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gavmur [86]
3 years ago
13

The back of Monique's property is a creek. Monique would like to enclose a rectangular area, using the creek as one side and fen

cing for the other three sides, to create a corral. If there is 580 feet of fencing available, what is the maximum possible area of the corral?
Mathematics
2 answers:
umka2103 [35]3 years ago
7 0

Answer:

Maximum possible area of the corral will be 42050 square feet.

Step-by-step explanation:

Monique wants to enclose a rectangular area, using creek on one side and other three sides by the fence.

Let the length of one side of the rectangular area is 'x' feet and other side is 'y'.

Length of the fencing has been given as 580 feet.

Therefore, (2x + y) = 580

2x + y = 580

y = (580 - 2x)-------(1)

Area of the rectangular area = length × width

A = xy

Now we replace the value of y in the area

A = x(580 - 2x)

   = 580x - 2x²

For the maximum area of the corral, we will find the derivative of the area A with respect to x and equate it to zero.

\frac{dA}{dx}=\frac{d}{dx}(580x-2x^{2}) = 0

580 - 4x = 0

4x = 580

x=\frac{580}{4}

x = 145 feet

From equation (1)

y = (580 - 2×145)

  = 580 - 290

  = 290 feet

Maximum area covered of the corral = xy

= 145×290

= 42050 square feet

Therefore, maximum possible area of the corral will be 42050 square feet.

kondor19780726 [428]3 years ago
3 0
So long as the perimeters are the same, rectangles and squares share the same area. For example, a square that is 2m by 2m across is 4m squared. A rectangle of 4m by 1m across is still 4m squared.

Therefore all we want to do here is see how big we can make our “square” perimeter using the creek. We have three sides to spread 580ft across, therefore if we divide this by 3, we get 193.3ft of fencing per side. If we then square this figure, we will then get the maximum possible area, which comes to 37,377ft squared. (That’s a huge garden).
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The perimeter of this triangle is 25 units. An angle bisector divides one side of the triangle into lengths of 2 and 3. Find the
grin007 [14]

Answer:

   8 and 12

Step-by-step explanation:

Sides on one side of the angle bisector are proportional to those on the other side. In the attached figure, that means

  AC/AB = CD/BD = 2/3

The perimeter is the sum of the side lengths, so is ...

  25 = AB + BC + AC

  25 = AB + 5 + (2/3)AB . . . . . . substituting AC = 2/3·AB. BC = 2+3 = 5.

  20 = 5/3·AB

  12 = AB

  AC = 2/3·12 = 8

_____

<em>Alternate solution</em>

The sum of ratio units is 2+3 = 5, so each one must stand for 25/5 = 5 units of length.

That is, the total of lengths on one side of the angle bisector (AC+CD) is 2·5 = 10 units, and the total of lengths on the other side (AB+BD) is 3·5 = 15 units. Since 2 of the 10 units are in the segment being divided (CD), the other 8 must be in that side of the triangle (AC).

Likewise, 3 of the 15 units are in the segment being divided (BD), so the other 12 units are in that side of the triangle (AB).

The remaining sides of the triangle are AB=12 and AC=8.

7 0
4 years ago
This is due tomorrow and i have answers but i think ive done it wrong! please help
Eva8 [605]

Answer:  i) 1 - 9x² - 12x

               ii) 17 - 3x²

             iii) - 20 + 10x² - x⁴

<u>Step-by-step explanation:</u>

g(x) = 3x + 2       h(x) = 5 - x²

i) h(g(x))

  h(3x + 2) = 5 - (3x + 2)²

                  = 5 - (9x² + 12x + 4)

                  = 5 - 9x² - 12x - 4

                  = 1 - 9x² - 12x

ii) g(h(x))

   g(5 - x²) = 3(5 - x²) + 2

                 = 15 - 3x² + 2

                 = 17 - 3x²

iii) h(h(x))

   h(5 - x²) = 5 - (5 - x²)²

                 = 5 - (25 - 10x² + x⁴)

                 = 5 - 25 + 10x² - x⁴

                 = -20 + 10x² - x⁴

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AlexFokin [52]
Is it like this?

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In this problem, you have two different denominators, 2 and 4. If you multiply them all by 2, the term -3/4m will still be a fraction.

-3/4m * 2 = -3/2m

As a result, we must try 4 instead. This causes all of them to cease being a fraction. Even the one with 2 as a denominator.

-1/2 * 4 = -2

7 0
3 years ago
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