help-39

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[mellow pine]

my homies

[mellow pine]

I literally have no clue how to start this

I have been learning proofs, and I know how to do proofs given definitions & for universal and existential quanitifiers

but IDK what this is haha

any resources / advice on how to start would be much recocmended

[hallow cedar]

First of all Do you know what all of the symbols mean

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[hearty geyser]

can i solve it using this?

[pearl pond]

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@hearty geyser Has your question been resolved?

[gleaming heath]

that's literally the same thing yes, since P(X>=x)=1-F(x)

[hearty geyser]

thanks

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[pseudo dune]

Hi, i do not understand how it an can be zero för n=k^2 for some k

[pseudo dune]

i seems that if n is not a perfect square it can be zero. for example 3, does not a_n become -1/3?

how can even (-1)^n/n become 0?

oh i think i understand. So because x has the power n^2 they choose n^2=k therefore n=sqrt(k) and when you put it in an it becomes (-1)^sqrt(k)/sqrt(k) therefore any number that cant be finely squared is undefined as (-1)^sqrt(3) for example does not work, am i thinking right?

[tropic saddle]

it would help to write out the first few terms of the sum

you will notice that for example x^3 doesnt appear

[pseudo dune]

that is kind of what i was trying to write, i just did a variable swap

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[wary heron]

isnt removing the indeterminate form the same as a continuous expansion ? in limits

[wary heron]

as doing* a continuous expansion

[pearl pond]

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[wary heron]

<@&286206848099549185>

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[fallen cobalt]

I'm not sure what you mean. Is there an example you could provide?

[pearl pond]

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[woven lantern]

How do I solve this? a and b are arbitrary

Was doing a physics problem and got stuck here

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[hollow dew]

for this question, why did they make the values on the left side of the transformed functions table over 8

they keep doing different numbers for these examples and idk how i'm supposed to know what to make the denominator

[shy imp]

probably to make it easier to look at

if the fractions are = then ur answer is correct still

[shy imp]

they want all the values to be /6 so its easier to graph, operate on etc

thats what im assuming

[hollow dew]

hm alright!

thank you :)

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[dense sphinx]

I found it 35/12

[old geyser]

Show me your work

[dense sphinx]

Sec = 5/4 cosec = 5/3

[final bone]

Steps are required

show ur steps

[dense sphinx]

Knowing that tan 3/4 , hypotenus is 5 cos = 4/5 sin = 3/5 since sec and cosec equals reverse if them I replace with reverses I end up with 35/12

Any thoughts on this @final bone

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[wraith vine]

Can someone help me revise on how to do this, without software for graphing?

[final bone]

ys

[dense plank]

can you differentiate the function

[wraith vine]

yes, 12x^2 + 36x - 216

[dense plank]

what does the derivative tell us

[wraith vine]

rate of change

[dense plank]

correct

[wraith vine]

but should help, would it?

[dense plank]

so it means, whenever the derivative is positive, then the original function is increasing

and when its negative, the original function is decreasing

[dense plank]

this is for A. and B. btw

how do you think we can determine the intervals where the derivative is negative or positive?

[wraith vine]

nvm, makes sense, hold on, i am going to find values

[wraith vine]

understood

[dense plank]

alright

[wraith vine]

I simplify it first to x^2 +3x -18

[dense plank]

then

[wraith vine]

x = -3 and 6

so when x is 3 and -6

those changes

[dense plank]

the factored form is (x+6)(x-3)

while the roots are x=3 x=-6

meaning, when x is 3 or -6, the derivate touches the x axis

well if it doesnt touch the x-axis then it must be either positive or negative right?

[wraith vine]

so do I test between 2, 3 and 4, find if its increasing?

[dense plank]

not exactly

you will test between the infinities and the roots

[wraith vine]

that didnt sound right but thats where I logically would go

[dense plank]

so you have to test the intervals
(-inf, -6) and (-6, 3) and (3, inf)

[wraith vine]

so range between (-inf and -6) and (3 and +inf)

[dense plank]

yee

but also (-6,3) too

test if they are positive or negative

if its positive, then the original function is increasing in that interval

[wraith vine]

test what betweens these points, sorry I got lost

[dense plank]

oh i forgot to tell you

you have to choose a point in inside those interval

any point

as long its inside the interval

substitute it in the derivative

and check if its positive or negative

[wraith vine]

so -5. 0 and 4

[dense plank]

-5 isnt exactly between (-inf, -6)

[wraith vine]

nvm, yea, so -7

[dense plank]

0 is a good choice for the 2nd interval

4 is correct

for the 3rd

[wraith vine]

0 is just -216

[dense plank]

ye

how bout -7 and 4

[wraith vine]

sorry still calculating

[dense plank]

(-inf, -6)
test point: -7
result:
sign:

(-6,3)
test point: 0
result: -216
sign: NEGATIVE

(3, inf)
test point: 4
result:
sign:

youre basically doing this table

[wraith vine]

-7 = 120 POS
0 = -216 NEG
4 = 120 POS aswell?

[dense plank]

ye thats right

so what are the intervals on which the derivative is positive?

