help-41

[upbeat gyro]

Just to share my thought process, I thought by making a secant as I said, and make the second point as close as the first one, we could find the slope of the tangent.

you mean ax+b?

or a = delta y / delta x

[naive pivot]

Yea

you can estimate the the slope using two points x and y

[upbeat gyro]

Yeah indeed, we're now learning about f'(x)

[naive pivot]

Yep so estimated change in y/estimated change x
use 2 points from the graph

[upbeat gyro]

Hm alright, I now found the correct answer to be 1. So with a tangent that is linear we can basically just easily find it by using a = delta y / delta x

[naive pivot]

yep but some time points won’t be obvious so we use derivative for its case

[upbeat gyro]

I assume the derivative rules in those cases?

Since we've only really learned about f'(x) when we're given the function

[naive pivot]

If like given a function yea

[upbeat gyro]

Ah right, so that means that we could even find the function to the tangent?

y = x+b, put in the coords and we would be able to find out what b is

[naive pivot]

Yep

[upbeat gyro]

Ah awesome, that makes sense. I guess I was more so overcomplicating it here

Thanks 🙂

[naive pivot]

It’s ok

[upbeat gyro]

.close

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[split sail]

find (871, 299)

[split sail]

note: this is the complete question

[frank zealot]

do u mean gcd

[split sail]

yes

[frank zealot]

hm

[primal holly]

prime factorization didn't work?

[frank zealot]

euclidean algorithm seems good

[split sail]

well yesn't

[split sail]

I can't find prime factors

[frank zealot]

how can u not

[split sail]

they are big

[primal holly]

,w prime factorize 871

[grizzled pagoda]

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[split sail]

lol

[frank zealot]

oh

[primal holly]

13

not too big

[frank zealot]

Then use euclidean algorithm

[primal holly]

,w prime factorize 299

[grizzled pagoda]

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[primal holly]

also 13

also not too big

:/

[frank zealot]

13 is big tho

[split sail]

yes

so basically

[frank zealot]

pretty sure no ones first instincts is to go "13 it is"

[primal holly]

It's like the first number you check after 7

[primal holly]

I would chek 7 and then 13

[frank zealot]

11 should also be checked

[split sail]

yeah

[primal holly]

11 has an easy divisibility rule

[frank zealot]

im pretty sure 11 is the hardest in my country for divisibility

not too sure what u do in ur country

[split sail]

just do the division

[warm burrow]

just use synthetic division on the floating point polynomial

[primal holly]

You take the difference of the sum of alternating digits

[split sail]

it's not long

[primal holly]

And you can do it in all parts of the universe

[frank zealot]

i guess that could do it

[split sail]

we'll do 299 * 2 + 273 = 871
(299, 273) = (871, 299) = (26,273)

[frank zealot]

yes

eucludean algorithm

[split sail]

GG

so euclid's algorithm's the best?

[frank zealot]

yes

[split sail]

ok

tanks

[frank zealot]

better than prime factorization

in most of the cases

[split sail]

if a | b, then a is the gcd right?

[frank zealot]

gcd of a and b?

[split sail]

and the gcd of 'a' and 0 is 0 right?

[frank zealot]

yes

[split sail]

ok

thanks

[frank zealot]

thats not true

gcd(a,b) exists if a and b are non-nil

0 has no divisors

wait no

[split sail]

no

[frank zealot]

i messed it up

yes ur correct

[split sail]

oh

[frank zealot]

wait no ur wrong

[split sail]

LOL

[frank zealot]

i keep messing it up im sorry

i hate arithmetic

[split sail]

no problem

[frank zealot]

by definition, GCD function works for two strictly positive integers only

[split sail]

No?

I thnk the only thing is that gcd is always non negative

and not 0

[split sail]

it works for any pair of integers that are not both simultaneously zero.

[frank zealot]

actually hold

[split sail]

ok

[frank zealot]

are we spesking arithmetic in N or Z

[split sail]

Z

integers

[frank zealot]

ok then the GCD(0, a) is actually

|a|

[split sail]

mhm

ok

then

also

wait

[frank zealot]

.

[split sail]

(-a,b) = (a,b) = (a,-b) = (-a,-b)

is this correct?

[frank zealot]

it is

if Z ofc

[split sail]

k

.close

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[split sail]

I had this question not done on first try

[split sail]

halp

wait

[tulip tapir]

wut

[split sail]

Given

By induction

I had done all similar ones by induction.

[kind ravine]

Is that….what-

[coral wigeon]

[primal holly]

What exactly do you have to do

[split sail]

prove by induction that a_(2n) = a_n * b_n

simple

[primal holly]

What have you tried?

