help-10

[unborn canyon]

i looked at the answer key

[tardy epoch]

expand (1+h)^6

[unborn canyon]

but i dont wanna

whats the fast way to do it

or is that the only way

[tardy epoch]

write it as a derivative

lots of ways

whether or not you know the techniques is up to you

[unborn canyon]

so just use the limit defiinition of the derivative?

[warm canopy]

It could be sideways 8?

[tardy epoch]

yes. usually you should use the definitions

[tardy epoch]

lmao i didn't notice that at first. silly word doc

[unborn canyon]

ok

thanks

x^6 is the f(x) and 6x^5 is the derivative bc power rule right

or am i being dumb

[tardy epoch]

,w differentiate x^6

[warm shale]

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[unborn canyon]

yea im right

[tardy epoch]

[unborn canyon]

im so smart

but hows the answer 6

im still not following that

[stable rain]

woohoo

[tardy epoch]

find an $a$ such that it's equal to $f'(a)$

[warm shale]

riemann

[unborn canyon]

im so dumb 😭

i stlil dont get it

[tardy epoch]

also use this

[unborn canyon]

thts how i found the x^6

[warm canopy]

But there are no x's in the limit

[tardy epoch]

write your fraction as the fraction inside the limit

with your f(x) = x^6

[unborn canyon]

im failing this exam

[thick roost]

🤨

[unborn canyon]

(this is practice sheet)

[shadow oriole]

That's the wrong way to do the question. If you plug in 0, you get 0/0, which is an indeterminate form. Therefore you're going to need to apply l'Hopitals rule.

[tardy epoch]

no.

[unborn canyon]

what

[tardy epoch]

ignore them

[unborn canyon]

ik

[tardy epoch]

write this for f(x) = x^6

[unborn canyon]

brb

[tardy epoch]

as another hint: 1^6 = 1

[unborn canyon]

i give up

im expanding it

o

i got it now

thanks

.close

[obtuse pebble]

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[tardy epoch]

that was fast expansion

[obtuse pebble]

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[mild slate]

help

[warm shale]

cyanometer

[mild slate]

I would like help verifying the identity.

[fierce lagoon]

Break tan^2(x) into a fraction

[mild slate]

yeah, let me show you where I am at now.

[warm canopy]

Probs better to move sin to left and tan to right

[fierce lagoon]

They're trying to prove that the LHS = RHS

not solve for x

[warm canopy]

You do what I said and you can use pythagoras

Without having to mess around

[mild slate]

I feel like I might be making this more difficult than it is now.

$((1+cos2x)/2)-((1-cos2x)/1+cos2x)$

[warm shale]

cyanometer

[mild slate]

That looks bad but I canceled out the 1+cos2x.

I have no idea how to turn the lhs into the rhs using identities.

[fierce lagoon]

Actually

It's a lot easier to turn the RHS into LHS

2 = 1 + 1

Literally no joke

[mild slate]

hmm. let me try it out.

[fierce lagoon]

$$(1 - \sin^2(x)) + (1 - \sec^2(x))$$

[warm shale]

Umbraleviathan

[warm canopy]

The trick on the other side is 0=1-1

[fierce lagoon]

You're 1-1

[warm canopy]

Take it back

[mild slate]

lol

[fierce lagoon]

No, I won't

Fight me bitch

You're 1-1

https://tenor.com/view/squidward-spongebob-meme-dab-gif-24208634

[mild slate]

oh shit

[warm canopy]

@moderators

[mild slate]

I'm still lost. I'm missing some fundamental knowledge about how this works.

.close

[obtuse pebble]

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[placid nova]

why is my Y not 0? if x^2 + y^2 = Z^2

[placid nova]

and my Z which 7.5/cos(pi/3) is 7.5

so isnt Y = z-x

[timber fox]

how would y be 0 if there is angle theta?

[high lily]

and my Z which 7.5/cos(pi/3) is 7.5
how are you getting 7.5 for z from that

[timber fox]

lol

I see the problem

put your calculator in radian mode

and from now on always work in radians and make your life easy :))

@placid nova what do you get for z now

[placid nova]

oh

LOLL

[timber fox]

degree * pi/180 to convert from degrees to rad

otherway around for rad to degree

[high lily]

or don't use a calc for this at all

[timber fox]

true

[obtuse pebble]

@placid nova Has your question been resolved?

[obtuse pebble]

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[forest epoch]

whatup

[forest epoch]

how is what i did wrong?

[obtuse pebble]

@forest epoch Has your question been resolved?

[obtuse pebble]

@forest epoch Has your question been resolved?

[obtuse pebble]

@forest epoch Has your question been resolved?

[obtuse pebble]

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[primal idol]

How do i sketch #22?

I absolutely have no idea of how I should do it.

There's not one single example in my textbook.

[fierce lagoon]

Mfw spiral

[obtuse pebble]

@primal idol Has your question been resolved?

[primal idol]

Thanks..

I'll try

[obtuse pebble]

@primal idol Has your question been resolved?

[obtuse pebble]

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[timid silo]

<@&286206848099549185>

[zenith raft]

:c

[timid silo]

How would I figure out the function for this problem?

[zenith raft]

try not to tag helpers too early 😭

[timid silo]

:||

[hazy vine]

by looking at the table it is possible to see that this is a linear equation of form mx+b

do you get that?

[timid silo]

Nope

[solar hornet]

I guess in the context of this assignment you are talking about linear functions only?

[hazy vine]

ok so here you can see that every times x increments by 1 y decreases by 3

[timid silo]

Ohhhh

[hazy vine]

oh fair

[timid silo]

I’m not sure on how to write out the formula

[zenith raft]

it just says “a function” so maybe anything is ok but linear is the expected route

[timid silo]

Y= x+3?

[hazy vine]

alright I'll guess you never saw the formula to find the slope

[hazy vine]

nop

[timid silo]

Damn

[solar hornet]

aight. I wanted to give something more general, but it's usually best to consider the context of what that assignment is

maybe teacher expects a system of equations with standard formula for linear

[hazy vine]

so here lets start by finding m

so can you agree with me that for each x going up by 1 there's 3 y going down

[timid silo]

Yes I agree

And I see that

[hazy vine]

ok here so what would you multiply x with for it to match y decreasing every time

[timid silo]

Would it be a -3

[hazy vine]

yep

so this is what is commonly called the slope

and in our y=mx+b function would be the m

[timid silo]

1?

[hazy vine]

what is 1?

[timid silo]

Y=1x+-3

[hazy vine]

nop

[timid silo]

Oof

[hazy vine]

mb that was rude lol

[timid silo]

It’s okay lol

[hazy vine]

but basically you multiply x by -3 right?

[timid silo]

Yes

[hazy vine]

so it would be the m

[timid silo]

X

[hazy vine]

so we would have -3x+b

[timid silo]

Oh

[hazy vine]

now

the b would be the value of y when x =0

this is because suppose when x =0, y=-3*0+b

so y=b when x=0

[timid silo]

I’m slowly understanding

[hazy vine]

oh are you struggling with something that I've said before?

[timid silo]

No , I just gotta practice a bit more so I can really begin to understand

thank you so much 😊

[hazy vine]

alright well

let's finish this by finding what would be the value when x=0?

[timid silo]

B

[hazy vine]

yes

[timid silo]

Yay

[hazy vine]

but do it step by step

like

x=4 y=9, x=3 y=12...

until you get to x=0

[timid silo]

Okay

Interesting

[obtuse pebble]

@timid silo Has your question been resolved?

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[finite loom]

does anyone know the formula for finding a certain term of binomial expansion

[finite loom]

for example (x+y)^8

is the formula like

n choose k?

where k is 1 less than the 3rd number

and then (x)^blank and (y)^blank

where u would find blank by

thats where im confused how du u find blank 😭

ohhh wait

is it that the 7 term just equals what k is

[civic zealot]

,w binomial theorem

[warm shale]

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[finite loom]

i see

but

isnt that a diff formula frm the term

[civic zealot]

the sum on the right is the sum of the terms

starting with the first term at k=0

second term is k=1
third is k=2
.... etc

[finite loom]

yep

but im trying to find like

a certain term

not the sum

3rd term of (x+y)^8

[timid silo]

the (r+1)th term will be (n_C_r)x^(n-r)y^r

[civic zealot]

if you don't use the sum, those are the terms

every term of the binomial expansion is nCk * x^k * y^(n-k)

[warm shale]

What the hell am I doing here?

[finite loom]

so thats the formula

[pine sail]

This is term number k+1

[finite loom]

wait does

(a+b)^n = (n choose k) (a^(n-k)(b^k) work

[pine sail]

You're missing a sum symbol

[finite loom]

Why do i need a sum symbol if im trying to find a term 😭

[pine sail]

(a+b)^n is the sum of all such terms

[zenith raft]

the binomial theorem tells you all the terms, so if you want one in particular, it tells you it

[pine sail]

(a+b)^n isn't just one term

[finite loom]

yeah

but u can get 1 term out of it

right?

if u plug in the right numbers

like

[pine sail]

yes, like I mentioned above.

[finite loom]

wait so why do i need the sum symbol

[pine sail]

You only need the sum symbol if you need (a+b)^n

You don't need the summation, if you need one term out of (a+b)^n

But you wouldn't write (a+b)^n = that particular term

[civic zealot]

you don't. The theorem gives you the sum of the general formula for the terms
so each term of the binomial sum is given
if you just don't use the sum, and use the term formula shown you have the formula for term you want

[pine sail]

Rather (a+b)^n is the sum of all such terms.

[finite loom]

oh

so i would write'

term = (n choose k)(a^(n-k)(b^k)?

[pine sail]

yeah

[finite loom]

ah

okay so n represents

[pine sail]

Also, it matters which term you're talking about

[finite loom]

the number of term right

like the expansion term

in (x+y)^8 n = 8

[pine sail]

yeah.

If you need term number 5 then k = 4

Term number 9 would be k = 8.

You get it, right?

[finite loom]

yee i learnt that k is always 1 less than the term u need

[pine sail]

n doesn't represent the number of terms btw.

In the expansion of (x+y)^n you'll actually have n+1 terms. Not n terms.

[finite loom]

ok wth

Why is it N+1

i feel like sequences/series is trying to kill me

[pine sail]

The theorem tells us.
You've known it all along. (x+y)^2 isn't x^2 + y^2 is it? It's x^2 + 2xy + y^2

For n = 2, you've always used 3 terms. And now you're asking why is it n+1, strange.

[finite loom]

o

okay yeah i knew n doesnt represent the number of terms

but n represents

the expansion number right

in (x+y)^8 n =8

its always the expansion number

[pine sail]

Yep.

[finite loom]

aight

okay so

k = 2

because 3 is the term

but - 1 so 2

so

8 choose 2

x^6

y^2

ah i see

[pine sail]

sure.

[finite loom]

I had one more question 🙂

So

They want me to find the coeffecient which is a

of the term z^6

[pine sail]

let z^2 = x and -1 = y
z^6 = x^3
So you need the term when you have x^3
The Exponent of x is given by n-k. You know n already, so get x.

Then later, sub the values back.

[finite loom]

okay i understand z^2 = x

and -1 = y

im confused why ur substituting tho

is there a diff way

i feel like thats too complicated for me

[pine sail]

Yes you don't need any x or y. But either way you need to figure "k" out.

Since once you know that, you're essentially done.

[finite loom]

wait what formula am i using

for this problem

[pine sail]

You tell me?

[finite loom]

(n choose k) (a^(n-k)(b^k)?

[pine sail]

yep.

But you need k.

As you can see, z^2 is your "a"

And -1 is "b"

It's just comparison.

So that we know what terms to use, right.

[finite loom]

yes

i understood making it a and b

so does n equal 12 still

yeh

n is always that number

wait isnt k

just

the value of the b term exponent

ok nvm

excuse my dumbness

thats literally in the formula

umh

soo

so how do we find k lmao

[pine sail]

I told you how already.

[warm shale]

What the hell am I doing here?

[finite loom]

ahh im looking throug all the msgs

OH WAIT

BRO IM STUPID

okay so

the power of z

has to be equal to 6 right

so that must mean

it has to be 3

because theres already a square

okay so

n = 12

12 - k = 3

k = 9?

right?

[pine sail]

yep. Sure.

[finite loom]

Tysm

ahh i have an exam tmrw but i just am failing all this stuff somehow

.close

[obtuse pebble]

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[next reef]

I suppose it's the AM GM Inequality

[tender jewel]

what that

[zenith raft]

i think i see a simple proof by contradiction

[quaint glen]

either a>b or a<=b

Well we've had 3 different methods suggested which is probably confusing so I'll leave this one lol

[tender jewel]

its contradiction

[tender jewel]

can yo ueXPLAN IT PLS

[zenith raft]

lol

[tender jewel]

SORRY CAPS

sorry caps

[zenith raft]

honestly rather someone else explain so i can sleep, sorry :c

[tender jewel]

its all good

hey can anyone explain pls

[zenith raft]

@quaint glen wanna join again? 🙂

[thorn stag]

Assume it isnt, square both sides |a^2| /leq |ab| and divide by a. Do the same for the other side. a /leq b or b /leq a is true for all positive integers.

[tender jewel]

um whats /leq

[quaint glen]

$\leq$

[warm shale]

Sneaky

[tender jewel]

ohnok

[tender jewel]

oh i get it ok

but it wants me to use contradiactoin

[next reef]

It's a classic theorem in algebra that says arithmetic mean of n positive real numbers is greater than or equal to their geometric mean.

[thorn stag]

Yes then assume it isnt and prove that its a contradiction. If a and b are positive integers, then they can be equal or one can be greater than the other.

[zenith raft]

that argument is totally not valid

[tender jewel]

idk what to do tbh

[tender jewel]

can you be more specific plesae

[thorn stag]

$|a| \leq |b| \lor |b| \leq |a|$. Assume that this isnt true by taking the negation. $|a| > |b| \land |b| > |a|$. Let a=b so the negated inequality is a contradiction.

[warm shale]

SirMonkey

[zenith raft]

still no

but i’m going to bed fr this time

[tender jewel]

ok thanks i get it

[elfin dune]

Layla is right

[elfin dune]

this doesn't work

[tender jewel]

why doesit not work?

waht about this one thoug

[tender jewel]

this one

[elfin dune]

wait no

forget it

I'm sleepy, but trust Layla

[tender jewel]

but they went to sleep as well

😢

[novel knoll]

Lets prove a<=0 or b<=0 using his method: Assume it isn’t by taking the negation. a>0 and b>0. Let a=b=0 so the negated inequality is a contradiction.

