help-0

[fast raven]

triangle area = 1/2 base times width

[vagrant raptor]

we're not at triangle area yet.

use these measurements to figure out the sides of the triangles

[restive pilot]

5cm?

[vagrant raptor]

yes

and then do the same for the length

[restive pilot]

sorry, i dont quite understand

[vagrant raptor]

find the other side

[restive pilot]

5 aswell?

[vagrant raptor]

yes. but we must do the calculation to know that

[restive pilot]

28-18

[vagrant raptor]

perfect

now, what's the area of such a triange?

[restive pilot]

12.5

[vagrant raptor]

perfect

area of all 4 of those triangles?

[restive pilot]

so 12.5 x 4 = 50

[vagrant raptor]

yep yep

area of this big rectangle?

[restive pilot]

so now i have to add the area of all the triangles to the area of the polygon?

[vagrant raptor]

find area of big rectangle, subtract area of missing corners!

[restive pilot]

OHH that makes sense!!

okay

[vagrant raptor]

great!

[restive pilot]

398!

[pine kettle]

no

not 398!

398

[vagrant raptor]

hahaha it is 398 cm^2

not 398! cm^2

[pine kettle]

,w 398!

[vagrant raptor]

they're messing with you

[ocean seal]

Results provided by WolframAlpha

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

Requested by sills#9713

[restive pilot]

ohh LOL

[vagrant raptor]

there you see what 398! is haha

it's 398 * 397 * 396 * 395 * ... * 2 * 1

but you did it! good job

[restive pilot]

tysm!

just one more thing

[vagrant raptor]

sure

[restive pilot]

on my answer sheet it says "400 (398cm^2)"

is 398 cm sqared still right?

[vagrant raptor]

yes, 400 is rounded. because of significant figures.

398 is the exact answer

[restive pilot]

so that would give me full points, on lets say a test?

if i wrote exactly 398 cm squared

[vagrant raptor]

that's hard for me to say - significant figures and rounding is something teachers and such can be pretty strict about

it depends on if you assume the lengths in the figure to be exact, or to be approximate measurements

[restive pilot]

mhm okay! how do i round the number correctly then? or like the answer

[vagrant raptor]

https://www.youtube.com/watch?v=-X0amVzMKFo

this explains how to round to significant figures

[restive pilot]

ohhh okay!!

[vagrant raptor]

I'm sure there are other good videos on it

[restive pilot]

it was just an extra question, i got the method which is most imporant

[vagrant raptor]

yes! and both answers are technically correct

[restive pilot]

okay! thank you so much this really helped me out

[vagrant raptor]

im glad! good job

good luck and have fun :)

[restive pilot]

thank you again 😄

[lone heart]

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[iron swallow]

Sorry

[ruby summit]

Please close this help channel if you don't plan to ask a question.

[lone heart]

@iron swallow Has your question been resolved?

[lone heart]

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[hardy panther]

Need help solving this question

[hardy panther]

Here are my notes

Am I doing it right?

<@&286206848099549185>

<@&286206848099549185> '

<@&286206848099549185>

[vagrant raptor]

bro

don't do that

[hardy panther]

I need help ASAP

[vagrant raptor]

is it a test?

[hardy panther]

No it's not

[vagrant raptor]

what's the rush then

[hardy panther]

No one has responded in the last few minutes

[glass lichen]

oh no

Be patient.

All the help in the server is voluntary

[hardy panther]

I need help calculating the percentage change between the 2008 price and the 2018 price for a single family home

[glass lichen]

Ok, so wait for someone to help you

[hardy panther]

I followed the procedure and got what I thought was the answer

It was 18.3 or 18.4%

It was incorrect

Was there something wrong with the equation?

Hello?

[glass lichen]

Hi, still, be patient

[lone heart]

@hardy panther Has your question been resolved?

[hardy panther]

<@&286206848099549185>

[hardy panther]

<@&286206848099549185>

<@&286206848099549185> am I doing it right?

<@&286206848099549185> STUDENT NEEDS ASSISTANCE!

It's been 15 minutes already

<@&286206848099549185>

<@&286206848099549185>

[slate kayak]

Omg stop pinging so much!! Nobody is obligated to help you nobody is getting paid it is not their job.

[hardy panther]

With trail and error I managed to solve the question

All of you are useless

.close

[lone heart]

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[quasi nacelle]

I need help figuring this out (cal)

[buoyant kayak]

do you know what a secant line is?

[quasi nacelle]

the point passing through both points?

line*

@buoyant kayak

[buoyant kayak]

more or less, yeah

so given two points, how would you get the equation of a line?

[quasi nacelle]

i plugged it in desmos but idk how to get that line

wait

slope formula?

[buoyant kayak]

you would get slope with that, yeah

and what do you know that can get you the equation of a line using the slope and a point?

[quasi nacelle]

slope m = 1

yea

lol

[buoyant kayak]

so now that you have the slope, use one of those points to get an equation for the line

[quasi nacelle]

i plug in slope into the main formula?

[buoyant kayak]

what is the "main" formula?

[quasi nacelle]

the function given

[buoyant kayak]

no

you're trying to get the equation of a line

using the slope and a point

perhaps you know a formula that allows you to do this?

[quasi nacelle]

hmm let me think rq

point slope form?

or

[buoyant kayak]

yes

since you have a point and the slope

point slope form is applicable

[quasi nacelle]

y=x-4 got it thanks

[buoyant kayak]

👍

[quasi nacelle]

what about c?

i got b done

[buoyant kayak]

what did you get for b

[quasi nacelle]

1

[buoyant kayak]

alright

so c asks for the line tangent to the point x=1 for the function f(x)=x^2-x-12

[quasi nacelle]

plug it in?