[split sail]

I tried to do

couldn't do

I am stuck at the inductive step

[primal holly]

You need to show that $a_{2n} = a_nb_n \implies a_{2n+2} = a_{n+1}b_{n+1}$

[grizzled pagoda]

neon

[split sail]

yes

mhm

Maybe I can try

Simpliflying the latter equation

[primal holly]

Maybe try working with all fibonacci since

[split sail]

And work backwords?

[primal holly]

You know

Yeah sure

[split sail]

wdym

[primal holly]

Convert these Lucas numbers whatever they are into Fibonacci sequences

[split sail]

oh

yea

right

.close

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[fringe saddle]

Hi I need some help

[amber wasp]

@fringe saddle Has your question been resolved?

[hollow salmon]

What have you tried?

[fringe saddle]

Well the question doesn't make sense to me like what should I start with

[warm burrow]

surface area how do you calculate it?

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@fringe saddle Has your question been resolved?

[warm burrow]

bro just left lol

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[unique blaze]

.

[fringe saddle]

No sorry had an exam so had to go

[lavish zenith]

question

?

[fringe saddle]

I never did any of it so dk any formula I got it like a test to learn so I wanna know how to do this

[warm burrow]

i did it earlier you need to think a little hard

you need only pythagoras formula

[fringe saddle]

A^2+B^2=C^2?

[warm burrow]

Yeah

also $\frac{\sqrt{3}}{4}s^2$

[grizzled pagoda]

Steel

[warm burrow]

equal triangle area formula

[fringe saddle]

What does s^2 stand for?

[warm burrow]

side squared

and the regular a = bh 1/2

[fringe saddle]

So the mid point is 32?

[warm burrow]

the midpoint between what?

[frank zealot]

pretty sure they need to prove this

it isnt a straight forward formula

[warm burrow]

damn they gotta prove the triangle formula thats mad

[frank zealot]

well

the height is sqrt(3)/2 the side is a formula

they have to restate everything tho for the area

[fringe saddle]

I dont think it matters just need to show some steps can skip some if possible

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[thorny terrace]

hi need some help with part b

[thorny terrace]

so for part a, i just used the hint, and got to the u-sub: u = (2+i)t and du = (2+i)dt and just finished it out from there

and now for part b, i want to use similar reasoning, but i dont exactly understand how they changed the cos(t) to Re(e^it) in the hint

[pallid canopy]

there's a typo on the right integral

should read Re(int exp(2+i)t dt)

[thorny terrace]

heres euler's formula from my textbook

[thorny terrace]

oh hmm

[shadow stump]

yeah so we find that cos(b) is the real part of e^(ib), and sin(b) is the imaginary part of e^(ib)

[pallid canopy]

i think the idea is to write sin(4t) as Im(exp(4it)) * i. or maybe * -i

[thorny terrace]

ah i see

let's see

[shadow stump]

don't think you need to multiply by i here

[pallid canopy]

oh yes thanks

[thorny terrace]

$\int e^{2t} \sin (4t) , dt = \int e^{2t} Im(e^{4it}) , dt$

[thorny terrace]

does the Im have any effect on the integral now or can i just remove it and make \int e^(2t+4it)?

[woeful dagger]

But with Im outside the integral?

[pallid canopy]

you can just hand wave and say integral (imaginary part of a function )= imaginary part of (integral of a function)

since your teacher did it for the cosine

[thorny terrace]

ahhhh ok that makes more sense

[shadow stump]

https://math.stackexchange.com/questions/2139671/why-does-integral-and-the-imaginary-part-commute

[thorny terrace]

yea its like a nonrigorous ODEs class so i dont think we will go that deep

[thorny terrace]

thank you!

ill try to go from here, thanks for ur guys help riemann and cloud : )

.close

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[thorny terrace]

haha i think i kinda understood until i read banach

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[proper egret]

E(Y)=lambda

So surely only 4 is unbiased?

[pallid canopy]

yea you can prove it using linearity of expectation

[amber wasp]

@proper egret Has your question been resolved?

[proper egret]

So I am correct?

[pallid canopy]

no idea, the Y_i aren't defined anywhere

[proper egret]

exponential, but E(Y) = lambda is enough right?

[pallid canopy]

maybe the 5th one too

[proper egret]

does this not just give bias under intuition, as in it'll just choose the smallest number

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[proper egret]

.reopen

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✅

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[formal elk]

trying to do log rules but im kinda stuck here

solving for A

[dense meteor]

You want to get rid of the log

[formal elk]

how would i go about getting rid of it

ok so that puts it in a different format

[quick ridge]

is this supposed to be log base 10

[formal elk]

no

[quick ridge]

log base e?

[formal elk]

10log

[quick ridge]

brother

💀💀

[formal elk]

yea

[misty bobcat]

so u get 10 log701-10loga =76.6 right

[quick ridge]

i’m not talking about that 10

[quick ridge]

^

[formal elk]

bro i didnt write it

its 10 x log

cursed as hell

[quick ridge]

that’s not what i asked

[formal elk]

oh

[quick ridge]

$\log_b (x)$

[formal elk]

im sorry then i am missunderstanding

[grizzled pagoda]

knief

[quick ridge]

b is the base

[misty bobcat]

isnt it log base 10

if its log without anything

[quick ridge]

does your book use 10 or e when just writing log

[formal elk]

if nothing is wirtten yea i think its 10

10

[quick ridge]

some people use e for log

instead of ln

[formal elk]

thats ln isnt it?

not log

[misty bobcat]

yeah

[quick ridge]

it is, but some people have wrong opinions:)

[misty bobcat]

ln is e and log is 10

[dense meteor]

Theres a 10 multiplying it, you should divide both sides by 10.
Then, take "e" and raise the whole equation to e
e^log(x) simplifies to x
Then its pretty easy from there

[shadow stump]

the context that it's a decibel calculation means it's definitely base 10

[quick ridge]

not everyone follows that conventional

[formal elk]

so

[quick ridge]

good to know

[formal elk]

from here

[misty bobcat]

ye

very easy

cause log 701 is jsut a constant

[formal elk]

so you can move it away?

[misty bobcat]

ye so its loga= log701-7.66

[quick ridge]

$\log_b(x) = y \iff b^y = x$

[grizzled pagoda]

knief

[quick ridge]

do 10^ on both sides

[misty bobcat]

and to remove the log u just do power of 10

[dense meteor]

Brother in christ what are you telling him to do 💀 there was no need to apply difference of logs 💀

Not saying you cant

[formal elk]

sorry?

to both sides?

[misty bobcat]

ye

to get rid of the log base 10

[formal elk]

?

[misty bobcat]

nah its

10^(log701-7.66)

[formal elk]

ah

[misty bobcat]

cause u gotta hink abut the definition of a log

right

its the number that u have to raise 10 to, in order to get A

so u wanna raise 10 to that log to get A

and since they are equivelant u do that

[formal elk]

?

oh

no

[misty bobcat]

its 10 to the power of log701-7.66

[formal elk]

oh

[misty bobcat]

yee

[misty bobcat]

but not 701, log 701

[quick ridge]

$7.66 = \log\left(\frac{701}{A}\right)$

[misty bobcat]

yeye

[grizzled pagoda]

knief

[quick ridge]

$10^{7.66} = 10^{\log\left(\frac{701}{A}\right)}$

but we know

$b^{\log_b(x)} = x$

[grizzled pagoda]

knief

[formal elk]

bro

what are you doing

[quick ridge]

yes

[quick ridge]

explaining it

[formal elk]

theendisnevertheendisneverthend did it better

[quick ridge]

nah their way was using subtraction of logs

hence why you’re getting the wrong answer

and fucking up

this way is easier

[grizzled pagoda]

knief

[quick ridge]

$10^{7.66} = \frac{701}{A}$

[formal elk]

what is this step and why

[grizzled pagoda]

knief

[quick ridge]

to get rid of the log

[quick ridge]

using this rule

[formal elk]

i see

[quick ridge]

since log is log_10

[quick ridge]

from here just solve for A

$A = \frac{701}{10^{7.66}}$

[grizzled pagoda]

knief

[quick ridge]

none of this shit

[formal elk]

kk

omg

[misty bobcat]

wait my answer is wrong??

[formal elk]

it gave the same answer lol

[misty bobcat]

holup where did i fuck up

[quick ridge]

just way simpler

[formal elk]

no you didnt, the way he showed just did it a different way

[quick ridge]

splitting the log into two logs is utterly useless

and you clearly were prone to making more mistakes that way

[formal elk]

well thank you for the help :3

both of you

[misty bobcat]

Oh i see

[formal elk]

.close

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[fathom ruin]

ive been stuck on this for a while

i found the derivative to be: 3x^2 + p

[fathom ruin]

i made the derivative equal to the original function but i cant seem to prove that its -3q/2p

[fathom ruin]

x^3 - 3x^2 + px + q -p = 0

[quick ridge]

why the derivative?

[sharp depot]

ok good

for a double root

which well call alpha for now

p(alpha) = 0

and P'(alpha) = 0

[fathom ruin]

so you could make them equal to each other so you get this?
x^3 - 3x^2 + px + q -p = 0

[full orbit]

Hey knief 👋👋👋

[quick ridge]

Gigagoat

what’s good

[full orbit]

What’s going on here

[quick ridge]

mr chicken is helping

[sharp depot]

type shi

[full orbit]

Oh okay 👍

[balmy hedge]

:0

hi

[quick ridge]

hello

[sharp depot]

if u want to discuss

please move ot discussion

cahannel

[quick ridge]

@sharp depot are you going to help them?

