<span>68=-16t2+64t+8
16t2-64t+60=0
4t2-16t+15=0
(2t-5)(2t-3)=0
t=5/2 or 3/2
The object is at 68 feet after 3/2 secs, and returns to 68 feet after another second..</span>
Answer: have you tried to use photomath?
7 is digit for hundred thousands and 6 is for ten thousands so the answer is 800,000
Answer:
c
Step-by-step explanation:
The first step is to locate the coordinates of the vertices of the arrow. From the graph,
A = (2, 1)
B = (2, 3)
C = (1, 3)
D = (3, 5)
E = (5, 3)
F = (4, 3)
G = (4, 1)
If (a, b) is reflected on the line y = - x, the coordinates of the image is (- b, - a)
By applying this rule, the new coordinates after reflecting the arrow over the line y = - x are
A'(- 1, - 2)
B'(- 3, - 2)
C'(- 3, - 1)
D'(- 5, - 3)
E'(- 3, - 5)
F'(- 3, - 4)
G(- 1, - 4)