(-inf, -6)
test point: -7
result: 120
sign: POSITIVE

(-6,3)
test point: 0
result: -216
sign: NEGATIVE

(3, inf)
test point: 4
result: 120
sign: POSITIVE

[wraith vine]

inf to -6,
3 to inf
are pos intervals

[dense plank]

yes

[wraith vine]

-6 to 3 are neg Intervals

[dense plank]

meaning, the original function is increasing at these intervals

thats the answer for A

[dense plank]

while here, the original function is decreasing at that interval

thats for B

thats how you determine whether a function is decreasing/increasing

• Take the derivative
• Find the roots of the derivative
• Create intervals based on the roots
• Choose an x for each interval, and sub that x into the derivative

-if the result is positive, then the original function is increasing in that interval
-if the result is negative, then the original function is decreasing in that interval

[wraith vine]

what am I looking for with max and min, havent touched this in a while, hence rusty but assuming that when it goes from + to - is a max, and - to + is a min

[wraith vine]

thank u, that is very helpful for test prep and practice

[dense plank]

alright

lets proceed to C and D

to find the local minimum and local maximum, we need to take a look at the derivative again, and also the functions second derivative

the first derivative is

$f'(x)=12x^2+36x-216$

[wraith vine]

f'' = 24x + 36

[dense plank]

find the second derivative

alr

$f''(x) = 24x +36$

[jolly parrot]

Renz

[wraith vine]

when x is -1.5, f'' = 0

[jolly parrot]

Renz

[dense plank]

wait a min

ok first

what do you know when the first derivative of a function is 0?

[wraith vine]

rate of change, inflection?

[dense plank]

not quite

if the derivative is positive, then the graph is increasing

if negative, then decreasing

but how about when 0

[wraith vine]

it changes like inflection point?

[dense plank]

its a turning point

the point where a graph changes from increasing to decreasing and vice

think of a parabola

it turns around in its vertex right

[wraith vine]

when f'' = 0 ?

[dense plank]

when f' = 0

[wraith vine]

yep

[dense plank]

first derivative

so what are our zeroes in our derivative again?

[wraith vine]

-6 and 3

[dense plank]

from earlier

correct

it means

the original function has a "turning point" at x=-6 and x=3

but we dont know if its a maximum or a minimum

we only know that the graph turns around

right?

[wraith vine]

so f''(-6) and f''(3) indicates max and minima?

[dense plank]

nono ignore the 2nd derivative first

everytime the derivative is 0, there will be a turning point, but we dont know yet if its a max or min

[wraith vine]

but cant we put it back into f(x) to find which way it turns

[dense plank]

not in f(x)

but in f''(x)

so if f''(x) is found to be positive, then its a minimum
and
if f''(x) is found to be negative, then its a maximum

[wraith vine]

so f''(-6) and f''(3) will tell us about max and min

[dense plank]

ye

it will tell us if its a min or max

now test f''(-6)

[wraith vine]

f''(-6) = -108

so its a max

[dense plank]

correct

[wraith vine]

f''(3) = 108
so its a min

[dense plank]

correct

but those are x-value only right?

we need to find its corresponding y-value

sub -6 in the original function

[wraith vine]

so f(3) and f(-6)

[dense plank]

ye

-6 first

[wraith vine]

f(3) = -371, so its min
f(-6) = 1087, so its a max

[dense plank]

correct

make it a point tho

(-6, 1087) is a local maximum
(3, -371) is a local minimum

thats your answer for C D lol

[wraith vine]

it wants only an answer, so I would just put in the y values, but need to get in the habit of putting points

rather than values

[dense plank]

Finding local min and max

• Take the roots of the first derivative
• Substitute the roots to the 2nd derivative
• Test if the answer is positive or negative, if its positive then its a local minimum, if its negative then its a local maximum
  -Now you have x-values for these points, find the y-values by substituting the x-values into f(x) to find the y-value and make them into ordered pairs (x,y)

got it?

[pearl pond]

@wraith vine Has your question been resolved?

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[signal kettle]

angle of 2 sides is 60degree and on one side there is a point and from that point to other side length is 18cm what is distance between that point and an edge

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[quartz ermine]

to find the shortest distance between a line and a plane

[quartz ermine]

do i find the point where the line is perpendicular to the plane

[compact ridge]

There's a formula

[quartz ermine]

oh is there

[compact ridge]

But yes, to do it yourself you would find the normal vector of the plane (that's easy if you have the plane in ax + by + cz = d)

[quartz ermine]

yeah it's in that form

what do i do after i find that point

[compact ridge]

https://external-content.duckduckgo.com/iu/?u=https%3A%2F%2Ftse2.mm.bing.net%2Fth%3Fid%3DOIP.9xq3E-U9oXJIjJWljsXBNwHaCw%26pid%3DApi&f=1&ipt=28fa2217af130ad4f6177e2cc727c6db66150228a892dcabb9c1e85eeb5728e7&ipo=images

[quartz ermine]

ahhhh thank you

so my x0,y0,z0 would be the coordiante of the perpendicular point

between line and plane

[compact ridge]