[sharp depot]

im trynna

unless if u hvae teh answer

then go ahead

[quick ridge]

i mean i wouldn’t take this approach personally

but i’m curious

[full orbit]

Let knief cook 🗣️🔥🔥

[quick ridge]

so i’ll let you finish

[sharp depot]

nah lowkey i dont think derivative works

u get answer

but not in the form the q watns

[quick ridge]

what did you get?

[sharp depot]

i dont see it tho

cause even if i solve for x

which i got like root(-p/3)

i dont know if its plus or minus

i dont konw how to turn that into the form the q wants

[quick ridge]

:/

well

[sharp depot]

so id just do let roots be alpha alpha beta

and try solve for alpha

[quick ridge]

yep

[sharp depot]

using sumand produdcts

[quick ridge]

$P(x) = x^3 + px + q = (x-a)^2(x-b)$

[grizzled pagoda]

knief

[quick ridge]

$x^3 + px + q = x^3 - (b+2a)x^2 + (2ab+a^2)x - a^2b$

[grizzled pagoda]

knief

[quick ridge]

$p = 2ab + a^2, \quad b+2a = 0, \quad q = -a^2b$

[grizzled pagoda]

knief

[sharp depot]

aight yea hi got it

[quick ridge]

$p = -3a^2, \quad q = 2a^3$

[grizzled pagoda]

knief

[full orbit]

Another successful cook up by chef knief 🧑‍🍳🔥🔥

[quick ridge]

$\frac{p}{q} = -\frac{3}{2a}$

[grizzled pagoda]

knief

[quick ridge]

$a = -\frac{3q}{2p}$

[grizzled pagoda]

knief

[quick ridge]

🔥🔥🔥

😤

[full orbit]

Tuff

[fathom ruin]

sorry! how did you get this line?

[quick ridge]

just expand the two terms on the right

$(x^2-2ax+a^2)(x-b) = x^3 - bx^2 - 2ax^2 + 2ab x + a^2x - a^2b$

[grizzled pagoda]

knief

[fathom ruin]

umm

[quick ridge]

then just group the x^2 terms

and x terms

[fathom ruin]

OHHHH

[quick ridge]

yes

[sharp depot]

or you could use sums and product of roots if u know them

i think their called vieta's formula

or summin

[quick ridge]

true

[fathom ruin]

ohhh i get it now

thanks for explaining !

[quick ridge]

https://tenor.com/view/dr-pepper-soda-can-gif-10290138

you’re welcome

[fathom ruin]

lmao

.close

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[empty kraken]

is the sum for this right

[empty kraken]

can someone double check my work

i think i may have done something wrong

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@empty kraken Has your question been resolved?

[empty kraken]

<@&286206848099549185>

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@empty kraken Has your question been resolved?

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[distant scarab]

9

[distant scarab]

ASAP

<@&286206848099549185>

[final matrix]

don't ping before 15 minutes

thanks

[distant scarab]

Ok

[final matrix]

why do you need this asap?

[distant scarab]

I just want to answer a question

It’s related to this

[final matrix]

ok well

there's a particular theorem which indicatees angles P will be half of angle alpha unless im not mistaken

[distant scarab]

P?

I have written points A B C and D

And O

[distant scarab]

So is angle AOB = angle ADB ?

[final matrix]

angle ACB and ADB should be half angle AOB iirc

there's a particular theorem for it

[distant scarab]

So it’s correct

Ok

Thanks

.close

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[frail totem]

Let two circles (O1) and (O2) intersect at A and B. Through A, draw any secant line intersecting (O1), (O2) at E and P respectively. Through B, draw any secant line intersecting (O1), (O2) at F, Q respectively. Prove that EF//PQ

[frail totem]

help me pls

[frank zealot]

did u do a drawing

cause drawing this by hand looks pretty rough ngl

can u show what u drew

[amber wasp]

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

@frail totem Has your question been resolved?

[frail totem]

@frank zealot

[amber wasp]

@frail totem Has your question been resolved?

[frank zealot]

ok i got it @frail totem

its all about cyclic quadrilaterals and the sum of their opposite angles

just try to look at angle EFB and angle EAB

anf then look at angle BQP and angle BAP

js try to relate them using the rule: the sum of two opposite angles in a cyclic quadrilateral (means a quadrilateral where his four vertices lie on the same circle) is 180°

let that be ur start

[amber wasp]

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[obsidian glacier]

i’m not sure how to begin with question 16

[obsidian glacier]

like, do i take the surface area? do i take the volume??

[split sail]

Probably both, and come up with a solution, but cast it onto the increments formula.

Otherwise the solution would be wrong

[obsidian glacier]

do i write either in terms of radius?

hm

all i know is that

the radius expands by 0.05cm

i found it using the calculator

pi(10.05)^2 * h = pi(10^2)(20)

and just solved for height

which was 19.8

so it decreased by 2mm

but that’s too easy

is there a different way?

[amber wasp]

@obsidian glacier Has your question been resolved?

[obsidian glacier]

.close

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[neat dagger]

hi

[neat dagger]

can someone help me

i need this to be a half circle

[soft hearth]

can u like elaborate?

[fathom bramble]

1. Which point do you want the center of the semicircle to be?

2. If what you want is for the circle to “close“ the two lines under, either you need to swap the center AND radius of the semicircle or the semicircle just doesn't exist

[fair rain]

I assume you want the semicircle to connect the two lines?

[amber wasp]

@neat dagger Has your question been resolved?

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[neat dagger]

yes

[fair rain]

ok so first you need to consider where the centre of the semicircle would be, which is the midpoint of the two points you are connecting

[neat dagger]

hmm

(sqrt(2) + 1) / 2

[amber wasp]

@neat dagger Has your question been resolved?

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@neat dagger Has your question been resolved?

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[spice dagger]

how can I approximate value of 4π^2?

[spice dagger]

Any method or i multiply?

[alpine flower]

hm

[spiral zealot]

take your pick for the approximation of pi

[alpine flower]

(pi)^2 can be approximated to 9.8

[spiral zealot]

use 22/7, 3.14 or other reasonable approximations

results may vary depending on your choice

[spice dagger]

Thanks

I got 39.mm

.close

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[keen pawn]

how do I deal with the last case

[elfin vale]

If both x and y are negative then xy is positive

[robust isle]

you need to mention for the first case you did that x < 0 and y >= 0, or x >= 0 and y < 0

it is not sufficient that one of x < 0 or y < 0

[amber wasp]

@keen pawn Has your question been resolved?

[keen pawn]

Why not

[keen pawn]

Yes

[robust isle]

if one of x or y < 0, lets say x < 0

then i choose y < 0

now xy is positive not negative as you claimed

so your first case is wrong

[keen pawn]

Only one

[keen pawn]

But how does that help me

got it

thanks

.close

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[compact wagon]

for this question

i got +- 2sqrt3 from the second derivative of f as well

[patent raptor]

really

[patent raptor]

,w D[(x^3)/(x^2-4),{x,2}]=0

[grizzled pagoda]

Results provided by WolframAlpha

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

Requested by adoniszgks

[patent raptor]

x^2+12 has no real roots

[compact wagon]

oh

.close

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[patent raptor]

oh

[amber wasp]

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[winter dock]

can anyone help, plz explain how to solve these questions, and plz do one of them as sample plz.

[runic kraken]

have you learned about sin, cos, and tan ?

[winter dock]

yes

[tulip tapir]

!noans

[amber wasp]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[runic kraken]

okay lets go through a together

what can you tell about it

what is given

[tulip tapir]

actually i think it'd be more helpful to label the opposite and adjacent sides, along w/ the hypotenuse, first

then you can do some very basic rearranging to figure out what "x" is

[runic kraken]

yeah true

[winter dock]

we are givven with perendicular and hyp we can apply sin ? p/h

[runic kraken]

yes

first time i hear it being called perpendicular but yes you can apply sin

[winter dock]

we can also say that height

[runic kraken]

yeah use what you feel comfortable with

[winter dock]

okay,

can i solve 1

u can check

[runic kraken]

yeah go ahead solve it

[winter dock]

i tried but coudnt

[runic kraken]

what was the issue you ran into

[winter dock]

i confused with the questions like part a, we hv 1 angle, hight and what we hv to calculate is hyp, but HOW

[runic kraken]

recall the equations of sin cos and tan

you can also use soh cah toa

[winter dock]

what r they, soh cah toa

sin cos and tan?

[runic kraken]

yeah another way to write them that helps you remember what they are.

[winter dock]

plz help @runic kraken

[dire gyro]

u may use sin3x = sin60 to calculate the value of sinx but it will be a really weird cubic

[runic kraken]

soh is sin = opposite / hypothenus. cah is cos = adjecent / hypothenus. toa is tangent = opposite / adjecent

[winter dock]

ik

[runic kraken]

no thats too complicated for no reason

okay good you know it

[dire gyro]

yeah

[runic kraken]

so in the first 1 you are given the angle, the opposite and you need to find the hypothenuse

write down the equation, fill in the values

and see what you can do to solve it

should make it clear

[tulip tapir]

okay look

fairly simple

label the sides

[dire gyro]

yo mb dis a dumb ques

but

[tulip tapir]

what do you get?

okay what

[amber wasp]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[tulip tapir]

please delete that

[sick sandal]

Oh sorry, @runic kraken told me to post question here:(

[tulip tapir]

no

#help-30 is free

[runic kraken]

yeah my bad the chat got occupied

[tulip tapir]

@winter dock what question are you on

[winter dock]

part a

[sick sandal]

@runic kraken What should I do now? 😮

[runic kraken]

go on another help channel and paste your question

[tulip tapir]

dude.

