great
#help-0
long roost
vale wigeon
long roost
45 at the apex
adding up to 360
Next step is to identify the side length and try and find the apothem correct?
vale wigeon
no, this is incorrect.
you said you would draw lines from the center to the vertices, but instead you drew lines to what appear to be the centers of every side.
long roost
The vertex is the common point of two line segments correct?
vale wigeon
and the shapes your octagon was split into are not triangles. they are instead quadrangles, whose area is definitely harder to find than what we are going for.
yes, a vertex is a corner.
long roost
Ok you wanted me to draw the lines from the vertices of the shape and not the triangle
long roost
This, is true
I just want to know
where I am going wrong in my solving of this equation
I understand
I am not a A+ graph drawer
can somebody please explain how I solve this problem in some capacity so I can figure it out?
vale wigeon
I am not a A+ graph drawer
it doesn't take an A+ graph drawer to follow instructions properly
we have not yet set up any equations either
vale wigeon
as far as the drawing goes, I TOLD YOU WHERE YOU WENT WRONG.
vale wigeon
ok then allow me to show the next step in the drawing
actually draw the apothem in one of the triangles. doesn't have to be all, just one will suffice.
(in fact do NOT draw all the apothems unless you like your diagrams cluttered and unreadable)
the apothem is, at once, the altitude, the bisector and the median for this isosceles triangle. (they have this very nice property for the line drawn from the apex to the base.)
last ether
Oh dear
vale wigeon
do you want a list of steps or do you want to understand whats happening and why things are what they are?
long roost
yes.
long roost
Ok I think understanding what is happening and why could be useful
long roost
we don't truly know
vale wigeon
the reason we are doing this at all is thus:
we have split the octagon into 8 copies of the same triangle.
if we now find out the area of the triangle, we can then get the area of the octagon by multiplying the area of the triangle by 8.
do you understand this? yes/no
long roost
yes
vale wigeon
ok
one of the formulas for the area of a triangle is the half-base-times-height formula (for which i know of no better name other than this somewhat clunky but most descriptive name)
the base of our triangle (or rather, the side most natural to choose for the base) is the side of the original octagon.
we know its length; we are given it in the problem directly.
what we don't know yet is the height, which in the context of regular polygons is also called the apothem.
do you follow? yes/no
long roost
yes 12 is the base
last ether
You also know that the radii of the octagon are equal, so you'll have an isosceles triangle.
vale wigeon
You also know that the radii of the octagon are equal, so you'll have an isoscel...
this has been said already
vale wigeon
actually draw the apothem in one of the triangles. doesn't have to be all, just ...
now look at this diagram again.
the apothem is, at once, the altitude, the bisector and the median for this isosceles triangle.
because it is an altitude, it splits the isosceles triangle into two smaller triangles that are right.
because it is a bisector, the angle at the apex gets split exactly in half, hence my marking of the yellow angle as 22.5°.
because it is a median, the segment opposite the yellow angle is half the polygon's sidelength, which in your particular problem means it's 6 cm.
do you follow? yes/no
long roost
I follow on all of those except the 3rd/last one about the median
Wait no I get it
Splits the base in half
Sidelength is 6 instead of 12
Ok I am following
vale wigeon
allow me now to draw closer attention to this small triangle by drawing it alone
here a is the apothem (which we want) and s is the polygon's sidelength (so the side opposite the yellow angle is, as i said already, half that, so s/2).
now you mentioned right-triangle trigonometry at some prior point in our discussion.
this is where it comes in now.
namely: $\frac{a}{s/2} = \cot(22.5\dg)$.
ocean sealBOT
Ann
long roost
what the fuck is cot?
vale wigeon
cotangent.
long roost
isnt that cos
vale wigeon
no.
i said cotangent, not cosine.
writing out the ratio for cosine would involve the hypotenuse, but the hypotenuse is irrelevant to us right now.
long roost
I am not familiar with the term cotangent
vale wigeon
are you familiar with the term tangent, then?
long roost
It sounds familiar but I am unsure of its meaning
vale wigeon
$\frac{s/2}{a} = \tan(22.5\dg)$
ocean sealBOT
Ann
vale wigeon
will this invoke a similar revolted reaction from you?
long roost
No because tan is opposite over adjacent
vale wigeon
great. does that mean you understand the equation i just wrote?
long roost
Yes
vale wigeon
half base times height.
long roost
ok then multiply by n (number of sides)
vale wigeon
yes, as i said.
long roost
Ok I have found a hiccup
6/a = tan(22.5) in this instance
I am unsure how to isolate the variable because i cant multiply both sides by 6 because it is the numerator
last ether
Raise both sides to the -1 power, multiply by 6
long roost
I have never done that before.
