PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
8th Edition
ISBN: 9781337904285
Author: Larson
Publisher: CENGAGE L
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Chapter 3.1, Problem 74E

a.

To determine

To find the initial quantity of radium (when t=0 ).

a.

Expert Solution
Check Mark

Answer to Problem 74E

The initial quantity of radioactive radium was 25 grams .

Explanation of Solution

Given information:

The given statement is:

“Let Q represent a mass, in grams, of radioactive radium, whose half-life is about 1600 years. The quantity of radium present after t years can be found using Q=25(12)t/1600 .”

Calculation:

The initial quantity can be calculated by:

  Q=25(12)t/1600Q=25(12)0/1600  [substitute t=0]Q=25(12)0Q=251     [a0=1]Q=25

Hence, the initial quantity of radioactive radium was 25 grams .

b.

To determine

To find the quantity of carbon after 1000 years.

b.

Expert Solution
Check Mark

Answer to Problem 74E

The quantity of radium after 1000 years is 6.5 grams .

Explanation of Solution

Given information:

The given statement is:

“Let Q represent a mass, in grams, of radioactive radium, whose half-life is about 1600 years. The quantity of radium present after t years can be found using Q=25(12)t/1600 .”

Calculation:

The quantity of carbon can be calculated by:

  Q=10(12)t/1600Q=10(12)1000/1600  [substitute t=1000]Q=10×0.65Q=6.5

Hence, the quantity of radium after 1000 years is 6.5 grams .

c.

To determine

To sketch the graph over the interval t=0 to t=5000 .

c.

Expert Solution
Check Mark

Explanation of Solution

Given information:

The given statement is:

“Let Q represent a mass, in grams, of radioactive radium, whose half-life is about 1600 years. The quantity of radium present after t years can be found using Q=25(12)t/1600 .”

Graph:

The graph can be obtained by:

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 3.1, Problem 74E

d.

To determine

To determine when will the quantity of radium be 0 grams .

d.

Expert Solution
Check Mark

Answer to Problem 74E

The mass of radium is always decayed by half of its mass, so the mass of radium will be 0 grams in an infinite amount of time.

Explanation of Solution

Given information:

The given statement is:

“Let Q represent a mass, in grams, of radioactive radium, whose half-life is about 1600 years. The quantity of radium present after t years can be found using Q=25(12)t/1600 .”

The mass of radium is always decayed by half of its mass, so the mass of radium will be 0 grams in an infinite amount of time.