[buoyant kayak]

well what information would you get by plugging x into the original function?

[quasi nacelle]

not the tangent line

[buoyant kayak]

well let's think ahead

for the tangent line, you're trying to find the equation of a line again

so what might you want to use again?

[quasi nacelle]

point slope form

[buoyant kayak]

right

and for point slope form, you need a point (x and y) and the slope

you currently have x

so plugging x into the original function will give you what?

[quasi nacelle]

-12

[buoyant kayak]

well that's the value, but what is -12?

[quasi nacelle]

f(x)

[buoyant kayak]

in other words?

[quasi nacelle]

the function

[buoyant kayak]

can you use it in point slope form?

[quasi nacelle]

not sure

[buoyant kayak]

if you were to plot points for the function f(x)=x^2 and you plugged in x=2, you'd get 4. what does that 4 represent?

[quasi nacelle]

the y value

[buoyant kayak]

exactly

[quasi nacelle]

ohhhhhh

[buoyant kayak]

so -12 is..?

[quasi nacelle]

is y

[buoyant kayak]

right

so now you have x=1 and y=-12

you just need the slope

how do you go about getting the slope of a function at a point?

[quasi nacelle]

wait so we dont need x1 and y1?

[buoyant kayak]

that's what you just got

1 is x_1 and -12 is y_1

[quasi nacelle]

i mean x2 and y2

[buoyant kayak]

no, you're not getting the slope using $\frac{\Delta y}{\Delta x}$

[quasi nacelle]

oh ok

[ocean seal]

a disappointing son

[buoyant kayak]

you're taking a more... calculus approach

the slope at a point... any ideas?

[quasi nacelle]

deriv?

[buoyant kayak]

yep

[quasi nacelle]

ok wait so im taking the deriv of what exactly? with x = 1 and y = -12

[buoyant kayak]

forget about those points for now

[quasi nacelle]

ok

[buoyant kayak]

you want the derivative of your function

[quasi nacelle]

ok

2x-1

[buoyant kayak]

right

so that's the derivative, which tells you the slope at a point

you specifically want the slope at x=1

[quasi nacelle]

well if i set to 0

or wrong approach

[buoyant kayak]

wrong approach, yeah

don't overthink is

[quasi nacelle]

0k

[buoyant kayak]

you have f'(x)=2x-1 and you want the slope at x=1

where f'(x) tells you the slope at any point

[quasi nacelle]

plug it in

[buoyant kayak]

yep

[quasi nacelle]

ohh i was thinking that at first idk where my mind went

[buoyant kayak]

happens to everyone lol

[quasi nacelle]

so its 1

[buoyant kayak]

yep

so now you have the slope, x_1, and y_1

you can create the equation for your line

[quasi nacelle]

1-y1=1(1-x1)

or not slope form

[buoyant kayak]

you're plugging in the wrong things

in $y-y_1=m(x-x_1)$, $y$ and $x$ stay as $y$ and $x$

[ocean seal]

a disappointing son

[quasi nacelle]

y-1=1(x-1)

[buoyant kayak]

remember what y_1 is

it's not 1

[quasi nacelle]

-12

wait

oh it is

[buoyant kayak]

yes, that's right

[quasi nacelle]

y--12=1(x-1)

[buoyant kayak]

right, now solve for y

[quasi nacelle]

y=x-13

[buoyant kayak]

looks good

[quasi nacelle]

thanks a lot u a real one

anywhere i can praise u or vouch u

[buoyant kayak]

i store it in my heart 🥺

[quasi nacelle]

got it lol

i got one last question

if u dont mind

[buoyant kayak]

go for it

[quasi nacelle]

idk how to get second one

[buoyant kayak]

you know the formula you're supposed to use for the mean value theorem?

oh wait

you don't even need that

wait yes you do

i'm a mess

[buoyant kayak]

back to this lmao

[quasi nacelle]

yea

f(b) - f(a) / b - a

[buoyant kayak]

wait yeah i was right with thinking i was wrong

instantaneous velocity is the velocity at a point, with velocity being the derivative of position

and you already found the average velocity... so your goal is to just find where the average velocity = the instantaneous velocity

so how can you get from s(t) to v(t)

[quasi nacelle]

value theoreom?

[buoyant kayak]

not quite

how are position, velocity, and acceleration related?

[quasi nacelle]

deriv?

[buoyant kayak]

yeah

so what's your derivative

[quasi nacelle]

of s(t)?

[buoyant kayak]

yeah

[quasi nacelle]

wait i think ik what to do one sec

got it

i took deriv

and set it = to other one

[buoyant kayak]

👍

[quasi nacelle]

and solved for t

thanks

i added u

thanks dawg

do i close this channel?

how so

[buoyant kayak]

.close

[quasi nacelle]

.close

[lone heart]

Channel closed

Closed by @quasi nacelle

Use .reopen if this was a mistake.

[quasi nacelle]

thanks bud

[lone heart]

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[full pelican]

How can I change the base of 1 / \sqrt(2) ?

[full pelican]

that is " 1 divided by the square root of 2 " if it isn't obvious

[gray isle]

wdym by
"change the base"

[full pelican]

I need to set the base equal to the base on the other side of the equation

but I don't know how to work with a number such a 1/\sqrt(2)

This is for solving exponential equations

[gray isle]

can you show the full question

[full pelican]

nevermind, I was able to use the change of base formula to get a decimal which I think I can use.... but how can I convert .125 into a fraction?

