help-26

[vernal matrix]

In this case, we want (1, 1) in "new basis", but it's in standard

We have the image of (1, -1), the first "new basis" vector, given to us, but in standard basis rather than in "new basis"

But to make the matrix of the linear map, we need the coordinates of the image vector with respect to the new basis

So change the image of (1, -1), which is (1, 1), out of standard basis and into new basis

[agile tree]

this part?

[vernal matrix]

Yep, that bit

[agile tree]

okk!

i think i get it, i will try a few more problems and hopefully continue tomorrow :D

thank you for such elaborate and kind help i really really appreciate it

im #1 chartbit fan

have a great day ^^

[vernal matrix]

Awwww, you're so sweet hope you have a wonderful day too, shall catch ya around

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[keen harness]

Hello this might be a simple question but uh

In no.5 is it EWO or OWE?

[split cloak]

Ig it doesn't really matter

[keen harness]

I think there's only one answer but thanks still!

[ruby tree]

EWO if the vertices are ordered, but technically both refer to the same triangle, so... depends on your teacher

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[shut obsidian]

Shouldn't we get e.g. y(x) = x as solution too?

[odd pagoda]

we do because lambda=0 is a double root

[weary lodge]

can we please translate?

nvm you guys handle this

[shut obsidian]

How does that give us y(x) = x

[neon iron]

Okay so the thing is the solution set e^{0x}, e^{0x} and e^{3x} are linearly dependent.

To solve this issue, consider the particular solution e^{0x} = 1. To make the solution set linearly independent, multiply it by x and assume another particular solution of the form Ax, where A is some constant. This makes the solution set: 1, Ax and e^{3x} linearly independent, and you can find A now.

Yeah sorry

[shut obsidian]

Alright

Why are we allowed to just do that

[neon iron]

The goal is to make the solution set linearly independent, because a linearly dependent solution set leaves out some solutions to the DE

In this case, functions of the form y(x) = cx are left out

[shut obsidian]

We could've aswell multiplied by x^3, no?

[neon iron]

Sure, but you would have to verify if it satisfies the DE by substitution

[shut obsidian]

Well don't we have to do that with x too

[neon iron]

You have to, when you do it, you find a unique value of A or if all values of A satisfy the DE, then such a particular solution exists

If you aren't able to find a value of A (for example), then the assumed particular does not exist and you would have to think of another

Slight guesswork is involved here

[shut obsidian]

you mean for A in Ax, right?

[neon iron]

Yes

[shut obsidian]

Alright

[neon iron]

Another way you can deal with this problem is directly tackle the DE

[shut obsidian]

Hm, isn't this fulfilled for all A

y'''(x) - 3y''(x) = 0 holds for all y(x) = Ax

That's just 0 - 0 = 0

Or do you mean plug in e^(3x) + Ax + 1

[neon iron]

Apologies, edited my message

[shut obsidian]

Why is that an issue

[neon iron]

^

[shut obsidian]

So now we can just arbitrarily toss out one of those linearly dependant pieces and throw in anything we want (as long as it works out after plugging back)

[neon iron]

Kind of

[neon iron]

Do you remember that system of ODEs message in #odes-and-pdes ?

Of turning a higher order linear DE to a system of ODEs?

Turns out that this is the reason why some solutions are left out, because there are always n linearly independent solutions to a nth order linear ODE

[neon iron]

So you have y"' - 3y" = 0

=> (y' - 3y)" = 0
=> y' - 3y = a + bx
=> ye^{-3x} = -3ae^{-3x} - 3bxe^{-3x} + C
=> y = A + Bx + Ce^{3x}

I'll brb in 10 min, dinner

Back

[topaz sinew]

@shut obsidian Has your question been resolved?

[shut obsidian]

Ah

[shut obsidian]

Ok, so after we have these three solutions, we need to check if they are linearly independant

My instructor usually does this by picking some random x

And then getting multiple equations

Though those were two variable cases

I guess we need another equation

[neon iron]

Yeah that's the general way of checking linear independency/dependency.
If you find that LHS is 0 iff every alpha_n = 0, then the particular solutions are linearly independent.
This is just like checking linear independency of vectors

There's also another way to check this, by computing something called the Wronskian

[neon iron]

Here, you form three equations (since three variables (a1, a2,a3)) and solve for them

[shut obsidian]

Ah, alright

[shut obsidian]

How do you go from line 2 to 3

[neon iron]

I again apologise if I have caused any inconvenience

[neon iron]

Multiplied by integrating factor then integrated

[shut obsidian]

Absolutely not

[shut obsidian]

From [y' - 3y = a + bx] to [ye^{-3x} = -3ae^{-3x} - 3bxe^{-3x} + C]

What's the integrating factor

[neon iron]

y'e^{-3x} -3ye^{-3x} = (a+bx)e^{-3x}

=> (ye^{-3x})' = (a+bx)e^{-3x}

=> ye^{-3x} = int((a+bx)e^{-3x}dx)

[neon iron]

Uhh have you done exact and non exact des yet?

