help-26

[trail timber]

need help

[normal adder]

lhopital

[trail timber]

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[reef fjord]

conjug-

okay

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[tawdry tundra]

Hi can someone help me out on this

[tawdry tundra]

I want to understand how the domain and the intervals work with the inequalities

<@&286206848099549185>

[topaz sinew]

@tawdry tundra Has your question been resolved?

[tawdry tundra]

<@&286206848099549185>

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@tawdry tundra Has your question been resolved?

[tawdry tundra]

<@&286206848099549185>

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@tawdry tundra Has your question been resolved?

[sand egret]

let me look

nvm I don't think I can help with that at the skill level I have

[tawdry tundra]

Alr thanks for looking at it

[twin heath]

sorry what part of it do you need help on?

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[neon iron]

I don’t get how this works

[neon iron]

This is what I did but the bases make it hard to do quadratic

[vernal matrix]

What base is that other log supposed to be? Natural log, or?

If anything, base change that too, I guess?

[agile harness]

does this say $log_x(1000) - log(z^2) + (10^{-3})^0 = 0$?

[neon iron]

Yeah

[sharp birch]

oh wait thats a z???

[neon iron]

But x

[sharp birch]

mb

[neon iron]

No

X

[thorny flame]

knief

[sharp birch]

oh

[agile harness]

it looks like a z

[neon iron]

I ended up with this but I can’t just merge a logx and lgx together right?

[sharp birch]

then you messed up somewhere cause youre way off

[neon iron]

I triple checked

[sharp birch]

oh wait im being dumb

1s

[neon iron]

Ok

[agile harness]

so $log_x(1000) - log(x^2) + (10^{-3})^0 = 0$

[thorny flame]

knief

[agile harness]

and is log just log base 10?

[sharp birch]

its either log base 2 or 10

if nothing is given

[neon iron]

Isn’t it base 1?

[sharp birch]

its always assumed to be base 10 or base 2 unless given

[agile harness]

lol

[neon iron]

Otherwise they would’ve put lg

[sharp birch]

it depends on where you live

[agile harness]

base 1

😹😹

[neon iron]

Yeah

[agile harness]

sorry

[neon iron]

I don’t get it

[agile harness]

there is no such thing

..

[neon iron]

Oh

[sharp birch]

(btw base 1 would be 1 to the what power would be <something>)

[agile harness]

logs are the inverse functions of exponentials

tell me

[sharp birch]

(meaning its impossible)

[agile harness]

what is 1^any power

[neon iron]

So I can just turn logx into lgx?

[neon iron]

1?

[agile harness]

yes sir

[neon iron]

Yeah

[sharp birch]

start from here, sub in t for logx

solve from there

[neon iron]

Ok

[agile harness]

what’s lg(x)

you mean ln?

[neon iron]

No

[agile harness]

⁉️⁉️

[neon iron]

I definitely put my

LG*

[agile harness]

but what is lg

[neon iron]

log_10

[sharp birch]

are you shorttening to lg for log

:p

[agile harness]

then why did you write log for the other

[neon iron]

Idk

[sharp birch]

^anyways do this and itll work out, i just got the answer

imma go sleep now

gn sheeple

[neon iron]

So I can just put logx as lgx?

[agile harness]

go sheeple

[agile harness]

please don’t

[neon iron]

But you said assume the base is 10

[agile harness]

yes but write it as log

not lg

[neon iron]

Ok

[agile harness]

log is generally assumed to be the common log or base 10

some people use it as ln

[neon iron]

Ok got it

[agile harness]

or base 2

but given that the problem has 1000 and 10^-3

i’m going to assume it’s base 10

[neon iron]

Yeah

[agile harness]

do you know the change of base formula

[neon iron]

Yeah

[agile harness]

what is it

[neon iron]

log_a(b)=log_c(b)/log_c(a)

[agile harness]

right so we can change the log_x to log

how

[agile harness]

yes

but please

again

just write log(x)

[neon iron]

Ok

[agile harness]

no one writes lg(x)

[neon iron]

Oh…

Then it’s just me

[agile harness]

so our equation simplifies to $$\frac{3}{log(x)} - log(x^2) + (10^{-3})^0 = 0$$

[agile harness]

indeed

[neon iron]

Ok

Ik what to do from here Ty

[agile harness]

what is $(10^{-3})^0$?

[thorny flame]

knief

[agile harness]

ok

[neon iron]

1

[agile harness]

here you didn’t

[neon iron]

Oh

[thorny flame]

knief

[agile harness]

do you know how to solve it

[neon iron]

Yeah

0.1

[agile harness]

did you use a calculator

[neon iron]

No, I just got it down to 10^-1

[sharp birch]

There’s a second answer

[neon iron]

Oh yeah

31.6(3s.f.)

[sharp birch]

Ew 3 sig figs

Hate sig figs

[neon iron]

But that’s the way

[sharp birch]

Disgusting

[neon iron]

But that’s how I always do it

[sharp birch]

Alright now I’m actually gonna go to bed

[neon iron]

Ok

1/3 = 0.3333333333…

3/3=0.9999999999….

