help-33

[bold abyss]

Got the first one then :))

Anyways my issue with it is that i would usually just figure out the equation corresponding to the graphs and then put those equal to eachother and solve for x. Issue is i cant analyse the a, b and c values in the parable (arch like one) so idk what to do

[spark mason]

notice the unit squares

[bold abyss]

I have in fact been staring at those

[spark mason]

and where 2 functions intersect

i think you can work the coordinates out

[bold abyss]

Is that all? Theres no big calculation i gotta do?

[spark mason]

i mean you can if you want but

it's just right there isn't it

given no information whatsoever, that's all you have and all you gotta use i guess

[bold abyss]

Omg okay thank u so much!!!

[spark mason]

you're welcome 👍

[bold abyss]

Okay second issue

Ive done all this

And i dont realky get what a is? Like what does a and b determine and how do i know what a is?

Cus if a is the percentage with wich the grap declines its way too high

[marsh citrus]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[bold abyss]

Okay this is the original problem, translation will follow shortly

[twilit grove]

how did you get that number for b?

[bold abyss]

Basically it says: the data below show the development in childdeaths in bangladesh i the period 2000-2019. Child deaths are measured in number of deaths for kids under the age of 5 per thousend born living.

In a model the childdeatlyness f(x) can be described by a graph
f(x)=b•a^x
Where x is years after 2000

A. Choose(?) The numbers a and b by exponential regression on all the datas (i have a seperate document with more data)

I did the regreasion and it gave me f(x)=85,591•(o,946894)^x

[twilit grove]

oh I see. it wants you to do a regression

[bold abyss]

Exactly

But then im like this dont make no sence

[twilit grove]

what doesn't?

[bold abyss]

I dont really get what a and b determines in the graph and if either of them is the decline of the graph then how are they positive numbers?

[twilit grove]

b is the y-intercept

i.e., the number at time 0

it's been a long time since I did this kind of regression, but I think that the rate of decline is actually ln(a)

,calc log(.946894)

[elfin berry]

Result:

-0.054568124479122

[twilit grove]

so that would be a 5% decrease

[bold abyss]

Ooooh okay that makes a lot more sence

I have learned about log and ln so i get it nit thank you!!!

*now

[twilit grove]

yw

[bold abyss]

But then can i just put b=(0.946894)?

[twilit grove]

no, b is 85.591 or whatever

[bold abyss]

Sorry i meant a

Keep mixing them

Imma do it

A=(0.946894)

I got a thrid and final question tho two sec

This is the assignment

It says: a quadratic polynomial f is given by
f(x)=2x^2+8x+k

A. Find the two zero points for f, when k=-4,5 (which i did)

B. Find k, so that f has exactly one zero point

(Zero point is where f(x)=0)

I dont really know how im supposed to do assignemnt b

[spark mason]

do you know the delta

and its properties

ax²+bx+c=0
🔺 =b²-4ac

i'm pretty sure you've heard it

[bold abyss]

I do not

Only in relation to physics

No actually i do know it

Sorry we just call it d, or the discriminant

[spark mason]

that's the same thing yeah

so there's few things you gotta know

like when d>0, there's 2 real roots to that function (meaning it intersects the x axis at 2 different points)

and when d<0, the function doesn't intersect the x axis at all, leaving you with no real roots

can you guess what would happen at d=0

[bold abyss]

1 intersection 😎

[spark mason]

👍👍

[bold abyss]

How is this used i correlation with my issue tho?

[spark mason]

you basically have the same function as ax²+bx+c

it's just a=2, b=8 and c=k

[spark mason]

you can rewrite d=b²-4ac as d=(8)²-(4)(2)(k)

and set that to 0 👀

[bold abyss]

How can i set that to 0 tho? Like how is that possible?

Wait it makes sence

Sense

Thank you so much! This has all ben such a great help!

[spark mason]

you're welcome

[bold abyss]

.close

[marsh citrus]

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[lyric relic]

!help

[marsh citrus]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[marsh citrus]

Please don't occupy multiple help channels.

[lyric relic]

isnt this there are an infinte amount of solutions

4c + 6m = 100

8c + 12m = 200

first equation you manipulate it

but its gonna be

8c + 12m = 200
8c + 12m = 200

so what is the answer

ok

but that option isnt there lol

khan academy sucks

because you just told me the answer is

there is an infinite amount of solution

which is not the same thing as there is not enough information to determine the exact number of capulets and montagues

help please

[marsh citrus]

@lyric relic Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

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[lyric relic]

!help

[marsh citrus]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[lyric relic]

hi

i need help solving this systems of equation

the two equations are

f + p = 21
4f = p + 24

could i divide the bottom equation by 4 to solve for f?

[winter smelt]

that works, there are many ways to do it, as long as you can find an equation solely depending on one of f or p, and solve them individually, it's fine

[lyric relic]

Is there an easy way to do it

Fractions are hard

[twilit grove]

substitution would be easiest here, probably

[lyric relic]

Hi

I need help

How can I substitute

I am not allowed to to substitute cause f is not the same as 4f

[twilit grove]

you can solve for "f" right away from the first equation

use that in the second

[lyric relic]

How

4f isn't the same as f

[twilit grove]

right, it isn't

[lyric relic]

So how

I solve for f?

But that's what I did

Ok here is what I am trying to do

I'm trying to isolate f so I can use whatever f is and substitute it into the first equation

And I need help with fractions

[twilit grove]

I understand. I'm saying it's easier to solve for f from the first equation, and use that in the second

[lyric relic]

So the picture I sent in asking if I did it right

Oh

So f would be f = 21 - o

f = 21 - p?

[twilit grove]

yup

[lyric relic]

Or p - 21

Does the order matter

[twilit grove]

very much so

[lyric relic]

Ok so how do I know which order is correct

[twilit grove]

you have f + p = 21, right?

