help-39

[vast berry]

they just factorized it

"a" is the common factor

[dim timber]

ah

ok thx

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[ebon thistle]

I solved this problem with y=ux, but I thought this substitution would be easier and I’m not sure why this is incorrect. Can someone explain the issue with my process here?

[ebon thistle]

<@&286206848099549185>

[sonic patrol]

So just to check - is the solution $x = A y^2 - 2 y$ for some constant A?

[jolly parrot]

Pseudonium

[ebon thistle]

my solution isn’t simplified, but it’s ln(x) - y^2/2x^2 - y/x = C

[sonic patrol]

Hmm…

I didn’t get that

[sonic patrol]

I got this

Using y = ux

I also think your working is correct (up to a sign error in an intermediate step, but it cancels out for the last line)

And it reproduces the answer i got when you do e to the power of both sides

[ebon thistle]

sorry this is something completely different, I got confused

here’s what I got

[sonic patrol]

Right… i mean i would just do e^ both sides to get rid of the logs?

Possibly after multiplying by 2

So $2 \ln(\frac yx) - \ln(1 + \frac{2y}{x}) + \ln(x) = D$

[jolly parrot]

Pseudonium

[sonic patrol]

Really there should be absolute values here

So you get $\frac{y^2}{x^2} \frac{1}{1 + \frac{2y}{x} } x = B$ for some arbitrary real constant B

[jolly parrot]

Pseudonium

[sonic patrol]

This re-arranges to the form $x = A y^2 - 2y$ for a suitable A which I can’t be bothered to figure out

[jolly parrot]

Pseudonium

[sonic patrol]

Oh I think $A = \frac 1B$ probably

[jolly parrot]

Pseudonium

[ebon thistle]

yeah I pretty much see what you’re doing

does that mean I was on the right track with my work?

[sonic patrol]

So

Your work is correct up to $(-v-2) dy = -y dv$

[jolly parrot]

Pseudonium

[sonic patrol]

The next line is incorrect, it should read $- \frac 1y dy = \frac{1}{-v - 2} dv$

[jolly parrot]

Pseudonium

[sonic patrol]

So you inserted an extra minus sign

However the line after that and all subsequent lines are correct

And if you exponentiate your last equation, you get $\frac{y}{\frac xy + 2} = D$ for some arbitrary constant D

[jolly parrot]

Pseudonium

[sonic patrol]

Again, this rearranges to the form I gave before

Again with $A = \frac 1D$

[jolly parrot]

Pseudonium

[ebon thistle]

alright, I think I understand

I must have typed it in wrong or something

thanks for the help

[sonic patrol]

Np

[ebon thistle]

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[tawdry atlas]

I just need quick answear

Cuz I got confused atp

Xc = -j10 ohm
V = j10 V

I = ?

[wet swallow]

What?

[pearl pond]

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

@tawdry atlas Has your question been resolved?

[warm current]

V=IR

[pearl pond]

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[dim timber]

if anyone knows physics could you help please

[dim timber]

i have 2 unknown variables and i dont know what to do

only thing i can think that could be right is assuming to friction to find Ft then find a then find mew but idk

[proven arrow]

Hi @dim timber

[dim timber]

hi

[proven arrow]

like there is no need to consider acceleration

since both are in rest

and the question is dealing with static friction not kinetic friction

[dim timber]

so a = 0

[proven arrow]

yes

[dim timber]

ah

that makes sense

and after that it gives you the right answer

thx

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[proven arrow]

no problem!

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[round heart]

arent all choices wrong

[quiet tendon]

translation?

i'm assuming n is a nonzero natural number?

[pearl pond]

@round heart Has your question been resolved?

[round heart]

yes n is a nonzero natural number

[quiet tendon]

since this is a multiple choice, why not just plug in n = 1

and see which answer you get?

[round heart]

sorry for being late

you can do that and get -1

[round heart]

but now try plugging n=4k then you would get 7

where k is a non-zero natural number

[proven arrow]

hmm

[round heart]

so are all choices wrong

or am i doing something wrong

<@&286206848099549185>

[summer sundial]

(i)^n + (-1)^n + (-i)^n + (1)^n this repeates itself

so technically if its even or odd it becomes 0

but at last (i)^8n is missing so we have output as -(i)^8n = -(1)^4n = -1

[round heart]

repeats but without the last 1^n

[summer sundial]

yup

exactly

u add n subtract (i)^8n

[round heart]

so basically it is 2(i^n)+2(-1)^n+2(-1)^ni^n+1

[summer sundial]

no

[summer sundial]

this gives output 0

[round heart]

it doesnt

[summer sundial]

try any int

[round heart]

try n=2

[summer sundial]

it does

whats the prob?

[round heart]

ah yes this does

[summer sundial]

yeah exactly

[round heart]

but n=4 doesnt it becomes 4

[summer sundial]

no

[round heart]

+3 from the other terms that we didnt write

becomes 7

i^4+(-1)^4+(-i)^4+1^4+i^4+(-1)^4+(-i)^4=1+1+1+1+1+1+1=7

every natural number which is a multiple of 4 will give this

[summer sundial]

hm true

wait i used to solve like this

ig i am forgetting something

[round heart]

this alone proves that all of the given choices are wrong

this expression doesnt give a single result

depends on n

[summer sundial]

just a sec

i have solved it before

and it gives a definite output

[round heart]

ok take your time

[summer sundial]

Only natural number is the given number?

[round heart]

n is a non-zero natural number

idk french

but the one who sent the question told me what it says

[proven arrow]

like it has multiple answers

[round heart]

yes

[proven arrow]

I think maybe the question is asking the best result between these 3?

