help-42

[atomic rune]

it cant be 0 tho right?

[potent smelt]

x-1 being equal to 0 is the third case, and there are no solutions associated with it

[atomic rune]

so x cant be 1

[potent smelt]

Exactly

But what if x = 0?

[atomic rune]

but it can be either positve or negative

[atomic rune]

then it would be / by -1

cuz 0-1

[potent smelt]

Yes and you have an inequality

And you just multiplied both sides by a negative number, right?

[atomic rune]

OHHH

[potent smelt]

What do you need to do when multiplying by a negative?

[atomic rune]

but thats only if x is 0 tho

[potent smelt]

No, that's whenever x-1 is negative

So $x \in (-\infty, 1)$

[potent lotus]

OmnipotentEntity

[atomic rune]

bro what is this 😭

i think i get what you mean tho

any negative nummber - a nagitve number give another nagitve nummber

so x can be -1 or -2

and so on

[potent smelt]

Or -π

Or 1/π

[atomic rune]

what does π mean

is it the same

as

[potent smelt]

Pi

[atomic rune]

3,14?

ye

[potent smelt]

The circle constant

[atomic rune]

but why does that have anything to do with this?

[potent smelt]

Just giving more examples, because you've only given negative integers

[atomic rune]

oh alr

[potent smelt]

So I wanted to make sure you understand that any negative real, and some positive ones, cause x-1 to be negative as well

[atomic rune]

so any negative nummber or positive number can be x

[potent smelt]

The only requirement for x-1 to be negative is x < 1

So rolling back a little bit

[atomic rune]

oh ye its not all positve and nagitve nummbers that are x only some of them

[potent smelt]

What do you have to do when you multiply both sides by a negative number in an inequality?

[atomic rune]

flip it

[potent smelt]

Yes

[atomic rune]

right?

ye

wait give me a sec to try out what you just said

[potent smelt]

So let's review, you have multiplied both sides by x-1, and you have three cases, x-1 is positive, negative, or zero. If x-1 is zero then there are no solutions, so that's not interesting, otherwise you have two inequalities, what are they?

And when do we consider them?

[atomic rune]

so this is if its negative

and this is if its positive

is that right?

[potent smelt]

\begin{align*}
x^2 + 4x + 3 <& (2x+5)(x-1) && \text{if } x < 1 \
x^2 + 4x + 3 >& (2x+5)(x-1) && \text{if } x > 1 \
\end{align*}

[potent lotus]

OmnipotentEntity

[potent smelt]

A minor correction to parentheses

[atomic rune]

ye isnt isnt this right then

and the one above

[potent smelt]

(2x+5)(x-1) not 2x + 5 (x-1)

[atomic rune]

ohhhhh ye i forgot

[potent smelt]

You can see that the second is different, because the x-1 is being multiplied only by the 5

[atomic rune]

ik why you have to do it

[potent smelt]

Ok

[atomic rune]

now what?

[potent smelt]

Now you have a little bit of algebra

[atomic rune]

wait give me a sec to think

[potent smelt]

Expand the right, make everything into a simple quadratic greater than or less than 0

And then use the quadratic formula.

[atomic rune]

how do i get the x-1 in

[potent smelt]

????????

[atomic rune]

when its nagtive

i messed up

[potent smelt]

How did you arrive at that?

[atomic rune]

NO IK WHAT I DID WRONG LOL

wait give me a sec to fix it 😭

like this

i just said

minuse this with both sides

[atomic rune]

and i got this

is that right?

[potent smelt]

\begin{align*}
x^2 + 4x + 3 &< (2x + 5)(x + 1) \
x^2 + 4x + 3 &< 2x^2 + 2x + 5x + 5 \
x^2 + 4x + 3 &< 2x^2 + 7x + 5 \
0 &< 2x^2 - x^2 + 7x - 4x + 5 - 3 \
0 &< x^2 + 3x + 2 \
\end{align*}

[atomic rune]

i mean i just took this and said -((2x+5)(x-1)) with both sides

[potent lotus]

OmnipotentEntity

[potent smelt]

Does the above make sense?

[atomic rune]

uhh

1 sec

let me have a look

[atomic rune]

nope i dont get this

[potent smelt]

What step do you get lost?

[atomic rune]

so 2x*x =2x^2?

[potent smelt]

Yes

[atomic rune]

im not good when mutiplying () with eath other

[potent smelt]

Second line I FOIL the right side

This is definitely something you need to practice then, because you're going to continue needing to do it

[atomic rune]

i mean i know how to when its with nummbers but it becomes alot harder with a and x and so on

so you said 2x * (x+1)

and you got

2x^2 and 2x right?

[potent smelt]

That's First and Outer, yes

[atomic rune]

oh now i get it

ye and then you had 5x and 51 so you got 5x and 5

alr im with you now

[potent smelt]

Exactly

[atomic rune]

ye i understand the rest

[potent smelt]

Ok now what?

[atomic rune]

but theres 1 thing i dont get tho

[potent smelt]

What's that?

[atomic rune]

why didnt you say -(2x+5)(x+1))

and take everything to the right side

isnt that much easyer?

