help-10

[proven oar]

you will get A = area of square + area of rectangle

[sinful saffron]

that would leave the first sqare with w: x, l: x, then the second rectangle: w: 2x, l: 2x + 1

[proven oar]

yes

so then for the area you would get

[sinful saffron]

so w: 3x, l: 3x + 1

[proven oar]

is this

for the whole shape?

[sinful saffron]

yes

[proven oar]

how did you get this

[sinful saffron]

i added w and l together

[proven oar]

if you are trying to find the area

then you need to find the area of the individual shapes

[sinful saffron]

so the rectangle would be 4x + 1

[proven oar]

no

[sinful saffron]

or squared

[proven oar]

can you tell me what the formula for area of a rectangle is

and formula for area of a square

[sinful saffron]

a = l * w

[proven oar]

correct

so going back to what you said before

width of the rectangle is 2x+1

and length of the rectangle is 2x

[sinful saffron]

2x + 1 * 2x

[proven oar]

yes

[sinful saffron]

like terms: 2x * 2x

4x^2 + 1

[proven oar]

That isnt correct

when you sub the values into the formula you get

A = (2x+1)*(2x)

can you tell me what the expanded form of this would be

[sinful saffron]

uh 1 sec

2x(2x+1)

i entered it in my calculator thats what i got

[proven oar]

That is correct

however that is not entirely expanded

do you know how to expand brackets

[sinful saffron]

yeah

[proven oar]

ok

so how would you expand

[proven oar]

this

[sinful saffron]

2x^2+2x

[proven oar]

correct

so thats the area of the rectangle

how would you find the area of the square

[sinful saffron]

same formula

[proven oar]

oh wait

sorry

there is a mistake

[sinful saffron]

oh

[proven oar]

2x^2

this should be something else

[sinful saffron]

4x^2

[proven oar]

nice

[sinful saffron]

mb

[proven oar]

so now you have area of rectangle (4x^2+2x)

how would you get the area of the square

[sinful saffron]

x^2

[proven oar]

nice

so the total area of the shape would be?

[rain forge]

?

Hum not sure

This is not a square or am i wrong?

[sinful saffron]

6x^2+2x

[rain forge]

Like it's more x and x+1

[proven oar]

not quite

The total area would be

(x^2) + (4x^2+2x)

so collecting like terms

you would get

[rain forge]

Bro i don't think

Or am I wrong

[sinful saffron]

5x^2+2x

[proven oar]

Yes

nice

that should be the final answer

[sinful saffron]

idk why but i read it as 2x^2 mb

the answer sheet says 5x2+3x hmmmm

[rain forge]

I would say 5x² + 3x

Bro you get wrong@proven oar

[sinful saffron]

how did u get the 3x?

[rain forge]

It's not a square

It's 2x+1-x

So the length are x and x+1

[proven oar]

oh tru

mb

[sinful saffron]

oh yeah forgot about that 1

[rain forge]

Or you can do $3x*(2x+1)-x²$

[proven oar]

its a rectangle not a squarte

[warm shale]

phoestaclies

[proven oar]

So area instead would be x(x+1) + 2x(2x+1)

[rain forge]

Yes

[sinful saffron]

ah that makes sense now

[proven oar]

Anything else I can help with

[sinful saffron]

all good thanks tsym

phoestaclies ty for assisting

[rain forge]

Np good luck 👌

[sinful saffron]

ty

well i shall close this now

.close

[obtuse pebble]

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[open mango]

Hello everyone if someone know how to solve this question?

[obtuse pebble]

@open mango Has your question been resolved?

[open mango]

NO

@obtuse pebble nope

[timid silo]

Lable the diagram first

Would be easy to communicate if someone helps you out

[open mango]

Ok

[obtuse pebble]

@open mango Has your question been resolved?

[obtuse pebble]

@open mango Has your question been resolved?

[obtuse pebble]

@open mango Has your question been resolved?

[obtuse pebble]

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[brisk kayak]

How do I do question 6

[gleaming ridge]

What are you supposed to do? Differentiate?

[brisk kayak]

calculate for y prime

[thick gyro]

so differentiating it

[brisk kayak]

yeah

[thick gyro]

for question 6

do you know

what the diffferential of

inverse cos is

[brisk kayak]

1/cosx

[thick gyro]

no

that is arccos

as in, the actual inverse cos

[brisk kayak]

sin?

[thick gyro]

no

as in

if cos(pi/2) = 0

then cos^-1(0) = pi/2

[gleaming ridge]

Clearly she doesn't know

[thick gyro]

that was learnt in standard trig years before differentiation was taught

[gleaming ridge]

well, you can divide both sides by x and then take cos of both sides

cos will cancel with cos inverse

then differentiate using product rule and rearrange

[thick gyro]

i dont think the person knows how to do implicit differentiation

[gleaming ridge]

She won't be able to do question 8 without that

[thick gyro]

guess youre right

but honestly, she just needs to know what the differentiation of arccos is, and then can just use product rule normally

instead of

trying to derive it for xarccosx

would be less work

[gleaming ridge]

another option is to derive the derivative of arccos separately and then use it freely

[brisk kayak]

So how do I solve it?