If i raise something to the -1 power
I am just multiplying them by -1 no?
last ether
No
long roost
which would swap the fraction
last ether
You just flip the fractions around
long roost
but there negative now?
last ether
No
long roost
ok
ocean sealBOT
Umbraleviathan
vale wigeon
I am unsure how to isolate the variable
how come you said no to my offer to explain it
last ether
The pain never ends
long roost
Ok I got it wrong
I did something wrong
a/6 = 1/tan(22.5)
multiply both sides by 6
6tan22.5= a
6(side length) x 6tan22.5 = a
1/2 b x h x 8?
lone heartBOT
@long roost Has your question been resolved?
marsh sedge
Hi! I have a very simple question that i couldn't answer myself, i hope someone can help me. Thx!
i'm having problems understanding how to get the domain of f(x)=((3x+1)/(2x+1))^((2x+1)/(x-3))
in first place, x can't be -1/2 to avoid dividing by 0.
same case for x=3.
then i get confused. I know the answer is Df= (-inf; -1/2) U (-1/3; 3) U (3;+inf)
i figured that x can't be -1/3 because if the exponent is negative, ((3x+1)/(2x+1)) can't be 0, but i don't understand why x can't take a value within (-1/2 ; -1/3).
I hope i was clear enough. Thanks for helping, have a nice day!
waxen flame
$f(x) = \left(\frac{3x+1}{2x+1}\right)^{\frac{2x+1}{x-3}}$
ocean sealBOT
Kookiemon
waxen flame
Is that your equation?
marsh sedge
It is, thx
waxen flame
Where did x=-1/3 come from? Did you mean x=-1/2?
wary stream
To answer your question, it won't give a real number
marsh sedge
idk for sure. I used GeoGebra and other tools and i saw that the interval (-1/2;-1/3) wasn't part of the domain, so i thought that if the exponent was negative, the inverse of 3x+1/2x+1 should be considered, so (3x+1)/(2x+1) cant be 0
wary stream
,w ((3*-.34+1)/(2*-.34+1))^((2*-.34+1)/(-.34-3))
ocean sealBOT
wary stream
See, imaginary result
Because of the exponent part, you can't take the root of a negative number
marsh sedge
so how does that affect the domain?
wary stream
Because you can't have a negative number under the root
marsh sedge
aah, so (3x+1)/(2x+1) can't be negative either?
ocean sealBOT
wary stream
That's just the base part and I picked a number between -0.5 and -0.33, notice how it's negative
Then when you raise that to some power, you're going to be taking the root of that number
Hence not real number
,w ((3 * -.6+1)/(2 * -.6+1))
ocean sealBOT
wary stream
There's a number smaller than -0.5
marsh sedge
aah ok, i'll try to solve it considering that
avoiding the values where i get roots of negative values
wary stream
You were correct about the denominators can't be zero but you just forgot to take in consideration that the base can't be negative since you're taking a power/root
Hopefully that makes sense
marsh sedge
thanks, it was very helpfull
marsh sedge
Hopefully that makes sense
not entirely, but hopefully it will
is a very helpfull hint
wary stream
Do you know $a^{\frac{m}{n}} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$ concept?
ocean sealBOT
dldh06
wary stream
Because that's how exponents can be presented
And so, you can't take the root of a negative number
marsh sedge
i "knew it" but i've never considered it
wary stream
When you see an exponent, you should consider it
lone heartBOT
@marsh sedge Has your question been resolved?
marsh sedge
yes
oblique parcel
hello
oblique parcel
need help to figure 124.6315 ( need this number to 3 decimal places , to the nearest gram and nearest kilogram please thanks !
abstract fractal
What have you tried
oblique parcel
ive tried but not sure on how to round to nearest gram or kilogram
abstract fractal
What about the 3 decimal places?
oblique parcel
yh i know how to do that
all i need to know is what the 124.6315 is to grams and kilograms but just not sure hense why i am here
abstract fractal
Well, the mass is 124.6315 grams. What is this number to the nearest whole gram?
tacit arch
all i need to know is what the 124.6315 is to grams and kilograms but just not s...
1000 grams = 1 kilogram
0.001 kilograms = 1 gram
lone heartBOT
@oblique parcel Has your question been resolved?
raw whale
How do i get to the Right side
lone heartBOT
How do i get to the Right side
gilded citrus
t(n) is the sum of the first n+1 terms of a geometric series with ratio 2
and first term 1
raw whale
can u maybe send me ressources that explains this
gilded citrus
sick remnant
lone heartBOT
sick remnant
can i have help im confused
swift shore
lone heartBOT
@sick remnant Has your question been resolved?
shrewd vortex
Hi, I knew a websites that was good for checking number sequences for patterns.
