help-42

[brittle siren]

AE = 0.5 * EB

but we dont have ea or eb as r

so how can we say that

[brittle siren]

black line is R and blue line is?

[random adder]

Blue line is $\frac{r\sqrt{3}}{2}$

[potent lotus]

casework

[brittle siren]

why

im sorry im

i just dont get why

E O2 is r

other than that

there is no angels there

nor sides that can be helpful to say this

i dont see why

maybe im stupid but i cant see how

[random adder]

AEO2 is a 60 60 60 triangle

[brittle siren]

i see now

so

what how is it 60 6060

oh nvm

isosceles and

parallel

so ae = ao2 = eo2 = r

i see where u get r sqrt 3 fom

what about the division by 2

[random adder]

Thats the formula for the height of a 60 60 60 triangle

If you want to you can derive it

[brittle siren]

if AB = r sqrt 3

and AB also equals 2 r

does r = r * sqrt 3 / 2?

are u saying that the blud line is r?

oh wait

nvm its EB

got me confused there

ok nvm all i wrote

i just googled

and found the formula

i just didnt know the formula for this

[brittle siren]

so this is correct i agree

[mint verge]

@brittle siren Again the circle, again you

[brittle siren]

heheh

whats up

[mint verge]

not bad

[brittle siren]

nice

[mint verge]

Is there any problem

or

maybe you figured it out

[brittle siren]

yeah im trying to figure it out

casework here helps me get it but i still struggle

[mint verge]

i understand

[brittle siren]

well casework i can also say the big traingle under that small one is also 60 60 60

so same formula

?

just now that the EO1 is R

so the height would be h = Rsqrt3/2

myabe now i can get the height of the trapezoid

R sqrt3/2 - r sqrt3 /2

[mint verge]

what is this for

so what's the main idea?

[brittle siren]

find area of the trapezoid

[mint verge]

I read the question

[brittle siren]

ABCD

[mint verge]

Doesn't he ask if they meet at point E?

[brittle siren]

express it using r

yeah thats the first section

this is the second one

[mint verge]

okay let me look

understand

understand now

[brittle siren]

ok

height of trapezid is sqrt3/2 ( R-r)

AB is 2 r

[mint verge]

R is bigger and r is smaller

?

[brittle siren]

and DC is 2R

yes

so 2R+2r * sqrt3/2(R-r) /2

[mint verge]

(r+R)/2×height

[brittle siren]

2r + 2R

[mint verge]

yes

so r+R remains

[brittle siren]

yeah thats what i wrote here

[mint verge]

from where btw

[brittle siren]

i find the height of the trinagle EO2A and EO1D

and then subtract

they are both 60 60 60 traingles

[mint verge]

is they equaliteral?

[brittle siren]

yes

[mint verge]

why

[brittle siren]

ADC = 60

and we know DEC = 90 bc diameter

[mint verge]

ADC 60°

[brittle siren]

yes

[mint verge]

Do we know this?

[brittle siren]

we do

[mint verge]

okay

[brittle siren]

it is given

[mint verge]

understand

[brittle siren]

ok

so R ( sqrt 3 /2 ) - r (sqrt 3/ 2)

we can write

[mint verge]

okay easy then

[brittle siren]

sqrt3/2 (R-r) * R + r

[mint verge]

yes

[brittle siren]

this is area

but we need to put R in terms of r

so we have one x

not x,y

[mint verge]

okay understand

[brittle siren]

btw AB||DC

so we can do similar

maybe

[mint verge]

// you mean

?

[brittle siren]

parallel

yes

[mint verge]

Because it's already trapezoid

lol

understand it but

[brittle siren]

yes i say trapezoid because parallel

[mint verge]

I guess it's not easy

i understand

R/r is

ratio of similarity

[brittle siren]

yes

its gonna be R/r = R/r

i alr try that lol

[mint verge]

no length?

