help-27

[tender cobalt]

they are both acute angles

so since sin(A - B) = 1/2

and sin(30) = 1/2

you can see that A - B should be 30, right?

[restive river]

But sin a + b we got π/2

[tender cobalt]

thats for A + B remember?

A + B and A - B are not the same (!)

[restive river]

OH

[tender cobalt]

silly

[restive river]

Yess

[primal ferry]

factorial moment?

[tender cobalt]

its in parentheses specifically to cancel out any effect a factorial could have (...!)

[restive river]

Sin ( A-B) = sin (30)

[tender cobalt]

so we know A + B = 90 and A - B = 30

[restive river]

Yes

[tender cobalt]

now can you find A and B from this?

A + B = 90, A - B = 30 is called a "linear system of equations"

[restive river]

Can u give me a hint pls 😅

[tender cobalt]

heres a hint

try adding both equations together

do you know how to add both equations?

[restive river]

Okay!

[restive river]

Yes

2A = 120
A = 60

[tender cobalt]

nice

now since A = 60, what is B?

[restive river]

B = 30

[tender cobalt]

thats correct

now can you figure out tan A : tan B?

[restive river]

Omg

TYSM 😭❤️❤️❤️

I reaaaaaly appreciate you help

[restive river]

Yes boss

[tender cobalt]

np then

as soon as you figure that out, we can move onto 2 which should be easier

[restive river]

I got the answer 3:1

[tender cobalt]

thats correct

[restive river]

Got the second one, I tried changing the equation a little bit,
4 sin ²x = 3
Then, sin²x = 3/4
Sinx = √3/2
X = 60

Tan3(60)

Tan180

If idk what tan 180 is, can I do it like tan90+tan90? @tender cobalt

But than 90 is infinite...

[tender cobalt]

no you cant

[restive river]

Oh

[tender cobalt]

do you remember that tan(x) = tan(x + 180)?

[restive river]

What

Idk that 😭

[tender cobalt]

what do you know about tan?

[restive river]

Sinx/cosx

[tender cobalt]

now whats sin(x + 180)?

[restive river]

Sin ( 60+180) = sin 240

[tender cobalt]

thats not what Im saying

you need to remember that sin(x + 180) = -sin(x)

and cos(x + 180) = -cos(x)

[restive river]

Teacher haven't taught that

[tender cobalt]

what about that sin(x + 90) = cos(x)?

[restive river]

Do you mean the formula, sin a + b?

[restive river]

He taught , that sin (-x ) = -sinx and cos ( -x ) = cosx

[tender cobalt]

then remember that tan(x + 180) = tan(x)

so tan(0) = tan(180)

[restive river]

But x is 60

[tender cobalt]

but youre calculating tan(3x) not just tan(x) remember?

[restive river]

Oh yes

Tan 180 = 0

Sorry my bad

So the answer is... 0?

[tender cobalt]

yes

[restive river]

Oh it correct

How do I become math god like you omg 😭

Tysm for helping me out ❤️

[mystic scarab]

Another way to see it easily:

tan(180°) = sin(180°)/cos(180°) = 0/(-1) = 0
And you're done 😉

[restive river]

wow!

I'll keep that in mind ❤️

Wait

[mystic scarab]

Always keep definitions in mind, they can save or help you a lot of times

[restive river]

How can I find sin 180 then?

[mystic scarab]

This should be a well known value 😅

[restive river]

Dangit

[mystic scarab]

But if you want you can use sin(a+b) formula

[restive river]

Ok 👌

A 90 and b 90 in case of sin 180

Right?

[mystic scarab]

a=90° and b = 90°

[restive river]

Ok boss

@tender cobalt can u help me with a last single question

I cannot understand it

😭

For value of ∅ , 0 < ∅ < π/2, does sin∅+sin∅cos∅ attain maximum value

[tender cobalt]

do you have a screenshot

[restive river]

Sure

[tender cobalt]

how many options are there?

try to make sure you dont cut off anything

[restive river]

When I put these as the ∅

I got √3 for pi/3

2+√2/2√2 for pi/4

Pi/6 same answer as √3/2

[tender cobalt]

unfortunatel all of those are wrong

lets try this one step at a time

[restive river]

Ok

[tender cobalt]

first lets start with pi/2

[restive river]

But

[tender cobalt]

thats correct

so we can skip pi/2

moving on to pi/3

[restive river]

In the question it says ∅ < pi/2

[tender cobalt]

we need to calculate sin(pi/3) + sin(pi/3) cos(pi/3)

[restive river]

I got √3

[tender cobalt]

now do you remember what sin(pi/3) is

[restive river]

√3/2

[tender cobalt]

and what about cos(pi/3)

[restive river]

1/2

[tender cobalt]

so whats sin(pi/3) cos(pi/3)

[restive river]

√3/4

[tender cobalt]

so sin(pi/3) + sin(pi/3) cos(pi/3) looks more like

√3/2 + √3/4

right

[restive river]

Yes

[tender cobalt]

now whats 1/2 + 1/4

[restive river]

What?

