The half-life of a newly discovered radioactive isotope is 6.4 hours. initially, there are 56 grams of a sample of the isotope. a) how many grams of the isotope remains after 24 hours? show all of your work. round your answer to the nearest hundredth of a gram. b) how many hours will have passed when the sample of the isotope to reaches 6 grams? show all of your work.

A) Using the attached formula where "lambda" is natural log of 2 divided by half-life.  lambda = 0.69314718056 / 6.4 hours

lambda = 0.108304247  Then we use the formula:

Present Amount = Beginning Amount * e ^ (-lambda * time)

Present Amount = 56grams * 2.718281828459 ^ (-.108304247 * 24)

Present Amount = 56 * 0.0743254446

Present Amount = 4.1622248988

=4.16 grams (rounded)

B) Using the SECOND attached formula:

time = natural log (Present Amount / Bgng Amt) / -lambda

time = natural log (6 / 56) / -.108304247

time = -2.2335922219 / -.108304247

time = 20.623311493

time = 20.62 Hours

Step-by-step explanation:

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

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5050 i think

step-by-step explanation: