Answer:
• David
,
• 4 miles
Explanation:
In the graph:
The given locations are:
• Owen's House, A(11,3)
,
• David's House, B(15,13)
,
• School, C(3,18)
We determine both Owen's and David's distance from the school using the distance formula.
Owen's distance from school (AC)
![\begin{gathered} AC=\sqrt[]{(3-11)^2+(18-3)^2} \\ =\sqrt[]{(-8)^2+(15)^2} \\ =\sqrt[]{64+225} \\ =\sqrt[]{289} \\ AC=17\text{ miles} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20AC%3D%5Csqrt%5B%5D%7B%283-11%29%5E2%2B%2818-3%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B%28-8%29%5E2%2B%2815%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B64%2B225%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B289%7D%20%5C%5C%20AC%3D17%5Ctext%7B%20miles%7D%20%5Cend%7Bgathered%7D)
David's distance from school (BC)
![\begin{gathered} BC=\sqrt[]{(3-15)^2+(18-13)^2} \\ =\sqrt[]{(-12)^2+(5)^2} \\ =\sqrt[]{144+25} \\ =\sqrt[]{169} \\ BC=13\text{ miles} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20BC%3D%5Csqrt%5B%5D%7B%283-15%29%5E2%2B%2818-13%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B%28-12%29%5E2%2B%285%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B144%2B25%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B169%7D%20%5C%5C%20BC%3D13%5Ctext%7B%20miles%7D%20%5Cend%7Bgathered%7D)
We see from the calculations that David lives closer to the school, and by 4 miles.
The graph below is attached for further understanding:
Answer:
0.085
Step-by-step explanation:
1% is the same as 0.01, so 8.5% of 100 is 0.085
Answer:
Area=190.091 cm^2
Step-by-step explanation:
Area = 1/2(Pi x r^2) one-half because it's a semi-circle
Area=1/2(3.14 x 11^2)
11^2=121 so, Area=1/2(3.14 x 121)
Area=1/2(379.94)
Area=189.97 cm^2
adjustment:
Area=1/2(3.142 x 11^2)
Area=1/2(3.142 x 121)
Area=1/2(380.182)
Area=190.091
Answer:
y = 8
Step-by-step explanation:
8 + 24 = 32
32/4 = 8
The train traveled 60 miles per hour since 360 miles divided by 6 hours gives you 60 mph
