help-0

[gray isle]

sqrt(3)^2 + sqrt(5)^2 = 3 + 5 = 8

[cinder sundial]

wait you said the value of sqrt(3)^2 is 3

shoudnt it be 6

[gray isle]

no

[cinder sundial]

ohhh sqrt time sqrt is the number inside it

like √3 x √3 = 3

does it correct

[vale wigeon]

yes, sqrt(3) * sqrt(3) = 3.

literally the definition of square root.

[cinder sundial]

ohh im so smart

but 2x√3x√5 got me confused

x is multiply sign btw

in my understandin, 2x√3 the answer will be 3?

[vale wigeon]

no, 2 sqrt(3) is not 3

2sqrt(3) is just 2sqrt(3)

doubling ≠ squaring

[cinder sundial]

ok

then √3 x √5

what should i do with them

[vale wigeon]

you should know that $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$

[ocean seal]

uu∀

[cinder sundial]

ohhhh

i forgot it

so its √3*5

[vale wigeon]

if you're going to use the square root symbol, then use parentheses too

√3*5 is ambiguous. it could mean sqrt(3*5) or sqrt(3)*5

[cinder sundial]

(√3*5)

does it correct

[vale wigeon]

does it correct → is it correct

[cinder sundial]

thank you

[vale wigeon]

and no, it isn't

you should put parentheses like this: √(3*5)

[cinder sundial]

so 2 x √3 is two √3

then whats the value of two √3

[vale wigeon]

,calc 2 * sqrt(3)

[ocean seal]

Result:

3.4641016151378

[vale wigeon]

$2\sqrt{3}$ cannot be simplified, you should just leave it as-is

[ocean seal]

uu∀

[cinder sundial]

ok

two √3 x √5?

can we multiply them

[vale wigeon]

you can turn $2 \cdot \sqrt{3} \cdot \sqrt{5}$ into $2 \sqrt{15}$, but that's it.

[ocean seal]

uu∀

[cinder sundial]

then √5^2, the value is 5 right

[vale wigeon]

sqrt(5)^2 = 5, yes

[cinder sundial]

ohhh im so smart

[random token]

[cinder sundial]

how do i know if a question can be converted to an different form like this

[vale wigeon]

converted*

[cinder sundial]

thank you

[vale wigeon]

to a* different form

[cinder sundial]

thank you

[vale wigeon]

anyway... there's no universal rule for that

[cinder sundial]

like (a+b)^2 = a^2+2ab=b^2

i noticed he converted a super hard question to this

and find the answer easily

[sly mantle]

**+**b^2

[vale wigeon]

^ you made a typo there

well (a+b)^2 = a^2 + 2ab + b^2 is an identity that you should know

as you get more experience, you get better at noticing when it and other identities like it can be applied

[cinder sundial]

it seems like i can use the method

for this one to find the answer

it said x and y is rational numbers, and i have to find x^2+y^2=

[vale wigeon]

$x + y \sqrt{16 + \sqrt{252}} = x \sqrt{8-\sqrt{28}} + 5$

[cinder sundial]

what it means

[vale wigeon]

i'm trying to summon the bot

[cinder sundial]

ohh

[vale wigeon]

but it's not working for some reason

[random token]

лишний пробел

после первого доллара

[ocean seal]

uu∀

[vale wigeon]

@random token спасибо

@cinder sundial is this your equation?

[cinder sundial]

yes

exactly

[vale wigeon]

okay

i'll need a few minutes

[cinder sundial]

x and y is rational

it seems like i can use the [method(a+b)^2] to find the answer,but im not sure about the process

[vale wigeon]

it'll take some trickery, i'm sure of it

[cinder sundial]

what should i do in the first step

my understandin, i gotta get rid of those sqrt

[vale wigeon]

there is no step-by-step procedure for this kind of questions, sorry

[cinder sundial]

ohhh

[vale wigeon]

however, something that i came up with is to factorize the 252 and 28

[white obsidian]

there is... you have to isolate the sqrt

[vale wigeon]

...

you never "have to" do anything, seabird

[cinder sundial]

i have to square it in order to get rid of the sqrt inside of sqrt

[white obsidian]

for step-by-step of course

[vale wigeon]

and it turns out you can write $\sqrt{252} = \sqrt{4 \cdot 9 \cdot 7} = 2 \cdot 3 \cdot \sqrt{7}$

[ocean seal]

uu∀

[vale wigeon]

and you can also write $\sqrt{28} = \sqrt{4 \cdot 7} = 2 \cdot \sqrt{7}$

[ocean seal]

uu∀

[vale wigeon]

so from here:

$\sqrt{16 + \sqrt{252}} = \sqrt{16 + 2 \cdot 3 \cdot \sqrt{7}} \ = \sqrt{9 + 2 \cdot 3 \cdot \sqrt{7} + 7} = \sqrt{(3 + \sqrt{7})^2} = 3 + \sqrt{7}$

[ocean seal]

uu∀

[cinder sundial]

what is 2X3X sqrt7

shouldt it be 2x3x7

[vale wigeon]

no it should not

$\sqrt{252} = \sqrt{4} \cdot \sqrt{9} \cdot \sqrt{7}$

[ocean seal]

uu∀

[cinder sundial]

yes 4x9x7 is 252

[vale wigeon]

just to be clear, this is NOT the result of a strict, step-by-step procedure. this relies on being able to notice a lot of things. such as knowing to split 16 into 9+7 so that what's under the big root looks like a^2 + 2ab + b^2

it looks like the roots have been set up specifically to be simplifiable like this

[white obsidian]

y you're right, i somehow read a x within the sqrt. pardon.

[vale wigeon]

anyway

$\sqrt{8 - \sqrt{28}} = \sqrt{8 - 2 \sqrt{7}} = \sqrt{7 - 2\sqrt{7} + 1} = \sqrt{(\sqrt{7} - 1)^2} = \sqrt{7} - 1$

this follows a similar logic to the above

[ocean seal]

uu∀

[cinder sundial]

ohhh 2 is the factor of 4 and 3 is factor of 3, like 2^2 is 4

and 3^3 is 9

[vale wigeon]

3^3 isn't 9, it's 27. you meant 3^2 is 9.

[cinder sundial]

but you cant do nothing about 7 so sqrt 7 is what it is

yes

[vale wigeon]

2 is the root of 4.

[cinder sundial]

thats what i means

[vale wigeon]

3 is the root of 9.

you can't do anything about 7.

(and what you really mean here is that 7 isn't a perfect square)

[cinder sundial]

thats why sqrt 4 and 9 and 7 became 2x3x sqrt7

i noticed you rearranged those numbers

if i were you, i wont be able to rearrange those like you did, how you do that

[vale wigeon]

magic

i saw 2 * 3 * sqrt(7) and i saw that i could make it the 2ab in a^2+2ab+b^2

[cinder sundial]

is there sth you can look at to rearaange

2ab is 2 * 3 * sqrt(7) right

then turn the 16 to those a^2 b^2

[vale wigeon]

a^2 + b^2

that's important

that's why i split 16 into the sum of 9 and 7

[cinder sundial]

wait, a and b cant be different?

can a and b be the same number

i know the leftover number 16 gotta to be a^2 + b^2

and 2^3 + 2^3 will be perfect

ohhhh i noticed im wrong

[vale wigeon]

2^3 isn't a square, it's a cube

[cinder sundial]

yes

im so dumb

what number should i use instead?

