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Posted on 2016-09-06
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Can you please tell me how f(0), f(1), f(2).... is calculated?
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Question by:mustish1
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LVL 74

Expert Comment

by:sdstuber
ID: 41786944
substitute the value for the variable and complete the arithmetic

f(x) = -0.8x(x-9)

so f(0) = -0.8*0(0-9)= 0(-9) =0

f(1) = -0.8*1(1-9) = -0.8(-8) = 6.4

and so on
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by:mustish1
ID: 41786973
Can you please also explain the part B
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sdstuber earned 2000 total points
ID: 41787024
f(x) = g(x)

f(x) = -0.8x(x-9)
g(x) = 1.7x + 5.7

so  -0.8x(x-9) = 1.7x + 5.7

-0.8x^2 + 7.2x = 1.7x + 5.7

-0.8x^2 + 5.5x - 5.7 = 0

-10 * (-0.8x^2 + 5.5x - 5.7)  = -10 * 0  
       ----- this step what was done in your example except it wasn't shown

8x^2 - 55x + 57 = 0

use quadratic equation to solve for x

x = (-b +/- sqrt(b^2 - 4ac))/2a

a = 8
b = -55
c = 57

x = (55 +/- sqrt(55^2 -4*8*57)) / (2*8)

x = (55 +/- sqrt(3025 - 1824)) / 16

x = (55 +/- sqrt(1201)) / 16

x = (55 +/- 34.66) / 16

x = 89.66/16  or 20.34/16

x = 5.60 or 1.27
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Author Closing Comment

by:mustish1
ID: 41787035
Thank You.
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