.close

[obtuse pebble]

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[solid topaz]

how do I factor x^4 + 2x^3 - 16x^2 - 32x?

[heavy depot]

$x^4+2x^3-16x^2-32x$

[warm shale]

CrEpasPmkinPie

[heavy depot]

this?

[fickle turret]

Factor out an x and guess a root

[heavy depot]

first take the x out

[thick gyro]

you can use the rational root theorem

to see which roots to guess

[heavy depot]

$x(x^3 + 2x^2 - 16x -32)$

[warm shale]

CrEpasPmkinPie

[solid topaz]

ok how step by step guide pls

[heavy depot]

now you can split that cubic equation

$x((x^3+2x^2)+(-16x-32))$

[warm shale]

CrEpasPmkinPie

[thick gyro]

thats acc genius, but not much equations are that obviouss

[heavy depot]

and now take common factors from both parenthesis

and do some algebra to it

$x( x^2(x+2) + -16(x+2))$

[warm shale]

CrEpasPmkinPie

[heavy depot]

some rearranging shows the factors

$x(x^2-16)(x+2)$

[warm shale]

CrEpasPmkinPie

[heavy depot]

and even that x² factor could be split up

[solid topaz]

To be factored in a way that I could use for the graphing for the polynomialss

[heavy depot]

$x(x-4)(x+4)(x+2)$

[warm shale]

CrEpasPmkinPie

[heavy depot]

and those are your zeroes aswell

[thick gyro]

honestly, rational root theorem gives me all the roots in this case

-2, -4, 4

[heavy depot]

and 0

[thick gyro]

so i just know to split it up into (x+2)(x+4)(x-4)

although if it only gave me 1

id have to do polynomial long division

[heavy depot]

do synthetic instead

its a linear divisor

[thick gyro]

never heard of it

[heavy depot]

it's a lot easier than long division

take the polynomial

then take a linear factor

must be linear

and the polynomial has to have every degree

so put zeroes in as coefficients

then take the opposite of the constant in the factor

its funky

take this polynomial

[solid topaz]

Wait can u do the factoring step by step

[heavy depot]

sure

$x^4+2x^3-16x^2-32x$

[warm shale]

CrEpasPmkinPie

[heavy depot]

each of those terms has an x

so you can factor the x out

[solid topaz]

ohh ok

[heavy depot]

$x(x^3+2x^2-16x-32)$

[warm shale]

CrEpasPmkinPie

[heavy depot]

now that's a cubic polynomial

so split the two halves

$x((x^3+2x^2)+(-16x-32))$

[warm shale]

CrEpasPmkinPie

[heavy depot]

and now you can take common factors out of both of those

$x(x^2(x+2)+-16(x+2))$

[warm shale]

CrEpasPmkinPie

[heavy depot]

and the do some algebra (I don't remember what it is exactly)

[solid topaz]

Wait this is the part where it got me confused, from 32 to 2? Simplified?

[heavy depot]

ah

$(-16x-32)$

[warm shale]

CrEpasPmkinPie

[heavy depot]

that has a common factor of -16

so divide both terms by -16

$-16(x+2)$

[warm shale]

CrEpasPmkinPie

[heavy depot]

$\frac{-32}{-16}=2$

[warm shale]

CrEpasPmkinPie

[solid topaz]

or can u help me do it from the point where I need to graph it

like the behavior of the graph

[heavy depot]

$x(x-4)(x+4)(x+2)$

[warm shale]

CrEpasPmkinPie

[heavy depot]

those are the zeroes

${-4,-2,0,4}$

[warm shale]

CrEpasPmkinPie

[heavy depot]

The leading coefficient is also positive

so it will open upwards

[solid topaz]

upwards since its greater than 0?

[thick gyro]

donr use someonee elses chat

[solid topaz]

Since idk where to start, to be able to graph it

[solid topaz]

Where do I put the point for the imaginary root?

[heavy depot]

there is no imaginary root

[solid topaz]

there's another thing that Idk how to solve

The volume of a rectangular packaging box is 73dm^3 more than the volume of a cube
packaging box. The length of the rectangular box is 1dm longer than twice the edge of the cube.
Its height is 2dm shorter than the length of the edge of the cube. Its width is 1dm longer than the
cube’s edge. What are the dimensions of the two packaging boxes?

[obtuse pebble]

@solid topaz Has your question been resolved?

[solid topaz]

How do I answer this?

