Answer: the length is 6 feet. The width is 2 feet
Step-by-step explanation:
The garden pool is rectangular in shape.
Let L represent the length of the rectangular garden.
Let W represent the width of the rectangular garden.
The area of a rectangle is expressed as L× W. The area of the rectangular garden pond is 12ft^2. It means that
L× W = 12 - - - - - - - - - - 1
The length of the pool needs to be 2ft more than twice that width. This means that
L = 2W + 2 - - - - - - - - - - - 2
Substituting equation 2 into equation 1, it becomes
W(2W + 2) = 12
2W^2 + 2W = 12
2W^2 + 2W - 12 = 0
W^2 + W - 6 = 0
W^2 + 3W - 2W - 6 = 0
W(W + 3) - 2(W + 3) = 0
W - 2 = 0 or W + 3 = 0
W = 2 or W = - 3
Since the Width cannot be negative, then the width is 2 feet
Substituting W = 2 into equation 2. It becomes
L = 2×2 + 2 = 6 feet
Answer:
a) ∀x∃y ¬∀zT(x, y, z)
∀x∃y ∃z ¬T(x, y, z)
b) ∀x¬[∃y (P(x, y) ∨ Q(x, y))]
∀x∀y ¬ [P(x, y) ∨ Q(x, y)]
∀x∀y [¬P(x, y) ^ ¬Q(x, y)]
c) ∀x ¬∃y (P(x, y) ^ ∃zR(x, y, z))
∀x ∀y ¬(P(x, y) ^ ∃zR(x, y, z))
∀x ∀y (¬P(x, y) v ¬∃zR(x, y, z))
∀x ∀y (¬P(x, y) v ∀z¬R(x, y, z))
d) ∀x¬∃y (P(x, y) → Q(x, y))
∀x∀y ¬(P(x, y) → Q(x, y))
∀x∀y (¬P(x, y) ^ Q(x, y))
15a) 1.2 x 10^6
15b) 1.2 x 10^-4
Looking at the graph, we can see the domain to be from (0 , 2π).
Now we have to find one period that corresponds to cos(x).
The half-period of cos(x) for this graph appears to be pi/3 and adding another pi/3 gets us 2pi/3 to be our cosine period.
b = 2pi/3
a is the same range as cos(x). Range: (0,0)
y = [a] * cos ([b]*x)
y = [1] * cos([2pi/3]x)