[lyric relic]

Yes

[twilit grove]

and you subtract p from both sides

[lyric relic]

Yes

[twilit grove]

which gives you f + p - p = 21 - p

[lyric relic]

Oh so it's 21 - p

[twilit grove]

yup

[lyric relic]

Because 21 is first

It's the closest to the = sign

Ahh thx

Ok thx everyone

[twilit grove]

do you understand how to use that expression for "f" in the second equation?

[lyric relic]

Ho Ho Ho

Y'all are so smart

Yes I know how to do systems of equations

I guess I was just being blind

Anyways I got the answer right

[twilit grove]

no worries

[lyric relic]

P is 12 and f is 9

[marsh citrus]

@lyric relic Has your question been resolved?

[marsh citrus]

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[timid prairie]

Help pls

[timid prairie]

<@&286206848099549185>

[dawn niche]

well you know the area is $\pi {r_2}^2 - \pi {r_1}^2$

[elfin berry]

Potatomonke

[dawn niche]

where r_2 is the outer radius

and r1 is the inner radius

so $\pi {r_2}^2 - \pi {r_1}^2 = 10 \pi$

[elfin berry]

Potatomonke

[dawn niche]

when accounting for the circumferences

$2 \pi r_2 + 2 \pi r_1 = 14\pi$

[elfin berry]

Potatomonke

[dawn niche]

we are trying to find the width, which is r_2 - r_1

[timid prairie]

is {r_2}^2 - {r_1}^2 = 10?

[dawn niche]

yes

then r_2 + r_1 = 5

[dawn niche]

we can factor this to get $(r_2 + r_1)(r_2 - r_1) = 10$

[elfin berry]

Potatomonke

[dawn niche]

we have the value for r_2 + r_1, which is 5, so we plug that in

$5(r_2 - r_1) = 10$

[elfin berry]

Potatomonke

[dawn niche]

simplifies to r_2 - r_1 = 2

which is your answer

[timid prairie]

I see

thank you very much!

[dawn niche]

np

[timid prairie]

.close

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

Please don't occupy multiple help channels.

[fathom mountain]

Did I do smth wrong? How can I proceed?

[dapper bough]

(to make it easier to read)

[fathom mountain]

Thx

[fathom mountain]

U see smth wrong?

[dapper bough]

i'm not sure i understand the question. could you try to explain it to me, or provide me some more context, or perhaps a picture of a prompt or something?

[fathom mountain]

Find z

With z=x+iy

<@&286206848099549185>

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

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[ruby cloak]

In a game of chance, six cards numbered 1 to 6 are lying face down on a table. Two cards are selected without replacement and the sum of both numbers is noted. State the total number of outcomes.

[lilac siren]

well which values are possible to be presented as a sum of two values between 1 and 6?

clearly the minimum is 3

and the maximum is 11

is it possible to achieve every value in the range {3,...,11}?

@ruby cloak

[ruby cloak]

yes. all possible

[lilac siren]

yop, then you know the total number of outcomes 🦇

[ruby cloak]

sorry -- don't understand

[lilac siren]

well we know that 3 is the minimum outcome

11 is the maximum outcome

and all values between 3 and 11 are also possible outcomes

thereby we know 3 to 11 are the only possible outcomes

meaning there are 11-2 = 9 possible outcomes

3,4,5,6,7,8,9,10,11

[ruby cloak]

oh, okay

thank you!

[marsh citrus]

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[marsh citrus]

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[native tapir]

How would i do q13

[lilac siren]

firstly distributive law

like we did before

[native tapir]

Did i do this wrong

[lilac siren]

careful careful

it's the dot product

not multiplication

[native tapir]

Like dis?

[lilac siren]

everything right except it's 10a*a :)

and after that just remember that a*a = |a|²

[native tapir]

Oops...

[lilac siren]

b*b = |b|²

and "a is parallel to b"

so let's remember what the dot product means

we project a onto b and then multiply the lengths

but if a is parallel to b

then a projected is just a again

[native tapir]

Is magnitude squared just the original value

[lilac siren]

wdym

[native tapir]

So magnitude 6 squared is just 6?

[lilac siren]

|a|² = 6² = 36

[native tapir]

Ohhh u rite

[lilac siren]

[native tapir]

OOPS.

[lilac siren]

:D

nw

therefore since a and b are parallel: a * b = |a| * |b|

because a_projected = a

[lilac siren]

I don't have the dot symbol here, but these are two different multiplications just to have it clear

left is vector dot product

right is normal multiplication

so written in latex:

[native tapir]

How about da ab

is it magnitude ab?

[lilac siren]

but I wrote it here

[native tapir]

Ohhh

I SEE

[lilac siren]

as they're parallel

[marsh citrus]

@native tapir Has your question been resolved?

[marsh citrus]

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[pseudo aspen]

I'm kind of lost of what to do at this point, using partial fractions

[fervent rampart]

set that equal to the original function

[pseudo aspen]

okay

[rocky lark]

$x^2(.....)+x*(.....)+....$ you could try doing something like this

[elfin berry]

BobTheBuilder

[still temple]

you can find the value of A and c by substituting x equals 0 and -1

[pseudo aspen]

how did you know that's what i was supposed to do?

i'm confused there

[mighty gyro]

tf?

[mighty gyro]

now u compare coefficients in LHS and RHS

[still temple]

🤨

[rocky lark]

after having done this you should create the system with 3 equations, placing the corresponding known value to each

[pseudo aspen]

i have no idea what that means

[rocky lark]

What did you not understand?