[obtuse pebble]

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[timid silo]

Find all real solutions and justify your steps $1 < \ceil{3x+5} \leq 3$

[warm shale]

hibyehibye

[timid silo]

So I did this by just letting f(x) = 3x+5 and finding the preimage of (1,3]

but im not sure how I’m meant to actually solve it and justify all my steps

[rigid lintel]

thats partially correct

its correct for the <= 3 part

but not for the 1 < part

[timid silo]

hmm

[rigid lintel]

because of the ceiling function

[timid silo]

oh i meant f(x) = ceil(3x+5)

[rigid lintel]

oh alright then youre good

[timid silo]

whats the like justification for doing this?

[rigid lintel]

definition

[timid silo]

definition of what?

[rigid lintel]

$1 < \ceil{3x+5} \leq 3$ is equivalent to $x \in \mathbb{R}$ such that $f(x) \in (1, 3]$

[warm shale]

heavy

[rigid lintel]

with $f(x) = \ceil{3x+5}$

[warm shale]

heavy

[timid silo]

Tysm!!!

.close

[obtuse pebble]

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[dark surge]

how would i go about disctreticizing a partial derivative like this (from the euler equation) $\frac{{\partial u}{\partial x}}, where x is the grid spacing in the x direction and u is the velocity vector$

[warm shale]

D3U5

[dark surge]

wait

[timid silo]

stop the $ before u get into the text

[dark surge]

$\frac{\partial u}{\partial x}$

[kind hawk]

essentially just do it in both directions separately and then add together

[warm shale]

D3U5

[kind hawk]

do you know the multidimensional taylors theorem?

oh wait you don't have a mixed derivative

[knotty crow]

you can also use

[kind hawk]

I read too much

[warm shale]

Modus

$$\pdv{u}{x}$$

[karmic bison]

Can any one help me with a short assignment

[kind hawk]

read #❓how-to-get-help

[dark surge]

no, all this stuff is conpletely new to me and i dont know where to start, cuz im still in middleschool

[kind hawk]

then why are you doing it

[dark surge]

because i want to understand it

where do i even start with such stuff

[kind hawk]

well you are probably missing like 3-4 years of basics

[timid silo]

probably reading all the stuff preceding to it

[dark surge]

but how is this supposed to be done

[kind hawk]

by knowing the basics

what exactly do you want to hear here

[dark surge]

the ones with the polynomial equations i understand

its just that this is a vector and a grid spavincing

[kind hawk]

"the ones with the polynomial equations" ?

[dark surge]

where you have f(x,y) and you have to do $\pdv{f}{x}$

[warm shale]

D3U5

[dark surge]

and the normal ones too dx/dy

but i dont understand how we go from a function to having a vector

[kind hawk]

a vector is just a couple of functions stacked on top of each other

$u(x) = (u_1(x), u_2(x), \ldots, u_n(x))$ where all the $u_i(x)$ are "normal" functions

[warm shale]

Denascite

[dark surge]

oh

[kind hawk]

and you can treat each coordinate separately

[dark surge]

ah ok thanks, thats exactly the thing i need to know

.close

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[fringe dew]

x

[fringe dew]

Hello, i'm Mallow. I'm currently programming an isometric Chess Game for my school. And i'm a bit stuck with maths.
Has you see. I've got the "chess_board.png" scaled by 4 so it's 616x364. Using some rudes archaic paint skill. I've found the 4 coordinate of the corner :
A = [0, 132];
B = [386, 0];
C = [616, 232];
D = [232, 364];
And i got an x and y for the mousePointer on the image. So basically mouseX can be [0, 616] and mouseY [0;364].
I can't figure how to determine the x and y of one of the squares based on the mouse coordinate.
Can someone help me please ?

[royal basin]

is this image going to be hardcoded into the game, as well as the screen resolution?

one would think that you'd want to grab the coordinates of the mouse within the window rather than on the image.

maybe i am mistaken about this.

[fringe dew]

Yes, the image will always have the same resolution, but the screen size can change. Don't worry about that. The mouse value that will be forwarded to the calc fonction will be based on the image location. At image 0,0, cursor will be at 0,0

The problem is i'm trying to figure which square as been pressed. And for that i've figured to get the deplacement between this 2:

left corner, and the right corner of the first square

+50.875x -16.5y

(the 0;0 point is the top left corner of the rectangle image)

[obtuse pebble]

@fringe dew Has your question been resolved?

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[mild badger]

I need help on question 22

[naive fossil]

Have you tried using the Law of Sines?

[mild badger]

idk how to use it that way

i can only use it a certain way when i have 2 angles

[high lily]

consider law of cosines

which relates three sides of a triangle with an angle

[mild badger]

i havnt learned law of cosines

[obtuse pebble]

@mild badger Has your question been resolved?

[obtuse pebble]

@mild badger Has your question been resolved?

[obtuse pebble]

@mild badger Has your question been resolved?

[obtuse pebble]

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[dark tundra]

I got tripped up ont his problem. I dont understand why the largest integer is x+127

[dark tundra]

if its the sum of the first 127 integers

x+0 is an integer

and then (x+1)….(x+126) are the other 126 integers

so wouldnt the largest term be x+126

[timid silo]

I dont really get their working out but you’ll get the same answer.

127x + sum 0 to 126 = 27178
x = 151
151+124=275

[obtuse pebble]

@dark tundra Has your question been resolved?

[dark tundra]

thanks alot

.close

[obtuse pebble]

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[compact prawn]

Can anyone help me with this? I just basically need to know what case it is and then I can do it after that

Well I know it’s non homogenous case

Wait I’m dumb I can do this

[timid silo]

Have you figured it out?

[obtuse pebble]

@compact prawn Has your question been resolved?

[compact prawn]

I got this but I’m not entirely convinced it’s correct lol

I was not in class when this was taught and haven’t had the time to catch up so I’m not 100% sure what I’m doing 😅

[timid silo]

This is correct.

[compact prawn]

Oh perfect, thanks!

!close

!thanks

[timid silo]

.close

[obtuse pebble]

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[storm wing]

I have a question about one of the steps here. I think it's simple arithmatic

[storm wing]

Is there some basic math here that is making me miss how we are able to do this?

So I understand how sqrt(x+3) -sqrt(x) can be divisible by one

but i dont understand the rule that allows us to multiply it by sqrt(x+3) + sqrt(x) .sqrt(x+3) + sqrt(x)

Hopefully, that makes sense.

[supple granite]

Multiplied the conjugate of the numerator to rationalize the numerator

[storm wing]

Thank you for definining it for me. I can see how to do this online now. Thank you.

.close

[obtuse pebble]

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[worldly trench]

when I'm writing a math paper, how do I make it clear that the (1) I'm referring to is not part of the equation?

[worldly trench]

Like this, I want to reference it throughout my paper

[frosty spoke]

do people actually believe that it's in there?

[worldly trench]

Yes

[frosty spoke]

lord help us

[worldly trench]

😭 I thought it was clear enough

[frosty spoke]

you could maybe write (Equation 1) or some abbreviation

[worldly trench]

Do you have any suggestions for making it cleaer?

[worldly trench]

wdym abbreviation?

[frosty spoke]

like (E1) or (Eq 1)

[worldly trench]

Ah gotcha

[frosty spoke]

I honestly prefer the (1) though

didn't know people fell for that

[worldly trench]

What if it's not necessary an equation per se?

[timid silo]

use \begin{align*} instead of \begin{align}

if you're using LaTeX

[worldly trench]

I wrote that in google docs

[timid silo]

ah

[obtuse pebble]

@worldly trench Has your question been resolved?

[obtuse pebble]

@worldly trench Has your question been resolved?

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[open nimbus]

how do i do (b) ?

[supple granite]

Well you have the center and the radius

[open nimbus]

yeah im just fucking stupid and forgot everything in algebra 2 apparently i just figured it out im sorry

thank you tho! ❤️

[supple granite]

[open nimbus]

.close

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[livid agate]

How is that a?

[livid agate]

Oh crap

Yeah, just realize

Thank you

.close

[obtuse pebble]

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[zinc zodiac]

.reopen

hi?

[obtuse pebble]

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[zinc zodiac]

Does this make sense?

[zinc zodiac]

The x and y are for referencial use, that's why the subscripts

[obtuse pebble]

@zinc zodiac Has your question been resolved?

[tardy epoch]

yes it does

[zinc zodiac]

Could you please explain it to me? Like, i stays the same until j completes 4, and then i is 2 and the same happens, etc?

[tardy epoch]

https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-3/v/sigma-notation-sum

[zinc zodiac]

Alright, thanks

.close

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[wispy cave]

In a circle of radius 2 miles, the length of the arc that subtends a central angle of 5 radians is

[wispy cave]

I know how to do the problem, but I'm not sure what 5 radians exactly is

I know how to convert degrees to radians and radians to degrees. but is 5 radians 5pi?

or

[timid silo]

$radians \cdot \frac{180}{π} = degrees$

[warm shale]

pulse

[wispy cave]

Ahh I see

So since I'm using s=r0 (theta) am I using 900/pi for theta?

[timid silo]

if you want the angle to be in degrees, yes

[wispy cave]

Ohhh I see

It did want it in radians but I just converted my answer back to radians

appreciate it

.close

[obtuse pebble]

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[obtuse thunder]

anyone...

[obtuse pebble]

Please don't occupy multiple help channels.

@obtuse thunder Has your question been resolved?

[sullen raptor]

Reciprocal

[obtuse pebble]

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[clever temple]

I have a simple question

So

In that image

Sin(u) is set equal to x/sqrt of (1+x^2)

But why isn’t it just x?

Why do we have to divide by the hypotenuse?

[gleaming ridge]

cause the definition of sine in that it is ratio of perpendicular and hypotenuse

(sin is just short for sine) @clever temple

[clever temple]

Oh wait

Yeah i couldn’t think for a sec thanks so much

@gleaming ridge

Helped me solve the problem

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[gleaming ridge]

sure

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[red thistle]

Don't mind that answer in there

[red thistle]

How do I go about getting the coefficients that should go in front of the 'y=e^rt"

I have the derivatives and the system of equations

[floral canopy]

actually yea

Try it

[fathom flicker]

have you tried

y=Ae^(rt)

and then differentiating

and plugging in

and then solving for A

[floral canopy]

but A falls off anyways

in the eq. above

[red thistle]

hold on lemme take a pic of what i did

i was thinking linear algebra but the uh

extra kinda stumped me

gimme a sec

[fathom flicker]

you can try aswell

y=xe^(rt)

is this just undetermined coefficients?

[floral canopy]

how did you get to that

like the first line

[floral canopy]

just try e^(rt) you will get a solution

[red thistle]

just strictly e^(rt) without the added t in front of the e?

ill try that

[floral canopy]

yea

[red thistle]

the way i was taught like two days ago

was when you have the original differential equation

which will turn out to r^2(r-4)^2

when its squared you'll have the

repeated roots

and just throw in an added t at the front

actually heres an example of what he did

anyway i digress, gimme a sec

[fathom flicker]

uhh

is the y^4 meaning the fourth derivative

[red thistle]

mhm

[fathom flicker]

oh

well

that changes a lot

[red thistle]

ah

[fathom flicker]

yeah, so then you do what your professor did

find the characteristic polynomial

which would be

r^4 -8r^3+16y^2 =0

[red thistle]

well that turns out to the 2nd pic that i did

[fathom flicker]

,, r^4 -8r^3+16r^2 =0

and you solved for the roots of ^ that

[warm shale]

AustinU

[red thistle]

mhm

[fathom flicker]

what'd you get

[red thistle]

r^2(r-4)^2

[floral canopy]

why did you add a t before the e(-4t)

[fathom flicker]

so

it has two repeated roots

one at 0 , and one at 4

[red thistle]

i still dont know like the base answer to that, but because of the repeated roots

[fathom flicker]

so then the solution is..

[red thistle]

just throw in a t

Austin, as for what i got for trying, (albeit poor quality) was in the pic with the blue handwritten

maybe im getting lost

just

trying to figure out how to account for the uh

constants

given like

the third derivative that i found with the base form

y=e^rt

i dont know if i can solve for the constants

with linear algebra because of something like on the third derivative

48c4e^(4t)+64c4te^(4t)

he didn't show how to do it by hand either

just

"use technology"

trying to see if there is one by hand

[fathom flicker]

,, y=c_1+c_2t+c_3e^{-4t}+c_4te^{-4t}

this is what you got yes?

[red thistle]

no

idk why the third constant has c3e^-4t

is it simply because its a repeated root?

[fathom flicker]

so

[red thistle]

• or - 4

[fathom flicker]

here's the deal

[warm shale]

AustinU

[fathom flicker]

when you have repeated roots

it tells you there is a solution

e^(-rt)

at that root

so from that we get the C_1 and the c_3 e^(-4t)

which I assume you follow

but

because this is a fourth order ODE

we know that

when we know one solution y_1 , to the ODE , there is another solution y=y_1u

where u is a function of t

the way you can solve for u

is writing

,, y=e^{-rt} u(t)

[warm shale]

AustinU

[fathom flicker]

we will take the case of our first repeated root at 0

,, y=u(t)

[warm shale]

AustinU

[fathom flicker]

now differentiate

,, y'=u'(t)

[warm shale]

AustinU

[fathom flicker]

etc.... etc...

plug in once you have them all

solve for the fact that 1*u(t) must equal 0

and you will end up getting (most likely) u(t)=t

and repeat the same process for your other repeated root

starting with

,, y=e^{-4t}u(t)

[warm shale]

AustinU

[fathom flicker]

differentiating 4 times

plugging in

solving for u(t)

and then you get your new solution

[red thistle]

trying to process all this
will the same initial setup for each derivative be the same, just with

[red thistle]

that

or

above

[fathom flicker]

I'm sending you an example image

although it is only for a second order ODE

this is for the case where we find a solution y=e^(-at)

so you would do this process twice

for the two repeated roots you had

y=1 and y=e^(-4t)

the method for solving in the simplest way I am assuming would be to follow the same pattern outlined with the blue pen

multiply the first line of the equation by a^4

then the second line by 4a^3

then the third by 12a^2

[red thistle]

ah ok i see how that pic is set up now then

uhm

well let me try my hand at this

[fathom flicker]

I actually think

it may be unnecessary to even solve the system

and that the case always becomes

u ^ whatever prime = 0

so then u(t) will become

the polynomial that arises from integrating it

,w y''''-8y'''+16y''=0

[warm shale]

Results provided by WolframAlpha

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

Requested by AustinU#0146

[fathom flicker]

so we get our ^ e^(4x) and e^(4x) * x type solutions along with our constant and constant * x solutions

[red thistle]

going back to this

are the equations multiplied after we have found the derivatives then adding them together?

