to get the number that when multiplied by -5, gets -40
#help-36
muted coral
me omw to explain every aspect of algebra to you
tranquil pine
is there anything special about the - 5
tranquil pine
me omw to explain every aspect of algebra to you
lol
muted coral
about the negative?
tranquil pine
yeah
muted coral
negative numbers work the same way as positive
tranquil pine
yeah
muted coral
since b is -5
but basically no it would work whether that was -5 or +5
tranquil pine
im just wondering how you got b
muted coral
for example
1x^2 + 8x + 15
find 2 numbers that have a product of 1*15 and a sum of 8
that's 5,3
x^2 + 5x + 3x + 15
tranquil pine
so this would give you b?
muted coral
b is 8
tranquil pine
ahhh
muted coral
for example
1x^2 + 8x + 15
find 2 numbers that have a product of 1*15 and a sum ...
would you feel comfortable factoring. this
tranquil pine
so in that situation the other situation b is 3
muted coral
yes
tranquil pine
but would that change anything
muted coral
no
tranquil pine
ok
muted coral
you can still solve by factoring (sometimes) or quadratic formula
not really much difference
muted coral
alr
muted coral
yw
final saddleBOT
Available help channel!
shadow quail
I found this problem on one of my books, but i have no ideia on how to solve it. Can someone please give me a hint or a direction on where to go please (Sorry if i have bad english, its not my first language)
determine all integer positive values for n, so that for any natural number m>=1, the following is true:
(1 + 2^n + 3^n + ... + m^n) = (1 + 2 + 3 + ... + m)^(n-1)
I tried separating the second part using x^(n-1) = x^n/x, and then tried creating a system to check for possible values of n, but i was unable to to it. Can someone help me please
final saddleBOT
I found this problem on one of my books, but i have no ideia on how to solve it....
shadow quail
I really just need a heads up, maybe a concept or a formula im missing
or a hint on how to continue it
vital crag
try plugging in different values of m and n
shadow quail
i tried solving it assuming the m was going to be the three, so i got (1+2^n+3^n)=(1+2+3)^(n-1)
vital crag
that's a good start
shadow quail
but i wasnt able to find a way to either isolate n, or to find an ax²+bx+c type equation
vital crag
you just need to check a few n
shadow quail
wich i believe should be the way to go to find multiple values of n
vital crag
"positive integer values for n"
start with the smallest one
then increase one by one to see if it's true
shadow quail
ok
lemme try that
i found that for m=1
n can be any integer value
for m=2 i found n=3
for m=3 i found n=3 as well
for m=4 also n=3
but is there any way of explaining why n=3 every time?
or at least it seems like 3 every tinme
time
@vital crag
vital crag
for m=2 i found n=3
you're working backwards. the question says to fix an n, then show equality for all m
what you're doing is fixing an m, then finding an n
shadow quail
i see
let me try that way
i wasnt able to put a fixed n
actually
i was
but the thing im able to show equality for all m
just plugging different values of m the equality is always right
if i use n=3
vital crag
but the thing im able to show equality for all m
Write out mathematically what you mean exactly and give an example
vital crag
you're working backwards. the question says to fix an n, then show equality for ...
FOR ALL m
vital crag
and i found m=4
Not just a single one
shadow quail
im sorry
but i dont understand how in one equation
i can represent all m
can you give me an example?
maybe?
vital crag
can you give me an example?
(1+2+...+n) = n(n+1)/2 for all n
Change that to m
If you're looking to prove something false, you just need one value as a counterexample
vital crag
(1+2+...+n) = n(n+1)/2 for all n
E.g. if this were instead
(1+2+...+n)=n(n-1)/2 for all n, you just need to check small n to see that it's wrong
shadow quail
(1+2+...+n) = n(n+1)/2 for all n
how did you get to thaفًّ
that?***
i dont think i understand it quite well
im sorry
lethal estuary
Gauss spoke it to him in a dream
vital crag
how did you get to thaفًّ
does it matter?
you asked for an example
this is the simplest one
shadow quail
im sorry
yes it doesnt matter
what im trying to understand is how can i do this for my problem, and explain it
shadow quail
because this was translated, it comes from a textbook that asks to write in discursive and explain your answer
vital crag
,w define discursive
soft zealotBOT
shadow quail
so if i say that, for example, for m=3, the only value of n that works is 3, it proves that in all scenarios the only value that works for n is 3, since i can use a counterexample from the value of m=3 and n being equal any other number, that it doesnt work, so 3 has to be the only value?
is that what you were trying to explain to me?
shadow quail
but if fix an n, how do i prove the other values of n wrong?
like, if i fix the n, i prove all values of m correct for a certain n
but then how can i prove to him that in other values of n, none of those values of m work, in a simple manner
vital crag
but if fix an n, how do i prove the other values of n wrong?
you fix one n at a time
vital crag
like, if i fix the n, i prove all values of m correct for a certain n
have you done this?
shadow quail
yes
vital crag
but then how can i prove to him that in other values of n, none of those values ...
again, ONE COUNTEREXAMPLE is enough
shadow quail
i proved that for n=3, all values of m work
from 1 to 10
what i thought of as an answer was
i have to leave, im sorry
.close
final saddleBOT
vital crag
from 1 to 10
you didn't do it correctly then
final saddleBOT
Available help channel!
white frost
final saddleBOT
white frost
struggling to convert the ellipsoid equation into cylindrical
or even if thats the correct approach
my first step is using z = 0 as a lower bound and z = ??? as the higher bound, ??? being the cylindrical version of the ellipsoid equation
final saddleBOT
@white frost Has your question been resolved?
white frost
<@&286206848099549185>
final saddleBOT
@white frost Has your question been resolved?
mortal sluice
regal depot
I pinged u
final saddleBOT
@white frost Has your question been resolved?
white frost
<@&286206848099549185>
final saddleBOT
@white frost Has your question been resolved?
harsh cypress
or even if thats the correct approach
@white frost
The highest point along the z-axis that the ellipsoid touches occurs when x and y are 0. So from the ellipsoid equation, this happens when z^2 = 16 or z = +/- 4.
So that's already the final integral taken care of with bounds from z=0 to z=4.
Next just plug in the cylindrical conversions for x and y and you should be able to work out what "functions" for the other two integrals are the bounds.
Sorry if unclear, a bit busy atm
broken scroll
|x-3| + |x+2| ≥ 8
final saddleBOT
|x-3| + |x+2| ≥ 8
broken scroll
|x-3| + |x+2| ≥ 8
I'm new to modulus to I need help with problems like this
What I did was opened |x-2| first which gave 2 cases
lethal estuary
What I did was opened |x-2| first which gave 2 cases
is it x - 2 or x + 2?
broken scroll
|x-3| ± (x+2) ≥ 8
broken scroll
> |x-3| ± (x+2) ≥ 8
-> |x-3| + x + 2 ≥ 8
|x-3| + x ≥ 6
lethal estuary
the way I'd do it is split into 3 cases: when both expressions are negative, when one is negative and one positive, and when both are positive
broken scroll
-> |x-3| + x + 2 ≥ 8
|x-3| + x ≥ 6
Now two more cases arrived
-> ±(x-3) + x ≥ 6
broken scroll
the way I'd do it is split into 3 cases: when both expressions are negative, whe...
So you're saying that
(x-3)-(x+2) ≥ 8
and
(x+2)-(x-3) ≥ 8
Are the same case?
broken scroll
first think: where will *both* of those expressions be less than 0?
I don't understand
lethal estuary
when are both (x - 3) and (x + 2) negative?
broken scroll
when are both (x - 3) and (x + 2) negative?
In a case when we take both of them as negative ig
lethal estuary
no, I mean for what values of x
broken scroll
x ≥ -(7/2)
For values greater than negative 3.5 I suppose @lethal estuary
lethal estuary
greater than? so if you plug in x=0 to both of those, they'll both be negative?
broken scroll
greater than? so if you plug in x=0 to both of those, they'll both be negative?
One of them would be positive
Is that your point?
lethal estuary
no, I asked when will both be negative
broken scroll
By both do you refer to |x-3| and |x+2|
lethal estuary
yes...
broken scroll
They'll be negative if |x-3| is taken as +(x-3) and |x+2| as -(x+2)
lethal estuary
no, forget the absolute value signs for a moment
lethal estuary
when are both (x - 3) and (x + 2) negative?
@broken scroll
broken scroll
when are both (x - 3) and (x + 2) negative?
when x is lesser than -2
lethal estuary
yes
broken scroll
makes sense
lethal estuary
so case 1, x < -2: since you know the absolute value expressions will be negative, you can write your inequality as (3 - x) + (-x - 2) >= 8
broken scroll
so case 1, x < -2: since you know the absolute value expressions will be negativ...
What exactly do you mean by absolute value expression
I'm not really familiar with the term
lethal estuary
if x < -2, then |x + 2| = -(x + 2), and |x - 3| = -(x - 3) = (3 - x)
broken scroll
if x < -2, then |x + 2| = -(x + 2), and |x - 3| = -(x - 3) = (3 - x)
correct
understood
so we have values for x<-2 in that case
lethal estuary
well now you can solve the inequality for that case
then you look for another interval of x-values for which one of them will be positive (x+2) and one will be negative (x - 3)
and then a 3rd case will be the x values for which both are positive
frank folio
is there any way to use the unit circle to find arctan(2)
final saddleBOT
is there any way to use the unit circle to find arctan(2)
royal gust
Like, just a picture of the unit circle? No not really
frank folio
alr just making sure
It was on one of my problems that said no calculator
and i was so lost lol
royal gust
Ah gotcha. Yeah that's a difficult angle.
royal gust
Go for it
royal gust
,w graph r = sin^2(theta)
soft zealotBOT
royal gust
They must mean one of the loops
frank folio
how did they find the bounds tho
royal gust
You only need to traverse between 0 and π to cover a loop
I personally would find it by graphing this:
,w graph y = sin^2(x)
soft zealotBOT
royal gust
And seeing where it comes back to 0
That's exactly where y = sin(x) comes back to 0, though
frank folio
so its just the bounds of the squeeze therum?
royal gust
Just the roots of sin(x)
royal gust
If you're in a tight situation, your best bet might be to actually graph it
I'm pretty sure this one is just a circle to the right of the origin, though
Like, plug in a few θ into your calculator, get a few r out, and you can draw a graph on the fly
still tangle
can someon explain how you would solve this
final saddleBOT
Available help channel!
zinc slate
final saddleBOT
zinc slate
for b) shouldn't it be $\vec{AB}$ = $\vec{AO}+\vec{OB}$?
soft zealotBOT
yomiko
half linden
Do you know the parallelogram rule of vector addition?
