help-33

[jovial lance]

i can see that you're right, but what i want to know is where I'm wrong in my explanation

[serene spoke]

P = VI

[jovial lance]

P=V^2/R

[serene spoke]

we will use this ^

[jovial lance]

but why???

[serene spoke]

Except we will also have to include R

[serene spoke]

Because Z = 2r

X = R

Got it ?

[jovial lance]

its fine ima get some more sleep than check again its 4 rn😩

thanks man

[serene spoke]

Wth bro

[serene spoke]

Welc.

[jovial lance]

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[marsh citrus]

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[marsh citrus]

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[sour bolt]

How do I solve sin 2x + 1 = x?

[sour bolt]

solve for x?

i changed the equation to sin 2x = x - 1

but I do not know what to do next

[calm harbor]

What’s the goal? Find how many intersects? Or find the coordinates of the intersect?

[sour bolt]

i think thats the answer

but im not sure how to show it algebraically

1.38

[crystal wraith]

You can't show it algebraically, with sin(2x) over there...

[sour bolt]

so would it be no solution

?

[marsh citrus]

@sour bolt Has your question been resolved?

[amber birch]

No there still is a solution

[sour bolt]

1.38?

[amber birch]

yes

If you sub in x = 0 or x = 2, both sides aren't equal

[sour bolt]

its like 1.37

when you sub 1.38 into sin 2x + 1

but yea ig it makes sense

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[terse basin]

can someone tell me why it should be log of 0.7(0.3) is larger than or equal to x?

[tender sparrow]

https://math.stackexchange.com/questions/2849125/why-does-taking-logs-of-base-0-b-1-for-exponential-inequalities-flip-the-i

this is better

mb

[terse basin]

alr

[tender sparrow]

the other explanation, (if this doesnt work for you)

basically take log base 10 on both sides

[terse basin]

yea

[tender sparrow]

so x log(0.7)>= log(0.3)

[terse basin]

oh so make everything log

[tender sparrow]

use this

buuuut

when u divide u gotta be careful

cuz log base 10 (0.7)

is a negative number

so when u divide by negative numbers of both sides of an inequality

the inequality flips

[terse basin]

what do i do after this

[tender sparrow]

divide

both sides

by log(0.7)

but flip the inequality sign

[tender sparrow]

this bit

[terse basin]

so basically any base that is, for example, 0<x<1, then i need to flip the inequality sign?

[tender sparrow]

thats the first method

from the stack exchange one

yeah

you flip it

[terse basin]

whats the reason behind if you times by a negative number or something, the inequality sign is flipped?

[tender sparrow]

take an example

say we had

-2<-1

which is true

[terse basin]

yep

[tender sparrow]

then we divide both with -1

if we didnt flip

it'd be

2<1

which isnt true

implying that we had to flip the sign

to make it true

[terse basin]

ohhh

alr i get it

[tender sparrow]

👍

[marsh citrus]

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[native tapir]

can i have help w question ci

[dusky viper]

yo

usually with show that questions, u just work on one side of the equation

so i'd start by expand sin(x+45) using sum of angles

[native tapir]

i did dat

[winter smelt]

everything looks alright, do you know what sin(45) and cos(45) are

[native tapir]

sin45 = sqrt(2)/2 and cos45 = 1/2

wait

sin45 =

root2/2

[dusky viper]

for this instance 1/root(2) looks better

[winter smelt]

what about cos(45)

[native tapir]

cos45 = 1/2

wait

bno its not

thats 30

45 is root 2/2

oh wait

[winter smelt]

that's not true, cos(30) = sqrt(3)/2

[native tapir]

oh................

ur right.................

[winter smelt]

sin(30) = 1/2

[native tapir]

oki

and den wat

[winter smelt]

but cos(45) = sqrt(2)/2 is right here

[native tapir]

yeee

[dusky viper]

distribute out root 2

[native tapir]

i dont get it i got this

[winter smelt]

why is it cosx * 1/2 in the second bracket in the second last line

we found that sin45 = 1/sqrt(2)

[native tapir]

ohhhhhh 45 not 30

okok

ok thz

[marsh citrus]

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[marsh citrus]

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[raw quest]

hi i have no clue how to do this using binomial theorem

i could do it w induction but uh

💀

could someone give a heads up?

wait

wrong question

[marsh citrus]

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[raw quest]

huh

.close

[marsh citrus]

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[raw quest]

hi i have no clue how to do this using binomial theorem
i could do it w induction but uh
💀
could someone give a heads up?

[hasty pumice]

u can use induction tbh more easy

[vivid chasm]

Wait this is from the hsc lol

[cunning jackal]

5=4+1

They instructed to use the binomial theorem so let's make a binomial 😄

[hasty pumice]

base:
n= 0 so we get 5^0 + 3 = 4 and 4 is divided by 4

now assume that 5^n + 3 divided by 4 and prove that 5^(n+1) + 3 is divided by 4

5^(n+1) + 3 = 5*(5^n) + 3
5*(5^n) + 3 = 4*(5^n) + 5^n + 3
4*(5^n) is divided by 4 and according to what we assume 5^n + 3 is also divided by 4 now if we add 2 numbers that each of em dividied by 4 then the sum of em also divided by 4

[raw quest]

wait so in this case

r we writing it in

sigma notation or ?

