Answer:
Follows are the solution to the given point:
Step-by-step explanation:
In point a:
¬∃y∃xP (x, y)
∀x∀y(>P(x,y))
In point b:
¬∀x∃yP (x, y)
∃x∀y ¬P(x,y)
In point c:
¬∃y(Q(y) ∧ ∀x¬R(x, y)) 
∀y(> Q(y) V ∀ ¬ (¬R(x,y)))
∀y(¬Q(Y)) V ∃xR(x,y) )
In point d:
¬∃y(∃xR(x, y) ∨ ∀xS(x, y))
∀y(∀x>R(x,y)) 
∃x>s(x,y))
In point e:
¬∃y(∀x∃zT (x, y, z) ∨ ∃x∀zU (x, y, z))
∀y(∃x ∀z)>T(x,y,z) ∀x ∃z> V (x,y,z))
Answer:
x=sqrt{22}
Step-by-step explanation:
Answer:
12 units²
Step-by-step explanation:
The area (A) of a trapezoid is calculated as
A = 
h (b₁ + b₂ )
where h is the perpendicular height and b₁, b₂ the parallel bases
Here h = 4 ( perpendicular distance between the bases ) and
b₁ = SR = 2, b₂ = TA = 4 , then
A = × 4 × (2 + 4) = 2 × 6 = 12 units²
Answer:
17
Step-by-step explanation:
f(x) = 3x + 2
f(5) = 3(5) + 2 = 17
<span>The topic of decimals, and patterns of decimals, seems to be of slightly greater interest to GMAC in the GMAT OG13e than in previous editions. What decimals terminate? What decimals repeat? In this post, we’ll take a look at these questions.</span>