[round heart]

thats why i am saying the choices are wrong

i dont think thats the case

[proven arrow]

hmm

[round heart]

because that wasnt the intention of any mcq that i saw eve

r

[proven arrow]

oh

[summer sundial]

yeah i see so it depends on n congruence of mod 4

[round heart]

what did you get

[summer sundial]

it builds a step function

so technically what u said is right it does depend on n

[round heart]

what are the possible outcomes

[summer sundial]

.t tex If 𝑛 ≡ 0 ( m o d 4 ) n≡0(mod4), the sum is 7 7. If 𝑛 ≡ 1 ( m o d 4 ) n≡1(mod4), the sum is 0 0. If 𝑛 ≡ 2 ( m o d 4 ) n≡2(mod4), the sum is − 1 −1. If 𝑛 ≡ 3 ( m o d 4 ) n≡3(mod4), the sum is 0 0.

[jolly parrot]

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

[summer sundial]

lil bad tex

[round heart]

np i understood the content

but how did you reach this

how 1 like if you plug n=2 dont you get -1

[summer sundial]

well i just divided the cases

tried plugging in some values

and as i^n depends on n from Nx(1,2,3)

[round heart]

$2i^n+2(-1)^n+1+2(-1)^ni^n$

[jolly parrot]

pirateking0723

[round heart]

this is the expression that we are trying to simplify

according to the values of n

its hard to proceed from here

@summer sundial

someone solved it in a handwavy way

for the cases where n isnt a multiple of 4

this sum is a geometric series with common ratio i^n

add and subtract 1

you will get -1+ a sum from k=0 to 7 with k being the number multiplied by n in the exponent of i

so this becomes $-1+\sum_{k=0}^7i^{kn}=-1+\frac{1-(i^n)^{7+1}}{1-i^n}=-1+\frac{1-(i^n)^8}{1-i^n}=-1+0=-1$

[jolly parrot]

pirateking0723

[summer sundial]

we can sort it as every 4th interval becomes 0

for i^n

thats for sure

[round heart]

wdym

[summer sundial]

wait mb

i was taking n from 0 to 7

forgot its a var

[round heart]

ohhh

anyways tysm

sorry if i wasted your time everyone

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[round heart]

.reopen

[pearl pond]

✅

[round heart]

but if you take it as a geometric series

then you need to reject all n of the form n=4k

but you cant do that

because it wasnt initially there

so you find the value of the sum for other values of n

and then do the n=4k by manually substituting that in the expanded sum

and then evaluating

am i right

[summer sundial]

Even for geometric series there are diff outputs

[round heart]

no how

i took any arbitrary n under the condition that it is not of the form n=4k where k is a natural number

[summer sundial]

Summation will be i^n(i^6n-1)/(i^n-1)

[round heart]

by doing this

[round heart]

but if you add a term to that sum

and then subtract it

[summer sundial]

Hm?

[round heart]

add and subtract 1 as here

leave the -1 that results

then group the other terms together

[summer sundial]

I dont get it ig

[round heart]

they will form a geometric series

[summer sundial]

Hm they could

But is summation dependent on only 1 i?

[round heart]

initially what you want to evaluate is $\sum_{k=1}^7i^{nk}$ now this $=-1+1+\sum_{k=1}^7i^{nk}=-1+\sum_{k=0}^7i^{nk}=-1+\frac{1-i^{8n}}{1-i^n}=-1+\frac{1-1}{1-i^n}=-1$

[jolly parrot]

pirateking0723

[round heart]

wdym by that

[summer sundial]

Hmm I see

[round heart]

again this is true for all n diff than 4k where k is a natural number

otherwise the num and denom will be 0

[summer sundial]

Well however it could lead to multiple outputs

[round heart]

how

[summer sundial]

It still gives -i,+i and -1

So 3 diff forms still

[round heart]

for which values of n does the sum give -i

[summer sundial]

I m talking abt i^n

In the result

[round heart]

so you are saying each i^{kn} alone

[summer sundial]

...

[round heart]

not the whole sum?

[summer sundial]

Yes

It will lead to variation however u write

[round heart]

wait

let me try n=1

for n=1 the sum will be i-1-i+1+i-1-i=-1 not 0

[round heart]

so if n is congruent to 1 mod 4 it doesnt give 0

also if 2 mod 4 then it will give -1

[summer sundial]

Yeah

[round heart]

same thing when n is congruent to 3 mod 4

so it is 2 cases in fact

when it is cong 0 mod 4 and when it is cong something else mod 4

tysm and sorry if i wasted your time

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[muted shale]

!status

[pearl pond]

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[fallen tartan]

1

[pearl pond]

@fallen tartan Has your question been resolved?

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[mortal marten]

What does it mean by 16C

Is it asking to form an equation?

Is it asking for the slope?

[pearl pond]

@mortal marten Has your question been resolved?

[pearl pond]

@mortal marten Has your question been resolved?

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[reef plinth]

Hey I have calculated the fourrier transfomations of cos(t) and got the following result
Got stuck on how to use it to calculate the integral under it

[pearl pond]

@reef plinth Has your question been resolved?

[pearl pond]

@reef plinth Has your question been resolved?

[reef plinth]

<@&286206848099549185> if anyone's free

[reef plinth]

.close

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[finite jolt]

What

[pearl pond]

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[ancient spruce]

Where did he get the square from in this equation? howd 3/x = x/24 turn into x squared

[sharp quest]

multiply both sides by x to get rid of the denominator

[ancient spruce]

oh is it because theres 2 x's

[ancient spruce]

ah ok lol stupid question

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[silent magnet]

in desparate need of help 😂 gotten nowhere

[muted shale]

!status

[pearl pond]

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[silent magnet]

1

[sweet forge]

Hemisphere is basically half of sphere

[midnight haven]

hemi means half

Since the formula of sphere's total surface area is 4πr^2

Then half of it i.e. formula for hemisphere will be 2πr^2

[silent magnet]

so would i be right to do 2 x π x 11^2?

