document.write( "Question 1087074: The depreciated value VV of a machine is a linear function of time t in years. A machine that is purchased for $100000 today will be worth approximately $67681 in 33 years. Write the linear function V(t) that represents the value of the machine at a given time t.\r
\n" ); document.write( "\n" ); document.write( " My work---> 100,1000---->67,681 3yrs
\n" ); document.write( "67,681v+3=100,000 this is what I got but I'm not sure.
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Algebra.Com's Answer #701359 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
I think you meant to say \"The depreciated value V\" instead of \"The depreciated value VV\". I changed the VV to V. \r
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\n" ); document.write( "\n" ); document.write( "The value today is V = 100000 dollars when the time is t = 0, which is the starting time value.
\n" ); document.write( "The value in 33 years, when t = 33, is V = 67681 dollars\r
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\n" ); document.write( "\n" ); document.write( "So we have two ordered pairs (0,100000) and (33,67681). I'm going to treat t as x, and treat V as y. So each ordered pair goes from the form (t,V) to (x,y). Using the (x,y) form we can use the slope formula\r
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\n" ); document.write( "\n" ); document.write( "First point = (x1,y1) = (0,100000)
\n" ); document.write( "Second point = (x2,y2) = (33,67681)
\n" ); document.write( "\"m+=+%28y%5B2%5D+-+y%5B1%5D%29%2F%28x%5B2%5D+-+x%5B1%5D%29\"
\n" ); document.write( "\"m+=+%2867681+-+100000%29%2F%2833+-+0%29\"
\n" ); document.write( "\"m+=+%28-32319%29%2F%2833%29\"
\n" ); document.write( "\"m+=+-10773%2F11\"
\n" ); document.write( "\"m+=+-979.3636\" Use a calculator here. The decimal form is approximate (the '36' portion repeats forever)\r
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\n" ); document.write( "\n" ); document.write( "So the slope is roughly \"m+=+-979.3636\"\r
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\n" ); document.write( "\n" ); document.write( "The y intercept is \"b+=+100000\" because this is the starting value (at t = 0)\r
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\n" ); document.write( "\n" ); document.write( "So we go from \"y+=+mx%2Bb\" to \"y+=+-979.3636x%2B100000\"\r
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\n" ); document.write( "\n" ); document.write( "The last thing to do is replace x with t and replace y with V(t) to get \"V%28t%29+=+-979.3636t%2B100000\"\r
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\n" ); document.write( "\n" ); document.write( "Therefore the value function is \"V%28t%29+=+-979.3636t%2B100000\"\r
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\n" ); document.write( "\n" ); document.write( "-------------------------------------------------\r
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\n" ); document.write( "\n" ); document.write( "Let's check the function. Plug t = 0 into the function to get
\n" ); document.write( "\"V%28t%29+=+-979.3636t%2B100000\"
\n" ); document.write( "\"V%280%29+=+-979.3636%2A0%2B100000\"
\n" ); document.write( "\"V%280%29+=+0%2B100000\"
\n" ); document.write( "\"V%280%29+=+100000\"
\n" ); document.write( "So that matches with the fact the initial value is $100,000\r
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\n" ); document.write( "\n" ); document.write( "Now plug in t = 33
\n" ); document.write( "\"V%28t%29+=+-979.3636t%2B100000\"
\n" ); document.write( "\"V%2833%29+=+-979.3636%2A%2833%29%2B100000\"
\n" ); document.write( "\"V%2833%29+=+-32318.9988%2B100000\"
\n" ); document.write( "\"V%2833%29+=+67681.0012\"
\n" ); document.write( "\"V%2833%29+=+67681.00\"
\n" ); document.write( "Which matches with the value after 33 years. So the answer is confirmed\r
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\n" ); document.write( "\n" ); document.write( "Side Note: if you use the fraction form for the slope then you won't run into rounding errors. The fact that we're rounding to 2 decimal places means that we don't have to worry about precision too much as long as the slope is expressed to 3 decimal places or more. You're probably wondering what the slope means? If so, then the slope is simply the rate of value decay or drop. In this case, the slope -979.3636 means the value V(t) is decreasing by $979.36 each year. The actual drop is a bit more than that but you can only round to the nearest hundredth for money problems like this.
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