[pseudo aspen]

literally everything, i don't know that x^2(...)+x*(...) is

[pseudo aspen]

okay

i basically don't know how i handle 5-x

[pseudo aspen]

all the examples we did had a x^2 if there was a x^2 in the Ax^2 place

oh so Ax^2+Bx^2, since there's no x^2 in the 5-x numerator is 0?

we didn't do any examples like that so i'm flabbergasted

[pastel granite]

yes the coefficient on the x^2 term must be 0

[pseudo aspen]

that gives me A=5, B=-5, C=-6

then i just do 3 integrals with those values

[pastel granite]

yup you can resubstitute into this and solve the integral

[pseudo aspen]

that gives me that, right?

[pastel granite]

be careful with the last integral

[pseudo aspen]

ah u sub

it's saying this is wrong???

without and without the bars

god i hate this software so much

[late geode]

don't need () when you have | |

[pseudo aspen]

it's the software being a dumbass, i needed to use their specific bars

I couldn't type in my own

the class made us pay $130 for this software and it doesn't even work

love it so much

[pastel granite]

that is upsetting it seems to be correct

[marsh citrus]

@pseudo aspen Has your question been resolved?

[pseudo aspen]

have i done something wrong here?

i can't figure out a way to find B or C

[pastel granite]

subbing in -1 for A, you can setup B in terms of C or vice versa

[pseudo aspen]

so I do -1 + B + C = 0 giving me B = 1 - C?

[pastel granite]

yes

[pseudo aspen]

giving me C = 2

then i just use -1 + B + 2 = 0

and B = -1

got it, that makes sense

simple algebra that i forgot lol

this whole process seems like it's more algebra based than calculus ngl

[rocky lark]

if you did the previous calculations correctly everything is correct for now

[pseudo aspen]

yeah, got it right!

for this problem, i would need to use the division stuff?

in my notes i have something like this

[rocky lark]

but if you do the calculations you see that the numerator has a lower degree than the denominator, so no

You do the division only if the numerator has a degree greater than or equal to that of the denominator

[pseudo aspen]

ahh, i see

[marsh citrus]

@pseudo aspen Has your question been resolved?

[pseudo aspen]

just going to keep this open for a little bit longer

if i see no help channels available, i'll close it though

[pseudo aspen]

.solved

[marsh citrus]

Channel closed

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[marsh citrus]

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[lethal bridge]

I don’t really understand what’s going on here

[spark mason]

thats not maths

discord.gg/physics

[lethal bridge]

so what

i can still ask it here

[spark mason]

great for you

[lethal bridge]

???

[spark mason]

We are a mathematics server but we are a part of a larger network of science-centered servers. If you are interested in other scientific topics, check out the servers listed in #old-network.
‌

this quite literally the first thing written in #info

[lethal bridge]

why are you minimodding for no reason

[spark mason]

whatever i give up

[lethal bridge]

dont know why youre making such a big fuss if youve been in the server long enough youd see that this isnt unusual

[crystal wraith]

So as you may or may not know, the photo current originally (before applying a potential difference to stop it) was because of the electrons that were ejected from the metal surface.

[lethal bridge]

right

[crystal wraith]

The fastest of those electrons had kinetic energy that you find out using the equation hυ = Φ + KE

[fathom ridge]

hu

[crystal wraith]

It's not u?

[lethal bridge]

hf

[crystal wraith]

υ

[lethal bridge]

hu?

whats u

[crystal wraith]

$\nu$

[elfin berry]

! What the hell am I doing here?

[lethal bridge]

nu?

u?

[crystal wraith]

frequency

f

God damn it.

sure, f

[lethal bridge]

in my book frequency is denoted as f

[crystal wraith]

Fair enough.

[lethal bridge]

so what hapens after the photoelectrons are emitted?

it turns into photocurrent?

[crystal wraith]

Yes, it normally would have.

But those electrons are now stopped because of the potential applied.

[lethal bridge]

this is the voltage source im guessing?

[crystal wraith]

Yes.

[crystal wraith]

So as it turns out, to stop those electrons the potential applied has to be equal to the kinetic energy these electrons would have normally had because of the light source.

That's the idea.

[lethal bridge]

by potential applied are u talking about like the voltage?

[crystal wraith]

Sure, that.

[lethal bridge]

so if the voltage is stopping the photoelectrons from running through the wire, how does it reach the collector?

[crystal wraith]

It'd have reached the collector had there been no voltage.

And the voltage applied only acts after the electrons reach the collector.

They'd have had kinetic energy still but this voltage now acts against the electrons' motion.

[lethal bridge]

is this the direction in which this is happening?

so it goes to the collector first?

this is more like it didnt mean to add the extra red line

or is it the other way around

[proper zodiac]

Other way, conventional current (+ to-) opposes electron flow. Increasing voltage makes more electrons "build up at the collector" meaning the photoelectrons don't want to be there more

[lethal bridge]

O

so if in the question it says "all emitted photoelectrons are gathered by a collector" if the voltage is stopping the photoelectrons from flowing in the wire, how does the collector ever gather it?

[proper zodiac]

At the start of the experiment the voltage is low enough that all photoelectrons that want to get to the collector, do so

As you increase the voltage less and less reach the collector until none do

[lethal bridge]

ohhh

Im confused why increasing voltage would reduce the current detected? I thought it would do the opposite no? from V=IR voltage is directly proportional to current

[cursive raptor]

Yo you need help here still? I think I get it

[lethal bridge]

yeap

[cursive raptor]

Mk what do you want me to explain?

[lethal bridge]

this

[cursive raptor]

The voltage is essentially travelling in the other way to the photoelectron current

[lethal bridge]

oh

[cursive raptor]

This lowers the current of the photoelectrons emitted; as the collector becomes more "negative" the photoelectrons don't want to travel there as much, they're being resisted (repelled)

[lethal bridge]

ahh okay

so this reduces the kinetic energy of photoelectrons and makes it move through the wire more slowly?

[cursive raptor]

Yeah, so the voltage is known as the stopping voltage, and is the voltage that resists the photoelectrons

[lethal bridge]

ohhhhhh alright that makes sense

[cursive raptor]

It reduces it to make it not reach the collector

Only the electrons that have enough KE can bypass this to reach the collector

[lethal bridge]

so thats why less current is picked up on the ammeter?

[lethal bridge]

i see

[cursive raptor]

Yep, less photoelectrons have the KE required to overcome the input voltage (stopping voltage) so it is less

I'll admit I have no idea if the stopping voltage is a current, I dont think it is

[lethal bridge]

wait so the negative side of the voltage source is connected to the collector or something, is that why the voltage is fighting against the photo current?

[cursive raptor]

Dude get lost

[cursive raptor]

I'm still trying to find sources for this, I don't think this is right

The stopping voltage makes the collector more negative which increases the amount of KE required for a photoelectron to make it to the collector

I don't know how the voltage is applied...

Let me cook

[lethal bridge]

alr lmao

[cursive raptor]

I'm confused because if that symbol is a battery then wouldn't the stopping voltage induce a current on the ammeter?

I'm learning this topic rn btw

[lethal bridge]

idk its just a voltage source im not sure if thats the same thing

[cursive raptor]

Wait hang on its not closed, theres no current

[cursive raptor]

Yeah batteries are a voltage source

[lethal bridge]

the metal here acts as a current source

[cursive raptor]

Nvm, I was thinking about the battery, I get it

[cursive raptor]

Yes the negative side here makes the collector negative so that it creates an opposing electric force to oppose the photoelectrons

[lethal bridge]

wait so im kind of confused, does the voltage source create electrons? or is it just a electrical force that guides the electrons

[cursive raptor]

It's impossible to create electrons, it essentially gathers them

And having more electrons at a point makes that point more negative, which would be the collector in this case

[lethal bridge]

so it just provides the force for electrons to flow in the circuit

?

ohh ok

i think i get it now

so if this was negative to positive, there would be no force the photocurrent has to fight against?

[cursive raptor]

Yea, it would be attracted

Because the collector would be positive

[lethal bridge]

kk

i get it now

thanks

[cursive raptor]

Hope that helps

If you want to mess around with the experiment I found this to be pretty good: https://javalab.org/en/photoelectric_effect_2_en/

[lethal bridge]

[marsh citrus]

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[marsh citrus]

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[open kayak]

#1195800467880558672 message

[marsh citrus]

Please don't occupy multiple help channels.

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[vernal forge]

not sure what answer do you want to get, it's like asking what's the difference between an orange and an apple

they are two distinct rules

sine rule
a/sin(A) = b/sin(B) = c/sin(C)

cosine rule
c^2 = a^2 + b^2 - 2*a*b*cos(C)

[marsh citrus]

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[plain crescent]

if $y=x^2-3x$, then $\frac{dy}{dt}=\frac{dx}{dt}$ at x = ?

[elfin berry]

Joeller

[plain crescent]

$\frac{dy}{dt}=2x\frac{dx}{dt} - 3\frac{dx}{dt}$ = $\frac{dx}{dt}(2x-3)$

[elfin berry]

Joeller

[plain crescent]

and from this

$\frac{dy}{dt}=\frac{dx}{dt}(2x-3)$, then $\frac{dx}{dt}=\frac{(\frac{dy}{dt})}{2x-3}$ = $\frac{dy}{dt}\frac{1}{(2x-3)}$

[elfin berry]

Joeller

[plain crescent]

and since dy/dt=dx/dt so we'll just remove them so we'll end up with

$2x-3=\frac{1}{2x-3}$

[elfin berry]

Joeller

[plain crescent]

$4x^2-12x+9-1=0$ divide by 4, $x^2-3x+2$

[elfin berry]

Joeller

[plain crescent]

$x=1,x=2$

[elfin berry]

Joeller

[plain crescent]

and I got both of them in the choices

• the answer was 2

[plain crescent]

dx/dt*

[still temple]

yea both satisfy actually

[plain crescent]

Yeah I thought so

its crazy that they have them both in the choices

I'll have to ask my teacher about this

.close

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[distant lichen]

what role does n=2 play in this

[distant lichen]

if im finding convergence or divergence

[marsh peak]

No role, it could start at any index greater than -1

[distant lichen]

oh ok

thanks

[latent coral]

it shouldn't really make a difference, n = 1 produces a finite term so if this sum diverges then the sum would diverge even if n started at 1, and likewise if this converges then adding the n = 1 term will still lead to a finite sum

[distant lichen]

this sum coverges tho

[tired oxide]

yeah but it’s the same reasoning

[distant lichen]

ohhhhh ok

[tired oxide]

if the whole sum converges then starting one term earlier doesn’t make a difference 👍

[distant lichen]

alright thanks

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[twin ether]

How do I solve b)

[still temple]

no clue i'm restarted

[twin ether]

a) is 7

[dawn niche]

Average rate of change formula is $\frac{f(b) - f(a)}{b-a}$ when $a < x < b$

[elfin berry]

Potatomonke

[twin ether]

$\frac{f(x-0.01)-f(x)}{0.01} = 7$

[elfin berry]

rynite

[twin ether]

I think this is what I should be solving

but idk if its possible

[merry pewter]

have you learned power rule yet?

[twin ether]

nope

[pseudo flax]

derivative based math is taught after functions based math

[twin ether]

so what do I do now

[boreal rose]

derivative based math?

[dawn niche]

which part of the problem is this meant for

[twin ether]

b)

in the book, one possible answer given is (4.5, 3)

idk how they got that tgho

tho*

[dawn niche]

its right between 4 and 5

its the midpoint of that line

[twin ether]

im not sure if I understand

[pseudo flax]

are you familiar with solving for the instantaneous rate of change at a point

[twin ether]

yes

[pseudo flax]

if you were to find a point **common ** to the instantaneous rate and the average rate , what could you do?

think in terms of x values

[twin ether]

no idea, don't you need 2 points for average?

not one

[pseudo flax]

think about the equations for each

[versed pagoda]

Is this a cubic?

