precalculus

[echo plaza]

well are you used to the one for limits to real numbers

the one as x tends to infinity is slightly different

[tropic crown]

im not too sure lightning what u meant by that

and idk if this is exactly delta

it has an epsilon sign, but i havent done any questions like this before

that has epsilon - N

only epsilon - delta

[echo plaza]

it's mostly the same

[tropic crown]

do i use the same procedure?

to prove?

[echo plaza]

it's just a different measure of "closeness" if you want, because it's a limit as x tends to infinity

mostly the same

given some arbitrary $\ep>0$ just find some $N$ satisfying this

[obsidian monolith]

CaptainLightning:

[echo plaza]

symbolically ofc

[tropic crown]

ill take ann suggestion and move the convo to calculus

but ill try something

and show my work

in calc soon

[echo plaza]

just as a hint

there's a more helpful way of writing this

[tropic crown]

uhh ;lol

what would that be?

[royal gull]

2= 0.5 + 1.5

[willow bear]

maybe consider $\frac{x+2}{2x+1} - \frac12$ instead

[obsidian monolith]

Ann:

[tropic crown]

so half angle formula for cos^2x is 1/2(1-cos(2x))

but if i have something in x, like, cos^2(2x)

what would the half angle look like

would it be 1/2(1-cos(4x)?

[willow bear]

indeed it would

[tropic crown]

so to confirm, cos^2(4x) would turn into 1/2(1-cos(8x)?

were just multiplying whatever is inside the intial ( ) by 2?

[willow bear]

well like

first off, +

not -

and second, yes, $\cos^2(4x) = \frac12(1 + \cos(2 \cdot 4x))$

[obsidian monolith]

Ann:

[tropic crown]

oh right, okay

thanks!

[fervent lagoon]

yo could someone help me out with this problem?

i was thinking about law of cosines but idk how to approach with two unknown values

[proud raven]

Im trying to figure out what the geometry trick is

which was never my strong suit

[grizzled orchid]

yah same lol

[proud raven]

im wondering if like

you can rewrite the d/b angles

but it doesnt look nice

[grizzled orchid]

yeah eugh

hmm

[viscid thistle]

Im wondering if the 180 in the 2nd triangle applies for all 3 sides

Its not an equilateral triangle right?

[proud raven]

i doubt it

[viscid thistle]

Got it

[grizzled orchid]

if we know perimeter equals 140 we can label AD and BC as x and y and say x + y = 280

[proud raven]

if it was equilateral

the condition that the two unknown sides are different wouldnt be true right

or maybe it would

640*

[grizzled orchid]

regardless we can't just assume : p

[proud raven]

640-(180*2)=280 youre right sorry >.>

[grizzled orchid]

yah

[proud raven]

so one is 280-x other is 280-y

[grizzled orchid]

ye

[proud raven]

then uh

[grizzled orchid]

ooh wait

or

hmm no dice

nvm

ugh why can't geo just be like calc

ez pz

[viscid thistle]

Theres too many tricks in geometry

[proud raven]

i wanna ping helpers

lol

[viscid thistle]

Or you could ping the mods

[proud raven]

admins yea

help we dont understand geometry

it also looks like the asker evaporated

[grizzled orchid]

yes

they dissipated

[proud raven]

bummer

guess ill never know

geometry is a nonsense anyways

which is to say im too stupid to get it

anyways

[grizzled orchid]

[fervent lagoon]

im lost too guys 😦

no clue where to go

[native sequoia]

7/9, you have answer so I can check my answer?

oh wait I can check it while writing the solution

[valid flint]

thats what i get too, 7/9

[native sequoia]

didn't realise they moved to #geometry-and-trigonometry rip

[viscid thistle]

Hey John Doe smith

[spark cliff]

This is C, correct?

[pale cedar]

yes

[mystic dagger]

Please help simplify this would be a huge help

[uncut mulch]

start with factorisation

[mystic dagger]

did gcf then factored am i on the right track @uncut mulch

[uncut mulch]

you've reached the solution

[mystic dagger]

thank you @uncut mulch really appreciate it

[rugged ice]

having a hard time reading this one

[uncut mulch]

assuming ur trying to find the eqn?

[rugged ice]

just the equation for it

[willow bear]

can you post the entire problem

judging by the graph alone, this looks like a circle centered somewhere around (-1, 6) with radius around 3.5

[rugged ice]

That's what I had figured but I wasn't completely sure, thanks

[mystic dagger]

Anyone know where to begin?

[river kindle]

What do you do when you divide fractions?

[harsh spoke]

https://cdn.discordapp.com/attachments/489120026423722015/604755835754774544/unknown.png

i think this is precalc

idk its summer homework for calculus

anyway how would i go about solving it

[mystic dagger]

Recripical got it

[harsh spoke]

i got u-v by just guessing for a

idk howto do the others though

[serene heath]

ok what's 0.125 as a fraction

[harsh spoke]

lmao i got all of those finally

except the last

cuz log (10^1/2

so i thought it would be 1/2

but it needs u or v

and idk how to put u or v in there

[limber bone]

log_10(10^1/2) = 1/2log_10(10) = 1/2

[wraith idol]

$\log_{10} (10^{\frac{1}{2}}) =\frac{1}{2} \log_{10}(10)=\frac{1}{2}$

[obsidian monolith]

Whoever:

[mystic dagger]

Where do I even start?

[pale cedar]

can cosecant be written as a more friendly trigonometric function?

[mystic dagger]

1/sin

[pale cedar]

so if you substitute that in, can you rearrange to get sin theta = constant?

[mystic dagger]

Yes do I graph from there?

Not too sure

[pale cedar]

you can do that; tho this happens to be a nice angle

[mystic dagger]

So I just solve from there getting theta alone @pale cedar

[pale cedar]

yea at that point you do have theta alone (if thats what ur asking? im not sure haha)

[mystic dagger]

I get this as an answer but how do I write that in radians in terms of pie?

Or am I still wrong?

[pale cedar]

ur fine; you found a solution.... since it seems ur calculator is in radians (which means 0.7854 = x * pi) just divide 0.7854 by pi

(also i will recommend you sketch the graph; these things sometimes have more solutions!)

[mystic dagger]

That’s where I am a bit confused is there something I can look up because I am not sure what this topic is

[pale cedar]

the radians part or the graph part?

[mystic dagger]

Graph per

Part

[pale cedar]

how familiar are you with sine?

[mystic dagger]

Haven’t done this in months so def need a refresher. Any thoughts on what to look up

I appreciate the help as well but I would hate to waste your time on something I can look up

[pale cedar]

you could google sine? or pull up a graphing calculator (i recommend desmos) and plug it in?

[mystic dagger]

ok thank you

[pale cedar]

😄 gl!

[mystic dagger]

Please help for final solution

[willow bear]

what is giving you trouble here

[mystic dagger]

taking the 80 out of rad @willow bear

[willow bear]

well, how do you normally simplify radicals with large constants?

[mystic dagger]

nvm i got it thank you

[mystic dagger]

ik its trig identity i am just having trouble subtracting the sin over for substitution

do i subtract 2sinx and leave the 2?

cos^2x - 2sinx = 2

is this correct?

<@&286206848099549185>

[surreal mesa]

Wait 15 minutes before tagging helpers

[wraith idol]

15 min

[tawny nacelle]

cos^2 = 1 - sin^2

and then its a quadratic in sin

[grizzled orchid]

reee

[mystic dagger]

Thank you and sorry for not waiting 15

[viscid thistle]

So given two sequences that satisfy the recurrence relations a(n)=Aa(n-1)+Ba(n-2) and b(n)= Cb(n-1)+Db(n-2) where A,B,C,D are positive real constants, what's the limit as n goes to infinity of their ratio?

[frozen needle]

You should be able to fully determine those two sequences

[viscid thistle]

do you mean i need to solve the recurrence equation?

[frozen needle]

that would be a way

[viscid thistle]

ok say i solve the recurence relations what then

[frozen needle]

then you calculate the limit with usual techniques

it'll probably be expressed in term of a(0), a(1), b(0) and b(1)

and A, B, C and D of course

[viscid thistle]

ok so i can't do the limit

is there another simpler way

[late pewter]

i feel like it'd be hard to solve the recurrence

isnt that the fibonacci sequence

[viscid thistle]

well if A=B=1 then a(n)=a(n-1)+a(n-2) which is the recurrence that the fibonacci sequence satisfies

[late pewter]

whats the closed form for fibonacci sequence though?

[viscid thistle]

it's kind of ugly

[patent beacon]

It's fibonacci-like

[viscid thistle]

the closed form is on the wikipedia page for fibonacci numbers

under mathematics

Relation to the golden ratio

[late pewter]

they really used the 2 most similar looking symbols

ok yeah i think i remember this

but I imagine deriving it for the general equation a(n) = Aa(n-1) + Ba(n-2)

wouldnt be that easy

[viscid thistle]

maybe it can be done without solving the recurrence

[late pewter]

yeah thats what im thinking

[viscid thistle]

https://www.youtube.com/watch?v=dTWKKvlZB08

[patent beacon]

Move the terms up to
a(n + 2) = Aa(n + 1) + Ba(n)

Then assume the solution is Cλⁿ. This brings us to
λ² = Aλ + B

Let the solutions of λ to that be r and s.

Then the solution to the whole thing is
a(n) = C1 rⁿ + C2 sⁿ

[viscid thistle]

like on numberphile kind of

well wolfram alpha can solve the recurrence

[patent beacon]

There's a few problems here, that r and s could be complex. As well, C1 and C2 depend on your starting point

[late pewter]

what can't wolfram alpha solve 😫

@patent beacon how'd you get λ² = Aλ + B

[frozen needle]

(-A)²-4(-B) > 0, r and s are real

[viscid thistle]

the best rational approximations to root 2 are kind of like this

like the numerator terms

satisfy the recurrence a(n)=2*a(n-1)+ a(n-2) where a(0)=2 and a(1)=2 but like half of this sequence

and the denominators

b(n)=2*b(n-1)+b(n-2) ,b(0)=0 b(1)=1

and the limit of their ratio is root 2

any ideas?

<@&286206848099549185>

So given two sequences that satisfy the recurrence relations a(n)=Aa(n-1)+Ba(n-2) and b(n)= Cb(n-1)+Db(n-2) where A,B,C,D are positive real constants, what's the limit as n goes to infinity of their ratio?

Repost

anyone?