[raw quest]

wait waht do u exaclty mean by this

do i split it up so its like

[hasty pumice]

u can also do this:
5^n = (4 + 1)^n

[raw quest]

yeahhh

[hasty pumice]

then u can use binom

[raw quest]

and then...

uh

[hasty pumice]

induction is more ez bro

why u need binom

[raw quest]

no ik induction

[hasty pumice]

lol

[raw quest]

is easier

but

thats what the quesiton wants

💀

[hasty pumice]

oh

[raw quest]

yeahhh

[hasty pumice]

then use binom 🙂

[raw quest]

thats why

wait can u explain tho

again

like

[cunning jackal]

I mean what I have written

[raw quest]

😭😭

NO BUT

i dont get what to do next..

[cunning jackal]

Do you know the binomial theorem?

[hasty pumice]

u know what (x+y)^n = ?

[raw quest]

yes...

[cunning jackal]

Apply that to 4 and 1

[hasty pumice]

do x = 4 and y = 1

[raw quest]

do i do it in sigma notation ?

[hasty pumice]

(4+1)^n

[raw quest]

or

[cunning jackal]

Upto you

[hasty pumice]

u can with sigma

[raw quest]

oh ok

um

if possible could one of u write it out and ill try it myself ifrst

jsut so i can confirm if i did it right

[hasty pumice]

I already did

[hasty pumice]

..

[raw quest]

no i mean like

wait but thats not the full proof

is it

or

bruh im so confused

wait ok so

its

$(4+1)^n + 3$

[elfin berry]

✯✡ρєввℓєѕ✡✯

[raw quest]

and then if u expand it out its

$nC0 * 4^n + nC1 * 4^(n-1) * nC2 * 4^(n-2) ... + nCn + 3$

[elfin berry]

✯✡ρєввℓєѕ✡✯

[raw quest]

uh

thats not writing out

nC0 * 4^n + nC1 * 4^(n-1) * nC2 * 4^(n-2) ... + nCn + 3

anyways its this

so where do i go

after this

[cunning jackal]

good what is nCn

[raw quest]

oh

1

OH

[cunning jackal]

yes

[raw quest]

OHHHH

and since everything else is times by 4 to the power of smth

it has to be divisble

by 4

[cunning jackal]

yes

[raw quest]

okok

.close

[marsh citrus]

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[hasty pumice]

dont forget that ur adding 3 in the end

@raw quest

[cunning jackal]

he didnt, he got that nCn+3=1+3=4

[hasty pumice]

aight

[marsh citrus]

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[silent prawn]

would you use a Angle sum and difference identity?

<@&286206848099549185>

[shadow ibex]

Yes

...probably

......maybe

[silent prawn]

thanks

very helpful

https://tenor.com/view/write-this-down-taking-notes-take-notes-gif-17975847

[shadow ibex]

It's 5 in the morning, I'm in the dark, and not being paid. Why don't your try your idea

[silent prawn]

im quite literally in the same position

why do you think im asking for help

you didn't give me a straight answer

[shadow ibex]

Use double and triple angle formulas (it will work out that way if you use general cosine angle addition formulas). Use the format that only involves cosine. You will get a polynomial

(Disregard second line)

[silent prawn]

yes, i got the same conclusion

thank you very much melvin

have a great morning in a few hours

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[narrow estuary]

riley

bro

[still temple]

v

stop bothering people

in their channels

[narrow estuary]

😭yep im cooked bro

[still temple]

what did you try

so far

try making a free body diagram

and find the best point

about which you can write torques due to the forces

and balance them since its equilibrium

[faint knoll]

uh i tried taking moments about the intersection of the rod and the disc

i let that point be X and i got 3/4a as that length

[still temple]

no not that good you will have to find hinge force for that which will be a difficult task

do it about the point of suspension of rod

so you wont have to deal with torque due to hinge force

[faint knoll]

does A work?

im confused what to do with the friction

[still temple]

yes thats what i meant by point of suspension

[still temple]

let friction be = f

and about A you can write its torque as f(a)

[faint knoll]

im so confused omg i dont do mechanics i have a major exam in three days

so i consider the weight, the circle intersection and the particle?

[still temple]

how do you not do mechanics and have a major exam

[still temple]

what circle intersection

consider weights and frictions

only

[faint knoll]

uhh a bunch of shit happened in my life and im learning everything in a week instead ...