[shut obsidian]

Oh, nope, we haven't

Yeah, we haven't had this method yet

[neon iron]

Oh, it will pop up there

[shut obsidian]

But seems interesting

[neon iron]

It's a factor that turns a non-exact de into an exact de

[shut obsidian]

Do we need to do this if the exercise asks to "determine the solutions"?

[neon iron]

If you want, you can solve this as usual, obtain general homogeneous solution, then a particular solution, then taking their linear combination as the general solution

[shut obsidian]

I mean testing for linear independancy at the very end

[neon iron]

Well whenever you get particular solutions to the DE, you have to check their linear independency yes or else you would not obtain all the solutions

[shut obsidian]

Ah, true, same reason as above, where we missed out on x

[neon iron]

It's not that hard to make out, if you just spot two solutions being multiples of each other, the solution set is linearly dependent

[shut obsidian]

Yeah but if it's linearly independant

Then you need to make a system of equations

[neon iron]

Another trick is computing the ratios of particular solutions and see if you get a constant

[shut obsidian]

Alright, thank you!

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[shut obsidian]

$y''''(x) - 4y'''(x) + 3y''(x) + 4y'(x) - 4y(x) = 0$. \ With $y(x) = e^{\lambda x}$ we get [\lambda^4e^{\lambda x} - 4\lambda^3e^{\lambda x} + 3\lambda^2 e^{\lambda x} + 4 \lambda e^{\lambda x} - 4 e^{\lambda x} = 0]

So in this case, it's a bit larger

We should guess, right?

After factoring out e^(lambda x)

Then polynomial long division

Since that's a quartic

[neon iron]

Yeah you can try that, if it doesn't work, try and check if it factorises into multiplication of two quadratics

Guessing works here

[shut obsidian]

Alright

Thanks

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[shut obsidian]

I got 1, -1 and 2 twice

Again a solution that occurs twice...

So we need to replace one $e^{2x}$

[neon iron]

Yes check if Axe^{2x} satisfies the DE

[shut obsidian]

How did you come up with that

Why not e^(3x)

[neon iron]

This is like the usual way of doing it

Because multiplying by x gives xe^{kx} that has nice derivatives

If you go like way too general and assume a particular solution of the form u(x)e^{kx}, then you again come back to the same DE

(k is a root of the CE)

[pastel salmon]

Homogeneous linear equations with constant coefficients always have fundamental systems based on analogous solutions, so there is no need to waste time to always check independence

[shut obsidian]

Alright, thanks

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[shut obsidian]

Do you have any online reference to this maybe

.reopen

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✅

[neon iron]

https://youtu.be/xAZ-hdBJ8u8?feature=shared

[shut obsidian]

Do you also know of a website rather than a video?

[neon iron]

No sorry, but let me search if I find any

[shut obsidian]

Oh, I think this is basically this https://en.m.wikipedia.org/wiki/Method_of_undetermined_coefficients

in example 2

[neon iron]

Yeah examples section

I am not sure if you have covered variation of parameters yet but that is also another useful way of solving such DEs (to be specific, nth order linear non-homogeneous odes)

[shut obsidian]

We haven't

[shut obsidian]

Hm, but isn't this kinda different

They didn't obtain solutions and then have to throw one of them out and replace it

[edgy frigate]

I can suggest a book which will be far greater than any YouTube video or online resource on the matter if you want it.

[shut obsidian]

Sure, though I'm taking a course on it rn anyways, but go ahead

[neon iron]

Example 2

[edgy frigate]

It will layout precisely when to multiply by t or t^2, etc.

[shut obsidian]

Yeah

[shut obsidian]

Which one is it?

[neon iron]

I mean the example is different, but the core idea is the same, multiplying by x or its higher powers to make the solutions linearly independent

[edgy frigate]

The book is Ordinary Differential Equations by Morris Tenenbaum and Harry Pollard, you can probably view the book on Internet Archive. (Starts at Lesson 21A, page 221.)

[shut obsidian]

Thanks

[shut obsidian]

Is there any motivation for multiplying by x or is it just something that often works

[edgy frigate]

Wait, are you wondering why sometimes if you have one solution that you can get another by multiply by t for the complementary solution?

[shut obsidian]

No, I obtained lambda = 1, -1 and lambda = 2 (twice) from my characteristic polynomial

Using y = e^(lambda x)

So I have to replace one of the e^(2x) solutions or it won't be linearly independant

And hellothere suggested x * e^(2x)

[edgy frigate]

Ok. I directed you to somewhere unrelated here, sorry.

[shut obsidian]

Alright, np

[edgy frigate]

I think you might be best served by looking at something called the method of Reduction of order. In that same book it is shown (without reference to the name of the method) on page 214.

It all comes back to this for why we require n linearly independent solutions. It's essentially because (19.32) will be a true general solution, i.e., it will contain every possible solution.

