help-26

[raw horizon]

oh wait that was the guess and check equation

we're making a quadratic rn

right

ok

yeah it should be +5x

[gusty bane]

yes

then divide

polynomial long division or synthetic, or anything else

[raw horizon]

alright 4 and 3 are roots

so 0,-1,4,3

thank you so much

:)

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[spring quail]

Hi there how do i solve this im in high school and not very good at calculus

[spring quail]

$$\frac{d}{dx}(\frac{x}{(a^2 + x^2)^\frac{3}{2}})$$

[neon iron]

is z meant to denote a complex numbe

[spring quail]

Nope

Really sorry,I’ll change it to x so its not confusing

[thorny flame]

skygazer2239

[main shale]

you can either start with quotient rule, or change the 3/2 exponent to -3/2 and do product rule

one of the parts is just x, which is simple to deal with, the other will require some chain rule applications

[spring quail]

Im not smart enough to perform chain rule on the denominator thats what i need help with i should’ve mentioned sorry for that

[main shale]

I'll just focus on $(a^2+x^2)^{-\frac{3}{2}}$ for now then

[thorny flame]

Desync

[main shale]

so the inner function is $a^2+x^2$, and the outer is $(-)^\frac{3}{2}$

[thorny flame]

Desync

[spring quail]

Ok

[main shale]

what's the derivative of the inner function wrt x

[main shale]

(I missed a minus sign in the exponent there, it should be -3/2)

[spring quail]

2x

[main shale]

good

now, the outer?

[spring quail]

No problem i got that

[spring quail]

0

[main shale]

consider it as $x^{-\frac{3}{2}}$

[thorny flame]

Desync

[main shale]

what's the derivative wrt x

[spring quail]

Ooh ok

-3/2 x^-1/2

[main shale]

not quite

the exponent is already negative

[spring quail]

Sry

[main shale]

so when you subtract 1, it should become more negative

[spring quail]

Yea yea

-5/2 right?

[main shale]

in the exponent yeah

[spring quail]

Ok

[main shale]

so putting it together, we have $-\frac{3}{2}2x\left(a^2+x^{2}\right)^{-\frac{5}{2}}$

[thorny flame]

Desync

[spring quail]

A question

[spring quail]

Why did we do this?

[main shale]

you take the derivative with respect to the inner function

the chain rule tells you how the composition of two functions varies

you need to know how the outer function varies with respect to its input (i.e. wrt the inner function)

[spring quail]

Ohk

[main shale]

as well as how the inner function varies

[spring quail]

Is this like substitution?

[main shale]

uhh

chain rule makes substitution work, I guess, sure

[spring quail]

Ty

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[spring quail]

I need to clear this up wait a minute plz

So the whole derivative becomes this..?

@main shale

[main shale]

sure

[spring quail]

Tysm

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[vivid charm]

im unsure if it's true or false. my linear algebra intuition says true, but my geometric intuition says false

linear algebra: B spans V, so B spans W, so some subset of B must be a basis for W

[gusty bane]

is this for a test

[vivid charm]

no

an open-everything quiz

[gusty bane]

also, this is false

[whole geode]

@vivid charm this is false

[vivid charm]

oh what

why?

[gusty bane]

B must span W, yes, but it is not necessarily a basis

[vivid charm]

or well whats wrong with my linear algebra intuition

[whole geode]

No member of B has to be confined to W

For instance consider a general plane in 3d space

This plane is a subspace

[gusty bane]

e.g., B = {(1,0), (0,1)}, and V = R^2
if you say W = span{(1,1)}, no restriction of B will be a basis for W
but B still spans W

[whole geode]

But might not contain any the standard basis

[vivid charm]

yea i have a counter example too, but how is #help-26 message not true

[gusty bane]

well W can be lower dimensional

so why should B be a basis

B will certainly span some set containing W

but it won't necessarily have the correct number of vectors

[vivid charm]

*mainly the first one

oh

i see

ok ty

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[shut obsidian]

If $f: A \to B$ and $A_1, A_2 \subseteq A$ with $B_1, B_2 \subseteq B$, show that $A_1 \subseteq A_2 \implies f(A_1) \subseteq f(A_2)$.

[shut obsidian]

So, let $x \in A_1 \subseteq A_2$. Then $f(x) \in f(A_1)$, but also, since $x \in A_2$, it is in $f(A_2)$.

[sterile geyser]

Good enough! Just end with "so f(A1) \subseteq f(A2)" and you're done

[shut obsidian]

Is this really all?

[sterile geyser]

yup! you could be more explicit by saying since x \in A1 and A1 \subseteq A2, this implies x \in A2

[shut obsidian]

Yeah, this is probably better, because I could change up the proof a bit and end up with

So, let $x \in A_1 \subseteq A_2$. Then $f(x) \in f(A_2)$, but also, since $x \in A_1$, it is in $f(A_1)$. Thus $f(A_2) \subseteq f(A_1)$.

[shut obsidian]

And with this argumentation, this won't work anymore

Thank you

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[honest mantle]

I found the gradient but not sure how to find the directional derivative

[honest mantle]

The directional deriv I had to look up to find

[topaz sinew]

@honest mantle Has your question been resolved?

[honest mantle]

No, I don’t know how to find the directional derivative in this situation, I had to look it up to get the solution

[sterile geyser]

If you have the gradient, the directional derivative is just the gradient dot u

in this situation its asking for the directional derivative in the direction of maximum rate of change

that's just the gradient again so you get

gradient dot gradient i.e. the magnitude of the gradient squared

[honest mantle]

Thank you

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[candid grotto]

Can someone teach me how to solve this manually

[topaz sinew]

Please don't occupy multiple help channels.

[candid grotto]

If it's of any help it's part of another integral going from zero to infinity

[raven sparrow]

u-sub should be good here. Have you given it a try?