Notice a problem?

[thorn stag]

Literally no

[tender jewel]

cus if a=b then the original euqation works

[thorn stag]

Yes but in the proof by contradiction, it doesnt work so the original statement is true.

[tender jewel]

yerah i was agreeing with u

[thorn stag]

Yeah ik im just referencing the other post

[thorn stag]

"Do you notice a problem with the proof by contradiction creating a contradiction?"

[novel knoll]

You haven’t shown anything just like I haven’t shown a<=0 or b<=0

[thorn stag]

Notice how in this example, a or b leq 0 while in the question a leq b or b leq a.

[obtuse pebble]

@tender jewel Has your question been resolved?

[novel knoll]

So prove a<=0. So multiply both sides by 0 so 0<=0

This is true so we done

[obtuse pebble]

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[novel knoll]

Don’t assume what you want to prove

[obtuse pebble]

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[obtuse pebble]

@half veldt Has your question been resolved?

[half veldt]

No

[obtuse pebble]

@half veldt Has your question been resolved?

[half veldt]

<@&286206848099549185>

[obtuse pebble]

@half veldt Has your question been resolved?

[devout jolt]

All you gotta do is work out the volume of the large cube(VLarge) and the volume of the small cube(VSmall) and subtract VSmall from VLarge

You have the height width and depth of the large cube, and the height width and depth of the small cube

(8x11x11)-(5x5x8) where x means multiply

[obtuse pebble]

@half veldt Has your question been resolved?

[obtuse pebble]

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[copper spade]

Won't this be vector C x vector A?

[cedar mantle]

yes

[copper spade]

Oki

[cedar mantle]

are those allen notes?

[copper spade]

Nope

Ncert

[cedar mantle]

oh lol

okie

[copper spade]

Hehe

.close

[obtuse pebble]

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[hybrid tartan]

for Q2(a)(v) what type of probability would it be, independent or dependent?

As i need to find P(AnB)

Some context on what im trying to do

<@&286206848099549185>

[obtuse pebble]

@hybrid tartan Has your question been resolved?

[obtuse pebble]

@hybrid tartan Has your question been resolved?

[obtuse pebble]

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[trail musk]

Let $a_0$ be any real number. Then define the sequence $a_n$ such that $a_{n+1} = \sqrt{a_n}$. It's easy to see that for any positive real, this quantity goes to 1, as it is an attractive fixed point.

Consider now the sequence $b$, with $0<b_0<1$, and $b_{n+1} = (b_n)^2$. It's easy to see that this sequence always goes to 0.

What is the algebraic difference between the equations $x=\sqrt{x}$ and $x=x²$? Why is 1 the attractive fixed point for rooting but 0 the attractive fixed point for squaring? My guess is due to the concavity of the functions but I'm not too sure how to make this a rigorous argument.

[warm shale]

Mr. Gamer

[trail musk]

Also a more general question: when a function has multiple fixed points, how do you determine which one is attractive over what interval?

[gilded needle]

sqrt has the property that it makes numbers greater than 1 smaller, and it makes numbers less than 1 bigger

squaring has the opposite property

[trail musk]

Yeah true. How do you determine which fixed point of a function is attractive tho?

[gilded needle]

well you can argue as follows

for square root, the property i noted above implies that if you start with a number less than 1, and repeatedly take square roots, you get an increasing sequence

that sequence is bounded above (by 1) so it converges

similarly if you start with a number greater than 1, and repeatedly take square roots, you get a decreasing sequence

that sequence is bounded below (by 1) so it also converges

so no matter where you start, you get a convergent sequence

to find the limit (call it $L$), start with $x_{n+1} = \sqrt{x_n}$ and take the limit of both sides as $n \to \infty$ to obtain $L = \sqrt{L}$ (we have used continuity of sqrt)

[warm shale]

Bungo

[gilded needle]

squaring both sides yields $L^2 = L$, or $L^2 - L = 0$, or $L(L-1) = 0$, hence $L$ is either 0 or 1

[warm shale]

Bungo

[gilded needle]

it can't be 0 because of the previously observed properties about increasing and decreasing sequences

so L=1

you can make a similar argument for squaring

the difference being that in the case of squaring, if you start with a number larger than 1 and repeatedly square it, you get an increasing unbounded sequence, so that doesn't converge

whereas if you start with a number less than 1 then you get a decreasing sequence, which you can show converges to 0 using the same method i used above

[trail musk]

Right, that's a good way to reason about it. But consider for a moment a process more general than squaring or rooting. Let's say we have a function with some number of fixed points. How do you determine which of those are attractive over what intervals?

[gilded needle]

well the banach fixed point theorem is one tool

in general i'm not sure if there is a single universally applicable method that will tell you all the fixed points

[trail musk]

I don't think you quite understand what I'm asking here

If the fixed points of a function are given, then which of those are attractive over what intervals?