Ohh

4+2/8 = 3/4

[tender cobalt]

btw when you add fractions, its ideal if you have the denominator be as small as possible

so you could just do (2+1)/4 instead of (4+2)/8

anyways yea 1/2 + 1/4 = 3/4

so √3/2 + √3/4 is?

[restive river]

3√3/4

[tender cobalt]

correct

moving on to pi/4

sin(pi/4) + sin(pi/4) cos(pi/4)

whats sin(pi/4) and cos(pi/4)?

[restive river]

Looks like I did calculations wrong b4

[restive river]

1sec

[tender cobalt]

alr take your time

[restive river]

2+√2/2√2

[tender cobalt]

you are forgetting parentheses

[restive river]

1/√2 + 1/√2 x 1/√2

1/√2 + 1/2

When 1/1 x 1/1 , we don't cross multiplication

What how

[tender cobalt]

read this closer

[restive river]

Ithats how

[tender cobalt]

PEMDAS would say that you typed:

[restive river]

1?

[tender cobalt]

hold on there husky

youre getting too excited

let me finish here first ok?

[restive river]

Ok

[tender cobalt]

first

[tender cobalt]

you forgot to type parentheses here

if you look closer, it looks like you typed:

$2+\frac{\sqrt2}2\sqrt2$

[woven radish]

mtt

[tender cobalt]

this is not what you meant

[restive river]

Yea

[tender cobalt]

so where would you want to put parentheses?

clearly this + shouldnt be done last

[restive river]

So I add first then multiply

But what about BODMAS then?

[tender cobalt]

PEMDAS and BODMAS both mean the same thing

[tender cobalt]

this is not what I meant

again dont take guesses

lets try this a bit slower

you typed 2 + sqrt(2) / 2 sqrt(2)

under PEMDAS or BODMAS, the + sign would be calculated last

[tender cobalt]

and you would end up with this

which is wrong

you now need to type parentheses (or as you call them brackets)

to make sure that the + sign is not calculated last

because that is not what you meant when you typed 2 + sqrt(2) / 2 sqrt(2)

[restive river]

( 2+√2/2√2 )

[tender cobalt]

thats not it

thats just parentheses around the entire thing

the problem is inside there somewhere

do you need a reminder of how parentheses/brackets work?

[restive river]

Yes pls 😭

[tender cobalt]

parentheses or brackets allow you to select an operation to be done first

for example

(2+sqrt(2))/(2sqrt(2))

you can see here the parentheses forced the 2 + sqrt(2) and the 2 * sqrt(2) to be done first

this would mean the division is left last

[restive river]

oh

[tender cobalt]

so this would be seen as $\frac{2+\sqrt2}{2\sqrt2}$

[woven radish]

mtt

[tender cobalt]

and you can see in how its written, that the + and * is done first before the /

which matches up with (2+sqrt(2))/(2sqrt(2))

does that make sense?

[restive river]

Yea

[tender cobalt]

ok thats good

[tender cobalt]

you correctly calculated sin(pi/4) + sin(pi/4) cos(pi/4) to be this fraction

moving on to pi/6

sin(pi/6) + sin(pi/6) cos(pi/6)

try doing this one again

[restive river]

√3

[tender cobalt]

that is not correct

whats sin(pi/6) and cos(pi/6)?

[restive river]

1/2+1/2x √3/2

[tender cobalt]

so thats $\frac12+\frac12\cdot\frac{\sqrt3}2$

[woven radish]

mtt

[tender cobalt]

now how is this simplified?

[restive river]

1/2 + √3/4 = (4+2√3)/(2)

[tender cobalt]

(4+2sqrt(3))/2 is not correct

right now you have this

how would you add these two fractions together

[restive river]

Cross multiplication

[tender cobalt]

look at the denominators

what would you need to multiply 1/2 by to get the denominators to match?

[restive river]

Multiply denominator and numerator of 1/2 by 2

[tender cobalt]

yep

so now we have three numbers to compare together

from using x=pi/3, x=pi/4, x=pi/6,

we have these three numbers

and we need to figure out which one of these is the biggest

[restive river]

But we can simplify it even more right?

[tender cobalt]

no we cant

[restive river]

Oh

[tender cobalt]

theres something nice we can do to cross out one of these numbers

[restive river]

3√3/4 seems to be the biggest

[tender cobalt]

are you using a calculator to say that?

[restive river]

No

[tender cobalt]

did you guess?

[restive river]

Yea

[tender cobalt]

I wouldnt guess if I were you

Id prefer if you said how you managed to guess that one

[restive river]

🥲

[tender cobalt]

lets consider something else

[restive river]

3√3 is bigger than 2√2

That's how I guess

[tender cobalt]

just seeing bigger numbers isnt good enough you know

but lets try something else out

[restive river]

Okie

[tender cobalt]

we can multiply all of the numbers by 4

after all we just need to figure out which one is the biggest

and that doesnt change if we multiply them all by 4

do you understand?