[candid grotto]

yeah but it just keeps getting more complicated

[gray mulch]

if you substitute t = u^(10/49) , what do you get for dt/du?

[candid grotto]

(at least if I do integration by parts) but I don't know any other way

[stone verge]

maybe u^(10/49) as u^(-39/49 + 49/49) idk

[candid grotto]

-e^-t? not sure tho?

[gray mulch]

you also have to substitute the du by dt, thats why you need the derivative dt/du

[candid grotto]

oh man

I´ll try 😞

[gray mulch]

think of it as t(u) = u^(10/49) , then t'(u) = dt/du = ...

[thorny flame]

svc gkawpogklawg

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[merry phoenix]

For what values of a parameter m the function f(x) is increasing in the interval ]-3,0[?

[thorny flame]

Zachy

[vernal vale]

what have you tried?

[merry phoenix]

So first I took the first derivative

and plugged in -3 and 0 for x in f(x)>0

[vernal vale]

,w D[m x^4 - (m+2) x^2, x]

[merry phoenix]

and I got the interval that m <= 12/102

[thorny flame]

Results provided by WolframAlpha

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Requested by janniku

[merry phoenix]

and then I tried graphing it and I got that the anwser is [-2, 12/102], but I don't know where that -2 comes from

[vernal vale]

its so weird because it seems like the answer would be way lower

oh -3 0

1s

[vernal vale]

i can explain where the -2 comes from

but i dont understand how you got the 12/102

[merry phoenix]

we can trade hahaha

but my overall apprach was this:

1. prove that the gradient for argument -3 is non-negative
2. prove that the gradient for argument 0 is non-negative
3. prove that there are no points of inflection in between

so this is how i got 12/102

[vernal vale]

well heres my reasoning

say you want the derivative to be 0

$2mx^3-(m+2)x=0$

[thorny flame]

jan Niku

[vernal vale]

lets assume that we aren't looking for a root at x=0

$2mx^2-(m+2)=0$

[thorny flame]

jan Niku

[merry phoenix]

(we dont have to assume, since x=0 is not included in the interval ]-3,0[ )

[vernal vale]

this implies that $x=\pm \sqrt{ \frac{m+2}{2} }$

[thorny flame]

jan Niku

[merry phoenix]

quadratic formula, right?

[vernal vale]

yes

[merry phoenix]

i got that too, but scrapped the idea

[vernal vale]

well the important part is the discriminant

[merry phoenix]

i see where the -2 comes from

[vernal vale]

yea

[merry phoenix]

yea, i meant discriminant

[vernal vale]

you can think of this as a bifurcation if its helpful

maybe its not

but lemme find something for you here

[merry phoenix]

nope it is not

[vernal vale]

you can see the two boundaries here

@merry phoenix

the horizontal axis is m

the vertical axis is f*, places where f is fixed (where f' = 0)$

so the black lines are boundaries between places where f' is positive or negative (red and blue)

if thats nonsense dwai

but you can clearly see the appearance of additional roots of f' on the left and right, which is cool

[merry phoenix]

yea im gonna choose to not worry about it

cool, but way beyond comprehesable to me lol

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[shrewd atlas]

Object A contains 30 g of salt in 500 ml of water. And object B is unknown then concentration of salt water when these two substances are mixed is 2/35 then what is concentration of B

[shrewd atlas]

I got negative value for B when I tried to set 30/500 + B = 2/35

[cerulean basalt]

you need to apply mass balance

what is the total volume of the solution?

[shrewd atlas]

540

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[outer portal]

how many mL of water does object B have

[shrewd atlas]

700

[outer portal]

edit your question to include this information

if the problem tells you that object B has 700mL of water, then you have to put that in

[outer portal]

this is also not correct

if you add Object A + Object B together, you would then ahve to add 500mL of water + 700mL of water = 1200mL of water instead of 540mL

@shrewd atlas please take a picture or to take a screenshot of the question

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[cerulean basalt]

tried the second one, unsuccessful

[cerulean basalt]

but almost there

in the solutions they do this. which is quite confusing to me. Is my way acceptable too? at least for the first part of the exercise

basically my question is, do I really need to learn all of these?

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[ember dune]

So I took this linear algebra test a while ago and I wasn't able to solve this question, now I'm trying again and I think I solved it but I'm quite rusty and these types of problems always confuse me. I'm not sure if my answer is correct [TRANSLATION: "given g the linear transformation defined as... Define if possible, a linear transformation f:R^4->R^4 such that..."]

[ember dune]

my answer:

I'm pretty sure I got it because I tried to show that f fulfills all the requested conditions but I just wanna make sure

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[proven grove]

gallia please enlighten us on what’s troubling you

[ember dune]

I just want to know if my answer is correct

I checked myself but with these types of problems I'm never sure

[proven grove]

oh i see your questions above

yeah that’s too long for me🙏

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[empty jay]

could someone please explain how to answer part B of this question to me? i know my answer has to include what side the distribution of each employees sales is skewed to, or if the distribution is symmetrical, but i don’t really understand how to figure out skewness in stem and leaf plots

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[mortal mirage]

i know the formula and the ans . then what do i do it says calculate the total amount

yes

206.045

[wooden osprey]

pretty sure this is correct

[mortal mirage]

it is

bt

but

it said calculate the total amount

this is thw ans

but u just said that 206.045 alone is correct.

ok

y=kx

?

so what is it 0 or x+2

cubed

o

oh

lemme try to solve

nope im lost

then what

yep

then wha

[serene spade]

k=4

[mortal mirage]

no 8

how 4

?

4

i thought he said 2 cubed is 4

so then we divide both side by 32 or

or both side by 8

oh

k= 4

now i will find y

i got 108

i got k= 2 and y= 512

am i corrrect

..