Wait actually

I think the Banach theorem might be what I'm looking for?

Let me read through it lol

[gilded needle]

in general i think the behavior can be quite complicated. check out the wikipedia page for "attractor / basins of attraction", see this example in particular:

https://en.wikipedia.org/wiki/Attractor#Basins_of_attraction

[trail musk]

Woah

Ok yeah this looks fun

Strange how I can't find much on it though, maybe I'm not using the right terminology

[gilded needle]

yeah i'm not sure about what terminology to search for either, i don't know much about dynamical systems

i think the concept you're after is "basin of attraction of a fixed point", searching for that shows some interesting MSE posts at least

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

how

[timid silo]

we

pass

from

-cot2x

to

tan

to

tan(-pi/2+2x)

from -cot2x to tan(-pi/2+2x)

$-cot2x$ to $tan(-pi/2+2x)$

[warm shale]

Kizima

[timid silo]

how we do this

pls

[fierce lagoon]

-cot(2x) or -cot^2(x)?

[timid silo]

-cot2x

[fierce lagoon]

Try using the tan(2x) identity

cot(2x) is the reciprocal of tan(2x)

[timid silo]

u mean 2tanx/1-tan²x ?

[fierce lagoon]

Well whatever the identity got tan(2x) is

[fierce lagoon]

Oh yeah, but then raise that to the -1 power since cot(2x) = 1/tan(2x)

[timid silo]

what

@fierce lagoon wdy

wdym

[fierce lagoon]

I hope to god you know how to raise things to the -1 power

[timid silo]

i dont understand english

[fierce lagoon]

Translate it then

[timid silo]

it didn't translate good

you mean to do ^-1 ?

[fierce lagoon]

Yes

cot(2x) = (tan(2x))^-1

[timid silo]

so its 1/tan2x

[fierce lagoon]

Yeah but use the tan(2x) identity

[timid silo]

1/2tanx/1-tan²x

[fierce lagoon]

Simplify that

[timid silo]

by doing sin/cos or

[fierce lagoon]

I gtg but hopefully someone can take over

[timid silo]

i found the answer

tan(2x-pi/2)=-sin(pi/2-2x)/cos(pi/2-2x)=-cos(2x)/sin(2x)=-cot(2x)

[fierce lagoon]

1/a/b = b/a

[timid silo]

without parentheses wouldnt that be (1/a)/b= 1/(ab)?

yep

,w 1/a/b

[warm shale]

Results provided by WolframAlpha

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

Requested by 13part#1101

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[hexed agate]

Divisor reciprocates to multiplier

[raven iris]

I have no idea what that means

[cold thistle]

You know about reciprocals of fractions?

[raven iris]

No im lost

[hexed agate]

https://youtu.be/4lkq3DgvmJo

[cold thistle]

A fraction's reciprocal is basically just itself upside down

Reciprocal of 4/3 is 3/4

5/18 and 18/5

So on

[raven iris]

Alr i got that

[cold thistle]

Dividing a number by some fraction is the same as multiplying the number by the reciprocal of the fraction

$2 \div \frac34 = 2 \times \frac43$

[warm shale]

lexitorius

[cold thistle]

Example ^

It even works with fractions: $\frac54 \div \frac67 = \frac54 \times \frac76$

[warm shale]

lexitorius

[raven iris]

So its like switching it aroujd?

[cold thistle]

Yup

If you want to switch the sign (from divide to times, or time to divide), you need to also switch one of the fractions

As such, your problem gives us 2 divided by 17/5

And then asks us to rewrite that as 2 times []/[]

2 time something over something

[obtuse pebble]

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[obtuse pebble]

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[blissful nimbus]

Anyone knows why -3+4i = 4i²+4i+1 ?? Is there a trick to it?? Because in the right form i can factor it and use the square root of it..

[pliant compass]

4i^2 is just -4

[narrow vault]

it's just completing the square, you halve and square 4i

[blissful nimbus]

ok but is there some sort of algorithm to come from -3+4i to 4i²+4i+1? or do i need to see something like that?

[pliant compass]

You’re talking about i as in the imaginary number right?

[blissful nimbus]

yes

[pliant compass]

You just have to know that it is the square root of -1

So squaring it gives -1

[dark mango]

complete the square

[dark mango]

.

[pliant compass]

Am I missing something

Why does he need to complete the square?

[dark mango]

thats how u get from left to right

[pliant compass]

Oh he’s going backwards

[dark mango]

which was his question

[pliant compass]

I see

Mb

[blissful nimbus]

ok is this using Gaussian Prime Factorizationß

?

[narrow vault]

it has nothing to do with gaussian primes

do you know the complete square technique to factoring quadratics? you're doing the same thing

[blissful nimbus]

I only know how to complete the square in the form of x²+bx+c . I dont have a square to begin with

i only have -3+4i

or do I just add 4i²on both sides?