[alpine sable]

Breh $\frac {.125}{1} = \frac {125}{1000}$

[ocean seal]

CatHashira

[alpine sable]

Just multiply the numerator and denomitaor by 1000

[full pelican]

is there a way to get it into the form 1/x?

[gray isle]

simplify the fraction

[alpine sable]

Ya

[full pelican]

Ah, 1/8. Thank you!

.close

[lone heart]

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[real maple]

Need help with this Khan academy question

[real maple]

here the pics

from bottom to top

top 3 are answer choices

Need help with this Khan academy question

<@&286206848099549185>

.close

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[lapis tendon]

can somone help me

[lapis tendon]

plz

[lone heart]

@lapis tendon Has your question been resolved?

[lone heart]

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[somber mango]

How can i find the probability for a discrete random variable? is there are an equation?

[pliant current]

iirc for discrete random variables, you can find the probability mass function

then use the pmf to find probabilities for specific cases

this might help

https://ocw.mit.edu/resources/res-6-012-introduction-to-probability-spring-2018/part-i-the-fundamentals/MITRES_6_012S18_L05.pdf

[somber mango]

tbh i don't know what this is because my teacher didn't say anything about this so i am not sure if i am supposed to use this. My quesiton is based on a homework question and i am a little bit confused about where to start. I know the random variable but to answer the question i need the probability of the outcome for the random variable. Can i DM you the question so you understand what i am talking about?

[somber mango]

Thanks for the link. I will see i this will help

[pliant current]

as much as i would like to help, not sure DMing me would be in your best interest because i havent taken statistics in a while

https://www.youtube.com/watch?v=d2STHFVHGAg

perhaps this will be more relevant?

[somber mango]

i just finished watching this and in my case i dont have P(X)

this is the confusing part for me

[somber mango]

ohhh I see np. I am new to statistics too.

[pliant current]

https://www.radfordmathematics.com/probabilities-and-statistics/discrete-probability-distributions-discrete-random-variables/discrete-random-variables-and-distributions.html

this should show you how to construct a table like khan's

[somber mango]

thank you

[pliant current]

yep

[lone heart]

@somber mango Has your question been resolved?

[somber mango]

hey so i watched it but in the question it did not give a PDF function.

[pliant current]

https://youtu.be/pC2p2yXxgYQ

there's two videos in that page

second one talks about constructing a distribution table

[somber mango]

got it, i will take a look at this

[pliant current]

sorry i didnt specify lol

[somber mango]

np

[somber mango]

the thing is that in this case the question is given the probability of X. If i had that i would have solved the question long time ago

[pliant current]

can you send the question you're working on so i have a better idea?

[somber mango]

this question is giving me headache

[lone heart]

@somber mango Has your question been resolved?

[lone heart]

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[hollow cairn]

Anyone know how to do this

[lone heart]

@hollow cairn Has your question been resolved?

[hollow cairn]

Nvm I got it ty anyways

[mortal hazel]

how did you got it?

[lone heart]

@hollow cairn Has your question been resolved?

[steady echo]

@hollow cairn didn't you say you got it, if that's right, pls close this channel

[lone heart]

@hollow cairn Has your question been resolved?

[lone heart]

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[foggy acorn]

if my bottom row of the matrix is [ 0 0 (2-a^2) | a-2], what values of a will provide no solution, infinitely many solution, and unique solution?
if this were a^2 - 2 i wouldnt be asking this question but this time it's 2-a^2

[foggy acorn]

i think for unique solution it would have to be a =/= +- sqrt(2) right?

but how about for inf many and no solution?

[oak perch]

no solution: a=+/-sqrt(2)

The number of solutions depends on the coefficients of x and y…

[foggy acorn]

for inf many ill have to find a value for a that will make the value of (2-a^2) and a-2 then?

[oak perch]

No

[knotty spire]

you only have infinite solutions if the last line becomes 0 = 0 in essence

[oak perch]

You only know that if a=+/-sqrt(2) then it has no solution

When a=/=+-sqrt(2), the situation depends on coefficients of x and y

[foggy acorn]

would a =/= +-sqrt(2) mean that 2-a^2 = 0 in which i can divide the 3rd row by 2-a^2?

so that 1 = a-2/2-a^2 which would be a unique solution assuming each row having a leading 1

[oak perch]

I told you it doesn’t only depend on z. You only know that if a=+/-sqrt(2), then no solution.

When a doesn’t equal +/-sqrt(2), the situation depends.
Let B=
(a_11 a_12 b_1-a_13(a-2)/(2-a^2)
a_21 a_22 b_2-a_23(a-2)/(2-a^2)
Let A=
(a_11 a_12
a_21 a_22)
If rank(A)=rank(B)=2, then unique solution, if rank(A)=rank(B)=1 or 0,then infinity many solutions, if rank(A)<rank(B),still no solution

[foggy acorn]

umm so for example a^2 - 2 = a - 2,

no solution if a = -2, inf many if a = 2, and unique solution if a =/= +-2 right?

[oak perch]

You didn’t listen at all…

You are talking to yourself…

[knotty spire]

https://en.wikipedia.org/wiki/Rouché–Capelli_theorem

[oak perch]

That’s basically what I told him

[knotty spire]

I know

[oak perch]

Here

[knotty spire]

but maybe he'll read it

a second time

and understand

[oak perch]

Yeah wish he will

[foggy acorn]

we havent gone through this yet

so far we've only had Gauss elimination and gauss-jordan

[knotty spire]

Then read this

http://sites.science.oregonstate.edu/math/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/gauss/gauss.html

They have examples of solving a system where there are no solutions or infinitely many. Notice what the last row becomes.