[fathom flicker]

that was the method there yes

^ although I think it is unnecessary

that was probably just the first proof in my notes of how it ended up giving the result

[red thistle]

hhmm

[fathom flicker]

See how there, the end result was u'' = 0 , and then integrating would give u=c1 t + c2

I think that you will always reach an end result of

something like

u'''=0

and for us it will be the fourth derivative

which will give

u=c1t^3+ c2t^2 +c3t+c4

[red thistle]

sorry i-

so with that u(t)

do we use that in tandem with something else to find the constants?

cuz if

well if t=0

[fathom flicker]

In order to solve for the initial values we first need to get the entire solution of our ODE, which we can do once we find the other two solutionst that aren't y=c1 and y=c2e^(-4t)

[red thistle]

agh
im feeling really dumb right now
Im not sure where to continue off with that right now

so to find the other two solutions

uhm

[fathom flicker]

Don't feel dumb , this is a struggle for me aswell

[red thistle]

hhm im still a bit lost with this new method

[fathom flicker]

lol

I think you just attach a t to the previous two solutions

[red thistle]

so then we'd have y=c1 y=c2e^(-4t) y=c1t and y=c2te^(-4t)?

would that last two be two different constants, however?

[fathom flicker]

I want to say yes, but also I am afraid I might not be leading us in the right direction

I don't want to confuse you or get you a wrong answer

[red thistle]

is there a way we can solve that IVP with wolfram?
if not i want to figure out something we can do with that way you showed me, but idk how to continue off where i left in the pic with blue ink

[fathom flicker]

so

I'll point out

,w y''''-8y'''+16y''=0

[warm shale]

Results provided by WolframAlpha

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

Requested by AustinU#0146

[fathom flicker]

if you can't tell from this

wolfram gives a weird answer

but

grouping constants

gives you

,, y=c_1e^{4x}x+c_2e^{4x}+c_3x+c_4

[warm shale]

AustinU

[fathom flicker]

and I can ^ explain how I got that if you need/want

and you can solve for those constants, based on the solution from wolfram

and the solution from wolfram

is so so close to what we were going to get

I don't understand why it is to the power of positive 4x

instead of -4x

otherwise that is what our solution was exactly

[red thistle]

i think i can see how we get this

to solve for the constants now

[fathom flicker]

just expand through their solution and group constants if you need to

To solve for the constants

the method will be

differentiate Wolfram's solution 3 times

[red thistle]

oh an then

[fathom flicker]

evaluate the derivatives and original solution at 0 (the initial values provided are all at x=0)

[red thistle]

do the, mk

[fathom flicker]

and then setup a system of equations

[red thistle]

i will try that rq

[red thistle]

oh crap thats upside down

[fathom flicker]

, rotate

[fathom flicker]

, rotate

[red thistle]

uhm

anyway

i need to recheck

[fathom flicker]

so uh, I personally can't really check that work for you without doing it all over again myself lol. I can tell you that if you used the method I detailed above that you will be able to solve the constants correctly

[red thistle]

i got the coeffs wrong, i need to go through again

[fathom flicker]

ah dang

btw

I am trying to figure out why wolfram is getting a positive exponent

so hopefully I'll get back to you with some news on that

[red thistle]

mk

unless i...

derivative of c1xe^(4x) is 4c1x^(4x)+c1e^(4x)

right?

unless i forgot the 4

[fathom flicker]

@red thistle

I figured out why I (or we?) had the signs swapped

the solution is e^(-at) for when the repeated roots are (r+a)^2

I assumed that it was for (r-a)^2

which is not the case

so that is why the signs for wolfram got swapped

the correct answer has positive exponents like wolfram says

[red thistle]

Alright, thats good

I went afk for a bit while i was trying to solve it

so im almost done with that

[fathom flicker]

nice

[red thistle]

epic

imma need to commit that u(t) method to memory

and if i cant

just hope that i dont have to deal with higher order differentials like this on an exam

Thank you so much for your help, I was really bamboozled with this one

[fathom flicker]

yeah no problem

[red thistle]

its time for me to get to bed as its 1:30 in the morn, have a good one

.close

[obtuse pebble]

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[obtuse pebble]

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[peak hamlet]

I don’t know how to do any of the question 4 questions

Like for example Q4 a.)

