Let $f(x)$ be a quadratic polynomial such that $f(-4) = -22,$ $f(-1)=2$, and $f(2)=-1.$ Let $g(x) = f(x)^{16}.$ Find the sum of the coefficients of the terms in $g(x)$ that have even degree. (For example, the sum of the coefficients of the terms in $-7x^3 + 4x^2 + 10x - 5$ that have even degree is $(4) + (-5) = -1.$)

The sum of the coefficients of the terms in (-1.5·x² + 0.5·x + 4)¹⁶ that have even degree is  -4273411167.501

Step-by-step explanation:

The parameters given are;

f(-4) = -22

f(-1) = 2

f(2) = -1

g(x) = f(x)¹⁶

The function f(x) is presented as follows;

f(x) = a·x² + b·x +c

We have;

-22 = a·(-4)² + b·(-4) +c

-22 = a·16 - 4·b +c (1)

2 = a·(-1)² + b·(-1) +c

2 = a - b +c(2)

-1 = a·(2)² + b·(2) +c

-1 = 4·a + 2·b +c(3)

Solving the equations (1), (2), and (3) by using an online linear systems solver, we get;

a = -1.5, b = 0.5, c = 4

Therefore, f(x) = -1.5·x² + 0.5·x + 4

f(x)¹⁶ = (-1.5·x² + 0.5·x + 4)¹⁶ which gives the coefficients of the even terms as follows;

656.841 - 19267.331 + 248302.054 - 1772904.419 + 6735603.932 - 2868054.635 - 119602865.901 + 750783340.827 + -2542435585.611 + 5338903756.992 - 6048065910.25 -1031335136 + 17223697920 - 32238338048 + 32107397120 - 17716740096 = -4273411167.501.

x = 25

step-by-step explanation:

a^2+b^2=c^2

7^2+24^2=c^2

49+576=625

sqrt625=25

x=25

126.66 m²

step-by-step explanation:

the pertinent area formula is a = πr², where r is the radius.

here, a = π(6.35 m)² = 40.32π, or approximately 126.66 m²