[still temple]

OH MY

i am sorry

i read the question incorrectly

[faint knoll]

i do a buch of pure math and algebra stuff but im cramming a bunch of mechanics and stats right now im so screwed

[still temple]

you can't do it about A because then you won't be able to deal with friction i thought friction was acting between the disk and the ground

[still temple]

thats cool

ill do it too next year as i graduate hs

[faint knoll]

yea im trying to be done w hs rn so if i get this exam done im done. no clue how im making that happen tho

ok i found the mark scheme i just have no clue what its saying

[still temple]

huh

i don't know this shit

look you have some forces and

and theyre all in equilibrium

so apply 1) newtons laws

and write moment equation

thats it

there shouldn't be so many of them

just 2-3 equations should do all the work

[faint knoll]

im just confused how much friction there is

[still temple]

also just a reminder you cant cram mechanics

[still temple]

the question clearly says

its limiting friction

[faint knoll]

yes but then how much is limiting friction

i have the coefficient but how much force is acting normal to the surface

[still temple]

and maximum static friction is (coefficient of static friction)(N)

[faint knoll]

why not lmao

[still temple]

idk

maybe its just me

lol

i dont think i can cram it

you can cram math

[faint knoll]

yea idk math has always come super easy to me like i learnt calculus and trig and stuff in a couple days and i took the ap calc exam without studying but this doesnt feel like math i hate it so much

[still temple]

use newtons law for this

[faint knoll]

second law?

[still temple]

yes

ob

vi

[still temple]

same lol

i learned it in few days

aswell

it seemed brainless work

but it was never same with physics

[faint knoll]

idk ive gotten through like a half of the question types in the past three hours. might be possible. but i can barely lose any marks if i want an a* so im stressed out

anyway how tf do you do this im still so confused

[still temple]

balance the forces acting on rod

[faint knoll]

yes but about what point

and where does the frictional force come into play

and the ball too is it an upwards force?

[still temple]

any point its equilibrium

forces about all points must balance out

literally

[faint knoll]

i still dont get what forces are acting at each point

[still temple]

weight of rod

will act

on the centre of mass of rod

so at 3a/2 from A

[faint knoll]

yes that parts easy

and the weight of the particle im fine with too

im just trying to figure out the disc and the friction between the rod and the wall

[still temple]

hmm join the line O and point of intersection of rod

and circle

to find distance between A and point of intersection

[faint knoll]

3/4a

[still temple]

now balance forces about point of intersection

[faint knoll]

thats what i tried in the first place

but theres the friction against the wall i have no clue

[still temple]

assume friction to be some f acting upwards at A

write moment due to friction aswell in your equation

and moment due to rods weight

discs weight

particles weight

wtf why is discsweight not given

fuck this

[faint knoll]

lmao idk who to ask anymore i homeschool so i dont even have a teacher or tutor im so fucked

[marsh citrus]

@faint knoll Has your question been resolved?

[marsh citrus]

@faint knoll Has your question been resolved?

[marsh citrus]

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[stoic needle]

Yo

[stoic needle]

Anyone know how to solve this?

[merry pewter]

what class is this

[stoic needle]

Algebra 2

Idk how to solve it

[amber birch]

Right so it's a cubic function

It has the form y = a(x - h)^3 + k, where (h, k) is the point of inflection

the middle of the flat bit

[stoic needle]

Ohh

[oblique elm]

Cant we write it like this X-4=y?

[stoic needle]

I can try hold up

Nah that just makes a linear equation

[stoic needle]

What would x be?

[amber birch]

x and y are variables

Like it wouldn't make sense to ask what x is in say, y = 2x - 3

[stoic needle]

Holy shit I think that was the right answer

Ty

[amber birch]

No worries!

yeah you need another point on the graph to find a, where you can sub in

[stoic needle]

Sounds good thanks

[marsh citrus]

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[stoic needle]

Yeah

[marsh citrus]

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[brazen flame]

I need to find a and b

but I don't really know what to do from here

[rich bough]

Elimination maybe

The last 2 equations are literally the same

Just factored

I feel like squaring the last one would help

[valid cape]

seems like a simul equation with 2 variables

just looks a bit scary, but isnt that hard

solve for one variable in terms of another, sub that in another equation and solve for it

[tender lily]

I would jus isolate for a in the third equation and then substitute it into the first equation and then solve

[valid cape]

yeah thats what i said

[tender lily]

Yeah

[brazen flame]

wait

maybe there is a firework I can do or something

in the exercise, a and b are mentioned to be rational numbers

a sum of rational numbers can't be equal to an irrational number

I hope

so

[serene spoke]

if a+√b = c +√d
then a = c and b = d, there was a theorem like this about surds

Subtract equation 3 from eqn 1

I mean eqn 1 - eqn 3, then a will be eliminated

[serene spoke]

No sry

All the equations are same !!!

[vast shoal]

It's like a single equation but simplified

[serene spoke]

b(1+2√2) + a(3-√2) = -√2
Multiplying with 3+√2
b(1+2√2)(3+√2) + 7a = -√2(3+√2)
Or b(7 + 7√2) + 7a = -3√2 - 2
Or 7b + 7a + 7√2 b = -3√2 - 2
or [7(a+b) = -2
And[7√2b = -3√2

@brazen flame ?

