document.write( "Question 164053: A company's assets have a value of $30,000 at the beginning of the year 1999. By the beginning og 2001 the assets has depreciated to $26,000.
\n" ); document.write( "1) Assuming that the depreciation is linear, write the linear equation rating the value V of the assest in terms of the years passed since the beginning of 1999.
\n" ); document.write( "2)Find the value of the asest in the year 2006
\n" ); document.write( "3)Find after how many years the value of the asset will be $7,000.
\n" ); document.write( "4)Find after how many years wil the asset be worthless(value=0) \n" ); document.write( "
Algebra.Com's Answer #121479 by MRperkins(300)\"\" \"About 
You can put this solution on YOUR website!
ok, the first thing we need is to find the formula:\r
\n" ); document.write( "\n" ); document.write( "The problem says that the formula is linear so let's use the slope-intercept form or y=mx+b where m=slope and b=the y-intercept.\r
\n" ); document.write( "\n" ); document.write( "Remember that the slope is simply how much y changes for each unit of x. So in this situation when x moves from 1999 to 2001 the y changes from 30,000 to 26,000.
\n" ); document.write( "Formula: slope=(change in y)/(change in x)
\n" ); document.write( "or
\n" ); document.write( "\"slope=%28y1-y2%29%2F%28x1-x2%29\"
\n" ); document.write( "or
\n" ); document.write( "\"slope=%2830000-26000%29%2F%281999-2001%29\"
\n" ); document.write( "\"slope=4000%2F-2\"
\n" ); document.write( "\"slope=-2000\"
\n" ); document.write( "So in this case x changed 2 and y changed -4,000, so to find out the change per year instead of per 2 years: divide -4000 by 2 and you get the change in y divided by the change in x or -2000
\n" ); document.write( "now we know that y=-2000x+b
\n" ); document.write( "the y-intercept or \"b\" is the y value when x=0 and x is the number of years passed since the beginning of 1999 so when x=0 y=30,000
\n" ); document.write( "therefore our formula is \"y=-2%2C000x%2B30%2C000\"\r
\n" ); document.write( "\n" ); document.write( "Part 2
\n" ); document.write( "Find the value of the asset in the year 2006:
\n" ); document.write( "2006-1999=7
\n" ); document.write( "so find the value of y when x=7
\n" ); document.write( "y=-2000(7)+30000
\n" ); document.write( "y=-14000+30000
\n" ); document.write( "y=16000\r
\n" ); document.write( "\n" ); document.write( "Part 3
\n" ); document.write( "Find after how many years the value of the asset will be $7,000.
\n" ); document.write( "let y=7000 and solve for x
\n" ); document.write( "7000=-2000x+30000
\n" ); document.write( "subtract 30000 from each side
\n" ); document.write( "-23000=-2000x
\n" ); document.write( "divide each side by -2000
\n" ); document.write( "11.5=x
\n" ); document.write( "so the value of assets will be $7000 after 11.5 years\r
\n" ); document.write( "\n" ); document.write( "Part 4
\n" ); document.write( "Find after how many years wil the asset be worthless(value=0)
\n" ); document.write( "let y=0 and solve for x
\n" ); document.write( "0=-2000x+30000
\n" ); document.write( "subtract 30000 from both sides
\n" ); document.write( "-30000=-2000x
\n" ); document.write( "divide each side by -2000
\n" ); document.write( "15=x
\n" ); document.write( "so the value of assets will be worthless after 15 years
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "I hope this helps
\n" ); document.write( "PayPal donations accepted through justin.sheppard.tech@hotmail.com
\n" ); document.write( " \n" ); document.write( "
\n" );