G(x) = 2x² - 5x + 2 = 2x² - 4x - x + 2 = 2x · x - 2x · 2 - 1 · x - 1 · (-2)
= 2x(x - 2) -1(x - 2) = (x - 2)(2x - 1)
g(x) = 0 ⇔ (x - 2)(2x - 1) = 0 ⇔ x - 2 = 0 or 2x - 1 = 0
x = 2 or x = 0.5
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
Answer:
As in Decimal: 0.86602540378
Angle Tan=Sin/Cos
30° 1 √3 = √3 3
45° 1
60° √3
Step-by-step explanation:
Answer:
c
Step-by-step explanation:
reason for is because when you add all the like terms you end up with c