4^2

can a and b both be 4^2

not

then its will be 32

i cant find an number that multiply itself will get 16

[vale wigeon]

i can't find a number that, when multiplied by itself, will give 16

[cinder sundial]

thank you

whats the number that when it multiplied by itself will give 16

[vale wigeon]

4 * 4 = 16

[cinder sundial]

ohh i made a little mistake here, 16 have to be the combination of a^2 and b^2

am i right

16 is the leftover

[vale wigeon]

well... yes

splitting 16 into 9+7 is something you just have to notice, i guess.

[cinder sundial]

so sqrt9 and sqrt7 is the answer?

sqrt9^2 and sqrt7^2

[vale wigeon]

we aren't anywhere close to solving the entire problem yet

but if you wish... then yes

sqrt(9) is just 3 though

[cinder sundial]

ohhhh

i think were close to the answer

we just have to put them in place

like get them into the 2ab stuff

[vale wigeon]

once we get through this thing with the radicals, we go back to the original equation, which will look like this:

$$x + y(3+\sqrt{7}) = x(\sqrt{7} - 1) + 5$$

[ocean seal]

uu∀

[cinder sundial]

but we find the x and y is 7 and 9?

[vale wigeon]

no??

we never even talked about x and y until now.

all i did was mess with the big roots to make them look less heavy.

it was all just preparation

[cinder sundial]

ohhh

whats the next step, what should i do

i know (9+7)^2 is 16x 16

[vale wigeon]

that's irrelevant.

i want to make sure that you understand this:

\begin{itemize}
\item $\sqrt{16 + \sqrt{252}} = 3 + \sqrt{7}$
\item $\sqrt{8 - \sqrt{28}} = \sqrt{7} - 1$
\end{itemize}

[cinder sundial]

no i cant, it beyond my understanding

[ocean seal]

uu∀

[vale wigeon]

mmm

[cinder sundial]

do you simplify them

[vale wigeon]

yes! that's what i've been doing all along!

[cinder sundial]

ohhh i have to simplify

so when we simplified them we get sqrt7-1

[vale wigeon]

that's the second root. the second root is sqrt(7) - 1.

[cinder sundial]

the top and bottom are same thing right,same question but you simplied

[vale wigeon]

they're not the same thing, they're different roots

the simplification follows (almost) the same logic though.

[cinder sundial]

oh ok

why sqrt 8- sqrt28 = sqrt7 - 1

how you simplify it to sqrt 7 - 1

[vale wigeon]

sqrt**(8- sqrt28)**

ok, let me repeat myself

$\sqrt{8 - \sqrt{28}} = \sqrt{8 - 2 \sqrt{7}} = \sqrt{7 - 2\sqrt{7} + 1} = \sqrt{(\sqrt{7} - 1)^2} = \sqrt{7} - 1$

[ocean seal]

uu∀

[vale wigeon]

this is how

[cinder sundial]

i got stuck in first part to the second, why 28 became 2 sqrt 7

[vale wigeon]

$\sqrt{28} = \sqrt{4 \cdot 7} = 2 \cdot \sqrt{7}$

[ionic jewel]

have you seen problems like that before or did you just find it? i never would have thought about splitting the 8 @ ann

[ocean seal]

uu∀

[vale wigeon]

i have done problems like that before yeah @ionic jewel i guess i just picked up some intuition

[cinder sundial]

why 2 * sqrt 7 is 28

[vale wigeon]

look here.

[cinder sundial]

sqrt 7 is 14?

[alpine sable]

This is correct

[cinder sundial]

i know everythin she said is correct, but in my understandin i cant multiply sqrt 7 by 2

ohhhhh

i can but 2 * 7 is 14

or you simply them so 28 became 14

[ionic jewel]

$\sqrt{28} = \sqrt{4 \cdot 7} = 2 \cdot \sqrt{7}$

[ocean seal]

ʎuunq

[ionic jewel]

consider reading the middle step of Ann's work here

it shows how you get the 2 sqrt 7

[vale wigeon]

sqrt 7 is 14?
no, where did i say sqrt(7) = 14???

[crisp night]

Suggestion- when you write like that in steps i suggest going down a line every time you do a step, ex:
$\sqrt{28} =\ \sqrt{4 \cdot 7} =\ 2 \cdot \sqrt{7}$

[ocean seal]

ƃuǝɹo_ʎuᴉɐɥS

[ionic jewel]

a) it's not mine
b) i think that looks worse

[crisp night]

Btw how can i not include tye text in the latex?

[vale wigeon]

tye?

[ionic jewel]

you can't not include the text

afaik

[crisp night]

The*

$\sqrt{28} =\ \sqrt{4 \cdot 7} =\ 2 \cdot \sqrt{7}$

[ocean seal]

ƃuǝɹo_ʎuᴉɐɥS

[crisp night]

This is what i meant

[ionic jewel]

okay i definitively would put the equals at the beginning of the line if j was doing that but i prefer 1 line

for something this simple

if it was more complex i can see splitting it

[latent thistle]

can someone help me out with the parametrisation of this?
the curve from one is the first linked pic

[vale wigeon]

you start at (0,1), then follow the black semicircle, then the blue line, then the purple arc, then the red line

[latent thistle]

i got that i just dunno how to word it

like the anto clockwise thing

wait ill show what i have

i have all the parametrisations i think

$(sin(\frac{\pi \times t}{2})), cos(\frac{\pi \times t}{2}))
\break
-1\le t\le 0$

[ocean seal]

74ʞ∀_oɯlƎ

[latent thistle]

$t, - (\frac{1}{\sqrt{3}+1 }) \cdot(t+1)
\break
-1\le t\le \sqrt{3}$

[vale wigeon]

ouch

this tex

[ocean seal]

74ʞ∀_oɯlƎ

[latent thistle]

im sorry im learning

[vale wigeon]

it's s o w i d e

can you write it out on paper

i think i can alreay see something wrong

your parameterizations don't have unit speed

[vale wigeon]

yeah, only one of these is unit speed

[latent thistle]

which one is that?

[vale wigeon]

and also they're kinda... not connected to each other

the green one

[latent thistle]

yea they are all over the place

i just dunno how to make it clockwise

thats the graph it prodcues

[vale wigeon]

ok so...

that purple arc

is it an arc of a circle of radius 2 centered at the origin

[latent thistle]

no its a circle

with parts cut off to fit

[vale wigeon]

hhhhhhh

please

read my question

this does not at all answer my question

if you continue like this i will have to stop helping you

[latent thistle]

yes

[latent thistle]

oh sorry about that, im thankful for your help

[vale wigeon]

okay

okay

so

since your parameterizations all have to be unit speed

[tiny carbon]

Hmm... How can I know the average speed without its net displacement?

[vale wigeon]

@tiny carbon channel busy please move

[tiny carbon]

Oh ok

[vale wigeon]

@latent thistle since your parameterizations have to be unit speed, it means that the time taken to move along a curve is equal to the length of the curve

the black arc has length pi/2

the blue line has length sqrt(5+2sqrt(3))

the purple arc has length 2pi/3

the green line has length sqrt(3)

these are named in the order that they'll be traversed

do you understand what i've said so far Y/N

[latent thistle]

i think so

Yes

[vale wigeon]

okay

great

so

i'll make the following abbreviations so as not to have to write out pi's and roots and whatnot a thousand times:

$t_1 := \frac{\pi}{2}, \ t_2 = \frac{\pi}{2} + \sqrt{5+2\sqrt{3}}, \ t_3 = \frac{\pi}{2} + \sqrt{5+2\sqrt{3}} + \frac{2\pi}{3}, \ t_4 = \frac{\pi}{2} + \sqrt{5+2\sqrt{3}} + \frac{2\pi}{3} + \sqrt{3}.$

the idea is that the particle arrives at the four "breaking points" in your curve at times t1, t2, t3 and t4 respectively.

do you understand this? Y/N

[ocean seal]

uu∀

[latent thistle]

Y

[vale wigeon]

okay

great

now

for the circular arc

we will want the parameterization (-sin(t), cos(t)) for 0 ≤ t ≤ t_1

do you understand this

[latent thistle]

yes

[vale wigeon]

okay

for the line

i'll write the parameterization in terms of $t-t_1$, since this is how much time the particle spends moving along the line (not just $t$, mind you)

[ocean seal]

uu∀

[fossil oriole]

One people is ok for help me

[vale wigeon]

@fossil oriole this channel is occupied, please move.

[fossil oriole]

Ok

[vale wigeon]

$\paren{-1 + \frac{\sqrt{3}+1}{t_2 - t_1} (t - t_1), -\frac{1}{t_2 - t_1}(t - t_1)}, \quad t_1 \leq t \leq t_2$

[ocean seal]

uu∀

[vale wigeon]

@latent thistle this is the parameterization for the blue line. do you understand this? y/n

[latent thistle]

y

[vale wigeon]

great

now for the purple arc

we could use (2cos(t), 2sin(t)) or some variant thereof, but this would have speed 2 rather than 1, and we don't want that.

so we will need to divide the argument by 2.

and get the parameterization $$\paren{2\cos\paren{-\frac\pi6 + \frac{t - t_2}{2}}, 2\sin\paren{-\frac\pi6 + \frac{t-t_2}{2}}}, \quad t_2 \leq t \leq t_3$$

[ocean seal]

uu∀

[vale wigeon]

do you understand this? y/n

[latent thistle]

y

[vale wigeon]

great

and now all we have left is the green line, which is very easy

$(\sqrt{3} - (t-t_3), 1), \quad t_3 \leq t \leq t_4$

[ocean seal]

uu∀

[vale wigeon]

since we are going to the left along it

[latent thistle]

ah yep

[vale wigeon]

$(x,y) = \begin{cases} (-\sin(t), \cos(t) & 0 \leq t \leq t_1 \ \paren{-1 + \frac{\sqrt{3}+1}{t_2 - t_1} (t - t_1), -\frac{1}{t_2 - t_1}(t - t_1)} & t_1 \leq t \leq t_2 \ \paren{2\cos\paren{-\frac\pi6 + \frac{t - t_2}{2}}, 2\sin\paren{-\frac\pi6 + \frac{t-t_2}{2}}} & t_2 \leq t \leq t_3 \ (\sqrt{3} - (t-t_3), 1) & t_3 \leq t \leq t_4 \end{cases} $

[ocean seal]

uu∀

[vale wigeon]

putting it all together

[latent thistle]

omg thankyou so much

[tiny field]

Decide K so that the parabola has its highest point in x=-2 f(x)=kx^3+6x^2

Could someone give me some tips for that one

[vale wigeon]

f(x) = kx^3 + 6x^2 is not a parabola

[tiny field]

It said so in the problem

But you are probably right, if we leave out the parabola thing, how do i proceed?

The answer is also supposed to be 2

[jagged imp]

unfortunately that never has a highest point for any value of k. are you sure the question is correct?

its not kx^2+6x or something?

[tiny field]

This is the question

Roughly translated it says: Decide K so that the parabola f(x) has its top in the point x=-2

[jagged imp]

oh its looking for a k where the graph has a local maximum at x=-2

[tiny field]

yea

[jagged imp]

weird wording to call it the top or whatever but who cares

so, do you know derivatives?

[tiny field]

Yup the course is derivates so I know im supposed to do something with derivates

[jagged imp]

yeah, so at a local maximum(what the question wants), the derivative is always 0. note that the derivative being 0 doesn't necessarily imply a local maximum however. Since you want a local maximum at x=-2, you should set f'(-2)=0 and find what value of k satisfies that

(and in some cases you will have to check if this is a local maximum or something else, but the wording of this question kinda guarantees its a local maximum in this case)

[tiny field]

I get the wron answer so there got to be something wrong in there, do you see it?

wrong*

[jagged imp]

3*4=7?

quite the leap

[tiny field]

yup there it was, i added them togheter insted of multiplying, thanks

[jagged imp]

nwnw

[tiny field]

Do you have time to kickstart me with another one to?

[jagged imp]

perhaps

[tiny field]

Decide the equation for a tangent to the curve

[jagged imp]

at a certain point?

[tiny field]

When k is 4

Direction coefficient = 4

[jagged imp]

does that mean gradient/slope?

i understand that might be hard to answer cuz language barrier but is direction coefficient the m in y=mx+b in other words?

[tiny field]

yes

google says slope

xD

[jagged imp]

sure, so the slope of a tangent to f(x) at a point x=k is f'(k)

so we can find out what point its tangent to the curve at by setting f'(k)=4 and solving for k

[tiny field]

is that he same as f'(x)=4?

or is it f'(4)=4

[jagged imp]

yeah the first one just with a different variable name

there was no real reason why i chose to use k there over x

[tiny field]

Yup, when I do that i get that x =1

Should i then calculate f(1)

[jagged imp]

yeah

[tiny field]

then i get 2

[jagged imp]

so after you do that, you have a point on a line and the line's slope?

are you aware of point-slope form?

[tiny field]

y=mx+b

m=4 and b =2

[jagged imp]

it looks something like $y-y_1=m(x-x_1)$

[ocean seal]

ʎʞɐǝuS

[tiny field]

Is that how I should think

[jagged imp]

oh you can do it from y=mx+b if you aren't familiar with what i just posted

are you familiar with what i posted?

[tiny field]

Yea I will use the one you posted that is the one we are using in the course

Thanks for the help I understand what im supposed to do now

[jagged imp]

awesome

[alpine sable]

Find tan (A) when a = 8 and b = 4.