[brazen flame]

🖐️

[serene spoke]

is this correct bro ?

b = -3/7 and a = 1/7

[brazen flame]

yea

that's correct

[serene spoke]

You got it ?

[brazen flame]

how did you think to multiply with the conjugate of (3-sqrt(2))?

[serene spoke]

To simplify the equation

[serene spoke]

I mean its intuition

K bye have a good day

[brazen flame]

have a nice day

!

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[marsh citrus]

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[marsh citrus]

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[shadow osprey]

hello I have a question

[broken dome]

hello I have an answer

(maybe)

[shadow osprey]

I've got the x integrand and I can't find the y integrand

i've tried the shell method but idk im stumped

[broken dome]

Try drawing a sketch of the situation

it might help you

[shadow osprey]

i did in desmos

I tried formatting it as this

2pi(-5)(y^1/8)

but says its wrong

im going insane

[broken dome]

how did you figure out a=0, and b=1

[shadow osprey]

I checked on desmos, its the intersection between y = 1 and y = x^8

and for this integral they want the dy integrand

[broken dome]

Oh nvm I missread the question

Well we could rewrite y=x^8 in terms of x first

[shadow osprey]

x = y^1/8

[broken dome]

and then the integral of this from 0 to 1 dy

gives us the surface of the area

well half of it

[shadow osprey]

i don't need to find the area

[broken dome]

since +- shenanigans

[shadow osprey]

i need to find the integrand expression

[broken dome]

and if you know the area

[shadow osprey]

i already have the area

• volume

[broken dome]

and you know the axis of rotation

you could derive the volume from that

[silk sable]

Integrating for volume parallel to the axis of rotation requires the Shell method.

[shadow osprey]

yeah i tried something like this

cuz -5 is the radius and y^1/8 is the height

[broken dome]

feels like you're missing a square somewhere

[shadow osprey]

shell method doesn't need a square afaik

[silk sable]

Your radius is incorrect.

[shadow osprey]

is my height correct?

[silk sable]

2y^(1/8), so yes.

Your radius should be 6-y.

[shadow osprey]

got it

thank u

have a swell day

[marsh citrus]

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[marsh citrus]

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[heavy raven]

im confused how this works

[heavy raven]

prove both equations btw

ok number 36 i can

i am completely loss on number 36 sorry

i can kind of do 37 but im stuck

[stark sandal]

Nice to see you lmao

[heavy raven]

oh hello agian

lol

same trig stuff theres a test tmrw

do you know how this works?

[stark sandal]

for the first you can again prolly just use double angle for sin

the second I dont think I like

this is a proof?

[heavy raven]

yes

[stark sandal]

well we can begin with the first

use double angle for sin

$\sin(2x)=2\sin(x)\cos(x)$

[elfin berry]

Percy

[stark sandal]

Last time we had cos(2x) lmao

[heavy raven]

oh ok

actually it ehlps knowing the name of the formula so i can search i jsut realize theres a name to it

ok

$\frac{3}{4}(2\sin(a)\cos(a))^2$

did i mess something up?

[stark sandal]

we have double angles, half angles, compound angles, sum to product, product to sum identities, usually

that's about it

[stark sandal]

the space between end and $

also use () for the argument of the trig fxns

[heavy raven]

ok this is it right

what do i do about that ^2

[stark sandal]

just square it lmao

[heavy raven]

is that gonna goa t the back?

oh ok

[elfin berry]

veryhuman

[stark sandal]

mmm

[heavy raven]

so that should just be

$3\sin^2(a)\cos^2(a)$

[stark sandal]

no?

how did you get two 3's lmao

the 3 isnt squared

also it's \sin

[heavy raven]

i thought it

oh wait yeah i should treat this as AB

sorry thought about it like A + B

[elfin berry]

veryhuman

[heavy raven]

i removed the 4s

ok so the front part

[stark sandal]

mmmm

[heavy raven]

what do i do about that $\sin^6(a) + cos^6(a)$

[stark sandal]

it's sin^6

[elfin berry]

veryhuman

[heavy raven]

typo yeah

[stark sandal]

write the whole thing down

see if it seems familiar

[heavy raven]

$\sin^6(a) + \cos^6(a)+3\sin^2(a)\cos^2(a)$

this is what we got sofar

it looks like

that one rule

[stark sandal]

\cos

does it seem familiar?

[heavy raven]

this one

[elfin berry]

veryhuman

[stark sandal]

yeah no

[heavy raven]

no?

[stark sandal]

this shouldd seem like an identity

it should feel familiar

hold on ill assist

$\sin^6(a) + \cos^6(a)+3\sin^2(a)\cos^2(a)\left(\cos^2(a)+\sin^2(a)\right)$

now?

[heavy raven]

OH WAIT

YEAH

wait i dont get how this fits

[stark sandal]

btw we use \left( stuff \right) for big brackets

[stark sandal]

what are you thinking of

[heavy raven]

im thinking of making it $(\sin^2(a)+\cos^2(a))^3$

[stark sandal]

close it

[elfin berry]

veryhuman

[stark sandal]

mmm

why does that not work?

[stark sandal]

I just realised I used x lol.

[elfin berry]

Percy

[heavy raven]

i thought that was on purpose

[stark sandal]

i mean it's 1 anyway but lol

[heavy raven]

bc thatsjust

1^3 yeah

$1+3\sin^2(a)\cos^2(a)$

[elfin berry]

veryhuman

[stark sandal]

hm?

what

we already added the 3 sin^2 cos^2

[heavy raven]

yeah this is just like my thought process ik something is wrong bc i did $\left(\sin^2(a) + \cos^2(a)\right)^3+3\sin^2(a)\cos^2(a)$

[elfin berry]

veryhuman

[heavy raven]

ok ok

i think

theres is suppose to be another instance of $3\sin^2(a)\cos^2(a)$

[elfin berry]

veryhuman

[heavy raven]

that is negative

[stark sandal]

This is $\left(\sin^2(a)+\cos^2(a)\right)^3$, no?

[heavy raven]

but from where im not sure

[elfin berry]

Percy

[heavy raven]

i assume the first 2 are?

is the $\left(\cos^2(a)+\sin^2(a)\right)$ a part of the equation?

[elfin berry]

veryhuman

[heavy raven]

bc if it is i dont get where it coems from

[stark sandal]

NO??

jeez.

$(a+b)^3=a^3+b^3+3ab(a+b)$

[stark sandal]

That's just one.

I added it to make it apparent that it's the same as the above identity

[heavy raven]

oh this is its own formula my b

[stark sandal]

which I butchere.

[elfin berry]

Percy

[stark sandal]

how're you doing trig and not konw (a+b)^3 xD /lh /nbr

[heavy raven]

i was forced to do math my math is very bad 😭 !!!!!!

[stark sandal]

its not that hard lmao

just some practicee

[heavy raven]

i know thats how math works from what ive seen

i have realized like from a non math expert that math seems pretty similar to learning a language

[stark sandal]

yesss it does

[heavy raven]

i see people who great at math just finish questions they never see

just like how people can figure out new vocab from the context its used in

ok so continueing on im still sorry kinda confused

if i

$\left(\sin^3(a)+\cos^3(a)+3\sin(a)\cos(a)\right)^2$

[elfin berry]

veryhuman

[heavy raven]

with your formula

should look like

[stark sandal]

no

[heavy raven]

$\left((\sin+\cos)^3\right)^2$

[stark sandal]

not squared

[elfin berry]

veryhuman

[stark sandal]

it's THIS expession

[heavy raven]

oh just?

[heavy raven]

OHHHHHHHHH OH OH OH OH OK

THEN

OHH

OH IT ALL CONNECTS

OW I GET IT

sorry i switched the placement

yes

[stark sandal]

okay lmao

[heavy raven]

this is supposed to be
$\left(\sin^2(a)+\cos^2(a)+3\sin(a)\cos(a)\right)^3$

[elfin berry]

veryhuman

[stark sandal]

no its not

[heavy raven]

its not?

[stark sandal]

you can just distribute powers

$(a+b+c)^2$ is ACTUALLY $a^2+b^2+c^2+2ab+2bc+2ac$ iirc.

[elfin berry]

Percy

[heavy raven]

oh yeah it is

[stark sandal]

compare this

[heavy raven]

so when i take the ^3 out

[stark sandal]

with this

tha's what you should be doing

[heavy raven]

$3\sin^2(a)\cos^2(a)$ dissapears?

[elfin berry]

veryhuman

[stark sandal]

just compare them jeez

[heavy raven]

ok sorry

OH WE ASSUME

A IS COS^2

NOT JUST COS

OK

sin too

so when SO

ok

yeah i 100% get it actually now i think

ty

i think we can move to the enxt one which is this

so ive reached tot he point where its

$(\sin^2(x)-2)-(\cos^2(x)-2)$

[elfin berry]

veryhuman

[stark sandal]

huh

[heavy raven]

but this doesnt end up with cos 2x

[stark sandal]

so that's about it then lmao

[heavy raven]

wait what

[stark sandal]

yes it does

$=\sin^2(x)-2-\cos^2(x)+2$

[heavy raven]

$sin^2-cos^2$

[elfin berry]

veryhuman

[heavy raven]

is what i got

[elfin berry]

Percy

[stark sandal]

I think you fucked the sign up

it should be $cos^2(x)-sin^2(x)$

[elfin berry]

Percy

[stark sandal]

how did you get here

[heavy raven]

well this i multiply $(\cos^2(x)-2)$ with -1 since theres a negative infront of it

[elfin berry]

veryhuman

[heavy raven]

making the cos -

how did the sin become negative?