Or the triangle law of vector addition
If you don't it's like this @zinc slate
Can you make out why it's like that from this?
zinc slate
you can do it like that??
surely that means that there is multiple answers
since both would get to the same point?
or because they asked for distance so its the nearest i guess
half linden
or because they asked for distance so its the nearest i guess
It is as you said, it should be AO + OB, just like in the triangle rule. However the vector given in the solution is not AO but OA
(-OA) + (OB) = OB - OA
final saddleBOT
@zinc slate Has your question been resolved?
final saddleBOT
final saddleBOT
Available help channel!
gilded marten
what is my f(x) and g(x)?
tranquil pine
<@&286206848099549185>
For A, the question wants you to practice the product rule $h'(-1) = (f(x)\cdot g(x))_{x=-1}' = f'(-1)\cdot g(-1) + f(-1) \cdot g'(1)$ and read the values of the functions and the values of their derivatives from the graphs shown.
soft zealotBOT
Landau08
tranquil pine
The left picture in you screen shot shows the function f(x). For x=-1 the blue curve has the value y=-2.
At x=-1, the function f has the slope f'(-1) = 2. Do you know how to determine the slope of a line graphically?
gilded marten
derivative?
tranquil pine
slope = rise / run
gilded marten
yeah
gilded marten
1.5?
i dont get this at all
all i know is product and quotient rule
so my f'(-1) = 2
tranquil pine
Using the product rule, you see that once you know the four numbers f(-1), f'(-1), g(-1), and g'(-1) you can calculate h'(-1)
tranquil pine
You read it off the picture on the left that shows f(x).
gilded marten
is f(-1) the same thing as y = -1?
tranquil pine
right
gilded marten
so 0.5?
tranquil pine
what about 0.5?
tranquil pine
No, actually at x=-1 the blue line goes through the point (-1, -2) so f(-1) = -2
gilded marten
i thought it was the same thing as y = -1
tranquil pine
gilded marten
so g(-1) = -2
tranquil pine
no the picture shows f
tranquil pine
right, you need the slope of g at x=-1.
slope = rise / run. In this case, when you go 1 to the right (run=1) you go up by roughly 1.2 (rise = 1.2).
So g'(-1)=1.2/1 = 1.2
gilded marten
isnt it a little bit more than 1.2
wait
no
nvm
so they are both equal to 1.2
and f(-1) f'(-1) = -2
gilded marten
oh so not negative
gilded marten
so f(-1) = 2, f'(-1) = -2, g(-1) = -1.2, g'(-1) = 1.2
gilded marten
yes
gilded marten
so h'(-1) = 2(1.2)+(-2)(-1.2)
gilded marten
f(-1) = 2,
f'(-1) = -2
g(-1) = -1.2
g'(-1) = 1.2
tranquil pine
right
but you wrote h'(-1) = 2(1.2)+(-2)(-1.2)
it should be h'(-1) = 2(1.2)+(-2)(1.2)
gilded marten
that equals 0?
tranquil pine
I think that's the solution to part A
gilded marten
the website says its wrong
tranquil pine
ah I see, you did another copying mistake
Again: h'(-1) = f'(-1) * g(-1) + f(-1) * g'(-1)
when you insert
f(-1) = 2,
f'(-1) = -2
g(-1) = -1.2
g'(-1) = 1.2
you get 2 * (-1.2)+(-2)(1.2) = -4.8
(the minus in the first bracket was missing)
tranquil pine
You can check it again, but that should be roughly right (the 1.2 might be a little bit imprecise).
For example, if the 1.2 was rather a 1.25 then the answer would be -5. I'm not sure how precise the answer must be
Is -5 correct?
gilded marten
i tried all single digit numbers
its wrong
idk
i also change the number after the decimal point
tranquil pine
I found another of your copying mistakes.
I wrote f(-1) = -2 and f'(-1) = 2, right. You made that into f(-1) = 2 and f'(-1) = -2.
h'(-1) = f'(-1) * g(-1) + f(-1) * g'(-1)
when you insert
f(-1) = -2,
f'(-1) = 2
g(-1) = -1.2
g'(-1) = 1.2
h'(-1) = 2 * (-1.2) + (-2) * 1.2
Fixing that error doesn't lead to a different solution.
So you tried both -4.8 and -5?
I'll try to get the exact result.
tranquil pine
i tried all single digit numbers
The solution for B would be:
k'(1) = (f'(1)*g(1) - f(1)g'(1))/(g(1))^2
f(1)=0
f'(1)=2/3
g(1)=4/3
g'(1)=4/3
k'(1) = (2/3 x 4/3)/(16/9) = 1/2
tranquil pine
i tried all single digit numbers
in the graph for g it looks like, it goes through the point (1.5|2) which mean the -1.2 should actually be rather -4/3.
The final result (using this more precise reading) for A should then be -4*4/3 = -16/3
Can you try -16/3 ?