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[fervent furnace]

hey can anyone help me with this problem:

[fervent furnace]

The radius and height of a cylinder are in the ratio 5:7 and its volume is 550cm cube. find its radius?

[fossil mist]

ah

do you know the formula for volume of a cylinder

[fervent furnace]

i think its pi R spquare H

[fossil mist]

$\pi r^2h$

[thorny flame]

flurry

[fossil mist]

yep

[fervent furnace]

yea

[fossil mist]

however we know that the ratio of radius and height is 5:7

let 5x and 7x be the radius and height respectively

[fervent furnace]

ok

[fossil mist]

this means that

$\pi \times (5x)^2 \times 7x$ is the volume

[thorny flame]

flurry

[fossil mist]

which is equal to the given volume

can you do it now?

[fervent furnace]

i need to find its radius

[fossil mist]

when you equate this volume to the given volume

you find 'x'

and then 5 times x is the radius

[fervent furnace]

ok let me try

22/7 times 25x^2 times 7x

hey idk how but i somehow got 12100 = 25x^2 times 7x now what should i do

hwlllo?

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[fervent furnace]

noooooooooooooooo

hello any genius here

nvm

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[fallow chasm]

Can someone please help me with this FFA question. Am I on the right track and if so what should I do next? Thank you in advance!

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[radiant marlin]

I'd post this in #advanced-analysis

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[sharp cobalt]

Using 4 colours, in how many ways we can colour the vertices of this graph such that every two vertices that are adjacent have distinct colours?

[wary tulip]

call the colors a,b,c,d.

if you color, for example, 5

what does that force upon 6,7,8?

[sharp cobalt]

C(6),C(7),C(8) are not C(5)

[wary tulip]

yea, but can you say something even stronger than that

[sharp cobalt]

But C(8) is not C(6) and not C(7)

[wary tulip]

so the inner 4 all need to be distinct

how many ways are there to make the inner 4 distinct?

[fossil mist]

if you struggle to approach your problem
you can try a simpler problem
the answer is 36

it has a similar approach and ig you can build a base with this

[sharp cobalt]

Oh i see

4! ?

[wary tulip]

yep

now if you fix a coloring on 5,6,7,8, how many ways could you color 1,2,3,4?

[sharp cobalt]

Hmm

[wary tulip]

this might help

[sharp cobalt]

Yes i did that on the paper:))

I need to fix a colour for 1

And then take cases?

[wary tulip]

yea i would prolly just think about each of the three cases for ways to color 1

[molten raven]

two of them are the same (yellow and purple), but i think casework is the best way here

[sharp cobalt]

And then i will have subcases i think

But the trick for inner vertices is good

[wary tulip]

[sharp cobalt]

I started with 1 and then i got in so many cases ://

[molten raven]

yeah without the two diagonal edges for the inner vertices the problem becomes significantly harder (or you can just use wolfram alpha to find the chromatic polynomial for a cubical graph)

[long stirrup]

i only see two cases

the outer ring has 4 colors, or the outer ring has 2 colors

ok it could be 3 i guess

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[summer patio]

given a point in Rn x_k = (xk1, xk2,...,xkn) will converge to x = (x1,x2...,xn) iff xki all approaches xi as k tends to infinity for each i

[summer patio]

now a hint is given such as

"important observation" is |x_j| <= ||x|| (1<=j<=n) for any x in Rn

how is that possible? even for R the real number line, xk while converging to x can be bigger or smaller anytime wont they?

[wary tulip]

pay attention to | | and || ||

the hint basically says the length of x cannot be less than the lengths of any of its components

if you want to imagine it in R^2 or R^3 or whatever

[summer patio]

isnt the norm and the mod value for x would be same for the euclidian distance of the real number line? is distance from origin

i mean when I am talking about only the R not the R^2 yet

[wary tulip]

yea sure

|x_j| = ||x|| for 1<=j<=1 in that case

still agrees with the hint

[summer patio]

that is hwat i am saying that x_j is only a sequence

while x is a fixed vector

x_j can sometimes be greater sometimes be smaller depending upon j

[wary tulip]

o ok i see the confusion

treat the notation in the hint independently from the question

the x_j's are the components of x

[summer patio]

awww okay that is a well explanation

but still it will still bug me that if it is possible for all n, it should be possible for n=1

wait no no uhh my bad

x_j is the individual component not a sequence as per my question

x_j ie the jth component my bad

thanks @wary tulip

[wary tulip]

[summer patio]

.close

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[versed island]

Sort of a programming question but it's not related to any language so I thought I'd ask here.

When working in a coordinate grid, you can use a ceiling function or a floor function to get which "square" of the grid you are in. Something I was wondering is how you would go about doing the same but for a hexagonal 2d grid.

[gusty bane]

part of the localization when you do it on a grid is that there is a natural way to number the grid squares by row/column

how do you identify the hexagons in the grid?

[versed island]

is that a rhetorical question?

[gusty bane]

mo

no

when you take a point and ask for the grid, the output is a tuple (row, col)

to identify the square

what kind of output are you looking for here

[versed island]

ok gotcha

basically for a given coordinate on a plane, with a hexagonal grid that is much larger than the scale of the coordinates on the plane, I'd want to be able to get the coordinate of the centre of the current hexagon

for a square grid I just used a floor division function to get a whole number for the x and y coordinate of the square, but I can see that that would be harder with a hexagonal grid

e.g. if the coordinate is at x 81 and the size of the grid is 20 units, the current square would be x 4

though I think that obtaining the coordinate of the hexagon's centre would be sufficient

the important thing is that each hexagon has its own unique x and y coordinate so I don't think it really makes a difference if a normalised coordinate is used or if the real coordinate is used