[narrow vault]

well it's the same form, halving 4i then squaring it gets you the expression on the right

[blissful nimbus]

ahh now i got it. ty so much

how can I close the channel?

[narrow vault]

type .close

[blissful nimbus]

close

[narrow vault]

.

[blissful nimbus]

.close

[obtuse pebble]

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[obtuse pebble]

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[primal grove]

Hi

[primal grove]

I don't understand linear equations

And i need to study for an upcoming exam

[wary vigil]

the number of people seated is 120

and for example 4 seat tables take 4 away for every y

and 2 seat tables take 2 away for every x

so 120 = 2x + 4y

[primal grove]

Oh

[wary vigil]

but without more info idk if you can solve for either

you can express one in the other

but solving i think isn't possible without more info

[primal grove]

Ok thanks i will figure the rest out

.close

[obtuse pebble]

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[rapid relic]

can someone please help me with this

I got to

ln2sinx=3ln(cos2x+2)

This

Idk what to do next

[tardy epoch]

how'd you get to this? change of base formula?

[rapid relic]

yeah

[tardy epoch]

lots of options

one could be to use log(xy) = log(x) + log(y) and double angle for cosine

[rapid relic]

I'm not suppose to use a calculator also

[tardy epoch]

nothing i said involves a calculator

[rapid relic]

is the formula used above a viable formula too or should is the log x +log y one better

Sorry I just wanted to make sure I said that because the hw question is suppose to be solved without it

[tardy epoch]

you can get rid of logs in the first step by bringing the 3 into the exponent

[rapid relic]

3^2

[tardy epoch]

xlog(y) = log(y^x) then you set the arguments of the log to each other

[rapid relic]

so you're saying 2log3(2sinx)=log3(cos2x+2)

Wait is what is written correct then or did I accidentally mess up

[tardy epoch]

not at all

[tardy epoch]

i assume this is correct

[rapid relic]

I think it should be fine

I used change of base and stuff

[tardy epoch]

right hand side can be written as 3log(stuff) = log(stuff^3)

[tardy epoch]

application of this

[rapid relic]

log(2x+2)^3

Sorry I haven't used that log change rule before

[rapid relic]

is there a chance I can solve that problem with ln or is it easier to do with logs

[tardy epoch]

i use log for natural log

[rapid relic]

oh ok

ln2sinx=ln((cos2x+2)^3)

so is that what you mean then?

[tardy epoch]

yes

so you have if log(stuff1) = log(stuff2) then stuff1 = stuff2

[tardy epoch]

but since cubing cosine is excruciating, this is more likely to be simple

[rapid relic]

ok so ln(2sinx)=ln((cos2x)^3)+ln(2^3)

ad then would I be able to divide everything by ln

[tardy epoch]

nah

you have some algebra mistake somewhere

[rapid relic]

wait so what am I missing out on

[tardy epoch]

(x+y)^3 is not x^3 + y^3

,calc (1+1)^3

[warm shale]

Result:

8

[tardy epoch]

,calc 1^3 + 1^3

[warm shale]

Result:

2

[rapid relic]

ok so ln(2sinx)=ln(cos2x)^3+ln(2)^3

Wait mb is that correct

[tardy epoch]

$3\log(\cos(2x)+2) = \log((\cos(2x)+2)^3)$

[warm shale]

riemann

[tardy epoch]

write out (a+b)^3

[rapid relic]

oh so we have to do (a^3+3a^2*b+3ab^2+b^3)

mb

ln(2sinx)=ln(cos2x^3)+ln(3*cos2x^2*2)+ln(3*cos2x*2^2)+ln(2^3)

is that what we have to do

how would we expand cos2x^3

would it be cos8x

[obtuse pebble]

@rapid relic Has your question been resolved?

[obtuse pebble]

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[rotund crest]

Dicothomy algorithm i dont understand why if its 7 the number you receive when the algorithm goes on until the end without finding x is 3 etc

[obtuse pebble]

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[floral tendon]

How would I find the missing sides (visually)

[hybrid gull]

You can’t get the missing sides visually, you’ll need to use what the tick marks tell you

[smoky slate]

Each mark is x

[floral tendon]

Oh

Its that simple?

0-0

Thanks

.close

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[obtuse pebble]

Please don't occupy multiple help channels.

[wide aurora]

.close

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[lofty yacht]

i don't understand how to relate this one

the one equation i have is

1331 = xy

[cedar mantle]

okay

so whats the perimeter in terms of x and y

[lofty yacht]

2(x+y)?

[cedar mantle]

yeah

and we now substitute 1331/x for y

so now our function is f(x)=2(x+1331/x)

[lofty yacht]

ok

i'll try that

[cedar mantle]

and we want to minnimize the perimeter so we are finding the local minnimum of f(x)

[lofty yacht]

first derivative test?

[cedar mantle]

yeah

[lofty yacht]

can i multiply by x first?

or i can leave the 2 outside of the derivative

2*(1-1331/x^2)

that's my derivative

it's zero at 36.48

what now?

