A 2nd baseman throws a baseball toward the 1st baseman 60 feet away. The path of the ball is given by f(x)=-0.004x^2+0.3x+6, where f(x) is the height of the ball (in feet) and x is the horizontal distance from the 2nd baseman (in feet). The 1st baseman can reach 8 ft high. Can the 1st baseman catch the ball without jumping?

The 1st baseman can't catch the ball without jumping.

Step-by-step explanation:

To know if the 1st baseman can catch the ball without jumping we need to find the height reached by the baseball by solving the following equation:

We know that x (the horizontal distance) = 60 feet, so we have:

Since the 1st baseman can reach 8 ft high and the height reached by the baseball is 9.6 ft, the 1st baseman can't catch the ball without jumping.

Therefore, the 1st baseman can't catch the ball without jumping.

I hope it helps you!                             

hello from mrbilldoesmath!

answer:   x^2 + y^2 = 1024

discussion:

the equation of a circle with center (h,k) and radius "r" is

(x-h)^2 + (y-k)^2 = r^2.

in our case

(x-0)^2 + (y-0) ^2 = 32^2 = 1024

you,

mrb