[fading zephyr]

gonna need a bit more context

do you have a picture that accompanies this?

[alpine sable]

nope

[fading zephyr]

how do a and b relate to A?

[alpine sable]

It might be these - the assignment is a bit unclear

[normal cloud]

lol the one on the top left is different to the rest

[alpine sable]

yep

[normal cloud]

it’s impossible to say the answer without knowing which side is which

[fading zephyr]

indeed

[alpine sable]

Yh lmao what is this question even

[normal cloud]

if there really is no context you should probably ask your teacher

[fading zephyr]

is it part of another question?

show the whole page where the question is

[alpine sable]

Yes

I think they mean like this

[alpine sable]

Ok I'll do that

hosliggende katete = adjacent

[fading zephyr]

the whole page, you still cut out some stuff at the bottom

[alpine sable]

it just continued

continues

nothing there

[fading zephyr]

i'm pretty sure questions 2 and 3 use the triangle in that second pic

question 1 uses the 4 triangles at the top

[alpine sable]

Probs

[fading zephyr]

that's still kind of a guess, though. this assignment wasn't designed very carefully

[alpine sable]

indeed

[fading zephyr]

you should ask your teacher to confirm, as the other person suggested

[alpine sable]

well

let's say it's like that

[fading zephyr]

the notation this thing is using is that side a is opposite to angle A

and similarly for the other letters

so put the info they give you about the sides on that triangle, and then use the SOH CAH TOA

[alpine sable]

okay

Like this?

[fading zephyr]

yeah that looks ok

then, from soh cah toa, tan (A) = opposite/adjacent

[alpine sable]

okay

tan(v) = 4/8

[fading zephyr]

careful there

[alpine sable]

v = tan^-1(0,5)

[fading zephyr]

what's v

[alpine sable]

the angle

[fading zephyr]

which of the 3

[alpine sable]

A

[fading zephyr]

you got it backwards there

with respect to angle A, the opposite side is a = 8

the adjacent is b= 4

[alpine sable]

huh

[fading zephyr]

side a is not touching the angle A

it cannot be adjacent to it

[alpine sable]

how's it like then

[fading zephyr]

as i just said

the opposite is a, the adjacent is b

adjacent means it is right beside

[fading zephyr]

no, this is wrong

[alpine sable]

no

wait

[fading zephyr]

the drawing was already correct

[alpine sable]

oh okay

[alpine sable]

yes?

[fading zephyr]

i have indicated there the hypotenuse, the adjacent side, and the opposite side with respect to A

tangent A = opposite/adjacent

[alpine sable]

8/4

[fading zephyr]

there you go

[alpine sable]

what now

[fading zephyr]

well, that's what they asked you for, wasn't it?

tan A

question 2. a

[alpine sable]

i thought they meant i had to calculate something

[fading zephyr]

you did

8/4 = 2

[alpine sable]

no

nvm that's the next one

oh

[fading zephyr]

those are 2 different things

[alpine sable]

so the answer is basically 8/4= 2

[fading zephyr]

to find the angle A you need the inverse tangent, as you did before

yes

[alpine sable]

Find sin (A) when a = 8 and c = 10.

[fading zephyr]

follow the same procedure

put the sides on the triangle, then SOH -> sin A = opposite/hypotenuse

[fading zephyr]

nope

[alpine sable]

well

i do like

a is the smallest, b is the second smallest and c is the largest side

i've got to eat, brb

[fading zephyr]

i would say to just ignore the shape of a triangle, that's just a visual aid... that is making it worse here

[cinder sundial]

can someone have a look at #help-4 , its an easy one

[drifting thorn]

A cubic function is in the form y = kx^3 + c. Find it's equation if it passes through: (1,-4) and (-2,33).

[subtle jetty]

the coordinates are in the form (x,y)

so you can substitute their values in the equation

and you'll be left with a linear system

[drifting thorn]

Oh, yes

I can see it

Thanks mate @subtle jetty

[elfin wasp]

Anyone help with this?

[elfin wasp]

<@&286206848099549185>

how do I do this one. I suck at stats

[woven whale]

the distribution for the next 7 christmas holidays would be X~B(7,0.65)

since you're investigating whether the rate has fallen you would do something liek this

you add up the bars from 0 to 2 and see if it exceeds 5%=0.05

my calculator shows that it adds up to 0.0556 which is greater than 0.05

therefore you can reject the hypothesis at the 5% level

[trail isle]

Hmm while doing a question i messed around and found this to be true for all numbers i pressed into the calculator. Is this thing i have here true or false

Where y-x > 0, a>0, a is not equal to 1

[woven whale]

well a^log_a(x) = x so yeah

[trail isle]

Ooo interesting

[trail isle]

Can you explain why this works to me a bit more? I find this to be quite interesting

[woven whale]

alright so log is a bit of an interesting function

log_a(x) would be equivalent to asking "what is the value of c if a^c = x?"

for example log_2(8) would be equivalent to "what is c if 2^c = 8"

it would be 3 in this case

[trail isle]

Oh wait i get this now

[woven whale]

and if you take 2^3 you get 8 back

it's really cool

exponentials and logarithms are inverse functions and they're pretty fun to mess around with

[trail isle]

Yeah its really interesting!

[spring harbor]

For how many three-digit integers is the middle digit the average of the other two digits?

[woven whale]

you set up a summation with the middle digit i guess

or just count; it's just 3 digits

[spring harbor]

I've tried to count and got to 13

But I am not sure how to continue

The answer is not even close to 13

[woven whale]

if you have -0- then you can't form one

but say you have -1-

[spring harbor]

111

[woven whale]

then you could have a common difference of 0

yeah there's one

and then for -2-

you could have a common difference of 0 or 1

[spring harbor]

222 333 444 555

that's what i've found

[woven whale]

that would get you 123 222 or 321

well yeah those always work

my strategy would be to find the maximum "common difference" between the digits and then count

[spring harbor]

hmmm

[woven whale]

for example if you have 3 as the middle digit you could have a maximum common difference of 3 (630) but not the other way round(036)

[spring harbor]

840 420 210

[woven whale]

yep those

if you need to set up a summation then separate into two parts, one excluding numbers with 0 and one just for those

[spring harbor]

But I guess there is no real way to immediately calculate it, because I am bound to miss one or two by doing it thiss way

[tired creek]

you could note that the sum of the first and last digit has to be even, so they both have to be either even or odd, then count the cases

[spring harbor]

yeah thats true

i

ill look into it thanks

[trail isle]

Lol im still feeling so amazed, its so cool, never had math do this to me so much

[fathom vortex]

How do I show where <v,w> is the dot product of vectors v and w, if <Av,Aw> = <v,w> and A is invertible, the equation is also true for the inverse matrix of A?

