X=10 and x=0
Steps:
5x=x^2/2
10x=x^2
x^2-(10x)=0
(x-10)(x+0)=0
Solve for factors
x-10=0
x=10
x+0=0
x=0
Answer:
Left
Step-by-step explanation:
(x,y)
For (2), start with the base case. When n = 2, we have
(n + 1)! = (2 + 1)! = 3! = 6
2ⁿ = 2² = 4
6 > 4, so the case of n = 2 is true.
Now assume the inequality holds for n = k, so that
(k + 1)! > 2ᵏ
Under this hypothesis, we want to show the inequality holds for n = k + 1. By definition of factorial, we have
((k + 1) + 1)! = (k + 2)! = (k + 2) (k + 1)!
Then by our hypothesis,
(k + 2) (k + 1)! > (k + 2) 2ᵏ = k•2ᵏ + 2ᵏ⁺¹
and k•2ᵏ ≥ 2•2² = 8, so
k•2ᵏ + 2ᵏ⁺¹ ≥ 8 + 2ᵏ⁺¹ > 2ᵏ⁺¹
which proves the claim.
Unfortunately, I can't help you with (3). Sorry!
Diagonals of the parallelogram are congruent.
