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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Concept explainers

Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
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Chapter 3.2, Problem 109E

a.

To determine

What was the average score on the original exam (t=0) ?

a.

Expert Solution
Check Mark

Answer to Problem 109E

The average score on the original exam was 80.

Explanation of Solution

Given:

Students in a mathematical class were given an exam and then tested monthly with an equivalent exam. The average scores for the class are given by human memory model f(t)=80−17log10(t+1), 0≤t≤12 , where t is the time in months.

Calculation:

To find the average score on original exam, we will substitute t=0 in our given formula.

  f(0)=80−17log10(0+1)f(0)=80−17log10(1)f(0)=80−17⋅0f(0)=80

Therefore, the average score on the original exam was 80.

Upon graphing our given function, we will get:

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 3.2, Problem 109E , additional homework tip 1

We can see that at t=0 the value of our given function is 80.

b.

To determine

What was the average score after 2 months?

b.

Expert Solution
Check Mark

Answer to Problem 109E

The average score after 2 months was 71.89.

Explanation of Solution

Given:

Students in a mathematical class were given an exam and then tested monthly with an equivalent exam. The average scores for the class are given by human memory model f(t)=80−17log10(t+1), 0≤t≤12 , where t is the time in months.

Calculation:

To find the average score after 2 months, we will substitute t=2 in our given formula.

  f(2)=80−17log10(2+1)f(2)=80−17log10(3)f(2)=80−8.11106133f(2)=71.88893867f(2)≈71.89

Therefore, the average score after 2 months was approximately 71.89.

Upon graphing our given function, we will get:

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 3.2, Problem 109E , additional homework tip 2

We can see that at t=2 the value of our given function is 71.89.

c.

To determine

What was the average score after 11 months?

c.

Expert Solution
Check Mark

Answer to Problem 109E

The average score after 11 months was 61.65.

Explanation of Solution

Given:

Students in a mathematical class were given an exam and then tested monthly with an equivalent exam. The average scores for the class are given by human memory model f(t)=80−17log10(t+1), 0≤t≤12 , where t is the time in months.

Calculation:

To find the average score after 11 months, we will substitute t=11 in our given formula.

  f(11)=80−17log10(11+1)f(11)=80−17log10(12)f(11)=80−18.346081182809f(11)=61.653918817191f(11)≈61.65

Therefore, the average score after 11 months was approximately 61.65.

Upon graphing our given function, we will get:

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 3.2, Problem 109E , additional homework tip 3

We can see that at t=11 the value of our given function is 61.65.

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Chapter 3.2, Problem 108E
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Chapter 3.2, Problem 110E

Chapter 3 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

Ch. 3.1 - Prob. 1ECh. 3.1 - Prob. 2ECh. 3.1 - Prob. 3ECh. 3.1 - Prob. 4ECh. 3.1 - Prob. 5ECh. 3.1 - Prob. 6ECh. 3.1 - Prob. 7ECh. 3.1 - Prob. 8ECh. 3.1 - Prob. 9ECh. 3.1 - Prob. 10E
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Prob. 32ECh. 3.6 - Prob. 33ECh. 3.6 - Prob. 34ECh. 3.6 - Prob. 35ECh. 3.6 - Prob. 36ECh. 3.6 - Prob. 37ECh. 3.6 - Prob. 38ECh. 3.6 - Prob. 39ECh. 3.6 - Prob. 40ECh. 3.6 - Prob. 41ECh. 3.6 - Prob. 42ECh. 3.6 - Prob. 43ECh. 3.6 - Prob. 44ECh. 3.6 - Prob. 45ECh. 3.6 - Prob. 46ECh. 3 - Prob. 1CRCh. 3 - Prob. 2CRCh. 3 - Prob. 3CRCh. 3 - Prob. 4CRCh. 3 - Prob. 5CRCh. 3 - Prob. 6CRCh. 3 - Prob. 7CRCh. 3 - Prob. 8CRCh. 3 - Prob. 9CRCh. 3 - Prob. 10CRCh. 3 - Prob. 11CRCh. 3 - Prob. 12CRCh. 3 - Prob. 13CRCh. 3 - Prob. 14CRCh. 3 - Prob. 15CRCh. 3 - Prob. 16CRCh. 3 - Prob. 17CRCh. 3 - Prob. 18CRCh. 3 - Prob. 19CRCh. 3 - Prob. 20CRCh. 3 - Prob. 21CRCh. 3 - Prob. 22CRCh. 3 - Prob. 23CRCh. 3 - Prob. 24CRCh. 3 - Prob. 25CRCh. 3 - Prob. 26CRCh. 3 - Prob. 27CRCh. 3 - Prob. 28CRCh. 3 - Prob. 29CRCh. 3 - Prob. 30CRCh. 3 - Prob. 31CRCh. 3 - Prob. 32CRCh. 3 - Prob. 33CRCh. 3 - Prob. 34CRCh. 3 - Prob. 35CRCh. 3 - Prob. 36CRCh. 3 - Prob. 37CRCh. 3 - Prob. 38CRCh. 3 - Prob. 39CRCh. 3 - Prob. 40CRCh. 3 - Prob. 41CRCh. 3 - 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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