Your answer for rounding 2.8497 x 10^3 is correct: 2.85 x 10^3.
350.0 is not correct because it has 4 sig figs. The proper rounding would be simply 350. with not additional zeros.
Answer:
t≈8.0927
Step-by-step explanation:
h(t) = -16t^2 + 128t +12
We want to find when h(t) is zero ( or when it hits the ground)
0 = -16t^2 + 128t +12
Completing the square
Subtract 12 from each side
-12 = -16t^2 + 128t
Divide each side by -16
-12/-16 = -16/-16t^2 + 128/-16t
3/4 = t^2 -8t
Take the coefficient of t and divide it by 8
-8/2 = -4
Then square it
(-4) ^2 = 16
Add 16 to each side
16+3/4 = t^2 -8t+16
64/4 + 3/4= (t-4)^2
67/4 = (t-4)^2
Take the square root of each side
±sqrt(67/4) =sqrt( (t-4)^2)
±1/2sqrt(67) = (t-4)
Add 4 to each side
4 ±1/2sqrt(67) = t
The approximate values for t are
t≈-0.092676
t≈8.0927
The first is before the rocket is launched so the only valid answer is the second one
Find the common denominator of 6 and 4.
6: 6, 12...
4: 8, 12...
We can use 12 for this example.
5/6 to ?/12.
12 / 6 = 2, therefore do 5 x 2.
5 x 2 = 10. 10/12.
1/4 to ?/12.
12 / 4 = 3, therefore do 3 x 1.
3 x 1 = 3. 3/12.
13/12. 1 1/12.
I believe you ran out of space or didn't put answer D. I researched this question and concluded that the other choice is 1 and 1/12.
The answer is 1 and 1/12.
66 is the anwer I think I'm not for sure I hope this helps and you get it right
