Answer:
<u>The solutions to this quadratic equation are 10 and 3.</u>
Step-by-step explanation:
1. Let's recall the formula for solving this type of equations, called quadratic equations:
x = ( - b +/- √ b² - 4ac)/ 2a
The equation given is x2 – 13x + 30 = 0,
where a = 1, b = - 13 and c = 30
Now replacing with the real values, we have
x = [ - (-13) +/- √ (-13)² - 4 * 1 * 30] / 2 * 1
x = [ 13 +/- √ 169 - 120] / 2
x = [13 +/- √ 49]/ 2
x = [ 13 +/- 7 ]/ 2
<u>x₁ = 13 + 7/2 = 20/2 = 10</u>
<u>x₂ = 13 - 7/2 = 6/2 = 3</u>
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
Idk tbh wat is a great game for me so I can play with the kids and play it for a while and then play me and play me when I get home I wanna Play is time
The interquartile range is 12-6=6
