I think after the 9th bounce it will lose 90% of its max height
times 40 by 0.9 which will give you 36
then take 36 away from 40 leaving you with 4.
answer = 4ft
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
If the machine can pack 50 boxes a minute, that would mean you have to multiply 50 x 25 which gives you your answer of 1,250
The answer is true trust just had this
The value of x is 4. The steps to solving the equation are to use distributive property first so that the equation then looks like this: 10x+20=5x+40
Next you would combine like terms and then the equation would look like this: 5x+20=40
Then you would use the subtraction property of equality to subtract 20 from 40 so your equation will now look like this: 5x=20
Then you use the division property of equality to divide 20 by 5 to get x=4