[stark sandal]

uh

[stark sandal]

how did you get to this from the problem

show your work

[heavy raven]

ok so it begins with me turning the cos in the left to sin and the sin in the right to cos

[stark sandal]

okay

how

[heavy raven]

$\sqrt{\sin^4(x)+4(1-\sin^2)}- \sqrt{\cos^4(x)+4(1-\cos^2)}$

i keep forgetting to clsoe it my b

oh

[elfin berry]

veryhuman

[heavy raven]

there we go

right

[stark sandal]

alrighttttttttt

[heavy raven]

after that i turn it to $\sqrt{(\sin^2-2)^2} -\sqrt{(\cos^2-2)^2}$

wait wait

[stark sandal]

huh

[heavy raven]

right? then i remove the

[stark sandal]

how?

[heavy raven]

oh

i sqrt everything inside the brackets to get a ^2 out

but wait how did i lose the 2sin

OH WAIT MY B

I FORGOT THAT IT GOES $A^2 + B^2 + AB$

[elfin berry]

veryhuman

[stark sandal]

im really lost rn

[heavy raven]

i basically factorize is taht the word?

sicne its just

$x^4-x^2+4$

[elfin berry]

veryhuman

[heavy raven]

sorry forgot to change the -

[elfin berry]

veryhuman

[heavy raven]

yeah if you count it together it should come out like that

yeah still the sin isnt negative

but you turne dit negative somehow?

[stark sandal]

right

sigh lemme just write this

im so tired

[heavy raven]

ty

[stark sandal]

god i get 0 somehow

oh i forgot 4

[marsh citrus]

@heavy raven Has your question been resolved?

[marsh citrus]

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[shadow osprey]

imma need some more help

[broken dome]

Specify what you need help with

[broken dome]

State the nature of your mathematical emergency

[shadow osprey]

for part d, why isn't my answer correct? im doing the larger radius minus the smaller radius but it keeps showing as wrong

this is the associated graph

someone please help i feel so stupid rn

[shadow osprey]

which part of my integrand in d is wrong?

[marsh citrus]

@shadow osprey Has your question been resolved?

[shadow osprey]

bro where is everyone

<@&286206848099549185>

[pastel ice]

k dot i gotchy

lemme help u out

is this alg 1?

[shadow osprey]

thank u

this is calc

[shadow osprey]

i got part d wrong and idk why

[pastel ice]

im in bc too im a freshman

hs

[shadow osprey]

i tried a diff answer as well

[pastel ice]

uhhh

it seems right

idk

[shadow osprey]

im tweaking rn

[pastel ice]

yo ima ping more help

i prolly woulda put that answer too

<@&286206848099549185>

[shadow osprey]

can u help me with this one other question

ive been stuck on these two questions for a while

[marsh citrus]

@shadow osprey Has your question been resolved?

[marsh citrus]

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[fathom mountain]

What method shall be used?

[fathom mountain]

Im kinda lost at the starting point, ive tried hopital but didnt give me a better psition

[balmy rain]

hint: taylor

[fathom mountain]

Why use taylor?

[balmy rain]

just do it

you'll see why

[balmy rain]

(TM)

[fathom mountain]

Ok it worked thx but why

I mean why know use taylor

[balmy rain]

because it works? XD

i cant think of a better explanation atm

[fathom mountain]

But how did u know

I must know how to tell or i wont be able to solve these

[balmy rain]

sinusoidal functions don't have vanishing derivatives at any order so you're kinda forced

[fathom mountain]

For hopital u mean?

[balmy rain]

yep

[fathom mountain]

So if theres a sinx or cosx unless its a pre set one like sinx/x u must taylor?

[balmy rain]

i mean there's no hard rule but usually yes

[fathom mountain]

Thx

.close

[marsh citrus]

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[marsh citrus]

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[wooden tartan]

Hello! This is a previous brazililian exam translated to English. I am not understanding how am I supposed to solve this

[marsh citrus]

Please don't occupy multiple help channels.

[trim quest]

Hint: Draw OC

and see if you can find the measure of angle BOC

[wooden tartan]

Ok, I was seeing a video ans I discovered that the central angle is always the double of the inscribed angle, so I made like this

I didn't understand very well why it is like it's and from now I don't know

The guy in the video is saying this

[trim quest]

Do you understand how you got here?

[wooden tartan]

Yes, the central angle is the double of the inscribed angle, so 72 * 2 = 144, the interior angles of the triangle are always 180º

So 180 - 144 = 36 / 2 = 18

[trim quest]

Yes

Then use triangle BDE

You already know two angles in BDE are 18 degrees and 90 degrees, so you can find the third

[wooden tartan]

woww

I understood now

[trim quest]

Nice 👍

[wooden tartan]

Awesome, aprecciate man, have a great day!

[trim quest]

You too

[wooden tartan]

.close

[marsh citrus]

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[marsh citrus]

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[tall sun]

is this correct?

[desert dirge]

,calc 25000*0.75^3

[elfin berry]

Result:

10546.875

[desert dirge]

yuh

[tall sun]

noice

thankyew

.close

[marsh citrus]

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[marsh citrus]

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[hushed minnow]

how do I know whether a rational function solution is extraneous

[crystal lintel]

[remote grove]

Flip a coin

[desert dirge]

extraneous to what

[crystal lintel]

what is a “rational function solution”

[lethal bridge]

slaaaaaaaayyyyyyyyylaaaaaa

[crystal lintel]

water beam

[lethal bridge]

FTMB when

[crystal lintel]

[lethal bridge]

YESSSSS

it’s backkk

[marsh citrus]

@hushed minnow Has your question been resolved?