[potent smelt]

You could do that, it's equally easy

You still have to FOIL

[atomic rune]

wait ima try rq

[potent smelt]

Left side?

[atomic rune]

like thos

and that >0

forgot to add that

is this right so far?

oh its not - its + 5x and + 5

[potent smelt]

Oh haha I made a mistake in my work

It is -

[atomic rune]

fr

😭

i thought i missed up fr

😭

[potent smelt]

\begin{align*}
x^2 + 4x + 3 &< (2x + 5)(x - 1) \
x^2 + 4x + 3 &< 2x^2 + 2x - 5x - 5 \
x^2 + 4x + 3 &< 2x^2 - 3x - 5 \
0 &< 2x^2 - x^2 - 3x - 4x - 5 - 3 \
0 &< x^2 - 7x - 8 \
\end{align*}

[potent lotus]

OmnipotentEntity

[potent smelt]

Corrected

[atomic rune]

like that?

0

[potent smelt]

No.

[atomic rune]

bruh

wait ima try to fix it 😭

[potent smelt]

Remember everything you moved to the left becomes negative

[atomic rune]

like this?

and cant i just multiply both sides by -1

and then i can just use the ax*bx+c

[potent smelt]

Be careful, you're making arithmetic mistakes

[atomic rune]

AAAAA

is it cuz theres a 0 on the other side

and i cant multipy by 0?

[potent smelt]

You never multiplied by zero though

You just added things wrong

[atomic rune]

cant i just pluse 1 on both sides and then multiply by -1

and add one again

[potent smelt]

This is correct here

[potent smelt]

This is not

[atomic rune]

is this wrong?

[potent smelt]

That is wrong

[atomic rune]

huh

doesnt all of them turn to -

bc its -()

[potent smelt]

+s become -s and -s become +s

[atomic rune]

so they flip?

[potent smelt]

Yes

[atomic rune]

wait ima look at it again then

like this?

[potent smelt]

Not to be rude or anything. But these concepts are things you should be familiar with and need to be familiar with before trying a problem as complicated as this one

No

Not like that

[atomic rune]

i havent started learning about this is school yet i maid this my self 😭

[atomic rune]

i just thought it would be fun to learn how to solv one that looked like that

[potent smelt]

\begin{align*}
x^2 + 4x + 3 - (2x^2 + 2x - 5x - 5) &= x^2 + 4x + 3 + (- 2x^2 - 2x + 5x + 5) \
&= x^2 + 4x + 3 - 2x^2 - 2x + 5x + 5
\end{align*}

[potent lotus]

OmnipotentEntity

[potent smelt]

Like this ^

[atomic rune]

wait so i just remove the ()??

[potent smelt]

No worries. It's fine to want to try new things, just you might find that you lack the background knowledge to successfully address the problem.

[potent smelt]

You need to pay careful attention to what it is I actually did, I didn't just remove the ()s

[atomic rune]

true ngl 😭

[atomic rune]

alr give me a sec

ahhh i see

i understood it right before i just missed up when i did it

you said -2x^2 bc 2x^2 was positve

and you made the other 1 negative

[potent smelt]

Yes, I think

[atomic rune]

and the last 2 positve

i think i get it

[potent smelt]

Anyway, it is 5PM here and I need to go afk.

[atomic rune]

oh 😭

anyway thx for the help aprichat it

and i fr gotta work on my speeling

.close

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[red locust]

am i right with B here?

<@&286206848099549185>

[quaint plinth]

what do you know about end behavior?

[red locust]

sorry i mistyped i mean D

[quaint plinth]

yeah that looks right

[red locust]

thank you so much

[quaint plinth]

yw np

[red locust]

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[red locust]

am i right with B here?

<@&286206848099549185>

[red locust]

this is the table for the question

[quaint plinth]

what is the y intercept?

[red locust]

the y intercept is (0,0)

[quaint plinth]

no from the table

when x=0, what is h(x)

[red locust]

wait 6 then?

[quaint plinth]

yeah

so what equation has 6 as the y intercept

hope that helps

[red locust]

wait so A?

[quaint plinth]

yep!

[red locust]

thank you

[quaint plinth]

yw

[red locust]

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[gray saddle]

Would anyone be willing to help me make sure that I am doing this problem correctly?

Here is the problem:

This is what I based my answer off of

<@&286206848099549185>

[vernal ore]

I don't understand it but that looks interesting

[calm coral]

@gray saddle Has your question been resolved?

[gray saddle]

<@&286206848099549185>

please 😭

[naive estuary]

seems like it is okey

i do not understand the specifics of the subject but it looks right

[calm coral]

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[gray saddle]

.reopen

[calm coral]

✅

[gray saddle]

<@&286206848099549185>

[storm hearth]

how do they or you get ln(2)?

[gray saddle]

to be completely honest i have no clue i dont rmemeber the teacher going over this

[storm hearth]

💀

[gray saddle]

i think they got ln(2) from another example

let me show u one sec

@storm hearth

i think they got ln2 from here but tbh i did not understand it

[storm hearth]

oh i see

[storm hearth]

so it seems that they are using the model
W(t) = W_0 e^(-lambda t)
where W_0 is the initial amount, t is the time, and lambda is a constant

then they determine that the half life, when W(t) = 1/2 W_0 (half of the initial amount), then
1/2 W_0 = W_0 e^(-lambda t)
-> 1/2 = e^(-lambda t)
-> ln(1/2) = -lambda t
-> t = -1/lambda * ln(1/2)
t = ln(2)/lambda

[gray saddle]

so is it right to use ln (2) as i did in the work i did?