(and assuming that your problem was initially a square matrix of coefficients, which you haven't stated)

[zenith compass]

Is that in RREF?

Umm at least REF i mean

[foggy acorn]

the REF would be as follows:
1 2 1 | 2
0 -6 1 | -3
0 0 (2-a^2) | a - 2

[zenith compass]

Alright then it is nice

[oak perch]

Great, rank(A)=rank(B)=2

So no solution or unique solution

[zenith compass]

If the last row is something like 1=1, 2=2, you actually find out that the last row is just telling you useless information

[oak perch]

That’s what I was talking about, you can’t only observe the third row

[zenith compass]

But if the last row is talking about sth like 0=1, 3=4, you can see it is totally nonsense, which implies no matter what you put x,y,z to be, 0is never 1 and 3 is never 4

[foggy acorn]

can i ask about the similar example by teacher gave us that deals the same questions? the main difference is that instead of 2 - a^2 it's a^2 - 4

[zenith compass]

Umm idk my explanation is good. Probably not rigorous but hope you can understand

[foggy acorn]

just in case cus our teacher's solution for that went by pretty quickly

[knotty spire]

This might not be rigorous, but I think you can if you have the row echelon form, and the only "variable" row is the last one.

[zenith compass]

But is he going to give a proof?

[oak perch]

That’s the case where rank(A)=rank(B)=0 like I said

[foggy acorn]

but we didnt get taught anything about the rouche capelli theorem yet

[zenith compass]

Forget that if you haven’t learn

Just rmb “no solutions” means something nonsense in the last row in this question

[foggy acorn]

our teacher's example was like this:
1 1 -1 | 2
0 1 2 | 1
0 0 a^2 - 4 | a -2

then she said that no solution if a = -2, inf many solution if a = 2, and unique solution is a =/= +-2

[knotty spire]

With no argumentation as to why?

[foggy acorn]

for the no sol. only wrote that 0 0 0 | 4 which isnt true

for inf many wrote that 0 0 0 | 0 if a = 2

[zenith compass]

Consider the square system of linear equations $Ax=b$ with $A=\begin{pmatrix}1&1&-1\0&1&2\0&0&a^2-4\end{pmatrix}, b=\begin{pmatrix}2\1\a-2\end{pmatrix}, x=\begin{pmatrix}x_1\x_2\x_3\end{pmatrix}$ where $a$ is an unspecified real parameter.

[ocean seal]

Trenton

[zenith compass]

Alright tell me, what happens when there is no solution

[knotty spire]

This is actually an application of the theorem, you just don't know it yet

but yeah, this works if you don't know the theorem

[foggy acorn]

our teacher said that the unique sol if it's a =/= +- 2

for her example that is

[foggy acorn]

no solution of a = -2

[zenith compass]

Yes do you know why

[foggy acorn]

because if you let a = 2 then the bottom row would be 0 0 0 | 4 which is false

[zenith compass]

Exactly

So how about infinitely many solutions

[foggy acorn]

inf many if a = 2 as it will make the bottom row 0 0 0 | 0

for unique solution she said that a would be =/= +- 2

in which the third row will be divided by a^2 - 4 because if a =/= +-2 then a^2 - 4 =/= 0

where the row changes to 0 0 1 | 1/(a+2)

[zenith compass]

Well this is only true when $a+2\not=0$.

But if $a+2=0$, this step is illegal which leads to no solutions.

Kinda different intuition yet still works.

[ocean seal]

Trenton

[foggy acorn]

does this imply that our teacher missed a step for finding values of a that provide unique solution?

[knotty spire]

No, but what she did do is leave out mentioning that the two upper rows have no bearing on the amount of solutions

In the more general cases, you can't assume that.

[foggy acorn]

so for my problem where
1 2 1 | 2
0 -6 1 | -3
0 0 (2-a^2) | a - 2

the two upper rows will have bearing then?

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[ocean sluice]

can someone show me how is this

[ocean sluice]

the same as this

[lone heart]

@ocean sluice Has your question been resolved?

[little drum]

exponent law

(a^b)^c = a^(bc)

[floral pawn]

$\left(\frac{1}{\frac{h}{x}}\right)\frac{1}{x}=\frac{1}{h}$

[ocean seal]

R31_

[ocean sluice]

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[frail grove]

How can we know that an angle is positive or negative if we can add and substract 2pi?

[frail grove]

Is it by finding the reference angle?

[glass gale]

I don't think positive or negative angles are really a distinction since you can use an angle of 270 degrees instead of -90 degrees

and achieve the same rotation

it only matters if you're also counting full rotations

which for things like sin and cos you aren't, so there's no difference

[frail grove]

I needed it to find the phase shift between two phases

[vale sapphire]

You reduce it to be between -π and π

Actually no

If we're talking phase and you have something like -3π, the phase is negative that's that

[glass gale]

I think there was a convention of sign depending on if the first wave is leading ahead of lagging behind the second, not entirely sure.

[frail grove]

Can't we add 2pi to - 3pi?

[vale sapphire]

Actually, let me rethink this once more

[frail grove]

What about in this case

[vale sapphire]

what's u_c?

[frail grove]

Vectors

[vale sapphire]

okay forget about it

So is φ calculated relative to u_c?