[restive river]

Why we multiply by 4 specifically? Is it bcuz 2 of them have 4 as denominator

[tender cobalt]

yep

we'll make the numbers easier to read

[restive river]

Okay

[tender cobalt]

afterwards we get this:

in the top and bottom, the / 4 and * 4 cancel out

in the middle, you have 4 / (2 sqrt(2)) which is sqrt(2)

[restive river]

Yes

[tender cobalt]

expand the parentheses to then get this:

are you with me so far

[restive river]

Yes

[tender cobalt]

now consider the following

we know that 2 + 2sqrt(2) and 2 + sqrt(3) only differ in that 2 sqrt(2) and sqrt(3), right

[restive river]

Yes

[tender cobalt]

thats good

now 2 sqrt(2) is the same thing as sqrt(4) sqrt(2)

or sqrt(8)

right?

[restive river]

Yes

[tender cobalt]

yep

so the second number can be written like this

[restive river]

The difference between 2 and 3 is of √ 5 that means

[tender cobalt]

nope thats not correct

sqrt(8) - sqrt(3) is not sqrt(5)

you cant just ignore the sqrt when you add or subtract

,calc sqrt(8) - sqrt(3)

[woven radish]

Result:

1.0963763171773

[tender cobalt]

,calc sqrt(8 - 3)

[woven radish]

Result:

2.2360679774998

[tender cobalt]

these numbers are different

[restive river]

Oh

[tender cobalt]

in general you dont ignore the symbols

[restive river]

Brackets are really imp I see

[tender cobalt]

yea

youre correct though that sqrt(8) is bigger than sqrt(3)

so that means the second number is bigger than the third

that lets us cross out the third number

so its just between these two now

[restive river]

Yes

If we take squares, we get 27 and 12

[tender cobalt]

thats still not correct

[restive river]

😭

[tender cobalt]

you need to know how things work

you cant just ignore square roots and squares and stuff

first

lets take (1 + 2)^2 for example

youll notice (1 + 2)^2 = 3^2 = 9

and also 1^2 + 2^2 = 1 + 4 = 5

so (1 + 2)^2 is not the same as 1^2 + 2^2

second

the real way you can calculate (1 + 2)^2 is with distributive property

(1 + 2)(1 + 2)

is then 1 * (1 + 2) + 2 * (1 + 2)

is then also 1 * 1 + 1 * 2 + 2 * 1 + 2 * 2

[restive river]

That's 9

[tender cobalt]

yep thats 9

in general, (x + y)^2 is x^2 + y^2 + 2xy

you forgot about the + 2xy

[restive river]

Sorry my bad

I forgot about that

[tender cobalt]

now try doing (2 + sqrt(8))^2 again

[restive river]

Ok

12+2√8+2√8

12+ 4√8

Am I correct so far?

[tender cobalt]

yep

good job

[restive river]

I don't think we can add 12 and 4 since 4 is with √8

[tender cobalt]

thats correct, you cant do that

what you can do though is write sqrt(8) as 2 sqrt(2)

so 4 sqrt(8) is 8 sqrt(2)

[restive river]

Yes

Now? Taking squares?

[tender cobalt]

we already took squares

before we continue, lets make things easier

lets subtract 12

[restive river]

So we remain with 8√2

[tender cobalt]

and 27 - 12 is?

[restive river]

Where did the 27 come from

Oh wait

Nvm

15

[tender cobalt]

yep

we're very close now

now what next?

[restive river]

Squares again?

[tender cobalt]

sure

[restive river]

So we can get rid of the root

[tender cobalt]

do you remember 15^2?

if you dont, its 15^2 = 225

[restive river]

Ohkkk

For the second one, I got 128

[tender cobalt]

yep

[restive river]

64x2

Oh god this one was so tricky bcuz of the root sign

Tysm you still managed to solve this one too 😭❤️

[tender cobalt]

np

we figured out the first option was the biggest

so pi/3

it took a while to just compare three numbers together

[restive river]

Any tips on how can I solve any problem like u?

[restive river]

Yea 😨

[tender cobalt]

you look for an easy way out

you start with problems you already know are easy

you get used to those problems so that you know you can do them, you recognize them

then as you move on to bigger problems, youll see theyre only made up of those easier problems you could always do

but this will only work on some problems

most problems are connected by some form of intuition

thats where it all begins

[restive river]

😯

[tender cobalt]

thats what you need to practice

and its not easy to practice

[restive river]

Practice makes man woman perfect

[tender cobalt]

if you get too used to doing a procedure, you no longer will be practicing intuition

no

practice makes persistent

practice cannot perfect anything

[restive river]

That's true...

[tender cobalt]

practice makes procedure

[restive river]

If I practice same type of sum, I will get used to it and cannot solve sums of other types

😨

[tender cobalt]

as you get more and more used to math, your intuition will grow

youll see how things fit together

then you can figure out how to do other things on your own

like I did

[restive river]

Tysm ❤️❤️

[tender cobalt]

np

[restive river]

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[restive river]

.reopen

[devout snow]

✅

[restive river]

@tender cobalt I tried solving this out, answer didn't match, can u point out my mistake?

