Question

Question about polynomial function. :)

Answer 1

I dont understand should I just calculate it according to the answers? o.O

Answer 2

Well thats a weird number -_-

Answer 3

you will have to make a system of equation to solve for a standard cubic equation :)

ax^3+bx^2+cx+d :D

Answer 4

lets plug in those numbers given
a*(x)^3+b(x)^2+c(x)+d=y
a*(x)^3+b(x)^2+c(x)+d=y
a*(x)^3+b(x)^2+c(x)+d=y
a*(x)^3+b(x)^2+c(x)+d=y
-----------------------------

a*(1902)^3+b(1902)^2+c(1902)+d=1.959
a*(1924)^3+b(1924)^2+c(1924)+d=6.373
a*(1946)^3+b(1946)^2+c(1946)+d=7.29
a*(1982)^3+b(1982)^2+c(1982)+d=15.817

we know have to solve the system of equations for a b c and d

Answer 5

Can we just choose two of them??

Answer 6

i'd do all of them

a~~0.000087222, b~~-0.507058, c~~982.61, d~~-634733.

http://www.wolframalpha.com/input/?i=a*%281902%29%5E3%2Bb*%281902%29%5E2%2Bc*%281902%29%2Bd%3D1.959%2C+a*%281924%29%5E3%2Bb*%281924%29%5E2%2Bc*%281924%29%2Bd%3D6.373%2C+a*%281946%29%5E3%2Bb*%281946%29%5E2%2Bc*%281946%29%2Bd%3D7.29%2C++a*%281982%29%5E3%2Bb*%281982%29%5E2%2Bc*%281982%29%2Bd%3D15.817

Answer 7

now plug into

ax^3+bx^2+cx+d =y

Answer 8

y=0.00008x^3-0.5x^2+982.61x-634733

Answer 9

o.O

Answer 10

a~~0.000087222, b~~-0.507058, c~~982.61, d~~-634733


0.000087222x^3-0.507058x^2+982.61x-634733 =y

Answer 11

but its not in the choices 0.0

Answer 12

looks like i did it wrong hmmm :(

Answer 13

Lol maybe its the calculation? XD

Answer 14

I did not take into accoutnt let x be the number of years since 1902 :O

Answer 15

NEW SYSTEM

a*(2)^3+b*(2)^2+c*(2)+d=1.959,
a*(24)^3+b*(24)^2+c*(24)+d=6.373,
a*(46)^3+b*(46)^2+c*(46)+d=7.29,
a*(82)^3+b*(82)^2+c*(82)+d=15.817

Answer 16

a~~0.000087222, b~~-0.00989259, c~~0.403068, d~~1.19174

Answer 17

D WORKS NOW :DDDDDDDDDD

Answer 18

I did not take into accoutnt let x be the number of years since 1902 :O
What does that mean?? :O
okay let me figure this out

Answer 19

it means before I just plug in the year directly but it is wron
a*(1902)^3+b(1902)^2+c(1902)+d=1.959
a*(1924)^3+b(1924)^2+c(1924)+d=6.373
a*(1946)^3+b(1946)^2+c(1946)+d=7.29
a*(1982)^3+b(1982)^2+c(1982)+d=15.817
^ ^ ^ years
-----
a*(2)^3+b*(2)^2+c*(2)+d=1.959,
a*(24)^3+b*(24)^2+c*(24)+d=6.373,
a*(46)^3+b*(46)^2+c*(46)+d=7.29,
a*(82)^3+b*(82)^2+c*(82)+d=15.817


the correct way is that the instruction says. Let x= years since 1900

so you take the year and subtract 1900 and that is your x :D

Answer 20

oHHHHHHH!!!!
I got it lol
wait..Mmm

Answer 21

http://www.wolframalpha.com/input/?i=a*%282%29%5E3%2Bb*%282%29%5E2%2Bc*%282%29%2Bd%3D1.959%2C+a*%2824%29%5E3%2Bb*%2824%29%5E2%2Bc*%2824%29%2Bd%3D6.373%2C+a*%2846%29%5E3%2Bb*%2846%29%5E2%2Bc*%2846%29%2Bd%3D7.29%2C+a*%2882%29%5E3%2Bb*%2882%29%5E2%2Bc*%2882%29%2Bd%3D15.817

Answer 22

c!
XD

Answer 23

Thanks smarty
XD