dagger]

So what do you get after solving this

[balmy sequoia]

i have an idea

of how to do it

so i will try it out'

and come back with an answe

[sage dagger]

You haven’t even solved the first question yet

[west sierra]

Just try out but first you need to answer prev question

[sage dagger]

what’s 5 • 1sqrt(2)

[balmy sequoia]

i dont know

what does the . represent

[west sierra]

It's multiplication

[balmy sequoia]

so then it would be 2√5

[west sierra]

You can either represent multiplication like that or by cross sign and sometimes by bracket

[balmy sequoia]

wait

[sage dagger]

Correct

[balmy sequoia]

ok

so the actual answer woould be 2√5 +√6

[west sierra]

Yes

Now try this one which you intended to

[balmy sequoia]

i actually have no clue

[west sierra]

Just try it

You solved similar question like this just now

[balmy sequoia]

ive done the first part

[west sierra]

a(b-c) on expanding is ab- ac

[balmy sequoia]

√24 - 2√16

[west sierra]

Yeah it's correct but do you know you can simplify that further

[balmy sequoia]

turning √24 into something like 6 and 4

[west sierra]

Yeah it's like √4*6

[balmy sequoia]

then 2√6

[west sierra]

Yes

[balmy sequoia]

and -2√16 can be like -2x4

then 2√6-8

[west sierra]

Yes

Exactly

[balmy sequoia]

wowzers

these one

i know its

[west sierra]

Do you have to rationalize it?

[balmy sequoia]

3/√2 x √3/√3

[balmy sequoia]

yes

[west sierra]

Yeah if you know you can do this

[west sierra]

Wrong

[balmy sequoia]

wait not √3 but √

√2

[west sierra]

Yes

3/√2 x √2/√2

[balmy sequoia]

so it would be 3x√2 / √2 x √2

[west sierra]

Yes

[balmy sequoia]

and it would be 3√2/ 2

[west sierra]

Yes

[west sierra]

You solved it

[balmy sequoia]

this ome

[west sierra]

Same process

You multiply √3 up and down

[balmy sequoia]

do i multiply the same

ok

[west sierra]

Because √3 is irrational

[balmy sequoia]

so i always multiple the botoom?

[west sierra]

Only the bottom irrational

[balmy sequoia]

oh ok

[west sierra]

2 is not irrational but rational for example

[balmy sequoia]

so its √6 / 6

[west sierra]

Yes

[balmy sequoia]

so if √2 was at the botoom it will be irrational

[west sierra]

Yeah

[balmy sequoia]

ok

[west sierra]

Also be careful in identifying irrational

Because √4 is rational as it's 2

√4, √9, √16 etc are rational even thou they have root

[balmy sequoia]

because they perfect squares?

[west sierra]

Yes

Only non perfect squares in root are irrational

[balmy sequoia]

so perfect squares and 2 is rational

in root

[west sierra]

Yes

[balmy sequoia]

oh alr

[west sierra]

Perfect sq in root are always rational because you can get rid of root from them

[west sierra]

Oh in this you have to simplify?

[balmy sequoia]

12a^7b^10

[balmy sequoia]

yes

[west sierra]

B's one is wrong

[balmy sequoia]

oh wait

[west sierra]

b's

[balmy sequoia]

my hand writig is attrocious'

[west sierra]

Power is written wrong for b

[balmy sequoia]

2 became a 7

[west sierra]

It's okay

[west sierra]

Lol what?

[balmy sequoia]

12a^7b^5

[west sierra]

Yes

[karmic whale]

good job

[balmy sequoia]

this one

also i know

[karmic whale]

remember :

[west sierra]

You substract power in division if base is same

[karmic whale]

$\frac{x^a}{ x^b}=x^{a-b}$

[west sierra]

Oh that's ratio sign

[balmy sequoia]

i got 3x^2y^6

[west sierra]

$x^a/x^b=x^{a-b}$

[warm shale]

Calamity

[karmic whale]

correct

[karmic whale]

ik

[karmic whale]

$(x^a)^b=x^{ab}$

[balmy sequoia]

27x^6

[west sierra]

It's like (ab)^2 is a^2*b^2

[warm shale]

JeremyGaming

JeremyGaming

[karmic whale]

yes

[karmic whale]

you nailed it

[karmic whale]

same

[karmic whale]

this

[balmy sequoia]

a^8 / b^4 c^12

[west sierra]

Yes

[balmy sequoia]

final one before a perimiter question

hmm

[west sierra]

Lol no it's same as previous but little tedious in writting

[west sierra]

You mean the above image, that's not perimeter question

[balmy sequoia]

before

[west sierra]

Yeah

[balmy sequoia]

so this is that last one

then there is a perimiter question

[west sierra]

Wait if it's final one how comes it before

[west sierra]

I see

[balmy sequoia]

im stuck on the question

[west sierra]

Which question

[balmy sequoia]

this one

[balmy sequoia]

.

[west sierra]

Just give a try and write answer. If there is mistake we tell you

[balmy sequoia]

30x^-3y / 45xy^-4

[west sierra]

Well on x it's -4 not -3

[balmy sequoia]

anything else

[west sierra]

On y it's 5

[west sierra]

Wait it's right

[balmy sequoia]

this is the soloution

[west sierra]

Yeah

[balmy sequoia]

i was way off

[west sierra]

You didn't write the last answer

Yes

[west sierra]

You weren't way off

[balmy sequoia]

how do i get to the 1x2 part

tho

[west sierra]

You just had to simplify

[balmy sequoia]

how

much

wtf

[west sierra]

30/45 is 2/3

[balmy sequoia]

oh

wth

[west sierra]

And like previous question you substract power of same base in division

[west sierra]

Chill, it's easy

[balmy sequoia]

ok final one then i will sleep

there is another question after

this

might aswell

just finish it

[west sierra]

Which one did you try

a or b

[balmy sequoia]

??

[west sierra]

Formula of perimeter of rectangle is 2(length + breadth)

You just have to put values in formula for answering a part

Because the answer is being asked in expanded form(see question a) you will have to expand after putting length and breadth into formula

[balmy sequoia]

i could do like 2(√2+3+2√3)

[west sierra]

Yes

But you have to expand now

Because see question a, answer is asked in expanded form

[balmy sequoia]

would it 2 effect the roots or not?

[west sierra]

No

[balmy sequoia]

or just like 2√2+6+4√3

[west sierra]

if it was √2 then yes √2 times √2 will be √4

But it's 2 only

[balmy sequoia]

is that the final answer?

[west sierra]

Should be

These 3 are unlike terms

So you can't simplify further

[balmy sequoia]

ok part b

[west sierra]

Do you know area of rectangle

[balmy sequoia]

wxh

width x height

[west sierra]

Ye

Length x breadth is what we call in my country but it's same

[balmy sequoia]

therefore just 2√3(√2+3)

[west sierra]

Yes

[balmy sequoia]

√6+6?

or no

[west sierra]

Where is 2

[balmy sequoia]

or

wait

[west sierra]

2√3 * √2 is not √6

[balmy sequoia]

its 2√6+6√3

[west sierra]

Yes

[balmy sequoia]

how come the 2 and 3 not effecyed even tho 2 is

outside bracket

[west sierra]

It's the case of like and unlike terms. Like term is multiplied like roots but not unlike terms

[balmy sequoia]

ok last one then i will seep

sleep

[west sierra]

Do you know formula for area of cuboid

[balmy sequoia]

width x height x length

[west sierra]

Yeah

So there goes your a part

[balmy sequoia]

i find base first

so part a

[west sierra]

What do you mean

What is base

[balmy sequoia]

a is find base the bottom

like the base of the box

[west sierra]

Oh

Sorry I missed the base

[balmy sequoia]

3√5(10+√3)

and we only multiply the roots

right?

or this time we do whole lot?

even numbers

[west sierra]

Yeah for base you do length * weidth

[west sierra]

Yes

That's how we kept doing

Practice more because you seem to forget math we been doing till now

[balmy sequoia]

ok

[west sierra]

So what's the expanded form

For a

[balmy sequoia]

isnt it 30√5+3√15

[west sierra]

Yes correct

[balmy sequoia]

good

[west sierra]

Now go to b part

[balmy sequoia]

volume

[west sierra]

Volume of box

[balmy sequoia]

width x heigh x length

[west sierra]

Yes

[west sierra]

When I asked you before the area of cuboid you told me volume of cuboid

[balmy sequoia]

oh

read it wrong

after this i will sleep and wish for the best

[west sierra]

Area of cuboid is 2(length times width + width times height + height times length)

Anyway what's your answer for b

[balmy sequoia]

so 3√5(10+√3+2√2)?