Do you know the website?
wet plover
How can i calculate the first and the second derivatives of this function?
heady pollen
1/N is a factor
you can pull taht our
out
and since Sigma is a sum, the derivative will be easy
because
d(a+b+c)/dx=da/dx+db/dx+dc/dx
same for the second derivative
@wet plover
wet plover
Thank you man!
heady pollen
you're welcome
wet plover
Do you know why the author compares the first derivative to zero while the second to 2/N?
I expected that will be the same result that you had found in calculus.
heady pollen
well he doesnt say that the first derivative is 0
he puts it to 0 i think
this means that the derivative is 0 if the sum of all x-u_i is 0
this is because 2/N is never 0
also i forgot the last step here 🙂
wet plover
also i forgot the last step here 🙂
Nice man. Thank you so much. I 'll think about it.
heady pollen
have a great day^^
lone heartBOT
Available help channel!
slate kayak
lone heartBOT
slate kayak
Is this correct? I used quotient rule
tacit arch
,w differentiate 2e^(-x-1) / x^2
ocean sealBOT
tacit arch
looks good
lone heartBOT
Available help channel!
final ether
Since there are no solutions given for this, can anyone help me with graphing this?
raven rover
Just start with one bullet at a time - of course before everything else sketch an x-y plane
final ether
I have done this so far,
I do not know if I am correct. I tried doing all the ones except the last bullet. I really do not know what f(0)=0 means. Is it just like a ssingle dot?
placid zinc
Red boi been drinking
median oar
First look at the first one
As x goes towards 3 from the negative side it gets bigger
minor needle
I do not know if I am correct. I tried doing all the ones except the last bullet...
yes, means it passes through (0,0), basically origin
placid zinc
Not a single dot. Function passes through there.
final ether
yes, means it passes through (0,0), basically origin
Aaaaa so then this ?
median oar
As x goes towards 3 from the negative side it gets bigger
First on an empty x y plane
Draw this first
minor needle
Aaaaa so then this ?
now last part is satisfied, but think about the rest huh
median oar
There are too many things wrong with this so let’s start from scratch
final ether
this means it approaches positive infinity right
median oar
Which means a dotted line
median oar
Asymptote should be at x=3
final ether
should one be at -1 too?
median oar
So draw the asymptote and also the part where it goes to +inf from the negative side
median oar
should one be at -1 too?
I don’t know we haven’t gotten there yet
final ether
median oar
No no do them 1 by 1
final ether
But how am I supposed to insert
f (x)=1
median oar
Don’t confuse yourself by doing too many things at once
final ether
will the x values at the asumptotes be holes?
median oar
Do you know what asymptotes look like?
weak prairie
imagine the Tan function graph
median oar
This is an example of a graph that gets bigger as you approach x=3 from the negative side
So for now let’s just draw the approaches +inf part of the graph
final ether
sry forgot to write x=3 and x=-1 for the asymptotes.
weak prairie
does your yellow line need to be like that, try to find a way to connect your three lines that preserves the definition of a function i.e. one input x and one output y
final ether
I see, I could then connect the two functions red and yellow
and make it look like a polynomial
median oar
Do you want my help or nah
weak prairie
yes and a green dot
final ether
I do want your help guys, sry that I am impatient. I thought I could do this.
median oar
It’s ok, take it step by step
Don’t do everything at once
That’s the whole point of starting from scratch
Start from scratch so we can satisfy the conditions one by one
final ether
ok
median oar
Now lastly we know f(1) = 1
final ether
median oar
You’re skipping ahead again
final ether
Sorry, I will not connect them then
median oar
What does the 3rd line mean?
final ether
Is it not a function?
like no matter where we are on the function the lim is 1
so it is like k=k a line?
final ether
Does that mean that we have a function that from both ends has the lim 1?
Like what I mean is from left and from right it approaches 1
final ether
and then put a dot on the (0,0) for the last
median oar
Right
final ether
then connect them
median oar
Yeah basically
lone heartBOT
median oar
Remember to label your axis
lone heartBOT
Available help channel!
median oar
Both the axis and asymptotes should have arrows to indicate they are not line segments @final ether
final ether
Both the axis and asymptotes should have arrows to indicate they are not line se...
yes
especially since they go to + and - infinity
median oar
Yes
final ether
thank you sir!
median oar
👍
final ether
u rly helped me 🙂 ❤️ !
median oar
.close
summer berry
How is integral x^3 | 2 above 1 = 7 shouldnt be 7/x^3?
lone heartBOT
How is integral x^3 | 2 above 1 = 7 shouldnt be 7/x^3?
last ether
What
last ether
How is integral x^3 | 2 above 1 = 7 shouldnt be 7/x^3?