[torpid wing]

yo guys

[earnest panther]

Trevor, I believe that the idea is to use the Z-score given the probabilities of the two benchmarks.
To find it out (or, at least, to get a good estimate), you can look up the probability of an entry to be within 0 and z standard deviations from the mean under normal distribution. Once you find the appropriate z score for the given probability, you can say that the given benchmark is µ + [z score]·σ, where µ is the mean and σ is the standard deviation.
You can use the chart at the bottom of this page:
https://www.calculator.net/z-score-calculator.html
As an example, given the first piece of data that the probability of running more than 36 miles is 58%.
This means that this benchmark is slightly smaller than the mean. We will attempt to find its Z-score.
You need to find some Z-score such that the probability of being between the mean and the Z-score is 58% - 50% = 8%.
Using the chart, you can see that this is a Z-score of about 0.02, which means the given 36 mile benchmark is about µ - 0.02σ. (the subtraction is because the Z-score is negative, as the benchmark is lower than the mean)

You can perform a similar operation on the other piece of data. Notice that since there's a probability of 7% of being smaller, it translates to 93% of being larger, and so you can perform a similar operation.

Then, you have two equations with two unknowns, allowing you to figure out the mean and the standard deviation.

Awynn, are you told that this is true for any two vectors v and w? If so, consider the following:
Given two vectors v and w, what vectors can you choose to apply the <Av, Aw> = <v, w> equality on in order to achieve the desired equality <A^(-1) v, A^(-1) w> = <v, w>?

[crisp pollen]

@earnest panther Got u. Thank you sm

[fathom vortex]

Yes, the equation is true for all v,w in R^n.

So I was thinking we have the identity matrix I where we trivially have that Iv = v and Iw = w. So <v,w> = <Iv, Iw>. We could make a substitution <A(A^-1)v, A(A^-1)w>, which by our assumption regarding A preserving the dot product is equal to <(A^-1)v, (A^-1)w>. Does that seem right?

[earnest panther]

I believe you're correct. Indeed, if the equality applies to all vectors, then in particular, for each $v$ and $w$, you can apply the inequality on $A^{-1} v$ and $A^{-1} w$ and receive: $\$
$<v, w> = <A (A^{-1} v), A (A^{-1} w)> =_{\text{known}} <A^{-1} v, A^{-1} w>$.

[ocean seal]

uoɯpɐʞᴉoɹ

[fathom vortex]

Alright, cool. Thanks!

[fresh wigeon]

hey guys

can i get help in q2

[somber spoke]

how do u get the values of tan sin and cos from -360 < theta < 360

i remember there being a way where u drew the graph and solved it by symmetry

do u know that method?

i cant quite remember how to do it

ok ty

[coral pilot]

Explain pls

[fresh wigeon]

for my last problem on my hw, this is right? or is it the second option

<@&286206848099549185>

should i just assume its the third one

its been like 10 min

[earnest panther]

You are correct.
Indeed, if you approach x = 2 from the left, then the limit you get is according to x² - 5, which will give you 2² - 5 = 4 - 5 = -1.
However, if you approach x = 2 from the right, then the limit you get is according to 2x - 3, which will give you 2·2 - 3 = 4 - 3 = 1.
Since the two one-sided limits differ, then the limit at x = 2 does not exist.

As to the other answers:

1. This is incorrect, as we have demonstrated that not only does the limit not equal the value at the point, but the limit doesn't even exist, since the two one-sided limits are different.
2. This is incorrect. The value x = 2 is defined under the function, since the x² - 5 description applies to values of x in the domain x ≤ 2, which x = 2 is included in.

[fresh wigeon]

thank you so much, you are good at explaining things

[sturdy grail]

so in partial integration You would have fg-integration(f()'g)

but they have integrated f meaning x to x^2/2 why?

[earnest panther]

From what I understand, one of the functions in the product in the original integral must be chosen as $u$ and one of them as $v'$. In this case, $x$ was chosen as $v'$ and $\ln (x)$ was chosen as $u$. In this case, we get: $\$
$v' = x \Rightarrow \int v' ; dx = \int x ; dx = \frac{x^2}{2}$

[ocean seal]

uoɯpɐʞᴉoɹ

[earnest panther]

In retrospect, in case I misunderstood the question and you meant to ask why pick $x$ as $v'$ and not $\ln (x)$: This is because picking $\ln (x)$ as $v'$ would require integrating it, which is more complicated than integrating a polynomial function. Moreover, if you pick $\ln (x)$ as $u$, then inside the integral of $\int u v' ; dx$, you get to differentiate it, turning it into a power of $x$, which makes integrating it a matter of using the power of $x$ formula, which is relatively easy compared to having $\ln (x)$ in there.

[ocean seal]

uoɯpɐʞᴉoɹ

[sturdy grail]

but

v prime is derivate

so if we gonna derivate x we will get 1

not

x^2/2

oh man

they juust switched it

yeah

man

im stupid

sorry

btw why would you switch ?

[earnest panther]

By 'switching', do you mean making the: $\$
$\int\limits_{a}^{b} u v' ; dx = [uv]|{a}^{b} - \int\limits{a}^{b} u' v ; dx$ $\$
transition to begin with, choosing another function as $u$?

[ocean seal]

uoɯpɐʞᴉoɹ

[sturdy grail]

yes

like how do you know which one to choose

I forgot that part

I know it has to do with in the integral in that it will cncel out if you chose one over another

[earnest panther]

Typically, you will choose $u$ as the function that you want to differentiate in order to make integration easier. For example, in the integral $\int x \cdot e^x ; dx$, you'd like to "differentiate away" the $x$, which you can do by choosing it as the $u$ and the other function, $v'$, as $e^x$, and get: $\$
$\int x \cdot e^x ; dx = x \cdot e^x - \int 1 \cdot e^x ; dx$ $\$
In this case, we wanted to "differentiate away" the $\ln (x)$, so we chose it as $u$ and the other function as $v'$.

[ocean seal]

uoɯpɐʞᴉoɹ

[tulip sedge]

theres this rule called ilate . search it up on google itll help

[clever pulsar]

if you know a*b and b*c, can you find a b and c?

[earnest panther]

Unfortunately, you can't. Since you have only 2 equations, this is insufficient to find 3 variables.

[sturdy grail]

wait

sorry

[clever pulsar]

damn... would this be a good server to ask about least squares method for system identification? or should i ask in a control server?

[sturdy grail]

so you choose the part that is easier to differentiate

[sturdy grail]

but why didn't they do that with x?

we could have removed it, how is lnx better suit for it?

[earnest panther]

would this be a good server to ask about least squares method for system identification? or should i ask in a control server?
I am only familiar with the least squares method as a way to find the line that best represents pieces of data. In which case, I believe this would go under #calculus or #linear-algebra.

so you choose the part that is easier to differentiate
I meant differentiating the function that you want to "differentiate away" such that your second integral would be easier. In this specific case, integrating x · ln (x) is difficult, but integrating 0.5x² · (1/x) is relatively easy. We "differentiated away" the ln (x) so that we could get an easier-to-integrate expression inside the integral.

If you chose x as function to "differentiate away", then you would have to integrate ln (x), which is relatively complicated compared to integrating powers of x.

[sturdy grail]

integrate ln x is 1/x

[eager orbit]

-n=40, i dont understand how to get the answer?