[proud raven]

hmm

write out some terms

[viscid thistle]

of a(n)/b(n) you mean?

i think this is done the same way this is done

Prove that the limit a(n)/b(n) as n goes to infinity = sqrt(2) where a(n)=a'(n)/2, a'(n)=2a(n-1)+a(n-2), a(0)=a(1)=2 and b(n)=2b(n-1)+b(n-2), b(0)=0 ,b(1)=1

so a(n) is half of this:

https://oeis.org/A002203

and b(n) is the nth pell number

and this is definitely true

[proud raven]

hmm

i think thats over solving it

should be 1, right?

or maybe some variable ratio

[viscid thistle]

the limit of their ratio is definitely the square root of 2

not of the general sequences i gave

but the other ones

[proud raven]

that doesnt make sense

[viscid thistle]

they give the best rational approximations to the square root of 2

[proud raven]

the only differece between the two are variable constants

[viscid thistle]

exactly

that's why i think it's solved the same way

[proud raven]

theyre not numerical in the context of this problem

how do you get a numerical answer

idk you might be right

i doubt it

[viscid thistle]

if we use the general sequences then of course the result will depend of A,B,C,D and maybe a(0),a(1),b(0),b(1) as well

but since in that more special case A,B,C,D and the rest have numerical values

the result will also have a numerical value

[proud raven]

ill take a look when im home

[viscid thistle]

cool thanks\

[late pewter]

idk if this is helpful at all but I got:
a(n) = (fib(n - 1)A + fib(n - 1)B) * a(1) + (fib(n - 2)A + fib(n - 3)B) * a(0)
where fib(n) = nth fibonacci number

[viscid thistle]

hmm interesting

[late pewter]

i might be wrong

but i just expanded a(n) a bunch

and noticed the pattern

obviously the same applies to b(n)

[viscid thistle]

yeah that

and maybe now the limit's easier?

[late pewter]

yeah, maybe you could use the closed form for fib() but it seems like you would get a very nasty answer

let me double chcek that expression I gave is even correct

oh yeah wait thats totally wrong lmfao

i forgot to plug in the A and B

🤦

[viscid thistle]

so maybe focusing on the more special case will help cause we know that the ratio of half of the companion pell numbers and the actual pell numbers is the square root of 2

and thos sequences have that form

those

a(n) = 2a(n-1) + a(n-2),

with different a(0) and a(1)

i thought that the proof would be kind of like the one in that numberphile video

although that one was for the limit of the ratio of consecutive terms of the same sequence

i believe it's close enough but idk

[proud raven]

i need to sit down and check if its telescoping

[viscid thistle]

i thought that term was used in series' only

like sums and stuff

wdym?

is there such a thing as a telescoping limit or smth?

[proud raven]

a telescoping series

[viscid thistle]

yeah but what series?

[proud raven]

no i mean terms go away

man i wish i had paper

[viscid thistle]

anw if you make any progress you're free to DM me or just post it on here and make a mention

anw

[harsh spoke]

can anyone help me with the last one

[proud raven]

@viscid thistle its not telescoping

i would post your question in #calculus

and then ping helpers if you dont get a reply

because im not sure but it seems fundamental so maybe its important

[harsh spoke]

they told me here yesterday lmao

i’ll move it

[proud raven]

no not you @harsh spoke

i was talking to another guy

@harsh spoke I don't think theres a way to write it without the square root

[harsh spoke]

shit

[proud raven]

so then its just a matter of simplifying the radicand, or maybe you don't even need to do that

[harsh spoke]

i can write with square roots

just no logs

i got 10rad(10)

if i simplify

[proud raven]

$\sqrt{55225*2}$?

[obsidian monolith]

jan Niku:

[proud raven]

has a nice prime factorization

[harsh spoke]

oh!

[proud raven]

you can express 10rad(10) in terms of your logs too

i think?

[harsh spoke]

3v+3u inside rad

[proud raven]

maybe I'm crazy 😄

you can definitely express 10

[harsh spoke]

it didnt work lmao

i tried sqrt(3u+3v)

maybe instead of using the square root i could use fractions as the

uhh

coeficcient

[proud raven]

$log(a)+log(b)=log(a\cdot b)$

😄

[obsidian monolith]

jan Niku:

[harsh spoke]

ye i know that one

[pale cedar]

can you write the stuff inside the log as 10^number

[proud raven]

and know that 1000 has a prime factorization into 5's and 2's

so if you can write 1000 in terms of 5's and 2s

[harsh spoke]

but how does the square root work

[proud raven]

you can write log 1000 in terms of log5's and log2's

just put it under the radical, i think

[harsh spoke]

oh

[proud raven]

its not a nice number, so i dont know how you would do it without that

[harsh spoke]

what

[proud raven]

unless there's a special way

[harsh spoke]

it turns into 3/2

[proud raven]

like this

[pale cedar]

yea thats correct

[harsh spoke]

thats like the answer

but i need it in terms of u and v

[pale cedar]

so what do you currently have?

[proud raven]

[spring thunder]

i mean you can write 3/2 (u+v)

but fuck that's useless

[pale cedar]

^

thats actually the answer ur lf i think

[harsh spoke]

it is but it worked LMAO

thank you all

[spring thunder]

stoopid webassign

[proud raven]

it should say

express your answer in terms of u's v's and 3/2's

[spring thunder]

express your answer in terms of u, v and numbers

[proud raven]

'express your answer'

[harsh spoke]

lol

and 3/2's

[young ridge]

hey guys how would I solve this inequality. 2x - 3 < x + 4 < 3x - 2

[willow bear]

solve 2x-3 < x+4, then solve x+4 < 3x-2, then take the intersection of the resulting solution sets

[young ridge]

Thank you @willow bear

[young ridge]

How do I solve this one: | x + 3 | = | 2x + 1|

[frozen needle]

split cases

[young ridge]

?