[fervent hornet]

what is the value of sec(2pi/3) (*hint its not 0)

[restive river]

2π/3 is 270°

[restive river]

Cos 270 is 0, sec is 1/0 that is 0

[fervent hornet]

yes but thats not sec(2pi/3)

,w sec(2pi/3)

[woven radish]

Results provided by WolframAlpha

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

Requested by convergence0091

[restive river]

I don't get it

[fervent hornet]

@restive river do you get it now?

[restive river]

No 😭

Wait...

-2

😨

Answer is 4

[fervent hornet]

yes

[restive river]

Tysm 😭😭

[fervent hornet]

:)

[restive river]

@fervent hornet can u explain me how the -2 came

Plz

[fervent hornet]

Cos(2pi/3)=-1/2
As secx=1/cosx
This means that sec(2pi/3)=1/(-1/2)
Thus sec(2pi/3)=-2

Sorry that I was late

[restive river]

How did the -1/2 come

😅

Oh.

Tysm

[fervent hornet]

:)

[restive river]

Lol wot xD

Still 2 years

Hope I can cover many things till then and become math pro like you

[fervent hornet]

It was not meant for you but still all the best

[restive river]

❤️

[fervent hornet]

Practice you can do it

[restive river]

Yes

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[restive river]

Hi

.close

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[wooden sage]

Translation: when Nasa trained the moon travelers, they used an inclined plane to simulate that the weight on the moon is only 1/6 of the weight on earth. What should the angle of inclination of this plane be?

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[open ether]

can i ask here?

[open ether]

how do i find the ratio here?

there is no e at the end btw

the question is to find the sum

[feral agate]

Looks like an infinite GP

[open ether]

yea it is infinite

i know that if r is smaller than 1 i can find the sum

but idk r

[covert root]

geometric series

[open ether]

yes

[covert root]

What do you think r is?

[open ether]

is it this whole thing?

[covert root]

no

Take the sequence 1, 3, 9, 27 what is r?

[open ether]

3

[covert root]

no

yes 3

[open ether]

my bad

so the first result i got 3/4 and second 24/15

do i need to make a1/a2?

[covert root]

You can do it that way. a2/a1 = r

[open ether]

i just forgot about the a2/a1 lol

[covert root]

By definition the sequence of geometric series is a, ar, ar^n-1

[open ether]

thanks for reminding me

[feral agate]

That is not necessary

[open ether]

32/15 ratio?

[feral agate]

Rewrite the sum as $\sum_{n=0}^\infty a\cdot r^n$

[woven radish]

kheerii

[covert root]

You most likely made arithmetic error

[open ether]

4/7 after given some thought

[feral agate]

Yes it is 4/7 indeed

[open ether]

thank you! , you can close this thread now

what was the command

.close

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[proven jackal]

I have a question on probability. So we know that P(A) = sum of P(A|Hi)P(Hi) if {H1,...} are a partition of probability space. Does that mean P(A) = sum of P(A, Hi), because P(A|Hi) = P(A, Hi)/P(Hi) so P(Hi) cancels out?

[topaz axle]

yes of course

[proven jackal]

Thank you, it was really stressing me because it's so obvious but we never used it on our exercises lol. If you know by any chance anything about poisson processes, we need to do an exercise on it: "On a warm summer evening, you open a window, and mosquitoes and flies start flying into the room, each type independently as its own Poisson process. Suppose that within one hour, at least three insects flew in, including at least one mosquito. What is the probability that within two hours, at least five insects will fly in, including at least three mosquitoes?" Mosquitos and flies together are insects, and i tried to do this exercise with total probability by spliting our condition into many parts (if number of insects within first hours is 3, 4 or >= 5, and then each of these even further to how many of those insects are mosquitos). Would this be an okay approach?

[topaz axle]

no idea

i know what it is, but this looks too hard

[proven jackal]

It's okay. That's all i needed to continue thank you:)

[topaz axle]

you can certainly split any probability like that and find the sum though

[proven jackal]

Another quick question, can we apply the same to conditional probability so that P(A|B) = sum P(A|B,Hi)P(B,Hi) = sum P(A,B,Hi) (since P(A|B,Hi) = P(A,B,Hi)/P(B,Hi)

[topaz axle]

you get P(A,B) from that

P(A|B) = sum P(A,Hi|B) i think

[devout snow]

@proven jackal Has your question been resolved?

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[glacial agate]

is my answer correct? Solving it my way I would just do basic algebra and get the answer ive selected

[scarlet sequoia]

it's not

how did you get 2 - 2 sin(t) ?

square roots don't work that way

[glacial agate]

substitued 2sin t and removed the squared and took square root of the 4

[woven radish]

Modus

[scarlet sequoia]

that's not valid

use the identity sin^2x + cos^2x = 1

[glacial agate]

oh it thought it did for a second, im just refreshing my math memore before college for fun

[glacial agate]

how would i use this in my problem, i dont see it being applied here?