[earnest panther]

integrate ln x is 1/x
Not quite. 1 / x is the derivative of ln (x), not its integral. The integral of ln (x) would be x ln (x) - x.

-n=40, i dont understand how to get the answer?
In case I understood your question correctly, you can get n by multiplying both sides of the equation by -1, which gives:

(-n) · (-1) = 40 · (-1)
n = -40

[alpine sable]

Can someone help me please

[sturdy grail]

yeah integrate of lnx is integrate(lnx *1)

through the rule of ilate

we get

lnxx-integral(1/xx) = lnx*x-x

ilate rule is the best

@tulip sedge thanks

where in ilate does cos and sin belong to?

[gray isle]

t for trig

also liate not ilate

[clever condor]

How would I go about solving this?

So far I've defined u as 4x - 2

du = 4 dx

[gray gorge]

Try u = x² - x + 2 instead

[clever condor]

Uhh, okay so now what,
du = 2x - 1 dx

[gray gorge]

Can you write 4x - 2 in terms of 2x - 1?

[clever condor]

ya, 2du

[gray gorge]

Yes

So now what integral in terms of u do you have?

[clever condor]

Uhh, 2u?
I'm not sure I understand your question, math in english is hard haha

[gray gorge]

Uhh what's your native language

[clever condor]

danish

I hardly remember all the math terms in danish, let alone in english

But if I were to say 2du, wouldn't I have to multiply the rest with 2 as well?

[gray gorge]

No

[clever condor]

Am I completely wrong here

Oh oops wait

[edgy cape]

when you do integration by substitution how do you know what function to chose from to let u = . for example: ∫[1-sin(x)]*cosx_ dx_ , is there a quick way for me to know which to make u = instead of using trial and error

[alpine sable]

Know derivatives very well so you know what to expect

[edgy cape]

ok ty

[alpine sable]

$1-\sin{x} = u\$
$-\cos{x} dx = du\$
So you get $- \int u du$

[ocean seal]

ɯƎ

[alpine sable]

It also sort of becomes obvious once you solve a lot of them

[thorn storm]

wouldnt it be u*sec(x)?

[alpine sable]

We dont want to have an x if we substitute

cosxdx is in the original integral, and when we find the derivative of our u sub we get that -cosxdx is du

So cosxdx is -du

[crisp pollen]

anyone know how to do this?

[manic quail]

Do you have any idea on how to approach this problem, @alpine sable ?

At first you have to calculate the function at x=3 and x=8

then you just calculate the difference between those in percent.

why +8?

+13

so y(3)=22 and y(8)=37, right?

then 22 + how many percent is 37?

and solve for x

ahm no

that would be equal to 35.98, not 37

it is close though

,w 22 plus how many percent is 37?

forget the equation I wrote I fucked up xD

$22 + x \cdot \frac{22}{100} = 37$, this should be correct now

[ocean seal]

uosɹǝdʎddɐɥʎɹǝʌ

[manic quail]

around 69, it is not exact

If you solve that equation, you get x=68.182

,w solve 22+ x*22/100 = 37

[ocean seal]

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[alpine sable]

Can we not multi post

[stuck gate]

i know since it stretches directly we need to use y=kx but i am confused with the variables given in the problem

rn i have d=40/3*5

[sly mantle]

@zinc tundra you've multiposted right after being asked to not do that. knock it off, also don't spam questions

[fading zephyr]

you should use the given info to find the spring constant first

you know that F = kx, yes?

they have given you the distance x, and you can find the force exerted by the mass of 3kg assuming this is happening on earth

[opal sentinel]

Hey guys, so Im solving this ODE. And I have a complementary solution of y_h = e^4x+e^x

The particular integral I got is y_p = -e^2x/3 +(e^4x)*x

So when I was gonna on check my answer on symbolab they omit the -e^2x/3 of the particular integral

there is a screenshot

Would really appreciate your help on this 🙂

[spare fern]

What's the ODE

[opal sentinel]

ordinary differential equation

[spare fern]

I know what ODE stands for

[opal sentinel]

Oh Im so sorry

[spare fern]

I'm asking what's the given ODE

[opal sentinel]

I missread your message

this one here

[sly mantle]

how'd you get y_p

[opal sentinel]

I use this expression

W is the wronskian and y1 and y2 the complementary solutions

h(x) would be 3e^4x

[sly mantle]

can you show work in using this formula

[opal sentinel]

sure

there you go

[sly mantle]

https://i.gyazo.com/9f1068cab6f1556fb8f35256507bc71b.png bad algebra

[opal sentinel]

im such an idiot

cant belive I didnt pick up on that

Thanks man

owe you a big one

[sly mantle]

no prob

[alpine sable]

just for fun how do i create a equation that's like:
everytime it hits a new number on the x axys on this 2d graph its y axis limit goes up by 1(while the x axis limit is infinite)
but also so the line travels at a 25 degree angle.

[pearl epoch]

<@&286206848099549185>

[jade wadi]

in 3:4 ratio
70 degrees
28 units
32 units

[stuck gate]

i can solve this by using the quadratic formula right?

[jade wadi]

yes

[agile compass]

quadratic formula can solve any quadratic equation

[stuck gate]

@jade wadi can you walk me through this one? i have some of the steps down

[jade wadi]

sure

[jade wadi]

well, you can also factorize this one

[stuck gate]

i accidently posted my work in #help-1 @jade wadi

[jade wadi]

np

let me see

[jade wadi]

you are doing it right as in the last step, it would be 16+-4/6

[stuck gate]

According to my calculator I should be getting 10/3 and 2 @jade wadi

[jade wadi]

yes is the answer you are getting not matching??

16+4/6=10/3

and

16-4/6=2

see they are the same answers @stuck gate

[stuck gate]

Oh I am stupid

the / was throwing me off 😂

It’s a fraction

[jade wadi]

yes

shall i write it and send?? if you are getting confused

[stuck gate]

No it’s okay

I understand now

[jade wadi]

okay np

[jade wadi]

factorizing such equations would be easier

[stuck gate]

@jade wadi like for this one. When I check my work I see that it should be -11 and 7 but I have the revered

It’s always the last steps that mess me up 😭

[jade wadi]

huh, your way is different let me send my way

sorry for the best image quality

[alpine sable]

First person is right cause he's doing complete the square method

U did the other method forgot what it's called

[jade wadi]

i don't use that method lol

[jade wadi]

idk i have practiced this only

[stuck gate]

@alpine sable thank you

Ya we have to complete the square

[jade wadi]

i see

i don't know how to do the completing the square method efficiently as i have not practiced it, but the way I did, i am good at that

[jade wadi]

in the last step you messed up, just do -9-2 and do +9-2 to get the answers

[agile pelican]

Any math gods care to explain how to solve this complex equation?

[manic quail]

$210 \cdot 291 \cdot 10^6$

[ocean seal]

uosɹǝdʎddɐɥʎɹǝʌ

[manic quail]

,w 210 \cdot 291 \cdot 10^6

[ocean seal]

Results provided by WolframAlpha

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Requested by veryhappyperson#1152

[manic quail]

Must I elaborate or is this enough, @agile pelican ?