what does that mean xD

[sour plinth]

|x+3| = x+3 for x>-3 or -x-3 for x<-3

and |2x+1| = 2x+1 for x>-1/2 or -2x-1 for x<-1/2

i think it's easier to understand if you plot it

[frozen needle]

look at what happens on $]-\infty,-3],~[-3,-1/2],~[-1/2,+\infty[$ separately

[obsidian monolith]

Tuong:

[fiery wave]

Another way to see it if you don't like restricting x values like Tuong suggests (but still a good way)
You can consider the 4 equations:

1. x+3 = 2x+1
2. x+3 = -(2x+1)
3. -(x+3) = 2x+1
4. -(x+3) = -(2x+1)

As you allow solutions to be different signs due to absolute values. Therefore you have all the possibilities up there^
But you should notice equations 1 and 4 are the same, and
equations 2 and 3 are the same too

So you should find all possible solutions by solving the 2 (remaining) equations

[young ridge]

Thank you guys. I understand now

[frozen needle]

🍻

[serene heath]

just sketch it tbh

[rugged ice]

Having a hard time finding the missing variable for this parabola related question, someone mind?

A company produces headlights for cars which are parabolic. If the equation for the headlights is 220xy, how far from the vertex should the bulb be placed?

Yikes that didn't format correctly

the equation is x^2 = 20y

[mystic dagger]

How should I go about this?

[rocky bison]

oof

okei

So first bit of mega useful info

dividing by a fraction is the same as multiplying by it's reciprocal

The reciprocal of a fraction is the same fraction just turned upside down

@mystic dagger

So that'll take us from that

To

[mystic dagger]

ok ill try do that thank you

[rocky bison]

,$ \frac{14c^3d}{3ab^2}\cdot\frac{6b^3}{7c^5d^4}

[obsidian monolith]

Pseudo:

[mystic dagger]

so thats the same as dividing or same equation from above @rocky bison

[rocky bison]

Yes

But instead of dividing by the fraction we're multiplying by it's reciprocate

Because that's easier

[mystic dagger]

ok thank you

[rocky bison]

Got it from there?

[mystic dagger]

yea

[rocky bison]

Awesome

[mystic dagger]

@rocky bison just would have to simplify and multiply across correct?

[rocky bison]

Yep

Just multiply top by top

bottom by bottom

Then just work by cancelling terms out a bit at a time

[mystic dagger]

ok got it

[silk sequoia]

why is this C and not B?

i tried to figure this out by coming up with all the possibilities by hand and only came up with 6

2314 2341 1234 4231 1423 4123

[rugged ice]

How do I find the direction angle of a vector?

[short sorrel]

@silk sequoia B is correct

whatever answer key you're using is wrong

maybe they meant, "if the digit 2 and digit 3 are always adjacent" or something?

@rugged ice draw the vector as a right triangle, and consider what trig ratio allows you to find the angle

this can be generalized

[rugged ice]

I don’t have any way to visually represent it

[short sorrel]

what's the vector in question

?

[rugged ice]

v = i - 2j

[short sorrel]

oh whoops

so this

that should be i - 2j in the bottom-right

the length of the left-right vector i is 1

the length of the up-down vector -2j is -2

so with some trigonometry

$\tan \theta = \frac{-2}{1}$

[obsidian monolith]

Namington:

[short sorrel]

solve for theta

[rugged ice]

Ah I thought it was i/j

that's why I had it wrong, lol

[short sorrel]

other way around

[rugged ice]

thanks

[grizzled orchid]

can't believe i need to ask this but: how can i rewrite $\frac{\sin ax}{\sin bx}$ in terms of $\frac{\sin x}{x}$

[obsidian monolith]

hegel:

[serene heath]

use a sub

oh theres a sin at the bottom

[grizzled orchid]

ye

[serene heath]

multiply by x/x

then

[grizzled orchid]

ah

[serene heath]

then it should be fine

[grizzled orchid]

hmm

ngl im gonna wolfram alpha this

[serene heath]

is this for a limit

[grizzled orchid]

yes

[serene heath]

oof

that's nasty

[grizzled orchid]

rewrite in terms of $\alpha$ where $\alpha = \frac{\sin x}{x}$

[serene heath]

x to 0 yea?

[obsidian monolith]

hegel:

[grizzled orchid]

ye

sorry

where $\alpha = \lim_{x \to 0} \frac{\sin x}{x}$

[obsidian monolith]

hegel:

[serene heath]

yea

I would split it up

and expand that cos

cos is bad here

[grizzled orchid]

smh its just 1 ree

[serene heath]

lol

[grizzled orchid]

oh god are you shitting me

this can't be right

$\cos (2bx) = 2 \sin (\frac{\pi}{4} - bx) \sin (\frac{\pi}{4} + bx)$

🤢

[obsidian monolith]

hegel:

[serene heath]

erm

y tho

[grizzled orchid]

i wolfram'd it

[serene heath]

double angle

writing it in terms of sin should help

[grizzled orchid]

[serene heath]

NO CRYING IN MATHS

nvm I'm crying too

cos(2bx)=1-2sin ^2(bx)

this helps alot

[grizzled orchid]

yeah i kn

i ddi that

[serene heath]

so what u got now

[grizzled orchid]

im at $\frac{\sin (ax) \sin (bx)}{\sin^2 (bx)}$

[obsidian monolith]

hegel:

[grizzled orchid]

this limit is grossss

[serene heath]

yea so that simplifies a bit

to sinax/sinbx

u had that

[grizzled orchid]