After finding the intersection of the line passing through your point and the plane (it's in parametric form and just sub into x, y, z in the plane equation)
Then just use the distance formula for your point and the perpendicular point

[compact ridge]

Nope, it's just your point

[quartz ermine]

ok thank you for the help :)

[compact ridge]

Yeah no worries

[quartz ermine]

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[quaint forum]

17 no question pls

<@&286206848099549185>

[compact ridge]

Please don't instaping

Only ping helpers after 15 minutes

[compact ridge]

There's a trick for this question actually

[quaint forum]

ok sorry

[compact ridge]

The minimum value of the LHS is the same as the maximum value of the RHS

[quaint forum]

what

[compact ridge]

Because the axis of symmetry of both the quadratics under the square root on the LHS is x = -1 (-b/2a)

And so is the axis of symmetry of the RHS also x = -1

(And the RHS is concave down, LHS is concave up)

[quaint forum]

ok i understand a little but for my classes purposes can u solve in the genral method too

like without the trick'

[compact ridge]

Squaring everything and expanding would be a terribly inefficient way of approaching this question

[quaint forum]

yes

[compact ridge]

Symbolab can do that for you

Question 17 is marked with a star cause it's not supposed to be solved using the normal methods

[quaint forum]

i tried but i did not give the steps

[compact ridge]

Yeah like try to understand what I wrote: come back to it later if you need to

[quaint forum]

actually i put that star on it

[compact ridge]

And also it helps to plot both equations on Desmos

[compact ridge]

Lmao

[compact ridge]

This is how I discovered there's a trick

And then I wrote the reasoning myself

https://www.desmos.com/calculator/6zumrcu6wj

[quaint forum]

ok

[compact ridge]

Of course coming up with the proper reasoning is going to be challenging, so in this case understanding > doing it yourself at least for now

[quaint forum]

ok

thx

[compact ridge]

No worries

[quaint forum]

.close

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[glacial swift]

i have find out the values of x and y

i dont know how to show that line L intersects the curve C only once

x= 4.8 and y=6.4 btw

if anyone can help mention me

[pseudo oxide]

if they intersect

they must have the same x and y

correct?

[glacial swift]

yes

correct

[spare crypt]

there are 2 ways

[glacial swift]

enlighten me

[spare crypt]

any observations on the equation of curve?

its a special type of curve

[glacial swift]

uhh...no but i did check it out on desmos and its a circle

[spare crypt]

yup

well any equation in the form

x^2 + y^2 = r^2
is a circle

anyways , if a straight line intersects a circle

and it only intersects once

what is it?

[glacial swift]

tangent

[spare crypt]

yup

tangent has a special property you know

[glacial swift]

that is? it intersects at one point?

[spare crypt]

another one

[glacial swift]

straight?

[spare crypt]

its about the angle with the center

[glacial swift]

oh

90 degs angle

[spare crypt]

yup

[glacial swift]

from the centre of the circle

[spare crypt]

so are you getting any ideas now?

[glacial swift]

yea..but how do i show it

like prove it

on paper

[spare crypt]

aight

so tangents from 90 degree

you have the equation of tangent

from question

you can find gradient/slope

and since you know point of intersection and center of circle , you can also find the slope/gradient of the radius connecting point of contact to center

if 2 lines have 90 degree between them, then the product of slope is -1

so if you get -1 , you can state , they only intersect once and it is a tangent

else it is a secant

[glacial swift]

holy, i did NOT learn this at school

but yea i understand

imma research it a bit more

[spare crypt]

other than that , i don't think so any method algebraically proves (maybe idk )

[glacial swift]

okay

thanks

[spare crypt]

np

[glacial swift]

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[glacial swift]

.reopen

[pearl pond]

✅

[glacial swift]

@spare crypt

yo

this Marking Scheme/ answer

says this

th is this

[spare crypt]

while solving

did you get any quadratic equation?

in end

[glacial swift]

to get the x and y?

yea

[spare crypt]

yup

[glacial swift]

yea

did

[spare crypt]

ahh

yeah this is also possible

this is pretty much that both the lines have only 1 solution

hence they intersect only once

i thought they were asking more than this

they are proving that the quadratic has no other solution by finding the discriminant

and D = 0 , hence equal roots ie 1 solution

the idea i told you would usually work only for circles

this idea is for any curve

rather any 2 equations in general

point of intersection is a solution, so more than 1 intersection points = more than 1 solution so yeah

[glacial swift]

oh

i see

wait so if i just prove

that both the equations have the same x and y

that should do it?

[spare crypt]

nope

you are proving here that both the equations don't satisfy for any other x and y other than the one you found

[glacial swift]

no, see usually, quadratic equations have two answers, this is case it was one

[spare crypt]

quadratic equation can have upto 2 answers

[glacial swift]

yea

thats what im saying

[spare crypt]

yup

quadratic has 3 cases

have you studied quadratic?

[glacial swift]

yea

[spare crypt]

so did you learn about the cases of quadratic

for different values of D

[glacial swift]

as in?

im giving gcses

[spare crypt]

do you know the quadratic formula?

