Share something pointless about your life!

me too jyxi!

Quote:

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Originally Posted by antalara

According to the model used by a specialist in influenza epidemics, the most recent outbreak of an unknown strain spreads on a city in a circular way where from center to "x" km point of the outbreak, the number of patients at the rate changes of:

F(x)= sin^3(x)
-----------------
---------- x

for x ∈ [1 , 2]
*

The health measures indicate that if the number of patients in an area where the model is valid, it is at least 4, is required to declare a quarantine and evacuate the mentioned area.

a. Provide an integral that calculates the number of patients to be in the mentioned sector ranging from 1 km. up to 2 km. the extent of the outbreak.

Justify the provided integral, to partition the range and significance of each of the factors, the Riemann sum corresponding to the context.

b. Get the number of patients who will be in the area referred to using the Fundamental Theorem of Calculus, and tell us if the quarantine si declare or not.


(probably some typos are into the big paragraph, not my native language, i speak the language i have created -ill-call-it-'rescrove')

---

ok fine, no one accepted

lemme try it

you can the people who are gonna be infected (patients) using the following

i can divide for x ∈ [1 , 2] on a N sub-intervals
every sub-interval has a - Δx wide -
i can have some sample points on the middle of every sub-interval ..and those receive the name of - Xi* -

knowing a bit of area and circles i can get some new data like this one

http://i.imgur.com/MDxYF.png

i can assume F(x) (mentioned on the quoted post) as a constant on every sub-intervals i have made..you can get how many patients i can get on the selected area [1,2]

so patients are "P"

P = π [(Xi)^2 - (Xi-1)^2] F(x)

P = π [(Xi + Xi-1) (Xi - Xi-1)] F(x)

having this i -probably- can get the exact number of patients
on this way

-------------n
Lim---------Σ π (2Xi*) F(x) Δx
n->∞-----n=1

this (finally) give us the integral

2
∫ π (2x) (sin^3(x)) dx
1 --------------------
---------------x


simplifying

2
∫ 2π sin^3(x) dx
1

---2
2π ∫ sin^3(x) dx
---1

I=∫ sin^3(x) dx

I=∫ sin^2(x) sin(x) dx

I=∫ [1-cos^2(x)] sin(x) dx

calling u= cos(x) ___ -du= sin(x)

so

I=∫ sin^3(x) dx = -∫ [1-u^2] du = -u+(u^3 / 3) + C

c is any constant (you can call it monkey or bear or sally or banned or what you want)

we have mentioned u=cos(x)
so the solution is

I=∫ sin^3(x) dx = -cos(x)+[cos^3(x)/3] + C

this give us the answer for a indefinite integral but ours is a definited one (so the constant will say bye this time)
-------------------------------------2
P= 2π [ -cos(x)+[cos^3(x)/3] | ]
-------------------------------------1


P is---

P = 2π [ ( -cos(2) + (1/3)cos^3(2) ) - ( -cos(1) + (1/3)cos^3(1) ) ] = 6


around 6 patients we can get in the mentioned area and 6>4 so .... yes, the quarantine declares



now you can sleep on my serious (and probably pointless) post ...zzz...

(probably some typos are into the big paragraph, not my native language, i speak the language i have created -ill-call-it-'rescrove')
black letters are my secret...lies!!! but i dont care
now i wonder if i made a mistake around..... feeling nervous

All of that for 6.
Psh.

Quote:

---

Originally Posted by antalara

ok fine, no one accepted

lemme try it

you can the people who are gonna be infected (patients) using the following

i can divide for x ∈ [1 , 2] on a N sub-intervals
every sub-interval has a - Δx wide -
i can have some sample points on the middle of every sub-interval ..and those receive the name of - Xi* -

knowing a bit of area and circles i can get some new data like this one

http://i.imgur.com/MDxYF.png

i can assume F(x) (mentioned on the quoted post) as a constant on every sub-intervals i have made..you can get how many patients i can get on the selected area [1,2]

so patients are "P"

P = π [(Xi)^2 - (Xi-1)^2] F(x)

P = π [(Xi + Xi-1) (Xi - Xi-1)] F(x)

having this i -probably- can get the exact number of patients
on this way

-------------n
Lim---------Σ π (2Xi*) F(x) Δx
n->∞-----n=1

this (finally) give us the integral

2
∫ π (2x) (sin^3(x)) dx
1 --------------------
---------------x


simplifying

2
∫ 2π sin^3(x) dx
1

---2
2π ∫ sin^3(x) dx
---1

I=∫ sin^3(x) dx

I=∫ sin^2(x) sin(x) dx

I=∫ [1-cos^2(x)] sin(x) dx

calling u= cos(x) ___ -du= sin(x)

so

I=∫ sin^3(x) dx = -∫ [1-u^2] du = -u+(u^3 / 3) + C

c is any constant (you can call it monkey or bear or sally or banned or what you want)

we have mentioned u=cos(x)
so the solution is

I=∫ sin^3(x) dx = -cos(x)+[cos^3(x)/3] + C

this give us the answer for a indefinite integral but ours is a definited one (so the constant will say bye this time)
-------------------------------------2
P= 2π [ -cos(x)+[cos^3(x)/3] | ]
-------------------------------------1


P is---

P = 2π [ ( -cos(2) + (1/3)cos^3(2) ) - ( -cos(1) + (1/3)cos^3(1) ) ] = 6


around 6 patients we can get in the mentioned area and 6>4 so .... yes, the quarantine declares



now you can sleep on my serious (and probably pointless) post ...zzz...

(probably some typos are into the big paragraph, not my native language, i speak the language i have created -ill-call-it-'rescrove')
black letters are my secret...lies!!! but i dont care
now i wonder if i made a mistake around..... feeling nervous

---

http://sarahsavvy.files.wordpress.co...9ca1e755e3.jpg

Now I'm sure that there are many jobs that require you to know how to do very complex math...But seriously :|

Quote:

---

Originally Posted by antalara

ok fine, no one accepted

lemme try it

you can the people who are gonna be infected (patients) using the following

i can divide for x ∈ [1 , 2] on a n sub-intervals
every sub-interval has a - Δx wide -
i can have some sample points on the middle of every sub-interval ..and those receive the name of - xi* -

knowing a bit of area and circles i can get some new data like this one

http://i.imgur.com/mdxyf.png

i can assume f(x) (mentioned on the quoted post) as a constant on every sub-intervals i have made..you can get how many patients i can get on the selected area [1,2]

so patients are "p"

p = π [(xi)^2 - (xi-1)^2] f(x)

p = π [(xi + xi-1) (xi - xi-1)] f(x)

having this i -probably- can get the exact number of patients
on this way

-------------n
lim---------Σ π (2xi*) f(x) Δx
n->∞-----n=1

this (finally) give us the integral

2
∫ π (2x) (sin^3(x)) dx
1 --------------------
---------------x


simplifying

2
∫ 2π sin^3(x) dx
1

---2
2π ∫ sin^3(x) dx
---1

i=∫ sin^3(x) dx

i=∫ sin^2(x) sin(x) dx

i=∫ [1-cos^2(x)] sin(x) dx

calling u= cos(x) ___ -du= sin(x)

so

i=∫ sin^3(x) dx = -∫ [1-u^2] du = -u+(u^3 / 3) + c

c is any constant (you can call it monkey or bear or sally or banned or what you want)

we have mentioned u=cos(x)
so the solution is

i=∫ sin^3(x) dx = -cos(x)+[cos^3(x)/3] + c

this give us the answer for a indefinite integral but ours is a definited one (so the constant will say bye this time)
-------------------------------------2
p= 2π [ -cos(x)+[cos^3(x)/3] | ]
-------------------------------------1


p is---

p = 2π [ ( -cos(2) + (1/3)cos^3(2) ) - ( -cos(1) + (1/3)cos^3(1) ) ] = 6


around 6 patients we can get in the mentioned area and 6>4 so .... yes, the quarantine declares



now you can sleep on my serious (and probably pointless) post ...zzz...

