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DedPeter [7]
3 years ago
6

being the first grandson, your grand parent decided to give a rectangular field for your coming wedding. If you are given 210 m

wires of fencing, what dimensions would you choose to get the maximum area? a.list all the possible dimensions of the rectangular field b. make a table of values for the possible dimensions c. compute the area for each possible dimension d. what is the maximum area you obtain? e. what are the dimensions of the maximum area you obtained?
Mathematics
1 answer:
Airida [17]3 years ago
5 0

if you are given a certain length of wire the biggest area you can fend up is a circle, 2nd a square.  For the rectangle to have the biggest area,the difference between the 2 sides must be the minimum,  it must approaches a square

 then we have         2(l +w)   =    200

                                  l + w   =     100

the  dimensions   are    49 & 51

multiply    49 by  51  =  2499 sq. m, will give the biggest area, any other      combinations will give a smaller area if you considered decimal numbers between  

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Ainat [17]

Answer: n = 10.

Step-by-step explanation:

In the bag, we have n counters.

4 of the counters are red.

the rest are blue, then we have:

(n - 4) blue counters.

Now, the probability that Ross takes a blue counter from the bag is equal to the quotient between the number of blue counters (n - 4) and the total number of counters, n

Then the probability is:

p1 = (n - 4)/n

Now he draws another, and it must be blue again, then we can calculate the probability in the same way as above, but he already take a blue counter, so the number of blue counters is (n - 5) and the total number of counters is (n - 1)

The probability of this event is:

p2 = (n - 5)/(n - 1)

The joint probability (the probability that Ross takes two blue counters) is equal to the product of the individual probabilities, and we know that this is equal to 1/3, then we have the equation:

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Now let's solve this for n.

n*(n - 1)/3 = (n - 4)*(n - 5)

(n^2 - n)/3 = n^2 - 4*n - 5*n + 20

n^2 - n = 3*(n^2 - 9*n + 20)

n^2 - n = 3*n^2 - 27*n + 60

0 = (3*n^2 - n^2) - 27*n + n + 60

0 = 2*n^2 - 26*n + 60

The two solutions of this equation can be found with Bhaskara's equation:

n = \frac{-(-26) +- \sqrt{(-26)^2 - 4*2*60} }{2*2} = \frac{26+- 14}{4}

Then the two solutions are:

n = (26 - 14)/4 = 3

This is not an option, because we know for sure that we have 4 red counters, then this option can be discarded.

The other solution is:

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Answer:

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