We can find the n linearly independent solutions by any means necessary.
Normally if you found one (by the usual methods of y_(t) = e^(λt)), which was a guess all along anyway so you might be tempted to try to find another by forming y_2(t) = u(t) y_1(t).
In fact, it can be shown that this always yields a second solution (See Lesson 23B).

[shut obsidian]

Ok, so we need to plug in $axe^{2x}$ into $y''''(x) - 4y'''(x) + 3y''(x) + 4y'(x) - 4y(x) = 0$

[edgy frigate]

I'd say for a thorough understand, try substituting

$y_2(t) = u(t) e^{3t}$.

[thorny flame]

stabulo

[edgy frigate]

The method of just multiplying by t probably comes from the pattern recognition that if the root is repeated it just always results in u(t) = t.

I do wonder if you choose

$y_{2}(t) = u(t) y(t)$, where this $y(t)$ is any of the non repeated roots if you just start generating all the others and eventually have to keep doing it till you get all the others.

[thorny flame]

stabulo

[edgy frigate]

The problem will quickly arise that finding y_2'. y_2'', y_3''', y_2'''' generally use the product rule a lot will quickly become very long to do by hand.

[shut obsidian]

Ok I did

We get 0

0 = 0

Great

[shut obsidian]

Oh, ok

[edgy frigate]

Resources will just probably show it true for the second order case and get you to notice the pattern.

[shut obsidian]

What do you mean by this, do we never need to check independence?

[pastel salmon]

Becasue every fundamental system of solutions of such euqations like yours are based on exp(ax)sin(bx), exp(ax)cos(bx), you never find other solutions if it comes to homogenous linar equations with constant coeff

fundamental syste, means its wronskian is not equal to zero

[pastel salmon]

hence if you find roots of the characteristic polynomila, like yours, 1, 1, -1, 2, then

fundamental system is:

{ exp(x), xexp(x), exp(-x), exp(2x) }

hence

the general solution is the lieanr combination of thsoe fours solutions i write

and that is end !

it lasts 3 minutes )

or less

and the appropriate theorem guaranttess

those solutisn are independent,

hence i said, do noto waste to much times on it

it is not worth

such equaitns belive me or not, can be solved in mind

[shut obsidian]

oh, my instructor in her example checked if theyre independant

[pastel salmon]

no, no need unleess there;'re equations with non - constant equations

but so far, you re showing only with contants coeff

[shut obsidian]

Thanks

[pastel salmon]

yw

[shut obsidian]

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[eager willow]

5 only answer

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

[old ether]

A linear equation can be constructed with the same number of data points as dimensions. This one is two dimensional, so you need two data points. Could you give some? Preferably one with x=0.

[eager willow]

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[glacial acorn]

you can solve it with an integral

[keen venture]

I think they're trying to go for Riemann sums, but that's not what the sum is

[eternal mauve]

can i try and do it with the trapezium rule?

[glacial acorn]

That looks like a Darboux sum to me

[keen venture]

They probably do intend an integral solution

As written, the sum is just:
0 + 0 + 0 + 0 + 0 + 0

[eternal mauve]

you guys cook but pls help me out

🙏🏾

[glacial acorn]

The simplest solution is to write an integral and solve it

Do you know how to do that?

[eternal mauve]

integral for root x?

and define limits using the width?

[glacial acorn]

yes

[eternal mauve]

i mean i can write the integral of root x

but without a numerical value for the limits

how would i be able to

oh wait-

do i do smthg with 4 and 9

[glacial acorn]

Whats the integral of root x?

[eternal mauve]

2x^3/2 divided by 3

plus c

[glacial acorn]

thats correct

Now mark that as F(x)

And calculate F(9) - F(4)

[eternal mauve]

38/3

[dusky socket]

why divided by 3

[eternal mauve]

wdym

[dusky socket]

ohhh nvm

mb

[eternal mauve]

x^n+1 divided by n + 1

[dusky socket]

sorry

[eternal mauve]

nah its cool lol

ntd

[glacial acorn]

And that is your solution

[eternal mauve]

wha-

i DID NOT KNOW it was this ez

im usually good at calc but the weird symbol and sigma threw me off

[glacial acorn]

Thats a sum sign

its used to note when you add multiple things at once

You can look up Riemann or Darboux integrals.

[glacial acorn]

these are two ways of formally defining integrals

You basically approximate a function with narrow vertical blocks and add up their areas

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[sterile cedar]

hello

[sterile cedar]

Show that the points form the vertices of the polygon.
isosceles triangle:
(2, −1), (3, 3), (−1, 4)
distance between (2, −1) and (3, 3) d1 =

distance between (3, 3) and (−1, 4) d2 =

distance between (2, −1) and (−1, 4) d3 =

hi @neon iron

it keeps saying my answer is wrong....

4.12
4.12
5.83

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@sterile cedar Has your question been resolved?

[neon iron]

oops

let me see

[sterile cedar]

u good bro

[neon iron]

Did you do the Pythagorean theorem for each one?