[glacial swift]

so i might not know this

[glacial swift]

yes

[spare crypt]

state it

[glacial swift]

-b (plus minus) square root b squared....

i know it

[spare crypt]

aight i will teach you rn, but just take a look online later on , you will understand

[spare crypt]

yup so

[glacial swift]

yea

[spare crypt]

quadratic equation has 3 cases

when it comes to solution

if you observe the equation , you will see that getting 2 solutions or 1 solution depends upon b^2 - 4ac

b^2 - 4ac is also called the discriminant

if the value of discriminant is greater than 0

then your formula becomes

-b ± k/ 2a

since you have a value after ±

it is bound to have 2 different solutions

but if b^2 - 4ac = 0

then the formula becomes -b/2a

that is when the quadratic has 1 solution

don't get confused with solutions and roots btw

quadratic still has 2 roots , but both the roots here are equal

example, (x-1)^2

opening it , you get x^2 + 1 - 2x
now if you find b^2 - 4ac for this equation, the discriminant will come out to be 0

but in reality it is (x-1)(x-1) , still 2 roots ( 1, 1) , but one solution (1)

got it?

[glacial swift]

let me process it

[spare crypt]

third case is when D is less than 0 , then you have complex solutions because root of negative no. is imaginary

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

watch this actually

he explains it pretty well

[glacial swift]

hmm, brb have to piss, imma watch this vid and hopefully i understand

brb

[spare crypt]

aight

[glacial swift]

ahhh okayyyyyyyy

i see

[glacial swift]

i undersstand

[spare crypt]

nice

[glacial swift]

im really weak at math

ngl

[spare crypt]

you understood how this idea was used in that question right?

[glacial swift]

yes

[spare crypt]

good

[spare crypt]

its fine

everyone isn't supposed to be good at something

you maybe are good at something else

just work on it daily and you will get better

[glacial swift]

tysm

until next time (which might be quick cuz im solving some past papers and i suck at math)

.close

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[spare crypt]

sure

[spare crypt]

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[pearl pond]

@restive pike Has your question been resolved?

[worthy perch]

weird for 3 but it's correct probably

[restive pike]

what are the reasons for the 1/2?

[worthy perch]

The inscribed angles or angles between the tangent and the secant are equal to half of the arc

[pearl pond]

@restive pike Has your question been resolved?

[restive pike]

would saying it's an isosceles be wrong?

even though it's not given

[worthy perch]

Bro, I proved it in the second point

[restive pike]

referring to the 1st

[worthy perch]

and?

https://tenor.com/view/adalfarus-okbb-okbt-okbu-okbr-gif-23339359

[restive pike]

RPQ = 70 (isosceles)

You proved it's an iscoscles

I'm just asking if giving isosceles as a reason without proof other than it being obvious would be wrong

[worthy perch]

huh?

If you are saying exactly that about an isosceles figure, then this is generally the third point, which is fully proven.

The remaining points imply elementary logical chains. I have no groundless evidence in my decision

[restive pike]

Alright I'll try and research about that 1/2 thing later as I don't fully understand it

[worthy perch]

oh

[restive pike]

We might use a different formula over here cause I've never seen that

[worthy perch]

bro, its theorem

[restive pike]

I meant that we go about things different worded or whtever

[restive pike]

how does that go into what you said though?

[worthy perch]

k, probably -_-

[restive pike]

I lack sm with these type of qstns

[worthy perch]

oh, i find

[restive pike]

sorta like this then?

[pearl pond]

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[sullen flare]

Hello, hope your having a good day, can anyone help me with this

[fluid wing]

hy

i can try to help you but i'm not sure of my answer

so if $T_a= 310 F \ T_0= 72F \ t= 1.5hours in second \ T = 122 F$ so you just remplace and resolve it

[jolly parrot]

pomale

[fluid wing]

so you find k and after you remplace another time with k for search T with t=6h in second

do you understand ?

[sullen flare]

yes

[fluid wing]

nice

but one question bc i not come from usa you use impérial units for the question ?

[sullen flare]

yes

[fluid wing]

oh ok

and for time you use second of hours

[sullen flare]

hour

i believe

so it would be 122 = 310+(72-310)e^-k(1.5)

and you solve for k

[fluid wing]

ok bc i come from belgium and we use second and degrees for this equation

[sullen flare]

not sure abt that

[fluid wing]

yes i think

[sullen flare]

hold on let me calculate

[fluid wing]

the surrounding object it's the oven right ?