[brittle siren]

im not sure

[mint verge]

well, maybe we can take advantage of the location of the small circle

tangent and halfway seated

[brittle siren]

is AE/DE = AB / DC ?

i think so

AE / DE = r / R

AE = r

sp DE = R?

oh we already know

equailateral

nevermind

[mint verge]

i guess

i found something

[brittle siren]

bro this is hard

really?

what is it

[mint verge]

I'll be back soon bro

[brittle siren]

ok

[mint verge]

not available right now

[brittle siren]

all good

[mint verge]

okay so

in here

there are two r's

EO2 and

O2-anonymus

[brittle siren]

yes

[mint verge]

let's say X

[brittle siren]

okk

[mint verge]

EO2 and

O2X

now let's draw EX

here we have

some angles

in the triangle

let's get to the wide one

we know DEC 90°

we draw EO1 and there is 60-60-60

[brittle siren]

in traignle EO1X?

[mint verge]

AO2X focus on the angle AO2X

no EO2X

this is more important

[brittle siren]

120

[mint verge]

but look at the AO2X angle right now

90°?

[brittle siren]

oh yeah nvm

its 90

[mint verge]

now lets look at

EO2A angle

[brittle siren]

wait

[mint verge]

it is 60 i think right?

[brittle siren]

AO2A is height for traoezoid

no?

[mint verge]

this is angle

EO2A

[brittle siren]

no wait

[mint verge]

We just said that it is equilateral

[brittle siren]

O2X

[mint verge]

60° must be

yes

[brittle siren]

O2X = r

[mint verge]

yes

[brittle siren]

but is also height for trapzeoid

we prove

already

sqrt3/2 (R-r) = r?

[mint verge]

what

Was it that easy?

[brittle siren]

i dont know hahha

[mint verge]

Come on, i was just going to get into the sine theorem

lol

[brittle siren]

bahahah

so r * R X sqrt3/2 (R-r)?

[mint verge]

r+r√3/2

= R√3/2

i think

and some things are cancelled

2r+r√3 = R√3

and r(2+√3) = R√3

R/r = (2+√3)/√3

we found

[brittle siren]

but we need area of ABCD

without R

[mint verge]

doesnt matter

it is

R = r(2+√3)/√3 then

[brittle siren]

ok i will try to do algebra

[mint verge]

let's find EDC-EAB

area edc area eab

[brittle siren]

yes

EDC is

R√3/2*2R/2

=

R^2√3/2

and for EAB

[mint verge]

yes

lets replace then

[brittle siren]

its r^2√3/2

[mint verge]

r√3/2×2r/2

yes

[brittle siren]

R^2√3/2 - r^2√3/2

= √3/2(R^2-r^2)

[mint verge]

r²(7+4√3)/3

is R2

^2

lets check it

√3/2(r²(((7+4√3)/3))-1)

= √3/2(r²((4+4√3)/3))

4r²√3/2•(1+√3)/3

2r²√3•(1+√3)/3

am i wrong or there is np?

[brittle siren]

is this final solution?

[mint verge]

yes i think

I want to do this with what i see

R/r if the ratio is the same, there is no need to

try to solve

r²+r²-2r²×cos(150)

2r²-2r²×(-√3/2)

= 2r²+√3r²

= r²(2+√3)

r×√(2+√3)

sin60/(r×√(2+√3)) = sin75/ED

sin75..

yes it is a little difficult

but it can be done

[brittle siren]

its ok i think its enough

[mint verge]

yes

[brittle siren]

i ask for proffesor tomorrow

[mint verge]

okay

[brittle siren]

can u help me last question?

its other

[mint verge]

okay

[brittle siren]

and then i go because i have to sleep

[mint verge]

let me see

[brittle siren]

ok

[mint verge]

Dude i guess you're having trouble with circles

[brittle siren]

circle with center O. radius is called "R". and another circle with center E. radius is called "r". OA and OB are tangents to circle E,

[mint verge]

lol

[brittle siren]

R=3r

[mint verge]

okay

[brittle siren]

yeah....

a little

bit

prove AOB is equilateral

[mint verge]

okay t is good

it is

we did this

you know

lol

so it's very similar

[brittle siren]

how we did

we have BO = AO

[mint verge]

okay yes

and AB = AO too

and BO too of course

?

