Take the vector u = <ux, uy> = <4, 3>.
Find the magnitude of u:
||u|| = sqrt[ (ux)^2 + (uy)^2]
||u|| = sqrt[ 4^2 + 3^2 ]
||u|| = sqrt[ 16 + 9 ]
||u|| = sqrt[ 25 ]
||u|| = 5
To find the unit vector in the direction of u, and also with the same sign, just divide each coordinate of u by ||u||. So the vector you are looking for is
u/||u||
u * (1/||u||)
= <4, 3> * (1/5)
= <4/5, 3/5>
and there it is.
Writing it in component form:
= (4/5) * i + (3/5) * j
I hope this helps. =)
Answer:
1048576
Step-by-step explanation:
calculator
Answer:
(D) 
Step-by-step explanation:
Answer:
f=
x2−2x+1
x3+2x2+x
Step-by-step explanation:
Let's solve for f.
fx=
(1−x)2
(1+x)2
Step 1: Multiply both sides by x^2+2x+1.
fx3+2fx2+fx=x2−2x+1
Step 2: Factor out variable f.
f(x3+2x2+x)=x2−2x+1
Step 3: Divide both sides by x^3+2x^2+x.
f(x3+2x2+x)
x3+2x2+x
=
x2−2x+1
x3+2x2+x
f=
x2−2x+1
x3+2x2+x
There are infinitely prime numbers "p", but p+2 and p+4 are not primes.
Example: Suppose p = 7 (prime) if you add 2, it becomes 9 (Not a Prime).
Same thing if p = 11(prime), if yo add 4, it becomes 15 (Not a prime number)
