SOLUTION: I would appreciate assistance with how this problem is completed.
In a study of worker efficiency at Wong
Laboratories it was found that the number of components
assembled pe
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Test
-> SOLUTION: I would appreciate assistance with how this problem is completed.
In a study of worker efficiency at Wong
Laboratories it was found that the number of components
assembled pe
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Question 292578: I would appreciate assistance with how this problem is completed.
In a study of worker efficiency at Wong
Laboratories it was found that the number of components
assembled per hour by the average worker t hours after
starting work could be modeled by the formula
N(t)=-3t^3+23t^2+8t.
a) Rewrite the formula by factoring the right-hand side
completely.
b) Use the factored version of the formula to find N(3). Found 2 solutions by nerdybill, Theo:Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! Laboratories it was found that the number of components
assembled per hour by the average worker t hours after
starting work could be modeled by the formula
N(t)=-3t^3+23t^2+8t.
a) Rewrite the formula by factoring the right-hand side
completely.
N(t)=-3t^3+23t^2+8t
N(t)=t(-3t^2+23t+8)
N(t)=t(-3t^2+24t-t+8)
N(t)=t((-3t^2+24t)-(t-8))
N(t)=t(3t(-t+8)-(t-8))
N(t)=t(-3t(t-8)-(t-8))
N(t)=t(t-8)(-3t-1)
.
b) Use the factored version of the formula to find N(3)
N(t)=t(t-8)(-3t-1)
N(3)=3(3-8)(-3(3)-1)
N(3)=3(-5)(-9-1)
N(3)=3(-5)(-10)
N(3)=(-15)(-10)
N(3)=150
Plug these into your original equation to see if they are good.
Your original equation is:
-3t^3+23t^2+8t
When t = 0, this equation becomes 0.
When t = -1/3, this equation becomes -3*(-1/3)^3 + 23*(-1/3)^2 + 8*(-1/3).
This becomes .11111111 + 2.55555555 - 2.66666666 which becomes 0.
When t = 8, this equation becomes -3*8^3 + 23*8^2 + 8*8.
This becomes -1536 + 1472 + 64 which becomes 0.
All 3 roots are good.
Our factored equation is:
N(t) = t * (3t+1) * (-t+8) = 0
When we substitute 3 for t, we get:
N(3) = 3 * (3*3+1) * (-3+8) which becomes:
N(3) = 3 * 10 * 5 which becomes:
N(3) = 150
Plug 3 into the original equation of -3t^3+23t^2+8t and you get -3*(3^3) + 23*(3^2) + 8*3 which becomes -3*27 + 23*9 + 24 which becomes -81 + 207 + 24 which becomes 150.
The factoring is good and the answer is N(3) = 150
I determined the factoring as follows:
I knew that 1*8 = 8 or 2*4 = 8
I tried 1*8 first.
I knew that 3*8 = 24 and -1*8 = -1 so I figured that the factors had to be something like (3t+1) * (-t+8)
When you multiply these together, you get:
(3t+1) * (-t+8) =
(3t * (-t+8)) + (1 * (-t+8)) =
(-3t^2 + 24t) + (-t + 8) =
-3t^2 + 24t -t + 8 =
-3t^2 + 23t + 8 which makes them the correct factors.
