## Prealgebra (7th Edition)

Substitute each given value and resolve. $16t^2=16(1)^2=16\times1=16$ $16t^2=16(2)^2=16\times4=64$ $16t^2=16(3)^2=16\times9=144$ $16t^2=16(4)^2=16\times16=256$