[sullen flare]

the temperature surrounding the object will be the room temperature which is 72 degrees

i think

idk

[fluid wing]

72 F is for the turkey

[sullen flare]

i think the oven temperature is the initial temperature

im confuse

[fluid wing]

i go watch the equation in my language i'm coming

for me is Ta the heat of the oven

T0 is the turkey

if the ambiant air is 72 F the turkey is also 72F

[sullen flare]

this is correct

[fluid wing]

sorry for my bad english

[sullen flare]

its ok

so i solve for k first

hold on

[fluid wing]

ok

i solve it also in my corner

[sullen flare]

ok

i got 0.15721914

k=0.15721914

[fluid wing]

2 sec

[sullen flare]

alr

take your time

[fluid wing]

i have the same

good job

now the temperature of the turkey after six hours

[sullen flare]

i believe the next step is to plug in k

[fluid wing]

you juste need to remplace k by the value you find and change the time so you have T

[sullen flare]

yes

[fluid wing]

i have 217.338244 F

[sullen flare]

let me calculate on my end

just to make sure

[fluid wing]

take your time no problem

[sullen flare]

T=310-188e^-0.15721914(6)

right

?

[fluid wing]

310-238

it's 72-310 = -238

[sullen flare]

so yea

[fluid wing]

just one question: what grade are you in?

[sullen flare]

i got 207.338

[sullen flare]

sophomore

[fluid wing]

so you are 20 years old ?

[sullen flare]

?

[fluid wing]

and i retry the calcul and i tell you what

[sullen flare]

wym i have 20 years

[sullen flare]

?

[fluid wing]

sophomore it's like second grand of university no ?

[sullen flare]

no

high school

[fluid wing]

oh sorry

[sullen flare]

dont worry

[fluid wing]

so i retry and i have 217.338244

i write this

[sullen flare]

its wrong

its 227

[fluid wing]

$310+(72-310)e^{-(0.1572191406)(6)}$

[jolly parrot]

pomale

[sullen flare]

it was 227

lol

but don't worry

i appreciate you helping me

[fluid wing]

if you have the correct answer all is good

[sullen flare]

i got it incorrect and the correct answer was suppose to be 227

[fluid wing]

no problem

[sullen flare]

thank you

appreciate you

[fluid wing]

have a good day

me too

[sullen flare]

you too

have a nice evening

[fluid wing]

thks

[sullen flare]

bye

.close

[fluid wing]

bye

[pearl pond]

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[arctic mortar]

When it says that they're vertical does it means that they're vertical to each other (i.e. perpendicular) ?

[quiet goblet]

No

Usually vertical means only y axis

[arctic mortar]

oh so they're parallel to the y axis then

[quiet goblet]

That is either they are parallel or anti parallel

Yeah

[arctic mortar]

I see, thx

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[unkempt violet]

Can someone explain to me how to do this

The second question

<@&286206848099549185>

[pearl pond]

@unkempt violet Has your question been resolved?

[unkempt violet]

<@&286206848099549185>

[pearl pond]

@unkempt violet Has your question been resolved?

[unkempt violet]

<@&286206848099549185>

[pearl pond]

@unkempt violet Has your question been resolved?

[shrewd mist]

what part of it don't you understand?

Looking at the dominant terms, its n^3/3^n

exponentials outgrow polynomials, and so the limit approaches 0

There are a bunch of ways to prove it at this point, with the less elegant one being L'hopitaling the shit out of it.

[unkempt violet]

??

Well yea I see that but I don’t understand how you get the answer using the comparisons test

I thought I could compare it to (1/3)^n but I can’t

[keen rain]

in the image it just depicts a geometric comparison

since the factor 1/3 in (1/3)^n is less than 1, the limit is 0

if you have any limit limn->inf., the geometric comparison is looking at x

if |x| < 1, the limit is 0

[unkempt violet]

Since it’s less than 1 I said it’s convergent

[keen rain]

yop

[unkempt violet]

But my teacher said that’s wrong

So I’m confused on how I should prove it

[keen rain]

which part do they refer to as wrong

[unkempt violet]

They didn’t specify just marked it one

Prob should’ve asked though

[keen rain]

you can further simplify but otherwise I don't see much of an issue

you mainly just need to mention that the growth of an exponential function outweighs a polynomial one

which is why you can ignore the n³ term

[unkempt violet]

Yea I understand I just find it odd how he said the comparison is wrong

Are you sure it’s not divergent or sum

[keen rain]

no if the factor is less then one absolutely, then it converges to 0

so they probably just missed the above mentioned growth comparison

because then it's just the limit of lim[n³/3^n * 3!] = 1/6 * lim[n³ / 3^n] = 1/6 * lim[1 / 3^n] = 1/6 * 0 = 0

[pearl pond]

@unkempt violet Has your question been resolved?

[pearl pond]

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[tidal otter]

Any ideas?

[somber locust]

try simplifying first and see where it gets you

[compact ridge]

My first instinct would be to do $x^3 (1 + x) + y^3 (1 + y) + z^3 (1 + z) \ge \frac{3}{4}(1 + x)(1 + y)(1 + z)$

[jolly parrot]

south

[somber locust]

forgot ^2 on them

[compact ridge]

No?

Multiply the first fraction by (1 + x)/(1 + x) and so on

[somber locust]

x^3 * (1 + x) + y^3 * (1 + y) + z^3 * (1 + z) = 3/4 * (1 + x)^2 * (1 + y)^2 * (1 + z)^2

[patent yew]

i dont think there's any squares

[dense goblet]

yeah no it's just common denominating

[compact ridge]

https://approach0.xyz/search/?q=OR content%3A%24\sum \frac{x^3}{\left(1%2By\right)\left(1%2Bz\right)}%24&p=1

Approach0 is super useful

You should use it before you ask cause your problem already exists in 99% of cases

[somber locust]

multiplying each denominator doesn't give (1 + x)^2 * (1 + y)^2 * (1 + z)^2 ?

