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otez555 [7]
3 years ago
9

A King in ancient times agreed to reward the inventor of chess with one grain of wheat on the first of the 64 squares of a chess

board. On the second square the King would place two grains of​ wheat, on the third​ square, four grains of​ wheat, and on the fourth square eight grains of wheat. If the amount of wheat is doubled in this way on each of the remaining​ squares, how many grains of wheat should be placed on square 25​? Also find the total number of grains of wheat on the board at this time and their total weight in pounds.​ (Assume that each grain of wheat weighs​ 1/7000 pound.)
Mathematics
1 answer:
Sveta_85 [38]3 years ago
3 0

Answer:

That's 2635249153387078 pounds or  1317624576693.5 tons

Step-by-step explanation:

The series is 2^(n-1) where n=1,2,3,4,...,62,63,64

 

We can adjust the index and write it as 2^n where n=0,1,2,3,4,...,61,62,63

 

The sum of the geometric series is:

 

 a1 * (1 - r^n)

------------------  

         1-r

 

where r is the common ratio in this case 2,

a1 is the first term, in this case 1,

and n is the number of term, in this case 64

 

 

1 * (1 - 2^64)

----------------  = 18446744073709551615

 1 - 2

 

Dividing that by 7000

That's 2635249153387078 pounds or  1317624576693.5 tons

Hope this helped!

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Answer:

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Step-by-step explanation:

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11 + 8 equals 19 not greater than 19.

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hi your question options is not available but attached to the answer is a complete question with the question options that you seek answer to

Answer:  v = 5v + 4u + 1.5sin(3t),

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Step-by-step explanation:

u" - 5u' - 4u = 1.5sin(3t)        where u'(1) = 2.5   u(1) = 1

v represents the "velocity function"   i.e   v = u'(t)

As v = u'(t)

<em>u' = v</em>

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so v' = 5v + 4u + 1.5sin(3t)

d/dt \left[\begin{array}{ccc}u&\\v&\\\end{array}\right]= \left[\begin{array}{ccc}0&1&\\4&5&\\\end{array}\right]  \left[\begin{array}{ccc}u&\\v&\\\end{array}\right] + \left[\begin{array}{ccc}0&\\1.5sin(3t)&\\\end{array}\right]

Given that u(1) = 1 and u'(1) = 2.5

since v = u'

v(1) = 2.5

note: the initial value for the vector valued function is given as

\left[\begin{array}{ccc}u(1)&\\v(1)\\\end{array}\right]  = \left[\begin{array}{ccc}1\\2.5\\\end{array}\right]

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