[west sierra]

No

Remember you seem to have already calculated length*width in a

Just multiply height to it and you will get your answer

[balmy sequoia]

oh so 2√2(10+√3+3√5)

wait

[west sierra]

It will be 3√5 * (10+√3) * 2√2

[balmy sequoia]

oh

[west sierra]

Length * width * height

[balmy sequoia]

cant i just multiply the high by area of base

[west sierra]

That's what I said above

Multiply height to answer of a

[balmy sequoia]

oh sorry

[west sierra]

Because a is length * width

[west sierra]

It's k

[balmy sequoia]

60√10+6√30

[west sierra]

Yes

[balmy sequoia]

ok

now c

i have no clue on where to start

[west sierra]

The c part is confusing

[balmy sequoia]

oh wait

[west sierra]

Because earring box in itself a cuboid

[balmy sequoia]

it says the base of the box

so we just have to

[west sierra]

Area of base/area of earring box

That would give number of boxes required

[balmy sequoia]

i dont even know

[west sierra]

What there is not to know

[balmy sequoia]

i think thats it then

[west sierra]

If you have to find the number of rooms of 1 sq foot that can be built in your 10 sq foot house, it's 10/1

[west sierra]

Yes

So what's your answer for c

[balmy sequoia]

what

i dont know

[west sierra]

Im giving example of similar question

[balmy sequoia]

but this is so different

[west sierra]

How so

Just divide

[balmy sequoia]

oh wait

tru

so 30√5 + 3√15 / 2√5

[west sierra]

You become very complicated sometime when in actuality the question is easy

[balmy sequoia]

now what

[west sierra]

Yes

To be exact

so (30√5 + 3√15) / 2√5

Remember to put brackets when you're typing online

[balmy sequoia]

ok

what next

[west sierra]

Didn't you practice questions like these

[balmy sequoia]

i havent seen a question like this in class if im being honest

[west sierra]

If (a+b)/c can't you distribute denominator over numerator?

Like a/c + b/c

If denominators are equal you can add numerators so the converse is also true

[balmy sequoia]

im too tired ill do this tmr

gn

[west sierra]

Lol okay gn

In fact it's just the last question's last part you would do before sleeping

But since you're tired, gn then

[obtuse pebble]

@balmy sequoia Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[true talon]

I understand the trig derivative for arctanx but I'm confused about solving the integral for this function

[true talon]

the answer being this

[elfin burrow]

factor out a 4

[true talon]

what about the 4 on the bottom?

[elfin burrow]

sorry i meant from the denominator too

[elfin burrow]

yeah

and note that [\frac{1}{4}x^2 = \left(\frac{1}{2}x\right)^2]

[warm shale]

tushar

[true talon]

aha

What about the 4/2 my teacher got?

[elfin burrow]

well you have a chain rule factor of 1/2

[elfin burrow]

take the derivative of this

and you'll see

[elfin burrow]

yeah

[true talon]

so i need a 4 on the top instead of 2

[elfin burrow]

so what you have is twice the derivative of arctan(x/2)

so your answer is 2arctan(x/2)

[true talon]

awesome I think I get it

[elfin burrow]

taking the derivative of this just multiplies your expression by 2

gets rid of the 1/2

[true talon]

.close

[obtuse pebble]

Channel closed

Closed by @true talon

Use .reopen if this was a mistake.

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[scenic fjord]

May I know how to solve this?

[scenic fjord]

Answer given is this

[jolly ginkgo]

use the second line

[scenic fjord]

I know we need to use that

But how to use?

And why we need to use that?

[jolly ginkgo]

substitute

[scenic fjord]

Like this?

[jolly ginkgo]

it does makes the pde easier

[jolly ginkgo]

yeah further

[scenic fjord]

Then sub again in second line?

[jolly ginkgo]

Nice reverse sub

[scenic fjord]

?

[jolly ginkgo]

you are doing great, go ahed

[scenic fjord]

I already don’t know how to continue

[jolly ginkgo]

oh

what is p?

what did we consider as p?

[jolly ginkgo]

since we integrate with respect to partial y the c will be function of x

[scenic fjord]

You see the point is now involve x and y

And I don’t know why there f(z) and g(y)

[jolly ginkgo]

wdym?