Let t: p3→p3 be the linear transformation such that t(â’2x2)=3x2+4x, t(0.5x+3)=â’3x2+3xâ’4, and t(2x2â’1)=â’3x+4. find t(1), t(x), t(x2), and t(ax2+bx+c), where a, b, and c are arbitrary real numbers.

It looks like we're told that

We use the fact that is linear to find . First, we notice that

We also have

So once we find , we can determine and . We have

and using this we find

Then

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40