[compact ridge]

No

[somber locust]

am I missing something?

[compact ridge]

The denominator for all of them will be (1 + x)(1 + y)(1 + z)

[compact ridge]

Yes

[compact ridge]

Multiply the second fraction by (1 + y)/(1 + y)
And the third fraction by (1 + z)/(1 + z)

[somber locust]

ahh

i took the longer route, and just noticed my mistake

[compact ridge]

Okok

IMO ShortList 1998, algebra problem 3

Most approaches use Muirhead or Holder

[pearl pond]

@tidal otter Has your question been resolved?

[tidal otter]

Thank you, I did not know about this

[compact ridge]

No worries!

[tidal otter]

.close

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[chrome flame]

How should I deal with this?

[compact ridge]

Check each of the answer options

It might be exact

[chrome flame]

What if it is exact?

It is not exact

[compact ridge]

RIP that's why I said this

Good luck finding the right integrating factor

There are so. many. cases.

[pearl pond]

@chrome flame Has your question been resolved?

[chrome flame]

@compact ridge

[compact ridge]

Ah nice one

Yeah I totally forgot you needed to divide by x^2 to have that form

I thought the form was just x dy - y dx

No worries

[pearl pond]

@chrome flame Has your question been resolved?

[pearl pond]

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[pearl pond]

@manic bison Has your question been resolved?

[pearl pond]

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[dense plover]

Can someone help me figure out what I did wrong?

[dense plover]

The answer is pi/32

I did integrations by substitution

[sterile tusk]

$\int \frac{du}{u^2+16}$is not inverse hyperbolic sine

but rather $\frac{1}{4} \arctan(\frac{u}{4})$

[jolly parrot]

y0shi

y0shi

[pearl pond]

@dense plover Has your question been resolved?

[dense plover]

oh ok

thank you for the help

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[dusk otter]

Hi i was wondering if anyone could help me verify this question

[dusk otter]

Was confused if the interval is right

Also since it says find all solutions should I be plugging numbers into n and solving ?

[quartz citrus]

Why do you restrict n from 0 to 3

[dusk otter]

I can't seem to remember, I did this 2 weeks ago. Should it be 0 to 2

My friend and I were debating about this

Like the possible solutions

[quartz citrus]

Cis is periodic so we can allow all integer values of n

[dusk otter]

Ohh I see

[quartz citrus]

cis(theta) = cis(theta + 2npi) doesn't only hold true for n between 0 and 3 right

Also, the reals is the left and right direction and the imaginaries are the up and down

You solved for when the LHS is 1, not i

[dusk otter]

ohh okay so I should redo the question?

from the start

[quartz citrus]

Yeah

[dusk otter]

okay thanks

[quartz citrus]

One last thing, you want to consider z^(3/4) but you only considered z^3

[dusk otter]

but wouldn't it be like if isolate for z I get i^4 which is equivalent ot (-1)(-1) which is 1

[quartz citrus]

Oh I see, yes. Myb I misinterpreted your work

[dusk otter]

yeah in anycase I think the intervals are still wrong

[quartz citrus]

nvm hang on

Are you familiar with using e^(itheta) instead of cis?

[dusk otter]

Nope we haven't done it like that

This is what my prof did in class for a similar quesiton but it was with a real number

[quartz citrus]

wdym? the values n is allowed to take?

[pearl pond]

@dusk otter Has your question been resolved?

[dusk otter]

mhm

oh wait I think we already covered that above right?

[quartz citrus]

yeah

n can be any integer, and you're done

[dusk otter]

So like this?

[quartz citrus]

yeah looks good

i would just simplify 1^(1/3) as 1 tho and you can drop it in the last line

[dusk otter]

yup good point thanks : )

[pearl pond]

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[timber matrix]

This is eigenvalues and eigenvectors

Can there be multiple answer for eigenvectors

So I used x2 as my free variable

And for eigenvalue - 1 I got the eigenvector -1 and 2

But my prof got 1 and - 2

[fluid axle]

yes

if you scale an eigenvector, you still have an eigenvector

[timber matrix]

So is mine right

-1 and 2

[fluid axle]

it's -1 times what your prof has

[timber matrix]

Yes

[timber matrix]

What’s the reasoning behind this?

That both works

[fluid axle]

you're just looking for a basis of your eigenspace here

you just have to pick enough linearly indepedent vectors to do so

the specific choice doesn't matter

[timber matrix]

If I picked 0 wouldn’t the vector be 0

0 for x2

[fluid axle]

I said linearly independent

[timber matrix]

Oh ok

[fluid axle]

you can't have 0

[timber matrix]

So anything other than 0 works l?

