The land bridge (Bering strait) and the crossing from Canada
Answer:
u = 5/4
Step-by-step explanation:
to evaluate for the value of u we would simply open the bracket and then evaluate for the value of u by collecting the like terms together.
solution
3=7(4 - 2u)-6u
3 = 28- 14u - 6u
collect the like terms
3 + 14u + 6u = 28
20u = 28 - 3
20u = 25
divide both sides by the coefficient of u which is 20
20u/20 = 25/20
u = 5/4
In order to find height from where ball is dropped, you have to find height or h(t) when time or t is zero.So plug in t=0 into your quadratic equation:h(0) = -16.1(0^2) + 150h(0) = 0 +150h(0) = 150 ft is the height from where ball is dropped. When ball hits the ground, the height is zero. So plug in h(t) = 0 and solve for t.0 = -16.1t^2 + 15016.1 t^2 = 150t^2 = 150/16.1t = sqrt(150/16.1)t = ± 3.05Since time cannot be negative, your answer is positive solution i.e. t = 3.05