[scarlet sequoia]

after the given substitution you should obtain (under the square root):

4 - 4sin^2(t), right?

[glacial agate]

yes

correct

[scarlet sequoia]

then take 4 out of the bracket to get

4(1 - sin^2(t))

and now:

[woven radish]

Modus

[glacial agate]

ohh because of pythagorean identites

so then i get this

5(2sint)/4(cos^2t) ?

[scarlet sequoia]

almost

[woven radish]

Modus

[glacial agate]

ah yes the root

and sin over cos is just tangent

so 2sin/2cos

is just tangent

so

[scarlet sequoia]

• 5

[glacial agate]

5tan t?

[scarlet sequoia]

yes

[glacial agate]

dang i cant believe i forgot how to do this

i used to be so good,

guess thats what 2 months of summer does to you

haha thank you bro

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[winter torrent]

okay pal

please don't ping helpers before 15 minutes

and especially don't ping helpers without actually, like, writing out your question

[devout snow]

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[restive bear]

help

[restive bear]

how to do basic qn2

[frozen stump]

.rotate

a

[frozen stump]

you know the addition and subtraction of logs with the same base right?

[devout snow]

@restive bear Has your question been resolved?

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[jaunty abyss]

Hey y’all
I got SSA for my triangle but that’s not one of the options..

[lusty sapphire]

SSA is not a triangle congruence theorem

[jaunty abyss]

Oh

It’s not?

You’re right

[jaunty abyss]

So what else would I do

To prove

Helloooo

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[honest meadow]

can someone help me with #3 for part i and j

ik the formula for P(A given B)

but im struggling on how to apply it because it's saying that the 2nd card is given as green

and is asking for the probability of the 1st card

i have that P(G1 and G2) = 25/64 for with replacement

but then when i do P(G2)=2/7 my result answer is over 1

im not sure if the question meant P(G2 given G1) though because the answer logically is 5/8 for both since the 2nd card happens after the 1st

[jade oak]

y u say 2/7?

[honest meadow]

for the one without replacement

the question itself is faulty i think

P(G1 given G2) is just 5/8 right?

because G2 happens after G1

i think they meant P(G2 given G1)

which would just be a tree diagram

[jade oak]

I dont think so

It seems like that

But the fact that you got a green on the second pull does give you some information about the 1st pull

[honest meadow]

does it?

you're already guaranteed the 2nd pull will be a green card

with replacement should be 5/8

because there's no difference between the pulls

with replacement is where im getting confused assumign this question isn't just written wrong

[jade oak]

Lets say we have 4 red cards and 1 green card. If you got a green card 2nd pull you know 100% you wouldnt have a green card on the 1st

Same logic here

[honest meadow]

then how do you calculate with replacement

because the formula doesn't really make sense

[jade oak]

What do u get fir G1 and G2 with out replacement

[honest meadow]

20/56 is without replacement

5/8 * 4/7

[jade oak]

Ok yeah we don’t know this for sure

[honest meadow]

cause even with a tree diagram

i get 5/8

[jade oak]

It could be 3/7 or 2/7 depending on if u pull green or red first

Actually isnt it 5/7 or 4/7 anyway

[honest meadow]

yeah that's what i get with the diagram

[jade oak]

Also I just realized it said yellow not red lol my bad

[honest meadow]

all g lol

[jade oak]

U got pic?

[honest meadow]

given that the 2nd card is green

the probability the first card is green is just 5/8

even if you replace it or don't

personally i think it's supposed to be g2 given g1

but i still would like to know how g1 given g2 works out

and for the 4 red 1 green example wouldn't that just be 0?

because if you draw a tree diagram out you cant draw a green card 1st if you got the green card 2nd

this is what i have for a 4 red 1 green scenario

if you draw the green 2nd then you can never draw the 1st card as green

ty for the help

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[small solar]

When taking the limit of the function, I have multiple terms being added to one another. The first term requires L'hoptials rule to find the limit, however the second term being added to it does not require L'hopitals.

[small solar]

Do I just leave it?

Or am I supposed to derive that term as well

[lusty sapphire]

!show

[devout snow]

Show your work, and if possible, explain where you are stuck.

[small solar]

one second

@small solar

@lusty sapphire

let me take a better pic

[lusty sapphire]

,rccw

[small solar]

basically from where 2. is

[lusty sapphire]

Where are you getting ln?

[small solar]

im taking ln of the whole thing

do you see in the first line

it starts off like this

[lusty sapphire]

I do not see

[small solar]

(-1)^n * (n^4-1)

over

[stable storm]

why are you taking the ln of the whole thing tho?