[brittle siren]

we need to prove that this is

we dont know

[mint verge]

okay im asking for it

already

[brittle siren]

well if we need to prove then

yes

it will be equal after

maybe we can work with 90 degree

from angels with tangents

in small circle

[mint verge]

okay

[brittle siren]

tell me

if we know OX is 90

can we say OC is 90?

[mint verge]

what is X

[brittle siren]

X is here

becuase its all on the same radois

straight line

maybe?

[mint verge]

OC is 90 already

[brittle siren]

why?

oh yeah

bc isosceles

and tangents

[mint verge]

AO = BO

[brittle siren]

yes but also

CO is angle bisector

bc of that OC is 90

[mint verge]

yes

[brittle siren]

ok

[mint verge]

I'm looking to see if there is any similarity

If we go around the back of the circle, we get equilateral without changing the measurements

[mint verge]

like this

This is what I meant when i said we've done it before

angles are equal

[brittle siren]

how is this equailateral

[mint verge]

3r we have

DO

AD = AB

divided by ratio of 1/2

[brittle siren]

AD=AB?

[mint verge]

yes

they are always equal anyway

Even if they come to my house

[brittle siren]

oh ok

[mint verge]

tangent must be equal

[brittle siren]

why is it 60 60 60 again

[mint verge]

.

Dividing 1/2 from the center of the smaller circle

i.e. it is center of gravity

also the center of the circle

[brittle siren]

ok

so?

[mint verge]

also bisector

angle and side

[brittle siren]

and

i dont understand

[mint verge]

but the circle is tangent to every side of the triangle

here's the thing

An ordinary isosceles can also do the rest i mentioned

but it can't do that

must be special

[brittle siren]

but how is it equilateral

[mint verge]

let me think a little

there must be something

im sure

[brittle siren]

ok

bro

maybe you want to do another exercise

i have a question

it looks eqsier

[mint verge]

I think I found it

I will tell you

what is this

[brittle siren]

what is it

@mint verge

[mint verge]

wait let me check

yes

i think i found it

@brittle siren

[brittle siren]

ok

what we do

?

[mint verge]

this is important

[brittle siren]

i cant see anything

[mint verge]

me was too

[brittle siren]

ok

[mint verge]

Now we know that the perpendicular line we drew to the left divides the side of the triangle into two

OD

[brittle siren]

yes

AD=DB

[mint verge]

Now let's look at where the small circle touches a point inside the triangle

this is right angle too of course

But if we show that it divides into two, we understand that it is the median

[brittle siren]

yes

[mint verge]

I draw a segment from point D to the tangent point from above

let's say X here

DX

[brittle siren]

why not DE

[mint verge]

E is center

as i see

[brittle siren]

yes

my bad

[mint verge]

np

[brittle siren]

ok DX

[mint verge]

if it comes with the same angle BOA

ODX = BOA

i mean

then it becomes isosceles and becomes a super tree

I don't know what they say there

[brittle siren]

?

[mint verge]

as angle

ODX = BOA

[brittle siren]

why

[mint verge]

When this is achieved, it is good if they intersect where the circle is tangent

i forgot

wait

a little

ahah

i remembered

there is 3r length between

D and O

If there is a tangent point when 1.5r the tangent point is the vertex of the triangle

this is important

So this is what i'm talking about

You and i are 6 meters apart

I raised my arm up alpha degree

You also raised your arm up at the "same angle"

Our arms would meet after traveling 6m/2 horizontally

Because if you draw a line segment perpendicular to the base from an isosceles triangle, it will divide it into two equal parts

[brittle siren]

each one?

