Based on the distance of the pitcher's mound from the home plate, the path of the ball, and the height the ball was hit, the distance the outfielder threw the ball is C. 183.0 ft.
<h3>How far did the outfielder throw the ball?</h3>
Based on the shape of a mound, the law of cosines can be used.
The distance the ball was thrown by the outfielder can therefore be d.
Distance is:
d² = 60.5² + 226² - (2 x 60.5 x 226 x Cos(39))
d ²= 33,484.42ft
Then find the square root:
d = √33,484.42
= 182.9874
= 183 ft
Find out more solving for distance at brainly.com/question/10739348
#SPJ1
Answer:
here
Step-by-step explanation:
15x-20x= -82+12=>
=> - 5x= -70=>
= x= 14
Answer:
See explanation below
Step-by-step explanation:
Find the value of x in each of the following
1) x-1+5√x-1+3=0
5√x-1 = -x -2
Square both sides
25(x-1) = (-x-2)^2
25(x-1) = x^2+4x + 4
25x-25 = x^2+4x + 4
x^2 +4x + 4 - 25x + 25 = 0
x^2 - 21x + 29 = 0
x = 21±√21²-4(29)/2
x = 21±√441-116/2
x = 21±√325/2
x = 21±18.03/2
x = 21+18.03/2 and 21-18.03/2
x = 19.515 and 1.485
4) x^4-64=0
x^4 = 64
![x = \sqrt[4]{64}](https://tex.z-dn.net/?f=x%20%3D%20%5Csqrt%5B4%5D%7B64%7D)
x = 2.83
Answer:
y=-15x+b
Step-by-step explanation:
Just move it around and use a calucator
The answer is 64!!
Hope this helps!