Can you enter fractions or do you have to enter a decimal like -5.3333333
Does it work?
gilded marten
in the graph for g it looks like, it goes through the point (1.5|2) which mean t...
yes
f(-1) = -2,
f'(-1) = 2
g(-1) = -4/3
g'(-1) = 4/3
gilded marten
The solution for B would be:
k'(1) = (f'(1)*g(1) - f(1)g'(1))/(g(1))^2
f(1)=0
f'...
k'(1) was also correct
thanks
tranquil pine
So the problem was, that one had to be more precise when graphically reading off the values of f and g...
gilded marten
So the problem was, that one had to be more precise when graphically reading off...
yeah its pretty dumb
.close
carmine plaza
final saddleBOT
azure trellis
You can come think of complex numbers somewhat like vectors
Can you tell me a vector property that will you get you the line joining two other vectors
carmine plaza
Can you tell me a vector property that will you get you the line joining two oth...
vector property?
carmine plaza
Between them
pythagoeran?
azure trellis
Not really
If the vector method is not working for you, you could always just use the distance formula
carmine plaza
well yea the pythagoeran theorem is the same as the distance formula
but like
that didnt work out so idk how to
what vector property?
azure trellis
well yea the pythagoeran theorem is the same as the distance formula
No not really
carmine plaza
yeah but the distance between two points forms a right triangle
carmine plaza
what vector property?
anyway idk
azure trellis
What kind of operations are you familiar with with vectors
azure trellis
You literally missed the most important one
carmine plaza
i probably know what its just vague where we're going here
uhh
well adding is the same as subtraction
theres component, trig and linear form
projections
azure trellis
well adding is the same as subtraction
You should treat them as if they are two separate things
azure trellis
yeah but the distance between two points forms a right triangle
Distance formula is a specific use case of the Pythagoras. Just referring to the method as Pythagoras can invoke wrong ideas
So take two vectors and subtract them
carmine plaza
yeah
carmine plaza
4 +12i
azure trellis
No geometrically
carmine plaza
geometrically?
final saddleBOT
@carmine plaza Has your question been resolved?
split geyser
geometrically?
do yk how to plot complex numbers on a graph
carmine plaza
so uh i just like did this
(2 − 5i + 6 + 7i)/2 = 4 + 1i
and distance formula for length
idk how to do this one doe
split geyser
you can just work it out normally by substituting z1 and z2
carmine plaza
yep i got it right
carmine plaza
now for parametric eq
i have to split this into x and y right and set up separate equations?
split geyser
uh i believe so
carmine plaza
.close
tranquil pine
.open
raven bridge
i'm not sure how to solve this, i have to check if W is a vector subspace
raven bridge
main problem is that i'm not sure what this R_2 is
tulip coyote
main problem is that i'm not sure what this R_2 is
Believe polynomials of degree at most 2?
tulip coyote
Yep yep, pretty much (or equivalently $c = -a$, given some $p(x) = ax^2 + bx + c$)
soft zealotBOT
@tulip coyote
tulip coyote
If you take linear combinations of such polynomials, do they retain the same form is the question now 
final saddleBOT
final saddleBOT
Available help channel!
novel sonnet
I don't understand how to do this
final saddleBOT
I don't understand how to do this
haughty bay
I don't understand how to do this
what theorems did you learn about triangles ?
novel sonnet
what theorems did you learn about triangles ?
my teacher just sprung this on us randomly, we havent done this since last year so i dont really remember sorry
sonic hamlet
Intercept theorem works to find the height of the tree, but not the angle A
and the assignment says to use trig ratios
novel sonnet
or alternatively, this figure ?
yeah i think ive seen that before
haughty bay
ohhh, mackwell is right, i didn't read properly, this is trig
okay, what do you recall about trigonometric functions ?
novel sonnet
like sin, cos and tan?
haughty bay
yup
haughty bay
Well, if you recall, you can find the angle from trig formulas as long as you know two lengths in the triangle
in which triangle do you know two lengths ?
novel sonnet
the one with the person?
haughty bay
yup
novel sonnet
1.6/3?
haughty bay
that's tan(A)
haughty bay
I think it's worth explaining where those formulas come from once we're done
novel sonnet
ok thanks
haughty bay
The relevant trig identity here is than tan(A) = opposite/adjacent
So in this case tan(A) = 1.6/3
novel sonnet
alright i got that part, but i dont know where to go from there to find A
haughty bay
Trig functions have (up to some details that aren't important here) inverses
sin, cos, tan have respective inverses arcsin, arccos, arctan (also written asin, acos, atan)
for example tan(arctan(x)) = x
So if you take A = arctan(1.6/3) you will find the angle such that tan(A) = 1.6/3
there are some asterisks to what i said, but in your case they don't matter
in the end, your solution is A = arctan(1.6/3), which you need a calculator for
,w arctan(1.6/3) in degrees
soft zealotBOT
haughty bay
so that's for finding A
novel sonnet
thanks
haughty bay
Now that you know one of the angles, you only need to know one side to deduce any other side, which allows you to do question (ii)
alright, now for my visual take on what trig is all about, and a way to not have to learn formulas by rote
let me get some paper
novel sonnet
ok
haughty bay
Here we are. This is a circle (close enough) of radius 1, and I drew a right triangle inside.
sine and cosine have very nice interpretations in terms of this unit circle, specifically the point in the top right
Definition : the point on the circle corresponding to angle θ (top right on the diagram) has horizontal position cos(θ) and vertical position sin(θ).