[mint verge]

yes

[brittle siren]

that makes no sense

[mint verge]

3m and 3m

[brittle siren]

even if we dont raise our arm

we meet

after 6m/2

raising our arm means our arm touch before 3m

[mint verge]

No, think of our arm as line

This is impossible but an example

Imagine that you can extend infinitely

[brittle siren]

my fingers travel less distance than my elbow

bc of x between fingers and elbow

elbow travel say 3m, fingers travel 3-x

after 2(3-x) fingers meet

if we have same fingers lol

[mint verge]

Think of the right and left lines as our arms

[brittle siren]

what are we trying to prove

again

on geometry

[mint verge]

They are alpha degree above the ground

anyway, they are isosceles after all

and when you draw a perpendicular line, two equal parts are formed

at the base

[brittle siren]

ok

how is that equilateral

[mint verge]

Assume the angles i did are equal

[brittle siren]

why are they equal

[mint verge]

I will show that they meet at the tangent point

ADO is 90°

If you draw it from a 90° angle towards the middle of the base, it will be equal to the parts it divides into

like this

I drew badly

But they all make 1 radius

1 radius = the others

because they are in the circle

and they are radius

dude we need to figure this out@brittle siren

I think it's going bad

[brittle siren]

lets try a different exercise

?

[mint verge]

good for me

i already solved it

[brittle siren]

i just need time

[mint verge]

but I would like to tell you

explain to you

[brittle siren]

ok

[mint verge]

no problem

[brittle siren]

i will isten

[mint verge]

We can look again tomorrow

[brittle siren]

ok

agreed

[mint verge]

ok

[brittle siren]

lets do this exercise

ithout circle

[mint verge]

okay

[brittle siren]

without

[mint verge]

lol

[brittle siren]

AE = EB

DF = FN

prove that BAL = BCL

[mint verge]

It's like we need to draw a circle

lol

[brittle siren]

cyclic quads?

[mint verge]

there is no joke

[brittle siren]

maybe

[mint verge]

maybe

[brittle siren]

look at AKF

[mint verge]

bro

lol

[brittle siren]

its 90

[mint verge]

yes

[brittle siren]

and ACB is 90

this is cyclic quad

AKNC

[mint verge]

ok then

[brittle siren]

i will try to drwa

[mint verge]

okay

[brittle siren]

because AE=EB

AE=EB=CE

CE is median to AB

i drew

[mint verge]

okay yes

[brittle siren]

wait

idk if its bisector

on A

say its 2a

[mint verge]

I think it's not bisector

[brittle siren]

yes its not

[mint verge]

what about

AFB

tirangle

triangle

[brittle siren]

what about it

[mint verge]

I think i saw it wrong

[brittle siren]

look my guess

[mint verge]

okay

[brittle siren]

we need to work with triangle KBD and CDB

[mint verge]

maybe

how

[brittle siren]

they are both right and share base DB

so maybe

[mint verge]

yes

wait yes

DB is diameter

[brittle siren]

also AN

[mint verge]

But are the circles equal?

[brittle siren]

esat

look

[mint verge]

yes

[brittle siren]

at what we need to prove

we need

BAL = BCL

if you llook

[mint verge]

yes

[brittle siren]

they both on BL

[mint verge]

yes

i saw thay

[brittle siren]

can u see circle with BL?

[mint verge]

I was looking at it

maybe ABCL

it must be ABCL

no way

because ALC and ABC must also be equal

when these are equal

triangles back to back

they already share an angle

both of them

[brittle siren]

but is ABCL cyclic quad?

[mint verge]

must be something

Does something need to happen?

only ABCL should be above a circle

[brittle siren]

why did they give AE=EB and DF = FN

we think on this

[mint verge]

yes

[brittle siren]

pro tip

[mint verge]

what it is

[brittle siren]

rotate your head 90 degree to the right

[mint verge]

AFB seems important

okay

i didnt see

lol

[brittle siren]

its supposed to be traingle ABD

where BC is 90 with AD

[mint verge]

yes

this is height

and okay i think

ALB must be 90°?

[brittle siren]

this is what i mean

[mint verge]

okay then

AE = EB. DF = FN

[mint verge]

yes i see

but what to do

ALB should be right angle i think

[mint verge]

yes

i see

isosceles?

maybe

[brittle siren]

ABC is also a

BC=AC?

no

nvm

ABC is not a

bro i open again on a new chanel ok?

so other can help try

[mint verge]

ok

[brittle siren]

.close

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Use .reopen if this was a mistake.