[tired sierra]

do you know how to derive

[untold flax]

differentiate using the chain rule

[peak hamlet]

Ok

[fluid nymph]

yes

[peak hamlet]

-(1-x^2)^-1/2

Sorry it should be -x in the front

I just realised

[untold flax]

ye

other than that it's correct though

[peak hamlet]

So what shall I do now?

[untold flax]

now plug in x= 1/2

for the gradient

and you're done

[royal basin]

bad notation

[untold flax]

this is true

[royal basin]

you should not be conflating a function with its derivative

third line should begin with y' = or dy/dx = but not just = (which implies y =)

[peak hamlet]

Ohh yeah I was writing like that to be quicker

[royal basin]

do not

your fingers will not fall off from one extra second spent to write the symbols y and prime

[peak hamlet]

True, I’ll use full notation in the future

The answer is 1/2(1-1/4)^-1/2

I think

[tired sierra]

simplify it

[peak hamlet]

I think I did something wrong

(-3/8)^-1/2

[tired sierra]

no thats ok?

[peak hamlet]

Wait what

But how do I got further

[tired sierra]

wait -3/8

go through how you got -3/8

[tired sierra]

you can't do that

[peak hamlet]

I did though

Why not?

[tired sierra]

because its still within the bracket

ok look at it like this

the -1/2 is a sqrt so

$-1/2-sqrt(3/4)$

ok that does nothing

uh hol horse

it would technically look like that

can you multiply -1/2 with 3/4?

[peak hamlet]

I think so

[tired sierra]

no you can't because its in a root

[peak hamlet]

Oh yeah

You’re right

[tired sierra]

so redo that

[peak hamlet]

So that would be the answer

[tired sierra]

... has your teacher not taught you this?

[peak hamlet]

Wait no that is not the answer

[peak hamlet]

If they did I was not there

[tired sierra]

ok so whatever x is

that is the gradient of the tangent

[peak hamlet]

Ok

But from where we are right now how do we get to finding its value?

[tired sierra]

wait did you sub in -1/2 or 1/2

[peak hamlet]

1/2

[tired sierra]

ah kk

so most of the time for me it'd give me a whole number or like a fraction

but do they want it in a decimal or exact form?

because you can kinda leave it like that if you want it in exact form or turn it into a decimal

[peak hamlet]

The answer is a fraction

-(1/squareroot 3)

According to my textbook

[tired sierra]

ok give me a minute

ok I got it

so our answer is right

now we just simplify

-1/2 / sqrt3/4

ok?

then we reciprocal sqrt 3/4 so we can get

sqrt 4/3

[peak hamlet]

Ohhh

[tired sierra]

since sqrt 4 is 2

= 2/sqrt3

so -1/2 * 2/sqrt3

and I think you can do the rest

[peak hamlet]

Yes

Thank you very much I just got the answer

.close

[obtuse pebble]

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[obtuse pebble]

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[raven portal]

Can someone help me with this

[knotty crow]

it's not clear

[raven portal]

what do you mean

[knotty crow]

is base 1 + sqrt(2)?

[raven portal]

no

[knotty crow]

so it's denom?

[raven portal]

no the (1+2) is the base

[knotty crow]

so it's just 3, right?

[raven portal]

no

(1+root of 2)

[knotty crow]

when I asked you said it wasn't the base

but nvm

okay

[raven portal]

oh my bad

[knotty crow]

what is the number we take log of?

[raven portal]

ohhh i forgot to change it

3+2sqrt(2)

[knotty crow]

and some power there?

[raven portal]

3

[knotty crow]

okay, now we can do it

[warm shale]

Modus

[raven portal]

my bad i forgot to translate the numbers

[knotty crow]

firstly I'd expand (3 + 2sqrt(2)) ^3

o no

notice that (1+sqrt(2))^2 = 3 + 2sqrt(2)

so it becomes

[warm shale]

Modus

[knotty crow]

and I guess you can proceed from here

[raven portal]

yeah the answer is 6

but i kinda didnt undestand

[warm shale]

Modus

[raven portal]

why 1+sqrt(2)^2 tho

sorry my math is bad

[knotty crow]

use formula (a+b)^2 = a^2 + 2ab + b^2

to expand that

[raven portal]

oh

so at first we xpand the (3+2sqrt(2))

[knotty crow]

no, we do (1+sqrt(2))^2

(we want to show it is equal to 3 + 2sqrt(2))

[raven portal]

ok

is there any other way to solve this question?

i think i would understand better if u show it step by step if you dont mind

can i ask another question?

[knotty crow]

ye

[raven portal]

What's this

[knotty crow]

find 'a' firstly, then subtitute it into 4a+1

[raven portal]

2^3/2 ithink

oh

[knotty crow]

that's 2sqrt(2)

yeah

[raven portal]

yeah

[warm shale]

Modus

[raven portal]

Yeag

[knotty crow]

now notice that 4^a = 2^(2a) and compare exponents or apply logs

[raven portal]

So its 4^2^3/2 + 1 right?

Oh my bad

So a is 3/4

[knotty crow]

yea

[raven portal]

So 4×3/4+1 right?

And that would be 4

[knotty crow]

right

[raven portal]

And i forgot to add at the end of the questions it says with the base of 4

Soo 4^4?

[knotty crow]

rather 4^1 = 4

for 4a + 1 obv

[raven portal]

Answer is 1 nice

But i still don't understand the first question

why we say they are equal

[raven portal]

can you explain it again

What about this

Sqrt(8) and 4 are base

I thought the answer was 7/12 but it wasn't

Aren't we supposed to to make the power of. Base upside down

<@&286206848099549185> sorry

i think i should make a new one

.close

[obtuse pebble]

Channel closed

Closed by @raven portal

Use .reopen if this was a mistake.

[nocturne dove]

.reopen

[obtuse pebble]

✅

[raven portal]

.reopen

huh?

did i do something wrong?

[high lily]

show work

[raven portal]

what?

[high lily]

I thought the answer was 7/12 but it wasn't
show all your work leading to that value

[raven portal]

ok let me write them in english

[raven portal]

Here

[grizzled pagoda]

addition mistake... last step

1/4 + 2/3

will be 11/12

[raven portal]

ohhhh

im so dumb

why did i do this

ok

i have two other questions

[raven portal]

first one is this ididnr quite understand it

And this

[high lily]

did you copy down the question correctly

[raven portal]

yeah

[grizzled pagoda]

From your second eqn (with log)
Use the sum rule, find relation between x and y. ||you get x=2y||
Then substitute back in the first eqn, take 2^y as another variable and solve for cubic... but uh i think it wud end up in non real roots...

[raven portal]

the answer is 4/5

[high lily]

,w 2^(4/5) +8^(2/5)

[warm shale]

Results provided by WolframAlpha

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

Requested by ramonov#0921

[high lily]

doesn't sound right

[raven portal]

hmmm ny math teacher kinda speed runed this part so

idk how to solve it

but he said the answer is 4/5

[sweet edge]

teachers can make mistakes too

[raven portal]

what about this one

[raven portal]

maybe

iknow this is basic log im just dumb

what hould i do

[obtuse pebble]

@raven portal Has your question been resolved?

[raven portal]

yeah i can make another one

yeah

[obtuse pebble]

Channel closed

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[obtuse pebble]

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• Show your work, and if possible, explain where you are stuck.
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[edgy acorn]

hello, how do you solve this integral without using the reduction formula? instead i'm supposed to reduce the power by using integration by parts.

[median dome]

what do you mean by the reduction formula?

[edgy acorn]

so you're not allowed to memorize that formula, then plug in, instead you need to get to the reduction formula solution by integrating by parts, so you kind of are still using it.

[median dome]

how would you do it by integration by parts?

[edgy acorn]

if you were to write a proof on the reduction formula, it would involve integration by parts. But I don't know how to do it

[median dome]

have you learned this formula in your class/course?

[edgy acorn]

no

[median dome]

well

you should use trig sub for this anyway

but idk about the reduction formula i have never come across it

[edgy acorn]

he won't let us use trig sub either

it's np

.close

[obtuse pebble]

Channel closed

Closed by @edgy acorn

Use .reopen if this was a mistake.

[median dome]

what

why

[edgy acorn]

bad luck on the prof I guess

[obtuse pebble]

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[floral canopy]

Yo I have a question.
Which set is [2, 3, 4) ∩ ℤ

[floral canopy]

like what does this [) mean

[kind hawk]

half open interval

no wait

that would be if it didnt include the 3

[2,4)

uhm

not common notation

[floral canopy]

what if the 4 wasnt there

was it just [2]

[kind hawk]

that would just be {2}

[floral canopy]

but why

oh I get it

tysm

.close

[obtuse pebble]

Channel closed

Closed by @floral canopy

Use .reopen if this was a mistake.

[kind hawk]

you should look up where that notation is defined in your book

it could also be that [2] = {1,2}

[obtuse pebble]

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[raven portal]

hi

[royal shard]

hello^^

[royal shard]

let me decifer it haha

$log_{y}x=1/\mu$

[warm shale]

~Martin

[raven portal]

where did that p came frpm

[royal shard]

looks like you didnt write mu, but i dont know how to write that weird letter haha

which p?

[raven portal]

p idk how write it

[heavy depot]

$\mu$

[raven portal]

3.14

[royal shard]

ah

[warm shale]

CrEpasPmkinPie

[heavy depot]

that?

[raven portal]

thre isnt any p

its not p

[royal shard]

i dont know how to write this in latex so i used mu

[raven portal]

its 3

[heavy depot]

pi

[raven portal]

3

i forgot to change to 3