I think you'll likely need to solve the cubic and then check where df(x)/dx = 7

[pseudo flax]

that is further on in functions , limits are taught before derivatives

[versed pagoda]

because intermediate value theorem isn't enough here

[versed pagoda]

But the question is asking for derivative clearly...

[pseudo flax]

@twin ether try to set the equations for instantaneous and average equal to each other

like this

[versed pagoda]

is f specified though?

[pseudo flax]

average has already been found

[versed pagoda]

I understand

but f isnt given...

[limber hearth]

we can find it with the graph if needed

[versed pagoda]

well not necessarily....you're assuming its a cubic but it could be a crazier function

anyway considering its a cubic we can do it quickly enough

[twin ether]

$\frac{f(x-0.01)-f(x)}{0.01} = \frac{f(5)-f(4)}{5-4}$

[elfin berry]

rynite

[twin ether]

like that?

[fervent yarrow]

you can, but that's not how u're supposed to solve this

[versed pagoda]

so anyway cubic is (x^2 -4)(x-2) = x^3 - 2x^2 -4x + 8

derivative is 3x^2 - 4x -4 which we want to be 7

3x^2 - 4x-11=0

just solve it

🤷‍♂️

[twin ether]

im confused asf

[versed pagoda]

ok

have you learned derivatives?

[twin ether]

no

[versed pagoda]

hmmm

but you've seen limits?

[twin ether]

no

[versed pagoda]

so what has been taught in calculus?

[twin ether]

im not taking calculus

[versed pagoda]

which grade are u in

[twin ether]

11

its a part of my advanced functions

course

[pseudo flax]

it’s advanced functions

[twin ether]

a single chapter

[twin ether]

yea

[pseudo flax]

limits are taught first

[versed pagoda]

ok you shouldve seen limits

but like ney said

[twin ether]

hm I didn't

[versed pagoda]

she gave you the def of limits

anyway ok

so let's do it slowly

assume the function is a cubic

what are it's roots

they're specificed?

from the graph

[twin ether]

yeah uh

-4 ,2, 4

[versed pagoda]

ok

so the function can be inferred right?

[twin ether]

i found it

[versed pagoda]

(x-4) x (x+4) x (x-2) right

[twin ether]

yea

and a is around 0.26

[versed pagoda]

ok its this x^3 - 2x^2 -4x + 8

right?

[twin ether]

yep

[versed pagoda]

ok

so you're looking to find an x such that

[f(x+h) - f(x)]/h

with h being super tiny (we'll quantify)

[versed pagoda]

so that this becomes 7

[twin ether]

yeah I understand that

[versed pagoda]

ok

so keeping x and h as variables

compute f(x+h) - f(x)

[fervent yarrow]

do you know this ?

[versed pagoda]

no he doesnt know derivatives

[twin ether]

not sure how

[twin ether]

no

[versed pagoda]

f(x) is x^3 - 2x^2 -4x + 8

f(x+h) is (x+h)^3 - 2(x+h)^2 -4(x+h) + 8

[twin ether]

oh

[versed pagoda]

open up the (x+h) powers

and then subtract it with f(x)

[twin ether]

alright

sec

[twin ether]

3x^2h + 3xh^2 + h^3 - 4x^2 - 4hx - 2h^2 - 8x - 4h + 16

[versed pagoda]

no that's not right

the constant should disappear

[twin ether]

what constant?

[versed pagoda]

16

8 and 8 cancel

you're subtracting not adding

[twin ether]

ok k

x^(3)-2x^(2)-4x+8 -(x+h)^(3)-2(x+h)^(2)-4(x+h)+8

is that what I should be solving

[versed pagoda]

the other way

f(x+h) - f(x)

[twin ether]

ah right

[twin ether]

-(x+h)^3 - 2(x+h)^(2)-4(x+h)+8 - x^(3)-2x^(2)-4x+8
this right now?

oops

too many variables

the 8s still dont cancel out

[twilit grove]

you have to distribute. it's minus all of f(x)

[twin ether]

ah

i forgot about that 😅

3x2h+3xh2+h3−4hx−2h2−4h

is that right

now

$3x^2h + 3xh^2 + h^3 - 4hx - 2h^2 - 4h$

clearer

[versed pagoda]

no its not

[elfin berry]

rynite

[versed pagoda]

3x^2 shouldnt exist

ok

[twin ether]

yeah mb

so what now

do i equal that to 7

and solve

[versed pagoda]

sounds reasonable

now divide this by h

[twin ether]

alr

ill do 0.01

wait you mean h

like the value itself

or literally "h"

[versed pagoda]

no the variable h

[twin ether]

ok

[twin ether]

$h^2 + 3hx + 3x^2 - 2h - 4x - 4$

[elfin berry]

rynite

[twin ether]

right?

[versed pagoda]

yeah

so now take h to 0

you should be left with 3x^2 - 4x - 4

[twin ether]

why 0

[versed pagoda]

thats what it means to have instantaneous change

you'll learn that in limits

[twin ether]

oh ok

[twin ether]

do i equal to 7

[versed pagoda]

yea

[twin ether]

got -1.36

and 2.69

is that right

[twilit grove]

yeah looks very close

[versed pagoda]

I mean did we make a mistake

because while the roots look fine

-1.36 and 2.69 on the graph are decreasing points

I think we got the function wrong

[twin ether]

is it because we didn't include the vertical shift/comp

[twilit grove]

yes, the function was wrong

[versed pagoda]

its (x-4) * (x+4) * (x-2)

[twilit grove]

the zeros are -4, 2, 4

[versed pagoda]

so (x^2 - 16) * (x-2)

[twin ether]

oh

[versed pagoda]

mb sorry

x^3 -2x^2-16x+32

[twin ether]

ight sec

[versed pagoda]

do the same thing as before

good practice 🙂

you should get 3x^2 - 4x - 16 = 7

to be solved

[twin ether]

i get 0.6666 and 1.5986

also -1.5986

[versed pagoda]

no that's not it

roots are 3.51, -2.18

did u figure it out?

[marsh citrus]

@twin ether Has your question been resolved?

[twin ether]

yeah sorry for late reply, I did something wrong

i got ur answer now

[versed pagoda]

ok cool

good luck

[twin ether]

I have a question though, I don't understand why we didn't consider a in the equation, vertical shift/comp

[versed pagoda]

whats a?

[twin ether]

~0.26

[versed pagoda]

im confused where are u getting -0.26 from?

[limber hearth]

can't be negativ

[versed pagoda]

no it can be

-2.18 is a valid solution

[limber hearth]

i think it's approximately 0.26 what he wanted to say

with the ~ symbol

[versed pagoda]

ah ok

thought it was a - lol

[marsh citrus]

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[spiral merlin]

How do I express this equation?

[dawn spear]

take 3^p+2 common

from the 3 terms

[spiral merlin]

Hmm how

[dawn spear]

acc to you, what should be left after you take 3^p+2 common from the first term?

[spiral merlin]

Factorise 3?

[dawn spear]

wdym factorise 3

[spiral merlin]

Like take out 3 from the 3 terms?

[dawn spear]

first just take out 3^p+2 from the first term

and tell me what do you get

[spiral merlin]

1^1

[dawn spear]

no

[spiral merlin]

Hmm I don’t get it 😭

[marsh citrus]

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[ripe python]

Is this correct

[ripe python]

But in book says its x= 1/12

I mean key

Are you sure

Okay

Thank you

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[tacit narwhal]

can someone help me solve this

[night turtle]

Try consider the general expression for each individual term

[tacit narwhal]

general expression ???

[night turtle]

Well you know the binomial theorem right?

[tacit narwhal]

yep

[night turtle]

$(x+y)^n = \sum^n_{k=0} {n \choose k} x^{n-k} y^{k}$

[elfin berry]

Afi

[night turtle]

The general expression is ${8 \choose k} (\frac{x^3}{2})^{8-k} (\frac{a}{x})^{k}$

[elfin berry]

Afi

[night turtle]

Since we are looking for the constant term, we want all the x to cancel out

And so we want to find k that satisfies

[versed pagoda]

look at terms of the form $C^8_k x^{3k}/2^k a^{8(8-k)}/x^{8-k} = C^8_k x^{4k-8} a^{8(8-k)}/2^k$

n is 8

[elfin berry]

tyler

[versed pagoda]

so k=2 is the valid term for the constant

[tacit narwhal]

this is the way my teacher did it but i want to know why x=1

[versed pagoda]

so 8C2 x a^{8 x 6}/2^2 = 7 a^48 = 5103

so a^48 = 729

729 = 3^6

so a^48 = 3^6 => a^8 = 3

[night turtle]

Its not that x=1
but x^(24-3r)/x^r is 1

[versed pagoda]

a = 3^{1/8}

[night turtle]

Since we are looking for the constant term here
There wouldn't be any x

[tacit narwhal]

why is it 1

[night turtle]

Is 5x a constant?

[tacit narwhal]

no i dont think si

now im getting confused

???

[night turtle]

If the term has x in it, then it is not a constant right?

[tacit narwhal]

tthe question is asking for the a and r vlue

[night turtle]

e.g. 510, -0.888, e are all constants
5x, 99/x^5, x^2 are not constants

Looking back to your question
We know that the constant term is 5103 (which has no x)

Meaning that the expression for the constant term $\frac{x^{24-3r}}{x^r}$ must be a constant, i.e. 1

[elfin berry]

Afi

[tacit narwhal]

ok

thnanks

bye

.close

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[drifting oar]

How do I find the highest common factor of s to the power of 8 and s to the power of 6

[tired oxide]

it's easy in this case since s^6 divides s^8

[still temple]

its s to the power 6

[tired oxide]

it's just like trying to find the HCF of 2 and 8, well it's 2 because 2 divides 8

[drifting oar]

Would it be okay for you to break that down for me?

So I can understand how it divides etc?

[tired oxide]

yeah i can, if you're trying to find the HCF of 2 and 8, we know that 2 is a factor of 8 and also of 2, but also there are no bigger ones since 2=2, so nothing higher will be a factor of 2

similarly, S^6 is a factor of both S^6 and S^8, but if you try anything higher, it will no longer be a factor of S^6

[drifting oar]

How is S^6 a factor of s^8?

[tired oxide]

because S^6 * S^2 = S^(6+2) = S^8

using our exponent rules

[drifting oar]

Got it

So the highest common factor would be s^6?

[tired oxide]

yep

[drifting oar]

Thanks! and how would I find the highest common factor of h^8 z^5 and h^7 z^6?

[tired oxide]

you would take the lower of the two powers in each and combine them. think "the max capacity for h's in h^8z^5 is 8" and for h^7z^6 it's 7, so we try to fill both with h's we can only fit 7 until one of them can't fit any more

and do the same for the z's

so how many z's go into each term?

[drifting oar]

Im sorry, I dont completely understand

[tired oxide]

that is okay, we should start at the basics actually, you are familiar with the term HCF? if not, we should start there.