[frail grove]

It's the angle between uc and v

[knotty spire]

$\cos \phi = \frac{\vec{u_c}\cdot\vec{v}}{|\vec{u_c}||\vec{v}|} = \cos (\phi + 2\pi k)$, where $k \in \mathbb{Z}$

[frail grove]

I understand why φ is negative since it's in the inverse way of w what i didn't understand is that if we have the value of φ how can i know if its positive or negative

[ocean seal]

Remavas

[glass gale]

I am trying to find on the internet what the convention for voltage-current phase shift is I think everyone just says the degree and then which lags behind which

here voltage lags behind current

aka. current leads ahead of voltage

[frail grove]

Yeah it's related to forced lcr oscillations btw

[glass gale]

I think relying on words rather than a sign is the smart thing to do here since there's no convention

and if there was it would be a hard convention to remember

since it would be rather arbitrary

[knotty spire]

Yeah, the two most often used conventions I've seen is either using -pi, pi or using no negative angles at all (0, 2pi)

[frail grove]

So whenever i come across an angle i need to verify that it's between - pi, pi in order to find out its sign

[knotty spire]

if you are using that convention

[frail grove]

Ok thnx guys

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[leaden tiger]

need help with trig

[leaden tiger]

i need to get this

into this format

[steady echo]

Have u tried firstly simplyifying sin 2theta and cos 2theta?

[leaden tiger]

i used the double angle formula to get the numerator as 2cos theta - 4sintheta costheta

[steady echo]

Alright great! How about the cos 2theta?

[leaden tiger]

i got the denominator as 1 - 4sin^2theta

usig this

which leaves me with this

[steady echo]

Aight got itt

Sorry for the delay

You're veyr closee!

[leaden tiger]

np

[steady echo]

Now you have to factor 2 cos from the numerator, and express the denominator in a diff way using (x-a)(x+a)

Ponder over this a little and see if you get itt

Hint: ||4 sin^2 (x) = (2 sin x)^2||

[leaden tiger]

ah

ok,

got it

just cancel out the common value from here

[steady echo]

Yes!

[leaden tiger]

thankyou for the help!

[steady echo]

Nppp

[leaden tiger]

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[languid furnace]

Can somebody help me with this problem? (financial applications)

[lone heart]

@languid furnace Has your question been resolved?

[languid furnace]

Bruh

[silver adder]

at is the number of days since the start

(a) wants you to plug in t=0 to the population equation

[languid furnace]

ah okk

[silver adder]

(b) wants you to plug in t=10, and equate P to 1606, and solve for k

[languid furnace]

i get the first two problems, but I'm not able to solve the problem for C

i set the equation as 1350 + 400e^-0.22314355t < 1500

but nSolve (t nspire calculator) isn't able to give me the answer

[silver adder]

youre suppose to solve it yourself

how did you solve for k?

[languid furnace]

1606 = 1350 + 400e^2k

400e^2k = 256

e^2k = 16/25

==> ln 16/25 = 2k

k = ln 16/25 / 2

= -0.22314355

[silver adder]

,calc log(16/25)/2

[ocean seal]

Result:

-0.22314355131421

[silver adder]

kk

cant you repeat for t?

1350 + 400e^-0.22314355t < 1500

[languid furnace]

wait

Damn

thank you

for part D would it be just 1350?

[silver adder]

lol

i mean yeah but its never going to reach it

the better answer is that there is no smallest population according to this equation

[languid furnace]

can you explain in a bit more detail please?

[silver adder]

the range of P is (1350,1750]

getting my shit mixed up a little

but yeah there is no minimum value

[languid furnace]

ah

Thank you for the help man. really appreciated

[silver adder]

you can say it has a lower bound of 1350

no problemo

[languid furnace]

ty ty

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[oak perch]

$=\frac{4}{|x^2-2x|}$

[ocean seal]

Cogwheels of the mind

[alpine sable]

.

[oak perch]

Yes

$=\frac{4}{x^2-2x}$ when $x< 0$ and $=\frac{4}{-(x^2-2x)}$ when $x> 0$

[alpine sable]

The greater that sign is looking weird in texit for some reason

[oak perch]

…

I wanted to say less than or greater than

Not equal to…

[ocean seal]

Cogwheels of the mind

[oak perch]

By definition…

Any positive M there exists δ,any |x|<δ,f(x)>M

It’s a quadratic function

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✅

[oak perch]

What are you doing…

No…

You supposed to show that any positive M, there exists d such that $=\frac{4}{|x^2-2x|}>M$ when $|x|<d$

It’s two quadratic functions of x

[ocean seal]

Cogwheels of the mind

[oak perch]

WLOG you can assume that $M>4$, you can take $d=min(\sqrt{1+\frac{4}{M}}-1,1-\sqrt{1-\frac{4}{M}})$ for example

[ocean seal]

Cogwheels of the mind

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[fossil mason]

Could I get help for last q

[oak perch]

ii C(5,5)C(2,1)+C(5,4)C(2,2)

iii ΣC(3,x)C(5,y)C(2,z) where (x,y,z) are (1,4,1) (2,3,1) (3,2,1) (1,3,2) (2,2,2) (3,1,2)

[fossil mason]

Ok it works!

Tysm

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[wise basin]

Fractions are the bane of my existence I have to make the bases equal to each other in order to solve for x

But

Yeah

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[alpine zephyr]

Factorise x^2+8x+4

[alpine zephyr]

how to to do it

its not quadratic

[abstract fractal]

Yes it is

[alpine zephyr]

can u pls tell how sud i split the mkiddle term?

[abstract fractal]

It's not factorable.