[gusty bane]

you could use a small amount of heurestics to do this fairly quickly

given an arbitrary point, you can find the closest 3 hexagon centers with a small amount of math, and then just compute which is the nearest of the three by taking their squared distances and finding the minimal one

a single formula derived from this method would be a piecewise function like a floor/ceil for the grid case, but would be pretty unwieldly

the general algorithm should be straightforward though

if that sounds useful lmk and i can expand, otherwise someone else will have to help cause im not sure i could derive a nice single formula for it

[versed island]

sounds useful for sure

worth understanding either way

[gusty bane]

alright well given a hex grid, adjacent centers form equilateral triangles

i shouldn't have started with that sorry

let me instead say that there are nicely defined rows and columns by the centers of the hexagons

sort of like this

[versed island]

I see

[gusty bane]

so given a point (x,y), you can find between which of these red and blue lines it lies

via the grid method you mentioned earlier

and in fact, given a single grid there is 2 closest centers, not 3 like i said earlier

[versed island]

yep that makes sense

[gusty bane]

all you'd need to do now is locate which grid you're in, then find the distance to the yellow corners for that grid, and take the minimal one

the actual numerics are a bit annoying but if you give me a minute i may be able to derive them, though hopefully you should too if you understand the picture im drawing

[versed island]

I had an idea but I'm not sure if i's the same as what you're getting at

[gusty bane]

real quick, are there any specifics as to the hexagons you're working with? does the grid have a hexagon centered at (0,0)? how are the hexagons defined?

as in, are you given their side lengths

[versed island]

it's a purely conceptual question atm but you would know the side lengths or the diameter of the hexagons yeah

[gusty bane]

though like you said most of this in normalized coords should be easily translatable

ok so supposing we have a hexagon centered at (0,0) with side length s, the distance between those red lines will be s * sqrt(3)/2, and the distance between the blue lines will be 1.5s

that means that if you want to find which of the grids you're in, you would take the point (x,y) and map it to (floor(2x / (s * sqrt3)), floor(y / 1.5s)), which should give you the bottom-left corner of the appropriate grid

well that's the index of those at least, so to get the actual corner you'd multiply the coordinates by s * sqrt(3)/2 and 1.5s respectively
so say we have a bottom-left corner (u,v), in order to get the other corners we simply take (u + s * sqrt(3)/2, v), (u + s * sqrt(3)/2, v + 1.5s), (u, v + 1.5s)

[gusty bane]

all that's left is to check which are actual yellow hexagon centers and which are "fake" centers

given the indeces we found earlier, (i,j) = (floor(2x / (s * sqrt3)), floor(y / 1.5s)), you should be able to just take the parity of i + j
if i + j is even, the bottom-left is a yellow center, otherwise it is a flase center

all that's left is to take the other yellow center and find the one with the smaller distance to the point. that will be the center of the hexagon the point lies in

[versed island]

yeah back here I was thinking you would have to check if you're on an even or odd rectangle to work out which corners to check

good to know I was on the same page

[gusty bane]

there may be some slight inacuraccies in the numbers i mentioned above, and some redundancy even, but the idea should be logical

[versed island]

yep

[gusty bane]

and if you really wanted to, this could be a single piecewise function, though that would probably be horrid

[versed island]

quick question, since I'm guessing this would be even easier if you know the length of one section of the red lines and one section of the blue lines, what ratio do those distances have to be in order for the rectangular grid to be "made into" hexagons?

if they're not at a specific ratio you would have a skewed hexagonal grid right?

[maiden cipher]

Right triangle where the red intersects the blue

If you know your trig functions, that's tangent

.. or it's just slope if you're not into trig yet.. rise / run

red / blue

[gusty bane]

i sort of derived the rectangular grid from the hexagons, so kind of opposite of what you said
the actual scale of the green arrows you drew doesn't really matter

[versed island]

yeah I get you it's kind of the opposite

[gusty bane]

in general, as long as the rectangular grid makes a right angle, the algorithm we discussed above should work to find which hexagon you lie in

but

if you were to care about a skewed grid (the red and blue one), all it would take would be to unskew the point, solve the problem there, then reskew

[versed island]

hmm

[gusty bane]

i also quickly wanna mention that while the algorithm above is a bit annoying and complicated, it is still O(1)

[versed island]

yeah that's good for sure

[versed island]

my thinking was that if you applied the algorithm to a rectangular grid of a slightly different length, the hexagons would be either stretched horizontally or vertically

[gusty bane]

thats right

[versed island]

so there must be a ratio of the distance between blue and red lines where the algorithm assumes regular hexagonal centers

[gusty bane]

ah i see. the ratio is sqrt(3) : 3

for an arbitrary side length s for a regular hexagon, the red lines will be s * sqrt(3)/2 units apart

the blue lines will be 1.5s apart

so their ratio would be sqrt(3) to 3

[versed island]

that makes sense, thanks for the help.

I mentioned originally that this was somewhat a programming question. So I think it would be even easier if you were to start with the distance between the blue lines, calculate the distance between the red lines, then for a particular coordinate (after working out which rectangle it is in) work out which two of the corners you need to compare the distance of to get the closest centre.

And it is definitely useful to know how to do it starting with the length of one of the sides of the hexagon, too.

[gusty bane]

well you can get the distance between the red and blue lines given the sides of a hexagon immediately

so they are essentially the same problem

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[mellow spindle]

can this be simplified

or should I just leave it as it is since it is such huge numbers

[worthy storm]

did you try writing it out in terms of factorials to see if you can cancel anything?

[mellow spindle]

no I haven't

i dont deal with probability often but the answer as I have is alright to display right?

it doesnt always hve to be super exact?

[worthy storm]

sure, unless someone asks you to simplify it

[mellow spindle]

okay

cause im looking at previous assignment solutions and like answers with binomial with like multiple powers on them are just left as it is

and I'm honestly 2 lazy to actually simplify if I dont have to 😭

thank you tho

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[cinder oxide]

did I forget something?