[muted shale]

yes

[silent magnet]

ive got 242π as my answer but says its wrong?

[muted shale]

you forgot one part

there is also a circle

look at the image

[silent magnet]

im confused

obviously im aware theres a circle

but i thought there was nothing else to do

like to calculate

[muted shale]

its a circle and a half sphere

you just calculated the area of the half sphere (hemisphere), but you still need to calculate the area of the circle

[midnight haven]

Sorry I din't read the question you are supposed to find the total surface area of hemisphere which is 2πr^2+πr^2=3πr^2

[silent magnet]

so i need to do this equation ?

[midnight haven]

You can just add πr^2 with this

[silent magnet]

so do

242π x 11^2

[midnight haven]

242π x 11^2*π

[silent magnet]

that gives me 289001.7561???

i need to keep it as π

so how can i do that

[muted shale]

3pi r^2

r=11, r^2=121

3pi r^2=363pi

[silent magnet]

oh, omg i just figured it out ty!!

thank you again for the help ❤️

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[cosmic raptor]

left and right limits

using table method

the table method or left and right limits?

for table method, you just keep plugging in values that get closer and closer to 3 and see what the limit approaches

yep

although you should go a bit closer like 3.01 and 2.99 just to be sure

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[pulsar stump]

is this defined? what would be the result?

[pearl pond]

Please don't occupy multiple help channels.

[autumn fossil]

That looks like {1} union {1, 2}

if im not mistaken

[pulsar stump]

i think so too\

but i have a feeling it might big $\big_cup{1,2}$

[jolly parrot]

Ayanokoji
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[autumn fossil]

Why?

[autumn fossil]

by definition, x is in this iff for some i (with appropriate restrictions), x is in {1, ... , i}

the appropriate restrictions being i is either 1 or 2

so x is in that iff x is in {1, ...., 1} or {1, ..., 2}

[pulsar stump]

so you're saying it's 2 sets in a set, and not just 1 set?

i mean 1 set with atomic elements

[autumn fossil]

it is 1 set with atomic elements

[pulsar stump]

then it has to be undefined right?

we can't do that

[autumn fossil]

No?

It's {1, 2}

[pulsar stump]

yeah

[autumn fossil]

iff x = 1 or x = 2. Hence the set must be {1, 2}

to complete that

[pulsar stump]

if it was {{1}, {1,2}}
then the generalized union would be {1,2},

[pulsar stump]

so you're saying

[pulsar stump]

this is the case?

[autumn fossil]

But it's not that

[pulsar stump]

then it generalized union on the set {1,2}?

[autumn fossil]

the first line is important

that tells us that it's just {1} union {1, 2}

[pulsar stump]

oh i see

so it is {1,2}

[autumn fossil]

Oh I get what you said here. If you meant that the union in your question is equal to the generalized union of {{1}, {1, 2}}, then you were correct

[pulsar stump]

but this is not the case

but rather the A1 intersection A2...intersection Ai

[autumn fossil]

those are actually equivalent notions

[pulsar stump]

oh

[autumn fossil]

generalized union of {A1, A2, A3...} is same as A1 union A2 union A3 ...

[pulsar stump]

that makes alot of sense

alright

i think i understand

thank you very much mate

[autumn fossil]

np

[pulsar stump]

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[eternal tulip]

do you know how to construct a secant line?

[grizzled dust]

what does the problem mean by use limits to solve this? you're not allowed to make use of basic derivative rules or something, it is expecting you to use definition of derivative?

[eternal tulip]

i would assume so

you use secant line definition. i.e. construct secant line at (-2,-1) to some other point on the line (a, 2/a)

well

do you know how to define a line based on just going through two points on 2/x?

so you know one of the points you want the line to pass through, (2,2/-2), which is just (-2,-1)

now what would a general other point on this curve look like

oh, it also helps to use point slope definition of line

i should say point slope form

[grizzled dust]

do you remember derivative definition? so just use this and plug in x = -2 then evaluate the limit

[eternal tulip]

its not just this

it does involve this

its the main focus yes

[grizzled dust]

that gets you the slope, then just plug it into a point \ slope equation

$\lim_{h\to0}\frac{\frac{2}{-2+h}-\frac{2}{-2}}{h}$

[jolly parrot]

∫oosh (lemonsaurus appreciator)

[grizzled dust]

so is this what you got?

do you know how to evaluate this limit?

what does the limit evaluate to?

this seems a bit more complicated than it needs to be

creating a common denominator between the top fractions is the right idea but don't really need to multiply the overall h denominator by everything too

so think about just multiplying the 2 / (-2+h) by -2 / -2 (which is just 2 / 2) and then multiplying the 2 / 2 by (-2+h) / (-2+h)

and after some small simplifcation youll get a lone h which should cancel with the overall h denominator

should probably work your way too but dunno why im not seeing where things are canceling

maybe you have a negative sign error

yep

so what is -1/2?

what does that represent?

slope of what

and where?

yes, and where specifically?

x = -2

so we have the slope of the tangent at some point on the curve, what point is this tangent going through?

how did you get that

what is this equation you are using?

so we have the equation f(x) = 2 / x

what is the point on the curve where x = -2 ?

im confused what exactly you are trying to do, why are you multiplying the slope of the tangent by stuff?

so im asking, how can we find a point on the tangent line?

the whole point (erm excuse the pun) of a tangent line is geometrically, it does what? it's meeting with our original curve f(x) with its exact slope at a specific point right?

so here is f(x) = 2 / x (in red) and what the tangent line would look like at x = -2, does that make sense?

