For the time period of the projectile at least 192 feet above the ground is between 2 second and 6 second
Since s=128t-16t2, we want to know the 2 times where 128t-16t2=192, as those two values will be the begin and end times of the interval in question. We can rewrite that equation in standard quadratic equation form:
16t2-128t+192=0 or, simplified, 8t2-64t+96=0
or t2-8t+12=0
since it is now in quadratic form, we can solve using the quadratic formula:
t = 2 and 6
So the time period from approximately 2 second and 6 second has the projectile above 192 feet.
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Answer:
8
Step-by-step explanation:
3/4 = 6/8
Answer:
-1
Step-by-step explanation:
h(t) = – t – 4
Let t = -3
h(-3) = – -3 – 4
= +3 -4
-1
Answer:
d=-4
Step-by-step explanation:
20 = –d + 16
Subtract 16 from each side
20-16 = –d + 16-16
4 = -d
Multiply each side by -1
-1 *4 = -1 * -d
-4 =d