[worthy storm]

you can see by taking the dot product of x and v that they are not perpendicular

[gusty bane]

when you take a cross product, the j should be multiplied by -1

[cinder oxide]

i thought it's just minor and cofactor expansion

so it should be [-10, 3, -1] ?

[worthy storm]

that's how cofactor expansion works

the signs alternate

[cinder oxide]

oh right

[cinder oxide]

so this is the answer?

[gusty bane]

take the dot product with the two original vectors

see if it is indeed orthogonal

[cinder oxide]

the dot product = -32

[gusty bane]

looks like you did 7 * -2 wrong

well you did 4 * -2 and 7 * -1 or something

oh never mind i was looking at k, my bad

[cinder oxide]

2(-1) + 7(-4) + 1(-2)

but what does -32 tell us?

[gusty bane]

oh

no i meant take the dot product of the vector you found

with u

and then the dot product of the vector you found with v

to make sure it is orthogonal

[cinder oxide]

oh, that equals 0

-10(2) + 3(7) + (-1)(1) for u

and 0 for v

-10(-1) + 3(-4) + (-1)(-2) for v

so both dot products = 0

so that means the correct answer is [-10, 3 -1], right?

if the dot product of two vectors is 0, the angle between them is 90 degrees

tyvm

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[gritty crater]

Can anybody give me a rundown on just how to do basic proofs with parallelograms? I'm in a geometry class and I am completely lost on how to do this stuff. Heres an example from my homework:

[topaz sinew]

@gritty crater Has your question been resolved?

[gritty crater]

<@&286206848099549185>

[vestal oak]

ok so you need help on proofs

do you know the theorems for them?

[gritty crater]

I mean, sorta? i know like the properites that make up quadrilaterls and stuff and all the angle and side proofs. in the assignment it already has them labeled, but on the test I have I would assume that they would require us to know by memory

[vestal oak]

ok so next step would be to use the midpoint theorm to prove that BC and AD are similar i think

[gritty crater]

how would that work?

also ik its both properties but would i do BC II AD or BC ~= AD

[vestal oak]

BC~= AD

[vestal oak]

Bruh

Sorry, it’s been a while since I’ve done geometry

[gritty crater]

all good haha

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[tall knot]

Just look at it bro

https://tenor.com/view/gigachad-chad-gif-20773266

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[neon iron]

How do i solve this question? my main doubt is what do I do with the variable n

[restive inlet]

work with it like you would another number

a good place to start would be to do some factorisation on the numerator

[neon iron]

If It was just another number I'd somate both and create a new expoent

[fading zephyr]

cant we write the 2.2^n as 2^n+1?

[restive inlet]

do you know your exponent laws

[fading zephyr]

i do?

[neon iron]

Would It become 2^4n @restive inlet ?

[restive inlet]

what's "it"

[neon iron]

Can you you solve just one of the ones that have a variable so I can see how would do It ?

[neon iron]

2^n+4

[restive inlet]

n+4 isn't the same as 4n

[fading zephyr]

that will be 2^n * 2^4

[neon iron]

Can you do only the First step of this question so I can see what do I do?

😭.

[restive inlet]

do you know your exponent laws?

[neon iron]

Multiplying potentiation you somate expoent, dividing potentiation subtração exponent

[restive inlet]

consult a list of exponent laws if needed, can you find another way to express
$$2^{n+4}$$
(this form will usually appear on the right side of the equation of such lists)11

[thorny flame]

ℝαμΩℕωⅤ

[neon iron]

Like this @restive inlet ?

[restive inlet]

yes

[neon iron]

But which one of those apply to this situation

[restive inlet]

lookcing at the right sides of the equation, what does $2^{n+4}$ resemble/look the closest too

[thorny flame]

ℝαμΩℕωⅤ

[neon iron]

Power of Power rule

[restive inlet]

no

on the right side of that equation/rule, there is a product in the power

which isn't what you have here

note that in the power here, you have: $n \red{+} 4$

[thorny flame]

ℝαμΩℕωⅤ

[restive inlet]

( a sum)

which of those rules has a sum / + in the power

[neon iron]

Power of quotient?

[restive inlet]

no

[neon iron]

Quotient of Power?

[restive inlet]

no

focussing on this section, which line has a + in the power

[neon iron]

Product of power

[restive inlet]

yes

apply that here

[neon iron]

Wouldnt that be true only If 2^n+4 was multiplying instead of subtracting 2.2^n?

[restive inlet]

wdym

equality works in both directions

[restive inlet]

starting from one side, you can get to the other

i.e. with 2^(n+4), you're starting from the right side of that product of powers rule

which will allow you to express it in the form on the left

[neon iron]

I thought that I could only use product of Power rule when they were multiplying each other

[restive inlet]

from the rule
$$x^{m+n} = x^m \cdot x^n$$
and apply that to
$$2^{n+4} = \what$$

[neon iron]

But there is a minha there between then

[thorny flame]

ℝαμΩℕωⅤ

[neon iron]

Aaaaa

I think i get it

Wait

Lets me try to solve the question by myself now, wait 5 minutes

This is what you were trying to explain me, right? @restive inlet

[restive inlet]

yes

try simplfying further now

[neon iron]

Can I simplify the numerator using the denominator

?

Like cutting -2 by 2 ? @restive inlet

[restive inlet]

wdym

show exactly what you'd be doing

[neon iron]

Like this

I cutted them

[restive inlet]

no, that will not be legal

you can't just cancel / erase expressions that happen to appear "somewhere" on the numerator and denominator

fraction simplification results from cancellation for common factors
$$\frac{\cancel{p}(q+r)}{\cancel{p}(s+t)} = \frac{q + r}{s + t}$$

[thorny flame]

ℝαμΩℕωⅤ

[restive inlet]

note that you have like terms on the numerator

[neon iron]

,rotate

[restive inlet]

some multiple of 2^n - another multiple of 2^n

[neon iron]

They were common factors though

2 and ,- 2

Hi Evry one

Or numbers with different signals arent common factors?

