help-10

[strange sapphire]

pls help me with this

[cunning sonnet]

I thought I had to place 48 into the first number I can fit it in , I couldn’t fit it in 2 or 26 so I went 265

[cunning sonnet]

Of course yea I think I see what I’ve done

Well I have one more question , why didn’t I have to move the decimal place back since I moved it 5 time to the right

[glossy basalt]

since we are actually doing this:

,rotate

[glossy basalt]

we move the digit on both numerator and denominator

that's we are actually expanding the fraction

while expanding the fraction, the value itself won't change

[cunning sonnet]

Ahhhhhh because a percentage of a fraction always remains the same no matter how much you multiply the top and the bottom!!

I seeee

Hmm so

[glossy basalt]

yea?

[cunning sonnet]

I will need to move it for multiplication addiction and subtraction

But not division

Im just thinking because this if for chemistry there’s lots of decimal places with division multiplication and addition

[glossy basalt]

because division works like fractions

we move the decimal on both dividend and divisor

[cunning sonnet]

But I’m saying

[glossy basalt]

yea?

[cunning sonnet]

When we had the answer we did not move the decimal like I do with subtraction or additiona

Once we had the answer

This only applies for division yeah? Since it a fraction?

[glossy basalt]

you can say that

[cunning sonnet]

I had to move the decimal back on this one

[glossy basalt]

like this?

[cunning sonnet]

Well I didn’t actually do that I moved them all the way over

[glossy basalt]

oh lol

i guess that's a way too

[cunning sonnet]

And then solved

And moved them back

[glossy basalt]

i see

[cunning sonnet]

That’s why I was confused When I tried with division

[glossy basalt]

if that's the case, ya, you'll have to move for +,- ,×

[cunning sonnet]

Yep awesome !

Well thanks again for saving the day biscuity

[glossy basalt]

no problem!

[cunning sonnet]

Gods speed.

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[timid silo]

Hey, Isn't this wrong?

[timid silo]

Because integration by parts formula is like this:

[pseudo swift]

What is your problem specifically with that ?

@timid silo

[timid silo]

Red line (They have taken the integration of uv part.

[pseudo swift]

I mean the formula you posted is for indefinite integrals

For definite integrals, you have to evaluate the uv yes

[timid silo]

I see. If possible then would you please share a formula for it?

[pseudo swift]

I'm on my phone right now so it's gonna be a bit difficult

But that's the only change pretty much

$$\int_a^b u\dd{v} = [uv]_a^b - \int_a^b v\dd{u}$$

[warm shale]

_aplatypus

[pseudo swift]

If you integrate (uv)' between a and b, you get [uv]_a^b

That's why it's there

@timid silo

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

I see. Thank you, @pseudo swift

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[slender sundial]

how did they jump from the first step to the next?

[slender sundial]

like why 17 * 17 divisors

[ashen dove]

It's a formula

The number of factors is product of the factors+1

[robust sleet]

<@&268886789983436800>

[slender sundial]

okay thanks whats it called

[slender sundial]

thanks dude, i tried looking for a derivation but couldnt find it, do you know how to prove it?

[candid yarrow]

Consider all the possible factors

the exponents of the primes must be smaller than or equal to those of the whole number

so the amount of possible combinations is this

[ashen dove]

It's easy to prove
The factors should be of the form
p⁰,p¹, p²,p³ ... p^a =(a+1)
q⁰, q¹, q²,q³ ... q^b=(b+1)
r⁰, r¹, r²,r³ ... r^a=(c+1)
And all their combinations
So the total number is (a+1)(b+1)(c+1)

[slender sundial]

ohhhhhh ok

makes sense

thank u

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[fair notch]

how does the sin(pi y/ 2) become 1 when substituting?

[ashen dove]

sin (π/2) is 1 only

[fair notch]

it gives 0.027 on my calculator?

[lusty olive]

Do you have it set to radians?

[fair notch]

no degrees

[lusty olive]

Theres your problem

The problem has it written in radians

Rule of thumb, if you see pi in trigonometry its almost always radians

[fair notch]

oh i see

thank you

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[lucid flame]

How to prove LHS = RHS?

[severe reef]

try expressing each ratio in terms of sin and cos

[lucid flame]

Then

[severe reef]

are you done with that?

[lucid flame]

Yep

[severe reef]

at what expression did u arrive

[lucid flame]

Sin A / 1-cos A - 1/sin A

[errant lark]

There's an easier way to do it:

Hint: Use cosec^2 A - cot^2 A = 1

[lucid flame]

I did this way

What can I do next

🤔

[errant lark]

Write sin^2 A as (1-cos^2 A)

[lucid flame]

Then

[errant lark]

Can you expand 1-cos^2 A somehow ?