[alpine zephyr]

ighjt

[gray isle]

its not factorable over the integers

you can do stuff like completing the square / apply quadratic formula to factorise over the irrationals

[alpine zephyr]

ohh ok

thnx

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[wet pulsar]

Hi

[wet pulsar]

So I’m just having an issue with understanding what an orthogonal function is

https://youtu.be/gAxt7XA7x6A

I was watching this video

And at 2:07 I just completely stopped understanding what the person was saying

And now I’m trying to understand if a function by itself can be orthogonal, or if a function can be orthogonal only in relation to another function

[oak perch]

It means it’s a set of functions, any two of them are orthogonal

Two functions being orthogonal means their inner product is zero, where the inner product is defined by an integral

[wet pulsar]

Ah okay so it’s a set of all the orthogonal functions

How would you know if your set was complete though?

And basically the concept is, given a complete orthogonal set of functions, you can express any other function that isn’t in the set as a linear combination of the complete orthogonal sets elements??

[oak perch]

If it’s complete, then any periodic function with the same period should be able to be written as a linear combination of them (possibly infinitely many, in terms of convergence)

[wet pulsar]

Same period as what?

And so only periodic functions can be described as a linear combination

[oak perch]

I mean if Those functions are cos (2πnx/T), sin(2πnx/T) then all functions with period T. Also some conditions need to be satisfied, not just periodic. I don’t remember the detail, long time since I saw analysis

If it’s not periodic, then maybe we should consider Fourier transform not Fourier series

Anyway I don’t study analysis, stein wrote a great book about Fourier analysis , but I don’t read it, I only know it’s recommended by many.

[wet pulsar]

Ahh okay that makes sense

Thank you I’m slowly understanding

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[snow flicker]

Please lhelp

Im new to calculus

[tacit arch]

Start with the limit definition of differentiable at a

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[supple tiger]

heyy, I don't understand the underlined notation
also if y'all have any pointers for a), I'd be grateful

[supple tiger]

meant to underline the left side of the equality in b)

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@supple tiger Has your question been resolved?

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@supple tiger Has your question been resolved?

[drifting hull]

@supple tiger $\max_{i,j} |a_{i,j}|=\max \lbrace |a_{1,1}|, |a_{1,2}|, \dots |a_{2,1}|, |a_{2,2}|, |a_{2,3}, \dots |a_{n,n}| \rbace$, so the maximum of the absolute values of any entry of the matrix

[ocean seal]

Alexander42
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[supple tiger]

I'm kinda confused, the subscript on the max means what tho?

i get max|a_i,j| would mean what you said, but is max_i,j|a_i,j| the same thing?

[drifting hull]

the subscript just means that you maximize over all possible values of i and j, so yes, it means the same.

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[leaden aspen]

Does this seem like a kind of wild question for the second example of implicit differentiation from a book?

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@leaden aspen Has your question been resolved?

[tight locust]

Personally I would find dx/dy

[leaden aspen]

I know it's kind of a theoretical thing

[tight locust]

Rotate the image by 90 degrees counterclockwise

Then the "rightmost point" just becomes a "minimum"

[leaden aspen]

Ahh that's super clever - nice

[tight locust]

Or I guess it would actually be a maximum

[leaden aspen]

Just getting introduced to implicit diff in calc 1 so I'm getting the basics of it

[tight locust]

Then set dx/dy = 0

[leaden aspen]

differentiating it isn't so tough it's kind of the 7 steps to get teh equation after

[tight locust]

Does what I'm saying make sense

[leaden aspen]

Yeah I think so - I'm not sure I've been introduced to that method of solving, is that from higher calculus?

[tight locust]

Sort of, but it doesn't require any actual math above calc 1

Just differentiate the eq with respect to y

And let x = x(y)

[leaden aspen]

Ok, does that skip right to the inverse if I'm understanding correctly?

[tight locust]

Yeah pretty much.

[leaden aspen]

Nice way to jump a step for sure thanks for pointing it out. I was originally just asking if it's a kind of crazy problem for the second example of Implicit differentiation i've ever seen

[tight locust]

Imo it sort of is

[leaden aspen]

true true I definitely like that way of solving, will bring it up with my teacher to see if she can walk if through next class

But yeah I'm kinda just looking at it like, might be good to build up to that level eventually haha. I appreciate the help thank you!

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[fluid dawn]

I am using 2 different algorithms to try and determine whether someone has diabetes. I have a dataset and margin that determines whether values are off is 0.4. i.e above 0.4 is positive and below 0.4 is negative. I want to show the deviation of the values that are on the wrong side of the margin and use that to compare which algorithm is better. What mathematical theory should I use for that eg. standard deviation or average error margin(by taking the average of the amount of error of each erroneous value) or something else.

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@fluid dawn Has your question been resolved?

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[fading nest]

guys

[fading nest]

A question

again

[raven rover]

Just ask your question

That should be your first message

[fading nest]

didn't understand how t.o factorize $ax^2-4a+3(x^2+4x+4)$

[ocean seal]

Xerral

[fading nest]

Oh sorry my bad

[raven rover]

np np

[raven rover]

Where did this problem come from? I'd be surprised if that was just given as a problem and told to factor that

[fading nest]

well an exercise asked me to factorize this

[raven rover]

Just send the original problem

A screenshot ideally

[fading nest]

ok

I'm italian so don't care about the language haha

letter d.

[bleak ridge]

Pane

[raven rover]

That's so weird - that's a strange question to ask

Don't think I've seen such a thing before

You can factor the (x^2 + 4x + 4)

[fading nest]

yeah

[raven rover]

What if you expanded everything

[fading nest]

for example?

