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ElenaW [278]
3 years ago
9

A leprechaun places a magic penny under a girl's pillow. The next night there are 2 magic pennies under her pillow. The followin

g morning she finds 4 pennies. Apparently while she sleeps each penny turns into two magic pennies. The total number of pennies seen under the pillow each day is the grand total; that is, the pennies from each of the previous days are not being stored away until more pennies magically appear. How many days would elapse before she has a total of more than $2 Billion?
Mathematics
1 answer:
dezoksy [38]3 years ago
5 0

Answer:

<u><em></em></u>

  • <u><em>32 days</em></u>

Explanation:

Simulate (build a table) the growing of the number of pennies for some nights to figure out the pattern:

First night:          1 penny = 2⁰

Second night:    1 × 2 pennies = 2¹

Third night:        2 × 2 = 2²

Fourth nigth:      2² × 2 = 2³

nth night:           2ⁿ⁻¹

You want 2ⁿ⁻¹ ≥ 2,000,000,000

Which you solve in this way:

  • 2ⁿ⁻¹  ≥ 2,000,000,000

  • 2ⁿ⁻¹ ≥ 2,000,000,000

  • n-1 log (2)  ≥ log (2,000,000,000)

  • n - 1  ≥ log (2,000,000,000) / log (2)

  • n - 1 ≥ 30.9

  • n ≥ 31.9

Since n is number of days, it is an integer number, so n ≥ 32.

Hence, she will have a total of more than $ 2 billion after 32 days.

You can prove that by calculating 2³² = 2,147,483,648.

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HELP ASAP PLZZZZ
Tcecarenko [31]
QUESTION 1

The given system of equations is

3d - e = 7...eqn(1)
d + e = 5...eqn(2)

To solve by linear combination, we add equation (1) to equation (2) to get,

3d  + d= 7 + 5


4d = 12


We divide through by 4 to obtain,


d =  \frac{12}{4}


d = 3


We put d=3 into equation (2) to get,



3+ e = 5


e = 5 - 3


e = 2


\boxed {The \: solution \: is  \: (3, 2)}



QUESTION 2


The given system is

4x + y = 5 ...eqn(1)

3x + y = 3 ...eqn(2)


To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.

This will give us,

4x - 3x = 5 - 3



This implies that,

x = 2


Put x=3 into equation (1) to get,

4(2) + y = 5

8+ y = 5


y = 5 - 8



y =  - 3

The solution is

(2,-3)



QUESTION 3

We want to solve the system;


a – 2b = –2 ....eqn(1)


2a + 2b = 14...eqn(2)

by linear combination.


We need to add equation (1) to equation (2) to eliminate b.


This implies that,

2a + a = 14 +  - 2




Simplify,

3a = 12



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a = 4
Put a=4 into equation (2) to obtain,



2(4) + 2b = 14


8 + 2b = 14
2b = 14 - 8


2b = 6


b = 3


The ordered pair in the form (a, b) is

(4,3)



QUESTION 4

The given system of equations is


11x + 4y = 18 ...eqn(1)

3x + 4y = 2 ...eqn(2)


We subtract equation (2) from equation (1) to get,


11x - 3x = 18 - 2


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x = 2


Put x=2 into equation (2) to obtain,


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The given system is ;

2d + e = 8...eqn1

d – e = 4...eqn2


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This implies that,

2d + d = 8 + 4


3d = 12



We divide both sides by 3 to get,


d = 4


We put d=4 into equation (2) to get,

4 - e = 4

- e = 4 - 4



- e = 0



e = 0


The solution is

(4,0)
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3 years ago
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4 years ago
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