[restive inlet]

in those terms yes, but you're cancelling 2 from a single term on the numerator
instead of a factor of 2 from the whole numerator

you can't do much clean cancellation of $p$ if you have stuff like
$$\frac{p+q}{p+s}$$

[thorny flame]

ℝαμΩℕωⅤ

[neon iron]

So I need to make a factorization first?

I'll try do one and show you, wait a second

I cant find a good vídeo about how I do factorization in this scenario 🥲

[restive inlet]

could just view it as combining like terms

how would you simplify something like
$$\red{7}\blue{x} - \violet{3}\blue{x}$$

[thorny flame]

ℝαμΩℕωⅤ

[neon iron]

Hi can any give me a perfect number
Non 6 or 28

[fossil mist]

!occupied

[topaz sinew]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[neon iron]

Are you sure ?

[restive inlet]

incorrect

[neon iron]

What do you think now? @restive inlet

Ohh

[restive inlet]

answer the question i posed above

simplifying 7x - 3x

[neon iron]

i dont know

[restive inlet]

if you had 7🍎 and someone stole 3🍎 from you, how many 🍎 would you have left?

[neon iron]

4 apples

[restive inlet]

yes

same principle applies to the simplification of 7x - 3x,
just instaed of apples, you have x there

and same principle also applies to the numerator of what you have,
just instead of apples or x, you have 2^n

$$\red{7}\blue{\text{apples}} - \violet{3}\blue{\text{apples}} = \underbrace{4}{\red{7}-\violet{3}}\blue{\text{apples}}$$
$$\red{7}\blue{x} - \violet{3}\blue{x} = \underbrace{4}{\red{7}-\violet{3}}\blue{x}$$
on the numerator you have
$$\red{2^4}\cdot \blue{2^n} - \violet{2}\cdot \blue{2^n}$$

[thorny flame]

ℝαμΩℕωⅤ

[neon iron]

Let me try do It again

What do you think now?

I'm on the correct direction ? @restive inlet

[restive inlet]

no

state the specific rules/laws are you applying to the numerator to get $2^n\cdot 2^3 \cdot 2^n$

[thorny flame]

ℝαμΩℕωⅤ

[neon iron]

Multiplying number with same base you sum the expoent

[restive inlet]

how exactly was that being applied to
$$2^4 \cdot 2^n - 2\cdot 2^n$$
or
$$2 \cdot 2 \cdot 2\cdot 2 \cdot 2^n - 2\cdot 2^n$$
to get
$$2^n \cdot 2^3 \cdot 2^n$$

[thorny flame]

ℝαμΩℕωⅤ

[neon iron]

I added the expoent of the 2^¹•2^¹•2^¹•2^¹-2^¹=2³

[restive inlet]

2^4 isn't being divided by 2

[neon iron]

Oh, só I wouldnt need to subtract the expoent?

[restive inlet]

wel you're not dividing..

division rule doesn't apply

[neon iron]

Let me do this again

[restive inlet]

what i'm trying to get you to do in this step is to simply combine like terms

it may be easier for you here to first use that 2^4 = 16

if you had 16🍎 and someone (me) snatched 2🍎 from you, how many 🍎 would you have left?

[neon iron]

,rotate

[neon iron]

What about now @restive inlet

[restive inlet]

no

when i asked you

if you had 7🍎 and someone stole 3🍎 from you, how many 🍎 would you have left?
you didn't say you'd have
4🍎🍎
did you?

[neon iron]

I did

[restive inlet]

you said just 4 apples

you will not end up with 4 alien weird hybrid double apples

you'd be left with just 4 normal apples

[neon iron]

3 apples

[restive inlet]

4

7 - 3 is 4

[neon iron]

16-2 is 14

[restive inlet]

yes

[neon iron]

So where did i do a mistake

[restive inlet]

you have 2^n twice

you will not end up with 4 alien weird hybrid double apples
you'd be left with just 4 normal apples

instead of apples, you have 2^n
so after subtraction you'd be left with 14 multiples of 2^n

and not 14 multiples of 2^n * 2^n

[neon iron]

So It would be 14 • 2^n+n

[restive inlet]

no

where's the extra + n coming from

[neon iron]

2^n • 2^n

[restive inlet]

i literally just said that's what you're NOT supposed to do

[neon iron]

Ohh

Like

[restive inlet]

if you had 7🍎 and someone stole 3🍎 from you, how many 🍎 would you have left?
you will not end up with 4 alien weird hybrid double apples
you'd be left with just 4 normal apples

[neon iron]

14(2^n)

[restive inlet]

yes

[neon iron]

But what happened to the n that were multiplied?

In this case they would be added to each other right?

[restive inlet]

wdym

like terms were combined

you seemed to understand what happened when apples were involved
exact same idea applies here

[neon iron]

I believe number of same base when multiplied you sum the expoent

[restive inlet]

this part/step was mostly irrelevant to that

[neon iron]

Why? I need to know because I have a Exam about that soon

[restive inlet]

the step i was getig you to perform wasn't really related to those exponent laws

[neon iron]

Why one 2^n would disappear and the other stay

[restive inlet]

but you were trying to improperly incorporate those laws in this step

[neon iron]

Ah

Get It

[restive inlet]

principle of combining like terms

you seemed to understand what happened when apples were involved
exact same idea applies here

[neon iron]

So we are Just combining terms now and nothing else

[restive inlet]

in that step, yes

[restive inlet]

yes

[neon iron]

I did the change but you need to explain me where the other 2^n ended when this end lol

Can I simplify all numbers by 2?

[restive inlet]

you should be able to see that 2^n is a common factor of both numerator and denominator

[restive inlet]

go back to the example i gave with apples

or if you have identical objects lying around in front of you, remove afew of them

in your numerator 2^n is that object

you have a certain multiple/number of that obect,
and a few of those are being taken away

leaving you with a lesser multiple of that object

16 bananas - 2 banans = 14 bananas
16 benjamins- 2 benjamins= 14 benjamins
16 cars- 2 cars= 14 cars
16 yachts- 2 yachts= 14 yachts

16 * 2^n - 2 * 2^n = 14 * 2^n

[neon iron]

So 16•2^n and 16•2^n•2^n have no different?

If thats true then 2^n have no value 😭

[restive inlet]

whut

$16 \cdot 2^n$ and $16 \cdot 2^n \cdot 2^n$ are very different

[thorny flame]

ℝαμΩℕωⅤ

[restive inlet]

i've been trying to say this the whole time

starting with 16cars and having 2 stolen from you doesn't result in you have 14 car cars

[neon iron]

This question is the Second case with 2^n 2^n

[restive inlet]

starting with 7 apples and having 5 stolen from you doesn't result in you having 2 apple apples

is this part of a different question now?

[neon iron]

It does result in 2 apples

[restive inlet]

yes, it results in 2 apples

NOT apple apples

[neon iron]

Ah

So

[restive inlet]

just 2 normal apples we normally see in the real world

[neon iron]

Did you cut the 2^n in the numerator with the one in the denominator?

Thanks for your help Man, I'm just stupid

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[high crypt]

is this right

[fluid belfry]

yea

[high crypt]

I can still keep going by doing factorization on (-x² + 16), right?

[half edge]

yep

[high crypt]

I gotta take out the - first though, right?

[half edge]

no need

(16-x^2)

[high crypt]

epic

thanks

oh hey steve

I didn't realize lol

[half edge]

Lol

hey

[high crypt]

tysm 👍👍

[half edge]

np

[high crypt]

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[neon iron]

Im confident I did it right?
Can someone see if I did it right?

For the Kite shown Angle EAB is
49 degrees determine the measure of the angles listed below in degrees

[neon iron]

If I didn't do it right can someone tell me a way I could do it? or a method

[azure panther]

AC and DB are perpendicular

this will help

[neon iron]

Ill return

[azure panther]

and AC is a bisector

[neon iron]

I will return in a few minutes with a new answer using the new given info

EAC is 24.5

Same with BAC

ACD is 90

[azure panther]

is EAD really EAB/2 ?

[neon iron]

Yeah
AC is a bisector so it divides EAB into two equal angles

[azure panther]

but EAB is not the angle that AC is bisector of

[neon iron]

But even if it isn't wouldnt it be those still?

[azure panther]

no ?

AC is bisector of DAB

not EAB

[neon iron]

I can't find anything else

[azure panther]

i cant help further than this sorry

[brave acorn]

Diagonals of kite intersect at 90 degrees

[brave acorn]

First one is right but not the other two

[neon iron]

ABE 49
AEB 41?

[brave acorn]

No not aeb

AEB is the angle between the two diagonals, so it will be 90

[neon iron]

Ah

I see

Im noting this info for the future

[brave acorn]

Second one is wrong

You know Angle EAB and you know angle AEB, use angle sum property of triangle and find Angle ABE

[neon iron]

so it's 90 also

?

If we do 180 - 49 + 41

[brave acorn]

180 = 90 +49 + ABE

[neon iron]

139

[brave acorn]

No I had written the - by mistake

[neon iron]

oh.

[brave acorn]

This one, do this

[neon iron]

41

[brave acorn]

Yeah

[neon iron]

Im having a hard time finding out why it's 41

But I aint that smart

[brave acorn]

You know angle sum property of triangle?

[neon iron]

yes

[brave acorn]

So Angle EAB + AEB + ABE = 180

Do you know what angle EAB and angle AEB are?

[neon iron]

I see now

thank u

[brave acorn]

Yeah

Your welcome

[neon iron]

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[sharp tapir]

from my understanding, when increasing the index of your sigma, the actual values inside should decrease adn when you decrease the index of your sigma, the values inside should increase. In the questioni above Im confused as to why theres another u^2 multiplied on the ouside

[sharp tapir]

feel like Im missing basic knowledge with these infinite series qs but I havent been able to find great resources 😥

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[radiant marlin]

you're right, but here they're not exactly shifting the indices because the (k-2)! at the bottom is the same

here it's just saying that (k-2)! doesn't exist for k=0 or k=1, since it's a negative factorial, and also in the step they pull out mu^2

it's always good to check if the first few terms of a sum make sense

[sharp tapir]

someone exaplined the factorial part a little bit ago, so I think I understand the fact that we can essentially ignore the factorial part because its invalid for certain values

but Im not sure what you mean by the fact that they pull u^2

[radiant marlin]

like the top equals the product of the bottom

[sharp tapir]

isnt the k-2 happening becuase of the sigma index shifting?

somethign like this?

[radiant marlin]

nah still no shifting just weird math simplifications, here some parts of the factorial are multiplied on top and cancel

[sharp tapir]

yea that makes sense

[radiant marlin]

usually for shifting you have to see all the k's move at the same time along with the sum bounds

[sharp tapir]

so if one of the k's isnt shifting, for example with the k-2! in this case, all of the k's dont shift?

so in this case, the ^k wont shift because the (k-2)! is holding it back in a way?

[radiant marlin]

yea that's what indicates it's not happening

[sharp tapir]

interesting

so if I wanted to shift 1 more, the (k-2)! would actually get efffected this time since it would lead to (3-2)!=1, would this then allow me to shift the ^k-2 to ^k-3

like this:

or would I still ignore it because x/1 = x either way?

