Answer:
y = 5/2x - 2 or C
Step-by-step explanation:
Use the rules of logarithms and the rules of exponents.
... ln(ab) = ln(a) + ln(b)
... e^ln(a) = a
... (a^b)·(a^c) = a^(b+c)
_____
1) Use the second rule and take the antilog.
... e^ln(x) = x = e^(5.6 + ln(7.5))
... x = (e^5.6)·(e^ln(7.5)) . . . . . . use the rule of exponents
... x = 7.5·e^5.6 . . . . . . . . . . . . use the second rule of logarithms
... x ≈ 2028.2 . . . . . . . . . . . . . use your calculator (could do this after the 1st step)
2) Similar to the previous problem, except base-10 logs are involved.
... x = 10^(5.6 -log(7.5)) . . . . . take the antilog. Could evaluate now.
... = (1/7.5)·10^5.6 . . . . . . . . . . of course, 10^(-log(7.5)) = 7.5^-1 = 1/7.5
... x ≈ 53,080.96
The initial kick is the first force applied to the ball. It sends the ball up into the air (at some angle). If gravity wasn't present, then the ball would go upward forever in a straight line. However, gravity is the second force pulling down on the ball. This explains why the ball hits some peak point or highest point before it is pulled to the ground. Overall, the path the ball takes is a parabolic arch.
In short, the two forces are the initial kick and gravity.
side note: technically air resistance (aka air friction or drag) is a force being applied since the air pushes against the ball to slow it down, but often air resistance is really complicated and beyond the scope of many math courses. So your teacher may want you to ignore air resistance.
Another note: the initial kick is a one time force that only happens at the beginning. Once the ball is in the air, that force isn't applied anymore. In contrast, the force of gravity is always present and always pulling down. It's probably incredibly obvious, but it's worth pointing out this difference.
Hope i understand it right
9y^2 -12y^4 = 3y^2 *(3 - 4y^2)