<@&286206848099549185>

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[broken rune]

please help

ive been trying different intervals but it never is right

<@&286206848099549185>

<@&286206848099549185>

pls help

anyone able to help

[obtuse pebble]

@broken rune Has your question been resolved?

[broken rune]

I unsderstood the right endppint is 2

but the left endpoint is incorrect

[obtuse pebble]

@broken rune Has your question been resolved?

[broken rune]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

i missed the 15 min intervals

but could you explain how to get left endopiint

[viral blade]

What did you put for left endpoint and why

[obtuse pebble]

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[gentle shard]

How do I sketch the graph of y=sinx+sqrt3 cosx

[obtuse pebble]

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[obtuse niche]

If we are given the vector field in the image, would taking the curl (which is zero) be enough to prove its conservative? Our book says that a vector field is conservative in star-shaped field if curl is zero, but that is not star-shaped since x or y cant be 0?

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[gentle shard]

How do I solve for x?

[pine sail]

If ab = 0 then either a or b or both are zero. In this case, 2e^x can't be zero. Meaning 1-2e^x = 0.
Solve from there.

[gentle shard]

.close

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[craggy stag]

can somebody explain how he got the formula here?

[craggy stag]

im watching an online lecture on series and he didnt quite xplain how he got the formula for summation n

[novel knoll]

Very standard result

Try proving it using induction for example

[craggy stag]

whats induction..?

[zenith raft]

there's a less mysterious to get it that might be better, especially if you don't know induction

[craggy stag]

how would one do that

[zenith raft]

i'll try my best here

we'll compute 2S_n:
1 + 2 + 3 + ... + (n-1) + n +
n + (n-1) + (n-2) + ... + 2 + 1

we just switched the order of the lists in the second row (each row is a copy of S_n)
now if you look at the sums vertically, they are
1 + n,
2 + (n-1),
3 + (n-2),
and so on

each of these are equal to n + 1

and there are n of them in total

so the whole sum is n(n+1)

and that's 2S_n

so S_n = n(n+1)/2

[craggy stag]

why is it 2S_n?

[zenith raft]

if we can compute that, we know S_n after

soooo why is it?

just cus it's something we can compute easier

[craggy stag]

but wouldnt that just be 2 + 4 + 6 ... and s on?

or like why wouldnt it be

[zenith raft]

yea it would

but it's also S_n + S_n

[craggy stag]

so, the 2S_n is more or less slapped in there for the sake of conveniency?

[zenith raft]

sure haha

[frozen vapor]

you could also do it with S_n, but the number of (n+1) terms would be equal to n/2 which just yields the same answer

[craggy stag]

how would the number of n+1 terms equal n/2?

[frozen vapor]

consider n = 5

1 + 2 + 3 + 4 + 5

(this is odd, even is easier to understand)

so keep the middle term, 3 aside for a while, you have 1 2 4 5

pair them up 1 + 5, 2+4

so there are two of these pairings

which can be understood that for n, there will be n-1/2 pairings (for 5, 5-1/2 = 2 pairs)

and each of these pairs sum to n+1, so the total pairings sum is (n+1)(n-1)/2

the middle term we kept aside is nothing but n+1/2

(n+1)(n-1)/2 + (n+1)/2

take n+1/2 common

which gives (n+1)/2 x (n-1+1)

which is the answer n(n+1)/2

for even terms, you dont need to deal with the extra middle term since the pairings nicely add upto n/2 number of terms

[craggy stag]

im a lils tuck on this part

i got to (n+1)(n-1)+(n+1)/2

[frozen vapor]

you're taking n+1/2 common so there will be n-1 in the first one and just 1 in the second

which sums upto n-1 + 1

[craggy stag]

(n+1)(n-1)+1/2?

[frozen vapor]

no i mean (n+1)/2 common

[craggy stag]

im not getting it

wdym common? i understand there 2 (n+1)'s but what exactly happens to them

[obtuse pebble]

@craggy stag Has your question been resolved?

[obtuse pebble]

@craggy stag Has your question been resolved?

[obtuse pebble]

@craggy stag Has your question been resolved?

[obtuse pebble]

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[steady notch]

hi

[steady notch]

i dont quite understand

the correct answer should be C meaning the number itself is a fraction… if say 1/3 is flipped how is it greater than four

it^

actually no not b. the only logical one would be C but i don’t understand how

[novel knoll]

But 1/3 is not in (0,1/4)

[steady notch]

wait

djdfjdjrjr you’re right

😭nvm

thanks for answering i forgot how fractions worked

.close

[obtuse pebble]

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[gritty carbon]

hey all

in need of some assistance with q19

[merry geode]

What have you tried?

