Answer:
pairs of numbers is 3,-3
Step-by-step explanation:
The computation is shown below:
Let us assume a and b are the pairs of the numbers whose difference is 6
i.e.
a - b = 6
a = 6 + b
So
f = ab
f = (6 + b)b is minimum
Now
d ÷ db (6+ b)b = 0
(6+ b) + b = 0
2b = -6
b = -3
Now
a = 6 + b
= 3
So pairs of numbers is 3,-3
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:
first is a) option and second is b) option
N=2p. Blah blah not part of the answer
