Answer:
a) t =2 sec
b) t = 0.5 sec
Explanation:
Given data:
a) set s=given equation to zero and solve for t:
s = 32t - 16t^2
0 = 32t-16t^2
32t = 16t^2
32 = 16 t
t = 2 seconds
b) set given equation equal to 12 and solve for t:
s = 32 t -16t^2
12 = 32t-16t^2
12 = 32t-16t^2
t^2 -2t + 0.75 = 0
t = 1.5, 0.5
hence t = 0.5 sec is right answer
Answer:
a) t =2 sec
b) 0.5s < t < 1.5s
Explanation:
Given data:
a) set s=given equation to zero and solve for t:
s = 32t - 16t^2
0 = 32t-16t^2
0 = t(32-16t)
t = 0
32 = 16 t
t = 2 seconds
Answer: The ball strikes the ground again at t = 2 seconds
b) set given equation greater than 12 and solve for t:
s = 32 t -16t^2
12 < 32t-16t^2
12 < 32t-16t^2
t^2 -2t + 0.75 < 0
0.5s < t < 1.5s
Answer: The ball is above 12 feet from the ground in the time interval of 0.5s < t < 1.5s
(25 POINTS) PLEASE HELP WILL GIVE BRAINLIEST, THANKS AND 5 STAR RATING!!!
5 QUESTIONS SHOW WORK!
[tex]PYTHAGOREAN THEOREM\\c^{2} = a^{2} + b^{2}\\ b^{2}=c^{2} -a^{2}\\ a^{2} =b^{2}-c^{2}[/tex]
1) x=9, because we can use the Pythagorean theorem, x^2 + 144 = 225, x^2 = 81
2) [tex]x=3\sqrt{8}[/tex], because 6^2 +6^2 = 72, [tex]\sqrt{72} = 3\sqrt{8}[/tex]
3) [tex]x=2\sqrt{3}[/tex], because [tex]x^{2} +7=19\\x^{2} =12\\x= 2\sqrt{3}[/tex]
hope this helps... load for images!