Answer:
C = (2,2)
Step-by-step explanation:
B = (10 ; 2)
M = (6 ; 2)
C = (x ; y )
|___________|___________|
B (10;2) M (6;2) C ( x; y)
So:
dBM = dMC
√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 6)^2]
(2-2)^2 - (6-10)^2 = (y-2)^2 + (x - 6)^2
0 + (-4)^2 = (y-2)^2 + (x - 6)^2
16 = (y-2)^2 + (x - 6)^2
16 - (x - 6)^2 = (y-2)^2
Also:
2*dBM = dBC
2*√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 10)^2]
4*[(0)^2 + (-4)^2] = (y-2)^2 + (x - 10)^2
4*(16) = (y-2)^2 + (x - 10)^2
64 = (y-2)^2 + (x - 10)^2
64 = 16 - (x - 6)^2 + (x - 10)^2
48 = (x - 10)^2 - (x - 6)^2
48 = x^2 - 20*x + 100 - x^2 + 12*x - 36
48 = - 20*x + 100 + 12*x - 36
8*x = 16
x = 2
Thus:
16 - (x - 6)^2 = (y-2)^2
16 - (2 - 6)^2 = (y-2)^2
16 - (-4)^2 = (y-2)^2
16 - 16 = (y-2)^2
0 = (y-2)^2
0 = y - 2
2 = y
⇒ C = (2,2)
Lets quickly run through the prime numbers and see what it is not divisible by
39 is not divisible by 2
39 IS divisible by 3
So lets stop right here.
What is 39 divided by 3? 13 !
And 13 is a prime number too!
So that is all we can do.
So prime factorization of 39 is <u>3 x 13</u><u><em /></u>
For each <em>x</em> in the interval 0 ≤ <em>x</em> ≤ 5, the shell at that point has
• radius = 5 - <em>x</em>, which is the distance from <em>x</em> to <em>x</em> = 5
• height = <em>x</em> ² + 2
• thickness = d<em>x</em>
and hence contributes a volume of 2<em>π</em> (5 - <em>x</em>) (<em>x</em> ² + 2) d<em>x</em>.
Taking infinitely many of these shells and summing their volumes (i.e. integrating) gives the volume of the region:
