Answer:
Step-by-step explanation:
1. -35, 2
2. -72, -6
3. -16, 0
4. 40, 13
5. -15, 2
6. -30, 7
7. 30, 11
8. 42, -13
9. -3, -24
10. -4, 20
11. 7, 1
12. 4, -24
13. -7, -70
14. 4, 9
15. 7, -3
16. 9, 3
17. -10, 11
18. 6, -5
19. 11, 6
20. 1, -9
21. -9, 4
22. -8, 7
23. 6, 4
24. -7, -5
25. 11, 4
26. 10, 3
27. -8, 5
28. -9, 0
![\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{64}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0\qquad \textit{from the ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20~~~~~~%5Ctextit%7Bin%20feet%7D%20%5C%5C%5C%5C%20h%28t%29%20%3D%20-16t%5E2%2Bv_ot%2Bh_o%20%5Cend%7Barray%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Cstackrel%7B64%7D%7B%5Ctextit%7Binitial%20velocity%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h_o%3D%5Cstackrel%7B0%5Cqquad%20%5Ctextit%7Bfrom%20the%20ground%7D%7D%7B%5Ctextit%7Binitial%20height%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h%3D%5Cstackrel%7B%7D%7B%5Ctextit%7Bheight%20of%20the%20object%20at%20%22t%22%20seconds%7D%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)
Check the picture below, it hits the ground at 0 feet, where it came from, the ground, and when it came back down.
20% of 124.95 is close to $125 so that is equal of 12.50 of 10% so multiply by 2 which is $25.00 so the estimate discount of the jacket will be $25.00
Answer:
No, not correct.
Step-by-step explanation:
Note that 8 * 4.3 can be broken up as follows:
8 * (4 + 0.3)
and that 4*8 =32, and that 0.3(8) = 2.4 (not 0.24).
Then 8 * (4 + 0.3) becomes 32 + 2.4 = 34.4