[lucid flame]

Yeah

Difference between 2 squares

[errant lark]

Good

Do that.

Then see if you can take something out as common in denominator.

[lucid flame]

?

Is there anybody

[errant lark]

Okay. Yes

[errant lark]

You are almost done.

Cancel out the terms. See what you get.

[lucid flame]

I get {(1+cos A) - 1}/sin A

[errant lark]

1 + cosA -1 is how much ? 😐

Shouldn't something get cancelled there ?

[errant lark]

No.

Can you write -1 before cosA ?

Like (1 -1 + cosA)

[lucid flame]

Yes

[errant lark]

Can you see what is that basically ?

[lucid flame]

1 - 1 closes

[errant lark]

1 -1 is zero. Right ?

[lucid flame]

Yes

[errant lark]

So, you only have cosx in the numerator.

So what do you get ?

[errant lark]

This is wrong.

[lucid flame]

What

[errant lark]

Wait...

[lucid flame]

You mean cos A/sin A

[errant lark]

You had:
$\frac {\left(( 1+\cos A) - 1 \right)}{\sin A}$

[errant lark]

Yes.

[lucid flame]

Then

[errant lark]

What the hell is this ?

[lucid flame]

Where is the answer

[warm shale]

.enemagneto

[lucid flame]

Yeah

[errant lark]

What is that equal to ?

[lucid flame]

cot A

[errant lark]

Yes.

[lucid flame]

But there is no answer at RHS

[errant lark]

Now, solve RHS and it'll come to be equal to cotA.

[lucid flame]

Can solve RHS too?

I thought I only can solve LHS

[errant lark]

Whole point is to show that both sides are equal.

If you show that LHS is cotA and that RHS is cotA.

[lucid flame]

Oh got it

[errant lark]

You are done.

[lucid flame]

Yeah I think I would be able to do it

I see these + and - marks only chnaged in both sides

Is this a way to identify that those are same?

[errant lark]

Not exactly. However, with more practise you'll come to see how those expressions are related.

[lucid flame]

Hum i see

[errant lark]

This was the way you were supposed to solve it.

[lucid flame]

I should try that way too

Thank you

👍

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[radiant frigate]

How exactly did we take 1/2 of the square root? Please explain.

[royal basin]

$\frac{x_1+x_2}{2} - x_1 = \frac{x_1+x_2-2x_1}{2} = \frac{x_2 - x_1}{2}$ and likewise for $y_i$ and $z_i$

[warm shale]

Ann

[royal basin]

then you have $\paren{ \frac{x_2-x_1}{2} }^2 = \frac{(x_2-x_1)^2}{4}$ and when you pull out the common factor $\frac{1}{4}$ outside the root it becomes $\frac{1}{2}$ again

[warm shale]

Ann

[radiant frigate]

Thank you!

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[pseudo prism]

ok so if i do 3/4 *8, i just multiply 8 and 4 right, the put it under 3. right?

[royal basin]

no

[pseudo prism]

oh

[royal basin]

that'd be if you were dividing 3/4 by 8, or multiplying 3/4 by 1/8.

$\frac{3}{4} \cdot 8 = \frac{3 \cdot 8}{4}$ perhaps.

[warm shale]

Ann

[timid silo]

Potentially lacking parentheses

Did you mean 3/(4*8) @pseudo prism

[pseudo prism]

wait but i thought if u wanted to get rid of the bottom number you just multiply the fraction by it

i mean this

D on question 1

[royal basin]

yeah sure

[pseudo prism]

ik its the right answer but unsure why

[royal basin]

$-\frac{3}{4} \cdot 8 = -\frac{3 \cdot 8}{4} = -3 \cdot 2$

[warm shale]

Ann

[royal basin]

does this explain it for you?

[pseudo prism]

yeah but if it was a 4 instead of an 8

it would just be reduced to -3

right?

[pseudo prism]

sorry im still confused

[royal basin]

sure, -3*4/4 would be -3 yes

[pseudo prism]

im so confused. but why

wait not devided by four

times four

[royal basin]

why what

[pseudo prism]

ok how does -3/4 *4= -3

like are you multiplying -3*3?

[royal basin]

i am not multiplying -3 by 3

[pseudo prism]

sorry wait

im so lost

[pseudo prism]

i mean -3*4

[royal basin]

do you agree that multiplying by a number and dividing by the same number takes you back where you started?

