Mathematics, Physics and Computer science : HOMEWORK HELP BY ME

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Salam walalo,

i am here to help anyone with maths,physics or CS problems.Ask away and hopefully our collective minds can answer your questions.
Let ² ≡ 1/N. Choose a number at random between 0 and 1. Choose a second number
between ² and 1 + ². Choose a third number between 2² and 1 + 2². Continue this
process, until you choose an Nth number between 1 − ² and 2 − ². What is the
probability that the first number you choose is the smallest of all the numbers?
Assume that N is very large, and make suitable approximations.
 
Let ² ≡ 1/N. Choose a number at random between 0 and 1. Choose a second number
between ² and 1 + ². Choose a third number between 2² and 1 + 2². Continue this
process, until you choose an Nth number between 1 − ² and 2 − ². What is the
probability that the first number you choose is the smallest of all the numbers?
Assume that N is very large, and make suitable approximations.
Assume a set:
{x_i} from 0=> N
b = 1/N
ib^2<x_0<1+ib^2
Assuming N => infinite
b=>0
hence:
0<x<1
Note then you pick N numbers independantly from a set of numbers between 0=>1
Hence
p(x) = 1/N~0 for Large N

HOWEVER

Note however your question is contradictive
1<N<2

thus b!= 0
 
Assume a set:
{x_i} from 0=> N
b = 1/N
ib^2<x_0<1+ib^2
Assuming N => infinite
b=>0
hence:
0<x<1
Note then you pick N numbers independantly from a set of numbers between 0=>1
Hence
p(x) = 1/N~0 for Large N

HOWEVER

Note however your question is contradictive
1<N<2

thus b!= 0
Can u write essays
 
Screenshot_20181221-191132.png
 
In any situation with two variables we do a trick.Take your question 5.

2x^2 + xy -y^2 +3x-3y-2=?

Now , when you witness an “xy” in a question it implies their is multiplication going on.The key to solve this question is by factorising my variables.

Let’s rearrange our question

2x^2 + xy -y^2 +3x-3y-2=2x^2 + xy +3x-y^2-3y-2=
A+B

Where we let A= 2x^2+xy+3x and B =-y^2 -3y-2=

Now we can factorise both A and B

x(2x+y+3) and (-y-1)(y+2)

Now since you are in grade 9 I would assume this would be sufficient.Thus for (5)=> x(2x+y+3)+(-y-1)(y+2)
 
Note However @abdiwarrior you could go further.

This is beyond scope but you should research it.

If of the form ax^2+bx +c=0

x=(-b(+\-)root(b^2-4ac))/2a

Note the solutions e.g(m,c)=>
(x-m)(x-c) is the factorisation.

You could also do a different method:

Let A=2x^2+3x-2 and B=-y^2-3y+xy

A=(2x-1)(x+2) and B = y(x-3-y)

(5)=> (2x-1)(x+2)+y(x-3-y)

If you need to factorise further use The formula.
 
@abdiwarrior

Ok so i actually havent seen much trinomials before so i found myself a challenge...for 20 mins.

I wont answer all your questions , instead i will teach you the process.

First of all lets do 5

2x^2+xy-y^2+3x-3y-2

Ok this is difficult!
lets get it to the form of ax^2 + bx + c : c,a,b are constant with respect to x

2x^2+(3+y)x-(y^2+3y+2)

note this is the form above, just looks a bit different.Now we want to factorise all portions:

=> 2x^2+(3+y)x-((y+2)(y+1))
you can verify this through expanding :)

Now that 2x^2 makes this tricky so instead of working through the maths , just try trial and error.

The solution will always be of the form:

(mx +/- (y+2))(gx +/- (y+1)) : g,m are random constants.

Then we get :
(2x-(y+2))(x+(y+1))

*note the 2 comes from the x^2 and the minus due to the constant(even though there are y's, y is a constant if x is our main variable)

try the others :)
 
Let ² ≡ 1/N. Choose a number at random between 0 and 1. Choose a second number
between ² and 1 + ². Choose a third number between 2² and 1 + 2². Continue this
process, until you choose an Nth number between 1 − ² and 2 − ². What is the
probability that the first number you choose is the smallest of all the numbers?
Assume that N is very large, and make suitable approximations.
What's ²?
Assume a set:
{x_i} from 0=> N
b = 1/N
ib^2<x_0<1+ib^2
Assuming N => infinite
b=>0
hence:
0<x<1
Note then you pick N numbers independantly from a set of numbers between 0=>1
Hence
p(x) = 1/N~0 for Large N

HOWEVER

Note however your question is contradictive
1<N<2

thus b!= 0

And what's b!?
 
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