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[remote mural]

asked to rationalize denominator

[remote mural]

my steps are

as follows

$\frac{2\sqrt{5}}{2\sqrt{5}+3\sqrt{2}}\cdot \frac{2\sqrt{5}-3\sqrt{2}}{2\sqrt{5}-3\sqrt{2}}$

[potent lotus]

mj

[remote mural]

$\frac{20-6\sqrt{10}}{20-18}$

$\frac{20-6\sqrt{10}}{8}$

[potent lotus]

mj

[remote mural]

$\frac{2(10-3\sqrt{10})}{2(4)}$

[potent lotus]

mj

[remote mural]

$\frac{(10-3\sqrt{10})}{(4)}$

[potent lotus]

mj

[kindred estuary]

How'd you get 12?

[rocky flower]

this step is incorrect, (3root2)^2=9*2=18 not 12

[remote mural]

oh ok mb

[potent lotus]

mj

[remote mural]

$\frac{20-6\sqrt{10}}{2}$

[potent lotus]

mj

[remote mural]

ah i see it now

$10-3\sqrt{10}$

[potent lotus]

mj

[remote mural]

okok thanks guys

🙂

.close

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[winter crypt]

I integrated the function from [0,t]

[winter crypt]

and I got an answer which wasnt correct

I got 764e^(0.02t) + C

[amber bolt]

did you drop +C ?

there's no plus see

[winter crypt]

no

[amber bolt]

764e^(0.02t)

[winter crypt]

.

?

[amber bolt]

and you said you didn't drop +C

that's the mistake

[winter crypt]

I tried it without the + C as well

it says its still wrong

[amber bolt]

okay

[winter crypt]

it says the variable isnt defined in this context

so im assuming theres no variable to represent the integration constant in the answer

but im not sure what else theyre looking for

since theyre just asking for the antiderivative

which represents the formula of the accumulation

of consumption

[amber bolt]

it doesn't make sense that after one year you consumed 764e^(0.02) when the rate is 15.28e^(0.02) at it's highest

i don't get it but i'm excited

[winter crypt]

<@&286206848099549185>

[amber bolt]

ok so the problem is that we need to subtract the value at 0

so it's 764e^(0.02t) − 764

but i still don't get it

[chilly heath]

The mistake is that you didn't find the constant. Taking out the constants from the integral and then using a substitution for u = 0.02t, it is very easy to apply integration rules and use the integral of e^u, which is e^u. Then, applying more basic integration gives you 764(e^u-1). Replacing u with 0.02t gives you the final answer: 764(e^0.02t-1)... which is 764e^0.02t - 764.

[winter crypt]

that makes sense

thank you so much

.close

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• 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.

[zenith timber]

just had a quick question for this. would the answer be the same for every function or would it vary? if so, how would u calculate the answer?

[pallid halo]

what is that in front of the 4x^4, is it a minus sign?

hard to read

[zenith timber]

yeah it is

[pallid halo]

ok then this is correct

[zenith timber]

would it be the same for every function? i didn't answer it, my teacher did actually

[pallid halo]

you mean would every function have limit -infinity as x-> infinity or -infinity?

[zenith timber]

would the answer be the same if the function changes but the question is the same?

[pallid halo]

well consider f(x) = 0

how does that behave when x->infinity?

[zenith timber]

there would be no answer since it goes forever

[pallid halo]

no

f(x) has limit 0 as x->infinity or x-> -infinity

[zenith timber]

so if the limit is 0, u would write the answer as infinity?

[pallid halo]

you would write it as 0

because 0 is 0, not infinity

[zenith timber]

oh i see, but im still a lil confused on why my teacher wrote -infinity then as the answer

[pallid halo]

because for your problem, the answer is -infinity

[zenith timber]

what are the steps to figure that out ? since you can't subsitute infinity into the function

[pallid halo]

the question isn't even worded correctly, it should be "as x approaches infinity and -infinity", not "as p(x) approaches infinity and -infinity"

[pallid halo]

isn't this covered in whatever class you're taking?

