Answer:
20 meters
Step-by-step explanation:
The track is circular so it means that after Patrick raced the entire track he is back at the starting point. In other words, every 440 meters he is back to the beginning.
So we would have that, if he races round the track twice, he would run 440(2) = 880 meters and he would be back at the starting point.
The problem asks us how far is he from the starting point at the 900 meter mark. If at 880 meters he is at the starting point, then at 900 meters he would be 
meters from the starting point.
Answer:
<h2><u>E</u><u>k</u>sponent</h2>
![\sf{ \large{ \boxed{ \red{ {a}^{ \frac{n}{m} } = \sqrt[m]{ {a}^{n} } } } }}](https://tex.z-dn.net/?f=%20%20%5Csf%7B%20%5Clarge%7B%20%5Cboxed%7B%20%5Cred%7B%20%7Ba%7D%5E%7B%20%5Cfrac%7Bn%7D%7Bm%7D%20%20%7D%20%20%3D%20%20%5Csqrt%5Bm%5D%7B%20%7Ba%7D%5E%7Bn%7D%20%7D%20%7D%20%7D%20%7D%7D)
![= \sqrt[3]{ {2}^{4} }](https://tex.z-dn.net/?f=%20%3D%20%20%5Csqrt%5B3%5D%7B%20%7B2%7D%5E%7B4%7D%20%7D%20)
![= \sqrt[3]{2 \times 2 \times 2 \times 2}](https://tex.z-dn.net/?f=%20%3D%20%20%20%5Csqrt%5B3%5D%7B2%20%5Ctimes%202%20%5Ctimes%202%20%5Ctimes%202%7D%20)
![= \boxed {\bold{\sqrt[3]{16}(c.) }}](https://tex.z-dn.net/?f=%20%3D%20%20%20%5Cboxed%20%20%7B%5Cbold%7B%5Csqrt%5B3%5D%7B16%7D%28c.%29%20%7D%7D)
Answer:
23. 0.4583 seconds
24. 0.0107 seconds
Step-by-step explanation:
The problem statement tells you how to work it. You need to convert speed from miles per hour to feet (or inches) per second.
90 mi/h = (90·5280 ft)/(3600 s) = 132 ft/s = (132·12 in)/s = 1584 in/s
__
23. The time it takes for the ball to travel 60.5 ft is ...
time = distance/speed
time = (60.5 ft)/(132 ft/s) = 0.4583 s
It takes 458.3 milliseconds to reach home plate.
__
24. time = distance/speed
time = (17 in)/(1584 in/s) = 0.0107 s
The ball is in the strike zone for 10.7 milliseconds.
D, its (x,y) and (8.2,5.7) are the only ones that match up