Chapter 3 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

Ch. 3.1 - Prob. 11ECh. 3.1 - Prob. 12ECh. 3.1 - Prob. 13ECh. 3.1 - Prob. 14ECh. 3.1 - Prob. 15ECh. 3.1 - Prob. 16ECh. 3.1 - Prob. 17ECh. 3.1 - Prob. 18ECh. 3.1 - Prob. 19ECh. 3.1 - Prob. 20ECh. 3.1 - Prob. 21ECh. 3.1 - Prob. 22ECh. 3.1 - Prob. 23ECh. 3.1 - Prob. 24ECh. 3.1 - Prob. 25ECh. 3.1 - Prob. 26ECh. 3.1 - Prob. 27ECh. 3.1 - Prob. 28ECh. 3.1 - Prob. 29ECh. 3.1 - Prob. 30ECh. 3.1 - Prob. 31ECh. 3.1 - Prob. 32ECh. 3.1 - Prob. 33ECh. 3.1 - Prob. 34ECh. 3.1 - Prob. 35ECh. 3.1 - Prob. 36ECh. 3.1 - Prob. 37ECh. 3.1 - Prob. 38ECh. 3.1 - Prob. 39ECh. 3.1 - Prob. 40ECh. 3.1 - Prob. 41ECh. 3.1 - Prob. 42ECh. 3.1 - Prob. 43ECh. 3.1 - Prob. 44ECh. 3.1 - Prob. 45ECh. 3.1 - Prob. 46ECh. 3.1 - Prob. 47ECh. 3.1 - Prob. 48ECh. 3.1 - Prob. 49ECh. 3.1 - Prob. 50ECh. 3.1 - Prob. 51ECh. 3.1 - Prob. 52ECh. 3.1 - Prob. 53ECh. 3.1 - Prob. 54ECh. 3.1 - Prob. 55ECh. 3.1 - Prob. 56ECh. 3.1 - Prob. 57ECh. 3.1 - Prob. 58ECh. 3.1 - Prob. 59ECh. 3.1 - Prob. 60ECh. 3.1 - Prob. 61ECh. 3.1 - Prob. 62ECh. 3.1 - Prob. 63ECh. 3.1 - 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Prob. 42CRCh. 3 - Prob. 43CRCh. 3 - Prob. 44CRCh. 3 - Prob. 45CRCh. 3 - Prob. 46CRCh. 3 - Prob. 47CRCh. 3 - Prob. 48CRCh. 3 - Prob. 49CRCh. 3 - Prob. 50CRCh. 3 - Prob. 51CRCh. 3 - Prob. 52CRCh. 3 - Prob. 53CRCh. 3 - Prob. 54CRCh. 3 - Prob. 55CRCh. 3 - Prob. 56CRCh. 3 - Prob. 57CRCh. 3 - Prob. 58CRCh. 3 - Prob. 59CRCh. 3 - Prob. 60CRCh. 3 - Prob. 61CRCh. 3 - Prob. 62CRCh. 3 - Prob. 63CRCh. 3 - Prob. 64CRCh. 3 - Prob. 65CRCh. 3 - Prob. 66CRCh. 3 - Prob. 67CRCh. 3 - Prob. 68CRCh. 3 - Prob. 69CRCh. 3 - Prob. 70CRCh. 3 - Prob. 71CRCh. 3 - Prob. 72CRCh. 3 - Prob. 73CRCh. 3 - Prob. 74CRCh. 3 - Prob. 75CRCh. 3 - Prob. 76CRCh. 3 - Prob. 77CRCh. 3 - Prob. 78CRCh. 3 - Prob. 79CRCh. 3 - Prob. 80CRCh. 3 - Prob. 81CRCh. 3 - Prob. 82CRCh. 3 - Prob. 83CRCh. 3 - Prob. 84CRCh. 3 - Prob. 85CRCh. 3 - Prob. 86CRCh. 3 - Prob. 87CRCh. 3 - Prob. 88CRCh. 3 - Prob. 89CRCh. 3 - Prob. 90CRCh. 3 - Prob. 91CRCh. 3 - Prob. 92CRCh. 3 - Prob. 93CRCh. 3 - Prob. 94CRCh. 3 - Prob. 95CRCh. 3 - Prob. 96CRCh. 3 - Prob. 97CRCh. 3 - Prob. 98CRCh. 3 - Prob. 99CRCh. 3 - Prob. 100CRCh. 3 - Prob. 101CRCh. 3 - Prob. 102CRCh. 3 - Prob. 103CRCh. 3 - Prob. 104CRCh. 3 - Prob. 105CRCh. 3 - Prob. 106CRCh. 3 - Prob. 107CRCh. 3 - Prob. 108CRCh. 3 - Prob. 109CRCh. 3 - Prob. 110CRCh. 3 - Prob. 111CRCh. 3 - Prob. 112CRCh. 3 - Prob. 113CRCh. 3 - Prob. 114CRCh. 3 - Prob. 115CRCh. 3 - Prob. 116CRCh. 3 - Prob. 117CRCh. 3 - Prob. 118CRCh. 3 - Prob. 119CRCh. 3 - Prob. 120CRCh. 3 - Prob. 121CRCh. 3 - Prob. 122CRCh. 3 - Prob. 123CRCh. 3 - Prob. 124CRCh. 3 - Prob. 125CRCh. 3 - Prob. 126CRCh. 3 - Prob. 1CTCh. 3 - Prob. 2CTCh. 3 - Prob. 3CTCh. 3 - Prob. 4CTCh. 3 - Prob. 5CTCh. 3 - Prob. 6CTCh. 3 - Prob. 7CTCh. 3 - Prob. 8CTCh. 3 - Prob. 9CTCh. 3 - Prob. 10CTCh. 3 - Prob. 11CTCh. 3 - Prob. 12CTCh. 3 - Prob. 13CTCh. 3 - Prob. 14CTCh. 3 - Prob. 15CTCh. 3 - Prob. 16CTCh. 3 - Prob. 17CTCh. 3 - Prob. 18CTCh. 3 - Prob. 19CTCh. 3 - Prob. 20CTCh. 3 - Prob. 21CTCh. 3 - Prob. 22CTCh. 3 - Prob. 23CTCh. 3 - Prob. 24CTCh. 3 - Prob. 25CTCh. 3 - Prob. 26CTCh. 3 - Prob. 27CTCh. 3 - Prob. 28CTCh. 3 - Prob. 29CTCh. 3 - Prob. 30CTCh. 3 - Prob. 1STPCh. 3 - Prob. 2STPCh. 3 - Prob. 3STPCh. 3 - Prob. 4STPCh. 3 - Prob. 5STPCh. 3 - Prob. 6STPCh. 3 - Prob. 7STPCh. 3 - Prob. 8STPCh. 3 - Prob. 9STPCh. 3 - Prob. 10STPCh. 3 - Prob. 11STPCh. 3 - Prob. 12STPCh. 3 - Prob. 13STPCh. 3 - Prob. 14STPCh. 3 - Prob. 15STPCh. 3 - Prob. 16STPCh. 3 - Prob. 17STPCh. 3 - Prob. 18STPCh. 3 - Prob. 19STPCh. 3 - Prob. 20STPCh. 3 - Prob. 21STPCh. 3 - Prob. 22STPCh. 3 - Prob. 23STPCh. 3 - Prob. 24STPCh. 3 - Prob. 25STPCh. 3 - Prob. 26STP
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