...

yeah

[serene heath]

yea so just multiply by x/x now

[grizzled orchid]

that's what i started with

[serene heath]

and split it up

[grizzled orchid]

ok

ah

[serene heath]

some juicy limit rules that allow such manipulation

[grizzled orchid]

heheh

im at $\frac{x \sin(bx)}{\sin^2 (bx)} \cdot \frac{\sin (ax)}{x}$

[obsidian monolith]

hegel:

[serene heath]

yup

simplify the right term a bit

then u can take the limit of the 2 terms separately

[grizzled orchid]

[serene heath]

*left term

not right

[grizzled orchid]

oh ok

[serene heath]

yea WA gets too excited sometimes

[grizzled orchid]

lol

i guess that's just x csc bx

or x / sin x

[serene heath]

yea

[grizzled orchid]

*sin bx

[serene heath]

so what's that limit

as x goes to 0

[grizzled orchid]

how is this in terms of sin x / x?

[serene heath]

x=(1/x)^-1

no?

[grizzled orchid]

: o

: O

[serene heath]

cheeky manipulation

[grizzled orchid]

hmm

but we still have all our sins as sin (ax) or sin(bx)

how do we get those a and b out?

[serene heath]

that's easily fixable

[grizzled orchid]

maybe im just dumb

[serene heath]

so u happy that we now got (sinbx/x)^-1 * sinax/x?

x goes 0

[grizzled orchid]

ye

[serene heath]

ok to deal with the a and bs

we just introduce a sub

[grizzled orchid]

: o

what do we say?

[serene heath]

for example for the sinax/x

if we let u=ax

x goes 0 so does u right?

[grizzled orchid]

yeh

ahhhh

i think i see

[serene heath]

yea?

that will allow you to right it interms of alpha

[grizzled orchid]

$\left. \frac{\sin (bx)}{x} .\right ^{-1} \cdot \frac{a \sin(u)}{u}$

uh

[obsidian monolith]

hegel:
Compile Error! Click the reaction for details. (You may edit your message)

[grizzled orchid]

uh ok

whatever u get the idea

[serene heath]

use round brackets

but yea I get u

[grizzled orchid]

and we can say that $\lim_{x \to 0} \frac{\sin x}{x} = \lim_{u \to 0} \frac{\sin u}{u}$

[obsidian monolith]

hegel:

[grizzled orchid]

right?

[serene heath]

yes

x is just a dummy var here

[grizzled orchid]

yu

*ye

doesn't x/(sin(bx)) = 1?

as x approaches zero

[serene heath]

not quite

[grizzled orchid]

hmm

[serene heath]

what does sinbx/x approach

[grizzled orchid]

...b?

i think

i kinda mentally lhopped it

[serene heath]

yea

it does

[grizzled orchid]

oh so it's 1 /b

[serene heath]

yer

yee

[grizzled orchid]

oh so the answer is

$\frac{a \cdot \alpha}{b}$

[obsidian monolith]

hegel:

[serene heath]

erm

should be an alpha from the sinbx term as well

[grizzled orchid]

why?

[serene heath]

(sinbx/x)^-1=(bsinu/u)^-1=(b alpha)^-1

[grizzled orchid]

oh

yah

ok

[serene heath]

hey cancel

so it diesnt even matter

[grizzled orchid]

spivak problems

[serene heath]

is this from spivak?

which chapter u doin

[grizzled orchid]

chapter 5

limits

[serene heath]

nice

did like 3 chaps and got bored

[grizzled orchid]

[cinder glade]

Is this right or is it 4pi

[shrewd urchin]

The sum of two force at a point is 16N. If the R[resultant] is normal to the smaller force & has value 8N find the two forces ?

R is normal to smaller forces means its perpendicular

R is perpendicular to small F ?

??????

anyone ?

[viscid thistle]

F1+f2=16N

[shrewd urchin]

i know that

[serene heath]

yes normal means perpendicular

[blazing parrot]

Try drawing a diagram. If you say there is a point O and that the resultant is OR, f1=OF, then f2=RF

then you know angle ROF=90 or ROF=pi/2, whichever u prefer

Now relate the 3 sides and solve the simultaneous equations

[shrewd urchin]

A body moving with speed 20m/s turns by 60 degree and continues with 20m/s. find the change in velocity?

I drew a diagram used the formula for resultant but got the wrong answer

$|R^2|=20^2+20^2-2 \cdot 20^2 \cdot cos(120)$

[obsidian monolith]

Krishna (An average mathy):

[shrewd urchin]

$R=20\sqrt{3}$

[obsidian monolith]

Krishna (An average mathy):

[shrewd urchin]

but the answer is 20

What went wrong

???????????????

[jolly silo]

huh

what is your question

[shrewd urchin]

scoll up

[jolly silo]

hmmm

The sum of two force at a point is 16N. If the R[resultant] is normal to the smaller force & has value 8N find the two forces ?
R is normal to smaller forces means its perpendicular
R is perpendicular to small F ?
??????

?

this is it

[shrewd urchin]

No not that boi

I done it a long ago

[jolly silo]

A body moving with speed 20m/s turns by 60 degree and continues with 20m/s. find the change in velocity?
I drew a diagram used the formula for resultant but got the wrong answer

[shrewd urchin]

A body moving with speed 20m/s turns by 60 degree and continues with 20m/s. find the change in velocity?
I drew a diagram used the formula for resultant but got the wrong answer
$|R^2|=20^2+20^2-2 \cdot 20^2 \cdot cos(120)$

[obsidian monolith]

Krishna (An average mathy):

[shrewd urchin]

YES

Wait i realised my mistake .