Do you mean $\int_2^1 x^3dx$?
ocean sealBOT
Umbraleviathan
last ether
That's what I thought
$$\int_1^2 3x^2dx$$
$$\=\left[x^3\right]_1^2$$
$$\= 2^3 - 1^3$$
$$\=8-1$$
$$\= 7$$
summer berry
yes
summer berry
alright
last ether
That isn't 7 though
summer berry
cant be
last ether
No it can't
summer berry
let me see
last ether
Cause this evaluates to 3.75
summer berry
I see that is my mistake
I had already finished most of it
so here on the 2nd part the x^4 is x^3
the initial thing was 3x^2
and than it became x^3
ocean sealBOT
Umbraleviathan
summer berry
but isnt it 2^3 / x^3 - 1^3/x^3
last ether
Why are you putting x^3 in the denominator
summer berry
Isnt that how it should be ?
last ether
No
last ether
**Umbraleviathan**
Quite literally this
last ether
That we are integrating?
summer berry
just a example
last ether
Well the primitive/antiderivative would have x in the denominator, yes
I'll brb I gotta drive my sister
summer berry
alright
summer berry
Well the primitive/antiderivative would have x in the denominator, yes
I think I got it if youre integrating and it there is something in the denominator than you also put it in the denominator if not than you dont
thanks for the help
.close
grand palm
how do i solve an inequality with an letter and a number in it like x + 5y > 10
grand palm
i just made that inequality up so not sure if it works
and how do i get the coordinate out of it
north needle
how do i solve an inequality with an letter and a number in it like x + 5y > 10
well I suppose you could re-arrange for the standard linear graph format
Y = mx + c
grand palm
OOOOOOOOOOOOOOOOOOOOOOOO
grand palm
what if its x
north needle
last ether
what if its x
Then it would be a function, x, with respect to y
Or in this case, an inequality with respect to y
Although it wouldn't really matter I'd think
humble remnant
I honestly don't know where to start going about solving this. If anyone has any ideas I'd love some help
tame willow
north needle
It’s an arrangement question
The total area of the rectangle is 45 but each tile is 3 in area
So 15 tiles
However those 15 tiles can be arranged in numerous ways
Sadly
humble remnant
yeah that's my problem
versed elm
ok this might not be the answer
but look, consider the 3 edge
there's only two way of filling that, you use 3 tiles with the 1 side
or you use 1 tile with the 3 side
now it's case work
you have x number of tiles with the 3 side filling the 3 edge
icy jewel
it’s mathin time
versed elm
and then do x from 1 to 15 lol
u can skip few cases because 15-x has to be divisible by 3
I think
north needle
It's 3 x 15
I read but didn’t comprehend
versed elm
for x=15 and x=0 we can only have 1 arrangement each
for x=12 we can only have 1 arrangement
actually
no nvm ignore what I just said
hmmmmm
this question is hard lol
humble remnant
yes I know
versed elm
I still think my way works, but it didn't
humble remnant
yeah I don't like these kind of problems
versed elm
because you have to fill the 3cm edge somehow
(let's just say the unit is cm, looks funny without unit when I type lol
oh right
nvm
mine works
I gotta keep doing case work
so
for x=12
we have a set of 3 tiles stacking up together
this thing can be placed in different places
in 13 positions
so that's 12 for x=12
now for x=9
there would be 2 sets of 3 tile stack
if you place one of the stack
then the other stack have 9 positions to be placed in
so 12*9 which is 108 positions
now x=6
we have 3 stacks
ahhhhhh
going back to x=9, order doesn't matter
I think I might got a few repetitions
humble remnant
I'm so confused
versed elm
right so if one of the stack has been placed in a position then when we keep counting we an't put the other stack in the same position (this sounds confusing Ibut I get my self lol
so it's actually like hand shaking
versed elm
I'm so confused
dw, I am confused too
I gotta draw it out
there is probably an easier way to solve this problem
but that is probably too hard for me considering my math skill (in tears
so I'll have to do case work
u wanna hop on a voice call?
humble remnant
can't
versed elm
ok, I'll type my thoughts here then
so for x=9, stack 1 in position1, 10
10+9+8+....+1
so 55
and then x=6
3 stacks, I am not happy
but
but
we might be able to compute from the x perspective for this case
humble remnant
how do we do that?