what we know so far is the slope of that green line (-1/2) which seems to make sense, that seems correct for its slope

but the important bit is that both the original curve (f(x), the red) and the tangent line (green) share the same point (the blue)

so what is that point?

yep

so now you know a point (-2, -1) and a slope (-1/2) now it's just a matter of using equation from basic algebra to write the equation of a line

find what y?

if you have a point and a slope that describes a unique line

reminder

so y - (-1) = -1/2(x - (-2)) and just simplify in terms of y

youre thinking of y = mx + b which is y intercept equation of a line, but use the point slope if you have slope and some random point

yes, probably simplified version would be y = -1/2 x - 2

whats the problem with that?

graphing in desmos, seems to look correct, the y intercept of the tangent line happens to be -2 yes

i suggest you review this algebra line stuff, calculus is going to eat you up if you aren't quite fluent with this stuff and just algebra basics in general

well, how does instant velocity relate to calculus concepts you've been learning?

have you only learned derivatives in terms of limit so far, you haven't learned any of the shortcuts or anything yet?

can you show the original problem statement, screenshot preferably for this?

ok so uh yeah the problem says position is s = 2t - t^2 whereas you said speed is that

that's why i asked lol

speed and position aren't the same 😄

but yeah part b) you would just do the exact same thing we did for last problem (well just need to find the slope of the tangent line not the whole equation unlike last problem), just different equation, using 2t - t^2 instead of 2/x

but just using this same thing

gotta go but good luck, if you are still having trouble for some reason feel free to open a new help channel if this one becomes inactive

[pearl pond]

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[pearl pond]

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[hollow pike]

can u solve this algebraically?

[quiet laurel]

yea

i believe so

or actually

maybe not

not without the lambert w function i don't believe so

[pearl pond]

@hollow pike Has your question been resolved?

[hollow pike]

im only in alg 2

[quiet laurel]

yea so def not lol

[hollow pike]

i just graphed it

[quiet laurel]

graphically is the way to go then

[hollow pike]

mar

[quiet laurel]

yupp

[hollow pike]

one quic

question

[quiet laurel]

mhm

hit me

[hollow pike]

how do i multiply sqrt root 10 by 3?

[forest lake]

Yes

[hollow pike]

in simplest radical form

[hollow pike]

how?

[forest lake]

Can I come to you after some time?

[hollow pike]

dm?

[forest lake]

Sure!

[quiet laurel]

its just $3\sqrt{10}$

[jolly parrot]

Mar the Marey

[quiet laurel]

thats the simpliest way to represent it

[quiet laurel]

oh my bad then sorry

[hollow pike]

wait how would -3/3sqrt root 10 be simplified?

[quiet laurel]

$\frac{-3}{3\sqrt{10}}$

[jolly parrot]

Mar the Marey

[quiet laurel]

is this what your talking about?

[hollow pike]

yea

-3/3

?

[quiet laurel]

yea so first lets simplify the -3/3 to -1

this gives

$\frac{-1}{\sqrt{10}}$

[hollow pike]

what if it was 3/3

[jolly parrot]

Mar the Marey

[hollow pike]

would it only be sqrt root 10?

[quiet laurel]

it just simplifies to +1

$\frac{1}{\sqrt{10}}$

[jolly parrot]

Mar the Marey

[hollow pike]

oh ok

thank you

[quiet laurel]

then do you know what rationalziing is

[hollow pike]

i dont think its needed

but ik what ur talking

about

multiplying

by sqrt root 10?

[quiet laurel]

yea

multiply top and bottom

[hollow pike]

i dont think we need that

[quiet laurel]

oh ok

then yea in that case this is simpliest form

[hollow pike]

take it easy

.close

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[lone nebula]

maybe its dumb question, but, is this correct

[lone nebula]

or no

[feral sedge]

Close

[lone nebula]

or just uh sinx=0 is x=kpi

[feral sedge]

It doesn't repeat every π

[lone nebula]

i see

so its just 0

[feral sedge]

sin(x) = 0 repeats every pi

[lone nebula]

okay

[feral sedge]

sin(x) = -1 doesnt

[lone nebula]

yeah so its 2kpi

[feral sedge]

remember pi is half a revolution around the circle

Yeah

[lone nebula]

okay

thanks

i got confused

[feral sedge]

Np

[lone nebula]

.close

[pearl pond]

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[feral sedge]

Usually there are 2 solutions that repeat every 2pi

[lone nebula]

.reopen

[pearl pond]

✅

[feral sedge]

For sin(x)=0 those solutions happen to be evenly spaced

[lone nebula]

yeah but

-1

[feral sedge]

For 1,-1 there's only one solution

[lone nebula]

i see

[feral sedge]

that repeats every 2pi

But anything else

sin(x)=1/2

[lone nebula]

2 solutions

yes

[feral sedge]

it's pi/6 + 2kπ and 5pi/6+2kπ

[lone nebula]

yup

just had that in exercise

thank u

.close

[feral sedge]

np

[pearl pond]

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[slate kindle]

ik what <, > sign mean but i have never seen its application in a equation

[slate kindle]

i just saw it being used in a solution

i want to know how does it work

[quick wadi]

<,> meaning a vector?

[slate kindle]

like less than greater than sign

[quick wadi]

oh i see

well x<x+2 is an application i suppose

[slate kindle]

whats an application

like use?

or its a math term

[quick wadi]

application means to use yes

[slate kindle]

ok

[quick wadi]

in a more mathematical setting, theres applied math

which is just this field of math that i guess "applies" math in daily settings if that makes sense

[slate kindle]

ok

im following

[quick wadi]

in terms of greater than and less than sign

in math terms we like to say some "inequality" holds

so x<x+2 would be an inequality

https://en.wikipedia.org/wiki/List_of_inequalities

[slate kindle]

whats meaning of inequality

[quick wadi]

heres a list of some inequalities

[slate kindle]

like literal

?