Change in velocity = v_2-v_1 not v_1-v_2

dumb me

No wait reeeeeee3

????

I mean it doesn't even mention directions

Does the question has less information

I can't figure out what's wrong

<@&286206848099549185>

[wraith idol]

[shrewd urchin]

Heck u . If you wanna help then help don't troll

[wraith idol]

Also \ before the cos

xD

[late pewter]

Can't you just subtract the vectors

[shrewd urchin]

Yes

But the answer I get is not correct

[late pewter]

What's the answer?

The correct one

[shrewd urchin]

20

I got 20*sqrt(3)

[late pewter]

Shouldn't the answer be a vector

[shrewd urchin]

The question lacks information I think

[late pewter]

Not really

Just assume it turns 60 degrees clockwise

[shrewd urchin]

I k

What answer do you get

[late pewter]

I get a vector

If the initial velocity is 20 along the x direction

[late pewter]

Oh wait

Is it asking you for the change in velocity along the positive x direction?

[shrewd urchin]

Q lacks information I guess

[late pewter]

Since it changes in velocity in the y direction as well

Are u sure the answer they give is 20

[shrewd urchin]

Yes

[late pewter]

That doesn't seem right in any world

[shrewd urchin]

Is my answer correct ?

[late pewter]

I'm honestly not sure what you're doing

I would think it's 10

Why do you square the numbers

[shrewd urchin]

@late pewter

[late pewter]

Can't really read that

[late pewter]

What's R

[shrewd urchin]

Resultant

[late pewter]

Resultant of what

Resultant of subtracting the vectors?

Oh I see

It uses law of cosines to get length of the resultsnt

Yyah 20 should be correct

I wasn't understanding the question usually vectors arent expressed in purely magnitudes

[shrewd urchin]

Laws of cosine ?

[late pewter]

Isn't that where it comes from

[shrewd urchin]

It's a parallelogram law of vector addition

[late pewter]

Well that comes from law of cosines

What you wrote is exactly the law of cosines

[shrewd urchin]

Oh kay

[late pewter]

Anyways to explain it simplistically

Since I don't feel like doing calculations

[shrewd urchin]

They subtracted the first vecotr from the second (the one after making 60°)

[late pewter]

If you take 2 vectors

Make a triangle from them

[shrewd urchin]

Wait

[late pewter]

The 3rd side is the resultant

Right

[shrewd urchin]

I know all that

[late pewter]

Yes

[shrewd urchin]

Bruh

[late pewter]

So if the triangle is an isosceles

With 60 degrees

Isn't that equilateral

[shrewd urchin]

They subtracted the 1rst with the second whereas change in velocity is v_2-v_1

[late pewter]

Ignore that

If it's an equilateral triangle

The resultant is 20

Right

[shrewd urchin]

Why

Calculations doesn't reveal so

[late pewter]

You agreed with me above

The resultant is a triangle

Lemme look at your calculations

But just from that logic alone it should be 20

I'm pretty sure you should be using cos60

Instead of cos120

[shrewd urchin]

So the subtract the first from second right

[late pewter]

I mean in the first screenshot you sent

It shows you subtracted 220cos120

Should be cos60

You should look into the law of cosines if you want to understand the formula you're using

[shrewd urchin]

@willow bear sorry to ping but they they took cos\theta and not cos(180-\theta) because the change in velocity is final velocity-intital velocity

[willow bear]

why would it be 180° - θ

[shrewd urchin]

Because final velocity- intial velocity

[prisma marten]

Angles == velocity tho

!=*

[shrewd urchin]

Angeles are formed by the velocity vector tho tho

Because of initial v-f v then final v vector will be opposite

[late pewter]

The basis of the formula is the law of cosines there's no way 180-theta makes sense

[shrewd urchin]

And if fv-iv then the iv will be opposite

[viscid thistle]

whats the question here

[late pewter]

the vectors will be opposite thats correct

If you add them head to tail the angle they form is still theta

[shrewd urchin]

I finally got what u said

No I don't

We were asked vector a-vecor b

So that's why 180-\theta

[late pewter]

The angle is still theta

[shrewd urchin]

We have to join tail to tail

[late pewter]

Ok then u make a parallelogram

[shrewd urchin]

Yeah

[late pewter]

This is why I'm saying look at the law of cosines because you cant solve for the resultant vector with 180-pi

[shrewd urchin]

180-theta

[late pewter]

The law of cosines is the way of calculating the length of a third side using the other 2 and the angle between them

If you use the law of cosines with 180-theta, you're calculating the length of the long side joining the two ends of each vector

You want to calculate the blue vector not the red one

If you use 180-theta you're calculating the red one

[shrewd urchin]

I can calc it

[late pewter]

Yeah, but not using 180-theta

That gives you the length of the red vector

[shrewd urchin]

Reading my physics textbook brought laurels in my life lol

[viscid thistle]

,rotate 90

[viscid thistle]

[silk sequoia]

how do you do this problem??

[pale kettle]

Is there another angle in the picture with the same measure theta?