[zenith timber]

it's an online class and he uploaded the document with the answers, unfortunately i missed the class and he doesn't record them

[pallid halo]

probably check khan academy then, that's a good place to learn/review stuff like this

[zenith timber]

alr ty

.close

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[calm coral]

Please don't occupy multiple help channels.

[pallid halo]

!status

[calm coral]

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

[remote mural]

i have begun but i got stuck midway

ln(x+3) = 2 - ln(2)

take natural log out

x+3 = e^2-2

[pallid halo]

ah no that step is wrong

[remote mural]

what did i do wrong?

[pallid halo]

e^(x-y) is not the same as e^x - e^y

[remote mural]

oh?

[pallid halo]

instead go back to ln(x+3) + ln(2)

[remote mural]

okay

[pallid halo]

this can be simplified into one log

[fringe fjord]

I'd first simplify ln(2)+ln(x+3)

[remote mural]

ln(2/3)?

ln(2/x+3)

[pallid halo]

ln(a) + ln(b) = ?

in general

[fringe fjord]

addition of logs!

[remote mural]

ln ab

[pallid halo]

yep

so use that here

[fringe fjord]

so ln2+ln(x+3)=? :]

[remote mural]

ln 2(x+3)?

i greatly appreciate the help!

[fringe fjord]

now you have ln(2x+6)=2, you can now simplify the logs

[remote mural]

take natural log

2x+6 = e^2?

then subtract 6

2x = e^2-6

then divide the whole thing by 2?

[fringe fjord]

🎉

[remote mural]

stumpman

are u a math major?

[fringe fjord]

not yet, want to be tho

still in high school

[remote mural]

ohh i see

oh damm fr?

this is uni pre calc

im horrible with logs

i have another problem, can i post it here?

[fringe fjord]

everything comes with practice, you'll get it eventually I got faith in your :]

[fringe fjord]

sure

[remote mural]

ill tell you what id do

and tell me if im wrong

subtract the log(x-2) from the right side

log(x+6) - log(x-2) = 2

log (x+6)/(x-2) = 2

(x+6)(x-2) = 9

so far so good?

[fringe fjord]

this is log_3 not log_10

otherwise looks good

[remote mural]

oh right

[fringe fjord]

also (x+6)/(x-2) not (x+6)(x-2), division sign is very important!

[remote mural]

but u subtracted log (x-2)

so using the rule that turns into division no?

oh

i see what u mean

(x+6)/(x-2) = 9

whats next?

idk

[fringe fjord]

multiply x-2 on both sides

now it just becomes a linear equation!

[remote mural]

got it!

can i add you my guy?

i got another problem!

[fringe fjord]

shoot!

[remote mural]

discord is lagging

x = 7 x =2

i got 2 solutions what do i do now?

[remote mural]

2 solutions

oh jusst plug it in and see which one gets negative

.close

[calm coral]

Channel closed

Closed by @thorn axle

Use .reopen if this was a mistake.

[fringe fjord]

yep

sorry was afk

log(x) is defined for x>0

so if a solution for x makes the inside of a log negative

it actually isn't a solution

[remote mural]

.open

.reopen

[calm coral]

✅

[remote mural]

idek where to begin

[remote mural]

new problem!

[fringe fjord]

sorry man I have to real son but here apply log to both sides and you get the answer very quickly :]

[remote mural]

thank you!

.close

[calm coral]

Channel closed

Closed by @thorn axle

Use .reopen if this was a mistake.

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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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[mild jetty]

For this problem, does chinese remainder theorem help to find the solution? Or does the theorem just guarantee a solution

One way I can see doing this problem is just using definition $x \equiv a \mod{n} \iff x = nk + a$

[potent lotus]

WhoTao

[mild jetty]

i see

This is what I’ve had so far. I don’t think it simplifies very nicely

wait what

[random swift]

This shit gives me flashbacks

[mild jetty]

it has been a while since i did number theory stuff. I didnt quite get how you get this 😅

[random swift]

Multiple by 2 then apply mod 5 in EQ 2

Then plug in k2

[mild jetty]

I see what you mean, like basically writing the first two equations like
$x = 3k_1 + 2$ and $x = 5k_2 + 3$. Then guessing?