[quick wadi]

thats a good question

I can't really define it direclty

[slate kindle]

ohk

np

[quick wadi]

but just think of them as some sort of relationship using < or >

so 1<2 is an inequality

3<2 is an inequality but it isn't true

[slate kindle]

so like ratio?

or iit only applies when these signs are used?

[quick wadi]

well not exactly a ratio, but there is ratios involved

[slate kindle]

uhuh

[quick wadi]

not sure what you mean by this

[slate kindle]

i thought like ratios have a relationship of being multiplied with same number to get the actuall number

so they should be considered too

thats why i asked

but nvm that

tell me about what u were saying

[quick wadi]

Sorry I’m not following this, but basically if you have the inequality 2<3 then it’s also true that 1<3/2

There’s certain things you can do to an inequality like you would with an equation

But just know inequalities are important I guess

[slate kindle]

like (80+3)/4 -3n/4 <0

and then iin solution -3n/4 was sifter to other side like we do with equal sign

and turned to 3n/4

so i have no idea what logic this follows

cuz this is was my first time seeing it like thiis

so can u tell me how did thiis work

like does transfering work with < sign

[quick wadi]

I see

So suppose we have a inequality 1<2 notice how adding or subtracting the same number from both sides still give us a true inequality

[quick wadi]

Think about adding 3n/4 to both sides

[slate kindle]

oh

i see

but what if i trasfer 80+/4

will it still work

[quick wadi]

What do you mean by transfer

[slate kindle]

cuz the lecture i was watching said dont do it with positives

[slate kindle]

i meant like we did with -3n/4

we added both sides right

[quick wadi]

Well you can subtract things from both sides so yes

I think the issue of doing that is then you get negative on both sides

So consider this

-1<2

If we multiply -1 to both sides

Not ice how it becomes 1<-2

Which is false right

[slate kindle]

ye

[quick wadi]

So when we deal with negatives, there’s a rule to flip the inequality sign

I think that’s why the teacher said to not do it the other way perhaps you haven’t convered that yet

[slate kindle]

well they never started it

it was for a arthmatic progression question

and the info he gave was we transfer it and it becomes positive and dont move positive there

so yea thx

i think i can do more question of that kind now

thank you

[quick wadi]

I see well I recommend some videos on YouTube

And if you have an library near by or the availability to purchase some math books that might help

Np

[slate kindle]

ohk

oh hey

incase i run into a situation where i gotta deal with negatives

i should just turn the sign to its opposite right

like < to > or visa versa

.close

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[peak steeple]

Using one of the quadratic identities to rewrite the equation 1x^{2}-13x+4=0. I know that I reach the right conclusion because I checked the assignments correct answer matched. But did I do it correctly and assigned the right signs in my calculation? Or was I just "lucky" to assign a minus sign twice (which would negate the first wrong sign and result in the final result being correct anyway).

[peak steeple]

$1x^{2}-13x+4=0\ \ \to\ \ x^{2}+\left(-\frac{13}{2}\right)^{2}-13x=-4\ \ \to\ \ x^{2}+\left(-\frac{13}{2}\right)^{2}-2\cdot1\cdot\frac{13}{2}x=-4+\left(-\frac{13}{2}\right)^{2}\ \ \to\ \ \left(x-\frac{13}{2}\right)^{2}=-\frac{16}{4}+\frac{169}{4}\ \ \to\ \ \ \left(x-\frac{13}{2}\right)^{2}=\frac{153}{4}\ \ \to\ \ \left(x-6.5\right)^{2}=38.25$

[jolly parrot]

DisplayName

[uneven kindle]

What is happening in this equatino

can you state the original problem?

[peak steeple]

$1x^{2}-13x+4=0\ \ \to$

$\ x^{2}+\left(-\frac{13}{2}\right)^{2}-13x=-4\ \ \to$

$x^{2}+\left(-\frac{13}{2}\right)^{2}-2\cdot1\cdot\frac{13}{2}x=-4+\left(-\frac{13}{2}\right)^{2}\ \ \to$

$\left(x-\frac{13}{2}\right)^{2}=-\frac{16}{4}+\frac{169}{4}\ \ \to$

$\left(x-\frac{13}{2}\right)^{2}=-\frac{16}{4}+\frac{169}{4}\ \ \to$

$\left(x-6.5\right)^{2}=38.25$

[jolly parrot]

DisplayName

[peak steeple]

I wish I knew how to put space vertically betwen the lines, that would help with better overview when looking at it

It's a second degree equation that I must rewrite to one of the quadratic identities. I was given 5 possible solutions, and I did get the right solution.

The right solution is $\ \left(x-6.5\right)^{2}=38.25$

[uneven kindle]

I dont get why there is the 1 in the 1x^2 and why there is a (-13/2)^2 comes from?

[jolly parrot]

DisplayName

[peak steeple]

Those who made the assignment put the 1 there. But of course 1x = x so you can just ignore it, just like I did. They do that to confuse students Im sure, which is good

[uneven kindle]

Looks correct

[peak steeple]

(-13/2)^2 comes from a rule about square completion. In the 2nd degree equation ax^2+bx+c=0 you can move c to the left side and get ax^2+bx=-c. And then you add half of b squared on both sides: ax^2+(1/2 * b)^2=-c+(1/2 * b)

So (1/2 * b) is (-13/2)^2

[uneven kindle]

Yea I saw it when i looked closer. It's a bit messy but i see the point

To create a perfect trinomial to make it into (something)^2

[peak steeple]

What would be a less messy calculation to get the quadratic identity?

