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 1.1, Problem 97E

  (a)

To determine

To Find: Linear equation giving the value V of the equipment for year t during its 10 years of use.

  (a)

Expert Solution
Check Mark

Answer to Problem 97E

  V=1011.5t+10995,0t10

Explanation of Solution

Given: Purchase price of equipment is $10,995 and price of equipment after 10 years is $880.

Formula used: Equation of line passing through two points (x1,y1) and (x2,y2) is:

  yy1=(y2y1x2x1)(xx1)

Calculation: We have given when t=0 , V=10995 and t=10,V=880.

So the corresponding points are (t1,v1)=(0,10995) and (t2,v2)=(10,880).

Equation of line passing through (t1,v1) and (t2,v2) .

  Vv1=(v2v1t2t1)(tt1)

  V10995=(88010995100)(t0)

  V10995=(1011510)t

  V10995=1011.5t

  V=1011.5t+10995

Also we have given t is years between 0 to 10 .

Therefore, 0t10.

Conclusion:

Therefore, linear equation giving the value V of the equipment for year t during its 10 years of use is: V=1011.5t+10995,0t10.

  (b)

To determine

To Graph: Graph the equation V=1011.5t+10995,0t10 using graphing utility.

To Verify: The answers algebraically.

  (b)

Expert Solution
Check Mark

Explanation of Solution

Given: V=1011.5t+10995,0t10

Graph:

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 1.1, Problem 97E

By tracing points on graph we get the following values:

    t012345678910
    V109959983.589727960.569495937.549263914.529031891.5880

Algebraic Verification:

We have: V=1011.5t+10995,0t10

Substituting 0 for t we get:

  V=1011.5(0)+10995=10995

Substituting 1 for t we get:

  V=1011.5(1)+10995=9983.5

Substituting 2 for t we get:

  V=1011.5(2)+10995=8972

Substituting 3 for t we get:

  V=1011.5(3)+10995=7960.5

Substituting 4 for t we get:

  V=1011.5(4)+10995=6949

Substituting 5 for t we get:

  V=1011.5(5)+10995=5937.5

Substituting 6 for t we get:

  V=1011.5(6)+10995=4926

Substituting 7 for t we get:

  V=1011.5(7)+10995=3914.5

Substituting 8 for t we get:

  V=1011.5(8)+10995=2903

Substituting 9 for t we get:

  V=1011.5(9)+10995=1891.5

Substituting 10 for t we get:

  V=1011.5(10)+10995=880

From these values table values are verified.

Conclusion:

Therefore, from graphical method and algebraic method we get the following table values:

    t012345678910
    V109959983.589727960.569495937.549263914.529031891.5880

Chapter 1 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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Prob. 59ECh. 1.6 - Prob. 60ECh. 1.6 - Prob. 61ECh. 1.6 - Prob. 62ECh. 1.6 - Prob. 63ECh. 1.6 - Prob. 64ECh. 1.6 - Prob. 65ECh. 1.6 - Prob. 66ECh. 1.6 - Prob. 67ECh. 1.6 - Prob. 68ECh. 1.6 - Prob. 69ECh. 1.6 - Prob. 70ECh. 1.6 - Prob. 71ECh. 1.6 - Prob. 72ECh. 1.6 - Prob. 73ECh. 1.6 - Prob. 74ECh. 1.6 - Prob. 75ECh. 1.6 - Prob. 76ECh. 1.6 - Prob. 77ECh. 1.6 - Prob. 78ECh. 1.6 - Prob. 79ECh. 1.6 - Prob. 80ECh. 1.6 - Prob. 81ECh. 1.6 - Prob. 82ECh. 1.6 - Prob. 83ECh. 1.6 - Prob. 84ECh. 1.6 - Prob. 85ECh. 1.6 - Prob. 86ECh. 1.6 - Prob. 87ECh. 1.6 - Prob. 88ECh. 1.6 - Prob. 89ECh. 1.6 - Prob. 90ECh. 1.6 - Prob. 91ECh. 1.6 - Prob. 92ECh. 1.6 - Prob. 93ECh. 1.6 - Prob. 94ECh. 1.6 - Prob. 95ECh. 1.6 - Prob. 96ECh. 1.6 - Prob. 97ECh. 1.6 - Prob. 98ECh. 1.6 - Prob. 99ECh. 1.6 - Prob. 100ECh. 1.6 - Prob. 101ECh. 1.6 - Prob. 102ECh. 1.6 - Prob. 103ECh. 1.6 - Prob. 104ECh. 1.6 - Prob. 105ECh. 1.6 - Prob. 106ECh. 1.6 - Prob. 107ECh. 1.6 - Prob. 108ECh. 1.6 - Prob. 109ECh. 1.6 - Prob. 110ECh. 1.6 - Prob. 111ECh. 1.6 - Prob. 112ECh. 1.6 - Prob. 113ECh. 1.6 - Prob. 114ECh. 1.6 - Prob. 115ECh. 1.6 - Prob. 116ECh. 1.6 - Prob. 117ECh. 1.6 - Prob. 118ECh. 1.6 - Prob. 119ECh. 1.6 - Prob. 120ECh. 1.6 - Prob. 121ECh. 1.6 - Prob. 122ECh. 1.6 - Prob. 123ECh. 1.6 - Prob. 124ECh. 1.6 - Prob. 125ECh. 1.6 - Prob. 126ECh. 1.6 - Prob. 127ECh. 1.6 - Prob. 128ECh. 1.6 - Prob. 129ECh. 1.6 - Prob. 130ECh. 1.6 - Prob. 131ECh. 1.6 - Prob. 132ECh. 1.6 - Prob. 133ECh. 1.6 - Prob. 134ECh. 1.6 - Prob. 135ECh. 1.6 - Prob. 136ECh. 1.6 - Prob. 137ECh. 1.6 - Prob. 138ECh. 1.6 - Prob. 139ECh. 1.6 - Prob. 140ECh. 1.6 - Prob. 141ECh. 1.6 - Prob. 142ECh. 1.6 - Prob. 143ECh. 1.6 - Prob. 144ECh. 1.7 - Prob. 1ECh. 1.7 - Prob. 2ECh. 1.7 - Prob. 3ECh. 1.7 - Prob. 4ECh. 1.7 - Prob. 5ECh. 1.7 - Prob. 6ECh. 1.7 - Prob. 7ECh. 1.7 - Prob. 8ECh. 1.7 - Prob. 9ECh. 1.7 - Prob. 10ECh. 1.7 - Prob. 11ECh. 1.7 - Prob. 12ECh. 1.7 - Prob. 13ECh. 1.7 - Prob. 14ECh. 1.7 - Prob. 15ECh. 1.7 - Prob. 16ECh. 1.7 - Prob. 17ECh. 1.7 - Prob. 18ECh. 1.7 - Prob. 19ECh. 1.7 - Prob. 20ECh. 1.7 - Prob. 21ECh. 1.7 - Prob. 22ECh. 1.7 - Prob. 23ECh. 1.7 - Prob. 24ECh. 1.7 - Prob. 25ECh. 1.7 - Prob. 26ECh. 1.7 - Prob. 27ECh. 1.7 - Prob. 28ECh. 1.7 - Prob. 29ECh. 1.7 - Prob. 30ECh. 1.7 - Prob. 