[raven rover]

You expanded the rightmost term in the polynomial

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[neat bane]

yo bro

do u know how to do this question

[flat blaze]

6sqrt2

[fluid eagle]

hint: this is a right triangle

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[rare salmon]

Can Someone Help Me?

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[wicked orchid]

Is this interval convergent or divergent and how do you justify it?

[pale kestrel]

I would consider bounding it

[vague coral]

^

[wicked orchid]

How so?

I'm actually so lost lmao

Also I posted the wrong thing

It should be

[pale kestrel]

my advice stands

Can you found the function inside

with something

[wicked orchid]

-1 and 1 I think?

[pale kestrel]

You could.

But then the integral of f(x) = 1 (or g(x) = -1)

from 1 to infinity

is infinite - doesnt converge

This bound does not give you much info.

[wicked orchid]

My brain hurts man

[vague coral]

|cos²(x)|<=?
|1+x^4|<= ?

[pale kestrel]

?

Just the numerator needs to be bounded

[vague coral]

yea i know

[wicked orchid]

Max value cos^2 is 1 isn't it xd

And 1+x^4 can go to +inf no?

[vague coral]

1+x^4 stays itself on (1,+oo)

[pale kestrel]

0 =< cos^2(x) =< 1
1 =< 1+x^4 < infty

So you should be able to bound that fraction

the denominator isnt of particular importance really...

its the numerator that you need to bound

$$0\leq\cos^2 x\leq1$$
$$0\leq\frac{\cos^2 x}{1+x^4}\leq\frac{1}{1+x^4}$$

[pale kestrel]

Think this is getting confusing unless you see ^

This was the process you needed to do to bound the integrand

You can divide by 1+x^4 because it is always positive

Now can you see the approach?

[wicked orchid]

Honestly no lmao

But it looks like I squeeze theorem

I'm not certain tho

[pale kestrel]

Can you not see how this chosen bound helps???

I don't know what integrals you have seen --- surely you should know int 1/(1+x^4) looks like it should converge

unless you have just started

[vague coral]

I think he gets the bounding, but dont know why we should bound

[wicked orchid]

I get that part I just don't get how this fits in the grand scheme

[pale kestrel]

The integral is the area underneath the blue curve

to the right of the purple line

Yes?

[wicked orchid]

Yeah

Converges at 0?

[pale kestrel]

?

[vague coral]

why 0 ?

[pale kestrel]

We just know it is the area under the blue curve.

We don't know anything about it - that's what the question is asking.

What we can do

Is show if there is a function (the red curve)

that bounds it

And we know the area under the red curve is finite

then we know the area under the blue curve is also finite.

(both functions are always positive --- so the lower bound for both areas is 0)

[wicked orchid]

Yeah...

[pale kestrel]

So we want the black area

We know the red area is greater than the black area

If we know the red area is finite

the black area must also be finite (ie. converges)

$$\int_1^{\infty} \frac{1}{x^n}\dd x$$

[pale kestrel]

Have you been show when this integral converges/diverges?

This is the usual goto when trying to bound integrals

[wicked orchid]

Mm mm

[vague coral]

a famous guy

[wicked orchid]

Convergent p > 1 divergent p <= 1

or n in this case

[pale kestrel]

This is a theorem for you?

I'm just checking

If yes, then you can use this exact result to help

[wicked orchid]

Yeah it is

[pale kestrel]

In that case we are almost there

[pale kestrel]

There is something small you can do to get exactly this

1/x^n

1/(1+x^4) <-- take this.

[wicked orchid]

Wdym

[vague coral]

To prove that the integral is converging, you need the theorem that Shuri stated

with 1/x^n

[pale kestrel]

Look, if we try to use the squeeze/comparison test right now

we get this

$$0\leq\cos^2 x\leq1$$
$$0\leq\frac{\cos^2 x}{1+x^4}\leq\frac{1}{1+x^4}$$
$$\int_1^\infty 0 \dd x\leq\int_1^\infty\frac{\cos^2 x}{1+x^4}\dd x\leq\int_1^\infty\frac{1}{1+x^4} \dd x$$

[pale kestrel]

The left integral is ok. That's just 0

But the right integral is not in this exact form

And you can't quite use that theorem.

Rather, before that there is one step to do.

[wicked orchid]

Yah get rid of the 1 in the denominator

just have no clue how xd

[vague coral]

a fraction is greater than an another if the denominator is less than the other denominator

don't know which grade it is

[pale kestrel]

,, 0<a<b\implies0<\frac1b<\frac1a

oops

[vague coral]

thats the inverse

[vague coral]

yes

[wicked orchid]

b = 1 + x^4
a = x^4
?

[pale kestrel]

good

So try writing the steps --- all the ideas are in place

[wicked orchid]

Left side is convergent, right side is convergent, middle is convergent as well then

Yeah?

[vague coral]

??

[pale kestrel]

You need another piece of information

(I'm thinking right now)

[vague coral]

it's pretty much done right now

1/x^4 is integrable so is cos²x/1+x^4

[pale kestrel]

This alone doesn't sound quite right - I think you need to fully state what is given

in the comparison test

So yes basically what you said, but justify it with 'By the comparison test'

[wicked orchid]

Okay, I think I can figure it out

Thanks Shuri

[pale kestrel]

^ and don't write what I did

Better to write it exactly as the test tells you

[wicked orchid]

Uhh

[pale kestrel]

You don't just 'integrate both sides of the inequality' really

You are applying the direct comparison test

$$0\leq\cos^2 x\leq1$$
$$0\leq\frac{\cos^2 x}{1+x^4}\leq\frac{1}{1+x^4}\leq\frac{1}{x^4}$$
By the Direct Comparison test (perhaps state explicitly why the conditions needed for the test are satisfied),
$$0\leq\int_1^\infty\frac{\cos^2 x}{1+x^4}\dd x\leq\int_1^\infty\frac{1}{x^4} \dd x$$

[wicked orchid]

Thank you ❤️

[wicked orchid]

.close

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[honest umbra]

determine equation of lines that are tangent to ellipse x^2+4y^2=16 and also pass through point (4, 6)

[raven rover]

I think there's only one such line.

