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:
x ≈ 1.4
Step-by-step explanation:
Using the sine ratio in the right triangle
sin50° = 
= 
= 
( multiply both sides by 1.8 )
1.8 × sin50° = x , then
x ≈ 1.4 ( to the nearest tenth )
Answer:
The dog weighs 16 pounds, while the cat weighs 8 pounds
Step-by-step explanation:
Let's call the weight of the dog d, and the weight of the cat c.
d+c=24
d=2c
Substitute the second equation into the first:
(2c)+c=24
3c=24
c=8
d=24-8=16
Hope this helps!
Answer:
hi
Step-by-step explanation:
have a nice day
