A baseball diamond is a square with side 90 feet. A batter hits the ball and runs toward first base with a speed of 24 ft/sec. At what rate is his distance from second base decreasing when he is halfway to first base?

9514 1404 393

  10.73 ft/s

Step-by-step explanation:

When the runner is halfway to first base, the triangle formed by his position and the first and second bases is a right triangle with legs of 45 ft and 90 ft. The angle between the runner's direction and the line from his position to 2nd base is ...

  angle = arctan(90/45) ≈ 63.43°

The "speed made good" in the direction of 2nd base is the product of the cosine of this angle and the runner's speed toward first.

  speed toward 2nd = cos(63.43°)×(24 ft/s) ≈ 10.73 ft/s

The rate the runner's distance to 2nd base is decreasing is 10.73 ft/sec.

it's d.

step-by-step explanation:

because: -1(4x + 1y) +2 (1x-3y) distribute each number into the parenthesis.

then you would get: -4x - 1y + 2x -6y combine like terms

and get: 2x -7y

hope my answer has you if not then i'm sorry.

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %