On the Cartesian plane in which each unit is one foot, a dog is tied to a post on the point $(4,3)$ by a $10$ foot rope. What is the greatest distance the dog can be from the origin?
The area the dog can go in is a circle of radius $10$ centered at the point $(4,3)$. The point farthest from $(0,0)$ in the circle would be the point on the circle's circumference which is on the same diameter as $(0,0)$ but on the other side of the center of the circle.
The distance from the origin to the center of the circle, by the distance formula, is $\sqrt{(4-0)^2+(3-0)^2}=\sqrt{16+9}=5$. As the radius of the circle is $10$, the distance from the origin to the point in the circle farthest from the origin is $\boxed{15}$.
15
Level 5
Algebra
Problem: On the Cartesian plane in which each unit is one foot, a dog is tied to a post on the point $(4,3)$ by a $10$ foot rope. What is the greatest distance the dog can be from the origin?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the greatest distance the dog can be from the origin, we first recognize that the dog is tied to a post at the point (4,3) and can move within a circular area defined by the length of the rope, which is 10 feet. The center of this circle is at (4,3) and the radius is 10.

Next, we calculate the distance from the origin (0,0) to the center of the circle (4,3). This distance can be found using the distance formula: 

Distance = √[(x2 - x1)² + (y2 - y1)²] = √[(4 - 0)² + (3 - 0)²] = √[16 + 9] = √25 = 5 feet.

Now, to find the maximum distance from the origin to any point on the circumference of the circle, we add the radius of the circle to the distance from the origin to the center of the circle. 

Maximum distance = Distance from origin to center + Radius = 5 feet + 10 feet = 15 feet.

Thus, the greatest distance the dog can be from the origin is 15 feet.<|END_OF_LATENT|> The final answer is $15$.