[pseudo prism]

sure yeah bc they're opposites of each other

[royal basin]

yeah well

-3/4 * 4

[pseudo prism]

ooohhh

i see

[royal basin]

you start with -3, divide by 4, then multiply by 4

[pseudo prism]

got it

but

doesnt that mean you multiyplayed 4 by the bottom 4

[royal basin]

... ???

no?

[pseudo prism]

yk what. maybe its a better idea for me to just turn it to a decimle

nvm that wouldnt really work in a graph

[royal basin]

you may want to review fractions

[pseudo prism]

that's the only bit im confused abt

oh wait nvm

you're multiplying it by the top and then deviding it

that makes sense

because a whole number is just it over 1

so it would be like 3/4 * 4/1

which would equal 12/4

which is 4

is that right @royal basin

[royal basin]

12/4 isn't 4

[pseudo prism]

fuck

then what is the logic behind it

are you not doing 3*4?

it was something else but it's gotten down to this

3/4*4= 3

but im not sure why exactly

[royal basin]

12/4 is 3 lol

you fucked up your arithmetic

[pseudo prism]

like where does the 4 plug in. shouldnt it have to be 3*4?

[royal basin]

idk what else to say like

we could go around in circles all day

it feels like we are going in circles here rn

there's no obligatory course of action with fractions like this

[pseudo prism]

what's the circle

yeah?

and 8*3 is 24

24/4 is 6

therefore thats the answer

yeah. so why does the mulitply by the top

but the 4 doesnt

if i did it times 4

instead of 8

why wouldnt it be 4*3 which is 12

and 12/3 is 4

wait no

right

biut

but*

3/4 *4 is 4

not 3

do you get my question here or no

it is?

[warm shale]

.general_jacob

[pseudo prism]

ooohhhh that's right

multyplying by the bottom number gets you just the top number

my b

i think i get it now

in school all they said is multiyply by the bottom number if you wanna get just the top number

[hidden compass]

What is 3/4 * 20? @pseudo prism

[pseudo prism]

it would by 3*20

which is 60

and 60/4 is 15

so 15?

[hidden compass]

Perfect

[pseudo prism]

got it. thanks guys! 😄

[hidden compass]

I think you got it 👍

[pseudo prism]

appricate it ❤️

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[quaint sierra]

i thought -90

will be answer since it is the smallest tens

[thick oracle]

-90 isnt a digit

[obtuse pebble]

@quaint sierra Has your question been resolved?

[quaint sierra]

how

<@&286206848099549185>

What

The answer is 001

Sir

@royal basin Help us

[thick oracle]

do you know what a digit is

you could try something like

10^2018 <= 7S < 10^2019

find lower and upper bounds

[quaint sierra]

Whattttt

Whyyyy

Where did you come from

How are you getting such conculsions

1 is unit

10^2018 <= 7S < 10^2019

How does this make sense

001

Yes

So how did you get it is the question

1

Sir

After some enlightenment

i have learnt that you were right

now show me your ways

so i can stop talking

oh great one

How did youget 1

[thick oracle]

7S has 2019 digits

[quaint sierra]

Now what

[quaint sierra]

Continue

[thick oracle]

mb

it should be

10^2019 <= 7S < 10^2020

cuz we know 7S has at least 2019 digits

and less tham 2020 digits

[quaint sierra]

So ans is 000

[thick oracle]

?

[quaint sierra]

You said it

I am wondering

[thick oracle]

?

[quaint sierra]

0 is divisible by 7

so why is this wrong

[thick oracle]

7 is also divisible by 7

[quaint sierra]

Yes but 0 is smaller

and leads to smaller number

[thick oracle]

we know S is divisible by 7, not the tens digit

[thick oracle]

yeah ur right

general

10^2019 <= 7S < 10^2020

just solve this

[quaint sierra]

so 10

[thick oracle]

10 what

[quaint sierra]

10.00

is ans

10^2019 <= 7S < 10^2020

[thick oracle]

no

[quaint sierra]

How can is solve this

art

10^2019 <= 7S < 10^2020

Why are there so many <<<<<<

and =

Are they to be solve simultaneously

???

<@&286206848099549185>

[thick oracle]

.

[hard igloo]

7S has 2019 digits

[mild swift]

0 is divisible by every natural number

[hard igloo]

So 7S must ve smaller than 10^2019 since that has 2020 digits

[quaint sierra]

Ok please lets get to the point

[hard igloo]

Ok

[thick oracle]

oh yeah ur right

10^2018<= 7S < 10^2019

[hard igloo]

What is the smallest number that has 2019 digits

1 and after 2018 zeros correct?

[quaint sierra]

-999999999999999999999......

[thick oracle]

<@&268886789983436800>

he keeps trolling

[quaint sierra]

That is not a troll

this is what i thought of

[hard igloo]

Wait your right

It was not said its positive

[quaint sierra]

and my initial answer

[thick oracle]

firstly -90 isnt a digit

[quaint sierra]

but it is something

[thick oracle]

what is it

[hard igloo]

So divide -9999999... by 7

[quaint sierra]

that is what i do not knwo how to do

[thick oracle]

even if 7s is negtive

the result is the same

[quaint sierra]

How

[thick oracle]

it will be the same result

but negative

[quaint sierra]

I do not have smarts in this area

i am simply trying to understand the question

[slim cove]

I'm not sure there's any trolling going on

[quaint sierra]

Please in layman's terms what is the solution

[thick oracle]

if you say so...

[slim cove]

Please don't ping random people to get help for your question.

[hard igloo]

You only need to focus on the -99 at the end

[quaint sierra]

how is answer 001

[hard igloo]

No

[quaint sierra]

-1

online

american book

questins

Why you divide

no s is

-999999999999......

Why no

demonstrate please

001

no

so 7S = -999999...9999999 --2019 digits

explain

the --

shouldn't s be -999999999999--2018

so s has 2018

help

7*7

49

Im assuming

wait please

The smallest s is -999999999...

why is so 7S = -999999...9999999 --2019 digits

correct

why

it did not say that 7s has to be the smallest

it said s has to be

hold

why 7s equals to so 7S = -999999...9999999

Why is it smaller

Why is 7s smaller

no

wait

so ans has to be negative

because c is smal number

that means c cannot be -11

because 9*-11 is -99

-99 are not digits

9C = -891
This is impossible if the two digits must be made

so what is ans

<@&286206848099549185>

for example

yesl

C must be the smallest integer such that 9C has 2 digits
how would you solve this

how would you find 10 digit

why you equate 9c to -99

ues

yes

meaning c is -99

why

so 7s must be equal to 2019

digits

but must be teh smallest

only way to change that is through variable s

so 7s =-999999999999.....

what then

then

but tens place is needed

how

but we are not solving that

we are dealing with infinite nines

explian patter thing

yr12

Year 12 highschool

grade 12

I am not in middle schol

i know integration by parts

calculus

circular fucntions

proof

That because my teacher said i should sove questions on my own

and when i need help meet use discord

beacuse he is busy

Please explain

Why

This is not normal math

Olympiad math

I am new to it

Please help i really have no other option

tHIS

[warm shale]

.general_jacob

[quaint sierra]

Ok

[warm shale]

.general_jacob

[quaint sierra]

What does this mean

Can you solve this without calculatore

how

yes

so why is ans 1

ok

ty

.close

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[obsidian wren]

i need help in creating two different two columnb proofs for two different questions

[obsidian wren]

<@&286206848099549185>

[obtuse pebble]

@obsidian wren Has your question been resolved?

[obsidian wren]

anyone?

[untold trail]

what triangles are ABd and adc?

[obsidian wren]

what question?

[crisp fiber]

Q2: <|Q = <|R , <|Q = PQ and <|R= PR => PQ = PR

Q1: well DC is tangent => AD perpendicular on DC => <| D = 90 also B = 90 degrees as AD is the diameter now we have B = D also A = A and there you have it

ABD ~ ADC

Not sure If I did it correctly. If you don't understand smth just tell me I'll clarify

[obsidian wren]

of first

[obsidian wren]

thanks ill check

[crisp fiber]

Okay!

tho it should be good from what I have learnt in 6th grade( im not 6th grade)

[crisp fiber]

lmk if its good

[obsidian wren]

ok mb had to do something ill look at it now

@crisp fiber Q2 is wrong

its arc pq and pr

not line

that we have to find

[crisp fiber]

yes

[obsidian wren]

and its in 2 column proof format

[crisp fiber]

yes ik

It shoudl be correct

they are equal

[obsidian wren]

ok but what theorems have you used?

[crisp fiber]

its bcs if u take it like that

No theoremes

[obsidian wren]

and postulates

[crisp fiber]

just simple math

[obsidian wren]

bro but i need help in making a two column proof

which requires a statement like you have provide and a reason for each

[crisp fiber]

Q = PR * 2
R = PQ *2

[crisp fiber]

oh

uh Ill try to research that rn

[obsidian wren]

ok thanks

so you understand what it is @crisp fiber ?

<@&286206848099549185>

anyonme

[crisp fiber]

Tbh I don't rlly know anymore

I can't help srry!

[obsidian wren]

oof

ok

<@&286206848099549185>

anyone anyone

@crisp fiber you there?

[untold trail]

lemme see

[obsidian wren]

THANK YOU PLEASE

[untold trail]

ok so first

do uk what a 2 column proof is

[obsidian wren]

yes yes

Q2: <|Q = <|R , <|Q = PQ and <|R= PR => PQ = PR

Q1: well DC is tangent => AD perpendicular on DC => <| D = 90 also B = 90 degrees as AD is the diameter now we have B = D also A = A and there you have it

ABD ~ ADC

Not sure If I did it correctly. If you don't understand smth just tell me I'll clarify. This is waht the other person rule rote

wrote*

[obsidian wren]

yes ik what it is

[untold trail]

Q1: well DC is tangent => AD perpendicular on DC => <| D = 90 also B = 90 degrees as AD is the diameter now we have B = D also A = A and there you have it

ok wait a moment

alright

so first we always put our given right

[obsidian wren]

ye

[untold trail]

so wahts our first statement and reason

[obsidian wren]

ok its gonna be

AD¯¯¯¯¯¯¯¯
is a diameter of circle O
and DC¯¯¯¯¯¯¯¯
is tangent to circle O
at D

[untold trail]

alr wait let me just create a table

[obsidian wren]

ok

[untold trail]

is that our statement or reason

[obsidian wren]

statement

reasonn is given

[untold trail]

yup

[obsidian wren]

ok now what

[untold trail]

did you understand the proof that @crisp fiber gave

he said "Q1: well DC is tangent => AD perpendicular on DC => <| D = 90 also B = 90 degrees as AD is the diameter now we have B = D also A = A and there you have it"

[obsidian wren]

yes i do understand it

just the reasons is the problem for me top match with the statements

you there?

[untold trail]

yes give me one moment

[obsidian wren]

ok sure

[untold trail]

what would be teh statement for teh second step

[obsidian wren]

Ad is perpendicular to DC

[untold trail]

we have to first define that DC is tangent to AD

[obsidian wren]

ok

so two different steps?

[untold trail]

yes

[obsidian wren]

is there a symbol for tangent?

or straightup DC is tanget to AD

??

[untold trail]

wait a moemnt, do u have a theorem sheet with u?

[obsidian wren]

yes actually

a pdf

[untold trail]

can u send that to me if u dont mind

[untold trail]

ok actually i think u can just say

DC is perpendicular to AD

sorry thats my bad

[obsidian wren]

ok all good

i should state incase though i think

dont ou think? whats there to lose out of it

[untold trail]

and then teh reason would be tangent line to circle theorem

[untold trail]

i made a mistake, DC isnt tangent to AD becaues AD is a line

and u cant be tangent to a line

[obsidian wren]

oh yea

should i say its tangent to circle in case?

[untold trail]

it already says taht in the given

[obsidian wren]

oh yea ok

ok lemme write down next step

[obsidian wren]

ok next step time

[untold trail]

alright what do u think the next statement is

[obsidian wren]

<D=90

[untold trail]

i think

at least when i did these last year

my teacher required me to say like (because definition of perpendicular lines does not say the angles are 90 degrees, but rather taht the angles are right angles)

<D is a right angle

and then m<D is 90 degrees

[obsidian wren]

ok

[untold trail]

so we can first say

<D is a right angle

and whats the reason

[obsidian wren]

ummmm thats what i was thinking

defintion of perpendicular lines?

[untold trail]

yes

the reason that we put <D is a right angle instead of m<D = 90 degrees

is because the definition of perpendicular lines states that they intersect each other at right angles not 90 degrees

[obsidian wren]

ok

[untold trail]

and then now we can say m<D = 90

and whats the reason behind that

[obsidian wren]

that i dont know

right angle=90

[untold trail]

whats the definition of a right angle

yeah

[obsidian wren]

defintion of right angle angles?

i was gonna say that

[untold trail]

yes ^

[obsidian wren]

ok

lemme write that now

[obsidian wren]

then m<B=90?

[untold trail]

first

1. which angle B
2. right angle or 90 degrees

[obsidian wren]

right angle

[untold trail]

right

and which angle B

[obsidian wren]

ok ima write the next two

[untold trail]

wait

which angle B

[obsidian wren]

?????

[untold trail]

<ABD or <DBC

there are two angles corresponding to angle B

you can't just say angle B

[obsidian wren]

DBC?

[untold trail]

also my bad i forgot to point it out you have to say <ADC because <D also corresponds to two angles 😅

[obsidian wren]

i think or wait

[obsidian wren]

oh ok

[obsidian wren]

mb

[untold trail]

so when we name angles

we can use the similarity statement as reference

[obsidian wren]

ok

[obsidian wren]

so is b angle DBA?

[untold trail]

yes, but i would write it as ABD

[obsidian wren]

ok

[untold trail]

so angle ABD is a right angle

and whats the reason

[obsidian wren]

def of perpendicular lines

then measure is =equal to 90 which is defintion of right angles

then felexive property of congruent <a=<a

reflexive*

and aa similairty theorem

damn

[untold trail]

wait no

[obsidian wren]

what what

[untold trail]

the reason is wrong

[obsidian wren]

for>?

[untold trail]

when <ABD is a right angle

you use this theorem's converse

and then u can use definition of right angle

and then AA

[obsidian wren]

ok

[untold trail]

can u weite out the proof

and send it to me

[obsidian wren]

why?

[untold trail]

ill recheck it

[obsidian wren]

ok

just one more proof after this and im done hopefully

[untold trail]

yup 😅

sorry i cant help rn 😭 i have a class

[obsidian wren]

please please

it will be quick

i n eed this i beg

we can go overn rn

it shouldnt take long

it looks simple

[untold trail]

okay

[obsidian wren]

THANK YOU

[untold trail]

whats ur general idea od how to prove it

[obsidian wren]

ok PR is equal to q

and pq is equal to r

to begin with right?

what theorem is that?

[untold trail]

PR is equal to Q?

what is Q

[obsidian wren]

no shoot

q and r are congruent

so sides pr and pq are congruent

[untold trail]

yes

and then

[obsidian wren]

idk where to go from there

thats where my brain stops

[obsidian wren]

so done

thats it right?

[untold trail]

yes tahts the general idea

[obsidian wren]

ok then i write that step

and done

that was quick

thanks a lot

[untold trail]

yup np

sorry i explain things weirdly sometimes lmao

[obsidian wren]

nah not at all

[untold trail]

gl on the rest of ur geo problems

[obsidian wren]

you were a great help

[untold trail]

ahh tahnk u sm
really glad to hear that

[obsidian wren]

thanks

have a good day

[obsidian wren]

so sry about that

[untold trail]

you too

[untold trail]

sorry ab wha

[obsidian wren]

i said man mistakenly

[untold trail]

theres nothing to be sorry about

HAHA NO WORRIES

i dont really mind

[obsidian wren]

ok cya!

[untold trail]

cya

[obsidian wren]

.close

[obtuse pebble]

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[timid silo]

Just gonna send this here because I don’t know what else do and I have no idea how algebra works 😓

[proven zephyr]

,rcct

,rctt

ah crap i forgot how

,rotate

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

😓

[timid silo]

No…

<@&286206848099549185> 🫡

[untold trail]

do you know what adjacent angles are

? @edgy valley did i do something wrong

[edgy valley]

No

It means I don't know

[untold trail]

oh

this orange angle is adjacent to the blue angle because they share common vertex A and common side AJ

[edgy valley]

Question 3 is therefore yes right?

4 is no

[untold trail]

yup

[edgy valley]

5 is yes

[untold trail]

@timid silo do u understand

[untold trail]

i would recommend color coding the angles like this

[edgy valley]

And color code the common side too

[untold trail]

i think color coding really helps with geometry and algebra

please keep unrelated bot use in #bots

[edgy valley]

O^K

[untold trail]

thank you!

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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

hello

[gray pulsar]

i need help

im loading smth rq

So

the equation 6^2+8^2 = 10^2 is true right?

but this equation DOES NOT follow the algebraic rules

because this is is similiar to a^2*b^2 = (ab)^2

but its adding

even though the equation is true, it does not follow the algebraic rules because this equation is the Pythagoream Theorm that makes the right triangle 6,8,10

am i right?

[nocturne minnow]

What algebraic rules are you referring to?

[gray pulsar]

the one i showed you up

[nocturne minnow]

It being similar does not mean it's the same

[gray pulsar]

yeah

its not the same

[nocturne minnow]

You can't apply a^2*b^2 = (ab)^2 to addition

[gray pulsar]

Yes

and the equation is asking me if the equation 6^2+8^2 = 10^2 is true

and if so how does it follow the algebraic rules IF IT FOLLOWS IT

so its basically a trick question

and it is true

but does not follow one of the algebraic rules

but instead its a pythagoream theorm equation

that makes the right triangle 6,8,10

[nocturne minnow]

As asked, what are the algebraic rules you are referring to?