[sterile cedar]

yep

i did it myself and then i ran it through chat gpt

same answer

[neon iron]

Can i see your work

[sterile cedar]

i used geogebra

so its gone now...

[neon iron]

These answers should be right

[sterile cedar]

cengage saying theyr wrong

[neon iron]

Weird

[sterile cedar]

yeahhhh i cba man

[neon iron]

yea idk then

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[fickle stirrup]

Determine the derivative of the following function:

[neon iron]

wtf is sen

[fickle stirrup]

Oh sorry the problem is in spanish

So sen is seno

But it is sin

[neon iron]

oh ok

lmao

gotcha

[fickle stirrup]

I know that sen2x/cos2x/1/csc2x

Which will end in 1/cos2x

If I am not wrong

[neon iron]

ye

[fickle stirrup]

But that is not the answer lol

[neon iron]

ye no

u didn't take the actual derivative yet

what is 1/cos(2x)

[fickle stirrup]

I know

[neon iron]

in other terms

[fickle stirrup]

The derivative is

-2sen2x/(cos2x)^2

But that is not the answer

[neon iron]

the derivative isn't the answer?

[fickle stirrup]

I mean

You have to simplify

[neon iron]

your question was to find the derivative

[fickle stirrup]

The answer is 2tan(2x)sec(2x)

But

My question is

How to get to that lol

That is the problem for me

I have been thinking how to get to that for 15 minutes

But I cant find a way lol

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[waxen niche]

just 27 for now

like x = 3cos pi/3 t

y = 2sinpi/3 t

i just plugged in 1 for t

and i got that the distance (pythag) is sqrt 21/2

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[waxen niche]

<@&286206848099549185>

[hidden zinc]

bro

for q 27

u have x = 3 cos(pi/3)t

so cos(pi/3)t = x/3

2 sin(pi/3)t = y

so sin(pi/3)t = y/2

now sin^2theta + cos^2 theta = 1

so x^2/9 + y^/4 = 1

now take lcm

4x^2 + 9y^2 = 36

this is an ellipse

[waxen niche]

im wondering why this doesn't work

i hav ethe solution

[hidden zinc]

ur supposed to find the reln b/w x and y

u dont plug values for t

it should be independent of t

do you know

how parameters work

[waxen niche]

no?

[hidden zinc]

exactly

parameters

are those points where

we take another variable and express x and y

to get the locus of the points

we must eliminate that variable

[waxen niche]

yeah but i don't get hwy testing values

wouldnt show u this relationship

cuz wouldnt it be constant

regardless

[hidden zinc]

no

t is not a constant

it can be anything

[waxen niche]

nah i mean

[hidden zinc]

it is just another variable

[waxen niche]

the relatinoship between x and y

like

[hidden zinc]

yeah?

[waxen niche]

lemme try smt

yeah like

ok

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[vapid notch]

unsure of 11

how would i calc shaded area?

ik the sector area is 39.26

do i need to solve the triangle to find triangle area? and sub that from sector area?

[whole geode]

Yes, it might also help to recognize that triangle as exactly one eighth of a regular octagon.

[vapid notch]

yes i did that

[whole geode]

But that depends on the formulas you have handy

[vapid notch]

law of sines

[whole geode]

It is also twice of a right triangle with one angle 22.5 degrees.

[vapid notch]

ill just use law of sines

[whole geode]

Fair enough

[vapid notch]

then find H with pythag right

make an apothem

and calc with b2 = c2 - a2

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[dreamy wolf]

need help

[dreamy wolf]

I have no idea how to solve some of these

thats what I did for the first two bits, pls see if thats right ill send the 3rd and 4rth but I cant figure out what im doing wrong or ho to do the 5 on wards.

<@&286206848099549185>

someone ?

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[void crow]

OH

I REMEMBER THIS

THE GRADE 10 QUESTION

AHHH

nostalgia

[thick inlet]

ye

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[grave bough]

why are we taking in terms of y

why not just x(f(X)dx

shouldnt shell height be sqrt(x)...

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@grave bough Has your question been resolved?

[grave bough]

oh boy im so fucked

even if we tookt he height as the red bar, it would be x = y^2

why 1-y^2....

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@grave bough Has your question been resolved?

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@grave bough Has your question been resolved?

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@grave bough Has your question been resolved?

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[fair thorn]

11:2::22:4

use that

[jolly patio]

let me try!

so pennies to nickles is 11:2

and nickles to dime is 4:3

and we need dimes to pennies

22:4 : 4:3 which is 22:3??

22/4 *4/3, 4 crosses out right so its just 22/3?? but thats not an answer

22 dimes and 3 pennies

which is 223 pennies

@fair thorn

[fair thorn]

yes

acutally no, 3:22

[jolly patio]

why 3:22

[fair thorn]

bc 22:3 is pennies to dimes

3:22 is dimes to pennies

[jolly patio]

oh wait

oh i see

so i know f is correct

bc it is 52 in total as well

isnt that the only answeR?

[fair thorn]

f is correct

but d is also correct

[jolly patio]

how do

[fair thorn]

same with e

3:22 again

[jolly patio]

d is 30 dimes and 220 pennies

which is 520 pennies

?

[fair thorn]

that does not matter

30 dimes : 220 pennies :: 3 dimes : 22 pennies

[jolly patio]

ah i see

so only d and f?

and e i assume

bc you can divide 24 by 3

and 176 by 22?

@fair thorn

[fair thorn]

yes

please dont ping me

[jolly patio]

ok

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[neon iron]

try n write out all the information you have first

[jolly patio]

<BCG is 10 degrees less than the measure of <CGD

[neon iron]

yeah

[jolly patio]

<DGE is 25 degrees more than the measure of <EGF

[neon iron]

represent that mathematically

what do u get

[jolly patio]

im not sure what to do next

this is DGE

[neon iron]

answer this

i want to see equations representing the sentences you mentioned

[jolly patio]

hm

maube

<CGD = <BGC-10

??

[neon iron]

i mean

if CGD = 50 and BCG = 40 does ur equation make sense

[jolly patio]

yes

[neon iron]

50 = 30?

[jolly patio]

bi

no**

that does not make sense

[neon iron]

yeah, so...

how do you correctly set it up

[jolly patio]

im not sure tbh

i havent taken geometry in 8 years

[neon iron]

50 - ??? = 40 what is ???

[tulip parrot]

<BCG=<CGD-10
<DGE=<EGF+25
<FGA?

[jolly patio]

<BGC=<CGD-10

so

[neon iron]

please let them figure out how to set it up at least

[jolly patio]

<DGE= <EFG+25

[tulip parrot]

ook i will go..

[neon iron]

but okay cool yes

[jolly patio]

ok

what should i do after

i haev to find FGA

i have two equations already

[neon iron]

those are your equations. Can you tell me what EGF is from the graph?

[jolly patio]

it seems like

its the same as

BGD

[neon iron]

no but like there is an x value associated with it

[jolly patio]

2X

[neon iron]

yes

[neon iron]

plug it in here what is DGE

[jolly patio]

why do i plug it into DGE?

[neon iron]

im saying to substitute in EFG = 2x

you know EFG is 2x from the graph

[jolly patio]

okay so

DGE = 2X+25

[neon iron]

yeah

keep that in mind

[jolly patio]

OK

[neon iron]

same thing for

what is cgd

[jolly patio]

X

[neon iron]

so sub it in yeah

[jolly patio]

bgc =x-10

[neon iron]

amazing

so now you have all the angles in term of x

so uh

sum of all angles = 360

find x

[jolly patio]

x-10+2x+25=360

[neon iron]

no

no

[jolly patio]

oh?

[neon iron]

ALL angles

[jolly patio]

hm

[neon iron]

so 2x + 3x + ...

[jolly patio]

6x+30

?

[neon iron]

yeah but also include those

[jolly patio]

can we consider BGC as x as well?

or no

6x+30+x-10+2x+35=360

[neon iron]

u already have bgc

[jolly patio]

which is 9x+55=360

[neon iron]

why is it 35

its not 35 tho

[jolly patio]

OH WAIT

25

6x+30+x-10+2x+25=360

SO

9x +45=360

which is 9x =315

which is 35

[neon iron]

yes

now

it wants fga which is 3x

so

whats 3 times 35

[jolly patio]

105

[neon iron]

yeah

[jolly patio]

ok so im going to be honest

im a little confus

like

I understand we found the equation of BGC and the equation of DGE

[neon iron]

yeah

[jolly patio]

oh wait i see

we found the missing ones

and now we add add the values

and equal to 360

i understand that

we got 35 cool

[neon iron]

ok

what's the issue

[jolly patio]

and then we did 3x bc we needed to find agf?

[neon iron]

yeah i mean thats what they say fga is in the graph

[jolly patio]

ig my tjing is why did we add all the values and equal to 360

[neon iron]

bcuz

[jolly patio]

and THEN find fga

[neon iron]

its a circle

like

a circle is 360 degrees

[jolly patio]

i see

[neon iron]

i mean it's just a way for us to recover x to get fga

[jolly patio]

ah isee

is it because we wanted to find x

and this is the only way hiw

????

and then we insert it into 3x

[neon iron]

probably not the only way how but probably the most straightforward one

[jolly patio]

ok i see

ty

[neon iron]

yeah

[jolly patio]

can you help me with another porblem

[neon iron]

sure

[jolly patio]

i can try first

[neon iron]

btw what are u even preparing for you do a lot of those

okay

[jolly patio]

its an exam called the GRE

graduate record examination

lol

[neon iron]

oh okay lmao

[jolly patio]

i take it in 10 days so

im trying to master is LOL

its still confusing LMAO

anyways

employee x is paid $19.50

and Employee Y is paid $18 an hour

for the first 40 hours

and then is paid 1.5 times the hourly rate for every additional hour worked

and then both employee worked the same number of hours and paid the same amount and we are trying to find how many hours they worked

ok so

i guess make two seperate equations??