[sharp birch]

Always go to 2 decimal points unless stated otherwise or in context (can’t have less than 1 of a human being)

[neon iron]

Ok

O = 360°
⬛️=360°
O=⬛️

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[spice sluice]

Would this mean the slope is zero when it says equal to zero

[gray ridge]

yes

[spice sluice]

👍

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[plain depot]

$$ \int \frac{1}{\sqrt{x^2-a^2}} ,dx $$

I'm confused regarding the x<-a case. I'm getting a different function, but the textbook doesn't clarify.

For theta in pi/2 to pi, tan theta becomes negative, so we factor with a negative.
But this is not done in the textbook

[thorny flame]

Anurag.[UdayS]

[plain depot]

This is my solution

But the textbook just ignored the second branch

<@&286206848099549185>

[topaz sinew]

@plain depot Has your question been resolved?

[gray ridge]

very interesting question indeed!

I'm not sure if this is the right approach, but for the results of integrals, it can be negative

while some books maintain positive around finding 'area' under curve, for integrals, we can consider area under x-axis as negative.

[thorny remnant]

ummm they both are the same..

[plain depot]

The second branch also gives some positive values

[plain depot]

How would that be? I'm unable to convert them. Kindly share.

[thorny remnant]

ok

sure

[gray ridge]

and i was checking with desmos lol

[gray ridge]

now that you've mentioned, haha

[plain depot]

There is a minus between one of them

I'm unable to get rid of it. I think convergence is going to show how

[thorny remnant]

$\frac{-1}{5}\ln|5x-1-\sqrt{5}\sqrt{5x^2-2x}|=\frac{1}{5}\ln|\frac{1}{5x-1-\sqrt{5}\sqrt{5x^2-2x}}|$

[thorny flame]

convergence

[gray ridge]

and then multiplying by the conjugate

[gray ridge]

thank you so much! that's hard to type

[plain depot]

Oh I think I see it now

But is this possible for the general x² - a² expression

Well I guess it's worth a shot

[gray ridge]

1/(x-a) * (x+a)/(x+a)

[thorny remnant]

it sure was..

[plain depot]

Damn

I feel so stupid I missed that

And spent an hour without taking the conjugate

[thorny remnant]

its alright

[gray ridge]

yea, log sometimes mess us up

[thorny remnant]

this happens alot of times

[gray ridge]

nice observation!

[plain depot]

The books tricked me actually

They straight up wrote root (tan²∅) = tan ∅

[thorny remnant]

me have a simple brain >sees - before log thinks of that property..

[gray ridge]

haha, for √ notation, we take a positive
e.g.
x²=4
x=±√2
but
√4=2, but not ±2

[plain depot]

Yeah, but root of a function, and that's as unsafe as C

[thorny remnant]

C?

[plain depot]

The Programming language

[thorny remnant]

oh i see

[plain depot]

It's unsafe cause there's bunch of things to consider

Same for square root of functions (at an elementary level)

[gray ridge]

note that tan²(theta) = (tan(theta))² is always non-negative where theta is in the domain of tan(theta)

[plain depot]

Yeah but sqrt (tan² x ) = |tan x|

≠ tan x

The book just ignored that

Anyways the problem is solved now. Thanks guys~

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[covert juniper]

how to prove this?

[opal vault]

hi, have you tried PFD?

or induction

[covert juniper]

what is that?

[opal vault]

partial fraction decomposition

[covert juniper]

I forgot induction

[covert juniper]

yeah

[opal vault]

well for example

when you have a rational function

so $P(x)/Q(x)$ where P and Q are polynomials

[thorny flame]

rafilou2003

[opal vault]

say you factored Q

so simplest example

$\frac{1}{(x-1)(x+1)}$

[thorny flame]

rafilou2003

[opal vault]

this is just an example

well you can decompose this into

$\frac{A}{x-1} + \frac{B}{x+1}$

[thorny flame]

rafilou2003

[opal vault]

where A and B are to be determined

[covert juniper]

I know PFD becuase I have done calc courses and before that I learnt it well but how can you use this thing to solve that problem?

[opal vault]

you'll see, there's a nice surprise

try PFD on 1/(k(k+1))

[covert juniper]

I get 1/k - 1/(k+1)

[opal vault]

yep

[opal vault]

so rewrite this left hand side

with that in mind

[covert juniper]

I get a different answer

I start from 1/(n+1) and end in 1/2

like decomposing the last term I get 1/n -1/(n-1) .As the second last term I get 1/(n-1)-1/n.

[opal vault]

?

the last term is 1/n - 1/(n***+***1)

[covert juniper]

I notice the first fraction of the last term cancels out with the last fractoin of the second last term

[opal vault]

second to last term is 1/(n-1) - 1/n

[covert juniper]

so I conclude that only the first and the last fraction would survive at the end

[opal vault]

which "first" and "last" fraction

[covert juniper]

which is 1/2 and -1/(n+1)

[opal vault]

and there's the problem

(1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/(n-1) - 1/n) + (1/n - 1/(n+1))

are you sure it's 1/2 that "survives"?

[covert juniper]

shouldn't the first term you used in that series be undefined? because the pattern tells that the first term should be 1/0.1 which is 1/0

[opal vault]

what

1/(k(k+1)) with k = 1

1/(k(k+1)) = 1/k - 1/(k+1)

[covert juniper]

ok I see

[covert juniper]

shouldnt the first fraction of first term be 1/(n-n) then?

the last one was 1/n+1 then 1/n then 1/n-1 then 1/n-2 and if we go like this we would end up in a fraction 1/(n-n)

[opal vault]

huh

no we stop at 1/(n-(n-1)) - 1/(n-(n-2))

[cedar wagon]

You can't have 1/(n-n) since it will divide by zero anyway

[silent siren]

The sequence, if it's any help, is $$\sum_{i=1}^n \frac{(i-1)!}{(i+1)!}$$

[thorny flame]

Shuba

[opal vault]

let me reassure you, it doesn't help

sorry xdd

$$\sum_{i=1}^n \frac{1}{i(i+1)} = \sum_{i=1}^n \left(\frac{1}{i} -\frac{1}{i+1}\right)$$

[thorny flame]

rafilou2003

[topaz sinew]

@covert juniper Has your question been resolved?

[silent siren]

You can prove it inductively.

Suppose $\sum_{i=1}^n \frac{1}{i(i+1)} = \frac{n}{n+1}$. Show $$\sum_{i=1}^{n+1} \frac{1}{i(i+1)} = \sum_{i=1}^n \frac{1}{i(i+1)} + \frac{n}{n+1} = \frac{n+1}{n+2}$$

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[thorny flame]

Shuba

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[cyan harbor]

How do you get range :{y/y≥-16} from f(x)=x²+6x-7?

[cyan harbor]

I'm stuck at a part

[frosty iron]

try to make f(x) into a perfect square, and you will be able to find f(x)'s minimum

[cyan harbor]

I'm gonna send my progress

Stuck here

[frosty iron]

try to put f(x) = (x + a)^2 + b

[cyan harbor]

How

[frosty iron]

well expanding this gets you x^2 + 2ax + a^2 + b

comparing this to the original equation, 2ax = 6x

then solve for a

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[mortal heart]

Guys pls help. The blue pen is my working

[restive inlet]

do you know the formulas for
area of a triangle
and trig formula for area of a triangle

[mortal heart]

Yess

I know base*height/2

And 1/2abSinC

[restive inlet]

apply the trig formula for the triangle

forgot to ask whether you knew about area of a sector

[mortal heart]

Yes i knoww

[mortal heart]

Sin cos tan?

[restive inlet]

?

you told me the relevant formula already,
apply that

[mortal heart]

ohh

but i don’t know what to do

i’ll try finding the area of the triangle and making it equal to the area of sector

[topaz sinew]

@mortal heart Has your question been resolved?

[mortal heart]

anyways thanks for ur help!

.close

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[distant spoke]

5. how is it an arithmetic sequence? and what is the common difference?

[distant spoke]

2. how do i find the common difference and 12th term?

[pseudo jetty]

you can check if it is an arithmetic sequence by computing the difference between adjacent terms

[distant spoke]

how about the common difference?

[pseudo jetty]

the common difference will be the difference between the adjacent terms when theyre all the same

i.e., common

[distant spoke]

so how do i compute the difference between adjacent terms? is there a formula i need to know

[pseudo jetty]

you just subtract the numbers

[arctic briar]

:0

[pseudo jetty]

the difference between two numbers is when you subtract them

[distant spoke]

even with fractions?

[pseudo jetty]

yes

[arctic briar]

deah explode moment

[distant spoke]

so something like 13 - 25/2?

[pseudo jetty]

yes

[distant spoke]

okay

so the common difference is -1/2? because when you subtract 25/2 and 12 it becomes 13/2, again with 13 and 25/2, it becomes 12/2

am i correct?

[arctic briar]

*13 = 26/2

*12 = 24/2

[distant spoke]

ohh

but is it still an arithmetic sequence since its multiplication?

shouldn't it be geometric?

@arctic briar

[arctic briar]

its geometric

[distant spoke]

okay got it

how about this?

[arctic briar]

nono wait wrong way around

geometric is multiplying,

arithmetic is adding

[distant spoke]

yeah

[arctic briar]

yea =

[distant spoke]

so what is it

arithmetic or geometric?

[arctic briar]

well how do u get from 12 to 25/2 and get from 25/2 to 13 etc

[distant spoke]

yeah exactly

[arctic briar]

do u multiply by a number or add by a certain number

[distant spoke]

forget it

[distant spoke]

how abt this?

[arctic briar]

ok lets go really basic

how do u write 1 in quarters

like when 4 is a denominater

then u can easily see the pattern

[distant spoke]

1/4

?

[arctic briar]

erm

[distant spoke]

4/4?

OHHH

okayy

[arctic briar]

yea

see the pattern now

[distant spoke]

so the common difference is?

[arctic briar]

6/4-5/4 or 5/4-4/4 is?

[distant spoke]

1/4

[arctic briar]

yea

there we go

[distant spoke]

okayyy tyty

[arctic briar]

np we got there

now off i go byebye

[distant spoke]

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[silk swallow]

I need help on how to get the answer (blue writing is the answer)

[slate radish]

90 means perpendicular to each other, so line KR and line RL must have relationship of gradient m1m2=-1

thereforce let R be (x,y), and find two m1 and m2 respectively that contain x and y

and just multiply them and =-1

[silk swallow]

okay lemme try brb

[slate radish]

also u learning STPM? i dont rmb SPM is this hard lol

or u r not malaysian

[silk swallow]

Huh

How uk im malaysian

Oh

Yea this is spm

They changed the syllabus

Nice i got the answer

Thank u

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[vestal hatch]

I answered 2 initially was told I was wrong. Can I have some help with this?

!status

[topaz sinew]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[vestal hatch]

3 - wrong answer

[topaz sinew]

@vestal hatch Has your question been resolved?

[lucid bridge]

(11-2x)/x+3
(-2x-6+17)/x+3
-2+(17/x+3)
in the form y=-2+(17/x+3)
y+2=17/x+3
(y+2)/17=x+3

i think it should be stretched by 17 lol

@vestal hatch

(the y is divided by 17)

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[grand trellis]

Isnt 18 just 1

(n) +(n+1) = 2021

So n will be 1010

[neon iron]

well yeah

but not only

it says number of sets

so u could also have (n) + (n+1) + (n+2) = 2021

and so

so on

[grand trellis]

Oh ;-;

How do I do it then?

<@&286206848099549185>

[topaz sinew]

@grand trellis Has your question been resolved?

[gray ridge]

interesting question!

you can try to see the pattern when there are 2, 3, 4,...., n consective integers

e.g.
when there are 2 consective integers, it's 1010+1011
when there are 3 consecutive integers, it's not possible (why?)
when there are 4 consecutive integers, also not possible
when there are 5 consecutive integers, also not possible
when there are 6 consecutive integers, .....
grrr
what i mean is you can just try to figure out the general case
when there are n consecutive integers

[grand trellis]

I see

[grand trellis]

Because n wouldn't give an integer?

[gray ridge]

so, do you have any idea on how to find it?

[grand trellis]

Hmm lemme think

[gray ridge]

yea, we can't find such integers

[grand trellis]

I don't really get it ;-;

I can't find a pattern

[gray ridge]

okay

when it's 2 consecutive integers
what we are doing is 2n+1=2021
right?

[grand trellis]

Yea

[gray ridge]

so for 3 consecutive integers,
we wanna find n such that
3n+1+2=2021
right?

[grand trellis]

Yea

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[grand trellis]

.reopen

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✅

[gray ridge]

and for 4 consecutive integrrs
we wanna find n such that
4n+1+2+3=2021

so, the pattern continues

[grand trellis]

Yep

How do we determine if n would be an integer or not?

Like we can't check manually

There has to be some limiting case or something?

[gray ridge]

so, for k consective integers,
we wanna find k such that
kn+1+2+3+...+(k-1)=2021

[grand trellis]

Yea

[gray ridge]

so, we just have to check all the numbers with 2021-k(k-1)/2 are positive and divisible by k

[grand trellis]

Ohhh

And the limit to where we check would be 4042 -k(k-1) >= 2k ?

But wouldn't this be too large?

Like there should be a smaller limit to where we check the value of k

[gray ridge]

well, i just use wolfram alpha to check, it's only 64 cases

[rigid schooner]

Try using this:

Since we know that $S_n = 2021$, we can therefore assume, that $n|4042$

[thorny flame]

Randel_

[grand trellis]

Oh

[gray ridge]

this is better @grand trellis

[grand trellis]

Yea I'm tryna understand that

[grand trellis]

I see

So n is a factor of 4042?

[rigid schooner]

Yep

Therefore the only options for n are 1, 2, 43, 47, 86, 94, 2021, 4042

[grand trellis]

Hmmm

I don't think this works

[grand trellis]

Ig the equation we get is not exactly of this form

[rigid schooner]

Yeah

$S_n = \frac {(a_1+a_n)} {2} * n$

[thorny flame]

Randel_

[rigid schooner]

$2*S_n = (a_1 + a_n) * n$

[thorny flame]

Randel_

[rigid schooner]

$S_n = 2021$

[thorny flame]

Randel_

[grand trellis]

This the formula for the sum of an AP i think

[rigid schooner]

Yes

[grand trellis]

This isn't exactly an AP ig

[rigid schooner]

it is

[grand trellis]

what about the kn?

[rigid schooner]

Okay let's try to rephrase this

$a_1 + a_2 + a_3 + ... + a_n = 2021$, where $a_n = a_{n-1} + d$ and $d = 1$

[thorny flame]

Randel_

[grand trellis]

Oh i see

Mb

[rigid schooner]

it's alr

[gray ridge]

👍

[grand trellis]

How are we not getting 64 values then?