#help-34

[solar gust]

Go to https://discord.com/channels/268882317391429632/488120190538743810

[tulip tapir]

rawkin post your question there

end of story

[dire gyro]

how would u solve

the a part

without calculator thou

[sick sandal]

Removing entries here, sorry guys had nowhere else to write 🙏🏻

[winter dock]

i hv calculator

[dire gyro]

yeah then just calculate the value of sin20

in part a

[winter dock]

okay wait

[dire gyro]

sin20 = P/H

[dire gyro]

= 10 /x

soo x = 10 / sin20

[runic kraken]

did you understand the following?

[winter dock]

other parts ?

[runic kraken]

you do them

we aren't going to give you solutions

[winter dock]

okay, im solving

[winter dock]

ik this chanel is to make u learn

[runic kraken]

if you need help understanding we'll try our best to explain

[winter dock]

thanks,

Ur a good guy

[runic kraken]

a good way to solve these is first to identify what is given, opposite, adjacent, hypothenus, the angle, then what you need to find. then write the appropriate equation cos, sin, tan in full form, then filling in what is given and rewriting the equation in a way where you can solve for what you need to find

[runic kraken]

perfect

[winter dock]

x = 10/20sin

but what to do next,

x= ?

@runic kraken

[runic kraken]

you can solve this on your calculator

[winter dock]

thats shoing a syntax error

[runic kraken]

ah yeah

its 10/sin(20)

[winter dock]

Jazak Alllah/Thanks

sorry im too dumb

[runic kraken]

nah you got this

just keep at it

practice it more and you'll get the hang of it

[winter dock]

Okay

[amber wasp]

@winter dock Has your question been resolved?

[winter dock]

how to solve if we hv to find angle is we r provided with all lengths or propotional length

[tender folio]

Not even my PhD’s helping u for this one lil bro 😭 🙏

[winter dock]

hv u done PHD in Maths

[runic kraken]

hur which one are you stuck on

[winter dock]

b

[runic kraken]

did you go through the steps?

[signal ermine]

Lemme see if I can help you

[winter dock]

yeah, sure

[signal ermine]

Spoiler alert:

[runic kraken]

where did you get stuck in the steps

naiver if u wana help go ahead

[signal ermine]

||use sin(35°)||

It's d/20

Plug sin(35°) into your calculator

Multiply that by 20

Bam

🥰

[winter dock]

but how

plz elaborate, and make me learn

@signal ermine

[signal ermine]

Sorry

So:

(I suck at explaining but bear with me)

[amber wasp]

Channel closed

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[runic kraken]

@winter dock type .reopen

[winter dock]

but still plz explain the steps @signal ermine

.reopen

[amber wasp]

✅

[signal ermine]

The sine of an angle is the opposite/hypothenuse

[tulip tapir]

"navier stroke"

[signal ermine]

And as you can see, d is the opposite and 20 cm is the length of the hypotenuse

[signal ermine]

I'm proud of this name XD

[tulip tapir]

xD

[winter dock]

yes so it would be d/20

[signal ermine]

So sin(35°) = d/20

And sin(35°) has a value that you can basically plug into the calculator (I can do a detailed explanation on what sine and cosine actually are but it's not relevant to the question)

[winter dock]

yes

they r not

[signal ermine]

So you gotta find out what sin(35°) is with the help of the calculator

And multiply it with 20

[winter dock]

= 0.57..

[signal ermine]

Since its the length of the hypotenuse

(brb)

[winter dock]

= ?

@signal ermine

[runic kraken]

brb = be right back

[winter dock]

but why we multiplied it with 20

[runic kraken]

write the equation down

and solve for x

as in make it x = ...

[winter dock]

teach me

how

[runic kraken]

solve for x

you want to isolate x

how can you do that

[winter dock]

oh, rule over the bridge

20 is deviding on right side so to make x isolated ill multiply sin35 with 20

[runic kraken]

perfect

you now have isolated x

and you can solve for x

[winter dock]

okaye.....

[runic kraken]

do you write down the equations like that on a paper?

