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Vilka [71]
2 years ago
13

Please help! You need to create a fenced off region of land for cattle to graze. The grazing area must be a total of 500 square

feet, surrounded by a fence, and in the shape of a regular polygon. Within this grazing area, the length of the apothem must be 10 feet long. Part I. Find the total perimeter of the grazing area. Part II. If the cost of the fence is $7.95 per linear foot, how much will it cost to place a fence around the entire grazing area? Part III. Suppose this grazing area is of an industrial grazing area for a major industry farm. What is the total cost to build a fence around the entire land of the large scaled farm? Include all of the necessary calculations in your final answers for Parts I, II, and III.
Mathematics
1 answer:
Svetradugi [14.3K]2 years ago
5 0

Answer:

multiply the price and max then divide it to get the total cost

Step-by-step explanation:

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For the binomial expansion of (x + y)^10, the value of k in the term 210x 6y k is a) 6 b) 4 c) 5 d) 7
Sloan [31]

Answer:

a) 6

Step-by-step explanation:

Expanding the polynomial using the formula:

$(x+y)^n=\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k $

Also

$\binom{n}{k}=\frac{n!}{(n-k)!k!}$

I think you mean 210x^6y^4

We can deduce that this term will be located somewhere in the middle. So I will calculate k= 5; k=6 \text{ and } k =7.

For k=5

$\binom{10}{5} (y)^{10-5} (x)^{5}=\frac{10!}{(10-5)! 5!}(y)^{5} (x)^{5}= \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5! }{5! \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } \\ =\frac{30240}{120} =252 x^{5} y^{5}$

Note that we actually don't need to do all this process. There's no necessity to calculate the binomial, just x^{n-k} y^k

For k=6

$\binom{10}{6} \left(y\right)^{10-6} \left(x\right)^{6}=\frac{10!}{(10-6)! 6!}\left(y\right)^{4} \left(x\right)^{6}=210 x^{6} y^{4}$

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