[rigid schooner]

It's important to note that $a_n = a_1 + (n-1)d$

[thorny flame]

Randel_

[rigid schooner]

No idea how we are supposed to get 64 values

[grand trellis]

Ye ik this

[grand trellis]

Oh ;-;

[grand trellis]

.

[gray ridge]

let's stick to randel's method for now👍

[grand trellis]

Alr

[rigid schooner]

So with this logic we only need to check for these values of n

I believe you can figure out the rest

[grand trellis]

Yea ig I'll have to try on my own ;-;

[grand trellis]

Ty tho

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[rigid schooner]

np

Have a nice one 👋

[grand trellis]

You too

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[potent crypt]

Find the value of x is the question

[potent crypt]

How do u solve b and d

[polar pollen]

You can apply addition of fractions

To solve them

[shut garden]

Hint : ||Multiply both sides by 6 to get rid of the fractions||

[potent crypt]

i dont really need hints i js wanna know how to solve them i havent opened math in so long😭

[potent crypt]

how?

[polar pollen]

Take the terms containing x one side

[shut garden]

Substracting fractions are exhausting.

[polar pollen]

And all others numbers another side

[shut garden]

So take my hint.

[polar pollen]

So is doing maths :))

[potent crypt]

and how do i subrtact 1 over 2 and 1 over 3 without calc

[polar pollen]

You don’t

[potent crypt]

what do i do

[polar pollen]

The only way to do it is with calculating it

[shut garden]

Bruh.

[potent crypt]

how do i add them without calculator

[polar pollen]

Oh u mean like that

[potent crypt]

i cant use calculator in these questions

[polar pollen]

Don’t you know addition and subtraction of fraction?

[potent crypt]

wait do i add or subtract

[polar pollen]

Here it will be x(1/2-1/3)=1+3

[potent crypt]

how do i subtract the fractions now

yes im this bad

[polar pollen]

Now what’s the least common multiple of 3,2

(The denominators)

[potent crypt]

huh

[polar pollen]

Except 0

[potent crypt]

what😭😭

[polar pollen]

Don’t u know LCM?

[potent crypt]

nope

[polar pollen]

Find a number which is a multiple of 3 and 2

[potent crypt]

one

[polar pollen]

And multiple it both sides

[potent crypt]

oh

six

[polar pollen]

It doesn’t come in the 3rd table or 2nd

[potent crypt]

six

[polar pollen]

Now you multiple it both sides

As the other one suggested

And now you should get it

[potent crypt]

so 6x(1/3X-1/2X)=4x6

?

[polar pollen]

Yes

Now multiple that 6 inside the bracket

[potent crypt]

so 6x1/3X-6x1/2X

[polar pollen]

Yes

[potent crypt]

so 6/3x-6/2

mb i was doing something

wait im dumb

[shut garden]

no $Umm$

[potent crypt]

2x-3x

oh my god im dumb

[thorny flame]

Roman_Garland

[potent crypt]

wait what

ye so x=24

wait

shouldnt i multiply them by 6 before putting the x in one side? @polar pollen

nvm

[polar pollen]

Anything works

[potent crypt]

tysm

[polar pollen]

X=-24

2x-3x is -x

Which is 4*6

So x is - 4*6=-24

[potent crypt]

how about d @polar pollen

same way?

[shut garden]

Brother, why didn't you just write it down on a piece of paper?

[polar pollen]

I’m on my bed

Going to sleep

[polar pollen]

Yea

Can’t respond now

[potent crypt]

alright

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[nimble spear]

How do i solve this crazy integral

[hearty tartan]

x=cos(2theta)

[nimble spear]

Huh

It is an integral

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[pallid harness]

i got this from math.stackexchange

but i don't really understand their answer

how does

Now just evaluate at x = 1.```

equate to n?

[vernal matrix]

From $\frac{x^n - 1}{x - 1} = x^{n - 1} + x^{n - 2} + \ldots + x + 1$

[thorny flame]

@vernal matrix

[pallid harness]

yes

i still don't really get it

[plush tulip]

or if you prefer take the limit as x --> 1 on both sides

[pallid harness]

huh?

[plush tulip]

do u know limits

[pallid harness]

yes i do

[plush tulip]

l'hopital

[pallid harness]

yes i do

no what i don't understand it how (1−α)(1−β)(1−γ)........... = n

is this a property?

[plush tulip]

what's ur question

[plush tulip]

ok do you agree with this

[pallid harness]

yes i do

i do agree with it

[plush tulip]

plug in x=1

on the rhs

[pallid harness]

that's just 1 then

[plush tulip]

no

[pallid harness]

oh i see it

it's 1*n

so n

[plush tulip]

yeah

[pallid harness]

1+1+1..... = n

[plush tulip]

or do l'hopital on the lhs

as x --> 1

[vernal matrix]

or just recognise that as the derivative of x^n at x = 1

[pallid harness]

so is it ok to say that x-α = x^(n-1)

and so on?

[plush tulip]

even better lol

l'hoptial is actually dumb cuz that's how u derive the derivative of x^n in the first place (at 1 at least)

[pallid harness]

?

[plush tulip]

no

where ru getting that from

[pallid harness]

this from odes not match the form (x−α)(x−β)(x−γ)... so i am just assuming there are equal somehow

because i have to prove that (x−α)(x−β)(x−γ)... is n

[plush tulip]

no ur getting mixed up

[plush tulip]

here it's a sum not product

[pallid harness]

it does seem so.

can we start from the begining

?

[plush tulip]

(x−α)(x−β)(x−γ) = x^{n - 1} + x^{n - 2} + ... + x + 1 is true yes

[pallid harness]

alright

[plush tulip]

no please 💀

[pallid harness]

and how do i prove that?

[plush tulip]

both sides are x^n-1/x-1

[plush tulip]

lhs from here

[plush tulip]

rhs

[pallid harness]

hmm

[pallid harness]

so given that the roots of a polynomial are such, is saying (x−α)(x−β)(x−γ).. = the original polynomial correct? is this a property?

[hearty tartan]

see x^2-3x+2=0
(x-2)(x-1)=0

[plush tulip]

if a is a root of p(x) then x-a divides p(x)

[hearty tartan]

so x^2-3x+2=(x-2)(x-1)

[pallid harness]

ooooooooooooooh i get it now

[hearty tartan]

where 2 and 1 are the roots

[pallid harness]

yea i get it

i get it now

[plush tulip]

what's p(a)?

[pallid harness]

0

[plush tulip]

yes

x-a divides p(x) so p(x) = (x-a)g(x) so p(a) = (a-a)g(a) = 0*g(a) = 0

[pallid harness]

yes i am aware of this

i think i get it now tbh

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[fading jackal]

Shouldn't the answer for this be A?

[fading jackal]

Per superposition principle:

Yp = Yp1 + Yp2
Yp1= A x e^2t x t^s (s=2 because of the homogenous solutions--->e^2t,te^2t)
Yp2= (Bt x C) x t^s (s=1 because of the homogenous solution-->1)

[sharp birch]

did it say that it wasnt A? cause it seems like it is

[fading jackal]

Well we got that question in the exam, the system corrected it as A but then the doctor rechanged it and announced the answer is NONE and the Yp would be this

from what i was taught, most general form an exponential ONLY function is = a constant multiplied by e^bt

A x e^bt

and then you add the t^s depending on the homogenous solutions

[sharp birch]

The non-homogeneous term is simply (t + e^{2t}). There's no need for a polynomial of degree two multiplied by (e^{2t}) and a separate polynomial of degree two.

there's also a lot of redundancy, terms like (Dt^2) and (Et) are already accounted for in the first part of the solution, (At^2e^{2t} + Bt + C).

yeah i got no idea what the instructor is on

did you ask your instructor>?

[fading jackal]

i dont blame him haha, he's the head of maths department in our uni but still

[fading jackal]

I sent an email to my doctor that taught me , waiting for a response

[sharp birch]

yeah probably want to just wait for that, im not the smartest but i dont think hes right saying its none

[fading jackal]

Yeh alright, thank you sir!

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[idle crypt]

So I understand the first part. I sorta just stumbled into the answer for part B.

I know it's tan-1(60/90) but Im not sure how to actually know whats the adjacent or the opposite.

[dense lily]

Tan-¹(y/x)

For inclination to the horizontal

So how can I help

[idle crypt]

Im not sure what that means either 😅

[dense lily]

Ok y is the total vertical forces

x is the total horizontal forces

Do you know how to resolve vectors into their components

@idle crypt

[idle crypt]

If thats what you mean then yes

[dense lily]

Yes

That's what I mean

[idle crypt]

Ah, ok I think I gotcha

[dense lily]

Ok so now replace y with j and x with i

[idle crypt]

OK, makes sense now so I can do others. tyvm 😄

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[tribal pewter]

hi, I have the following question: If I calculate the Eigenvalues and insert them into the Matrix A-lambda*I , I know that we will get the Eigenvectors. But which meaning does the span (which contains all Eigenvectors) has?? What´s that?

[tribal pewter]

Topic: Linear Algebra

[loud oasis]

the way we define eigenvectors is that they are the special vectors where the matrix acts like scalar multiplication:
Av = λv
to find them, we rearranged the equation:
Av - λv = Av - λIv = (A-λI)v = 0
so we see that eigenvectors are the null space/kernel of the matrix (A - λI)

but the actual utility and meaning of eigenvectors comes from the original equation, the bottom equation is just a convenient way to find them

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[idle crypt]

tan-1(21/7.5)

Thats the same way I did it for B but isnt working for D?

[somber carbon]

is there a pic or smth of a and b

[idle crypt]

Where I replied to

https://cdn.discordapp.com/attachments/903486479064522752/1280641676339052565/image.png?ex=66d8d200&is=66d78080&hm=59700cdfb7f21f2253dd0372e8b9a4d2d7e8e89c415ad3938538f0b590042562&

<@&286206848099549185>

[somber carbon]

my fault I forgot

[somber carbon]

u can kinda tell that a-b would point in the 3rd quadrant

so in ur tan-1, ur missing a negative somewhere

[idle crypt]

How do you tell that?

