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.

8 0

2 years ago

Find the vertex and length of the latus rectum for the parabola. y=1/6(x-8)^2+6

Ivan

Step-by-step explanation:

If the parabola has the form

$y = a(x - h)^2 + k$ (vertex form)

then its vertex is located at the point (h, k). Therefore, the vertex of the parabola

$y = \dfrac{1}{6}(x - 8)^2 + 6$

is located at the point (8, 6).

To find the length of the parabola's latus rectum, we need to find its focal length f. Luckily, since our equation is in vertex form, we can easily find from the focus (or focal point) coordinate, which is

$\text{focus} = (h, k +\frac{1}{4a})$

where $\frac{1}{4a}$ is called the focal length or distance of the focus from the vertex. So from our equation, we can see that the focal length f is

$f = \dfrac{1}{4(\frac{1}{6})} = \dfrac{3}{2}$

By definition, the length of the latus rectum is four times the focal length so therefore, its value is

$\text{latus rectum} = 4\left(\dfrac{3}{2}\right) = 6$

5 0

3 years ago