[silk sequoia]

no

[pale kettle]

are you sure

[silk sequoia]

wait

the angle opposite from theta

[pale kettle]

that's right

well that angle is in a nice triangle

if you can find the side lengths of that triangle, you can use the law of cosines to find theta

[silk sequoia]

i thought if you have 2 sides you need to know a 3rd angle to use law of cosines

or if we can find the third side on the bottom them im not sure how

[karmic topaz]

thats just the 2 roots of the functions

you can solve for that easy

[pale kettle]

you can find all 3 sides

[silk sequoia]

im not even sure how to find the length of the first 2 sides in the first place

trying to figure it out rn

[karmic topaz]

pythagorus

[silk sequoia]

that doesnt help

[karmic topaz]

just find the three points to get the distances

[silk sequoia]

the problem is that this graph doesnt give you exact values

you can tell that by when the graph intersects the y axis

[karmic topaz]

they give you the functions

[silk sequoia]

ok i see

is the answer E?

[karmic topaz]

no

[silk sequoia]

can someone explain how to do it

[pale kettle]

We've told you

[grizzled orchid]

: o

zoph can u ping me if you get really pissed off

i want it for bingo

[karmic topaz]

where a and b are perpendicular

if you found the height of the triangle, and the length of its base

just halve the base

[silk sequoia]

ahh arctan

mk i see now

ty

[karmic topaz]

the angle youll get is half of the actual angle

[silk sequoia]

ye

[viscid thistle]

3^(2n-4) -1

I mean

3^(2n-4) = 1

How do I solve for n?

[limber bone]

when is something ^ something =1

when that something is 0

the other something X

like the power it self

or u can say this

3^(2n-4) = 1 --> 3^(2n-4) = 3^0 ---> 2n-4 =0

n=2

[karmic topaz]

3^a = 1
means that a =0

dont just do it dude

[viscid thistle]

o shit

you opened my eyes

thank you

[karmic topaz]

you could use logarithms

if youve learned them

[viscid thistle]

In this area of study I'm not suppose to be using logs yet

so that first answer was bette

r

[karmic topaz]

kay

[viscid thistle]

thanks

[karmic topaz]

👍

[jolly silo]

this?

[young ridge]

How do I solve this: if sinx = 1/3 and secy = 5/4, where x and y lie between 0 and pie/2, evaluate sin(x+y)

[rocky bison]

Well first, what's sec?

[young ridge]

sec is 1/cos i think

hyp/adj i think

[rocky bison]

Ye so if sec(y)=5/4=1/cos(y)

What's cos(y)=?

[young ridge]

4/5

[rocky bison]

Now are you familiar with the double angle formula?

Or not double angle formula

[young ridge]

sin(x+y) = sinxcosy+cosxsiny?

[rocky bison]

Yep

[young ridge]

but how do i get cosy value

and siny

[rocky bison]

We've already got cos(y)

[young ridge]

cosx i mean

[prisma marten]

Draw a little triangle

With angle y, adj 4, hypotenuse 5

You can solve for opp

And do the same with the sinx

[rocky bison]

Yas

Tfw stolen

[prisma marten]

snipe

[young ridge]

what type of triangle? r-angle triangle?

[rocky bison]

Ofc

That's the only one where we can use soh-cah-toa

[prisma marten]

SOH CAH TOA, only works for r-angle

Oh

Counter snipe

Hehe

[young ridge]

so when solving for cosx i found the unknown side to be 3 (i hope) so is cosx = 3 then?

[willow bear]

i mean

disregarding the fact that you missed some parentheses

cos(x) is always between -1 and 1 for any angle

how can it ever be equal to 3

what are you doing rn

[young ridge]

yes u are a great help thanks ann. wheres the other guys

[willow bear]

can you show me the problem you're doing rn

i cbf to scroll up

[young ridge]

alright sec

How do I solve this: if sin(x) = 1/3 and sec(y) = 5/4, where x and y lie between 0 and pie/2, evaluate sin(x+y). im at the stage where im trying to find cos(x) and sin(y)

[willow bear]

pi

not pie

anyway, ok

[young ridge]

lol Jesus

[willow bear]

so can you show the triangle you made for x

and what side in it you found to be 3

[young ridge]

crap how am i gonna do that

let me paint

use paint

[willow bear]

did you do it on paper?

can't you take a picture and upload it here lol

[young ridge]

http://prntscr.com/omf38o

[willow bear]

ok, wait

[young ridge]

did i do it right

[willow bear]

yes, this is the correct triangle for y

but that's y, not x

[young ridge]

oh i need to make a triangle for x?

[willow bear]

i mean duh how else are you gonna find cos(x) given sin(x)

you're gonna need a triangle for x too

[young ridge]

thank you annalicious

i got it

[willow bear]

please don't call me annalicious

[prisma marten]

[young ridge]

sorry

[serene heath]

lmao

annazing

[shrewd urchin]

How can I find $\vector{A}-\vector{B}$ when the direction are not given just magnitude

[obsidian monolith]

Krishna (An average mathy):
Compile Error! Click the reaction for details. (You may edit your message)

[shrewd urchin]

I think the answer is -1m

Because I think it's the only value that difference of vectors A and vectors B can attain.

[pale kettle]

Can the magnitude of a vector be -1?

[shrewd urchin]

Bog woops

Nope

It's 1m

dumb me

Am I correct ? @pale kettle

[pale kettle]

1 is possible

there are other possible ones

[shrewd urchin]

Let me think

If the vectors were parallel then the magnitude will be 1 and if they were anti parallel then 7.

[pale kettle]

That's true

[shrewd urchin]

Yeah so these are the two values I can think of

Does there exist anymore?

[pale kettle]

yes

[brave topaz]

@exotic tide remember that vectors don't have to just be parallel (0 degree difference in vector angle) or anti-parallel (180 degree difference in vector angle). Also remember that the difference vector connects the end points of the two vectors being subtracted from each other, forming a certain geometric shape. Try visualizing the A and B vectors connected to a single origin point, and visualize them parallel, anti-parallel, perpendicular, and all sorts of angles in between, and think about what that means about the possible values of the difference vector's magnitude. You have already found two values, so see if this visualization helps you find any others

[shrewd urchin]

So the difference of the vectors can take on any value between 1 and 7 inclusive

[brave topaz]

Yes! Nice job! Now what other answer choice/choices does that lead you to pick?

[rugged ice]

A hot-air balloon is moving on a bearing of 291 degrees at 47mph. A wind is blowing with the bearing of 271 degrees at 14mph. Find the component form of the velocity of the balloon.

I don't know how to find velocity with a shift, can someone help?

Oh, I just add the value of the wind to the plane.

Alright, never mind

[rugged ice]

I have no idea what this is asking

[forest canopy]

isn't it the distance between the 2 points ?

[charred hull]

it's asking you to find the magnitude of the vector $\vec{GF}$, which would just be distance formula here essentially

[obsidian monolith]

çölórôdòrōdõ brøwn stæìn:
Compile Error! Click the reaction for details. (You may edit your message)

[shrewd urchin]

U are asked to find the magnitude of the vector

What are the direction cosines good for ?

in physics

$\cap{i}$

[obsidian monolith]

Krishna (An average mathy):

[shrewd urchin]

how do i denote a unit vector using latex

[willow bear]

$\hat{i}$

[obsidian monolith]

Ann:

[willow bear]

$\hat{\imath}, \hat{\jmath}, \hat{k}$

[obsidian monolith]

Ann:

[willow bear]

\imath and \jmath give you i and j but without the dot

[shrewd urchin]

Ok

$\overrightarrowhead{a}=acos(\alpha)\hat{\i]+acos(\gamma)\hat{\j}+acos(\phi)\hat{\k}$

whats wrong ?

[knotty spear]

\hat{\i]

imbalanced brackets

also \i

[obsidian monolith]

Krishna (An average mathy):
Compile Error! Click the reaction for details. (You may edit your message)

[knotty spear]

$\vec{a}=\acos(\alpha)\hat{i}+\acos(\gamma)\hat{j}+\acos(\phi)\hat{k}$

[shrewd urchin]

lol

[obsidian monolith]

BewareillEatYou:

[shrewd urchin]

[knotty spear]

[shrewd urchin]

Tank u

[willow bear]

alpha, gamma and phi

weird choice of letters

[hushed ember]

hello

is anyone willing to help me with some questions

[hard hornet]

[hushed ember]

ok let me send in the questions

[hard hornet]

and to clarify, we help, we don't give free answers

one question at a time

[hushed ember]

oh the question is

find the value of a b and c for the following polynomials using appropriate methods.

[hard hornet]

question b first?

[hushed ember]

yeah

[hard hornet]

what topic of math are you doing right now?

[hushed ember]

polynomials

[hard hornet]

it looks like questions that ask you to compare terms from both sides

$a(x-2) + b(x+3) = 5$

[obsidian monolith]

MemesPlease:

[hard hornet]

do we have to figure out x too?

[hushed ember]

it only says find the value of a b and c

[hard hornet]

o actually doesnt matter

[hushed ember]

im just confused what to do because theres only 5 on the other side

[hard hornet]

ok so the first question lemme c

i thought this question is a bit weird too

[hushed ember]

oh its not a equal sign its the three bar sign

[hard hornet]

but im just going to assume it's going to ask you to compare terms on both sides

so first question

i would recommend distributing everything, and comparing terms

let me know what you get when you distribute everything

im too lazy to look at your work LOL

[hushed ember]

alright

$ ax+bx+3b-2a=5$

[obsidian monolith]

woa:

[hard hornet]

correct

and now, we are going to do some comparing

on the left side, you can see that there are x terms, but on the right side, there are NO x terms

so

[hushed ember]

yeah

[hard hornet]

what is the first equation you can set up

its all about figuring out what is equal, based on what terms you have

[hushed ember]

i have no clue

[hard hornet]

ill start the first equatoin for you

ax + bx = 0

why is this true you ask?

because the right hand side has no x terms

[hushed ember]

oh because no x values

[hard hornet]

thus

[hushed ember]

ahhh

[hard hornet]

any x values in the left must be equal to any x values in the right

now set up the second equation

[hushed ember]

3b-2a=5?

[hard hornet]

correct

by finding the right values for (a,b), any x value, and i mean LITERALLY any x value would work with (a,b)

[hushed ember]

ohh do i do substition now

[hard hornet]

substitution, combination, whatever method you're most comfortable solving systems

Check your answer at your own discretion: ||(-1,1)||

[hushed ember]

ok is it ax+(5x+2ax/3x)=0

[hard hornet]

actually

[hushed ember]

i mean 3ax+5x over 3x =0

[hard hornet]

u know in the equation ax+bx = 0

[hushed ember]

wait that meakes no sense

ye

[hard hornet]

you can actually ignore the x's, and only take care of the coefficients

so you really get

a + b = 0

because we know the x terms have to be equal to the x terms

so we can take their coefficients

its like saying 100y + 1245246y = 200y, we dont relaly care what the y value is

we only care about the coefficients

tho please ignore that equation above, its complete nonsense

[hushed ember]

ohh

hold on

that is very strange

if a+b=0

3b-2a does not equal 5