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SOVA2 [1]
4 years ago
9

Pat wishes to fence in a rectangular area, using the wall of her garage as one side, so only three sides need to be fenced. If s

he has 80 ft of fencing available, what is the maximum area that can be fenced in?
Mathematics
1 answer:
natali 33 [55]4 years ago
8 0

Answer:

The maximum area is 800 ft².

Step-by-step explanation:

Lets call L the length of the side parallel to the garage and M the length of the other two sides. The total amount to fence is 2M+L and the area is L*M.

Since Pat got 80ft of fence, then 2M+L = 80, hence L = 80-2M. By replacing the value of L in the formula of the area, we get that

A(M) = M*(80-2M) = -2M² + 80M

The maximum of this function can be obtaining throught derivation, but since it is a quadratic with negative main coefficient, we know that the maximum is the vertex. The x-coordinate of the vertex is '-b/2a' = -80/2*(-2) = 20. The y-coordinate (which represents the maximum area),  as a result, is A(20) = -2*20²+80*20 = 800.

800 is the maximum area that can be fenced. Note that M = 20 and L = 80-2L = 40.

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3 years ago
Which ofnthe following shows the correct use of distributive property when solving 1/3(33-x)=135,2
madam [21]

Answer: OPTION C.

Step-by-step explanation:

The missing options are:

A. (33 - x) = \frac{1}{3} * 135.2\\\\B. (\frac{1}{3} *33) - \frac{1}{3} x = \frac{1}{3} * 135.2\\\\C.(\frac{1}{3}  * 33) - \frac{1}{3} x = 135.2\\\\D. (\frac{1}{3}  * 33) + \frac{1}{3} x = 135.2

In order to solve this exercise you need to remember the following:

1. The Distributive property states that:

a(b+c)=ab+ac\\\\a(b-c)=ab-ac

2. The multiplication of signs:

(+)(+)=+\\(-)(-)=+\\(-)(+)=-\\(+)(-)=-

In this case the exercise gives you the following equation:

\frac{1}{3}(33-x)=135.2

Since you need to solve for the variable "x" in order to  find its value, the first  step you must apply in the in the procedure is to eliminate the parentheses applying the Distributive property.

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3 years ago
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