versed elm
I'm still thinking
I wish my friends is here, maybe I should ping them
(lol me who can't solve math
ok back to actualy business
ok I think we should just do the 3 stack thing
there's only 9 cases for that
ah but 9 case is just too much
anyways let's go
1,2 near each other
7 for 3
1 gap 6 for 3
2 gap 5 for 3
until there's only 1 option
so
56
I guess.......
but if 1 is 1 square from the beggining
I am googling at this point
humble remnant
yea me too
versed elm
I'm asking my friend right now
okie
so my thinking at first was right
so for the 3 tiles stack
you can have 0-5 piles of these
now for 0 pile
there's just 1 arrangement
for 1 pile, there's 13 position to choose from
so 13 choose 1
for 2 piles
there would be 9 1 by3 tile (like the vertical ones), and then plus the 2 positon for the tile stack, in total there's 11 slots), you choose two of the 11 slots to put tile stack
so 11 choose 2
and for 3 piles
it's the same thing
6 vertical thingy, plus 3 tile stack
9 choose 3
and then 4 piles
3 vertical thingy, 4 tile stack
7 choose 4
then 5 piles is just everything in piles
so 1 arrrangement for that
in total there is 1+13c1+11c2+9c3+7c4+1=189 arrangement
thank you @atomic saddle
versed elm
no worries~ mostly solved by my friend (at least my initial case work thing was right..........lol
atomic saddle
Haha
humble remnant
oh god there's more problems
versed elm
oh gosh
humble remnant
thankfully they're different
versed elm
phew
humble remnant
I have like 66 hours to finish so it shouldn't be a problem
shadow pond
Right now, I'm having trouble solving this problem. I don't know why the negative goes away and I'm having trouble past the GCD part.
lone heartBOT
Right now, I'm having trouble solving this problem. I don't know why the negativ...
@shadow pond Has your question been resolved?
real gazelle
Right now, I'm having trouble solving this problem. I don't know why the negativ...
They shouldn't really take the negative away
That's kind of abuse of notation
Try solving -12x + 17y = -1 using the extended euclidean algorithm though
That should give you all solutions x to the original congruence equation
shadow pond
Okay, thank you.
icy jewel
it’s mathin time
fossil owl
lone heartBOT
normal ingot
What is your question
fossil owl
Solve for X I’m stuck
normal ingot
Make 1/a+1/b a single fraction
fossil owl
like substitute?
normal ingot
In the form A/B where A and B are some expressions
fossil owl
I don't understand
normal ingot
a/b+c/d=(ad+bc)/(bd)
fossil owl
Do you have like a youtube video that explains this?
normal ingot
fossil owl
.close
worn hinge
Say I have a function f(x, y) = x mod 10^y
If I have f(12345, 1) = 5
How would Improve this function so that
Instead of f(12345, 3) = 345 I will get f(12345, 3) = 3
rapid sundial
What
worn hinge
Mod = get the remainder
rapid sundial
Yes I know what mod is. I don't get your previous messages
worn hinge
Better question
rapid sundial
Is this computer science or math... sorry just slightly confused.
worn hinge
Cs
rapid sundial
Ok
worn hinge
Btw is mod sign in math %?
rapid sundial
yea
You could always do (x mod 10^y)/(10^(y-1))
That seems to work
At least it'd work in java
rapid sundial
oh
icy jewel
it’s mathin time
worn hinge
Say I have a function `f(x, y) = x mod 10^y`
If I have `f(12345, 1) = 5`
How wo...
It is already a for loop
rapid sundial
?
worn hinge
Maybe to get the previous unit I should just store it?
rapid sundial
Do you have your code?
rapid sundial
🤔
worn hinge
for (i=10; ...; i*=10);
That's how I increment by *10
What about just storing the previous i?
That might be good
.close
lone heartBOT
lone heartBOT
Available help channel!
small swift
lone heartBOT
https://gyazo.com/487662c4734f5d61912ee4eef01f5395
small swift
Hello,
Can someone take me through I can solve this?
I seem to get myself stuck in a loop by just taking y'' = y' and then y'' -y' = 0
vale wigeon
well surely you can solve the second equation for y itself, can't you? y' = y with y(0) = 8
lone heartBOT
@small swift Has your question been resolved?
forest dew
If I had a combination starting with aa0000. Goes to aa0001, which eventually rolls over to ab0000, and ends with zz9999, how many possible combinations are there?
forest dew
<@&286206848099549185>
vale wigeon
are you asking how many strings exist that have two letters followed by four digits?
forest dew
So aa0000 is one combination, and aa0001 is the second
So what number combination is zz9999
vale wigeon
surely that would simply be 26^2 * 10^4?
forest dew
I believe the answer is yes to what you are saying
vale wigeon
if you're considering aa0000 as combination #1, that is.
if you're instead considering it as #0, you'll need to subtract 1.
forest dew
if you're considering `aa0000` as combination #1, that is.