[storm hearth]

yeah

[gray saddle]

ok gotcha

[storm hearth]

yeah what you did is fine

[gray saddle]

and is my work correct, i wasnt sure if i did it wrong

the one i did in the picture

i think i understand it i just wanted to make sure

since all i did really was plug in the numbers into the example problem

[storm hearth]

it looks fine but uh

i got about 8133 rather than 8679

oh nah

i messed up i put

5370 instead of 5730 mb :skulL:

yeah your thing definitely looks correct xD

[gray saddle]

ok thx so much i was worried i did it wrong

i appreciate it 🫡

.close

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[exotic cosmos]

if a questions says a series of points are arranged on a number line, does that imply:

1. a < b < c

2. a <= b <= c

[potent smelt]

Not enough information. We would need a definition of "arranged," which I gather is intended to mean "placed on a number line in a monotonically increasing or nondecreasing manner," and also confirmation that a "series" of points refers to a, b, c, in order.

[exotic cosmos]

Ohh, so technically you could have 2 points coincident to each other (one on top of another)?

@potent smelt

sorry forgot to reply ping

i mean just reply 😭

[graceful dust]

[potent smelt]

That's what I'm trying to figure out, but based on what is written so far, it is impossible to determine, because the language used is either not specific enough, or I'm not aware of the specific definition.

[exotic cosmos]

hi

[graceful dust]

hi

[exotic cosmos]

oh, if one point is on top of another then isn’t that technically only one point

[potent smelt]

One point can have multiple names

[exotic cosmos]

what do you mean

[potent smelt]

If I say the point on the number line at 5 is named "a" there's nothing stopping me from also naming the same point "b"?

(After all, there is nothing to say that a sequence cannot repeat itself)

[exotic cosmos]

ohhh, i get it

[potent smelt]

https://oeis.org/A000012

[exotic cosmos]

okay thanks makes sense

[calm coral]

@exotic cosmos Has your question been resolved?

[calm coral]

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✅

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[exotic cosmos]

differentiate with respect to r on both sides

dV/dr = 4/3 * pi * 3 r^2

plug in r = 2 ft to get dV/dr

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[hidden ferry]

I find all primes p and q
such that pq | 5^p + 5^q. What I got: if p=q easy analysis p=q=5; now p≠q p=2: only (2,3); (2,5); p=3: only (3;2) it was before ; p=5 (2,5 it was before) and (5,313). Now p ≠ 2,3,5 and q ≠ 2,3,5. 5^p+5^q = 0 mod p => 5+5^q=0 mod p=> 5^(q-1) = -1 mod p Symmetrical for mod q 5^(p-1) = -1 mod q. 5^(p-1) = 1 mod p => 5^(q-p) = -1 mod p; symmetrical for mod q: 5^(p-q) = -1 mod q => 1/5^(q-p) = -1 mod q => -1 = 5^(q-p) mod q (that is same mod p) . maybe we can use orders here.

[calm coral]

@hidden ferry Has your question been resolved?

[calm coral]

@hidden ferry Has your question been resolved?

[hidden ferry]

<@&286206848099549185>

[dusk abyss]

what's the question?

[calm coral]

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[hidden ferry]

.reopen

[calm coral]

✅

[hidden ferry]

Find all (p,q)

[hidden ferry]

Do you have any ideas?

.close

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[abstract snow]

How'd I prove that the polynomial p defined by
P(x)=x^3-kx^2+2kx-4k has atmost one real root for k>0

[abstract snow]

P'(x)=3x^2-2kx+2k
This needs to have two roots by rolles theorem
D>=0
4k^2-24k>=0
k^2-6k>=0
k>=6
So for k<6 it doesn't work but I wanted to prove that it doesn't have roots for all positive values of k

Yes I am aware that the derivative being 0 at 2 points is a necessary condition but not the only one

[stoic crystal]

you got check on 0 too

k(k-6)≥0

it's union of both of answers

[abstract snow]

Note to myself:Why does it need two roots because if there were three roots a,b,c then f(a)=f(b) so

[abstract snow]

Please elaborate

[stoic crystal]

you edited ques?

I just realised

[abstract snow]

If you mean why I cancelled out the k in this inequality then it was given k>0

[abstract snow]

I edited the part that "no roots part" to "at most one root"

I didn't write it down properly so I am sorry for the inconvenience

Discord being laggy today

[stoic crystal]

yeah that changes a quite if things

[abstract snow]

I am once again sorry for the inconvenience

[stoic crystal]

well if you see the function is decreasing from 0 to 6 and then increasing from 6 onwards

[abstract snow]

(However the k>0 condition was already there when I wrote)

[stoic crystal]

yeah np

[abstract snow]

None of the functions mentioned by me are a function of k?