(probably some typos are into the big paragraph, not my native language, i speak the language i have created -ill-call-it-'rescrove')
black letters are my secret...lies!!! But i dont care
now i wonder if i made a mistake around..... Feeling nervous

---

tl:dr.....

Quote:

---

Originally Posted by AlaskanE

Spoiler: Show

Quote:

---

Originally Posted by antalara

ok fine, no one accepted

lemme try it

you can the people who are gonna be infected (patients) using the following

i can divide for x ∈ [1 , 2] on a N sub-intervals
every sub-interval has a - Δx wide -
i can have some sample points on the middle of every sub-interval ..and those receive the name of - Xi* -

knowing a bit of area and circles i can get some new data like this one

http://i.imgur.com/MDxYF.png

i can assume F(x) (mentioned on the quoted post) as a constant on every sub-intervals i have made..you can get how many patients i can get on the selected area [1,2]

so patients are "P"

P = π [(Xi)^2 - (Xi-1)^2] F(x)

P = π [(Xi + Xi-1) (Xi - Xi-1)] F(x)

having this i -probably- can get the exact number of patients
on this way

-------------n
Lim---------Σ π (2Xi*) F(x) Δx
n->∞-----n=1

this (finally) give us the integral

2
∫ π (2x) (sin^3(x)) dx
1 --------------------
---------------x


simplifying

2
∫ 2π sin^3(x) dx
1

---2
2π ∫ sin^3(x) dx
---1

I=∫ sin^3(x) dx

I=∫ sin^2(x) sin(x) dx

I=∫ [1-cos^2(x)] sin(x) dx

calling u= cos(x) ___ -du= sin(x)

so

I=∫ sin^3(x) dx = -∫ [1-u^2] du = -u+(u^3 / 3) + C

c is any constant (you can call it monkey or bear or sally or banned or what you want)

we have mentioned u=cos(x)
so the solution is

I=∫ sin^3(x) dx = -cos(x)+[cos^3(x)/3] + C

this give us the answer for a indefinite integral but ours is a definited one (so the constant will say bye this time)
-------------------------------------2
P= 2π [ -cos(x)+[cos^3(x)/3] | ]
-------------------------------------1


P is---

P = 2π [ ( -cos(2) + (1/3)cos^3(2) ) - ( -cos(1) + (1/3)cos^3(1) ) ] = 6


around 6 patients we can get in the mentioned area and 6>4 so .... yes, the quarantine declares



now you can sleep on my serious (and probably pointless) post ...zzz...

(probably some typos are into the big paragraph, not my native language, i speak the language i have created -ill-call-it-'rescrove')
black letters are my secret...lies!!! but i dont care
now i wonder if i made a mistake around..... feeling nervous

---

tl:dr.....

---

maybe the NEXT ONE should be a little bit shorter....huh? :D.. it took like 45 min to write all that post.. on paper and brains, you have like... 25 min to do that...(im thinking on a situation when you see this problem for the 1st time)..like a test

I have a boyfriend finally.
Go **** yourself society. <3

Quote:

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Originally Posted by Orphic

I have a boyfriend finally.
Go **** yourself society. <3

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well not sure to say .....but

Congratulations!..

Also found out:

Chafing.
Sucks.
So.
Much.

I thin i have like 12 different personalities

Im serious , i found the last one on the weekend

The ban is strong with this one.

i have a spot of some sort at the back of my neck and it's so akward to reach around.
who wants to pop it for me?

My breath smells so bad of garlic.

Not even my dog will kiss me.

Quote:

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Originally Posted by Arcayna

My breath smells so bad of garlic.

Not even my dog will kiss me.

---

A complete turn off.. good thing your good looks makes up for your bad breath. XD

Quote:

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Originally Posted by Marazhu

A complete turn off.. good thing your good looks makes up for your bad breath. XD

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i swear your gay?