Abaseball diamond is a square with sides that are 27.4 m. the pitchers mound is 18.4m from homeplate on the line joining hommeplate and second base. how far is it from the pitchars mound to first base

  19.4 m

Step-by-step explanation:

The law of cosines can be used to figure this. The angle between the line to the pitcher's mound and the first-base line is 45°, so the distance (d) from the mound to first can be found from ...

  d² = 18.4² +27.4² -2·18.4·27.4·cos(45°) ≈ 376.33

  d ≈ √376.33 ≈ 19.4 . . . . meters

It is about 19.4 meters from the pitcher's mound to first base.

14.3

step-by-step explanation:

add all the numbers together.

9 + 12 + 15 + 16 + 17 + 17 = 86

now, divide 86 by how many numbers we have: 6.

86 / 6 = 14.3

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204