[stoic crystal]

and from 0 to 6 it's value would be in -ve and then after 6 onwards the function is constantly increasing

well that kinda proves it will cut x-axis only once

[abstract snow]

Sir

P is a function of x

Not a function of k

[stoic crystal]

oh shit, you're correct

[abstract snow]

Sorry I gotta go

[stoic crystal]

i dont understand where are you facing difficulty here?

[calm coral]

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[calm coral]

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[true blaze]

How do I prove that the 4 thales circles of a convex quadrilateral cover the entire quadrilateral using the pigeonhole principle?
The only proof I was able to come up with was that Let x be a point in ABCD, then since ABCD is convex, one of the angles AxB, BxC,... is less than or equal to 90° and thus lies in the thales circle from (idk what word would be gramatically correct here) AB.

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[hidden ferry]

find all primes p and q such that pq | 5^p + 5^q. What I got: if p=q => p^2 |25^p=> p=5=q; now p≠q p=2: 2q | 25+5^q => 2q | 5(5+5^(q-1)) => 2q | 5(5+1) => q = 3 or 5; p=3: 3q | 5^3 + 5^q => For it to be divided by 3, 3 and q must have different parity => q = 2 ; p=5: 5q | 5(5^4 + 5^(q-1) => q | 625 +1 => (626 = 3132) q = 2 or 313. Now p ≠ 5 and q ≠ 5. 5^p+5^q = 0 mod p => 5+5^q=0 mod p=> 5^(q-1) = -1 mod p Symmetrical for mod q 5^(p-1) = -1 mod q. 5^(p-1) = 1 mod p => 5^(q-p) = -1 mod p; symmetrical for mod q: 5^(p-q) = -1 mod q => 1/5^(q-p) = -1 mod q => -1 = 5^(q-p) mod q (that is same mod p) . maybe we can use orders here.

[burnt heart]

bro what

[modern peak]

okay first off none of that makes sense

format it better and maybe we'll understand

[silver sonnet]

the "maybe" lol

[hidden ferry]

The initial task: find all primes p and q such that pq | 5^p + 5^q

The rest are my ideas on this task

[hidden ferry]

Changed it

[modern peak]

again.

forget it better

we are not reading through that jumbled mess

[dreamy wing]

What does this mean =?

pq = 5^p + 5^q ?

I don't understand what the" | "is indicating

[hidden ferry]

5^p + 5^q = 0 (mod pq)

[dreamy wing]

Wait then if that equals 0

Oh okay

[calm coral]

@hidden ferry Has your question been resolved?

[hidden ferry]

p and q are prime numbers

[dreamy wing]

Then there... isn't a solution.

how can p and q be prime and that = 0

[hidden ferry]

mod pq

It means that 5^p + 5^q is divided by pq

[dreamy wing]

OH WAIT SORRY I MISSED THAT

SILLY ME

But still I don't really understand all the conditions sorry!!!

[hidden ferry]

@modern peak @dreamy wing @silver sonnet I hope it's clearer now

[calm coral]

@hidden ferry Has your question been resolved?

[hidden ferry]

.close

[calm coral]

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[copper cedar]

approximate from ln(1-ax) to the other thing

x approaching 0

is a not all of the values

how is it not like

all of them

since x is approaching 0 value of 0 is negligible?

[ocean cypress]

let's define ~ rigorously

[copper cedar]

approximate to each other

[ocean cypress]

$$f(x) \sim g(x) \iff \lim_{x \to 0} \frac{f(x)}{g(x)} = 1$$

[potent lotus]

Mr. Gamer 🇵🇸

[ocean cypress]

does that make more sense?

[copper cedar]

yeah

but i still dont know how to get a

i knew the approximate thing already

[ocean cypress]

well use the definition of '~' i just provided

take the limiting ratio

of f(x)/g(x)

[copper cedar]

limiting ratio?

[ocean cypress]

for what values of a does that limit equal 1?

[copper cedar]

like just equate the equations above each other?

all of them like below infinity

[ocean cypress]

let me be a bit more explicit here

[copper cedar]

yeah sorry if im not following

[ocean cypress]

first, try to find:

$$\lim_{x \to 0} \frac{\ln(1-ax)}{\sqrt[7]{1+(a+8)x}-1}$$

[potent lotus]

Mr. Gamer 🇵🇸

[ocean cypress]

do you understand why we're doing this?

[copper cedar]

to try to find a value

of a

[ocean cypress]

but why are we taking the limit of a fraction?

[copper cedar]

caues its a limit of x

[ocean cypress]

we want to see if they are 'asymptotically equivalent', meaning that they have the same rate of change as x goes to 0

maybe you don't quite grasp the '~'

[copper cedar]

ah

so its always = at all values?

even at 430 for exmaple

[ocean cypress]

not quite

we only care about the behavior of the two functions as x approaches 0

here's an example

$$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$

[potent lotus]

Mr. Gamer 🇵🇸

[ocean cypress]

so we can say that at x = 0, $sin(x) \sim x$

[copper cedar]

ah

[potent lotus]

Mr. Gamer 🇵🇸

[copper cedar]

okay

i found a taylor series thing

to expand ln

ill try that

[ocean cypress]

just use lhopital's rule

taylor series will get really messy here

[copper cedar]

it works on the big thing?

