Step-by-step explanation:
Let the height above which the ball is released be H
This problem can be tackled using geometric progression.
The nth term of a Geometric progression is given by the above, where n is the term index, a is the first term and the sum for such a progression up to the Nth term is
To find the total distance travel one has to sum over up to n=3. But there is little subtle point here. For the first bounce ( n=1 ), the ball has only travel H and not 2H. For subsequent bounces ( n=2,3,4,5...... ), the distance travel is 2×(3/4)n×H
a=2H..........r=3/4
However we have to subtract H because up to the first bounce, the ball only travel H instead of 2H
Therefore the total distance travel up to the Nth bounce is
For N=3 one obtains
D=3.625H
<span>the first term 13m has TWO factors 13 and m
the second term 2n has TWO factors 2 and n</span>
There is a 2 of 6 percent chance that one them might land on three if you add one extra dice it would be 3 of 6 percent chance and so on.
Hope this Helped!
;D
Brainliest??
Step-by-step explanation:
1.
2.
3.
- 485 - 319 = 166 more people
4.
- 14.5x + 80 > 500
- 14.5x > 420
- x > 420/14.5
- x > 29
5.
- 1.25x + 3 ≤ 28
- 1.25x ≤ 25
- x ≤ 25/1.25
- x ≤ 20 miles
6. <em>This seems a typo... 460 is likely $60</em>
- 60x + 145 ≤ 625
- 60x ≤ 625 - 145
- 60x ≤ 480
- x ≤ 480/60
- x ≤ 8 days