[fluid axle]

yes

[timber matrix]

Okay makes sense

But the best options for x2 would be to get rid of the existing fractions right

[fluid axle]

sure

just pick what's the most convenient

[timber matrix]

Ok

[fluid axle]

@timber matrix other questions?

[pearl pond]

@timber matrix Has your question been resolved?

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

[rose ibex]

what is your problem here?

[real musk]

?

It is incorrect

[rose ibex]

well

did u divide by x first for each term?

so like ∫(-3x^2-2x+9-8/x) dx

[compact ridge]

Okay 9x - 8 ln(x) + C is correct

The rest isn't

Can you integrate -3x^2 and -2x using the integral power rule?

[real musk]

😵‍💫

-5x ^ 2

[compact ridge]

How do you get that?

Okay step by step

The integral of -3x^2 = -3 * (integral of x^2)

[real musk]

-3x - 2x

[compact ridge]

And also stop guessing
It really doesn't help

[real musk]

I’m not guessing 😭😭 I was adding like terms

-3x and the -2x

[compact ridge]

You have -3x^2 and -2x though

x^2 and x are not the same

[real musk]

I see I see, the /$ command helps In seeing that

(That’s where I got that from)

[compact ridge]

I hope you know basic algebra cause you're getting me worried

[real musk]

Messages like these make me dislike asking for help

I didn’t see x^2 at first of course so yeah you wouldn’t be able to simplify them together

[compact ridge]

Oh ok, I've just seen my fair share of "in calc 1/2/3 but can't do algebra" horror stories

I didn't mean to dissuade you from asking for help

[pearl pond]

@real musk Has your question been resolved?

[real musk]

OKAY

LAST PROBLEM

THEN I CAN START MY NIGHT

[real musk]

Okay, from here

What are we doing exactly

[compact ridge]

Integrate them both

[real musk]

Ok

-x2

And

[compact ridge]

-3 * x^3/3 or just -x^3

[real musk]

Yes I got the same

• C ofc

[real musk]

.

Am I just plugging those in? There’s no extra interval or anything

@compact ridge

[compact ridge]

Yes

You have +C cause there's no initial condition and no bounds

[real musk]

Hmm

Still incorrect

@compact ridge

[compact ridge]

Wait, try -8 ln |x|

[real musk]

Yup

Thanks

.close

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[untold lintel]

Guys what 😭

[untold lintel]

Linear Algebra Done Right - Sheldon Axler - Exercise 1A

[quartz citrus]

what are alpha, beta, and C?

[untold lintel]

This is linear algebra

[quartz citrus]

yes

provide definitions for alpha beta C pls

[untold lintel]

C is a set of complex numbers

Complex numbers are written as a + bi

So

[quartz citrus]

use the fact that z = a + bi where a, b are reals and use the fact addition is commutative in R

[ivory swallow]

this is just an axiom

can't really be proven per se

[untold lintel]

I’m trying to come up with something promising and intelligent to answer with but my brain is hurting

[untold lintel]

Kinda what i thought so I wrote “commutativity”

[ivory swallow]

so we have two complex numbers a + bi and c+di

by definition we add complex numbers component wise

so you must show that (a+c) + (b+d) i = (c+a) + (d+b) i

which just results from commutativity of addition in R

qed?

[untold lintel]

Vector addition huh? I leaned this in trigonometry/precalc it’s that easy?

But I have to come up with

The extra term c + di?

Or complex number I mean

I need to learn more math terms

[quartz citrus]

let $z_1 = a + bi, z_2 = c + di$ where $a,b,c,d \in\mathbb{R},$ then $z_1 + z_2 = a+bi +\mathbb{C} c+di = (a+\mathbb{R} c) + (b+\mathbb{R} d)i = (c +\mathbb{R} a) + (d+_\mathbb{R} b)i = (c+di) + (b+ai) = z_2 + z_1$

[jolly parrot]

M8 of 48

[untold lintel]

Whats the _Rc and the _Cc mean

[quartz citrus]

addition between reals, and addition between complex

i assume you have the axiom that addition under reals is commutative but addition under C isn't

i started omitting it throughout because lazy but it should be fully

[untold lintel]

Yes it’s right here

Or that’s in C

Nvm fuck

[quartz citrus]

$z_1 = a + bi, z_2 = c + di$ where $a,b,c,d \in\mathbb{R},$ then $z_1 + z_2 = (a+bi) +\mathbb{C} (c+di) = (a+\mathbb{R} c) +\mathbb{C} (b+\mathbb{R} d)i = (c +\mathbb{R} a) +\mathbb{C} (d+\mathbb{R} b)i = (c+di) +\mathbb{C} (b+ai) = z_2 + z_1$

[jolly parrot]

M8 of 48

[quartz citrus]

you can do an analagous expansion to show multiplication commutes in C as well

[untold lintel]

Okay

Gonna reread the first part again lmao

.close

.close

[pearl pond]

Channel closed

Closed by @untold lintel

Use .reopen if this was a mistake.

[pearl pond]

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[rustic viper]

hii i want to ensure i’m doing this question correctly

[rustic viper]

the highlighted is my answer

the answer i got

[ashen marsh]

(3.5, -3.0)

[rustic viper]

oh

where did i go wrong

[pearl pond]

@rustic viper Has your question been resolved?

