p^2-3
It is an inequality than can be solved for p: 5 ≥ p - 3 → p - 3 ≤ 5 → p ≤ 8 So any value less than, or equal to, 8 will do for p.
All points less than 3 distant from point 'P' comprise a circle, centered at 'P', with a radius of 3, but NOT including the line that is the circumference of the circle.
51 = 3*17
p-2
3p-5
p^2-3
It is an inequality than can be solved for p: 5 ≥ p - 3 → p - 3 ≤ 5 → p ≤ 8 So any value less than, or equal to, 8 will do for p.
All points less than 3 distant from point 'P' comprise a circle, centered at 'P', with a radius of 3, but NOT including the line that is the circumference of the circle.
p(a) = 1/3, p(b) = 1/2, p(a and b) = p(a)*p(b) = 1/6
As an algebraic expression, six less than a number p is simply: p-6
3(n-9)
51 = 3*17
p-2
p-6
p-15
p is less than or equal to 4