$1x^{2}-13x+4=0\ \ \to$

$\medskip$

$\ x^{2}+\left(-\frac{13}{2}\right)^{2}-13x=-4\ \ \to$

$\medskip$

$x^{2}+\left(-\frac{13}{2}\right)^{2}-2\cdot1\cdot\frac{13}{2}x=-4+\left(-\frac{13}{2}\right)^{2}\ \ \to$

$\medskip$

$\left(x-\frac{13}{2}\right)^{2}=-\frac{16}{4}+\frac{169}{4}\ \ \to$

$\medskip$

$\left(x-\frac{13}{2}\right)^{2}=\frac{153}{4}$

$\medskip$

$\left(x-6.5\right)^{2}=38.25$

Found out to make vertical space, Gives a bit better overview I think.

using \medskip

[peak steeple]

Thanks, Im not a native english speaker so my own languages terms for the math is a bit different. I need to learn the english equivalents too.

For example "x^2 + 6x + 9 is a perfect square trinomial expression, since it can be factored as (x + 3)^2, the square of the binomial (x + 3)."

[jolly parrot]

DisplayName

[pearl pond]

@peak steeple Has your question been resolved?

[pearl pond]

@peak steeple Has your question been resolved?

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[hybrid karma]

if r is the real part of a strictly complex solution of f, determine an interval I of length 1/2 so that r∈I

[pearl pond]

@hybrid karma Has your question been resolved?

[hybrid karma]

<@&286206848099549185> i'd like some help thanks :))

[warm current]

What tools can you use to solve this?

[hybrid karma]

i don't understand

do you mean which subjects i've studied?

[pearl pond]

@hybrid karma Has your question been resolved?

[pearl pond]

@hybrid karma Has your question been resolved?

[cyan ingot]

I think he's asking what techniques you can use

[hybrid karma]

i mean idk where to start, that's why i asked for help

i know that there are 2 complex solutions ( conjugates ) and one real solution so i guess if i could find the real solution that would be helpful but i don't know how to find it

[hybrid karma]

.closed

.close

[pearl pond]

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[neon blaze]

I got this answer as an integrals as no idea why it is wrong

[neon blaze]

Never mind as I see -2e^(-t/2)+C is the correct answer

[west sapphire]

no

your answer should be a number

it's a definite integral

[neon blaze]

So it is 2 then

Thank you as I was able to get the right answer for it

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[smoky saffron]

how do i solve

[smoky saffron]

the f(3) is all i haev so far

I dont know what to do with the f(u) stuff

[heavy snow]

first try writing $h'(x)$ in terms of $f(x)$, $g(x)$, $g'(x)$, and $f'(x)$

[jolly parrot]

esca (@ with reply)

[smoky saffron]

like this?

[heavy snow]

hmm are you using product rule? you wanna use chain rule

[smoky saffron]

ohhh ok

how come

[heavy snow]

yep exactly

[smoky saffron]

so

what do i do with the f(u)

[heavy snow]

solve for f'(u)

[smoky saffron]

im confsued tho

i dont nkow how to

[smoky saffron]

can you show me how

[heavy snow]

$f(u) = u^2 - 1$, so $\frac{\dd}{\dd u}u^2 - 1$ is?

[jolly parrot]

esca (@ with reply)

[smoky saffron]

2u

[heavy snow]

yep

so we have f'(u) = 2u

[heavy snow]

now you can solve for this right

[smoky saffron]

but the 3 and -1 are x's right?

how do the u and x relate

[heavy snow]

u is just a variable. you just sub in 3 for u in this case

[heavy snow]

i think your final answer is right but remember youre taking $f'(3)(-1)$, not $f'(-3)$

[jolly parrot]

esca (@ with reply)

[smoky saffron]

wait how come you cant multiply them?

[smoky saffron]

i dont understand how f'(3)(-1) would work

[heavy snow]

f'(3)(-1) = (2(3))(-1) = -6

[smoky saffron]

Ohhhhhhhh

ty!

.colse

.clos

.close

[pearl pond]

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[pulsar stump]

how to do in latek?

[rustic gate]

,, . \overline B_A

[marble bluff]

$\overline{B_{A}}$

[jolly parrot]

ekafeman

[pulsar stump]

$\underline B_A$

[jolly parrot]

Ayanokoji

[rustic gate]

disgusting

[pulsar stump]

do you have reference sheets you can reccomend? I hate having to google and sometimes being unable to find some things

like this one

[rustic gate]

i do not

i never use reference sheets

[pulsar stump]

I see

[rustic gate]

google detexify

[pulsar stump]

I did

it doesn't have it

[rustic gate]

learn more latex

[pulsar stump]

I know a sone

but the rest is just googling

I mean for signs in my courses

well thank you anyway mates

.close

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[open moth]

i made a triangle with the base having 3.5miles and the left angle being 26 degress and the right angle 42 degrees

so the last angle is 118 degrees

<@&286206848099549185>

this is my new question i got the last one wrong

<@&286206848099549185>

let me know if you guys know how to solve this

i cut the last one in half to make a right triangle but the answer was wrong

the angles add up to less than 180 if i cut it in half

do i use low of cosines instead?

law

but i have to find the height so that's why i thought i would cut it in half

[dusky ocean]

Try to draw a diagram

[open moth]

i drew a triangle with the base being 3.2 and the bottom left angle is 26 and the bottom right angle was 39

[quasi fern]

yeah that's correct

so what's the third angle going to be

[open moth]

so i thought the last angle would be 115

[dusky ocean]

You will see that the man will create a right triangle with the balloon and the woman will also create a right triangle with the balloon

[quasi fern]

ok good

[open moth]

so do i make it a right triangle after?

okay so i split it in half and half of 115 is 57.5

but 26+90+57.5 is 173.5 which is less than 180

so the answer wont be correct

am i supposed to split it differently

cause the answer wont make sense if the right triangle is less thatn 180 degrees right

[dusky ocean]

I think you are complicated thing

if we know two angles

then we can find the third angle

[open moth]

right so found the 3rd angle it was 115 degrees

[dusky ocean]

And then using law of sine

you can find two remaining sides of a triangle

If you draw a straight line from the balloon to the base of the triangle, it will create a right triangle, and from this, I think you can solve it using trigonometry.