[pearl pond]

@rustic viper Has your question been resolved?

[pearl pond]

@rustic viper Has your question been resolved?

[merry carbon]

I agree with the x coordinate of 11, but how did you get the y coordinate as -3.5?

[pearl pond]

@rustic viper Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

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Remember:
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[jolly parrot]

The Prophet Of The Damned

[pearl pond]

Channel closed

Channel closed due to the original message being deleted.
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[spice condor]

bruh

WA is giving me this shit

,w int x^4/((x^2-1)^2+1) dx

[jolly parrot]

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Requested by theprophetofthedamned

[spice condor]

which is not promising

[gilded temple]

fr

how do u get i in an integral

[rose ibex]

it is valid but instead

use polynomial long division

this is what u should get after the division

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

so the integral becomes of the form 1/(u^2+1)

which is obviously arctan

[north kelp]

i have a feeling it'd be bashy

very

[rose ibex]

it wont

[north kelp]

hmm

[rose ibex]

well its the only method i can think of

[north kelp]

$\int{\dfrac{2,{x}^{2}-2}{{x}^{4}-2,{x}^{2}+2}+1}{;\mathrm{d}x}$

[jolly parrot]

Singularity

[north kelp]

long div gives this

[wet osprey]

Have you tried completing the square

Oh it started as that form

[compact ridge]

They used partial fractions with imaginary numbers

(x^2 - 1)^2 + 1 = (x^2 - 1 - i)(x^2 - 1 + i)

[compact ridge]

Smort

[spice condor]

not particularly good

[north kelp]

THIS IS TOO BASHY

IM DONE 😭

[spice condor]

wait I wrote the question wrong

💀

[compact ridge]

RIP

[spice condor]

sorry mb guys

.close

[wild fable]

[spice condor]

bruh close

,close

how

[wild fable]

never

[rose ibex]

.close

[north kelp]

YOU'RE STUCK

[wild fable]

you cannot stop it

[spice condor]

bruh

[north kelp]

FOREVER

[compact ridge]

.close

[spice condor]

.close

[north kelp]

MUAHHHAHAA

[wild fable]

haha

[compact ridge]

WHAT

.close

[rose ibex]

its over

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[compact ridge]

Ohhhhh

[compact ridge]

It's already been closed

[spice condor]

ok helpers you can hangout here

[compact ridge]

.close

[pearl pond]

Channel closed

Closed by @compact ridge

Use .reopen if this was a mistake.

[wild fable]

What is your doubt?

[rose ibex]

LMAO

[compact ridge]

lmao.

[wild fable]

Can I help with a doubt sir?

[compact ridge]

I don't have a question

[spice condor]

doubt yo mama

[compact ridge]

I just happened to type something after the channel reopened

[north kelp]

i have a ques #help-9

its solved i think

but still

[spice condor]

in spherical coordinates what was x I forgor

x=rhosin(phi) then was it cos(theta) or sin(theta)

[compact ridge]

Doesn't matter actually

But usually it's cos(theta)

[spice condor]

aight

[wild fable]

sir typeclose when ur done

.coose

!coose

[spice condor]

coose

[north kelp]

choose

[compact ridge]

I typed closed already

[gilded temple]

.choose

[compact ridge]

Dude I'm a fricking helper

[gilded temple]

but this is ur help channel

[wild fable]

make sure to clear all ur doubrs sir

[spice condor]

what did the jacobian look like again

[gilded temple]

maximillian robespierre

[spice condor]

good one

[wild fable]

https://media.discordapp.net/attachments/917110295997063238/1209279645099233330/1192815796490080286.gif

[spice condor]

but I dont want my head off

[compact ridge]

lmao

[spice condor]

I need it to do math

[gilded temple]

lol

[spice condor]

this is now helper hangout channel

[verbal sedge]

help i fant cind out my ethnicity am i a pokemon or a sigma male

[rose ibex]

it appears you are a skibidi pokemale

[verbal sedge]

im real

[wild fable]

https://tenor.com/view/never-say-those-words-again-mason-kane-richard-madden-citadel-don't-ever-say-that-again-gif-17372024661020524727

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vivid pilot]

is this true or false?
im pretty sure its true, but not sure why

[vivid pilot]

and i have to explain why

[upper igloo]

i think its false

the converse is true tho

[wraith hare]

have you learnt any showing convergence/divergence methods

[vivid pilot]

yes

[wraith hare]

what methods did you learn

[vivid pilot]

geometric, p-series, integral, comparison, limit comparison, ratio, root, absolute value, alternating series

[wraith hare]

I think it's true

[wraith hare]

any guess to which one it could be

[vivid pilot]

absolute value

[upper igloo]

oh wait im actually braindead lol

[vivid pilot]

but that test says if |a_n| converges, than a_n converges

[upper igloo]

yeah this is true, the converse is the false one

[wraith hare]

mhmhmm so not abs val test

[vivid pilot]

comaprison?

[wraith hare]

yup

so how would you use it