[agile pelican]

Sorry I was afk @manic quail

[alpine sable]

how do i solve 2^x / 2^9 = 2^2x

[manic quail]

Do you have any idea how to approach this, @alpine sable ?

[alpine sable]

uhhh

probably

remove base 2

[manic quail]

but that doesn't work yet

[alpine sable]

this question is a lot easier if i write it on paper

well first thing u do is solve the exponents

cause the base is same

so u ignore it

2^2x means 2 to the power of 2x

not 2 and then x btw

[manic quail]

wait, let me write that real quick

[alpine sable]

kk

i can take a pic

or write it and then show me

[manic quail]

$\frac{2^x}{2^9}=2^{2x}$

[ocean seal]

uosɹǝdʎddɐɥʎɹǝʌ

[alpine sable]

yes

thats it

[manic quail]

you have to first do something about the fraction

what would you suggest

[alpine sable]

uh

2^x minus 9

and just remove the bottom 2

[manic quail]

that's one correct way: $2^{x-9}=2^{2x}$

[ocean seal]

uosɹǝdʎddɐɥʎɹǝʌ

[manic quail]

and then remove the 2, also correct

[alpine sable]

yup

[manic quail]

If you know how to do it anyway, why did you ask? xD

[alpine sable]

uh

cause i have to solve it

i had one problem

ill take a picture so its easier to explain

idk what to do after

this

[manic quail]

xDDDDD

you do minus x on both sides

[alpine sable]

o

so then

its

2 = -9

[manic quail]

$2x-x=x-x-9$

[ocean seal]

uosɹǝdʎddɐɥʎɹǝʌ

[alpine sable]

so its

[manic quail]

$x=-9$

[ocean seal]

uosɹǝdʎddɐɥʎɹǝʌ

[alpine sable]

wait

isnt it 2 = -9

[manic quail]

no, -9 is not 2

[alpine sable]

how

im pre sure my brain shut down lol

[manic quail]

you do minus x on both sides, then you get: $2x -1x =1x -1x -9$

[ocean seal]

uosɹǝdʎddɐɥʎɹǝʌ

[alpine sable]

OHHH

I FORGOT

X IS 1X

OHHH

crap im dumb

lol

[manic quail]

xD

[alpine sable]

ohh ok thank you man

😄

[manic quail]

No problem.

[patent burrow]

i'm a little stuck here because my answers aren't showing up as correct

so far i've done

sin(pi/4 - x) = -sqrt3/2

[manic quail]

,w solve sin(pi/4 -x) + sqrt3 = 0

[ocean seal]

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Requested by veryhappyperson#1152

[alpine sable]

@manic quail in exponential equations, what happens if both bases arent the same and u cant divide them by each other

so like one is 3 and another is 10

what happens there

how

[manic quail]

you use logarithms, @alpine sable

show me the question, maybe something else can be done

[alpine sable]

LOGARITHMS??

bro im in grade 8 what is that

learn it

[manic quail]

just show me the question -.- maybe in another channel

[alpine sable]

ok

learn logs rn

ez

well ill make one up ig

EZZZZZZ

<@&286206848099549185>

how

2^x / 3^9 = 2^2x

the one i just asked u to help me with lel

just with a different base

[manic quail]

you use logarithms here, the inverse function of exponentials

[alpine sable]

ok so what does that mean

[manic quail]

there is no way around them, I think you will learn them soon

[alpine sable]

so like reciprocal?

oh

is what ur going to tell me extremely difficult

[manic quail]

like $\sqrt{x}$ is the inverse function of $x^2$

[alpine sable]

ohhhh

[ocean seal]

uosɹǝdʎddɐɥʎɹǝʌ

[alpine sable]

wait thats not too bad

so i just

wait

whats the inverse function of x^3?

cubed...

[manic quail]

@alpine sable just use Pythagorean Theorem

[glass lichen]

x^2 with restricted domain*

[manic quail]

pfffffffffff

[alpine sable]

restricted domain?@??!

bro these terms lol

[manic quail]

nothing be restricted there are complex numbers

but anyway, don't worry about it

[alpine sable]

ok

[alpine sable]

its not a right triangle tho

[glass lichen]

x^2 doesnt have an inverse function if you dont restrict the domain, since x^2 isn't injective

[alpine sable]

crazy how changing one thing in the equation makes it completely different

[manic quail]

pffff, bijective, injective, same thing

[glass lichen]

they arent

[manic quail]

both has jective in it

@alpine sable you can make two though

[alpine sable]

@manic quail imma see if khan academy has a video on it

[alpine sable]

you can solve this without logs and without square root

if u didnt learn logs in class then u need to do it the way they want u to

u set t = 2^x

and then u have a quadratic equation (3^9)*t^2 - t = 0

[glass lichen]

yes it's a quad. in disguise, however unless you get t=obvious power of 2, you do need logs

[alpine sable]

what does quadrtatic mean

a quadratic is this: x^2 + x + 1 = 0

[glass lichen]

$ax^2+bx+ c$ is a quadratic in x

[alpine sable]

that ^

o

[ocean seal]

1llᴉɥsoɯ

[alpine sable]

ok wait

[patent burrow]

i think you missed the 2

[alpine sable]

let me show u the equation i want to solve

u can solve those by factoring, quadratic equation, completing the square, grouping, etc

let me take a pic of it

[manic quail]

@alpine sable do you see what I mean? You can make two right triangles out of that one and then use pythagoras

[alpine sable]

just changing the bottom to 3 instead of 2

,w (2^x)/(3^9)=2^(2x)

[ocean seal]

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[alpine sable]

i have never seen symbols like that in my life

[alpine sable]

but then i only have 2 sides so idk how to do

that big ass Z lol

[glass lichen]

No clue why you felt the need to post the answer @alpine sable

[manic quail]

You only need two sides for pythagoras

[alpine sable]

oh right

whats the problem with posting the naswer

ok thank u !

[glass lichen]

bad practice when helping

if you want to help, actually help

$\frac{2^x}{3^9}=(2^x)^2 \implies \frac{t}{3^9}=t^2, t=2^x$

[ocean seal]

1llᴉɥsoɯ

[alpine sable]

wait is log just the amount of times ur multiplying the base to get the desired number

is that basically it

thats not so bad lol

[alpine sable]

actually i still dont know what to do because i have 2 variables

[glass lichen]

$log_a{b}=c \iff a^c=b$

[ocean seal]

1llᴉɥsoɯ

[alpine sable]

what does t mean

[glass lichen]

I defined t. . .

[alpine sable]

oh yes

oops

[manic quail]

$h = \sqrt{(3x+7)^2 - (\frac{x-4}{2})^2}$, @alpine sable

[glass lichen]

$t^2-\frac{t}{3^9}=0\to t\left(t-3^{-9}\right)=0$

[ocean seal]

1llᴉɥsoɯ

uosɹǝdʎddɐɥʎɹǝʌ

[alpine sable]

how do I solve for x

[manic quail]

I might miss something, but I don't think you can. Don't you have any other information?