So for a triangle of hypothenuse 1 and one angle θ, the side angle has length cos(θ) and the opposite side has length sin(θ). Follow so far ?
novel sonnet
yep
haughty bay
Since θ can be anything, not only does this describe right triangles, this describes all right triangles
Now we focus on the right triangle proper
If we label our sides using the usual terminology of adjacent, opposite and hypothenuse, we see some familiar relations pop out.
Now, in the case of our triangle with unit hypothenuse they're really trivial, but the key observation is that if you scale the triangle by some constant factor, the side lengths will change, but the ratios will stay the same.
Which means that they hold in all right triangles and that's why they're true.
So in a sense, these relations are a direct consequence of the definition of sine, cosine and tangent
And if you're ever confused which one is which, you can always draw this prototypical right triangle (or even the unit circle) and the right relation will fall out if you look at it hard enough
final saddleBOT
haughty bay
@novel sonnet bonus :
final saddleBOT
Available help channel!
wicked cargo
Need help with this proof question. I don’t know where to start
final saddleBOT
Need help with this proof question. I don’t know where to start
wicked cargo
All I have written down is: assume if x is an irrational number, 2 + x is rational
gritty solar
You can assume it equal to some rational number p
onyx peak
I would start by assuming 2+x is rational
gritty solar
2 + x = p
x = p - 2
Oh no, but x is irrational, therefore this is incorrect
wicked cargo
Ok what does it mean when something is rational or irrational?
gritty solar
Irrational numbers cannot be expressed in the form of p/q where both p and q are integers
Rational numbers can
cedar ivy
Something more
onyx peak
So suppose 2+x=p/q, where both p and q are integers. And try to rearrange it to find out that x can also be written as ratio of 2 integers.
cedar ivy
q≠0
wicked cargo
So suppose 2+x=p/q, where both p and q are integers. And try to rearrange it to ...
Oh, and that means x is rational too which contradicts the negotiation statement
So I’ve come across proof questions which involve the square root of a prime number i.e prove by contradiction the square root of seven is irrational. I assume it is rational in the form m/n where m, n are integers. Apparently I have to also mention m/n is irreducible, why? Why is that necessary?
onyx peak
So I’ve come across proof questions which involve the square root of a prime num...
In this case it's not necessary. But in proof for irrationality of square roots, it's good assumptions to make. Can all fractions be written in irreducible form? Yes. So you can use this fact. Later you will show that the square root fraction is ALWAYS reducible, so it definitely cannot be fraction of 2 integers which means it must be irrational
By forming the assumption (x is rational) in most exact way (x can be written as ratio of integers p and q such that p and q are coprime), you heighten your chance to get to contradiction.
In this case it's not required, but in case of square roots it's quite useful
wicked cargo
In this case it's not necessary. But in proof for irrationality of square roots,...
So when something is reducible, its irrational?
onyx peak
ehmm not quite. I meant that every rational number can be written as irreducible fraction
but the square root fraction, is reducible
and cant be written in irreducible form
In assumption of the square root proof, you assume that $\sqrt{7}=\frac{p}{q}$. such that p/q is irreducible. And this assumption is completely fine, because if sqrt(7) was a fraction, then it could be written in irreducible form
soft zealotBOT
MathIsAlwaysRight
wicked cargo
so a rational number can be expressed in the form p/q where p and q are integers which can be always expressed as an irreducible fraction. But for square root of 7, p/q is reducible which cannot be the case for a rational number being expressed in its simplest form?
Ok thank you
I will take note of that
onyx peak
or rather, no matter on the p and q, it's always reducible. So it doesnt have irreducible form
wicked cargo
Ok
onyx peak
so it cannot be a fraction, since every fraction has irreducible form
wicked cargo
That makes sense
final saddleBOT
Available help channel!
still shuttle
can somebody please explain
final saddleBOT
can somebody please explain
still shuttle
why these are true and false
normal hill
find and identity the critical points of $f(x,y) = 10-3x^2-2y^2+8y+12x$
soft zealotBOT
Mortta
normal hill
wrote out all my deriivitves set them to 0 found the poitns to be 2,2
did the Hessian matrix
But i got my lambda as 2 negative numbers
which means its a maxiumum
but apperetly its a minimum
final saddleBOT
@normal hill Has your question been resolved?
normal hill
.close
final saddleBOT
final saddleBOT
Available help channel!
snow pendant
yes
stoic mural
!status
final saddleBOT
What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
clear hearth
!status
final saddleBOT
What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
you have a semicircle and a triangle
you know how to calculate their areas?
stoic mural
then show your work
we can see where you went wrong then
pearl wraith
yeah and I did
what did you get
clear hearth
Please be serious he's trying to help you...
snow pendant
i am
snow pendant
what did you get
like uh 2873.4
snow pendant
then show your work
we can see where you went wrong then
it was on paper
and i dont have my phone on me
atm
stoic mural
doesn't matter
how would you go about this?
