The nearest tenth of how fast a rover will hit Mars' surface after a bounce of 15 ft in height is 20.7ft/s.
<h3>What is the approximation about?</h3>
From the question:
Mars: F(x) = 2/3
Therefore, If x = 15
Then:
f (15) = 2/3
= 16/3
= 20.7ft/s
Hence, The nearest tenth of how fast a rover will hit Mars' surface after a bounce of 15 ft in height is 20.7ft/s.
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13. You add the 8 and the 5
Answer:
See in bold
1 x 1 = 1
1 x 2 = 2
1 x 3 = 2
1 x 4 = 4
1 x 5 = 5
2 x 1 = 2
2 x 2 = 4
2 x 3 = 6
2 x 4 = 8
2 x 5 = 10
3 x 1 = 3
3 x 2 = 6
3 x 3 = 9
3 x 4 = 12 Total 70
3 x 5 = 15
4 x 1 = 4
4 x 2 = 8
4 x 3 = 12
4 x 4 = 16 total 125
4 x 5 = 20
5 x 1 = 5
5 x 2 = 10
5 x 3 = 15 total 175
5 x 4 = 20
5 x 5 = 25 total 220
Total = 220
Algorithm Finding 142 = 220 - 78 = 142
-78 = 6 x 13 therefore 142six = 13
Algarithm Finding X 230six = X diagonal
X diagonal list on table shows;
1 x 1 = 1
2 x2 = 4
3 x 3 = 9
4 x 4 = 16
5 x 5 = 25
6 x 6 = 36
Total = 91
230/91 = 2.52747253 if diagonal algorithm
or 230 taken from the last 5 x 5 algorithm total = 220
1 x 6 = 6 220 + 6 = 226 x = 4+ then 1 x 6
2 x 6 = 12 = 18 220 + 18 = 38 x = 8 - then 2 x 6
as X230 = 8- 230 = 2 x 6
or X230 = 4+ 230 = 1 x 6
x = 60 - y
X is the amount of money left. 60 is how much Harris originally had. y is however much he spends.