[ocean cypress]

it's a 0/0 limit isn't it?

[copper cedar]

yeah

[ocean cypress]

then it works

[copper cedar]

differnetating bottom will be hard

actaully not really

nvm

w

ill try

ty for help

ill report back

[calm coral]

@copper cedar Has your question been resolved?

[calm coral]

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[remote mural]

My professor is marking this down as a tautology on the review sheet

[remote mural]

I imagine this is a mistake right

unless im missing something

[desert torrent]

i agree with you

a or (a implies b)) = T
so (a and b) must be true, which is false

[vagrant oak]

this table is correct though

so the correct answer is: Not a tautology

[eternal shard]

I think => and -> make a difference

[vestal dust]

isn't -> "determines"?

[remote mural]

well, even asking chatgpt, it spits out something similar

i appreciate the responses, but i think its safe to assume an error on my professors side, best i shouldnt get hung up on it, time is of the essence

.close

[calm coral]

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[hasty tree]

CAN SOMEONE HELP PLEASE, I JUST WANT TO KNOW IF THIS TRIANGLE IS "NO SOLUTION"

[maiden vigil]

are you trying to say that "There is not any possible value for C"?

[hasty tree]

yesh

[teal drift]

What would it mean?

[maiden vigil]

well that's not the case here

[hasty tree]

cus its cos^-1(3.201)

idk

isnt it supposed to be

between

-1 and 1

😭

[teal drift]

Yes correct, probably you've made some mistakes

[hasty tree]

o

can u

halp

😭

[teal drift]

Well, send your calculations

[hasty tree]

yw

ye

wait

[hasty tree]

Ye i send

[maiden vigil]

erm you are supposed to find the angle C

(even if you want to find angle B, the formula you used is incorrect)

[hasty tree]

o

can u send me formula

ples

[maiden vigil]

,tex .cosine law

[hasty tree]

i forgor magic word

[hasty tree]

o

thnx bru

me say thank you

🙏

[maiden vigil]

np

[hasty tree]

lols

[hasty tree]

wait but

the formula i used was corect eh

[maiden vigil]

not quite

[hasty tree]

where is the mistake 😭

can u ples tell

so i learn

[maiden vigil]

First of all you were supposed to find the angle C and not the angle B

[hasty tree]

o

but why C

can u tell 😭

its cus like

we can find all 3 right

cus there all all the degress

[maiden vigil]

read the question dude

[hasty tree]

😭

me so nub

oh..

oh

OH

OHH

OHHHHH

@maiden vigil do u mind sending the answer which u solved

[maiden vigil]

!noans

[calm coral]

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

[hasty tree]

eh

oki

[maiden vigil]

if you're stuck somewhere then you could send your own solution for verification

[hasty tree]

o

[calm coral]

@hasty tree Has your question been resolved?

[calm coral]

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[gleaming ivy]

Hi! Any attempts for this question would be appreciated)

[unkempt drift]

is there a previous part to this question?

[gleaming ivy]

I've expanded sin5x using De Moivre's theorem, then simplified so it is all in terms of cos so far

[unkempt drift]

ohhhh

[gleaming ivy]

No, just this

[gleaming ivy]

I'm not sure where to go from here)

[ebon epoch]

use taylor series fr

[unkempt drift]

that won't help

[gleaming ivy]

Sorta

I've got everyting in cos but no solution yet

Thanks for the quick replies guys :)

[unkempt drift]

no I mean Taylor series won't work, haha

[gleaming ivy]

Haha yup

[ebon epoch]

it can

but it will have long solution frfr

[unkempt drift]

Taylor series are only approximations

furthermore, you don't know what the solutions even are, so you don't have a centre of convergence to use

[ebon epoch]

fr

wait we need to solve it?

where is equals to sign in this Q

[unkempt drift]

yes

[ebon epoch]

wait nvm

my eyes rn fr, slept for 2 hours frr

[gleaming ivy]

Nice pfp btw

[ebon epoch]

thx

my fav scintist

schrodinger

[gleaming ivy]

Love schrodie

[ebon epoch]

uss litteraly i started to do add maths only due to him frfr

[unkempt drift]

oh I see you haven't joined the r/IGCSE server

[ebon epoch]

wdymm

[unkempt drift]

there's a server for IGCSE

and it also has active A level channels

[ebon epoch]

demnnn

i am at o level rn fr

[unkempt drift]

same thing, we have O level people there

[ebon epoch]

dem so how do i join

[gleaming ivy]

Look up that server I guess

[unkempt drift]

okay, so from what you did $\frac{\sin 5x \tan x}{\sin x} = \frac{\sin 5x}{\cos x}$

[potent lotus]

south's secret twin brother

[unkempt drift]

similarly, sin(x) * sin(5x) / (tan x) can be written in terms of cos

jesus yeah I think just convert everything to cos

no wait you should be using $\sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x$ instead

[potent lotus]

south's secret twin brother

[gleaming ivy]

Oh really?

But there's still sin5x?