[gritty carbon]

well i'm not sure how to approach proving the geometric series

because it seems to me as though z1 and z2 are the same

so how can i actually prove that they have any sort of order behind them

by that i mean, nothing in the question shows any difference between z1 and z2

all we know is that z1z2 is equal to (z3)^2

so i think i'm just getting lost in the wording

[merry geode]

For example, z1=1, z2=e^(i pi)

You will get a z3

[gritty carbon]

but with that example

you could also say

z1 = e^(i pi) and z2 = 1

and that would mess with the series wouldn't it

because they want it in the exact order

r1, r3, r2

i can show you the answers

[merry geode]

In this example, r1, r2 and r3 are 1

[gritty carbon]

not 100% certain on how it works

[merry geode]

So its still a series

Geometric

[gritty carbon]

it's still a series yes

but the order would change no?

and they want it in a specific order

one sec

uploading

not too familiar with what the series is in $\frac{r1}{r3} = \frac{r3}{r2}$ as well

[warm shale]

splooze

[merry geode]

They do not want the order. They only want you to show it is a geometric progression. Whether it is 9, 3, 1 or 1, 3, 9, they don't care.

[merry geode]

That would be the expression for both forward and reverse order.

[gritty carbon]

how is it a series though?

ah

silly question

jesus

okay

i'm a little confused with the last part

the bisect angle part

does bisect mean it will cut directly through the middle?

[knotty crow]

yes

[gritty carbon]

okay

yeah i think i got it, thanks guys

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[low frost]

Need help solving all equations. For question 1/4 where it says solve for the unknown values I believe y= 45 and x= 85. And for the rest I need a direction on how to solve them. I sort of understand how to solve question 3/4 but i need some direction on that one too. Question 4 I believe is 67° and I don’t know about question 2.

[gritty carbon]

the angles of a triangle add up to what?

[low frost]

180°

[gritty carbon]

so in question 3/4

if we know that all those angles add up to 180

write an expression that would state it clearly

an equation*

something along the lines of

(the sum of all this) = 180

[low frost]

Ok

[gritty carbon]

do you have an idea?

or need a bit more help

[low frost]

I think I can do it

[gritty carbon]

write down the equation here

[low frost]

79° + x + 43° + x + 66° = 180°

?

[gritty carbon]

bang on

there's no need to write the degrees symbol btw haha

[low frost]

Okay

[gritty carbon]

so

you have a nice equation there

now simplify it

collect like terms

all that good stuff=

[low frost]

188 + 2x = 180

[gritty carbon]

yep

do you know how to solve for x here?

[low frost]

Pretty sure x would be -4

[gritty carbon]

yep

looks good

[low frost]

[gritty carbon]

https://media.discordapp.net/attachments/903430332324405288/1044944252108541972/IMG_0013.png?width=1440&height=665

with this question

you wrote down 53 degrees for ACB

which is not correct

53 degrees is for which angle?

[low frost]

I rewrote it to 67° if that helps at all

[gritty carbon]

yes 67 is right

[low frost]

Okay

[gritty carbon]

https://cdn.discordapp.com/attachments/903430332324405288/1044944253090017300/IMG_0011.png

know what to do with this?

[low frost]

Not at all

[gritty carbon]

okay

this is the order of what we need to do

We want to find the angle of ABC first

[low frost]

Okok

[gritty carbon]

what do you think that is?

[low frost]

Hm

[gritty carbon]

here's a hint

see that blue line

what's the angle of a straight line? for example

if we had

imagine the blue line is straight

and the black line is a little circle

a perfect circle

what would be that angle

[low frost]

360°?

[gritty carbon]

here's an image of a compass

if we start at 0 degrees

and go 360 degrees

that would get us right back to where we started

[low frost]

Yes

[gritty carbon]

in this case

https://media.discordapp.net/attachments/903430332324405288/1044947327967842335/image.png

notice how it's only half a circle

[low frost]

Yes

[gritty carbon]

if a full revolution of circle is 360 degrees

[low frost]

So we would only go 180

[gritty carbon]

and that's half a circle

[low frost]

Degrees

[gritty carbon]

exactly

the most important thing for you to know is this

an angle on a straight line is 180 degrees

[low frost]

Okay

[gritty carbon]

so notice how those two angles add to 180 degrees

so in this case

We have a straight line

and we know part of the angle on that straight line

and with that information, we can find out the other part

because an angle on a straight line is how many degrees?

[low frost]

180

[gritty carbon]

right

so what do you think angle ABC is

[low frost]

Ummmmmm

[gritty carbon]

just to confirm

do you know what angle i'm talking about

when i say

angle ABC

[low frost]

The angle of B

[gritty carbon]

not quite

you need to be more precise

[low frost]

Uh

[gritty carbon]

angle ABC specifically points to that angle there

[low frost]

Okay

[gritty carbon]

What is the value of the angle CBE

?

[low frost]

120°?

[gritty carbon]

yes

okay

so lets get this question done

AE is a straight line yes?

[low frost]

Yes

[gritty carbon]

and the angle B lines on this straight line

so B's angle, in it's entirety, must add up to what?

[low frost]

180?