[open moth]

okay let me try that first

okay so i did the problem and the let me send a picture of my work

the hypotnuse is more than the bottom if i make it into a right triangle tho

i mean hypotnuese is less sorry

@dusky ocean

[dusky ocean]

Where does 1.75 come from

[open moth]

i divided 3.5 in half

[dusky ocean]

Ok I understand your mistake

[open moth]

cause u said make it into a right triangle right

i think i split it wrong im not sure

[midnight haven]

You have the angle which is 39 and you have the hypotenuse (m) so you can figure the distance from the balloon to the ground

[dusky ocean]

It is not simply divide 3.5 by half

[midnight haven]

yeah because the balloon is not in the middle

[dusky ocean]

yes

[midnight haven]

it's closer to the person whose angle is 39

[open moth]

ohh okay so how would i find out the base then

do i use law of sine again?

[dusky ocean]

law of sine or trigonometry

[midnight haven]

you don't need it for the height but it's easy

[open moth]

ohh so i can use the sin of 39 degrees with the height and sine of 90 with the hypotnuse right?

so i can just get the height i dont need the base

it says the answer is wrong tho its not 5625

so i have to find the base still?

[dusky ocean]

or you can use the definition of sine

[midnight haven]

SORRY FOR MY MOUSE WRITING

[open moth]

LOLL ur good

okay so i multiplied sin(39) but 1.6929

and then divided it by sin90 which is just one

and the multiples but 5280 cause if 5280 ft per mile

i got 5625 but it says my answer is wrong 😭

idk what to do LOL

[midnight haven]

let me check aha

[dusky ocean]

Have you converted it to feet?

[open moth]

ya i did

[midnight haven]

it's not 1.6929

sin(26)/(sin(115)/3.2) is approximately 1.5478

[open moth]

OH SHIT ITS 3.2

sorry sorry

[midnight haven]

mhm

1.5478*sin(39) = 0.9741 approximately

[open moth]

okay so 5143 ft

[midnight haven]

0.9741 * 5625 is (again approximately) 5143

yes

[open moth]

damn i rly use the wrong distance haha

thanks for the help

[dusky ocean]

Ok. This type of problem always has something to do with right triangle😂

[midnight haven]

Yw. it also helps cuz I have exams soon 😭

[open moth]

LOL i just finished im doing a summer class to get a credit done

it goes faster than the ones over the year so its harder to go so fast 💀

but thanks anyways good luck on ur exams

[pearl pond]

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[sharp smelt]

My first thought was to find $u_{n+1}$ in terms of $n$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

which would be $n-1 + \frac{1}{n-1}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

$\frac{n-1 + \frac{1}{n-1}}{2} \leq \sqrt \frac{{(n-1)^2 + \frac{1}{(n-1)^2}}{2}$

I guess AM<QM isn't the way to go here?

just yes/no please

wait, is it just sufficent to show $n-1 + \frac{1}{n-1} \leq \frac{3 \sqrt{n}}{2}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

like just solving this inequality is sufficent ?

[cosmic torrent]

You take u_n-1=n-1?

[sharp smelt]

$u_n = n-1 + \frac{1}{n-1}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[muted shale]

but then u_1 is undefined

[sharp smelt]

for all n>1

[muted shale]

since $u_1=1-1+\frac1{1-1}=\frac10$

[jolly parrot]

Flappie

[muted shale]

aha

[cosmic torrent]

You got the sequence by putting terms or what ?

[muted shale]

$u_2=2-1+\frac1{2-1}=1+\frac11=2$

[sharp smelt]

oh, vm

[jolly parrot]

Flappie

[sharp smelt]

$u_3= 2 +\frac{1}{2}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

oo

$u_4= \frac{5}{2} + \frac{2}{5}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

I think I ought to do hard problems when I'm fresh

.close

[pearl pond]

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[pearl pond]

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[red trail]

if for all vector v in subspace V: <Tv, v> = <Sv,v>
does T=S?

[fluid axle]

No

You don't have a counter example in mind I assume ? @red trail

[red trail]

yeah i just got one

if u take S as the transpose of T

when T is non symmetric

[fluid axle]

Yeah that works

[pearl pond]

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[midnight haven]

How would 50 be wrong. Is it suppose to be concave up and concave down inflection point?

[midnight haven]

To put it more clearly, an inflection point is where it goes from concave to convex or vice versa

Does it seem to you like that is the case at x=50?

Noo would it be 500? Since it's starting to go back up

I wouldn't say the inflection point is at x=500 but at least you seem to understand the reasoning behind it now

That is in fact how you spot in inflection point

Assuming we're going from left to right, before it was dipping further and further down and then it started to aim further and further up a little bit

Were that switch occurs (or the reverse), we have an inflection point

Personally I'd say the inflection point is before x=500

just at a glance

@midnight haven Does that all make sense?

Yes thank you!! This is making more sense I think I just need to look at more graphs and quiz myself. Thank you!!!

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[modern pilot]

Can someone walk me through this simplification process? y, w, and s are all bitstrings.

[spiral pivot]

Can you explain what the operations do?

[slow grove]

i believe the + with a circle is bitwise addition

[spiral pivot]

Circled plus is xor addition I assume

Dot is dot product with xor reduction?

What is exponentiation?

[slow grove]

ig dot is bitwise mult

so like and

[spiral pivot]

Is this over GF(256)?

