The ratio is 3:10. 18/6=3 and 60/6=10
First, an introduction: If the equation of the circle were x^2 + y^2 = 144, then the center would be at (0,0) and the radius would be 12. Note that the distance from the center to P(10,10) is 10sqrt(2), or 14.14. Thus, in this example, P would be OUTSIDE the circle (since 14.14 is greater than the radius 12).
Now let's focus on <span>(x-1)^2 + (y-2)^2 =144. Let x = 12 as an example; find the corresponding y: 9^2 + (y-2)^2 = 144, or (y-2)^2 = 63, and so y-2 is approx. -8 or +8. Then y (for x = 12) is either approx. -10 or 6: (12,-10) or (12,6). Are these inside the circle or outside?
A better way to address this would be as follows:
Find the distance from the center (1, 2) to the point P(10,10). If this distance is less than 12, the point P is inside C; if greater than 12, P is outside C.
This distance is sqrt( (10-2)^2 + (10-1)^2 ), or sqrt (64+81) = sqrt(145).
This is LARGER than sqrt(144). Thus, P is OUTSIDE the circle C.</span>
Answer:
I would choose to use the equation with feet per hour. Submarines reach such great depths that I feel this unit of measurements better represents the magnitude of those depths.
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given that the plane passes through (1,3,3)
Also the plane contains the line x(t)=3t,y(t)=2t,z(t)=4+3t.
This means all points lying in this line will also lie in the plane.
Find out two points on this line.
First point: Let t =0. Point is (0,0,4)
Next point : Let t = 1: Point is (3,2,7)
Now we have 3 non collinear points (0,0,4) (3,2,7) and (1,3,3) lying on the plane.
Equation of the plane is
![\left[\begin{array}{ccc}x-0&y-0&z-4\\3&2&3\\1&3&-1\end{array}\right] =0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx-0%26y-0%26z-4%5C%5C3%262%263%5C%5C1%263%26-1%5Cend%7Barray%7D%5Cright%5D%20%3D0)
Simplify to get
x(-2-9)-y(-3-3)+(z-4)(9-2)=0
i.e -11x+6y+7z-28 =0
11x-6y-7z+28 =0 is the equation of the plane.
