Answer:
last choice
Step-by-step explanation:
If we are looking at positive integers and we take out the odd ones, the only one's that are left are the even ones.
So last choice. (Also I assume AC meant the complement of A)
Answer:
Step-by-step explanation:
Question
Find the perimeter of a triangle with vertices A(2,5) B(2,-2) C(5,-2). Round your answer to the nearest tenth and show your work.
perimeter of a triangle = AB+AC+BC
Using the distance formula
AB = sqrt(-2-5)²+(2-2)²
AB = sqrt(-7)²
AB =sqrt(49)
AB =7
BC = sqrt(-2+2)²+(2-5)²
BC = sqrt(0+3²)
BC =sqrt(9)
BC =3
AC= sqrt(-2-5)²+(2-5)²
AC= sqrt(-7)²+3²
AC =sqrt(49+9)
AC =sqrt58
Perimeter = 10+sqrt58
F(x) = 1/(x+2) & g(x) = x/(x-3)
(f(x) + g(x) = 1/(x+2) + x/(x-3). Reduce to same denominator:
1/(x+2) + x/(x-3) =(x-3) + x(x-3)/(x+2).(x-3) ==> (x²+3x-3)/(x+2).(x-3)
C(-8,2) and M(0,0) , since M is at the origin. Let x₁ and y₁ be the
coordinates of S →s(x₁ , y₁)
C(-8,2) and S(x₁ , y₁)
The coordinates of M, the midpoint of CS are M(x₂ , y₂)
a) x₂ = (-8 + x₁)/2 , but x₂ = 0, then :
0 = -4+x₁/2 and x₁ = 8
b) y₂ = (2+y₁)/2 , but y₂ = o, then:
0 = 2+ y₁/2 and y₂ = -2
Then the coordinates of S are S(8 , -2)
