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igor_vitrenko [27]
3 years ago
7

Use the estimate in the text based on the Prime Number Theorem to give approximate values of the following. (a) The number of pr

imes between 1 and 1030. (b) The number of primes between 1 and 1029.
Mathematics
1 answer:
ozzi3 years ago
5 0

Answer:

a)148.47

b)148.34

Step-by-step explanation:

Let π(x) be the prime-counting function that gives the number of primes less than or equal to x, for any real number x. For example, π(7) = 4 because there are four prime numbers (2, 3, 5, 7) less than or equal to 7.

We can see that π(7)=π(10) because there are four prime numbers (2,3,5,7) less than or equal to 10.

A good aproximation for π(x) is x / log x, where log x is the natural logarithm of x.

a)The number of primes between 1 and 1030.

1030/log(1030)=148.47

b) The number of primes between 1 and 1029.

1029/log(1029)=148.34

 Extra: π(1029)=π(1030) because 1030 is not a prime.

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2 years ago
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C

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8 0
2 years ago
Simplify 6 to the 5th power over 7 to the 3rd power times 2
Anon25 [30]
My answer -

<span>1. Use symbols (not words) to express quotient
 

2. Use exponent symbol (^) to denote exponents

3. Just write out question number, question, and choices. No need for extra information (such as points). Also, don't leave blank lines between choices. This extraneous that we don't need just makes your whole question very very long, and means a lot of scrolling on our part.
 

4. You should only post 2 or 3 questions at a time.


1) (6x^3 − 18x^2 − 12x) / (−6x) = −x^2 + 3x + 2 ----> so much simpler to read !

2) (d^7 g^13) / (d^2 g^7) = d^(7−2) g^(13−7) = d^5 g^6 ----> much easier to read !

3) (4x − 6)^2 = 16x^2 − 24x − 24x + 36 = 16x^2 − 48x + 36

4) (x^2 / y^5)^4 = (x^2)^4 / (y^5)^4 = x^8 / y^20

5) (3x + 5y)(4x − 3y) = 12x^2 − 9xy + 20xy − 15y^2 = 12x^2 + 11xy − 15y^2

6) (3x^3y^4z^4)(2x^3y^4z^2) = (3*2) x^(3+3) y^(4+4) z^(4+2) = 6 x^6 y^8 z^6

7) 5x + 3x^4 − 7x^3 ----> Fourth degree trinomial

8) (5x^3 − 5x − 8) + (2x^3 + 4x + 2) = 7x^3 − x − 6

9) (x − 1) + (2x + 5) − (x + 3) = x + 1

10) (−4g^8h^5k^2)0(hk^2)^2 = 0 (anything multiplied by 0 = 0)
or.. (−4g^8h^5k^2)^0(hk^2)^2 = 1 (h^2 (k^2)^2) = h^2 k^4

Last question shows why it is so important to use proper symbols (such as ^ to indicate exponents). Without such symbols, I could not tell if the 0 was an actual number and part of multiplication, of if 0 was an exponent of the expression preceding it.


P.S

Glad to help you have an AWESOME!!! day :)
</span>
6 0
3 years ago
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