[unkempt drift]

yeah I don't think you can express sin 5x as powers of cos x

[gleaming ivy]

Hold on I can show you my working too

[unkempt drift]

otherwise you would need to use product to sum formulae with say cos(5x) and cos(3x)

idk

actually WA suggests using the half angle sub, t = tan(x/2)

[unkempt drift]

I think that's your best option

[unkempt drift]

do this then convert to half angle using sin x = 2t/(1 + t^2)

[gleaming ivy]

Ah, right

[gleaming ivy]

Makes sense

[unkempt drift]

https://www.desmos.com/calculator/manfie6lww

hopefully you get this

[gleaming ivy]

I've found a mistake, everything was not in terms of cosx :(

There was a sinx I didn't account for

But thank you I will try again using t formulas

[gleaming ivy]

Did you have to use graphical methods to solve it?

As in, is there no closed form solution you've found

[unkempt drift]

there is a closed form solution cause you get a big nasty polynomial, with neat roots

it's just hellish to get there

what kind of class are you even doing

also consider asking in #geometry-and-trigonometry if you want to follow up

cause help channels close and stuff

[gleaming ivy]

Will do

[gleaming ivy]

Alright, can I ask what you got here?

[unkempt drift]

I don't have it

[gleaming ivy]

Ah that's okay

[gleaming ivy]

But you got there this way?

[unkempt drift]

https://www.desmos.com/calculator/dhy3uhbwgq

I used Wolfram Alpha to simplify the expressions and you eventually get to this

=0

it's a miracle the 5 - 10t^2 + t^4 appears twice

[gleaming ivy]

Can't wait for that algebra 😅

[unkempt drift]

but then you still have to cross multiply

[gleaming ivy]

Ooh

[unkempt drift]

btw

[gleaming ivy]

That's saved me hours thanks XD

[unkempt drift]

basically use Wolfram Alpha to check each result

[gleaming ivy]

Oh, A-level Further Maths - and I got this question from some physics entrance exam prep

[unkempt drift]

I don't know how you can factor it like that either, I guess just keep guessing roots

[unkempt drift]

really?

there must be an easier way cause this is too hard for further maths

is that entrance exam from India by any chance

also I tried searching the question online and nothing came up

I'm hoping the sin 5x is a typo

[gleaming ivy]

Nah it's UK - prep questions given by a uni I'm applying to

[unkempt drift]

which uni

STEP is not this hard

[gleaming ivy]

I doubt lol

[gleaming ivy]

Er, might be disclosing too much there 😂

But thanks so much for your help

[unkempt drift]

oh you can DM if you want

I'm just insanely curious

and I think there's some super clever way that doesn't require this much algebra

np!

[gleaming ivy]

There probs is, as always XD

[calm coral]

@gleaming ivy Has your question been resolved?

[calm coral]

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[digital hearth]

How do i find the area of this whole thing

It confuses me because of the curved lines. There is no such tool to take the area of those elements

[calm coral]

@digital hearth Has your question been resolved?

[calm coral]

@digital hearth Has your question been resolved?

[calm coral]

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[shadow trout]

hii would this be washer or shell method? thank you in advance :)

[eternal shard]

washer

[shadow trout]

tyy

.close

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[manic scroll]

How come my professor's answer is different? (Incoming image)

[manic scroll]

Professor's answer

Only the first part is different

[rustic osprey]

$\int \frac{1}{f(x)} \dd{x} \neq \ln |f(x)|+C$ in general

[potent lotus]

Civil Service Pigeon

[manic scroll]

How to resolve it then? Regular substitution method?

[rustic osprey]

,w derivative of cot x

[potent lotus]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[rustic osprey]

Recall this

[manic scroll]

Is this correct??

How does this turn into his, though?

[rustic osprey]

Draw a right triangle with legs x and 3 and indicate the angle that is arcsen(x/3)

Then the cotg of that angle using the triangle

[manic scroll]

😵💀

I think I will stop here and hope that in an exam he would evaluate where I arrived as correct enough

.close

[calm coral]

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[supple ridge]

top question

[supple ridge]

How would i go about? im trying to do substitutions and get stuck when isolating K

[calm coral]

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

[supple ridge]

0 clue where to go

[rustic osprey]

I’d let the log be base e for convenience (aka ln)

So you have $\ln \frac{1}{10} =-k(0.01)$

[potent lotus]

Civil Service Pigeon

[supple ridge]

I never learned ln in school, is it fine to do the question without it?

Or is it just another way of saying something

[rustic osprey]

Suck it up and shove the whole thing in a calculator then

$k=\frac{\log \frac{1}{10}}{-0.01 \log e}$

[potent lotus]

Civil Service Pigeon

[rustic osprey]

(Ok it’s not that bad but I’m lazy as hell)

[supple ridge]

syntax error

@rustic osprey

#13, not sure where i messed up, but i tried 2 methods and keep getting 54

Answer key says 17, is there any mistakes u see

I also tried just making one log, but log_1.1^2500/14 also = 54

[calm coral]

@supple ridge Has your question been resolved?

[supple ridge]

<@&286206848099549185>

[remote mural]

Can you send the complete question

Nvm

[supple ridge]

The first variable is P

that was cut off

[remote mural]

Yeah figured it out

[remote mural]

C is the circumference

What they asked is the diameter

[supple ridge]

Ohh

Got it thanks

Waig

wait

But if my diameter is 54, wouldnt the circumference be around 200

Lemme plug it in

[remote mural]

Circumference =pi d

[supple ridge]

Oh i have the circumference, nvm i got it

.close

[calm coral]

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[tulip flint]

any1 know how to compute probability of F

im so lost rn

[calm coral]

@tulip flint Has your question been resolved?

