Find f(t – 3) for f(x) = 4x^2 – 8x + 4. question 11 options: a. 4t^2 – 32t + 64 b. 64 c. 4t^2 – 32t – 64 d. 4t^2 + 32t + 64

Step-by-step explanation:

Instead of x put (t - 3) in the equation of the function f(x) = 4x² - 8x + 4:

f(t - 3) = 4(t - 3)² - 8(t - 3) + 4

use (a - b)² = a² - 2ab + b² and the distributive property a(b + c) = ab + ac

f(t - 3) = 4(t² - (2)(t)(3) + 3²) + (-8)(t) + (-8)(-3) + 4

f(t - 3) = 4(t² - 6t + 9) - 8t + 24 + 4

f(t - 3) = (4)(t²) + (4)(-6t) + (4)(9) - 8t + 28

f(t - 3) = 4t² - 24t + 36 - 8t + 28

f(t - 3) = 4t² + (-24t - 8t) + (36 + 28)

f(t - 3) = 4t² - 32t + 64

i believe it is 0.1m because the measurement of 30.4m limits how precise you can be

you can only be as precise as the least precise measurement