Answer:
Step-by-step explanation:Since we want to know how much time elapsed since the object was launched until it ht the ground. At the ground, the height is 0.
Therefore the equation h(t) = 80t – 16t2 becomes 0= 80t – 16t2
Solving the equation for t gives 80t =16t2 or 16t=80 or t=5sec
The object hits the ground 5 seconds after it is launched.
Answer:
BE = FC = 3 inches, EF = 2 inches
Step-by-step explanation:
The sum of angles A and D is 180°, so the sum of their half-angles is 90°. That is, half of A plus half of B add to 90°, so the bisector from B intersects AE at a right angle. Call that point of intersection X.
Then angle ABX = angle EBX, so triangle ABX is congruent to triangle EBX. Sides AB and BE are corresponding sides of congruent triangles.
The same argument applies to sides DC and CF.
Thus we have BE = CF = 3 inches, and EF is the left-over distance, 2 inches.
1) 50x30
2) 2000+409
3) 500+800
Answer:
arc PQ = 124°
Step-by-step explanation:
The inscribed angle PRQ is half the measure of its intercepted arc PQ, so
arc PQ = 2 × 62° = 124°
Try this option:
way 1: using the picture, if x=-3 then y=-6.
way 2: the equation of this line is y=x-3; it means that y(-3)=-6.