[drifting oar]

yep

[tired oxide]

okay great

[drifting oar]

🙂

I havent done highest common factors of variables and exponents until today so def a learning curve 😂

Pretty sure ive got it nailed for variables now though 🙂

[tired oxide]

okay, that is fine. actually i want to start with an example for numbers, and it'll pretty much tell you how to do it for variables. I want to find the HCF of 18 and 24, how would you go about doing that?

i know this sounds simpler than what you want, but it's really instructive

[drifting oar]

Finding common multiples eg: 1,18 and 1,24 then working my way up till I find the highest one that goes into both

So in this one it would be 6

6x3 and 6x4

[tired oxide]

yes. okay, also do you know about prime factorizations (i promise i am getting somewhere with this)

[drifting oar]

Working on it, not completely there yet

[tired oxide]

okay, the point is that 24= (3)(2)^3 and 18 = (2)(3)^2

and we computed the hcf to be 6 = 2*3

so what i would do to find the hcf was to look at the minimum exponent of each of the primes (which in your case will be the variables) that divide it.

so we can go back to h^8z^5 and h^7z^6

i look and see "what is the minimum exponent of the h's" and i see that it's 7, and "what is the minimum exponent of the z's" and I see that it's 5

so we can just say that the hcf is h^7z^5

and in fact you can check for yourself that that's indeed the highest common factor

[drifting oar]

So the highest common factor is the minimum exponent of both variables in the equation?

[tired oxide]

yeah, you should verify my statement by trying to see if h^8z^5 is a factor of both, and then realizing that it's not a factor of the second h^7z^6

also be careful about using the word equation

there is no equation here!

[drifting oar]

So the highest common factor is h^7 and z^5?

[tired oxide]

it's the product of those

h^7z^5 is one expression

[drifting oar]

So what would I need to do to find the hcf?

[tired oxide]

that is the hcf, you basically found it, you looked for the min exponent in both variables, identified them correctly, and now the hcf is h^7z^5 (read: h to the seventh times z to the fifth)

[drifting oar]

Got it! So if I were to find the hcf of 35a^2 x^4 and 25a^3 x^6 I would do the same thing?

[tired oxide]

yes, this one is slightly different because you now have numbers out front, right? but do the variable part first, and then the number part after

[drifting oar]

Got it, so i would find the hcf of 35 and 25 which is 5?

[tired oxide]

yep

and then now do the variable part

[drifting oar]

and then find the minimum exponent for each variable?

[tired oxide]

yep

[drifting oar]

so a^2 and x^4

[tired oxide]

yrp

so now we combine them together

[drifting oar]

So would it be 5a^2x^4?

[tired oxide]

yeah

[drifting oar]

Thank you so much!

[drifting oar]

Would it be okay to get some help for other hcfs as well?

I have a few more which are a bit tricky

[tired oxide]

sure yeah go for it

I won't say much though, you pretty much have the technique down now

[drifting oar]

L(5-2x) and 3(5-2x)

[tired oxide]

okay so treat 5 - 2x as a variable

so we can do the numbers out front first, we have 3 on the right and just a 1 on the left (remember that if there is no number in front that just means there is an invisible 1)

[drifting oar]

So you mean 1L and 3?

[tired oxide]

and then go variable by variable. if there is ever not a variable in the either expression, then just leave it out, for example you won't have any L's in your HCF

[drifting oar]

Just making sure Im looking at the same part :laugh:

[tired oxide]

1 and 3

L is just another variable

i am assuming

[drifting oar]

So where did the 1 come from?

Looking at the L right?

[tired oxide]

the 1 is literally always there. L(5-2x) = 1* L(5-2x)

[drifting oar]

Got it

just making sure we are looking at the l

1x5 and 1x-2x for the left bracket?

or do we not expand and just find the hcf?

[tired oxide]

don't expand

you can directly find the hcf, just treat (5-2x) like a variable and do what we did before

actually every time you try to find the hcf you'll try to get them in a product of terms like this

so this is like the best case we could hope for

[drifting oar]

Sorry, ive got a bit of brain fog atm. would 5-2x be the minimum variable sincw its the same in both brackets?

since

[tired oxide]

if you're saying what I think you're saying then yes..

5-2x is the variable, min exponent is 1

[drifting oar]

So how would I find the hcf?

[tired oxide]

just like before. take hcf of the numbers, which is HCF(1,3) = 1, then the min exponent of the L's, which is 0, and then the min exponent of the 5-2x's, which is 1

so overall we just get (5-2x)^1

[drifting oar]

So the hcf is (5-2x)^1?

[tired oxide]

yeah

notice that (5-2x)^1 = 5-2x

so we can just say 5-2x

[drifting oar]

So the simplified varsion is 5-2x?

[tired oxide]

yea

[drifting oar]

Thanks! are you any good at factorising by chance?

[tired oxide]

yes, but i have to go to sleep, it is 4 am and i have to be up in the morning

[drifting oar]

ahh, sorry to be using your time like this!

[tired oxide]

other people here are also good at factoring though

no it's okay, i like doing this

i like doing math help in my free time it's fun.

I just get carried away sometimes and stay up too late :P

good luck factorizing though.

[drifting oar]

Thank you so much, Ive been very sick for the past term and havent been able to come to lessons! (+ we got a new teacher)

This has significantly helped me 🙂

[tired oxide]

lol yeah that can be really discouraging. hopefully you are not too far behind, i know that it can be rough in that situation

[drifting oar]

Not too far behind not that youve helped me!

Thanks so much again!

How do I factorise 14l^5 +8l^4 g^4

<@&286206848099549185>

Could you help break that down for me?

im sorry, i dont seem to be catching on. Could we try k(-6+y^2) +5a (-6+y^2)

is v^2-13v

trinomial

<@&286206848099549185>

Why is it trinomial?

would 2d^2+34 be the same then?

So thats also both quadratic and trinomial?

same with -6x+36?

but not with y^2?

<@&286206848099549185>

is y^2 quadratic and trinomial?

so all of those are quadratic and trinomial?

