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Musya8 [376]
3 years ago
13

The sixth number 20210A is prime for only one digit A. What is A?

Mathematics
1 answer:
Rainbow [258]3 years ago
6 0

9514 1404 393

Answer:

  A = 9

Step-by-step explanation:

Beyond the first few primes, all prime numbers end in one of the odd digits, 1, 3, 7, or 9. We can reduce this list by using a couple of simple divisibility rules.

The sum of the digits is 2+0+2+1+0+A = A+5. The number will be divisible by 3 if A+5 is divisible by 3. That will be the case for A=1 and A=7.

The sum of odd digits is A+1+0 = A+1. The sum of even digits is 0+2+2 = 4. When these sums are equal, the number is divisible by 11. That will be the case for A=3.

So, we have ruled out all digits except 9. If the number is prime it must be ...

  202109

  A = 9

_____

<em>Additional comment</em>

To make sure the number is prime, we would need to check divisibility by all primes less than √202109 ≈ 449.

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Answer:

7x, 7

Step-by-step explanation:

In each of the terms 3x 2 and 7x, x is a variable, and in both terms, we are multiplying a number by the variable. In the first term, 3x 2, 3 is being multiplied by the variable, so 3 is a coefficient. In the second term, 7x, 7 is being multiplied by the variable, so 7 is a coefficient.

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3 years ago
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Suppose that qequals30​, Lequals5​, and Kequals15 is a point on the production function qequals​f(L, ​K). Is it posssible for qe
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No

Step-by-step explanation:

If q=30​, L=5​, and K=15 is a point on the production function q=​f(L, ​K).

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The function N(t) = 100e^−0.023t models the number of grams in a sample of cesium-137 that remain after t years. On which interv
Andreas93 [3]

Options:

A. [1,10]        B. [10,20]      C. [15,25]     D. [1,30]

Answer:

A. [1,10]

Step-by-step explanation:

Given

N(t) = 100e^{-0.023t}

Required

Interval with the fastest rate of decay

The rate of change over [a,b] is calculated as:

Rate = \frac{N(b) - N(a)}{b - a}

So, we have:

A. [1,10]        

Rate = \frac{N(10)-N(1)}{10 - 1}

Rate = \frac{N(10)-N(1)}{9}

Calculate N(10) and N(1)

N(10) = 100e^{-0.023*10} = 100e^{-0.23} = 79.45

N(1) = 100e^{-0.023*1} = 100e^{-0.023} = 97.73

So:

Rate = \frac{79.45-97.73}{9} = \frac{-18.28}{9} = -2.03

B. [10,20]      

Rate = \frac{N(20)-N(10)}{20 - 10}

Rate = \frac{N(20)-N(10)}{10}

Calculate N(20) and N(10)

N(20) = 100e^{-0.023*20} = 100e^{-0.46} = 63.13

N(10) = 100e^{-0.023*10} = 100e^{-0.23} = 79.45

So:

Rate = \frac{63.13 - 79.45}{10} = \frac{-16.32}{10} = -1.632

C. [15,25]    

Rate = \frac{N(25)-N(15)}{25 - 15}

Rate = \frac{N(25)-N(15)}{20}

Calculate N(25) and N(15)

N(25) = 100e^{-0.023*25} = 100e^{-0.575} = 56.27

N(15) = 100e^{-0.023*15} = 100e^{-0.345} = 70.82

So:

Rate = \frac{56.27 - 70.82}{10} = \frac{-14.55}{10} = -1.455

D. [1,30]

Rate = \frac{N(30)-N(1)}{30 - 1}

Rate = \frac{N(30)-N(1)}{29}

Calculate N(30) and N(1)

N(30) = 100e^{-0.023*30} = 100e^{-0.69} = 50.16

N(1) = 100e^{-0.023*1} = 100e^{-0.023} = 97.73

So:

Rate = \frac{50.16 - 97.73}{29} = \frac{-47.57}{29} = -1.64

By comparing the values of the calculated rates,

-2.03 is the smallest, in other words; the fastest rate

Hence, the interval [1,10] has the fastest rate of decay

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Answer:

Step-by-step explanation:

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