[fluid axle]

the length of the bitstrings shouldn't matter here

GF(2^5456657) if you want

[modern pilot]

yes the dot is the dot product

the circle with the plus is the xor symbol

y, w, and s all have the same size

[modern pilot]

and the exponentiation is just normal exponentiation

[spiral pivot]

Ok, then this is pretty straightforward

First line to the second line is distribution in the exponent, then reverse distribution outside of it

[slow grove]

i think the two main things to realise are that $a \cdot \left(b \oplus c\right)) = \left(a \cdot b\right) \oplus \left(a \cdot c\right)$ because $\mathbb{Z}_2$ is a group so has the distributive property
and also that $\left(-1\right)^{a \oplus b} = \left(-1\right)^a \left(-1\right)^b$

[jolly parrot]

Acman

[modern pilot]

okay so because only the parity matters here we can do the reverse distrubution

right?

[spiral pivot]

(-1)^(y.w) + (-1)^(y.w + y.s)
(-1)^(y.w) + (-1)^(y.w) * (-1)^(y.s)

Then factor out the (-1)^(y.w)

[slow grove]

field*

[modern pilot]

okay so first line to second line works because how again?

I can factor out $(-1)^y$ easily

but how does $w$ also get factored out?

[jolly parrot]

Decoway

Decoway

[slow grove]

do you understand why $\left(-1\right)^{y\cdot\left(w \oplus s}\right)} = \left(-1\right)^{y \cdot w} \left(-1\right)^{y \cdot s}$

[modern pilot]

I understand this yes

[jolly parrot]

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

[modern pilot]

I get this

yes

[slow grove]

ok then you factor out the (-1)^(y.w) term

which is how you get to the second line

[modern pilot]

oh okay I see now how that works

[modern pilot]

actually wait I need to process this a little more

so the xor will be 0 if they are the same and then the dot product will sum the corresponding terms multiplied

oh I see

so if both are 0 then it will be (-1)^0 * -(1)^0 which is 1

and if both are 1 it will be(-1)^1 * (-1)^1 which is 1

so basically nothing happens

does that line of reasoning make sense?

if that makes sense, then I get how the first line goes to the second line

[slow grove]

say $a, b, c \in \mathrm{Z}2^n$ that is $a = \left(a_1, \dots, a_n\right)$ with $a_i \in \left{0, 1\right}$ and similar for b and c \
addition in $\mathrm{Z}2$ (explicitly written as $+2$ - though here ill continuet to use $\oplus$) works $\mathrm{mod} , 2$ so $1 + 1 = 0$, multiplication works like normal
\begin{align*}
a \cdot \left(b \oplus c\right) &= \begin{bmatrix} a_1 \ \vdots \ a_n \end{bmatrix} \dot \begin{bmatrix} b_1 \oplus c_1 \ \vdots \ b_n \oplus c_n \end{bmatrix} \
&= \sum{i=1}^n a_i \cdot \left(b_i \oplus c_i\right) \
&= \sum{i=1}^n \left(a_i \cdot b_i\right) \oplus \left(a_i \cdot c_i\right) \
&= \sum{i=1}^n \left(a_i \cdot b_i\right) \oplus \sum_{i=1}^n \left(a_i \cdot c_i\right) \
&= \begin{bmatrix} a_1 \ \vdots \ a_n \end{bmatrix} \dot \begin{bmatrix} b_1 \ \vdots \ b_n \end{bmatrix} \oplus \begin{bmatrix} a_1 \ \vdots \ a_n \end{bmatrix} \dot \begin{bmatrix} c_1 \ \vdots \ c_n \end{bmatrix} \
&= \left(a \cdot b\right) \oplus \left(a \cdot c\right)
\end{align*}

[jolly parrot]

Acman

LaTeX source sent via direct message.
```Compilation error:```! Extra }, or forgotten \endgroup.
\mathaccentV ...blethree \frozen@everymath {}$#5$}
                                                  \macc@tmp \ifnum \macc@dep...
l.58 \end{align*}
                 
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.```

[modern pilot]

okay I understand now

the first step to the second step

but then from the second to the third step, $(-1)^{y \cdot w}$ just disappears

[jolly parrot]

Decoway

[slow grove]

$\left|ab\right| = \left|a\right| \left|b\right|$

[jolly parrot]

Acman

[slow grove]

and also $\left|\left(-1\right)^n\right| = 1$

[jolly parrot]

Acman

[modern pilot]

okay that makes sense

[slow grove]

because $\left(-1\right)^n$ is either -1 or 1, either of which has absolute value 1

[jolly parrot]

Acman

[modern pilot]

right

and then step 3 to 4 is trivial

okay!

I got it!

thank you so much

you've helped me twice today 🙏

[slow grove]

got the latex to compile nicely

[modern pilot]

nice!!

[slow grove]

happy to !

[modern pilot]

[slow grove]

all the best

[modern pilot]

thanks!

you too

.close

[pearl pond]

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[pearl pond]

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[red trail]

i tried solving this and failed, can anyone help?

[tropic saddle]

I mean you dont even need the inner product structure here

[red trail]

yeah i figured that but it was given in the question for a follow up question

[tropic saddle]

can you give a few equivalent conditions for a map to be diagonalizable

[red trail]

that there exists a basis that the matrix represnting the map is diagonal

[red trail]

is there something i'm missing?

[tropic saddle]

there is a nice one with the minimal polynomial

of course you can also just find a matrix representation of the map and then find eigenvalues and eigenvectors of that

[red trail]

i can't recall something like that

[tropic saddle]

that is unfortunate

[tropic saddle]

then do that

[cursive wraith]

Never heard of the minimal polynomial to be split with multiplicities 1?

[red trail]

isn't that the charestersic polynomial?

[cursive wraith]

No, the minimal polynomial

[red trail]

alright