[gray pulsar]

a^2*b^2 = (ab)^2

actually no or

[nocturne minnow]

What are the other rules that are algebraic rules

I don't know what you mean by algebraic rules

[gray pulsar]

omg

such as

a*b = a(b)

or

a^a / a^b = (a^b-a)

those??

but what im saying is if my explaination is correct

[nocturne minnow]

Your definition of algebraic rules makes no sense, because a*b = a(b) is just a property of multiplication. a^a / a^b = (a^b-a) is a property of exponents

[gray pulsar]

but the question is asking

IF it follows that algebraic equations or it means RULES

algebraic rules

there

but the equation thats shown in the equation

does not follow any of the rules

[nocturne minnow]

And as I am asking, what are the algebraic rules you are referring to? Do you have a list from a book or your teacher?

The meaning of algebraic rules is too generic to understand what you mean

[gray pulsar]

the basic ones..

the ones that you started learning from pre-alg or alg 1

[quasi glen]

I think he talks about the multi platoon ones

Multiplication

Formulas

[nocturne minnow]

Not everyone learns it with the same terminology

[quasi glen]

Like (a+b)^2 and so on

[gray pulsar]

but the main point of the equation

[nocturne minnow]

I could literally say a + a = 2a as an algebraic rule

[gray pulsar]

i didn't came up in my head

but

the point is

and if so how does it follow the algebraic rules IF IT FOLLOWS IT
so its basically a trick question
and it is true
but does not follow one of the algebraic rules
but instead its a pythagoream theorm equation
that makes the right triangle 6,8,10

thats my explaination

And it says IF

it follows it

but no it doesn't

[nocturne minnow]

Your definition is broad, I don't know what you mean by algebraic rule.

[gray pulsar]

;-;

but generally if you know the topic

is it still correct?

[nocturne minnow]

It helps to know what rules you are referring to

[gray pulsar]

omg

[nocturne minnow]

Because you mentioned a^2*b^2 = (ab)^2 , that question doesn't follow that

But what other rules are there?

[gray pulsar]

HAVE YOU SEEN the equation a^2+b^2 = (a+b)^2 WORK?

CUZ 6^2 + 8^2 = (6+8)^2 DOES NOT MAKE 10^2

SO THE EQUATION IS CORRECT

but the ALGEBRAIC rules DOESNT FOLLOW in that equation, a^2*b^2 = (ab)^2

yk what

i'll solve it bymyself

.close

[obtuse pebble]

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[timid silo]

find the coefficient of x^2 in the expansion of (3+2x+x^2)(2-1/x)^4

how do i do a binomial expansion when one of the terms is a fraction? also, is there a shortcut to finding the coefficient of a specific term without having to expand and multiply by the other polynomial

[desert sinew]

Hint: Find the general term of the binomial first.

[high lily]

binomial expansion is the same

(a+b)^n = ...
doesn't matter if a,b are fractions

[timid silo]

what do you mean by this?

[high lily]

= 3(2-1/x)^4 + 2x(2-1/x)^4 + 3x^2(2-1/x)^4
consider 3* what type of term,
gives an x^2 term
identify that term in the expansion of (2-1/x)^4 from binomial theorem

repeat for the 2x and x^2

also did you copy down the question correctly?

as it's a bit strange to be set up like this

[timid silo]

ok so when i expand the terms i get many constants/x^n and when they are multiplied with the terms of (3+2x+x^2) i no longer have a fraction and get a term with x^2?

[timid silo]

yeah i think so

[high lily]

do you have a pic of the original problem

[timid silo]

yeah

give me a sec

[high lily]

don't expand (2 -1/x)^4 directly,
identify the specific terms needed considering what I typed above
and obtain those from binomial theorem

yeh, the question is a bit off. there's probably a typo somewhere.

[timid silo]

via 4C2, 4C1, 4C0?

[timid silo]

oh what makes you say that

[high lily]

you'll see following what I said

consider
3 * what type of term gives an x^2 term?

[timid silo]

x^2

[high lily]

and is there an x^2 term in (2-1/x)^4

[timid silo]

ok thanks, ill try

[high lily]

typo

again, don't expand

[timid silo]

just extract the terms using ncr?

