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
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The answer is 18, Hope you get it right!
The point slope form is y - 8 = 4(x + 6) and the slope intercept form would be y = 4x + 32.
In order to find this, we'll start with the base form of point-slope form.
y - y1 = m(x - x1) ----> Plug in the numbers
y - 8 = 4(x - -6) ----> Simplify
y - 8 = 4(x + 6)
Now to find slope intercept form, solve for y.
y - 8 = 4(x + 6) ----> distribute the 4
y - 8 = 4x + 24 -----> add 8 to both sides
y = 4x + 32
The distance between two points on the plane is given by the formula below
![\begin{gathered} A=(x_1,y_1),B=(x_2,y_2) \\ \Rightarrow d(A,B)=\sqrt[]{(x_1-x_2)^2+(y_1-y_2)^2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A%3D%28x_1%2Cy_1%29%2CB%3D%28x_2%2Cy_2%29%20%5C%5C%20%5CRightarrow%20d%28A%2CB%29%3D%5Csqrt%5B%5D%7B%28x_1-x_2%29%5E2%2B%28y_1-y_2%29%5E2%7D%20%5Cend%7Bgathered%7D)
Therefore, in our case,

Thus,
![\begin{gathered} \Rightarrow d(A,B)=\sqrt[]{(-1-5)^2+(-3-2)^2}=\sqrt[]{6^2+5^2}=\sqrt[]{36+25}=\sqrt[]{61} \\ \Rightarrow d(A,B)=\sqrt[]{61} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5CRightarrow%20d%28A%2CB%29%3D%5Csqrt%5B%5D%7B%28-1-5%29%5E2%2B%28-3-2%29%5E2%7D%3D%5Csqrt%5B%5D%7B6%5E2%2B5%5E2%7D%3D%5Csqrt%5B%5D%7B36%2B25%7D%3D%5Csqrt%5B%5D%7B61%7D%20%5C%5C%20%5CRightarrow%20d%28A%2CB%29%3D%5Csqrt%5B%5D%7B61%7D%20%5Cend%7Bgathered%7D)
Therefore, the answer is sqrt(61)
In general,

Remember that

Therefore,
Answer:
He made $108 in 9 hours
Step-by-step explanation:
I first multiplied 12 by 9 and got 108