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fredd [130]
3 years ago
15

Help with this pls?????

Mathematics
2 answers:
Svetradugi [14.3K]3 years ago
7 0

Answer:

83.74

Step-by-step explanation:

Radius = 6÷2 = 3in

Area = rectangle - circle

= (14 × 8) - (3.14 × 3²)

= 83.74 in²

Alenkinab [10]3 years ago
6 0

Answer:

83.74 in squared

Step-by-step explanation:

The area of the original rectangle is: 14 * 8 = 112 in squared.

The area of a circle is \pi r^{2}, where r is the radius. In this case, we know the diameter is 6, so the radius is 3. Then, the area of the circle is:

3.14 * 3^2 = 3.14 * 9 = 28.26 in squared.

The remaining area is just the total minus the circle area:

112 - 28.26 = 83.74 in squared

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Answer:

See deductions below

Step-by-step explanation:

1)

a) p(y)∧q(y) for some y (Existencial instantiation to H1)

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c) q(y) → r(y) for all y (Universal instatiation to H2)

d) r(y) for some y (Modus Ponens using b and c)

e) p(y) for some y (Simplification of a)

f) p(y)∧r(y) for some y (Conjunction of d) and e))

g) ∃x (p(x) ∧ r(x)) (Existencial generalization of f)

2)

a) ¬C(x) → ¬A(x) for all x (Universal instatiation of H1)

b) A(x) for some x (Existencial instatiation of H3)

c) ¬(¬C(x)) for some x (Modus Tollens using a and b)

d) C(x) for some x (Double negation of c)

e) A(x) → ∀y B(y) for all x (Universal instantiation of H2)

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g) B(y) for all y (Universal instantiation of f)

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b) P (x, y) ↔ ¬P (y, y) for some x, and for all y (Universal instantiation of a)

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But c) is a contradiction (for example, using truth tables). Hence the formula is not satisfiable.

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