tired hull
just split it into 2 shapes
stoic mural
we dont need to input any numbers
snow pendant
just split it into 2 shapes
I did
snow pendant
ye
tired hull
then just apply them
and add them together
(obviously divide the circle in half for semicircle)
tired hull
did you use radius
snow pendant
yeah
I did 48 squared
times pi
first I divided 48 by 2
to get 21
then I squared it
lethal estuary
48/2 does not equal 21
lethal estuary
what's the area of a semicircle
snow pendant
1/2 pi(radius) squared
lethal estuary
and what'd you get for the semicircle in this problem
snow pendant
576pi
lethal estuary
no, that'd be the area of a circle w/ radius 24
snow pendant
no, that'd be the area of a circle w/ radius 24
uh
tired hull
oops
tired hull
wait no
you wer eon the right track
so that was the area of the circle with radius 24 corrcect?
snow pendant
ya
snow pendant
divide by 2
snow pendant
<@1037102108291387484> ?
288
tired hull
288 pi
correct
so thats the area of the semi circle
to get the triangle you do B*H /2, with 64 as height, 48 as base
so whats the area of the triangle?
tired hull
np
thick wing
final saddleBOT
thick wing
Theta found is 7π/4 how?
dusty harbor
check the solutions for line 3
final saddleBOT
@thick wing Has your question been resolved?
final saddleBOT
final saddleBOT
Available help channel!
tranquil pine
How can I utilize the Lagrange error bound formula to find the error of a series?
final saddleBOT
How can I utilize the Lagrange error bound formula to find the error of a series...
vital crag
still haven't given any equation to help you with
tranquil pine
IM ASKING FOR HELP IM ASKING FOR SOMEONE TO EXPLAIN IT TO ME
vital crag
IM ASKING FOR HELP IM ASKING FOR SOMEONE TO EXPLAIN IT TO ME
calm yourself. there's nothing you've given to explain. what series?
tranquil pine
Ok for example, sinx to the 5th degree , find the lagrange error bound
vital crag
what part of the formula don't you understand
tranquil pine
The entire thing
vital crag
Pick a place to start
tranquil pine
x - x^3/3! + x^5/5!
vital crag
Ok for example, sinx to the 5th degree , find the lagrange error bound
$\sin^5(x)$ isn't a series. do you mean the taylor expansion of $\sin^5(x)$?
soft zealotBOT
riemann
vital crag
and what part of the lagrange error bound don't you understand
this formula right?
https://brilliant.org/wiki/taylor-series-error-bounds
vital crag
yes I dont get it
which part exactly
read one variable at a time. ask when you don't understand the first one
vital crag
do you understand the formula?
tranquil pine
no
vital crag
then do that first
vital crag
vital crag
I’m asking for help
no. you're asking how to use the formula. you can't use something you don't understand
faint locust
$R_{n + 1}(x) = \frac{f^{n + 1}(\xi)}{(n+1)!}(x-a)^{n + 1}$
vital crag
You’re causing hooplas
you're just complaining and whining. do some actual math
tranquil pine
no. you're asking how to use the formula. you can't use something you don't unde...
yes how can I utilize it, I will understand it when I know how to use it
faint locust
tranquil pine
you're just complaining and whining. do some actual math
I am
vital crag
https://www.youtube.com/watch?v=Q9y43_02vFw
this is a good start
tranquil pine
I’m not complaining and whining
vital crag
You’re causing hooplas
.
vital crag
And being difficult for no reason
.
faint locust
You want to maximize that function, n is the degree of your taylor polynomial and a the development point
tranquil pine
I’m unsettled because you’re being difficult for no reason
tranquil pine
.
Yeah you’re causing a hoopla
You’re being extra for no reason lol
How is that complaining
final tangle
we're not here to teach you from scratch
tranquil pine
I am asking for help, not for a math lesson
final tangle
view the recommended resource, show that you've put in effort on your end to try and understand
and point to specific parts that you need clarification on
vital crag
I am asking for help, not for a math lesson
you're still just complaining and whining.
tranquil pine
You want to maximize that function, n is the degree of your taylor polynomial an...
Thanks
tranquil pine
Thanks a lot
tranquil pine
you're still just complaining and whining.
You see how easy it was, but you just want to be unhelpful and cause problems and give me an inconvenience to deal with because you have nothing better to do
You’re not a good troll lol
faint locust
I just sent the formula and explained the parameters involved, lol
tranquil pine
It’s all I needed
vital crag
You see how easy it was, but you just want to be unhelpful and cause problems an...
kepe told you the same thing in the article i linked. you just weren't willing to do any work
final tangle
and the vid does everything in such more detail
tranquil pine
kepe told you the same thing in the article i linked. you just weren't willing t...
Why would I want to read a whole article over a summary of the error bound
faint locust
and the vid does everything in such more detail
Yeah, though I would recommend a book about analysis if you want to go more in depth
vital crag
it was in the first paragraph
tranquil pine
kepe told you the same thing in the article i linked. you just weren't willing t...