[marsh citrus]

@drifting oar Has your question been resolved?

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[drifting oar]

Is the factorised version of 25n^2-25 = 25 (n-1)(n+1)

[late geode]

yes

[drifting oar]

is the factorised version of 25q^2-f^2 = (5q+f)(5q-f)

[late geode]

yes

[drifting oar]

is the factorised version of 25t^2-25h^2 = (t+h)(t-h)

Is the factorised version of 4c^2n^2-49v^2 = (2cn+7v)(2cn-7v)

[marsh citrus]

@drifting oar Has your question been resolved?

[drifting oar]

is the factorised form of 14l^5+8l^4g^4 = 2l^4(7l+4g^4)

<@&286206848099549185>

[buoyant jetty]

,iamnot helper

[elfin berry]

Removed the Helpers role from you.

[buoyant jetty]

,w factorize 25t^2-25h^2

[buoyant jetty]

almost

,, (t-h)(t+h) \cdot 25

[elfin berry]

938c2cc0dcc05f2b68c4287040cfcf71

[buoyant jetty]

,w factorize 25q^2-f^2

[elfin berry]

Results provided by WolframAlpha

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Requested by __melomaniac

[buoyant jetty]

I think you get the idea

[buoyant jetty]

this is called difference of squares I think

[drifting oar]

kk

would the factorised version of 14l^5+8l^4g^4 = 2l^4(7l+4g^4)

[buoyant jetty]

,w factorize 14a^5+8a^4b^4

[elfin berry]

Results provided by WolframAlpha

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

Requested by __melomaniac

[drifting oar]

.close

[marsh citrus]

Channel closed

Closed by @drifting oar

Use .reopen if this was a mistake.

[buoyant jetty]

yes

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[celest harbor]

how would i intergrate y? so far im at $3 \int \frac{1}{u}dx => \frac{(4x+1)^{\frac{1}{2}}}du = dx => 3 \int \frac{(4x+1)^{\frac{1}{2}}}{2u} du$ but i feel like theres an easier way of doing this

[elfin berry]

morphine_addiction

[celest harbor]

hold on this latex is buggin

[naive fox]

ok... I think you forgot the square root or did you let your $u = \sqrt{4x+1}$?

[elfin berry]

᲼

[naive fox]

ok... suppose you let your $u = \sqrt{4x+1}$. What is your $du$?

[elfin berry]

᲼

[naive fox]

actually, you have it already

[celest harbor]

ye 2(4x+1)^(-0.5)

[naive fox]

one trick to make this simpler is to see that $u^2 = \sqrt{4x+1}$ then, $dx = \frac{u}{2} du$ but let's stick with what you have for now

[elfin berry]

᲼

[naive fox]

well, you will have a similar result anyways but with different bound

[celest harbor]

u mean u^2 = 4x+1?

[naive fox]

yes. it is the same with what you did

just write it in different way.

Anyways, I think you forgot square root on 3 and your solution should be fine

[celest harbor]

but like how do i carry on from where i left of bc the (4x+1) is still in terms of x

[naive fox]

well, you know that $u = \sqrt{4x+1}$

[elfin berry]

᲼

[naive fox]

then your $du = \frac{2}{(4x+1)^{\frac{1}{2}}} dx = \frac{2}{u} dx$

[elfin berry]

᲼

[naive fox]

do you agree?

[celest harbor]

yes

[naive fox]

Then, your integral will simplify to ...

[celest harbor]

1/2

[naive fox]

exactly

[celest harbor]

so 3x/2?

[naive fox]

well, close

not quite

[celest harbor]

i mean u

[naive fox]

still... really close

[celest harbor]

oh ye when i took out 3 it was supposed to be sqrt(3) so it would be sqrt(3)/u

[naive fox]

well, no. But yes, you should have $\sqrt{3}$

[elfin berry]

᲼

[naive fox]

it is $\left.\frac{\sqrt{3}u}{2}\right|_{a}^{b}$

[elfin berry]

᲼

[celest harbor]

hold on im buggin, 1/2(sqrt(3)*u

[naive fox]

yes.

Now, you need to find the boundary with respect to $u$ (or just substitute $x$ back) and you should be done

[elfin berry]

᲼

[celest harbor]

so b=9, a=1?

[naive fox]

yes, if you use $u$ that is

[elfin berry]

᲼

[celest harbor]

so $\frac{\sqrt{3}u}{2}\right|{1}^{9}$ is the same as $\frac{\sqrt{3}(4x+1)^{\frac{1}{2}}}{2}\right|{0}^{20}$?

[elfin berry]

morphine_addiction
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[naive fox]

yes.

[celest harbor]

how come my calculator gives different answers for both?

[naive fox]

left side, you are basically putting the $x = 0$ and $x = 20$ down already and the right side is you have not done anything yet.

[elfin berry]

᲼

[naive fox]

actually...

since $u=(4x+1)^{\frac{1}{2}}$, $x = 0 \rightarrow u = 1$

[elfin berry]

᲼

[naive fox]

and $x = 20 \rightarrow u = \sqrt{81} = 9$

[elfin berry]

᲼

[naive fox]

you should get the same result

[celest harbor]

ye i cnaged the bounds here and got different results

[naive fox]

well, you can't get different result when left side (u) is the second step of the right side (x)

so, for $u$, $\frac{9\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = \frac{8\sqrt{3}}{2} $

[elfin berry]

᲼

[naive fox]

for $x$, $\frac{\sqrt{3}}{2}(4(0) + 1)^{\frac{1}{2}} = \frac{\sqrt{3}}{2}$

[elfin berry]

᲼

[naive fox]

and $\frac{\sqrt{3}}{2} (4(20)+1)^{\frac{1}{2}} = \frac{9\sqrt{3}}{2}$

[elfin berry]

᲼