PLEASE HELP!!!!! And explain too! Thanks!!!

1 answer:

4 0

I don't know any slick way to do this. I would just list the prime numbers
between 40 and 50 and test them one at a time.

The prime numbers between 40 and 50 are: 41, 43, 47

Test 41:
-- add 1 . . . 42
-- prime factors of 42 . . . 2, 3, 7
-- sum of the factors . . . 12

Test 43:
-- add 1 . . . 44
-- prime factors of 44 . . . 2, 11
-- sum of the factors . . . 13

Test 47:
-- add 1 . . . 48
-- prime factors of 48 . . . 2, 3,
-- sum of the factors . . . 5

Maybe I'm not understanding how to list prime factors, but so far,
I haven't found any number that answers the question.

Maybe if I write the prime factorization of 44 like this:

44 = 2 * 2 * 11

Now the items in the prime factorization do add up to 15.
So the answer would seem to be 43.

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The radioactive isotope radium has a half-life of 1,600 years. The mass of a 100-gram sample of radium will be______ grams after

Olenka [21]

To solve this we are going to use the half life equation $N(t)=N_{0} e^{( \frac{-0.693t}{t _{1/2} }) }$
Where:
$N_{0}$ is the initial sample
$t$ is the time in years
$t_{1/2}$ is the half life of the substance
$N(t)$ is the remainder quantity after $t$ years

From the problem we know that:
$N_{0} =100$
$t=200$
$t_{1/2} =1600$

Lets replace those values in our equation to find $N(t)$ :
$N(200) =100e^{( \frac{(-0.693)(200)}{1600}) }$
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8 0

3 years ago

Multiplication of fractions 3/4×7/15

Pachacha [2.7K]

Answer:

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