[radiant marlin]

right that's what it'd look like

[sharp tapir]

nvm this is dumb

[radiant marlin]

the terms won't suddenly turn bad here no

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[radiant marlin]

in the original image it looks like they're effectively trying to make the summation look like a shifted version of the e^x series by factoring things out, just so they can plug in the value, so it's pretty confusing haha

[sharp tapir]

yea haha

in general, can you always factor out with series?

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[thick hound]

What is all other left and right truncatable numbers with ends on number 1? For example 11 is prime, and ends 1. Or 311 is a prime, and 11 is a prime, ends on number 1

[thick hound]

Or code to find this in python

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[thick hound]

<@&286206848099549185>

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[thick hound]

What is all other left and right truncatable numbers with ends on number 1? For example 11 is prime, and ends 1. Or 311 is a prime, and 11 is a prime, ends on number 1

[thick hound]

Or code to find this in python

[rigid ivy]

!nogpt

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[bitter hemlock]

i think he's asking where you discount the last step needing to be prime (1 not being prime)

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[oblique ivy]

I have the sketched graph for part A and the diff equation I came to is the second image. I plugged in 1 for x and -1 for y but got 0. Not sure what I did wrong

[oblique ivy]

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[remote agate]

yes

[zenith hollow]

no it was C

:/

[remote agate]

oh sorry reading comprehension

i am so blind

Wait

???

[zenith hollow]

???

[proven grove]

???

if it was C it would be moving 5 units to the left

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[tall plover]

For 7 h)

What should we get after we do u = 1 + e^x ?

i got like (u-1)²/u².

Is that right ?

[glossy fable]

make u=e^x

[tall plover]

it will be more complex i think

[glossy fable]

$\int{\dfrac{{e}^{2,x}}{\left({{e}^{x}+1}\right)^{2}}}{;\mathrm{d}x}$

[thorny flame]

prime

[glossy fable]

if you set u=e^x, du=e^xdx, and it cancels out with the top

$\int{\dfrac{u}{{u}^{2}+2,u+1}}{;\mathrm{d}u}$

[thorny flame]

prime

[glossy fable]

then factor the denominator, and use creative usub

$\int \frac{u}{(u+1)^2}du$

[thorny flame]

prime

[glossy fable]

$\alpha=u+1$ $d\alpha=du$ $u=\alpha -1$

[thorny flame]

prime

[glossy fable]

$\int \frac{\alpha-1}{\alpha^2} d\alpha$

[thorny flame]

prime

[glossy fable]

and now you can separate into 2 integrals due to linearity

then replace all the variables

[tall plover]

ty

[glossy fable]

the final answer should be $\ln|\left({e}^{x}+1\right)|+\dfrac{1}{{e}^{x}+1}+C$

[thorny flame]

prime

[tall plover]

yes it is that

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[kind plinth]

Does anyone have the proof of derivative of e^x?

[shrewd lance]

so do you know the derivative of a^x

[kind plinth]

Yeah

I mean I know the derivative of e^x is equals to e^x but I just need the proof

[shrewd lance]

or we can proof it too to get that and just substitute a=e

let y=e^x
take natural log both sides
ln(y)=xln(e) (lne=1)
differentiating both sides wrt x
1/ydy/dx=1
dy/dx=y (y=e^x)
dy/dx=e^x

[stone verge]

wouldn't that assume you defined the derivative of lny as 1/y already. I think that is a totally valid definition but idk

[shrewd lance]

uh. no that is the proof though and derivative of ln(anything var) wrt x is 1/(anything variable)*d(anything variable)/dx

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[glacial adder]

by definition of derivative $\lim_{h\rightarrow 0} \frac{e^{x+h}-e^x}{h}$

[thorny flame]

y0shi

[glacial adder]

$\lim_{h\rightarrow 0} \frac{e^x e^h - e^x}{h}$

[thorny flame]

y0shi

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[glacial adder]

we can factor out the $e^x$ and pull it outside the limit since it’s not being evaluated: $e^x\cdot \lim_{h\rightarrow 0} \frac{e^h-1}{h}$, the limit approaches 1 so the derivative of $e^x$ is just $e^x$

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[raven sparrow]

You can factor $\frac{1}{2^k}$ from the second and third term.

[thorny flame]

Azyrashacorki

Ava

[raven sparrow]

Hum more like $1 - \frac{1}{2^k} \left(1 - \frac{1}{2}\right)$

[thorny flame]

Azyrashacorki

[raven sparrow]

Well you can evaluate what is in the parentheses quite easily. You should be left with something already simpler than what you started with

[thorny flame]

Ava

Ava

[raven sparrow]

Yes

[thorny flame]

Ava

[raven sparrow]

$1 - \frac{1}{2^{k+1}} = \frac{2^{k+1}}{2^{k+1}} - \frac{1}{2^{k+1}} = \frac{2^{k+1} - 1}{2^{k+1}}$

[thorny flame]

Azyrashacorki

[raven sparrow]

a/a = 1

coconut / coconut = 1

So long as a or coconut isn't 0

Yeah it's probably not necessary to go the step further from the app

There's a minus sign because we're taking -1/2^k out of the two terms

And the rightmost term was meant to be positive

$- \frac{1}{2^k} + \frac{1}{2^{k+1}} = -\frac{1}{2^k} + \frac{1}{2^k}\cdot \frac{1}{2} = \frac{1}{2^k} \left(-1 + \frac{1}{2}\right)$

Well here I took -1/2^k as a factor.

But you can take 1/2^k as a factor doesn't matter

[thorny flame]

Azyrashacorki

[thorny flame]

Ava

Ava

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[languid dagger]

HELP HELP HEKP

exam is gonna end

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