[small solar]

3n^4-2n^+1

[stable storm]

dont think thats necessary

[small solar]

because there are two terms in the numerator being multiplied

and one in the denom

so by ln rules

you can seperate the top two

as term1 + term2

and subtract

by term3

so like

[stable storm]

you do know that ln(-1) is imaginary right

[small solar]

term1 + term2 - term3

[stable storm]

how about you ignore the (-1)^n term just for now

and try to compute the limit of the other bits

im guessing n is strictly a positive integer here

[small solar]

n goes to infinity

[lusty sapphire]

In answer to your first question, yes, you can take (-1)^n out of the lhopital

[small solar]

well

i make the n into 1/n under the ln(-1) to apply lHoptials

but the other terms im confused about

[lusty sapphire]

Unnecessary and incorrect

That's why you are confused

[stable storm]

try to compute $\lim_{n\to\infty}\frac{n^4-1}{3n^4-2n^2+1}$ first

[woven radish]

y0shi

[stable storm]

then think about what that (-1)^n is doing to the sequence

[small solar]

hold on

wait

so

i was watching this video by oct

https://www.youtube.com/watch?v=XdkoTb8PEG0

look at

26:23

he does it similarly to me

its only different because i have more terms

[stable storm]

well the thing is if you try to take the ln of both sides for just (-1)^n

you wont get the indeterminate form that is required for lhopital's rule

you're overcomplicating things here

intuitively think about what (-1)^n does for all n that are positive integers

[small solar]

well

it simply changes the positivity of the whole function depending on n

[stable storm]

exactly

it causes it to alternate

[small solar]

right

[stable storm]

so really if we just find what the limit of the other bit is

without this alternating bit

and then put this concept back at the end

itll be the same thing

[small solar]

so i can just pull it out?

[stable storm]

yeah

ignore it for now

[small solar]

okay

lemme try this out give me a second

so do i just solve for the rest of the limit

and then once i do

[stable storm]

mhm

[small solar]

do i just plug that back in or do i find the limit of it as well and then plug

[stable storm]

so the (-1)^n bit alternates it

if the limit of the other bits equals some finite number

depending on what n is

the limit as n approaches infinity could be positive or negative of that finite number

basically the limit then wouldnt exist

the only way the limit could exist is if the rest of the limit approaches 0

[small solar]

yeah the limit goes to infinity

okay

[stable storm]

since then the plus or minus would affect it

[small solar]

so the limit when i pulled out the first thing goes to 1.3

1/3

[stable storm]

yep

and now we put back the alternating concept

so the limit could be either -1/3 or 1/3 depending on what n is

hence it doesnt exist

[small solar]

but because its infinity, doesnt that make it (-1^inf)1/3

or am i confusing that part?

[stable storm]

well (-1)^n

no matter what n is

itll just either be -1 or +1

making the sequence alternate

[small solar]

right

[stable storm]

(-1)^infinity does the same

it could be -1 or +1

depending on what n is

[small solar]

for the sake of the sequence question, this would make it diverge right?

[stable storm]

if n is even then itll be positive, if not itll be negative

mm yeah

by the divergence test

since the limit doesnt equal 0

[small solar]

ohh okay

ngl

last week

i missed three classes and one got cancelled

[stable storm]

i see

[small solar]

i was on vacation and i talked to my prof,

so he understands but i thought i would have been able to study a bit more but i didnt get the chance

so i still dont understand the divergency test

for now ill put that it diverges because it alternates by infinity

i think hell be lenient enough to understand what i mean

hes chill

thank you a lot though

i really appreciate your help

[stable storm]

yw!

[small solar]

have a goodnight!

[stable storm]

you too!

[small solar]

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[vague obsidian]

So I tried raising all of them to 5

and I got

$$4\sqrt{2}, 3\sqrt[3]{9}, 5$$

[woven radish]

Shinutsi

[vague obsidian]

but idk what to do next

like without

manually calculating

idk the value of $$\sqrt[3]{9}$$

[woven radish]

Shinutsi

[winter patrol]

you'd want to consider lcms of the #root being taken

e.g. to compare the sizes of 3^(1/3) and 5^(1/5)
consider raising both to the power of the lcm of 3 and 5, i.e. 15

[vague obsidian]

ohhhh

why does 5 not work

like the greatest index

[winter patrol]

because raising 3^(1/3) to a power of 5 will leaving you with something with a fractional power

[vague obsidian]

ohhhh

[winter patrol]

from power rule
(a^m)^n = a^(mn)
you'd want to product for all stuff you're raising the power of to be integers

[vague obsidian]

thus the roots

i see

let me try

[winter patrol]

you "could" attempt to compare all 3 at once directly, but the numbers may be too tedious to calculate by hand

[vague obsidian]

i got 2^15, 3^10, and 5^6 by raising all of them to 30

is there a way to simplify this further?

[opal cloak]

he said u could do it this way

[vague obsidian]

oh

[winter patrol]

multiply that by hand

[opal cloak]

just stick to a comparison of 2

[winter patrol]

you could take a few shortcuts

[vague obsidian]

oh okay

[opal cloak]

i mean at that rate, u may aswell put it into a calculator

[winter patrol]

that's why i said "could" earlier

[vague obsidian]

got it now, thanks everyone

.close

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[lost laurel]

How do I prove $\lambda 1= \lambda$

[woven radish]

ƒ(Why am. I here)=I don't Know

[lost laurel]

I was thinking of writing lambda as a+bi

and then distributing it that it?