This. Thanks!
lone heartBOT
@forest dew Has your question been resolved?
fast smelt
I need to find the value of "?" in the following equation ----> ? + 50% of ? = 9912
fast smelt
I have no clue how to do that
remote heron
oh is each ? the same?
fast smelt
yeah
remote heron
or are they different numbers
okay
so
what is something plus half of itself
can you think of another way to say that
fast smelt
that's it, what is something plus half of itself that would equal 9912
remote heron
right but the
in mathematical language
to turn this into an equation we can solve
fast smelt
right but the
the waht
fast smelt
yeah
fast smelt
yeah
remote heron
so whats the eequation we want to solve
fast smelt
we want to know the x value
remote heron
okay, write your equation
fast smelt
x + (x/2) = 9912
fast smelt
Idk x + 0.5x ?
mystic flint
x + (x/2) = 9912
u also can multiply each with 2?
fierce prairie
Idk x + 0.5x ?
yes, now simplify it further
fast smelt
1.5x ?
pearl wraith
yeah
fast smelt
isolated the x right?
mystic flint
remove the fraction form, multiply everything by 2
only work with operation + and -
pearl wraith
why would you want that
fast smelt
idk I forgot I havent do algebra in forever
pearl wraith
you could have a beautiful equation like 3x/2 = 9912
mystic flint
simplify works but sometime beginner find it tough i guess
pearl wraith
right, so you’re at 1.5x = 9912, how can we solve for x now?
decimals 
fast smelt
doesn't I have to divide 9912 by 1.5 ?
fierce prairie
yeah but a fraction form would be easier if you don’t have a calculator
earlier, you could’ve done 1x + (1/2)x = (2/2)x + (1/2)x = (3/2)x
or just see that 1.5 = 3/2
mystic flint
yeah but a fraction form would be easier if you don’t have a calculator
of course, i will do the simplify method
its cleaner
then bring the denominator 2 to other side
fast smelt
.close
lone heartBOT
pearl wraith
(3x/2) = 9912
(2/3)(3x/2) = 9912 (2/3)
x = 6608
fast smelt
thank you so much
lone heartBOT
Available help channel!
silent scroll
1.9(3.2r - 7.5) = 38.6
lone heartBOT
1.9(3.2r - 7.5) = 38.6
lone heartBOT
@silent scroll Has your question been resolved?
old tangle
.ask
sly comet
lone heartBOT
sly comet
how do i do thjis?
lone heartBOT
@sly comet Has your question been resolved?
summer berry
How do you know with matrices if something is scalar multiplication or normal multiplication
naive valley
How do you know with matrices if something is scalar multiplication or normal mu...
do you have an example of what you mean?
marsh rapids
In general, you know whether you're multiplying by a number or by a matrix
lone heartBOT
@summer berry Has your question been resolved?
torn isle
lone heartBOT
torn isle
Hi, may I know how to show (b)
vale wigeon
$\nabla^2$ is the laplacian operator $\pdv[2]{x} + \pdv[2]{y} + \pdv[2]{z}$
ocean sealBOT
Ann
torn isle
Hmm… I did search in YouTube before for solving this, but his method I’m not so sure
Anyway, may I know how you solve that ?
Because I tried with this at first
vale wigeon
would you like me to walk you through part b
torn isle
would you like me to walk you through part b
Yea
ocean sealBOT
Ann
vale wigeon
$\dv{x} r^n = \frac{n}{2} (x^2 + y^2 + z^2)^{n/2 - 1} \cdot 2x = nx r^{n-2}$
ocean sealBOT
Ann
vale wigeon
skipping over a few steps here for brevity but does this make sense?
torn isle
so $r^n = (x^2 + y^2 + z^2)^{n/2}$
Erm, may I know it becomes this directly?
Is it because r=sqrt(x^2+y^2+x^2)?
vale wigeon
well $(x^2 + y^2 + z^2)^{n/2 - 1} = (x^2 + y^2 + z^2)^{\frac{n-2}{2}} = (r^2)^{\frac{n-2}{2}} = r^{n-2}$
ocean sealBOT
Ann
vale wigeon
Is it because r=sqrt(x^2+y^2+x^2)?
yes
torn isle
Why you know you start the question by substitute r^n at first and not do like what I had did at first?
torn isle
i don't understand your question.