[somber carbon]

idk if that helps

[idle crypt]

It does not xP

[somber carbon]

do you know what adding vectors look like

[idle crypt]

I guess not

I did it for Parts A/B but idk what it "looks" like

[somber carbon]

like this

U connect the two ig

So subtracting just think of it as

a + (-b)

so it’s the same but u just flip b around

[idle crypt]

Im lost

I have this on subtraction

but I cant find anything about direction

[somber carbon]

the -b flips the direction 180 deg

Like the minus part

[idle crypt]

so b is in Q3 now

[somber carbon]

maybe I’m not qualified to help you lol

[somber carbon]

Ya pretty much

• -b

[idle crypt]

Think Ill have to ask again later since I gotta go soon. Still not understanding.

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[lyric crypt]

I'm stuck on part e

It's asking for the definite integral of (2+f(x)) with upper bound of 6 and lower bound of 2

and the definite integral of f(x) with the same bounds = 20.5

I assumed it would be 20.5 + 2 but 22.5 was wrong

[fringe grove]

You can separate the integral of 2+f(x) into 2 integrals, do you know how to do that?

$\int_{2}^{6} (2+f(x))dx = \int_{2}^{6} 2 dx + \int_{2}^{6}f(x) dx$

[thorny flame]

Sevchenko

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[lyric crypt]

.reopen

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[lyric crypt]

yes i do

so that would make the answer 2x+20.5?

[scenic flare]

why 2x

[lyric crypt]

because the integral of 2 is 2x?

[scenic flare]

what does this equate to

[lyric crypt]

2

[scenic flare]

elaborate

[lyric crypt]

oh wait

i was thinking of sigma notation

[scenic flare]

kk

[lyric crypt]

im sorry i learned integrals a week ago

is it 1

[scenic flare]

no worries

exactly, so if we want to know what the integral from 0 to 6 of 2 is:

[lyric crypt]

2

[scenic flare]

you're right that the integral function of 2 is 2x, but that doesn't give you the interval over 0 to 6

[lyric crypt]

okay wait

[lyric crypt]

for this, why is this 1?

oh wait

because the integral of 1 is x

so it's 1-0

[scenic flare]

exactly!

you can also verify yourself with the visual interpretation

what's the area under the curve of the function f(x) = 1

from 0 to 1

makes sense that it's 1 since the area is just a square

[lyric crypt]

so for the integral of 2 is 2x, you would do 2(6) - 2(0)

[lyric crypt]

which is what you did here

because of the second fundamental theorem of calculus right?

[scenic flare]

yop

[scenic flare]

yessus

[scenic flare]

and for integral of 2 over [0,6]:

[lyric crypt]

okay that makes sense

also something i've been confused about

when im trying to find the integral of 1/x

is it just a law that 1/x = ln|x|

[scenic flare]

ehm nop

or do you refer to the integral function

[lyric crypt]

yes

When integrating 1/x

[scenic flare]

hm careful

[lyric crypt]

because you can bring x up to make it x^-1

[scenic flare]

the derivative of ln(x) is 1/x, BUT only for x>=0

because ln(x) is only defined for x>=0

[lyric crypt]

and then you do the reverse power rule and that gives you x^0/0

ohh okay

[scenic flare]

therefore yes, the integral function of 1/x is ln(|x|)

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[strong sable]

Assume ( f(x, y) ) is continuously differentiable in the region ( x^2 + y^2 \leq 1 ), and ( |f(x, y)| \leq 1 ). Prove that within the region, there exists at least one point ((x_0, y_0)) such that

[
\left[ \frac{\partial f(x_0, y_0)}{\partial x} \right]^2 + \left[ \frac{\partial f(x_0, y_0)}{\partial y} \right]^2 < 16.
]

[thorny flame]

miyo

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[feral charm]

cause it’s hard

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[wide wyvern]

we proceed to use ibp in the first line

however shouldnt the second term be negative ?

in the 2nd line i mean

[obsidian solstice]

The (x-t)^n becomes -(x-t)^(n+1)/(n+1) which changes the sign

[wide wyvern]

wait how we integrate it how could it change the sign

[obsidian solstice]

Because it has minus sign

[wide wyvern]

am i just dense or

[obsidian solstice]

dt

[wide wyvern]

the antiderivative doesnt have a neative sign

[obsidian solstice]

Not dx

[wide wyvern]

oh shit yea ur right

its to dt

were integratring

i gotta be careful of that thx

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[limpid plinth]

how do u do thie

[mellow arrow]

if x=secy

then can you represent coty in terms of x?

[sweet shard]

Drawing a picture might help

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@limpid plinth Has your question been resolved?

