Verified Answer :

MissPhiladelphia

For the answer to the question above, 
1 + nx + [n(n-1)/(2-factorial)](x)^2 + [n(n-1)(n-2)/3-factorial] (x)^3 

1 + nx + [n(n-1)/(2 x 1)](x)^2 + [n(n-1)(n-2)/3 x 2 x 1] (x)^3 

1 + nx + [n(n-1)/2](x)^2 + [n(n-1)(n-2)/6] (x)^3 

1 + 9x + 36x^2 + 84x^3 

In my experience, up to the x^3 is often adequate to approximate a route. 

(1+x) = 0.98 

x = 0.98 - 1 = -0.02 

Substituting: 

1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 

approximation = 0.834 

Checking the real value in your calculator: 

(0.98)^9 = 0.834 

So you have approximated correctly. 

If you want to know how accurate your approximation is, write out the result of each in full: 

1 + 9(-0.02) + 36(-0.02)^2 + 84(-0.02)^3 = 0.833728 

 (0.98)^9 = 0.8337477621 

So it is correct to 4

junioraltamiraoy1jqh

Answer:

a= 1 b= -0.02 c=5

Step-by-step explanation:

took it on edge

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