[neon iron]

ok ig lets start a step at a time

let x denote the time in hours

[jolly patio]

yes

so employee x =19.50x

[neon iron]

how do you represent "employee X is paid 19.50 an hour"

[neon iron]

yeah

for employee Y we have to set up a piecewise function

[jolly patio]

piecewise function?

[neon iron]

yeah

[jolly patio]

please provide an example of it

im not familiar with the term

[neon iron]

,, \begin{dcases*}
13x &if $x < 5$ \ 12-x &if $x\ge 5$
\end{dcases*}

[jolly patio]

ok i see

[neon iron]

so

you have two cases for employee Y

[jolly patio]

18x is x>40

[neon iron]

for two separate intervals of time

you mean < right?

[jolly patio]

oh yes

wrong sign haha

[neon iron]

yeah

so like

[jolly patio]

and then

[neon iron]

1.5 times that

[jolly patio]

27x if x >40

[neon iron]

yeah

exactly

so great we set our stuff up

the question now asks

[jolly patio]

ok perfect

how many hours did each employee work that week

thats what it asked^^

[neon iron]

yeah

[jolly patio]

ok what should i do next??

[neon iron]

so we need to find the hour x for which both are equal

[jolly patio]

so employee x to employee y?

[neon iron]

wait actually

Sorry i think i misguided you a bit. I guess i didn't fully read the rest of the question until now

we will rewind a bit

[jolly patio]

okay

[neon iron]

so

let x be the amount of time X worked and y the amount of time Y worked

[jolly patio]

ok

[neon iron]

from the question we know 19.50x

[jolly patio]

yes

[neon iron]

for employee Y, the "is paid $18 an hour for thr first 40 hours she works" is basically a constant. Because she MUST work that amount

so in reality thats just $18\240$

[jolly patio]

how do you know that?

that she HAS TO WORK

[neon iron]

because the question including an "additional" time is implicitly saying that those first 40 hours are going to be happening regardless

[jolly patio]

hm ok

so

720 (1.5-40)

[neon iron]

yeah uh so we are not done yet

[jolly patio]

oh

mb

[neon iron]

the second phase of the additional thingy

like you said its 27 the rate

[jolly patio]

yes

[neon iron]

but our variable is going to be y - 40

because like

it only starts after 40 hours right?

[jolly patio]

oh right

720 (27-40)

[neon iron]

no

[jolly patio]

oh

[neon iron]

720 is like

separate

you dont multiply it

[jolly patio]

hm ok

[neon iron]

its just added on top of the additional work

[jolly patio]

oh ok

continue, sorry for interrupting

[neon iron]

no its okay so the first 40 hours are 720

and then

you have 27(y-40) for the additional work

so what is ur equation

[jolly patio]

oh i see

so the eqyatuon is 19.50x =720+27(y-40)

[neon iron]

yeahh

exactly :D

so they want you to have x = y

and solge for x

[jolly patio]

so 19.50x=720-27y-21600

[neon iron]

substitute x for y

first

[jolly patio]

which is 19.50x=-20880-27y

x=20880-27y/19.50

[neon iron]

substitute x for y

[jolly patio]

why do we do that?

[neon iron]

the question says they worked the same number of hours

[jolly patio]

i see that

ig im not seeing why we are putting x into y

what is the x equation

and y equation

[neon iron]

x is the amount of time X worked

y is the amount of time Y worked

[jolly patio]

im sorry, im not sure im understanding wym

i get what x and y represent

but using the equation 19.50x=720+27(y-40) i dont know what number to replace into y

we didnt even find an x value to replace into y

[neon iron]

im not asking you to replace it with any number

[jolly patio]

if that makes sense

oh

I apologies

[neon iron]

im asking you to replace it with x

[jolly patio]

OH

[neon iron]

because we know it's equal to x

[jolly patio]

so 19.50x=720+27(x-40)

??

[neon iron]

yeah

now solve

[jolly patio]

but dont they mean different things?

why change it to x

[neon iron]

because

again the question's premise is saying "if they worked the same amount of hours how many hours did they work"

so

[jolly patio]

oh so they are the same

ok

[neon iron]

yeah

[jolly patio]

so -7.5x=720-1080

wait

[neon iron]

no

ok

thats rigjt now

[jolly patio]

48

[neon iron]

yeah

thats it

[jolly patio]

ok can we do another problemm together

[neon iron]

ok

[neon iron]

ok what do u think

[jolly patio]

this one, idk how to approach at all

tbh

are f g and h angles?

tbh i would say there's not enough info bc there's no numbers

and its not a specific triangle like equilatal or right so its kind of difficult to indicate the numebers

[neon iron]

yeah

[jolly patio]

what are m,n,p?

[neon iron]

angles as well

like

[jolly patio]

two angles in each corner?? that makes no sense

[neon iron]

whats the sum of the angles in a triangle?

[jolly patio]

180??