[alpine sable]

it was (2^x)/(3^9)=2^(2x)

right

if it was u can solve for x

how do we know when to

use logs

in a question

[glass lichen]

exponentials

exist

[manic quail]

no, the triangle, mr monkey

[alpine sable]

oh

maybe i can set up system of equations for 2 variable

with sohcahtoa

nvm idk which side would be adjacent to the 90 degree

[manic quail]

If you think about it, you only know "the proportions". There could be multiple triangles like this, just bigger or smaller, I think.

[alpine sable]

yea i am probably missing info, its a random screenshot i found

[sonic halo]

In Dirac brakets, can I do : -|f> = |-f> ? If yes, what property can I call this?

[alpine sable]

https://gyazo.com/5c9d4c311e73df140cd7e70102c38113

What am I doing wrong. Idk what to do for the next two

[alpine sable]

Answer pls

[manic quail]

pythagoras

[alpine sable]

what's 1+1

Can someone explain why it's 2

[alpine sable]

If S is a subspace of all polynomials that have degree at most 3 , and it is 2 dimensional and satisfies p(1)=p(2)=0, for those polynomials p(x). What would be the basis of S?

Would the set {[1 0], [2 3]} span S?

Those are column vectors

[solar cradle]

@alpine sable So you'll want "vectors" that give all possible polynomials of degree $$\le 3$$. These won't be vectors in the conventional way you're probably think about. Consider $$e_{1} = 1, e_{2} = at, e_{3} = bt^{2}, e_{4} = d t^{3}$$ and find the value of the coefficients that would span this space

[ocean seal]

snɹoʇuᴉlƆ ʎllᴉB

[alpine sable]

So basically solve a system

Plug in 1 and 2 for t?

[solar cradle]

Sort of, you're given $$p(1)=p(2)=0$$, and we're interested only in polynomials of degree less than or equal to 3. This tells us that all polynomials satisfying these two properties will be of the form $$p(t) = (t-1)(t-2)(at-b)$$. Expanding this polynomial out we'll find $$p(t) = bt^{3} - at^{2} + 3at - 3a$$. We want basis vectors that are linearly independent. Note the vectors $$e{1} = 1, e{2} = at, e{3} = bt^{2}, e{4} = d t^{3}$$ can span this whole space. That is $$p(t) = \alpha e_{1} + \beta e_{2} + \gamma e_{3} + \delta e_{4}$$. Now just find the coefficients $$a, b, d$$ and you'll have your basis. Hope this helps!

[ocean seal]

snɹoʇuᴉlƆ ʎllᴉB

[solar cradle]

@alpine sable

Oops, I made an error in the second expression of p(t), just expand out the first polynomial and you'll find the correct expressions. Then just apply the rest of the strategy.

[alpine sable]

Gotcha thanks

[wet gorge]

I'm trying to show that
$\int_0^\pi \frac{Rie^{i\theta}}{R^2e^{2i\theta}+a^2} d\theta = 0$
in the limit of $r \to \infty$, however I am using this to try to find an upper bound of 0 for the integral, and the arc is actually over a length of $R\pi$, so I really get

$R\pi \int_0^\pi \frac{Rie^{i\theta}}{R^2e^{2i\theta}+a^2} d\theta = 0$

[ocean seal]

sǝlƃuɐp

[wet gorge]

This integral has to be 0 for the answer to be right, but I can't seem to see how R^2/R^2 + a^2 = 0, or can i take that limit before multiplying the other R in?

That feels like cheating

a is real also

[solar cradle]

@wet gorge, are you certain that your set up is correct? Evaluating the last integral I get 2 \pi

[wet gorge]

Heres my idea, i have to evaluate that 1/x^2 +a^2 term using contour integration

So I set it up liket his, and it's also of course just a trig integral, so I plugged that in to check, and I get that the answer should be pi/a

But pi/a is simply the residue, so I am assuming that the \theta integral should be 0

cause that first dx integral is def = pi/a, and that's def the residue according to wolfram after checking

It is possible that I am messing up the theta integral though (or its setup in the first place).. I tried finding a way to use Jordan's lemma but couldn't

[solar cradle]

I see now, you don't need to multiply by R \pi in the original message. The integral that you have already describes the behavior over that piece of the contour. And that integral does tend to zero.

@wet gorge

[wet gorge]

Hmmm how come? I was trying to show that it goes to zero by taking

integral < max(integrand)*arc length

Is there a diff way to do that?

[solar cradle]

Your strategy doesn't seem to work here. It definitely sound and correct but will only work if max(integrand)*arc length goes to zero which it does not in this case. Any easier way to see that the integral goes to zero in this case is to majorize in the following way; $$ \vline \frac{Rie^{i \theta}}{R^{2}e^{2i \theta} + a^{2}} \vline \le \frac{R}{R^{2} + a^{2}}$$ Taking the limit as $$R \rightarrow \infty$$ then shows that the integral tends to 0.

[wet gorge]

Ohhhhh

[ocean seal]

snɹoʇuᴉlƆ ʎllᴉB

[wet gorge]

Soo the whole <max * arclength is really better for when the integrand doesnt already go to 0 itself

because it's more or less unecessary if the integrand itself already does?

[solar cradle]

Yes and no, if the max * arclength does tend to zero then the integral will go to zero. It's just that if you evaluate it and max * arclength = b, that result doesn't give you any information other than integral < b. Even in this case, you find that it is a valid statement, it just doesn't show that the integral tends to zero.

[wet gorge]

Right right that makes sense

awesome well thank you very much for your time, i appreciate it

[ocean seal]

˙ʎɹɹɐq

[indigo slate]

<@&286206848099549185>

[alpine sable]

For a right-angled triangle, ABC, https://imgur.com/a/vt1N7Z8 cos A = 4⁄5

Determine
a) cos B
b) sin (90°-B)
c) tan A

[untold warren]

real quick

dont think i entirely understand how I can go about this

[wheat rivet]

why is the answer to this negative

[cinder sundial]

i need an handsome person to solve me out the hot question i posted in #help-3

that person must be super handsome, so that hes able to solve the hot question

[wind hare]

What do you have against the aesthetically challenged?

[cinder sundial]

wdym, everyone is handsome and pretty, but the one that can solve the question i have posted in #help-3 gotta be the most handsome person in the world (today

[coral pagoda]

You've been around for a while. Don't do this. :(

[cinder sundial]

what you means @coral pagoda

[coral pagoda]

I think you know very well what I mean. Don't ask people to help from other channels, especially back to back.

[cinder sundial]

no, i thought you were talkin bout the joke i made, and im sorry

i thought i could do that like ask ppl check diffrent channel,

i wont do that again, sorry

[coral pagoda]

You're bribing unrelated compliments if ya get help. That also is a bit debatable

[unkempt viper]

i’m trying to find x

I can’t find x, i’ve tried it multiple times, please help