Compab=a.b/|a|
b=<0,1,−2√10>
a.b= 2√10
|a| = √10
a.b/|a|=2√10 / √10=2
Answer:
The points (x,y,z) that respond to Ir-r0I =1, are all that describes the form
with:
-1+x₀<x<1+x₀
-1+y₀<y<1+y₀
-1+z₀<z<1+z₀
Step-by-step explanation:
All points required in this problem came from applying the definition of modulus of a vector:
Ir-r0I =1.

Answer: f(x) = (x + 3)(x – 7)
Step-by-step explanation: Use "standard form" of the function and insert values given: vertex (2,-25) intercept point (7,0)
f(x) = a(x-h)² + k from vertex, h is 2 y is -25 from intercept, x is 7 f(x) is 0
to find a, 0 = a(7-2)² +(-25) 0 = a(7-2)² -25 add 25 to both sides
25 = a(5)² 25 = 25a 25/25 = a 1=a (seems useless but verifies implied "a"coefficient is 1)
f(x) = a(x-h)² + k solve to get the quadratic form
f(x) = (x-2)² -25 (x - 2)² is x² -4x +4
f(x) = x² -4x +4 -25 simplify
f(x) = x² -4x - 21 then factor
f(x) = (x + 3)(x - 7)