[neon iron]

its just a very concave angle

[neon iron]

yeah

[jolly patio]

ok

[neon iron]

so f + g + h is what

[jolly patio]

f+G+h=180

?

[neon iron]

yeah

[jolly patio]

ok

[neon iron]

so that's the numerator

now for the denominator

[jolly patio]

alright

which is m n and p

[neon iron]

i want u to look at, say, m

[jolly patio]

yes

[neon iron]

so here is the thing, that thing ALMOST looks like a full circle right?

[jolly patio]

yeah

[neon iron]

but its just missing a small degree that f has

[jolly patio]

yeah

[neon iron]

what did we say the degree of a circle is?

[jolly patio]

the full degrees is 360

but since its missing

[neon iron]

yeah

but take away f

[jolly patio]

maybe 270?

[neon iron]

what do u get

[jolly patio]

oh wait

i see

300

[neon iron]

no

[jolly patio]

wait

give me a secon

100

[neon iron]

you are not getting any number u dont know f

[jolly patio]

well if its 180 total

[neon iron]

im just asking you for the expression

[jolly patio]

i did 180/3

=60

and 360/3

=120

[neon iron]

ok but like you're assuming the triangle is equilateral with that

and we dont know that

[jolly patio]

ah i see

[neon iron]

we know nothing about the triangle

[jolly patio]

ok true

so

[neon iron]

anyways answer this

[jolly patio]

m=360-f

[neon iron]

yeah

exactly

do the same for all the others and sum them up

what do you have (give me an expression)

[jolly patio]

so 180/ 360-f+360-g+360-h

[neon iron]

yeah

so whats 360*3

[jolly patio]

ok

1080

[neon iron]

yeah

so what is Ur expression now

[jolly patio]

180/1080-f-g-h

[neon iron]

yeah

factor out - from f, g ,h

what do you get

[jolly patio]

like this?

f+g+h =180/1080?

[neon iron]

how did the entire thing flip lmao

you did like uh

[jolly patio]

oh wait

[neon iron]

some magic there because there is no = at all yet

[jolly patio]

oh alright

1/6 f-g-h

[neon iron]

no

like

factor $-x-y-z$

[neon iron]

how do you do it

[jolly patio]

-(x+y+z)

[neon iron]

right

exactly

so

[neon iron]

do this here with -f-g-h

[jolly patio]

so 1/6 -(f+g+h)

[neon iron]

no

[jolly patio]

oh

[neon iron]

,, \4a{a+c} \5r\ne \41{1+c}

[neon iron]

this is pretty much what you did

[jolly patio]

oj

oh

im not sure what to do tbh

[neon iron]

you have [
\4{180}{1080-(f+g+h)}
]

[neon iron]

right?

[jolly patio]

yes

[neon iron]

ok what did we say f+g+h was

at the beginning

[jolly patio]

180/1080 - 180/f+g+h

[neon iron]

no

you cant do that

you can't rip the denominator like that

,,\4a{b+c} \5r\ne \4ab +\4ac

[neon iron]

but anyways answer this

[jolly patio]

-(f+g+h)

[neon iron]

rewind a bit

right at the beginning

what did we say the sum of the angles of a triangle is

[jolly patio]

f+g+h=1080-m-n-p

?

[neon iron]

.

[jolly patio]

OH

yes

[neon iron]

substitute that in

[jolly patio]

oh 180/1080 - 180

[neon iron]

yes

[jolly patio]

180/900

[neon iron]

also use parentheses

[jolly patio]

1/5

[neon iron]

please

[neon iron]

yes

[jolly patio]

ok ty

i gtg i have class but ill come back in a couple of hours

tysm

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[neon iron]

alright good luck

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[junior linden]

is there Isosceles Trapezoid this formula a+b=2c

[junior linden]

i mean if isoscales trapezoid has this formula a+b=2c

[deft dew]

No

In an isosceles trapezoid a and b have the same lengths

a+b=2c suggests that the sum of the lengths of both sides equals twice the length of one side

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@junior linden Has your question been resolved?

[junior linden]

ok

.close

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[hasty bane]

can someone please help me understand where did the minus sign go?

[sweet shard]

$|c + b_q - b_p| \le |c+b_q + b_p|$ for all positive c

[thorny flame]

riemann

[sweet shard]

Nonnegative c

And the fact they give at the end

[hasty bane]

thanks

.close

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[neon iron]

The answer key says 25/2 is the answer for this but I'm getting 0

[golden mesa]

do you know how to integrate with absolute value

[neon iron]

No

[golden mesa]

if not, i would recommend sketching it out

looks like this

[neon iron]

The way we work solving it was by FTC

How would I solve using FTC

[neon iron]

Like yea I understand area under curve and all

[golden mesa]

the problem is that |2x-5| is different to the left and right of x=5/2, so you can't integrate it in one go
you have to split it into when it's 2x-5, and when it's 5-2x

[topaz sinew]

@neon iron Has your question been resolved?

[neon iron]

?