[autumn fjord]

what is lambda 1

[wicked turtle]

hmm maybe (a+bi)(1+0i) would be the most "formal" way to write it

[lost laurel]

$\lambda \cdot 1$

[woven radish]

ƒ(Why am. I here)=I don't Know

[autumn fjord]

oh like actual complex numbers

[lost laurel]

I see, thanks!

similarly I can take $a+0=a$ where $a \in \R$ as an axiom

[woven radish]

ƒ(Why am. I here)=I don't Know

[lost laurel]

right?

[fervent hornet]

Yes

[lost laurel]

thanks

[feral agate]

Morning wai

[lost laurel]

'afternoon

sorry

[feral agate]

It’s morning for me 😭😭

[lost laurel]

yeah, morning

[fervent hornet]

Lol

[lost laurel]

thanks

.close

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[covert furnace]

how is this answer 4374?

[covert furnace]

it doesn't give the initial so i wasnt sure what to do

i tried solving for the initial value but you need the output to find that

[devout snow]

@covert furnace Has your question been resolved?

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[covert furnace]

i need help with this sum question

[devout snow]

@covert furnace Has your question been resolved?

[covert furnace]

<@&286206848099549185>

this is the new version of the question i had a question on

[copper sonnet]

which one do you need help with?

[covert furnace]

the second one

[copper sonnet]

This is geometric series so general term is $6(\frac{13}{14})^n$

[woven radish]

Closer

[copper sonnet]

so you need to find n to make $6(\frac{13}{14})^n \leq 1$

[woven radish]

Closer

[covert furnace]

so how step would i do to fine that/

?

(13/14)^n <= 1/6

what would i do after

[copper sonnet]

using logarithm

[covert furnace]

so what would the base of the log be?

[copper sonnet]

base that cancels out exponent

which is 13/14

[covert furnace]

ohh okay

so log(13/14) and then what would i put inside

would i put 6(13/14)

or just (13/14)

[copper sonnet]

From this if you take log of both side, it become this: $\log_{\frac{13}{14}}(\frac{13}{14})^n \leq log_{\frac{13}{14}}\frac{1}{6}$

[woven radish]

Closer

[copper sonnet]

Do you know what to do next?

[covert furnace]

no i dont

how do you get n by itself from the log

[copper sonnet]

There is a property of log that will help you do it

which is: $\log_{b}b^n = n$

[woven radish]

Closer

[copper sonnet]

you are basically asking which power b needs to be raised to equal b^n

[covert furnace]

oh so log13/14 13/14^n = n

so just that answer itself is n

[copper sonnet]

Yea, this is often used to bring exponent down

[covert furnace]

so is n = 1?

[copper sonnet]

how did you get 1?

[covert furnace]

wait i dodn think i did it correctly

log13/14(13/14)

on the ti84

but how do you add the n on the calculator

[copper sonnet]

n is a variable that we are finding

[copper sonnet]

you just plug n into the inequality

[covert furnace]

so the previous property u told me about how do we use it if we dont even know n

[copper sonnet]

we use it to bring n down

[covert furnace]

it just told me the property

[copper sonnet]

meaning: $\log_{\frac{13}{14}}(\frac{13}{14})^n \leq log_{\frac{13}{14}}\frac{1}{6}$ to $n \leq log_{\frac{13}{14}}\frac{1}{6}$

[woven radish]

Closer

[covert furnace]

wait how did 13/14 become 1/6?

[copper sonnet]

we are solving inequality here

[copper sonnet]

You know how to solve equation, right?

We are doing the same here

This is the whole steps

[covert furnace]

ohh okay i get it now

i forgot about the 6 from before

[covert furnace]

so n is 24.177

or i mean 24.178 rounded

[copper sonnet]

wait

[copper sonnet]

you can plug it to the original: $6(\frac{13}{14})^n$ which is the distance the ball rebound after n time

because n should be an integer so you can test 24 and 25

which number give you the result that is < 1 is the answer

And look like 25 is the answer

[woven radish]

Closer

[covert furnace]

i searched up and someone had solved the same kind of problem for someone else

so for the log(13/96) / log(12/13)

what would they even use as the base

[copper sonnet]

It is convention in mathematics that log implies log base 10 and ln implies log base e

[covert furnace]

OH okay so ususllay you just do 10 unless it states to do a different number

oh e is 10

[copper sonnet]

Do you know how to calculate the total distance the ball travels before resting on the ground?

[covert furnace]

i did like sum infinity

ya

a1/1-r

[copper sonnet]

There is little subtle to this problem

because the ball bound up and then bound down

you need to double it

except the first 6 feet

[covert furnace]

a1 was equal to 112/9 and then i divided by 1/9 and i got 112 fr and then added 7 to it

so it got 119

[copper sonnet]

So, total distance will be: $6 + 2\sum_{k=1}^{\infty}6(\frac{13}{14})^k$

[covert furnace]

it said it was correct ig it accept infinity sum

[woven radish]

Closer

[copper sonnet]

I got 162

Do you have the answer?