You have a point, now you just need dy/dx at that point

[honest umbra]

there r 2 lines, one is x=4 and idk how to get the other one

[raven rover]

Oh - I didn't notice the vertical

[raven rover]

This - it'll help to draw a picture

[lone heart]

@honest umbra Has your question been resolved?

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@honest umbra Has your question been resolved?

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Channel closed

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[honest umbra]

.reopen

[lone heart]

✅

[ripe rain]

There must be 2 I believe

[lone heart]

@honest umbra Has your question been resolved?

[lone heart]

@honest umbra Has your question been resolved?

[honest umbra]

.close

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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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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[fading nest]

$x^7-x$

[ocean seal]

Xerral

[fading nest]

How do you factorize it?

Anyone on?

[alpine sable]

$x(x^6-1)$

[ocean seal]

Jezier

[fading nest]

can i continue?

cuz x^6 is pretty big

[buoyant kayak]

let u=x^2

[misty dagger]

of course u can

[fading nest]

how do I continue

idk how to do it

[buoyant kayak]

.

make a sub.. perhaps it'll be easier to see

[fading nest]

hmmm wdym

[misty dagger]

What is x^2-1?

[buoyant kayak]

i mean rewrite x^6 using exponent rules and let u=x^2, as a sub variable

[fading nest]

like $x(x^2-1)(x^4-1)$

[ocean seal]

Xerral

[buoyant kayak]

no

[misty dagger]

Uhhh

no

x^2-4 = (x+2)(x-2) For ex

[buoyant kayak]

$(x^a)^b=x^{a\cdot b}$

[misty dagger]

Since there x^6, What can you do to break down?

[ocean seal]

a disappointing son

[fading nest]

lemme think

[oak perch]

Try factor x^3+1 and x^3-1

[fading nest]

$(x^2-1)(x^4+1)$

[buoyant kayak]

please don't give answers...

[ocean seal]

Xerral

[misty dagger]

?? why you give him answer?

[buoyant kayak]

#rules #❓how-to-get-help

[oak perch]

Oh

[misty dagger]

Loll

[fading nest]

i didnt even understand with the answer

lol

[oak perch]

Edited

Hope he doesn’t remember it

[buoyant kayak]

i still advise rewriting x^6 and making a sub

what are the factors of 6?

[misty dagger]

that's correct.

Follow that step

[fading nest]

3 and 2

1 and 6

?

[buoyant kayak]

right, and 1 and 6 wouldn't be very useful in rewriting

[fading nest]

ok right

[buoyant kayak]

so try $(x^2)^3$ as $x^6$

[ocean seal]

a disappointing son

[fading nest]

hmm so

x[(x^2)^3-1]

[buoyant kayak]

right

now it'd be easier to see if you made a sub

you can pick any variable you want (other than x), i chose u because it's normally a sub variable

let u = x^2

[fading nest]

Ok so

$x(x^2+x)(x^3+1)$

[ocean seal]

Xerral

[fading nest]

Idk

[buoyant kayak]

i'm not sure what you did there

[fading nest]

Oh

[buoyant kayak]

all i'm advising is to let u = x^2 to make a sub

[alpine sable]

If $u=x^2$ what's $x^6$ in terms of u?

[ocean seal]

CatHashira

[alpine sable]

Is that what ur asking

[fading nest]

x^4?

[buoyant kayak]

1.) in terms of u
2.) exponent is wrong

[fading nest]

Oh so x^3

[alpine sable]

[fading nest]

Cuz 2•3 is 6

[buoyant kayak]

in terms of u

exponent is correct now tho

[fading nest]

x^3

No wait

[buoyant kayak]

[alpine sable]

[fading nest]

Wdym in terms u

Lol

[alpine sable]

Do you understand what "in terms of u" Means

[buoyant kayak]

there should be no x

[alpine sable]

Knew it

[buoyant kayak]

in terms of u means the expression you are telling me should have u as its only variable

no other variables

just u

[buoyant kayak]

you're very close with this... you're just not putting u lol

[fading nest]

(x^2)^3

[buoyant kayak]

[fading nest]

Lol

[buoyant kayak]

with this, where you see an x^2, replace it with a u

that is what a substitution is

replacing something with something else

[fading nest]

u^3

[buoyant kayak]

something simpler, so it makes things easier

yes

so now you have $x(u^3-1)$

[ocean seal]

a disappointing son

[buoyant kayak]

notice something special about the inside of the parenthesis?

[fading nest]

Okok

Lemme see

Difference of cubes?

[buoyant kayak]

yep

[fading nest]

So

Wait

$(u-1)(u^2+u+1)$

[buoyant kayak]

not quite, and remember we're in terms of u here

[fading nest]

Right

[buoyant kayak]

not sure where the 3x came from?

[fading nest]

Yeah got confused

Just x

[buoyant kayak]

all those x's should be u's

[fading nest]

Yeah ok

[ocean seal]

Xerral

[buoyant kayak]

right, now remember the sub

u=x^2

[fading nest]

Ok

[buoyant kayak]

so whenever you see a u, replace it with (x^2)