Question

Consider that the age, x, of a unicorn in human equivalent years can be given by the formula f(x) = - 0.003518x4 + 0.087326x3 – 1.3367x2 + 12.46x + 2.914. When a unicorn is 2.5 years old, what is its age in human equivalent years? What about when it is 12 years old?

Answer 1

So, by definition, a function gives you a "y" value for and y "x" value you plug in:
in this case the "y", or "range" is human years, and the "x" or "domain" is unicorn years

Answer 2

therefore, if you plug in the the age in unicorn years into the function, you end up with the equivalent age in human years

Answer 3

so you would just substitute in a 2.5 everywhere you see an "x" in the equation

Answer 4

(the unicorn age)

Answer 5

the answer you get will be the "y", or the age in human years

Answer 6

according to my calculator, the result is 27.20956563 human years for 2.5 unicorn years

Answer 7

did you get the same result?

Answer 8

double checking, I am getting 26.937

Answer 9

28.1789

Answer 10

I am getting 26.937 and 37.899, respectively

Answer 11

28.1789 and 40.6641

Answer 12

\[\small f(x) = - 0.003518x^4 + 0.087326x^3 – 1.3367x^2 + 12.46x + 2.914\]\[\small f(x) = \left(- 0.003518x^3 + 0.087326x^2 – 1.3367x + 12.46\right)x + 2.914\]\[\small f(x) = \left(\left(- 0.003518x^2 + 0.087326x – 1.3367\right)x + 12.46\right)x + 2.914\]\[\small f(x) = \left(\left(\left(- 0.003518x + 0.087326\right)x – 1.3367\right)x + 12.46\right)x + 2.914\]

Answer 13

I'll check my answer again but that's what I'm getting.

Answer 14

@UnkleRhaukus Is that last equation more efficient (in terms of algorithms)? (I think it is, seeing as how it only has n-1 multiplications as opposed to \[\sum_{x=1}^{n} n-x\]
multiplications, and the same number of additions)

Answer 15

\[\small f(2.5) = \left(\left(\left(- 0.003518\times2.5 + 0.087326\right)2.5 – 1.3367\right)2.5 + 12.46\right)2.5 + 2.914\]
\[\qquad= \left(\left(\left(-0.008795 + 0.087326\right)2.5 – 1.3367\right)2.5 + 12.46\right)2.5 + 2.914\]\[\qquad= \left(\left( – 1.3367\right)2.5 + 12.46\right)2.5 + 2.914\]\[\qquad= \left(-3.34175 + 12.46\right)2.5 + 2.914\]\[\qquad= \left(9.11825\right)2.5 + 2.914\]\[\qquad= 22.795625+ 2.914\]\[=25.709625\]

Answer 16

ops made a mistake going from line two to line three just now.

Answer 17

\[\qquad= \left(\left(\left(-0.008795 + 0.087326\right)2.5 – 1.3367\right)2.5 + 12.46\right)2.5 + 2.914\]\[\qquad= \left(\left(\left(0.078531\right)2.5 – 1.3367\right)2.5 + 12.46\right)2.5 + 2.914\]

Answer 18

that gives 26.937

Answer 19

28.1789 I know for sure is right. Answer without rounding

Answer 20

rounded to the nearest thousandth

Answer 21

\[\small f(2.5) = \left(\left(\left(- 0.003518\times2.5 + 0.087326\right)2.5 – 1.3367\right)2.5 + 12.46\right)2.5 + 2.914\]\[\qquad= \left(\left(\left(-0.008795 + 0.087326\right)2.5 – 1.3367\right)2.5 + 12.46\right)2.5 + 2.914\]\[= \left(\left(\left(0.078531\right)2.5 – 1.3367\right)2.5 + 12.46\right)2.5 + 2.914\]\[= \left(\left(0.1963275 – 1.3367\right)2.5 + 12.46\right)2.5 + 2.914\]\[= \left(\left(-1.1403725\right)2.5 + 12.46\right)2.5 + 2.914\]\[= \left(-2.85093125 + 12.46\right)2.5 + 2.914\]\[= \left(9.60906875\right)2.5 + 2.914\]\[=286.222157868798+2.914\]\[=286.222157868798+2.914\]\[=289.136157868798\]\[ \approx300\]

Answer 22

9.6 * 2.5 in NOT 286

Answer 23

oh good,

Answer 24

\[= \left(9.60906875\right)2.5 + 2.914\]\[=24.022671875+2.914\]\[\large=26.936671875\]

Answer 25

im in agreement with Mr Moose

Answer 26

I submitted the 28.17 and it was graded as correct so... oh well thanks anyways :)

Answer 27

/**
*no description
*/

import java.util.Scanner;

public class Function
{
public static void main(String[] args)
{
Scanner kboard = new Scanner(System.in);

System.out.println("x?");
double x = kboard.nextDouble();

double y = (((-.003518 * x + 0.087326) * x - 1.3367) * x + 12.46) * x + 2.914;

String temp = y + "";

System.out.println(temp);
}
}

says that whoever is grading has no clue

Answer 28

26.936671875 and
37.899280000 should be the correct answers