[hushed minnow]

ration equation

<@&286206848099549185>

explain how to know a solution is extra aneous when solving rational equations and what would this mean graphically

[sudden egret]

If the denominator evaluates as 0

so it would be an asymptote on a graph

[hushed minnow]

does it only have to be one denominator equaling 0?

[hushed minnow]

oh ok

[sudden egret]

yes

anything/0 is undefined, so yea

hope that helped 🙂

[hushed minnow]

very much so

wish me look on final

[sudden egret]

gl man 🙂

[marsh citrus]

@hushed minnow Has your question been resolved?

[marsh citrus]

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[compact sentinel]

stupid question (maybe), but how can i turn an equation like into a data table

[compact sentinel]

i have an equation:

$\sin(\frac{2\pi}{10}(x))$

[elfin berry]

j

[compact sentinel]

and i want a table of values between like x=1800 to x=1950 on excel or something

How would i go about doing this

[balmy rain]

you can just give the row/column of input values to the function

[quaint elm]

make your two columns, give them headers

put 1800 in the first column and then drag down to autofill

in the second column write =SIN( 2*PI() / 10 * A2) or whatever

and then drag that down to autofill

[marsh citrus]

@compact sentinel Has your question been resolved?

[compact sentinel]

Okay

[compact sentinel]

👍

Thank you very much both of you

[marsh citrus]

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[crude wren]

<@&286206848099549185>

[marsh citrus]

@crude wren Has your question been resolved?

[crude wren]

<@&286206848099549185>

[marsh citrus]

@crude wren Has your question been resolved?

[marsh citrus]

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[vague moss]

7 and 8. This has to do with permutations and stuff. I really am stuck with what to do because everything is so vague to me

[marsh citrus]

@vague moss Has your question been resolved?

[vague moss]

Solved this nvm

.close

[marsh citrus]

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[sly swallow]

can someone explain this problem, I don't quite get it. I currently have -1<=cos(1/x^4)<=1 because y value for cosx goes up to 1 and -1 (I think that's why) and I'm kinda stuck

[fervent rampart]

that's a good start, now try multiplying each side by sin(x)

[sly swallow]

okay that'd be -sin(x) <= cos(1/x^4)*sin(x) <= sin(x)

do i plug in after or?

[latent coral]

yeah, that expression in the middle is exactly what you want to find the limit of

what do you get when you take the limit of the left and right?

[sly swallow]

0

[latent coral]

so if the limit is both >= 0 and <= 0, what can you conclude?

[sly swallow]

oh

that its 0

[latent coral]

yes

[sly swallow]

right thats how limits work

gotchu

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[vestal rock]

Consider the rabbit pairs that illustrate the pattern in the Fibonacci sequence. These rabbits produce exactly 1 pair of new rabbits after reaching maturity at age 2 months. Imagine that the rabbits and all their offspring live forever. Also, imagine the field the rabbits live in can expand in size so that its side length is exactly equal to the number of pairs of rabbits living in the field. What is the side length of the field at the end of two years? Explain and show your work.

[vestal rock]

the side length of the field after 2 years = 75,025
1 month

Adult = 1 pair

Baby = 0 pair

after one month, adult pair will have a pair of baby rabbits

2 months

Adult = 1 pair

Baby = 1 pair

another month, those baby rabbits would become adults and the other pair of older rabbits would produce a new pair of baby rabbits

3 months

Adult = 2 pairs

Baby = 1 pair

again the pairs of baby rabbits would now become adults and the other 2 pairs of older rabbits would have 2 pairs of baby rabbits

4 months

Adult = 3 pairs

Baby = 2 pairs

Then, again, after a month, those 2 pair of baby rabbits would become adults and the other 3 pairs of older rabbits would again, have 3 pairs of baby rabbits.

@5 months

Adult = 5 pairs

Baby = 3 pairs

This would continue for 24 months

6 months

Adult = 8 pairs

Baby = 5 pairs

7 months

Adult = 13 pairs

Baby = 8 pairs

8 months

Adult = 21 pairs

Baby = 13 pairs

9 months

Adult = 34 pairs

Baby = 21 pairs

10 months

Adult = 55 pairs

Baby = 34 pairs

11 months

Adult = 89 pairs

Baby = 55 pairs

12 months

Adult = 144 pairs

Baby = 89 pairs

13 months

Adult = 233 pairs

Baby = 144 pairs

14 months

Adult = 377 pairs

Baby = 233 pairs

15 months

Adult = 610 pairs

Baby = 377 pairs

16 months

Adult = 987 pairs

Baby = 610 pairs

17 months

Adult = 1597 pairs

Baby = 987 pairs

18 months

Adult = 2584 pairs

Baby = 1597 pairs

19 months

Adult = 4181 pairs

Baby = 2584 pairs

20 months

Adult = 6765 pairs

Baby = 4181 pairs

21 months

Adult = 10946 pairs

Baby = 6765 pairs

22 months

Adult = 17711 pairs

Baby = 10946 pairs

23 months

Adult = 28657 pairs

Baby = 17711 pairs

24 months

Adult = 46368 pairs

Baby = 28657 pairs

​

46368 adult pairs + 28657 baby pairs

= 75,025 pairs

this was my answer, how did i do?

sorry its so long

[marsh citrus]

@vestal rock Has your question been resolved?