[viral skiff]

in limits,absolutes with variables have 2 different answers
a little unrelated but assume x -> 5/2+ (approaching from the right)
assume the value is very slightly higher than 5/2
if you put it this way,2x will be higher than 5 so it will leave the absolute at it's exact form

hence |2x-5| = 2x-5

if you approach it from the left side

the inside is a minus

because 2x will be smaller than 5

[neon iron]

Oh right

But like I don't know how to go about the second answer

[viral skiff]

on this particular problem I do not know integrals so I can't help you about it

but I do know limits

[neon iron]

.close

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[hasty bane]

how do I show that $$\text{exp}(x) := \sum_{k=0}^{\infty}\frac{1}{k!}$$ is continuous everywhere?

It's continuity can be reduced to continuity at $0$:
$$
|e^{x}-e^{x_0}| = e^{x_0} |\underbrace{e^{x-x_0}}{e^t \ \text{ as } t \text{ is close to } 0 }-\underbrace{1}{e^0}| < \varepsilon
$$

where $e^x := exp(x)$ for convenience

But idk how to show that $\text{exp}(x)$ is continuous at zero 🤔

[thorny flame]

Sweet Tea 🧋

[final jungle]

It's better to show that the sequence of partial sums uniformly converge to e^x. Continuity follows.

There's a typo in the summation btw

[hasty bane]

where?

[final jungle]

Should be x^n

[hasty bane]

Hmm, I don't remember that theorem... I thought uniform convergence => pointwise convergence, but what does continuity has to do with it?

[final jungle]

Uniform convergence along with each sequential function being continuous implies the convergent function is continuous

[hasty bane]

and either way, this theorem is discussed way before any mention of sequences of functions and their convergence

so I'd like to prove that w/o them

altho thank you for suggesting

[final jungle]

Yeah but that's the easiest way I could think of

[topaz sinew]

@hasty bane Has your question been resolved?

[rigid ivy]

Are you defining e^x as the limit sum?

@hasty bane

Also, what was your other question?

[hasty bane]

what other question? 🙃

[rigid ivy]

Hah it's okay.

[hasty bane]

that's it: continuity of e^x and multinomial expansion

[hasty bane]

yep

[rigid ivy]

Okay. So basically, you want to show that $\sum_{k=0}^{\infty} \frac{x^k}{k!}$ is continuous everywhere. Or at least continuous at $x=0$ for now.

[thorny flame]

SWR

[hasty bane]

correct

I suspect we have to bound the e^x - 1 by using either the expansion (or in our case the definition of e^x) or maybe the limit of e^x-1 over x as x goes to 0

like -1 term subtracts straight away the first term of the

$$
e^x := \text{exp}(x) := \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1+x+x^2/2+\ldots
$$

[thorny flame]

Sweet Tea 🧋

[rigid ivy]

I believe simply showing that the sum converges at every x would help show continuity? I think? If you can show that, you can show two neighborhoods are within some epsilon of each other.

But that's only my first guess and may not be helpful to you

[hasty bane]

hmmm. what if we assume, WLOG, that |x|<1

then e^x-1=x(1+x/2+x^2/3!+...)<=x(1+1/2+1/3!+...)=x(e-1)

[hasty bane]

definitely not

point wise convergence of functions and their continuity separately does not imply the function they converge to is also continuous

[rigid ivy]

Give me time after work to think this through. Not 100% sure I can solve this since I forget all my series analysis, but I'll at least take a good look

But I'll at least try to help

[noble laurel]

You need to show the series converges uniformly to e^(x)

Since all of the functions in the series are continuous

Their uniform limit will also be

[hasty bane]

: (

then (if I get this theorem on the upcoming exam) I'd have to prove this result as well to use it

and the thing is, even tho I'm slightly familiar with the theorem and followed a proof from a textbook (specifically not saying 'proved', because that assumes deeper understanding than I have atm 😅), we didn't cover sequences of functions

so looking for some more elementary methods

[noble laurel]

You haven’t covered sequences of functions ??

Yet your definition of e^(x) is an infinite series expansion

Can you show the original question

[hasty bane]

I've found a proof online, and it started with 'obviously |e^x-1|<2|x| if |x|<1'

but idk how to prove that

[noble laurel]

[hasty bane]

well, we simply defined e^x to be this series

and point wise convergence is elementary (with root test, for example)

so it's well defined

[noble laurel]

Sure

Okay I think I see

All you’re trying to do is prove e^x is continuous?

[hasty bane]

yes 🙂

exp(x) to remove any confusion

because u could have defined e^x as a supremum of something, with rational powers approaching your real number x

[hasty bane]

and specifically at the point x=0, since that would automatically prove its continuous everywhere else

[noble laurel]

I would do a more general epsilon delta proof

It’d be easier I think

Sure you can take c=0

Can you state then what you’re trying to prove in terms of the epsilons and deltas

[hasty bane]

given eps > 0 we can find delta > 0 st whenever |x|<delta it follows that |e^x-1|< eps

*with remark that e^x := exp(x) 🙂

[noble laurel]

Sure and now let c=0