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[sudden barn]

a particle moves in a straight line according to the relation s(t) = Ln (t^2-2t+e)

[sudden barn]

i need to find the velocity of the particle when it returns to its original position

i have some ideas but none of them worked

now i know
v (t) = (2t - 2) / (t^2-2t+e)

after differentiating

no idea what to do now though

is it v(t) = s(0)

[silent siren]

clearly the velocity is a function of time, so you'll first need to calculate the time at which it returns to the original position.

[sweet shard]

s(0) is a number

[opal vault]

you want to know v(t) when the POSITION is the original one again

can you translate mathematically the statement "the particle returns to its original position"

[sudden barn]

the original question is in arabic so its a bit hard to give the 100% accurate translation

oh you meant

ok nevermind im stupid

that means s(0) = 0 ?

or s(t) = 0

wait s(0)=0 doesnt work

[sudden barn]

its 1

[topaz sinew]

@sudden barn Has your question been resolved?

[sweet shard]

Yea find a t when s(t) is 1

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@sudden barn Has your question been resolved?

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[glacial pumice]

we just started probability statistics and teacher wrote this on his lecture

[glacial pumice]

Im just wondering if this is a mistake or something im missing?

cuz Im getting 31920

[prisma mesa]

,w (8 choose 2) * (20 choose 3)

[thorny flame]

Results provided by WolframAlpha

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

Requested by methisalwaysright

[prisma mesa]

you seem to be correct

[glacial pumice]

Thank you very much sir

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[delicate wharf]

heyy for part (g) and (l) i got a different answer than the book. for g, i got true and for l, i got false. but the book has the opposite answers for both. am i losing my mind or is the book wrong??

[eternal wing]

you're correct

g is true and l is false

[delicate wharf]

okay thank you 😭 i was like "this is so basic no way im tripping this hard"

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[spiral fulcrum]

The drag is posing a problem

The ln becomes ln(0) when I try to input the two values from the integral, which is not a number

[pearl osprey]

@spiral fulcrum your supposed to change the limits when u do the substitution

youre integrating v=0 to 69.44, so u = 2 to 2-0.00004(69.44)², youve got u=0 to 69.44

[spiral fulcrum]

Oh shoot
Thank you

Always the small things on these massive problems

[pearl osprey]

o7

[spiral fulcrum]

The upper limit becomes smaller than the lower limit, do I have to flip the integral sign?

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[night halo]

I need help with both the questions

[night halo]

For the first one I can’t figure out how to do it at all and the second one I can’t figure out how to create the integral for (1,2) to (4,2)

[wise salmon]

<@&286206848099549185>

Help pls

[night halo]

Bruh 😭 get ur own chat

[wise salmon]

Oh

Mb

[night halo]

U good

[wise salmon]

Help me tho

Pls

Daddy

🥺

Ik u more advanced

Plssss

[night halo]

P sure it’s just the 3rd option

That’s true

[night halo]

<@&286206848099549185>

[wise salmon]

I needa know how to do it

I’ll help u

@ them

PLEASEEEEE

U my only hope

[night halo]

Bruh odd function means the ends go opposite directions

[wise salmon]

Okay?

What does p(3) p(2) mean bruh

I’m so lost

[night halo]

Since it said p(3)=-7 and p(2)=0

[wise salmon]

Mhm?

[night halo]

It means at x=3 the y value is -7

And x=2 y=0

[wise salmon]

I thought p(x) means y

[night halo]

P is just the name of the function

[wise salmon]

💀

[night halo]

P(x) does mean y

Atleast in a graph

[wise salmon]

OHHH

SO WHEN X IS 3

[night halo]

U get it?

Yea

[wise salmon]

Y = -7

[night halo]

Yea

[wise salmon]

What’s the

Less than or equal to 0 when x is greater than or equal to 2 mean

🥲

[night halo]

And since it goes on forever. Limx->+infinity would be -infinity

[wise salmon]

Oh we haven’t got that far

Huh

So what statements r true

[night halo]

Basically it’s saying the y value can only ever be equal to or less than 0

And the x value can only ever be equal to or greater than 2

[wise salmon]

So it can’t be the second statement

[night halo]

The y value one is true whenever the x value statement is true

[night halo]

Yup

The first statement can be true but doesn’t necessarily need to

I don’t know what ur studying but take x^7 and x^3 and x^1 for example. If u graph them in say Desmos

[wise salmon]

Is precalc

🫤

[night halo]

U can see the higher the exponent is the more the 0s u get

[wise salmon]

mhm

[night halo]

And the ends are going in opposite directions bc it’s odd

Since we are told it’s odd

[wise salmon]

The degree correlates with the x int?

[night halo]

And we know when the x increases, the Y decreases

[wise salmon]

Yeah

[night halo]

That means the y must increase when the x decreases

[wise salmon]

MHM

[night halo]

So option 3 has to be true

[night halo]

Yea

Or atleast when the y-intercept is 0

[wise salmon]

Can u explain

The meaning

Of the signs

Like

[night halo]

Which signs 😭

[wise salmon]

lim and the stuff under

Idk what that indicate

And like what p(x) = infi is for

[night halo]

Oh basically the lim stands for limit and shows what ur approaching. So say for lim of a function that’s linear

Like y=x