Dimitristhym

I think anwer is G. 6 units long

Why f(1) = 1? 

\(f(\sqrt{2-1}) = f(\sqrt{1}) = f(1) \)

so \(f(1)= 1^10 + 2(1)^9 - 2(1)^8 - 2(1)^7 + 1^6 + 3(1)^2 + 6(1) + 1 = 1 + 2 - 2 - 2 + 1 + 3 + 6 + 1 = 3 + 6 + 1 = 10 \)

so \(f(1) = 10 \) not 1,where im wrong?

Thanks! 

I guess f(2)= f(1) + f(1) + 2(1)(1) = 4 + 4 + 2 =10 ,f(4) = f(2) + f(2) + 2(2)(2) = 10 +10 + 8 = 28 

f(8) = f(4) + f(4) +2(4)(4) = 28 + 28 + 32 = 88 

So i think f(8) = 88 

Hope it helps!

 f(f(f(x)))=a+b( a+b( a+bx))

 f(f(f(0)))=a+b(a+b(a))=27

 f(f(f(1)))=a+b(a+b(a+b))=29 

f(f(f(0)))=a+ba+b^2a=27

f(f(f(1)))=a+ba+ab^2+b^3=29

(1) a+ba+b^2a=27 -> f(f(f(1)))=(1)+b^3=29 so 27 + b^3 = 29 <=> b^3 = 29-27 <=> b^3 = 2 <=> b = \(\sqrt[3]{2}\)\(\)

after in f(f(f(1))) you are replacing and find the a!
 

Oof?

Dont post it again critical answer to you before - 6 2/9

\(3\sqrt{10} = \sqrt{3^2}\times\sqrt{10}= \sqrt{9}\times\sqrt{10}= \sqrt{9\times10} =\sqrt{90}\)

 The number under the radical sign is \(90\)

Hope this helps!

The polynomial is f(x)= 0x^7 + 8x^6 + x^5 +5x^4 + 2x^3 + 0x^2 + 0x + 2 (08152002)

so \(f(x)=8x^6 + x^5 +5x^4 + 2x^3 + 2\)

BTW we have the same day birthday but im in 1999

Hope this helps!

Here in Greece we wish to take 510$ per week!We take 350$ PER MONTH :(