[marsh citrus]

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[mint cape]

I need help

[mint cape]

I put right triangle but I am unsure

[sand kindle]

what makes you think it's a right triangle?

[mint cape]

uh i think 25 squred is equal to 7 squred plus 24 squared

[sand kindle]

yep

[mint cape]

wait odnt leave me pls

i have a couple more 💀

if thats fine with u

i think its obtuse

[sand kindle]

• If a^2 + b^2 < c^2, the triangle is obtuse
• If a^2 + b^2 = c^2, the triangle is right
• If a^2 + b^2 > c^2, the triangle is acute

[lucid zenith]

this comes from the cosine rule btw

[mint cape]

yo wait

i genuinly need help on tihs one

its not just clarification

im kinda lost rn

[sand kindle]

,w (2 sqrt(3), 4 sqrt(2), 3 sqrt(5))

[elfin berry]

Results provided by WolframAlpha

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

Requested by invariance.

[sand kindle]

those are in ascending order

so check (2 sqrt(3))^2 + (4 sqrt(2))^2 against (3 sqrt(5))^2

[mint cape]

uhhh

i think its obtuse again?

[sand kindle]

an easy way to check this by hand is to bring everything inside the radical

e.g. (2 \sqrt{3} = \sqrt{2^2 \cdot 3} = \sqrt{12})

[elfin berry]

Invariance

[sand kindle]

or check the order after you square them

[sand kindle]

what are the squares?

[mint cape]

uh

wdym

[sand kindle]

what are a^2, b^2, and c^2?

[mint cape]

uh i just added a squared and b squared in my calculator and i got 44

and c squred, i got 45

so since c is greater than a+b than obtuse?

well with everything squared

[mint cape]

uh am i wrong?

[sand kindle]

if a^2 + b^2 < c^2 then the triangle is obtuse

[mint cape]

uh

but maybe i put it in my calculator wrong?

[sand kindle]

you can double-check by doing it by hand

[mint cape]

uh i kinda have to finish this super quick

[sand kindle]

...do it by hand quickly?

or put it into the calculator again?

[mint cape]

uh

u think u could doulbe check me?

or

nuh uh

[sand kindle]

we're not here to do your exam for you

[mint cape]

its not an exam

[sand kindle]

assignment then

[mint cape]

its my homework 😭

dog diggity dang it

but it does say

if i make a mistake

u shoud politely point it out

[sand kindle]

it sounds like you've got it

[mint cape]

here i have another one and i will show u waht i did and maybe u can help me out according to the rules

[mint cape]

alr tysm

so i split the triangle in half to create two 30-60-90 triangles

to find the altitude, i used the shortleg, which is now 3, since i split it in half, and multiplied it my sqrt3

to get 3sqrt3 for the altitude

[sand kindle]

why multiply it by sqrt 3?

[mint cape]

uh thats like the rule for 30-60-90 triangles i think

cuz its a special right triangle

like this

[sand kindle]

oh that

yeah, that works

if you wanted to double-check you can use the Pythagorean theorem directly (3^2 + x^2 = 6^2)

[mint cape]

oh yea

ye thx i think its good

i used the same thing for this

i used these too

[mint cape]

And I did this

And added them up to get about 56.9

[sand kindle]

I gtg now, but you seem to be doing fine on all these

the ideas are right at least

[mint cape]

oh alr

thx for helpin g

[marsh citrus]

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[cosmic gust]

he finds it's dependence or independence and he predestined C_3 = -1 why he did that? didnt understand well

[marsh citrus]

@cosmic gust Has your question been resolved?

[cosmic gust]

<@&286206848099549185>

[marsh citrus]

@cosmic gust Has your question been resolved?

[marsh citrus]

@cosmic gust Has your question been resolved?

[still temple]

as a general rule, if there are more vectors than the dimension of the vectors, the set is linearly dependent. as to why he chose -1, i suppose it's because this system of equations has infinitely many solutions for arbitrary values of c_3 @cosmic gust

[cosmic gust]

thank you so much!

.close

[marsh citrus]

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Use .reopen if this was a mistake.

[marsh citrus]

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• 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.

[karmic shore]

Im done everything i can to find the answer but i cant understand it

[karmic shore]

One direction vector i found was (5,-2,3) and found a point (3,-2, 1) but i dont understand how to get the 2nd d vector

[marsh citrus]

@karmic shore Has your question been resolved?

[marsh citrus]

@karmic shore Has your question been resolved?