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Andrew [12]
3 years ago
14

There are 8 rows and 8 columns, or 64 squares

Mathematics
1 answer:
lawyer [7]3 years ago
5 0

Complete Question:

There are 8 rows and 8 columns, or 64 squares on a chessboard.

Suppose you place 1 penny on Row 1 Column A,

2 pennies on Row 1 Column B,

4 pennies on Row 1 Column C, and so on …

Determine the number of pennies in Row 1

Determine the number of pennies on the entire chessboard?

Answer:

255 in the first row

18,446,744,073,709,551,615 in the entire board

Step-by-step explanation:

Given

Rows = 8

Columns = 8

Solving (a): Number of pennies in first row

The question is an illustration of geometric sequence which follows

1,2,4....

Where

a =1 --- The first term

Calculate the common ratio, r

r = \frac{T_2}{T_1} = \frac{4}{2} = 2

The number of pennies in the first row will be calculated using sum of n terms of a GP.

S_n = \frac{a(r^n - 1)}{n - 1}

Since, the first row has 8 columns, then

n = 8

Substitute 8 for n, 2 for r and 1 for a in S_n = \frac{a(r^n - 1)}{r - 1}

S_8 = \frac{1 * (2^8 - 1)}{2 - 1}

S_8 = \frac{1 * (256 - 1)}{1}

S_8 = \frac{1 * 255}{1}

S_8 = 255

Solving (b): The entire board has 64 cells.

So:

n = 64

Substitute 64 for n, 2 for r and 1 for a in S_n = \frac{a(r^n - 1)}{r - 1}

S_{64} = \frac{1 * (2^{64} - 1)}{2 -1}

S_{64} = \frac{(2^{64} - 1)}{1}

S_{64} = \frac{(18,446,744,073,709,551,616 - 1)}{1}

S_{64} = \frac{18,446,744,073,709,551,615}{1}

S_{64} = 18,446,744,073,709,551,615

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Answer: 2m^2n^3 x ( -5m^2 + 4n^2 -3n^2 )

Step-by-step explanation:

<em>ok</em><em>,</em><em> so just let me</em><em>,</em><em> uh</em><em>,</em><em> make this a little easier on my eyes lol:</em>

-10m^4n^3+8m^2n^6-6m^2n^6 ----> (-10m^4)(n^3) + (8m^2)(n^6) - (6m^2)(n^6)

<em>I just put in parentheses. nothing changed. just parentheses. </em>

Since the leading coefficients, -10 , 8 , and 6 are multiples of two,

we can take out 2 from each... [-10/2 = 5 , 8/2 = 4 , 6/2 = 3 ]

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Looking at it now...

<em>we can take out m^2 from each... </em>m^4 , m^2 , and m^2

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= 2(m^2) x [ (-5m^2)(n^3) + (4x1)(n^6) - (3x1)(n^6) ]

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Finally, one more step...

<em>we can take out n^3 from each term.... </em>n^3 , n^6 , and n^6

n^3 [1 , n^2 , n^2] since n^3 /n^3 = 1 , n^6 /n^3 = n^2 , n^6 /n^3 = n^2

= 2(m^2)(n^3) [ (-5m^2)(1) + (4)(n^2) - (3)(n^2) ]

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= 2m^2n^3 x ( -5m^2 + 4n^2 -3n^2 )

ANSWER: 2m^2n^3 x ( -5m^2 + 4n^2 -3n^2 )

<em>keep this parentheses. seriously. </em>

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Answer: (a)  Job A , by approximately $69, 482

\\ (b)  Job A , by approximately $6,867

 \\(c) Job B , by approximately $ 767,362

  \\ (d) Job A

Step-by-step explanation:

JOB B

\\The starting Salary is $ 10,000

Since there is an increment of 25% at the beginning of each new year. The breakdown of the increment is as follow:

\\First year : 125% of $ 10,000 = $12,500

\\Second year : 125% of $ 12,500 = $15,625

\\Third year : 125% of $15,625= $19,531.25

\\Fourth year: 125% of $19,531.25 = $24,414.06

\\Fifth year : 125% of $24,414.06 = $30,517.58

\\Sixth year: 125% of $30,517.58 = $38,147.00

\\Seventh year: 125% of $38,147.00 = $47,683.75

\\Eight year: 125% of $47,683.75 = $59,604.69

\\Ninth year: 125% of $59,604.69 = $74,505.86

\\Tenth year: 125% of $74,505.86 = $93,132.32

\\Following the same procedure:

\\Eleventh year = $116,415.40

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\\Seventeenth year = $444,089.51

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\\Difference: Salary of Job A at the beginning of fifth year remains $ 100,000 while that of Job B resulted into $ 30,517.58, the difference implies

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\\(c) At the beginning of the twentieth year, the annual salary of A is still $ 100,000 while the annual salary of B is $ 867,362.32. Job B annual salary is greater by approximately $ 767,362

\\(d) If I were in Jane’s shoe I will take Job A and work for few years to gain more experience the look for a job that pays better. Waiting for many years in case of Job B is risky , market situation is uncertainty.

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