Erm, basically I don’t understand why you solve it that way? It looks more simple but I don’t understand why at first you directly use r^n and sub that in
vale wigeon
the reason i do things the way i do them is to not write more stuff than necessary
the more stuff you write, the higher the chance is that you screw up somewhere.
vale wigeon
i don't know what you mean by "apply r^n"
we are asked to find its laplacian, and that's what i'm setting out to do
i'm not done yet!
this is just the first step, in case that was not clear
torn isle
i'm not done yet!
Oh okay, I’ll see whether I can understand or not
Sorry for interrupting in the middle 😬
vale wigeon
you didn't interrupt. i asked you if what i'd done so far makes sense.
do i need to explain further why $\pdv{x} r^n = nxr^{n-2}$?
ocean sealBOT
Ann
torn isle
Ah, this I can understand, it just normal differentiate right?
vale wigeon
yes i am just taking the partial derivative with respect to x
only once, as of yet
my next step will be to do it again
torn isle
But I’m not sure why we need d/dx and not the others
torn isle
only once, as of yet
Oh
vale wigeon
But I’m not sure why we need d/dx and not the others
we'll get to that
okay
so now let's take the derivative wrt x again
$\pdv[2]{x} r^n = \pdv{x} nxr^{n-2} = nr^{n-2} + nx \cdot (n-2)x r^{n-4} = nr^{n-4} (r^2 + (n-2)x^2)$
ocean sealBOT
Ann
vale wigeon
does this make sense to you? yes/no
torn isle
Yes
vale wigeon
ok
so now
because r is symmetric in x, y and z, we can swap out the letters to get immediately, without any further calculations:
$\pdv[2]{x} r^n =nr^{n-4}(r^2 + (n-2)x^2) \ \ \pdv[2]{y} r^n =nr^{n-4}(r^2 + (n-2)y^2) \ \ \pdv[2]{z} r^n =nr^{n-4}(r^2 + (n-2)z^2)$
ocean sealBOT
Ann
torn isle
Oh I see, I was about to ask you am I going to do the same procedure for y and z just now
Just notice can apply what we got in x to y and z
vale wigeon
no need to do the same thing 3 times
now we add all these up and get $\nabla^2 r^n = nr^{n-4}(3r^2 + (n-2)(x^2+y^2+z^2))$
ocean sealBOT
Ann
torn isle
Until now I understood, but may I know how you show the last steps. Look complicated to me
vale wigeon
ok let me write it out more explicitly
$\nabla^2 r^n = nr^{n-4} (r^2 + (n-2)x^2 + r^2 + (n-2)y^2 + r^2 + (n-2)z^2)$
ocean sealBOT
Ann
vale wigeon
does this make more sense to you?
torn isle
does this make more sense to you?
I know how it simplify into the upper part
torn isle
I’m not sure how you convert back to n(n+1)r^(n-2) for the RHS
vale wigeon
x^2 + y^2 + z^2 = r^2, so you get nr^(n-4) (3r^2 + (n-2)r^2) = nr^(n-4) * (n+1) * r^2
vale wigeon
which part of my message are you questionmarking?
torn isle
which part of my message are you questionmarking?
No problem now
torn isle
which part of my message are you questionmarking?
Thank you so much
Guess the method that I solve initially is too long and there a mistake there too
errant cedar
i've solved sin(x+0.927)=1 (in rad)
x+0.927 = 0.64 -> this is correct
now it's asking me for a general solution.
I got x=pi/2 +2kpi - 0.927 where k is an integer
errant cedar
but the answer says this
i think it's the same since I've said k is an integer, and for their solution n might be in the naturals, but can someone confirm?
minor needle
i think it's the same since I've said k is an integer, and for their solution n ...
your answer is valid
I'd say
lone heartBOT
@errant cedar Has your question been resolved?
errant cedar
your answer is valid
ok thanks!
vivid acorn
lone heartBOT
vivid acorn
I'm stuck help please
harsh girder
,rotate
ocean sealBOT
ocean sealBOT
harsh girder
should equal to 2*sin(80°)
vivid acorn
The question is wrong?
vivid acorn
but what you're trying to do
Lhs=rhs
lone heartBOT
@vivid acorn Has your question been resolved?
final ether
Would the function be f(x)=500^t ?
final ether
.close
tribal oxide
lone heartBOT
tribal oxide
is it a) 0 b) 1 c)infinity(does not exist) d)-infinity(does not exist)
lone heartBOT
@tribal oxide Has your question been resolved?