[high lily]

if that's what you were thinking

you should be able to identify whether this will actually have an x^2 term or not without having to do anything extraordinary

[timid silo]

actually that's where i got stuck, because i didnt know how to extract an x^2 term out of 3 * 4C2 * (1/x)^2

[high lily]

overthinking

3 * 4C2 * (1/x)^2
isn't an x^2 term

is there any x^2 term in expansion of
(2 - 1/x)^4

do the full expansion if you feel the need for it

[timid silo]

is it just a trick question

16x^2 from the last term

[high lily]

yes

[high lily]

hence why I said there's probably a typo

if this question only has one part then

[timid silo]

yeah i think it was probably just taken out of context of a larger problem

[high lily]

they probably intended
(2x - 1/x)^4

[timid silo]

i see

[timid silo]

thank you

how do i account for this?

[high lily]

same principle

= x(x+y)^6 - y(x+y)^6
consider x * what type of term gives an x^2y^5 term

identify that term in the expansion of (x+y)^6 from binomial theorem

repeat for the y

[timid silo]

ok thanks

i'd like to see the thought process and methodology in solving this problem. can someone provide a step by step to explain their intuition when solving these problems?

i want to know how you can find r which gives a term in x^0 in a question like this

[obtuse pebble]

@timid silo Has your question been resolved?

[high lily]

again same idea

its pretty much always the same idea

consider x^2 * what type of term gives a constant

and extract that term from the expansion

[obtuse pebble]

@timid silo Has your question been resolved?

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[tidal mulch]

can someone please show me what did I do wrong here?

answer is supposed to be -104/3

[obtuse pebble]

@tidal mulch Has your question been resolved?

[tidal mulch]

<@&286206848099549185>

[obtuse pebble]

@tidal mulch Has your question been resolved?

[fair rapids]

@tidal mulch actually it should be |[(Y/2+9)^3/2]/(3/2)|0 to -16

[tidal mulch]

@fair rapids hi! are you able to write the working on paper?

[fair rapids]

ill try one sec

@tidal mulch sry i didnt get

[obtuse pebble]

@tidal mulch Has your question been resolved?

[tidal mulch]

tch it's upside down

ignore that in the upper left it's not relevant to this

then the continued ver

[obtuse pebble]

@tidal mulch Has your question been resolved?

[tidal mulch]

@stark geyser hi! please show me your working when you're available, thank you!

[tidal mulch]

differentiated the expression, brought it down-- that was what I was taught

[robust sleet]

Ummm

You can just use usub

@tidal mulch

[tidal mulch]

oogh i wish but idk it well enough to apply it

[robust sleet]

i mean i have no idea what you did, and usub is 100% easier

just u-sub y/2 +9 and you'll how see how easy it will become

[tidal mulch]

I'll try, but I'll close this channel for now and return when I still have issues

.close

[obtuse pebble]

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[obtuse pebble]

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

I was proving something, but I ended up proving every trapezium a parallelogram, I cant find out what I did wrong

[dense stirrup]

what were you proving?

[tranquil sonnet]

COB is not similar to AOD

[shell parrot]

ABCD is a convex quadrilateral in which AD= √3, ZA=60°, ZD=120° and AB+CD=2AD. M is the midpoint of BC. Then DM=​

I was thinking to prove this trapezium a rhombus first

[dense stirrup]

trapezium is not a rhombus

you know

[shell parrot]

did I use SAS criterion wrong?

[shell parrot]

uhm, a rhombus is a trapezium

[dense stirrup]

Please send a clear image of properly labelled construction

cuz i don't see any m in ur figure

[tranquil sonnet]

Evidently

[shell parrot]

the Z in here is for angle I think

[shell parrot]

could you explain why

I am struggling to understand

[tranquil sonnet]

They don’t have 2 sides that are congruent unless the trapezoid is isosceles

[shell parrot]

but the ratios of the other two sides are equal and the angle between them is equal

[dense stirrup]

they are proportional

[dense stirrup]

as proved before

[tranquil sonnet]

They aren’t

wait

I see what yiu are saying

Lemme see

[shell parrot]

ohk

I got it

gonna go jump off a cliff now

[tranquil sonnet]

Oh it’s because the ratios aren’t the same

[shell parrot]

thanks

[tranquil sonnet]

[shell parrot]

how do I close I forgot

nvm

.close

[obtuse pebble]

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[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

[halcyon hull]

2

[obtuse pebble]

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

[halcyon hull]

e i cantttt

someone please just HELP

I JUST CANT FIGURE IT OUTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

please bro

im actually begging someone to help

its been days...

this question alone has been halting my progress. I CANT FIGURE IT OUT MAN

[shy wasp]

Hey, the area under the v-t graph (the one that you have shown) equates to the displacement.

if we are considering the displacement at t = 0, to be 0. Then we just need to find t for which area under the graph from 0 to t is 0.