[winter dock]

yes i did

[runic kraken]

nice okay

keep writing it, it will make solving much easier

since you can see what you want to solve for

[signal ermine]

Yoooo

[winter dock]

how to solve if we have all the sides and want to find the angle, how to do that

part c

[runic kraken]

think about it first yourself

with cos sin and tan

[signal ermine]

@winter dock

[runic kraken]

for the angle a what does each length mean

[signal ermine]

I think you need some mathematical intuition around these concepts

[runic kraken]

which one is the hypothenuse, opposite, adjacent

[winter dock]

sin = P/h

=15/17 or 8/17

[runic kraken]

its one of them

i gotta leave rn good luck with the rest

[winter dock]

15/17 = 0.882 and 0.470 = 8/17

[amber wasp]

@winter dock Has your question been resolved?

[winter dock]

nope

[runic kraken]

You’re looking for the angle

[winter dock]

yes

[runic kraken]

What you found is the value for x in sin(x)

No

My bad

No thats not what you found

What you found is the result of sin(x)

Sin(x) = opposite/hypothenuse

Now you need to find x

[winter dock]

yeah

[runic kraken]

You’re supposed to know how to find x

[winter dock]

idk

[dire gyro]

sin-x

use calculator for it asw

no

i mean

[runic kraken]

What

Yeah

[dire gyro]

sin-0.470

[runic kraken]

Sin^-1

The inverse function of sin

Sin^-1 or arcsin

Both are the same

Some people use arcsin and some sin^-1

Both are correct

Both solve the problem

[winter dock]

oh so ans is 28.07

[runic kraken]

You then multiply both sides by sin^-1

I don’t have a calculator on me to check but it should be correct yes

Good job

I don’t want to see you asking questions for d

I know u can do it with what you have learned

[winter dock]

ill share my ans with u

Part d solved 🎉

[runic kraken]

yessirrr

Thats is correct

[winter dock]

yes

[amber wasp]

@winter dock Has your question been resolved?

[amber wasp]

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[queen dragon]

how can you find height in an isosceles trapezoid?

[shadow stump]

you can figure out the length of the base of that triangle with the information given

[queen dragon]

with only the hypotenuse?

🤷‍♂️

[shadow stump]

you don't need the hypotenuse for that

think about what would happen if you made a second right triangle to the left

[queen dragon]

I tried doing similarity with those two but

didn't work out

[shadow stump]

they aren't just similar

[hearty sequoia]

i agree

[queen dragon]

i'm struggling to see it

would diagonals help?

[shadow stump]

just the two right triangles is enough

[queen dragon]

yeah i still don't get it

[shadow stump]

can you show the diagram involving both right triangles?

[naive pivot]

How is the longer base 10 ?

[shadow stump]

you have that the angles of those triangles of the same, and at least one of their sides is. what does that tell is about those triangles?

[queen dragon]

oh

do you have to use the

postulate thing

🤷‍♂️

well

no idea why this works but

whatever

thanks for the help 👍

[shadow stump]

that does work

[queen dragon]

yea

https://www.youtube.com/watch?v=V_CeTUBl1VA&t=69s&ab_channel=GeometryConceptVideos

[crystal basalt]

because both hypothenuse are the same length

so it is an isocele triangle with a rectangle between it

[shadow stump]

the two right triangles are congruent so their base is the same

[queen dragon]

yea well

[crystal basalt]

yes that's what i meant

you expressed it better than me

[queen dragon]

I guess you can pull something like this only in isosceles trapezoids

[queen dragon]

so

if angles the same

atleast 1 side the same

their sides are also gonna be the same?

[shadow stump]

that's what the ASA postulate tells us

[queen dragon]

I see

alright, thanks for the help 👍

.close

[amber wasp]

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[normal bluff]

https://cdn.discordapp.com/attachments/1062063468225773569/1302394041853546516/IMG_5144.jpg?ex=6727f478&is=6726a2f8&hm=a8f80a6883740489294f5c8c4b8fa6156aee728f764617861ff51191e8559543&

[normal bluff]

im taking the second derivative implicitly

does this look right

i want to make sure im thinking through this correctly

[celest cove]

which is the original?

it doesnt look right in either case but itll help to know

[normal bluff]

sorry!

f(x) = 3x^2+y^2=5
f'(x)=6x+2y(y') = 0
f''(x) = the bottoim

[spare granite]

Well you have an ode on the bottom

[celest cove]

think youre missing a term

6x to 6

2y y' to 2y y'' + 2y' y'

[normal bluff]

wdym

[celest cove]

by the product rule d/dx (2yy')=2[dy/dx y'+ y dy'/dx]=2[(y')^2+yy'']

[normal bluff]

is the f'(x) i got correct? and im missing a term between that and f(''x)

[celest cove]

your f' is fine

im just saying your f'' isnt

[normal bluff]

i see

what u mean

i missed writing down

uyes

[celest cove]

you need to use the product rule on the 2yy'

but i dont think you did it quite right

[normal bluff]

I think this is right correct me if im wrong

[celest cove]

that one is right yeah

wait no

[normal bluff]

neat ok

so i solve for y''

orn ot

lmao

[celest cove]

well, unless the y' on the y''(2y') is just hidden

[normal bluff]

wait

shouldnt i be wy

2y

[celest cove]

nvm you right dont mind me

[celest cove]

did it fine here

[normal bluff]

f'(x)g(x)+g'(x)f(x)

[normal bluff]

So this should be correct?

[normal bluff]

what is ode

bro wtf i genuinely dont understnad

anymore

[celest cove]

sub in the expression for y'

[normal bluff]

y' has y' in itsself no

?

so wouldnt i tjust be a loop

of infinite plugging in y'

[celest cove]

no?

[normal bluff]

this is y'

[celest cove]

no it isnt

[normal bluff]

how

what

[celest cove]

its the first derivative

its not y'

because you havent made y' the subject

[normal bluff]

so what is y'

[celest cove]

rearrange it?

[normal bluff]

OH

yeah

thanks

appriciate u

@celest cove so this is a seperate problem but this shoulf be good yah?

[celest cove]

you should stop writing f and f' and f'', because it really makes little sense

[normal bluff]

Instead write what

[celest cove]

nothing

but yeah it seems okay

[normal bluff]

my professor will yell at me for not labeling stuff

And showing my work

[celest cove]

you are showing your work, if you really want to write something write 'first:' or something
youre defining a function when you write f somewhere

[normal bluff]

Oh and this isnt a function?

Cause its y and x

vertical line test shit

[celest cove]

like if you write f(x)=6x^2+y^2=7
youre then saying that
f'(x)=12x+2yy'=0
but then youre rearranging and saying f'(x)=-2x/2y
so then f'(x)=y'
so then y'=12x+2yy'=0
so then 12x=0 x=0

it makes problems if you really start to use it

[normal bluff]

Kk

Tyy

[amber wasp]

@normal bluff Has your question been resolved?

Channel closed

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• Be polite and have a nice day!

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[wheat elm]

Need help with these induction proofs. My brain is freakin blank on how to even start right now

[plush latch]

which one 1 or 2

[wheat elm]

both, but let's start with 1, and hopefully if we get through that 2 will seeem clearer

[plush latch]

ok so for 1

Your going to show the case where n = 1

which u can probably just do

next you want to say that when n=k you are going to assume the statement is true

and then u want to show that the statement is true for when n=k+1

[wheat elm]

sorry still working out the writing of it

[plush latch]

wym

[wheat elm]

I mean my brain is not coming up with the formula's notation

[plush latch]

do you know what a proof by induction is

[wheat elm]

Honestly, not really. was super sick for a bit

[plush latch]

oh right

[wheat elm]

The Latin E thing makes my brain explode.

[plush latch]

so basically the a proof by induction is where you prove a statement by assuming its true and then proving it for numbers bigger than it basically

[plush latch]

latin E thing?

[wheat elm]

this fucker

[plush latch]

oh thats an s

I mean u dont really have to use that here tho icl

[wheat elm]

Well good cause I hate the bugger

I think this is what the first part of the equation looks like right?

[plush latch]

there isnt an equation

you have to prove that 3^n-1 is divisible by 2 for all values of n

so for a proof by induction you split it into parts

first you show that the 3^n-1 is divisible by 2 when n=1

[wheat elm]

yup

Just turns into 3-1=2 /2

[plush latch]

then you assume that 3^n-1 is divisble by 2 when n=k

and then using that assumption show that 3^n-1 is divisible by 2 when n=k+1

[wheat elm]

OOkay

[plush latch]

ye then show that is divisible by 2

assuming that 3^k-1 is divisible by 2

[wheat elm]

It should be, since every full value exponent of 3 is odd, and minus one that becomes an even number

Do I do that by just plugging some random shit in or is there an algerbraic notation for htis

[plush latch]

you can show it algebraicly

[wheat elm]

okay, let's tart with that

[plush latch]

do you know how to do it or do u want me to do it

[wheat elm]

-sighs- the latter

Apologies

[plush latch]

ok so 3^(k+1)-1 = 3*3^k-1

3* 3^k-1=2* 3^k+3^k-1

2*3^k is divisible by 2 and 3^k-1 is assumed to be divisible by 2 therefore 3^(k+1)-1 is divisible by 2

[wheat elm]

Where did the two come from

[plush latch]

3* 3^k=2* 3^k+3^k

[wheat elm]

So what generates it? There's nothing that brings it from the first bit of your equation to the second

[plush latch]

your basically just doing 3=2+1

[wheat elm]

Ah