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densk [106]
3 years ago
10

A park ranger has 32 feet of fencing to fence four sides of a rectangular recycling site. What is the greatest area of recycling

site that the ranger can fence? Explain how you know.
Mathematics
2 answers:
wariber [46]3 years ago
8 0
The greatest area he can fence is 64 ft².

In order to maximize area and minimize perimeter, we use dimensions that are as close to equivalent as possible.  32 feet of fence for 4 sides gives us 8 feet of fence per side.  We would have a square whose side length is 8; the area would be 8*8 = 64.
Goshia [24]3 years ago
5 0

Answer:

A = 64\,ft^{2}

Step-by-step explanation:

The formulas for the perimeter and area of the rectangle are, respectively:

2\cdot x + 2\cdot y = 32\,ft

A = x\cdot y

After some algebraic handling, the formula for area is simplified into the following form:

A = x\cdot (16\cdot ft - x)

A = 16\cdot x-x^{2}

The maximum area is found by means of First and Second Derivative Tests:

A' = 16 - 2\cdot x

A'' = -2

According to the second derivative, the critical point leads invariantly to an absolute maximum. The value of the critical point is:

16-2\cdot x = 0

x = 8\,ft

The length of the other side is:

y = 8\,ft

The maximum area of the recycling site is:

A = 64\,ft^{2}

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Answer:

The shaded area is 23\ in^{2}

Step-by-step explanation:

we know that

The shaded area is equal to the area of the large square minus the area of the two smaller squares

so

A=6^{2} -(3^{2} +2^{2})\\ \\ A=(2*3)^{2} -(3^{2} +2^{2})\\ \\A=(2^{2})(3^{2}) -(3^{2} +2^{2})

Calculate the shaded area

Remember that

3^{2}=9\\ 2^{2}=4

substitute

A=(4)(9) -(9 +4)\\ \\A=36-13\\ \\A=23\ in^{2}

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kogti [31]

Answer:

The SSS theorem

Step-by-step explanation:

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1. Consider the following hypotheses:
Andrej [43]

Answer:

See deductions below

Step-by-step explanation:

1)

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

b) q(y) for some y (Simplification of a))

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)

f)  ∀y B(y) (Modus ponens using b and e)

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

h) B(x)∧C(x) for some x (Conjunction of g and d, selecting y=x on g)

i) ∃x (B(x) ∧ C(x)) (Existencial generalization of h)

3) We will prove that this formula leads to a contradiction.

a) ∀y (P (x, y) ↔ ¬P (y, y)) for some x (Existencial instatiation of hypothesis)

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

c) P (x, x) ↔ ¬P (x, x) (Take y=x in b)

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

7 0
3 years ago
Please help<br> What is the perimeter of PQR? Show your work.
9966 [12]
17 is the answer

3y-2=y+4
2y-2=4
2y=6
2y/2=6/2
y=3
5 0
2 years ago
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