[devout snow]

@covert furnace Has your question been resolved?

[devout snow]

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[solid osprey]

find the minimum of $$15x^2-14x+\frac{8}{x^2}+22$$

[woven radish]

Skill_Issue

[solid osprey]

no i dont know calculus

[restive river]

what have you tried

and do you want to learn it?

[solid osprey]

but sure what to even try

[solid osprey]

ye

[restive river]

I can teach you differentiation if you want

[solid osprey]

um AM-GM might work here

[restive river]

it will

but calculus is way faster

[solid osprey]

i dont think thats even allowed lmao

[restive river]

why not

[solid osprey]

im guessing so this is grade 9 olympiad

uh but sure ig

diffrentiation is the y' thing right?

[restive river]

in layman's terms yes

if y is a function

[solid osprey]

ok

[restive river]

so, we have $f(x) = 15x^2 - 14x + \frac{8}{x^2} + 22$

[woven radish]

45

[solid osprey]

the differentiation i know is the power rule and the $$f'(x)=\lim_{h->0}\frac{f(x+h)-f(x)}{h}$$

[restive river]

remove \ from f'(x)

but yeah

[woven radish]

Skill_Issue

[restive river]

that's the definition of the derivative

but we can use the power rule which states that the derivative of $x^n$ is $nx^{n-1}$

[woven radish]

45

[solid osprey]

ye i know that

[restive river]

perfect

differentiate f(x) then

and re-write $\frac{8}{x^2}$ as $8x^{-2}$

[solid osprey]

idk how to diffdfentiate the /

[woven radish]

45

[solid osprey]

oh

uhh ill try my best

[restive river]

I'll guide you if you get lost

[solid osprey]

$$15\cdot2x-14\cdot1+8\cdot{-2}x^{-2-1}$$
$$30x-14-16x^{-3}$$

[woven radish]

Skill_Issue

[solid osprey]

this?

[restive river]

yep

that's correct

[solid osprey]

cool

[restive river]

so there's a couple things you need to know

first off

when we set f' to 0, we obtain what are called the critical point(s) of the function

for now, all you need to know is that the maximum or minimum occurs when f'(x) = 0

we want the minimum

so in accordance with that, we have to make sure this is actually a minimum

and this happens when the second derivative is greater than zero for that point

so before we find out what value of x satisfies f'(x) = 0, we should also check what f''(x) looks like

in this case, we have $f''(x) = 30 + 48x^{-4}$

[woven radish]

45

[restive river]

what do you know about f''(x)?

[solid osprey]

what is f"(x)

[restive river]

this

[solid osprey]

idk what that is

[restive river]

oh

f''(x) is the derivative of f'(x)

[solid osprey]

oh

[restive river]

it's called the second derivative

[restive river]

.

[solid osprey]

oo ok

[restive river]

what do you know about this function?

[restive river]

.

[solid osprey]

um

[restive river]

can it be negative or 0

for x in R

[solid osprey]

the range or the r

[restive river]

real numbers

$\mbb{R}$

[woven radish]

45

[solid osprey]

i mean the range or the x

the range i think its >=0?

[restive river]

for any $x \in \mbb{R}$, is $f"(x) \leq 0$?

[woven radish]

45

[restive river]

can it be zero?

[solid osprey]

i mean >=30

[restive river]

great, so we've established that f''(x) is ALWAYS positive for ANY real number

this means that any critical point we have will ALWAYS be a minimum

[restive river]

.

in accordance with this

so, now we solve f'(x) (the first derivative) = 0

[restive river]

so we go back to this

$30x - 14 - 16x^{-3} = 0 \implies 30x^4 - 14x^3 - 16 = 0$.

[woven radish]

45

[solid osprey]

mhm

[restive river]

notice that 30 - 14 - 16 = 0

so x = 1 is a root

and this also coincides with the rational root theorem

so you've found that x = 1 is a critical point

and what you also need to know is that f(critical point) = minimum or maximum (we know this will be the minimum, since we checked using the second derivative)

[solid osprey]

do i try to find the other critical points by dividing by (x-1)?

[restive river]

you definitely could, for good practice

and in an exam, I'd advise you to do that

[solid osprey]

30x^3+16x^2

[restive river]

if you know synthetic division

[solid osprey]

wait what is this actually right

did i do synthetic wrong

[restive river]

you did

don't try to do it in your head

oh it might not be actually

[solid osprey]

30 -14 0 -16
30 16 16
30 16 16 0

[restive river]

nah it is

[solid osprey]

it is what

[restive river]

give me a second, I'll check your synthetic division work

[solid osprey]

oh

i know it

so 30x^3+16x^2+16x

what

no this is wrong

[restive river]

don't rush

it's $30x^3 + 30x^2 + 30x + 16$

[woven radish]

45

[restive river]

so now by the rational root theorem

[solid osprey]

question, where did i go wrong?

[restive river]

you used the format for a cubic not a quintic

you have a x^0 term as well

so there should be five terms

not four

[solid osprey]

o

[restive river]

anyway