[tulip flint]

@amber bolt sorry for the ping but you helped me yesterday if your free you able to help?

[amber bolt]

no i'm afraid of notation

or maybe of "proving"

[balmy kindle]

@tulip flint shouldn't it be $$j \neq i$$ for the first part instead of $$j\neq 1$$

[potent lotus]

hi im Bilel

[tulip flint]

yea mistake

[tulip flint]

its just using bayes rule

but i dont understand the bottom term and how to get there

which is this part

[tulip flint]

i understand that 1-alpha is the proability of not finding ball when in box i

but i dont get the multiplication of pi

[balmy kindle]

It should be $$P(E|F)=\frac{P(F|E)P(E)}{P(F)}$$

[potent lotus]

hi im Bilel

[balmy kindle]

yours has P(E) in the denominator @tulip flint

[tulip flint]

i forgot to change that mb

ur right it should have p(F) denom

[tulip flint]

i just dont understand how to get P(F)

[balmy kindle]

Find the complement. What is the probability that a search of box i discovers the ball?

what two things need to happen for this to occur

[tulip flint]

alpha

[balmy kindle]

not quite

[tulip flint]

why not

[balmy kindle]

alpha is the probability you find it, but the ball has to be in the box to begin with for it to work

so the probability is $$\alpha_i p_i$$

[potent lotus]

hi im Bilel

[tulip flint]

oh

[balmy kindle]

does this part make sense?

[tulip flint]

yea

[balmy kindle]

yes so this is the probabiliyt that the search of box i discovers the ball. but we want to NOT discover ball.

so what do we have to take?

[tulip flint]

the complement

[balmy kindle]

yes so that's where $$1-\alpha_i p_i$$ comes from

[potent lotus]

hi im Bilel

[balmy kindle]

is this part clear?

[tulip flint]

yea

[balmy kindle]

so that's P(F), you're correct about P(F|E) and P(E) is p_j, so the numerator is just 1*p_j

so the fraction is $$\frac{1\cdot p_j}{1-\alpha_i p_i}$$

[potent lotus]

hi im Bilel

[balmy kindle]

which is just $$\frac{p_j}{1-\alpha_i p_i}$$

[potent lotus]

hi im Bilel

[balmy kindle]

so that's the first part

lmk if u have questions

[tulip flint]

ok so for part 2

P[E} is just P_I

because J=i

i dont get how its 1-alpha for the prob of it being in box j given that the search box i does not find the ball

[balmy kindle]

so P(F)?

the probability that seaching box i discovers the ball is 1-P(F). in order to discover he ball, you need the ball to be in the same box that you're searching. the probability that the ball is in box i is P_i and the probability of you finding that ball is alpha_i, so you have to multiply both of those because you want both those things to happen.

[tulip flint]

no

[balmy kindle]

oh ok

[tulip flint]

this term which would be P(F|E)

[balmy kindle]

well you're searching box i and we know that the ball is in there

what's the probability that you find it?

[tulip flint]

alpha

[balmy kindle]

yes so what's the probability you don't find it

[tulip flint]

1-alpha

okay

[tulip flint]

is the P(F) for J =1 the same for J does not equal 1?

[balmy kindle]

you mean j=i?

[tulip flint]

oh yea

sorry

[balmy kindle]

yeah it's the same for both parts

[tulip flint]

yeah but is the reasoning

the same

[balmy kindle]

yes

all good?

[calm coral]

@tulip flint Has your question been resolved?

[calm coral]

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[halcyon ore]

i rlly dont have the unit circle memorized

[halcyon ore]

is there anyway

if i type like sin30 in my calc

i can get the fraction version?

i always get a decimal

if not i guess i memorize it tmr

ok

ik sin30 is 1/2 obv

like sin 60 i guess

[agile jetty]

It's best if you just memorize it. You'd have to use a scientific calculator or something for it to show as a fraction

[halcyon ore]

i have scientifc

cuz its like physics class

i guess its best

[agile jetty]

It is. You will probably keep using it in the future

Solving exercises is the best way to memorize it

[calm coral]

@halcyon ore Has your question been resolved?

[calm coral]

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[dark osprey]

how does that simplify to that

[prime yacht]

$9 - 6x - 21$ = $-6x - 12$ = $-6(x+2) $

[cedar kernel]

Wdym

[potent lotus]

Adarsh

[dark osprey]

but its + 21

oh wiat

ok

i see

not sure what i should be doing if i should be rationalizing

or factoring the highest power

[leaden thunder]

Just find the behavior of the highest powers in the numerator and denominator

[dark osprey]

got it

6/3

2

calculating f'(x)

i tried the quotient rule but my answer is wrong

[tacit moat]

what's the derivative of x sinx?

[dark osprey]

ohh i forgot about the x part of sinx

sinx + xcosx

[tacit moat]

yup

[dark osprey]

why does this not exist

instead of being negative infinity

is it becuase of the unit circle?

