Answer:
Step-by-step explanation:
in a parallelogram diagonals bisect each other.
so mid points are same.
Let coordinates of D be (x,y)
mid point of AC=midpoint Of BD
(-2+5)/2=(x+3)/2
x+3=-3
x=-3-3=-6
x=-6
(3-3)/2=(2+y)/2
y+2=0
y=-2
so D is (-6,-2)
54, 36, 24 are the 1st 3 element of a geometric progression with 2/3 as a common ratio: PROOF:
the 1st term is 54, (a₁= 54) the 2nd term a₂ = 24, then
(a₂ = a₁.r) or 36 = 54.r → r= 36/54 = 2/3. Same logique for the 3rd term.
So 2/3 is common ratio. We know that :U(n) = a.(r)ⁿ⁻¹. Then if a =54 and r = x (given by the problem), then f(x) = 54.xⁿ⁻¹
n, being the rank of any element of this geometric progression
9+12>18
12+18>9
9+18>12
So, yes. A triangle can be formed by the given angles.
Answer:
I dont know.
Step-by-step explanation:
