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babymother [125]
2 years ago
5

A farmer needs to enclose a section of land in the shape of a parallelogram by using one side of a barn for one side. the barn a

nd the side opposite of the barn will both be 13 feet long. one of the other sides will be 9 feet long. what is the length of the last side
Mathematics
2 answers:
Alecsey [184]2 years ago
6 0
The answer would be 9 feet.
alexgriva [62]2 years ago
3 0

Answer: 9 feet.

Step-by-step explanation:

Given: A farmer needs to enclose a section of land in the shape of a parallelogram by using one side of a barn for one side.

We know that in<em> the opposite sides of parallelogram are equal.</em>

<em />

Since, it is given that the barn and the side opposite of the barn will both be 13 feet long and one of the other sides will be 9 feet long.

Therefore, the last side must be opposite to 9 feet long side.

Hence, the length of the last side must be 9 feet.

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23.78

Step-by-step explanation:

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My granddaughter is having problems understanding math questions; Divide 360 / 4
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Question: \frac{360}{4}
There are two methods of doing this. One is via long division. It is long, and hard to type in a meaningful way, so I will not include that method. The other, which I often employ for dividing by small amounts such as 4, is to divide it in half 2 times. This is possible because 4 = 2 times 2. You now have the fractions \frac{360}{2*2} = \frac{180}{2} = \frac{90}{1} = 90
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What is the focus of the parabola y = 10x??
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A standard six-sided die is rolled twice and the top faces are observed. What is the probability that the sum of the numbers on
melisa1 [442]

<u>Answer: </u>

A standard six-sided die is rolled twice and the top faces are observed and the probability that the sum of the numbers on the top faces is 10 is \frac{1}{12}

<u>Solution: </u>

The sample space of two dice outcomes is given as follows:

{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3, 1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4, 1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5, 1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6, 1) (6,2) (6,3) (6,4) (6,5) (6,6) }

Total number of events n(s) = 36

Expected Outcomes = {(4, 6), (5, 5), (6, 4)}

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Probability (sum of numbers on the top faces is 10) = \frac{n(E)}{n(s)} = \frac{3}{36} = \frac{1}{12}

Therefore the probability that the sum of the numbers on the top faces is 10 is \frac{1}{12}

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