If I’m not willing to do any work I wouldn’t be here
vital crag
Why would I want to read a whole article over a summary of the error bound
you're spending 10x more effort into arguing than reading a paragraph of math
VERY FIRST PARAGRAPH
tranquil pine
you're spending 10x more effort into arguing than reading a paragraph of math
You’re spending 10x more effort being an inconvenience to me than just helping me with what I asked for
Instead you cause a hoopla and argument
For no reason too
robust sedge
What the hell is a hoopla?
tranquil pine
It’s like an unnecessary altercation or commotion
final tangle
We encourage self learning where possible
and can direct you to resources that would be useful to you
tranquil pine
I am already self learning
vital crag
You’re spending 10x more effort being an inconvenience to me than just helping m...
you're just lazy
final tangle
your argumentative behaviour indicates that you didn't even bother with clicking the link or viewing the vid
vital crag
can't help the lazy
tranquil pine
you're just lazy
So hypocritical
final tangle
as that would've given everything you asked for
tranquil pine
I’m not lazy lol
You’re judging me off of one experience because of a difference of opinion
tranquil pine
In your opinion im lazy, in my opinion you have no life
final tangle
get the formula, plug and chug
tranquil pine
Unwilling to work or use energy
?
How am I unwilling to work
I am in here speaking to you
Asking for help
You’re lazy
You’re avoiding helping me because it’s easier to argue and cause hooplas
If I was unwilling to work I’d be a discord mod who spends all their day arguing, not asking for help and doing math
I don’t have time for your shenanigans
.close
final saddleBOT
final saddleBOT
Available help channel!
wintry cedar
how would I go about starting this equation?
final saddleBOT
how would I go about starting this equation?
sterile cypress
basically asking which one is equal to cos(teta)
digital sable
I'm not really good with the unit circle.
do you know how transformations work
function transformations
wintry cedar
not really, no. I need a refresher.
wintry cedar
do you know how transformations work
would you want me to DM you?
digital sable
nah its ok
so if we have smth like
f(x-a), we transform this graph a units to the right
now; you should hopefully know what the graphs look like so see which one matches up
wintry cedar
would it be sin(0-pi/2) since all the values listed in the equation are positive?
@digital sable
wintry cedar
how come?
pi/2, pi, and 3pi/2 are all positive values. Since pi/2 could be all those values and the negative transformation would take it to positive is how I was thinking the answer could be right.
@digital sable (sorry for the mass pings)
<@&286206848099549185>
drowsy rapids
how would I go about starting this equation?
just put the given values instead of theta in the question and options and tally them if it is not easy
you can use sin(pi/2-theta) = cos theta
wintry cedar
what would cos(0) equal though? O?
drowsy rapids
no it equals 1
grim badger
Use the unit circle
drowsy rapids
cos zero = 1
i dont know how they teach it in the US
but in India we are made to remember values of all ratios of standard angles
cos zero equals one
wintry cedar
okay.
so for example
I did sin(0-pi/2)
for the first one
it gave me -1
so that would be wrong right?
wintry cedar
yeah
drowsy rapids
and that matches with cos 0
wintry cedar
right
another problem if you don't mind
it's a cos graph as the graph doesn't start at 0,0
drowsy rapids
sure
wintry cedar
the amp is 2
drowsy rapids
yes
wintry cedar
ohh right
drowsy rapids
2pi
wintry cedar
the graph doesn't really repeat
drowsy rapids
ohh right
all that happened is the graph skewed towards the right
it does repeat
use desmos
to plot cos(theta+pi/4)
it repeats
wintry cedar
,w plot sin(0+pi/4)
soft zealotBOT
wintry cedar
,w plot sin(x+pi/4)
soft zealotBOT
drowsy rapids
,w plot sin(x+pi/4)
soft zealotBOT
drowsy rapids
see it repeats
wintry cedar
I see
drowsy rapids
this is sin
wintry cedar
,w plot cos(x+pi/4)
soft zealotBOT
drowsy rapids
,w plot cos(pi/4+x)
soft zealotBOT
wintry cedar
this is sin
I just noticed 😆
drowsy rapids
see instea of reaching minima value (-1) at pi it reaches at it a little later at pi+pi/4
so you see adding pi/4 in the function just shifted the standard cos graph
wintry cedar
see instea of reaching minima value (-1) at pi it reaches at it a little later a...
right
drowsy rapids
pi/4 units left
if you subtract pi/4 it will shift to the right by pi/4
lets see
,w plot cos(x-pi/4)
soft zealotBOT
drowsy rapids
see it shifted to the left
wintry cedar
it shifted the graph to the left
drowsy rapids
yes
wintry cedar
because the swirl is to the the left
drowsy rapids
to pi-pi/4
wintry cedar
won't it shift to the right since it's negative?
wintry cedar
so would the answer be 2cos(x-pi/4)?
drowsy rapids
won't it shift to the right since it's negative?
for every value of theta you are decreasing pi/4 from it so where it used to reach a value at a certain x now it reaches it at x-pi/4
yes it is x-pi/4
drowsy rapids
go on
drowsy rapids
if theta is zero the height is thirteen
put zero in the function you end up with 11+2sinpi/2
which would be 13
wintry cedar
thanks.
drowsy rapids
np
final saddleBOT
final saddleBOT
Available help channel!
unreal juniper
If a 3-digit number has at least one '4and at least one '5as digits then, the number of such 3-digit numbers greater than 100, wherein the position value of each 5 is larger than the position value of each 4, equals
final saddleBOT
If a 3-digit number has at least one '4and at least one '5as digits then, the nu...