31ECh. 1.7 - Prob. 32ECh. 1.7 - Prob. 33ECh. 1.7 - Prob. 34ECh. 1 - Prob. 1CRCh. 1 - Prob. 2CRCh. 1 - Prob. 3CRCh. 1 - Prob. 4CRCh. 1 - Prob. 5CRCh. 1 - Prob. 6CRCh. 1 - Prob. 7CRCh. 1 - Prob. 8CRCh. 1 - Prob. 9CRCh. 1 - Prob. 10CRCh. 1 - Prob. 11CRCh. 1 - Prob. 12CRCh. 1 - Prob. 13CRCh. 1 - Prob. 14CRCh. 1 - Prob. 15CRCh. 1 - Prob. 16CRCh. 1 - Prob. 17CRCh. 1 - Prob. 18CRCh. 1 - Prob. 19CRCh. 1 - Prob. 20CRCh. 1 - Prob. 21CRCh. 1 - Prob. 22CRCh. 1 - Prob. 23CRCh. 1 - Prob. 24CRCh. 1 - Prob. 25CRCh. 1 - Prob. 26CRCh. 1 - Prob. 27CRCh. 1 - Prob. 28CRCh. 1 - Prob. 29CRCh. 1 - Prob. 30CRCh. 1 - Prob. 31CRCh. 1 - Prob. 32CRCh. 1 - Prob. 33CRCh. 1 - Prob. 34CRCh. 1 - Prob. 35CRCh. 1 - Prob. 36CRCh. 1 - Prob. 37CRCh. 1 - Prob. 38CRCh. 1 - Prob. 39CRCh. 1 - Prob. 40CRCh. 1 - Prob. 41CRCh. 1 - Prob. 42CRCh. 1 - Prob. 43CRCh. 1 - Prob. 44CRCh. 1 - Prob. 45CRCh. 1 - Prob. 46CRCh. 1 - Prob. 47CRCh. 1 - Prob. 48CRCh. 1 - Prob. 49CRCh. 1 - Prob. 50CRCh. 1 - Prob. 51CRCh. 1 - Prob. 52CRCh. 1 - Prob. 53CRCh. 1 - Prob. 54CRCh. 1 - Prob. 55CRCh. 1 - Prob. 56CRCh. 1 - Prob. 57CRCh. 1 - Prob. 58CRCh. 1 - Prob. 59CRCh. 1 - Prob. 60CRCh. 1 - Prob. 61CRCh. 1 - Prob. 62CRCh. 1 - Prob. 63CRCh. 1 - Prob. 64CRCh. 1 - Prob. 65CRCh. 1 - Prob. 66CRCh. 1 - Prob. 67CRCh. 1 - Prob. 68CRCh. 1 - Prob. 69CRCh. 1 - Prob. 70CRCh. 1 - Prob. 71CRCh. 1 - Prob. 72CRCh. 1 - Prob. 73CRCh. 1 - Prob. 74CRCh. 1 - Prob. 75CRCh. 1 - Prob. 76CRCh. 1 - Prob. 77CRCh. 1 - Prob. 78CRCh. 1 - Prob. 79CRCh. 1 - Prob. 80CRCh. 1 - Prob. 81CRCh. 1 - Prob. 82CRCh. 1 - Prob. 83CRCh. 1 - Prob. 84CRCh. 1 - Prob. 85CRCh. 1 - Prob. 86CRCh. 1 - Prob. 87CRCh. 1 - Prob. 88CRCh. 1 - Prob. 89CRCh. 1 - Prob. 90CRCh. 1 - Prob. 91CRCh. 1 - Prob. 92CRCh. 1 - Prob. 93CRCh. 1 - Prob. 94CRCh. 1 - Prob. 95CRCh. 1 - Prob. 96CRCh. 1 - Prob. 97CRCh. 1 - Prob. 98CRCh. 1 - Prob. 99CRCh. 1 - Prob. 100CRCh. 1 - Prob. 101CRCh. 1 - Prob. 102CRCh. 1 - Prob. 103CRCh. 1 - Prob. 104CRCh. 1 - Prob. 105CRCh. 1 - Prob. 106CRCh. 1 - Prob. 107CRCh. 1 - Prob. 108CRCh. 1 - Prob. 109CRCh. 1 - Prob. 110CRCh. 1 - Prob. 1CTCh. 1 - Prob. 2CTCh. 1 - Prob. 3CTCh. 1 - Prob. 4CTCh. 1 - Prob. 5CTCh. 1 - Prob. 6CTCh. 1 - Prob. 7CTCh. 1 - Prob. 8CTCh. 1 - Prob. 9CTCh. 1 - Prob. 10CTCh. 1 - Prob. 11CTCh. 1 - Prob. 12CTCh. 1 - Prob. 13CTCh. 1 - Prob. 14CTCh. 1 - Prob. 15CTCh. 1 - Prob. 16CTCh. 1 - Prob. 17CTCh. 1 - Prob. 18CTCh. 1 - Prob. 19CTCh. 1 - Prob. 20CT
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