[potent lotus]

WhoTao

[mild jetty]

theres gotta be a more mathematical way to formulate this stuff

[calm coral]

@mild jetty Has your question been resolved?

Channel closed

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[supple marsh]

The ome in the middle 813

Me neither

I think it's B too

I didn't understand the 'before' term

[idle marten]

B seems good

[supple marsh]

It says answer is A

[idle marten]

I think before is in front

Like you are standing before me

Cuz there are two ways that Rose can be positioned

I think

[supple marsh]

I don't have one 🥺

So I ll just skip it

And try this time to time

Thanks

.close

[calm coral]

Channel closed

Closed by @supple marsh

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[calm coral]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[remote mural]

Please help I have no idea where to start

[civic dirge]

Apply that identity to every term

[remote mural]

Ok so for 1/1*2 it’s 1/2??

[civic dirge]

And don't simplify

So you'll get:
(1/1 - 1/2) + (1/2 - 1/3) + (1/3...

[remote mural]

Yes

So I have to find the pattern here right?

[civic dirge]

Well, I see a -1/2, and a +1/2

Later, I see a -1/3 and a +1/3

[remote mural]

I’m confused I don’t know what the objective is

[civic dirge]

There's a nice simplification

[remote mural]

So 1/2-1*1/1-1/2 then we get 1/1-1/2

Same goes for the other ones right?

[civic dirge]

ye

[remote mural]

So we’re finding the bumbers that are in between 1/3* 4 and 1/9* 10?

[civic dirge]

Consider writing this out

[remote mural]

Yeah

Ohh

Wait uhhh

The next is 1/1*1/20??

Cuz 1/5-4*(1/4-1/5)

[civic dirge]

Don't simplify 1/4 - 1/5

The next is 1/4 - 1/5

[remote mural]

Yeah

[civic dirge]

I urge you to write the whole thing out on paper if you haven't yet.

[remote mural]

Ok

(1/1-1/2),(1/2-1-3),(1/3-1/4),(1/4-1/5),(1/5-1/6),(1/6-1/7),(1/7-1/8),(1/8-1/9)

[civic dirge]

what is , lol

[remote mural]

Um i also don’t know sorry

I was just writing the order

This is confusing

[civic dirge]

(1/1-1/2) + (1/2-1/3) + (1/3-1/4) + (1/4-1/5) + (1/5-1/6) + (1/6-1/7) + (1/7-1/8) + (1/8-1/9) + (1/9-1/10)

[remote mural]

Yeah

[civic dirge]

Any cancellations looking nice?

[remote mural]

Cancellation? I also have to cancel?

[civic dirge]

You get to cancel

[remote mural]

If I do that only 1/1 will remain right and 1/10

[civic dirge]

Lol okay there, you got it

Almost all of it goes away, you're left with
1 - 1/10

[remote mural]

Yeah

[civic dirge]

0.9, and we didn't even need a calculator

[remote mural]

Ohhhh

Wait how did 1-1/10 turn into 0.9? Isn’t it suppose to become 0/10? So 0??

[wheat marsh]

1 - (1/10)

[remote mural]

Ahh

So I convert (1/10) into a decimal first?

Thanks

.close

[calm coral]

Channel closed

Closed by @sick bolt

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[calm coral]

Available help channel!

Send your question here to claim the channel.

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

Is this correct so far?

I am not sure about the last part

This is my answer. Is it good?

[upbeat coral]

Well in the second line you should include cosine

You still calculated that bit correctly tho

Same thing for sine in first equality for x_1

Otherwise yes it's correct

[nimble field]

Yeah just forgot to write it

Thx

.close

[calm coral]

Channel closed

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[calm coral]

Available help channel!

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Remember:
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[latent zinc]

i normalised e and checked that the length is 1

not sure how to calculate the directional derivative of a scalar function at a point along the direction along span(v) ?

[calm coral]

@latent zinc Has your question been resolved?

[calm coral]

@latent zinc Has your question been resolved?

[latent zinc]

.close

[calm coral]

Channel closed

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[calm coral]

Available help channel!

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Remember:
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[trail flame]

In a GP, the t3 = root2 and t8 = 8. Find S8

[trail flame]

I missed out on a session today and I was given that for homework

Geometric Arithmitics

I know my AP's but I don't know anything about GP's

it seems like the channel keeps removing itself :/

.close

[calm coral]

Channel closed

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[twilit crag]

1. Parallelogram AXIS
2. Def. of parallelogram
3. Alternate Interior Angles Are Congruent
4. Line IA ≅ Line AI
5. Alternate Interior angles are Congruent
6. ASA congruence postulate

[twilit crag]

I wanna know if my no. 5 or 3 is correct lol

[calm coral]

@twilit crag Has your question been resolved?

[calm coral]

Channel closed

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[twilit crag]

.reopen

[calm coral]

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✅

@twilit crag Has your question been resolved?

[calm coral]

@twilit crag Has your question been resolved?

[calm coral]

Channel closed

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[sage cosmos]

What do we take time period for a bob of simple pendulum which is at rest?

[sage cosmos]

I'm confused between T= 0 or infinity

[pliant scarab]

t = 0?

it is at rest

so thats our initial measurement time

[sage cosmos]

Because if we go by the reasoning that a simple pendulum at rest completed infinite oscillation of amplitude= 0 then time period is 0

[pliant scarab]

firstly pendulum's time period has no relation w amplitude.

[sage cosmos]

Yes that's the same reason

[pliant scarab]

however long the amplitude is

it doesnt matter

[sage cosmos]

I see

[pliant scarab]

yeah

do you know

how the formula is derived

[sage cosmos]

2pi √l/g?

[pliant scarab]

do uk how it is derived

[sage cosmos]

No

I have not come to that part

[pliant scarab]

ok

but

idt it matters in our discussion here

i just want to assert that

the time when it is at rest

[sage cosmos]

What

[pliant scarab]

at the mean position

[sage cosmos]

Oh

Yes

[pliant scarab]

is considered as t = 0

in that case any object that is at rest which is dropped

should be considered as t = infinity

it is just a convention ig

[sage cosmos]

If it's at rest then no oscillation should happen then why is t=0

[sage cosmos]

Ohh

[pliant scarab]

t = infinity means

if u want more clarity

the oscillation never ends

[sage cosmos]

i see

[pliant scarab]

yeah

[sage cosmos]

In case of rest, oscillation never started though

So yeah it never ends

It's very confusing lol

[pliant scarab]

true

but nice thinking

[sage cosmos]

I guess I will go by convention

[sage cosmos]

Yeah

[pliant scarab]

u can ask ur teacher

worth asking

[pliant scarab]

😭 LMFAO

[sage cosmos]

If I could ask

But the teacher is online

[pliant scarab]

ohhh

[sage cosmos]

youtube

[pliant scarab]

who is it

[sage cosmos]

Name is rajwan

Indian teacher

[pliant scarab]

AREEE BHAI JEE

LMAO

[sage cosmos]

Ha

[pliant scarab]

HUM BHI JEE KE LIYE PADHTE HAIN

[sage cosmos]

Lol

Same

[pliant scarab]

i see

[sage cosmos]

Rajwan sir bole ki T= infinity

Usi k liye yaha Aya bhai

From oscillation chapter 11th

[pliant scarab]

ohhh

he said T = infinity

[sage cosmos]

Yes

[pliant scarab]

ok then he might be correct

let me think why though

[sage cosmos]

But you said T=0

Yes

[pliant scarab]

but

teacher is teacher

they are correct lol

[sage cosmos]

Yeah

lol

I came here for personal clarity

[pliant scarab]

wait

what reasoning did he give

[sage cosmos]

He said that g tend to infinity