On a baseball field, the pitcher’s mound is 60.5 feet from home plate. during practice, a batter hits a ball 195 feet at an angle of 32° to the right of the pitcher’s mound. an outfielder catches the ball and throws it to the pitcher. approximately how far does the outfielder throw the ball?

In this problem, we can imagine that all the points connect to form a triangle. The three point or vertices are located on the pitcher mount, the home plate and where the outfielder catches the ball. So in this case we are given two sides of the triangle and the angle in between the two sides.

With the following conditions, we can use the cosine law to solve for the unknown 3rd side. The formula is:

c^2 = a^2 + b^2 – 2 a b cos θ

Where,

a = 60.5 ft

b = 195 ft

θ = 32°

Substituting the given values:

c^2 = (60.5)^2 + (195)^2 – 2 (60.5) (195) cos 32

c^2 = 3660.25 + 38025 – 20009.7

c^2 = 21,675.56

c = 147.23 ft

Therefore the outfielder throws the ball at a distance of 147.23 ft towards the home plate.

3. parallelogram

step-by-step explanation:

we turn each of our sides into direction vectors

ab=b-a=(3 - 7, 1 - 0) = (-4, 1)

bc=c-b=(1,6)

cd=d-c=(4,-1)

da=a-d=(-1,-6)

we see how the opposite sides have negated direction vectors; this is characteristic of a parallelogram.   we have congruent sides ab=cd and bc=da with unequal lengths because 1^2+6^2 is different from (-4)^2+1^2.   so it's not a square or a rhombus.  

we check the dot product of direction vectors of adjacent sides.   we have

ab dot bc = -4(1) + 1(6) = 2

that's not zero so the sides are not perpendicular.   we've ruled out rectangle.

the direction vectors indicate two pairs of congruent parallel sides, so we choose parallelogram over trapezoid.

1/4 because there is 4 grades.