willow canopy
yea
fierce prairie
yea
stoic estuary
i think it should be 3/2, again i don't have a pen and paper so would have to confirm
you can use l'hospital rule
differentiate the numerator and denominator
tribal oxide
you can use l'hospital rule
we havent gone there yet, im supposed to be doing this with factorization
stoic estuary
ahh ok my bad
one sec
if that's the case, then it should be fairly simple enough as well, both the numerator as well as the denominator have t-2 as their factor
numerator is (t-2)(t^2+5t-2) and denominator is (t-2)(t^+2t)
lime cliff
,w 2+2=
ocean sealBOT
stoic estuary
remove the t-2 from both numerator and denominator, and that should give you 12/8
tribal oxide
remove the t-2 from both numerator and denominator, and that should give you 12/...
oh yes i was imagining something nvm, it is 12/8
lone heartBOT
tribal oxide
.reopen
lone heartBOT
✅
harsh girder
,w Limit[(t^3+t^2-5 t+3)/(t^3-3 t+2),t->1]
ocean sealBOT
final ether
Any idea how to evaluate lim when we have absolute values?
lone heartBOT
Any idea how to evaluate lim when we have absolute values?
marsh rapids
You're looking at 2+ anyways
vale sapphire
Yeah, the question is, what's the definition of "x → 2⁺"
marsh rapids
Always simplify|| if necessary by treating each case separately
final ether
I really do not know how to solve for absolute values, I forgot lol it has been forever since I did it
pulsar lark
The question ask to find the maximum and minimum value of y, but I'm stuck at sin²x, what does it mean? Sin x²?
minor needle
The question ask to find the maximum and minimum value of y, but I'm stuck at si...
no
$$\sin ^{2} x = (\sin x)^{2}$$
ocean sealBOT
Modus
vale wigeon
why does this say 3 - sin^2(x) - 4 and not just -1 - sin^2(x) 
minor needle
hint: think about range of sin^2(x)
pulsar lark
hint: think about range of sin^2(x)
I learnt that, is 1 to -1 right?
minor needle
I learnt that, is 1 to -1 right?
it's for sin(x)
well, think about a square of real number, can it reach a negative value?
pulsar lark
No
minor needle
so what the range of sin^2(x) should be then
pulsar lark
1?
pulsar lark
1?
minor needle
no, u can have sin(x) = 1/2 then sin^2(x) = 1/4
notice that for example if sin(x) = -1 then sin^2(x) = 1
so we cannot have negative values
pulsar lark
I get confused
last ether
Think about your important points of a sine wave
pulsar lark
Why the minimum of sin²(x) is not (-1)²
last ether
Because that's the same as 1^2
Recall that these are your 4 important points of a sine wave
Square each ordinate
last ether
Just square the ordinates (the y-values)
You want the range right? Just square the y-values
pulsar lark
I still not understand at all sorry
How to get this answer from range of sin(x) to this?
Or they are no relationship?
So y= -1-sin²x
The max.=-1-(1)²
=-2?
The min.=-1-(-1)²
=-2...?
Can you demonstrate 1 time please?
last ether
I still not understand at all sorry
You wanna square each y-coordinate
(-1)^2 = 1
0^2 = 0
1^2 = 1
fresh parcel
The maximum doesnt occur there lol
shut tendon
How do I use sigma
lone heartBOT
How do I use sigma
shut tendon
So?
chrome plank
This channel is not empty, ask again your question in another empty channel
brisk bay
Σ n is the maximum integer value at which the cumulative sum is taken, k is the first integer value that is taken of the sum
n is at the top
k is at the bottom
shut tendon
And the number that is in the sum
brisk bay
how do I explain this better
so an example would be
(sigma, n=3&k=0) k+1 which is equal to (0+1) + (1+1) + (2+1) + (3+1)
shut tendon
Ok
brisk bay
yes
shut tendon
Yes
shut tendon
If i wanted a multiplication i would have just done it
But i need it expressed as a sum
brisk bay
yes
but I don't see why you would need to use sigma notation
from what I understand
maybe I'm missing something
shut tendon
but I don't see why you would need to use sigma notation
Because i don't want to write 50+ numbers
brisk bay
you want the rate of change to be constant
shut tendon
Yes
brisk bay
2+2+2+...+2
ok
well
sigma(n=whatever max you want & k = wherever you want to begin) 0k + 2
I think that should do it
shut tendon
And where do i write how many 2s are there
shut tendon
So if i write n=50
shut tendon
So if i write n=50
Is that the total sum or how many numbers will there be
brisk bay
the way you find how many 2s there are is by determining how many numbers are between n and k plus 1
so for here, 50 minus one plus 1
50 2s
shut tendon
Oh