[calm coral]

@dark osprey Has your question been resolved?

[calm coral]

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[distant lagoon]

what would I do for k^x+1 = (square root of k)(k^2x)

[bleak ermine]

is it k^(x+1) or k^x

[distant lagoon]

k^(x+1)

[bleak ermine]

how can u re write

$\sqrt{k}$

[potent lotus]

Lebob

[distant lagoon]

would it be something like k^1/2

[bleak ermine]

yes

then simplify $k^{1/2} * k^{2x}$

[distant lagoon]

so its k^(x+1) = (K^1/2)+K^(2x)

[potent lotus]

Lebob

[bleak ermine]

ahh

no

they are multiplied remember

[distant lagoon]

Am I using the product rule?

[bleak ermine]

well when u multiply 2 numbers with indicies

u add the powers

[distant lagoon]

so it would be k^(1/2+2x)

[bleak ermine]

yea

then u just equate them

[distant lagoon]

so we would need to transpose next

[bleak ermine]

no not even

[distant lagoon]

ohh just skip to solving?

[bleak ermine]

u can do this

thats how exponential equations work if the bases are the same

[distant lagoon]

yep so just solving then

[bleak ermine]

if the bases arent the same the first thing you do is MAKE THEM THE SAME so u can do this

yea pretty much

[distant lagoon]

so x+1-1/2+2x

sorry typo

[bleak ermine]

alg

[distant lagoon]

x+1 = 1/2+2x

[bleak ermine]

yes

and solve

[distant lagoon]

which is x = -0.5 +2x

[bleak ermine]

no

x=1/2

oh wait

sorry ur still going

yep

[distant lagoon]

what would I do next?

unsure how to simplify the 2x bit

[gray smelt]

$(x+1) = \frac{1}{2}+(2x)$

[potent lotus]

Alaska

[gray smelt]

before subtracting simplify the right side first

[distant lagoon]

I cant transpose at this stage, can I?

[gray smelt]

1/2 +2x

[distant lagoon]

Not sure how to simplify the right side without transposing

How would I simplify 1/2+2x

[gray smelt]

$\frac{1}{2}+\frac{2x}{1}$

[potent lotus]

Alaska

[gray smelt]

to make the denominator same for the for 2x/1

multiply both numerator and denominator by 2

only for (2x)/1

[distant lagoon]

so 4x/2

[gray smelt]

yes

[distant lagoon]

1/2+4x/2

[gray smelt]

which is

[distant lagoon]

Would I do 5x/2 or 5/2 + x?

[gray smelt]

$(x+1) = (\frac{4x+1}{2})$

[potent lotus]

Alaska

[gray smelt]

keep the (4x+1) term in right side

multiply both sides by 2

[distant lagoon]

so 2x+2 = 4x+1/2

[gray smelt]

just 4x+1

on right side

because 2/2 = 1

[distant lagoon]

ohh bc of multiplying

yep

2x+2 - 4x+1

[gray smelt]

yes

[distant lagoon]

I mean = not -

[gray smelt]

2x+2 = 4x+1

[distant lagoon]

yep

[gray smelt]

isolate the variables

and numbers

[distant lagoon]

would I transpose them then?

swap sides?

[gray smelt]

yes

[distant lagoon]

so on one side I would have 2x-4x = 1-2

[gray smelt]

yes

[distant lagoon]

which is -2x = 1

[gray smelt]

-1*

-2x = -1

[distant lagoon]

which means -x = -0.5

yep sorry

x=0.5

[gray smelt]

yes

very good

[distant lagoon]

thank you!

Doing hours worth of math really does scramble my brain haha

[gray smelt]

remember to simplify fractions

before subtracting them

[distant lagoon]

so x =1/2

got it, thanks!

[gray smelt]

!done

[calm coral]

If you are done with this channel, please mark your problem as solved by typing .close

[distant lagoon]

.close

[calm coral]

Channel closed

Closed by @distant lagoon

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[calm coral]

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[mossy geode]

😭 why is this so hard

[paper haven]

what does number of points mean?, is it like the area, rounded down to an integer?

[calm coral]

@mossy geode Has your question been resolved?

[pearl skiff]

I'd just draw it out and count the points lol

[mossy geode]

coordinates

[mossy geode]

this is more of a pnc thing

[mossy geode]

i did draw

[pearl skiff]

it's like the points P(a,b) are on the lines y=kx for +ve integers k

they are in a really pattern 🙂

[mossy geode]

it says strictly inside and not on the line tho

[pearl skiff]

the points are on these y=kx lines 😛

[calm coral]

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[unreal flame]

Tried a lot of time, and still don't get why is my answer wrong 🥲

[sullen dock]

What answer

[jolly pilot]

Have you tried plotting the diagram for velocity wrt time?

It would make your problem clearer

Use that to plot for acceleration

[unreal flame]

So I just have to check the slope of the graph to find velocity right? But then I still don't understand why is my answer wrong 😭

[jolly pilot]

|| your velocity answers dont look wrong ||

[unreal flame]

Ye but it say I'm wrong idk why

[jolly pilot]

You need all 4 correct simulteneously

your other 2 are wrong

[unreal flame]

Oh