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GenaCL600 [577]
3 years ago
9

Consider U = {x|x is a positive integer greater than 1}.

Mathematics
2 answers:
iris [78.8K]3 years ago
5 0
2x will be an even number, which means it's immediately divisible by 2. Therefore it cannot be prime, so the second set is empty.
Irina18 [472]3 years ago
4 0
<h3><u>Answer:</u></h3>

Hence, the set that is empty is:

{x|x ∈ U and 2x is prime}

<h3><u>Step-by-step explanation:</u></h3>

We are given:

U = {x|x is a positive integer greater than 1}

i.e. U={2,3,4,5,6,7,.....}

The empty set among the following is:

{x|x ∈ U and 2x is prime}

As x ∈ U.

This means that x has to be an positive integer which is greater than 1.

and 2x will be an even integer i.e. {4,6,8,.....}.

Hence,

the term 2x can't be a prime as each of the number has more than 2 factors other than 1 and number itself.

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A coin bank that excepts only nickels and dimes contains $9.15. There are three more than twice as many nickels as there are dim
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45 dimes and 93 nickels were in bank

<em><u>Solution:</u></em>

Let "n" be the number of nickels

Let "d" be the number of dimes

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value of 1 nickel = $ 0.05

value of 1 dime = $ 0.10

<em><u>Given that There are three more than twice as many nickels as there are dimes</u></em>

Number of nickels = 3 + 2(number of dimes)

n = 3 + 2d ---- eqn 1

<em><u>Also given that coin bank that excepts only nickels and dimes contains $9.15</u></em>

number of nickels x value of 1 nickel + number of dimes x value of 1 dime = 9.15

n \times 0.05 + d \times 0.10 = 9.15

0.05n + 0.10d = 9.15  ---- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

Substitute eqn 1 in eqn 2

0.05(3 + 2d) + 0.10d = 9.15

0.15 + 0.1d + 0.10d = 9.15

0.2d = 9

<h3>d = 45</h3>

From eqn 1

n = 3 + 2(45)

n = 3 + 90 = 93

<h3>n = 93</h3>

Thus 45 dimes and 93 nickels were in bank

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