final saddleBOT
@unreal juniper Has your question been resolved?
final saddleBOT
final saddleBOT
Available help channel!
odd wren
$f(x) = \frac{x^{n}\cdot (1-x)^n}{n!}$
final saddleBOT
$f(x) = \frac{x^{n}\cdot (1-x)^n}{n!}$
soft zealotBOT
Nomad_InSearchfor_
odd wren
then for any integer k>=0, the k-th derivative
$f^{(k)}(0)$ and $f^{(k)}(1)$ are both integers, that is needed to be proved
!status
final saddleBOT
What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
Nomad_InSearchfor_
odd wren
$f(x)=\frac{(x-x^{2})^{n}}{n!}$
soft zealotBOT
Nomad_InSearchfor_
soft zealotBOT
Nomad_InSearchfor_
odd wren
$f^{2}(x)=\frac{(-2)^{n-2}}{(n-2)!}$
soft zealotBOT
Nomad_InSearchfor_
odd wren
someone ?
i have tried for different value of n , all i am getting is rational numbers and often only integers
i don't see how does this always produce integers,
NOTE: there are no conditions set for "n"
odd wren
i need to prove that k-th derivative of f(x) for x=0,1 are always integers
but see that i am getting rational numbers
woeful atlas
R^2?
odd wren
?
woeful atlas
what's the domain
odd wren
0 and 1
woeful atlas
do you know induction?
odd wren
nah
woeful atlas
is that sarcasm
odd wren
nah
woeful atlas
idk your text sounded sarcastic
odd wren
and this is also not a sarcasm
woeful atlas
not a sarcasm
odd wren
idk your text sounded sarcastic
do you have synesthesia
odd wren
i might know little bit from here and there about induction
woeful atlas
well you can solve for the nth derivative and then substitute x and then your inductive step will be the nth derivative with your x value subbed (1 at a time) and then you will induce your answer from the general form of an integer
but that's a mega ass pain
also this doesn't seem like pre-uni work lol
odd wren
i suppose that is quite treacherous
odd wren
sure
woeful atlas
great
final saddleBOT
@odd wren Has your question been resolved?
tranquil pine
final saddleBOT
tranquil pine
i know that this is the equation for it but whenever i replaced x=2 and inf=3 for order 3 for example the answer is wrong
final saddleBOT
@tranquil pine Has your question been resolved?
vital crag
i know that this is the equation for it but whenever i replaced x=2 and inf=3 fo...
order is degree polynomial expansion in the sum to include
,tex .maclaurin
soft zealotBOT
riemann
tranquil pine
ohhhhh
tranquil pine
5th and 6th degree (degree 6 term is 0)
so 4th degree would be 0 too right?
vital crag
yes, so 4th degree expansion is the same as 3rd degree as well
tranquil pine
And 5 and 6 is the same
because at 4 and 6 it equal to 0
i got it now
thank you @vital crag
i appreciate it!
have a nice day
.closw
.close
solid badger
How to do qk
final saddleBOT
How to do qk
strong atlas
How to do qk
Partial fractions
final saddleBOT
final saddleBOT
Available help channel!
granite oracle
If minus was before 7root3, would you need to change all signs (-+)? So all - would be + and all + would be -?
final saddleBOT
If minus was before 7root3, would you need to change all signs (-+)? So all - wo...
granite oracle
Also are there two ways to do this right?
1: multiply each number and then do the equation
2: do the brackets and then multiply?
final saddleBOT
@granite oracle Has your question been resolved?
granite oracle
<@&286206848099549185>
cyan kayak
@granite oracle it's not exactly clear what you're asking, but maybe you're driving at this:
[
\begin{align*}
&(7\sqrt{3} - 5\sqrt{3})(3\sqrt{2} + 4\sqrt{3}) \
&(-7\sqrt{3} - 5\sqrt{3})(3\sqrt{2} + 4\sqrt{3}) && \qqtext{If minus was before $7\sqrt{3}$} \
&= -1 (7\sqrt{3} + 5\sqrt{3})(3\sqrt{2} + 4\sqrt{3}) && \qqtext{Then you could pull out a $-1$} \
\end{align*}
]
All remaining signs are +s on the inside, so you'll get -1 (a + b + c + d), so -a - b - c - d.
soft zealotBOT
OmnipotentEntity
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
ashen geyser
You can bracket sqrt(3)
ashen geyser
Then calculate things in brackets, then multiply answer with sqrt(3)
Oh, wait
I missed sqrt(2) thing
Sorry
granite oracle
How do you know if a number is dividable by 4?
sturdy cypress
How do you know if a number is dividable by 4?
the rule is that it must end in 00, 08,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,92 or 96
so the last 2 digits is a number divisible by 4
it's not as easy as 3
cyan kayak
Look at the second to last digit, if it's even, then the last digit must be 0, 4, or 8. If the second to last digit is odd, the last digit must be 2 or 6.
It's the same as what frowny frog said, but a little easier to remember.
final saddleBOT
@granite oracle Has your question been resolved?
stiff harness
final saddleBOT
atomic mural
Do you know how to shift the coordinate axis?
stiff harness
nope
atomic mural
No no
stiff harness
neg 1